Previous Year Paper

DSSSB PGT Maths (Female) Official Paper (Held On: 10 Jul, 2021 Shift 1) (Previous Year Paper)

Held in 2021

300 questions180 minutes
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Mental Ability (20 questions)

Question 1

Choose the option that is to unrelated (or odd one out amongst) the given set of words.

  1. AOption (a)

    Oil

  2. BOption (b)

    Butter

  3. COption (c)

    Paneer

  4. DOption (d)

    Curd

View answer & explanation

Correct answer

Answer: (a)A.

Oil

Why this answer is correct

Butter, Paneer, and Curd are the products made from Milk.

While;

Oil is a thick liquid that comes from animals or plants and is used in cooking.

Hence, "Oil" is the correct answer.

Important PointsThere are may types of oil:

  • Crude Oils
  • Vegetable Oils
  • Animal Oils
  • Seed Oils etc

Question 2

Find the missing number in the series given below.

3, 8, 22, 63, ?

  1. AOption (a)

    185

  2. BOption (b)

    295

  3. COption (c)

    310

  4. DOption (d)

    285

View answer & explanation

Correct answer

Answer: (a)A.

185

Why this answer is correct

The pattern follows here is;

Hence, "185" is the correct answer.

Question 3

Pointing to the photo, Mamta said that the woman in this photo is the wife of the son of the only daughter of my father's wife. How is Mamta related to that lady?

  1. AOption (a)

    Aunt

  2. BOption (b)

    Sister

  3. COption (c)

    Mother

  4. DOption (d)

    Mother-in-law

View answer & explanation

Correct answer

Answer: (d)D.

Mother-in-law

Why this answer is correct

Preparing the family tree using the following symbols:

A possible tree diagram will be:

Hence, Mamta is the mother in law of a lady.

Question 4

In a certain code language, 'MAPED' is coded as '64' and 'RECK' is coded as '41'. How will 'NOSE' be coded as in that language?

  1. AOption (a)

    82

  2. BOption (b)

    51

  3. COption (c)

    63

  4. DOption (d)

    57

View answer & explanation

Correct answer

Answer: (d)D.

57

Why this answer is correct

The positions of the letters according to the English alphabet series:

The pattern for this code is as follows;

LOGIC: Sum of positional value + (Number of Letters)1or 2

If the number of letters is even then power is 1.

If the number of letters is odd then power is 2.

In MAPED:

Number of letter = 5 (Odd)

MAPED = (13 + 1 + 16 + 5 + 4) + 52

= 39 + 25

= 64

And;

In RECK:

Number of letter = 4 (Even)

RECK = (18 + 5 + 3 + 11) + 4

= 37 + 4

= 41

Similarly;

In NOSE:

Number of letter = 4 (Even)

NOSE = (14 + 15 + 19 + 5) + 4

= 53 + 4

= 57

Hence, "57" is the correct answer.

Question 5

Select the number from among the given options that can replace the question mark (?) in the following series.

A1Z , C3Y , E7X , G15W , ?

  1. AOption (a)

    A29G

  2. BOption (b)

    I31V

  3. COption (c)

    B29U

  4. DOption (d)

    O31V

View answer & explanation

Correct answer

Answer: (b)B.

I31V

Why this answer is correct

The positions of the letters according to the English alphabet series:

The pattern for this is as follows;

Hence, "I31V" is the correct answer.

Question 6

Preeti got more marks than Rahul. Yamuna scored as many marks as Divya. Lokita got less marks than Manju. Rahul got more marks than Yamuna. Manju got less marks than Divya. Who got the lowest marks?

  1. AOption (a)

    Yamuna

  2. BOption (b)

    Rahul

  3. COption (c)

    Manju

  4. DOption (d)

    Lokita

View answer & explanation

Correct answer

Answer: (d)D.

Lokita

Why this answer is correct

  1. Preeti got more marks than Rahul.

Preeti > Rahul

  1. Yamuna scored as many marks as Divya.

Yamuna > Divya 

  1. Lokita got less marks than Munj.

Manju > Lokita

  1. Rahul got more marks than Yamuna.

Rahul > Yamuna

  1. Manju got less marks than Divya.

Divya > Manju

After combining all the statements;

Preeti > Rahul > Yamuna > Divya > Manju > Lokita

Hence, "Lokita" got the lowest marks.

Question 7

How many triangles are there in the given figure?

  1. AOption (a)

    15

  2. BOption (b)

    13

  3. COption (c)

    14

  4. DOption (d)

    16

View answer & explanation

Correct answer

Answer: (a)A.

15

Why this answer is correct

The figures given below represents the number of triangles:

Hence, "15" is the correct answer.

Question 8

Identify the diagram that best represents the relationship among the given classes.

Book, Page, Pen

  1. AOption (a)

  2. BOption (b)

  3. COption (c)

  4. DOption (d)

View answer & explanation

Correct answer

Answer: (a)A.

Why this answer is correct

The correct Venn diagram representation is:

Hence, option (1) is the correct answer.

Question 9

In the question given, relations between different elements are shown in the statements. These statements are followed by two conclusions. Find out which of the given conclusions follow(s) the given statements and select the correct alternative from the given choices.

Statement:

L ≥ M = N < O, P < Q ≥ R = S ≥ L

Conclusion:

I. Q > M

II. N = Q

  1. AOption (a)

    None is true

  2. BOption (b)

    Only I is true

  3. COption (c)

    Only II is true

  4. DOption (d)

    Either I or II is true

View answer & explanation

Correct answer

Answer: (d)D.

Either I or II is true

Why this answer is correct

Statements:  L ≥ M = N < O, P < Q ≥ R = S ≥ L

On combining: P < Q ≥ R = S ≥ L ≥ M = N < O

Conclusion:

I. Q > M → False (Q ≥ R = S ≥ L ≥ M, Q ≥ M)

II. N = Q → False (Q ≥ R = S ≥ L ≥ M = N, N = Q)

Hence, Either I or II is true.

Important Points

If all three possible conditions (<, >, =) between any two entities are included in conclusions, and both the conclusion are individually false then it would be a case of ‘either-or’.

Question 10

Identify the diagram that best represents the relationship among the given classes.

Pandemic, Coronavirus, Diabetes

  1. AOption (a)

  2. BOption (b)

  3. COption (c)

  4. DOption (d)

View answer & explanation

Correct answer

Answer: (b)B.

Why this answer is correct

The correct Venn diagram representation is:

Hence, option (2) is the correct answer.

Question 11

Which answer figure will complete the pattern in the question figure?

  1. AOption (a)

  2. BOption (b)

  3. COption (c)

  4. DOption (d)

View answer & explanation

Correct answer

Answer: (d)D.

Why this answer is correct

Completing the incomplete pattern;

Hence, option (4) is the correct answer.

Question 12

Rahul starts walking in the south direction and walks a distance of 7 meters. Now he took a left turn and walk 6 m. Again he takes a left turn and walks 15 m and reached a point P. In which direction is Rahul from the initial point?

  1. AOption (a)

    North-West

  2. BOption (b)

    South-West

  3. COption (c)

    North-East

  4. DOption (d)

    South-East

View answer & explanation

Correct answer

Answer: (c)C.

North-East

Why this answer is correct

As per the given information;

Hence, Rahul is in North east direction from his starting point.

Question 13

Find the mirror image of the following figure.

  1. AOption (a)

  2. BOption (b)

  3. COption (c)

  4. DOption (d)

View answer & explanation

Correct answer

Answer: (a)A.

Why this answer is correct

The mirror image of this image is:

Hence, option (1) is the correct answer.

Question 14

Select the option in which the words have the same relation among themselves as the words in the pair of words given below.

Hospital ∶ Treatment

  1. AOption (a)

    Clay : Bricks

  2. BOption (b)

    Court ∶ Lawyer

  3. COption (c)

    Office ∶ Work

  4. DOption (d)

    University ∶ Employment

View answer & explanation

Correct answer

Answer: (c)C.

Office ∶ Work

Why this answer is correct

As Hospital is for Treatment.

Similarly, 

The Office is for work.

Hence, the related word pair is ‘Office ∶ Work’.

Question 15

In the question below are given two statements followed by two conclusion. You have to take the two given statements to be true even if they seem to be at variance with the commonly known facts and then decide which of the given conclusions logically follows from the two given statements, disregarding commonly known facts.

Statements:

All cars are tyres.

Some clutchs are tyres.

Conclusions:

I. Some cars are clutchs.

II. Some clutchs are cars.

  1. AOption (a)

    If only conclusion I follows

  2. BOption (b)

    If either conclusion I or II follows

  3. COption (c)

    If only conclusion II follows

  4. DOption (d)

    If neither conclusion I nor II follows

View answer & explanation

Correct answer

Answer: (d)D.

If neither conclusion I nor II follows

Why this answer is correct

The least possible Venn diagram for the given statements is as below,

Conclusions:

I. Some cars are clutchs → False (There is no direct relation between cars and cluths)

II. Some clutchs are cars → False (There is no direct relation between cars and cluths)

Hence, neither conclusion I nor II follows.

Additional Information

Question 16

Which two signs should be interchanged to make the given equation correct?

20 + 5 × 3 ÷ 3 - 1 = 14

  1. AOption (a)

    × and +

  2. BOption (b)

    ÷ and ×

  3. COption (c)

    ÷ and +

  4. DOption (d)

    × and -

View answer & explanation

Correct answer

Answer: (c)C.

÷ and +

Why this answer is correct

The logic followed is:

Checking options one by one;

1) × and + →

20 + 5 × 3 ÷ 3 - 1 = 14

After replacing the signs from left to right and using the BODMAS rule,

20 × 5 + 3 ÷ 3 - 1 

= 20 × 5 + 1 - 1

= 100 + 1 - 1

= 100 ≠ 14

2) ÷ and × →

20 + 5 × 3 ÷ 3 - 1 = 14

After replacing the signs from left to right and using the BODMAS rule,

20 + 5 ÷ 3 × 3 - 1 

= 20 + 5/3 × 3 - 1

= 20 + 5 - 1

= 25 - 1

= 24 ≠ 14

3) ÷ and + →

20 + 5 × 3 ÷ 3 - 1 = 14

After replacing the signs from left to right and using the BODMAS rule,

20 ÷ 5 × 3 + 3 - 1 

= 4 × 3 + 3 - 1 

= 12 + 3 - 1

= 15 - 1

= 14

4) × and - →

20 + 5 × 3 ÷ 3 - 1 = 14

After replacing the signs from left to right and using the BODMAS rule,

20 + 5 - 3 ÷ 3 × 1 

= 20 + 5 - 1 × 1

= 20 + 5 - 1

= 25 - 1

= 24 ≠ 14

All options are correct except option (3).

Hence, option (3) is the correct answer.

Question 17

Which oh the following conclusion can  be drawn on the basis of the given statements?

Statements 1: Some crystals are diamonds.

Statements 2: All emeralds are crystals.

  1. AOption (a)

    All crystals are emeralds

  2. BOption (b)

    Some crystals are emeralds

  3. COption (c)

    Some emeralds are diamonds

  4. DOption (d)

    All diamonds are emeralds

View answer & explanation

Correct answer

Answer: (b)B.

Some crystals are emeralds

Why this answer is correct

The least possible Venn diagram for the given statements is as below,

1) All crystals are emeralds → False (As All emeralds are crystal, then All crystal are emeralds not possible)

2) Some crystals are emeralds → True (As All emeralds are crystal, then some crystals are definitely emeralds)

3) Some emeralds are diamonds → False (There is no direct relation between emeralds and diamonds)

4) All diamonds are emeralds → False (There is no direct relation between emeralds and diamonds)

Hence, option (2) is the correct asnwer.

Additional Information

Question 18

From the given responses, find the missing letter cluster in the series.

ZXVU, VTRQ, ?, NLJI, JHFE

  1. AOption (a)

    KLMO

  2. BOption (b)

    RPNM

  3. COption (c)

    RSNP

  4. DOption (d)

    KLMN

View answer & explanation

Correct answer

Answer: (b)B.

RPNM

Why this answer is correct

The positions of the letters according to the English alphabet series:

The pattern for this is as follows;

Hence, "RPNM" is the correct answer.

Question 19

Four figures are given in the following question. All figures have something in common except one. Find the odd figure out.

  1. AOption (a)

  2. BOption (b)

  3. COption (c)

  4. DOption (d)

View answer & explanation

Correct answer

Answer: (a)A.

Why this answer is correct

Checking options one by one;

  1. Number of arrows = 5 (Odd)

  2. Number of arrows = 8 (Even)

3) Number of arrows = 4 (Even)

4) Number of arrows = 2 (Even)

Hence, option (1) is the correct answer.

Question 20

In the following question, select the related letter cluster from the given alternatives.

EDKX ∶ MLSF ∶∶ ACZO ∶ ?

  1. AOption (a)

    IKVG

  2. BOption (b)

    JKHV

  3. COption (c)

    IKHW

  4. DOption (d)

    JKIV

View answer & explanation

Correct answer

Answer: (c)C.

IKHW

Why this answer is correct

The positions of the letters according to the English alphabet series:

The pattern for this code is as follows;

Similarly;

Hence, "IKHW" is the correct answer.

General Awareness (20 questions)

Question 21

Who is the ex-officio chairman of NITI Aayog?

  1. AOption (a)

    Home Minister

  2. BOption (b)

    Vice President

  3. COption (c)

    President

  4. DOption (d)

    Prime Minister

View answer & explanation

Correct answer

Answer: (d)D.

Prime Minister

Why this answer is correct

The correct answer is Prime Minister.

Key Points

  • Prime Minister is the ex-officio chairman of the NITI Aayog in India.
  • NITI Aayog is an executive body, that was established by the government in 2015 through an executive resolution.
  • NITI Aayog replaced the erstwhile planning body- the Planning Commission of India.
  • The term NITI stands for National Institution for Transforming India.
  • NITI Aayog is a think tank that advises the government in policy formulation, unlike the PC which was empowered to allocate and distribute funds for the planning.
  • NITI Aayog's main focus areas are cooperative & competitive federalism, decentralised planning, scenario planning, capacity building etc.
  • Full-Time Organisational Framework:
  • Chairman: Prime Minister (Narendra Modi)
  • Vice Chairperson: Appointed by the Prime Minister. (Suman Berry)
  • Chief Executive Officer (CEO).
  • Full-Time members
  • Part-time members: maximum 2
  • Ex-officio members: maximum 4

Question 22

The Gateway of India was built to commemorate the visit of which of the following king?

  1. AOption (a)

    King Edward VII

  2. BOption (b)

    King George I

  3. COption (c)

    King William IV

  4. DOption (d)

    King George V

View answer & explanation

Correct answer

Answer: (d)D.

King George V

Why this answer is correct

The correct answer is King George V.

  • The Gateway of India was built to commemorate the visit of King George V.

Key Points

  • The early twentieth century saw the rise of the swaraj (self-rule) movement.
  • King George V (1910–36) and his wife, Queen Mary, carried out two royal tours of the Indian subcontinent in the hope of improving public perception of and loyalty to the British Crown.
  • Both fell in love with ‘dear beautiful India’.
  • Following their initial visit in 1905–06 as Prince and Princess of Wales, they returned as Emperor and Empress for a Coronation Tour in 1911–12.

Additional Information

  • Important facts about Gateway of India.
  • ​The Gateway of India was erected in honour of King George V and Queen Mary’s landing at Apollo Bunder on their visit to India.
  • However, the foundation stone was laid on March 31, 1911, by the then-Governor of Mumbai, Sir George Sydenham Clarke.
  • The design of the monument took three years to pass and finally got sanctioned on March 31, 1911.
  • The archway of the monument is 26 meters (85 feet) high and joined with four turrets and intricate latticework carved on stones.
  • The archway of the monument is built of Basalt and about 21 lakhs was spent to build the arch alone.
  • The structure is an adaptation of the Arc de Triomphe in Paris.
  • During the rule of the British government, this was used as the arrival point for the visitors from the west.
  • The monument is surrounded by another of Mumbai’s major tourist attractions, the Marine Drive, facing the vast Arabian Sea.
  • The last British ships set to sail in 1947 to leave the country from the Gateway of India making it a legend.

Question 23

Vaishali was the capital of which of the following Mahajanapadas?

  1. AOption (a)

    Panchal

  2. BOption (b)

    Vajji

  3. COption (c)

    Magadh

  4. DOption (d)

    Avanti

View answer & explanation

Correct answer

Answer: (b)B.

Vajji

Why this answer is correct

The correct answer is Vajji.

Key Points

  • Vaishali was the capital of Vajji Mahajanapada.
  • The kingdom was situated in Bihar to the north of the river Ganga.
  • It covered the districts of Vaishali and Muzaffarpur in Bihar.

Additional Information

MahajanapadaCapital
VajjiVaishali
PanchalAhichatra
AvantiUjjain & Mahismati
MagadhaInitially, Rajagaha was the capital of Magadha. Later, in the fourth century BCE, the capital was shifted to Pataliputra.

Important Points

  • The term "Janapada" literally means the foothold of a people.
  • There were 16 Mahajanapadas during 600 B.C. to 325 B.C. in the Indian Sub-continent.
  • There were 16 kingdoms or oligarchic republics that existed in Northern ancient India from the sixth to fourth centuries BCE during the second urbanization period.
  • The 6th–5th centuries BCE is often regarded as a major turning point in early Indian history; during this period India's first large cities arose after the demise of the Indus Valley Civilization.
  • It was also the time of the rise of sramana movements (including Buddhism and Jainism), which challenged the religious orthodoxy of the Vedic Period.

Question 24

The water holding capacity is the highest in which of the following soils?

  1. AOption (a)

    Silt Soil

  2. BOption (b)

    Clayey Soil

  3. COption (c)

    Chalk Soil

  4. DOption (d)

    Peat Soil

View answer & explanation

Correct answer

Answer: (b)B.

Clayey Soil

Why this answer is correct

The correct answer is Clayey Soil.

Key Points

  • Clayey soil:
  • The water holding capacity of soil is a very important agronomic characteristic.
  • Soils that hold generous amounts of water are less subject to leaching losses of nutrients or soil-applied pesticides.
  • The water holding capacity is highest in clayey soil because it is made up of very small tightly packed particles that do not allow water to percolate.
  • It is very sticky to the touch when wet, but smooth when dried.
  • Clay is the densest and heaviest type of soil that does not drain well or provide space for plant roots to flourish.
  • Hence The water holding capacity is the highest in Clayey soil. So, the correct answer is 'Clayey soil'.

Additional Information 

  • Soil is classified into four types:
  • Sandy soil
  • Silt Soil
  • Clay Soil
  • Loamy Soil.
  • Silt Soil: Silt has smaller particles compared to sand.
  • It is also made up of rock and other mineral particles.
  • Furthermore, its fine quality holds water better than sand.
  • Due to the above-mentioned features, it is also beneficial for agriculture.
  • ​Chalk soil: can be either light or heavy but always highly alkaline due to the calcium carbonate (lime) within its structure.
  • As these soils are alkaline they will not support the growth of ericaceous plants that require acidic soils to grow.
  • If chalky soil shows signs of visible white lumps then they can’t be acidified and gardeners should be redesigned to only choose plants that prefer alkaline soil.
  • Peat soil: is high in organic matter and retains a large amount of moisture.
  • This type of soil is very rarely found in a garden and is often imported into a garden to provide an optimum soil base for planting.
  • Loam soil: is a mixture of sand, silt, and clay that are combined to avoid the negative effects of each type.
  • These soils are fertile, easy to work with and provide good drainage.
  • Depending on their predominant composition they can be either sandy or clay loam.

Question 25

Which of the following district is the largest district in India by area?

  1. AOption (a)

    Mahe

  2. BOption (b)

    Bangalore

  3. COption (c)

    Mumbai

  4. DOption (d)

    Kutch

View answer & explanation

Correct answer

Answer: (d)D.

Kutch

Why this answer is correct

The correct answer is Kutch.

Key Points

  • Kutch is the largest district in India covering an area of 45,674 square km.
  • It is in the state of Gujarat.
  • Kutch has 10 Talukas, 939 villages, and 6 Municipalities.
  • The language Kutchi is used in this district.
  • The Kutch district is also known for ecologically important Banni grasslands with their seasonal marshy wetlands.

Additional Information

  • Mahe of Pondicherry is the smallest district of the country with an area of 9 sq km.
  • There were 640 districts in the 2011 Census of India.
  • Thane is the largest district in the country population-wise with a total population of 11060148.

Confusion Points

  • Kachchh Gujarat has an area of 45,674 sqkm, whereas Leh has 45110 sqkm area.

Question 26

Who was the judge who presided over the hearing of Mahatma Gandhi during the non-cooperation movement in the year 1922?

  1. AOption (a)

    Justice Henry Davison

  2. BOption (b)

    Justice CN Broomfield

  3. COption (c)

    Justice James Dewar

  4. DOption (d)

    Justice William McDonnell

View answer & explanation

Correct answer

Answer: (b)B.

Justice CN Broomfield

Why this answer is correct

The correct answer is Justice CN Broomfield.

Key Points

  • During the Non-Cooperation Movement thousands of Indians were put in jail.
  • Gandhiji was also arrested in March 1922, and charged with sedition.
  • The judge who presided over his trial was Justice C.N. Broomfield.
  • He made a remarkable speech while pronouncing his sentence.

Additional Information

  • The Programs of the Non-Cooperation Movement are:
  • ​Boycott of government colleges, schools and courts.
  • Boycott of foreign goods and language.
  • Boycott of Council elections to be held as per the suggestions of the Morley-Minto Reforms of 1919.
  • Resignation of Indian members from nominated seats in the local bodies.
  • Surrender of titles and honorary offices.
  • Refusal to attend functions organized by the government.
  • From the above, it is clear that attaining self-government through Constitutional means was not included in the Programme of Non-Cooperation Movement

Question 27

Who among the following persons is the author of the book "Anandmath"?

  1. AOption (a)

    Rabindranath Tagore

  2. BOption (b)

    Munshi Premchand

  3. COption (c)

    Dr. Rajendra Prasad

  4. DOption (d)

    Bankim Chandra Chatterjee

View answer & explanation

Correct answer

Answer: (d)D.

Bankim Chandra Chatterjee

Why this answer is correct

The correct answer is Bankim Chandra Chatterjee.

Key Points

  • Anandmath​:
  • The famous book ‘Anandmath' was authored by Bankim Chandra Chatterjee.
  • The book was originally written in 1882.
  • Anandamath takes the reader back to Bengal in the clutches of the famine, the backdrop of the pre-independence struggle, and the turmoil of human lives caught in this frame of time.
  • It is considered one of the most remarkable works of the Bangla language and Indian literature.
  • Through the book, Bankim also gave us the song ‘VandeMataram’which became the rallying call for rebels.
  • The first two stanzas eventually became the National song of India.
  • One of the gems of Indian Literature, ‘Anandmath’ carries a deep sentiment of nationalism which was the essence of the freedom struggle.

Additional Information

  • Bankim Chandra Chatterjee -
  • Bankim Chandra Chatterjee was born on 27 June 1838 in a Brahmin family in Kanthalpara village in Bengal.
  • Vande Mataram was composed in a mixed language (Sanskrit-Bengali) by Bankim Chandra Chatterjee.
  • This song was published in the year 1872 as an underlying song in his novel Anand Math.
  • In this novel, this song is sung by a monk named "Bhawanand".
  • Its tune was composed by Yadunath Bhattacharya. This song takes 65 seconds to sing.​

Question 28

Who among the following laid the foundation of Golden Temple?

  1. AOption (a)

    Guru Arjan Sahib

  2. BOption (b)

    Guru Amardas Sahib

  3. COption (c)

    Hazrat Mian Mir ji

  4. DOption (d)

    Guru Ramdas Sahib

View answer & explanation

Correct answer

Answer: (c)C.

Hazrat Mian Mir ji

Why this answer is correct

The correct answer is Hazrat Mian Mir Ji.

  • Hazrat Mian Mir Ji laid the foundation of the Golden Temple.

Key Points

  • Sri Harmandir Sahib, also known as Sri Darbar Sahib or Golden Temple.
  • The land for the site was bought by the Guru Ram Das Sahib on payment from the Zamindars (landlords) of native villages.
  • Guru Arjan Sahib got its foundation laid by a Muslim saint Hazrat Mian Mir Ji of Lahore on the 1st of Magh, 1645 Bikrmi Samvat (December, 1588).
  • The construction work was directly supervised by Guru Arjan Sahib himself and he was assisted by prominent Sikh personalities like Baba Budha ji, Bhai Gurdas Ji, Bhai Sahlo Ji, and many other devoted Sikhs.
  • Sri Harmandir Sahib is built on a 67ft, square platform in the centre of the Sarovar (tank).
  • The temple itself is 40.5ft. square.
  • It has a door each on the East, West, North, and South.

Additional Information

  • Guru Amar Das was the third of the Ten Gurus of Sikhism.
  • Sikhism was well established by the time of Guru Arjan, the fifth Guru.
  • Guru Arjan completed the establishment of Amritsar as the capital of the Sikh world.
  • And compiled the first authorized book of Sikh scripture, the Adi Granth.
  • The Ten Gurus in Sikhism
The First GuruGuru Nanak
The Second GuruGuru Angad
The Third GuruGuru Amar Das
The Fourth GuruGuru Ram Das
The Fifth GuruGuru Arjan
The Sixth GuruGuru Hargobind
The Seventh GuruGuru Har Rai
The Eighth GuruGuru Harkishan
The Ninth GuruGuru Teg Bahadur
The Tenth GuruGuru Gobind Singh

Question 29

Where was the first initiative of Mahatma Gandhi as a nationalist in India?

  1. AOption (a)

    Porbandar

  2. BOption (b)

    Kheda

  3. COption (c)

    Champaran

  4. DOption (d)

    Ahmedabad

View answer & explanation

Correct answer

Answer: (c)C.

Champaran

Why this answer is correct

The correct answer is Champaran.

  • Champaran was the first initiative of Mahatma Gandhi as a nationalist in India.

Key Points

  • Mahatma Gandhi was the leader who guided India towards Independence. India was under British rule for over 250 years.
  • Gandhi returned to India from South Africa in 1915 at the request of Gopal Krishna Gokhale.
  • Mahatma Gandhi's famous contributions to the Indian freedom movement:
  • Champaran Satyagraha (1917):
  • ​The Champaran agitation in Bihar was Gandhi's first active involvement in Indian freedom politics.
  • The Champaran farmers were being forced to grow Indigo and were being tortured if they protested.
  • The farmers sought Gandhi's help and through a calculated non-violent protest, Gandhi managed to win concessions from the authority.​

Additional Information

  • Kheda Satyagraha (1918):
  • ​When Kheda, a village in Gujarat, was badly hit by floods, the local farmers appealed to the rulers to waive off the taxes.
  • Here, Gandhi started a signature campaign where peasants pledged non-payment of taxes.
  • ​Khilafat Movement
  • ​Gandhi's influence on the Muslim population was remarkable.
  • This was evident in his involvement in the Khilafat Movement.
  • After the first World War, the Muslims feared for the safety of their Caliph or religious leader and a worldwide protest was being organized to fight against the collapsing status of the Caliph.
  • Gandhi became a prominent spokesperson of the All India Muslim Conference and returned the medals he had received from the Empire during his Indian Ambulance Corps days in South Africa.
  • His role in the Khilafat made him a national leader in no time.
  • Non-cooperation Movement(1920):
  • ​Gandhi had realized that the British had been able to be in India only because of the cooperation they received from the Indians.
  • Keeping this in mind, he called for a non-cooperation movement.
  • The ominous day of the Jallianwala Bagh Massacre triggered the non-cooperation movement.
  • Gandhi set the goal of Swaraj or self-governance, which since then became the motto of the Indian freedom movement.
  • Salt March(1930):
  • ​Also known as the Dandi Movement, Gandhi's Salt March is considered to be a pivotal incident in the history of the freedom struggle.
  • Gandhi started a Satyagraha campaign against the salt tax in March 1930.
  • He marched 388 kilometers from Ahmedabad to Dandi in Gujarat to make salt.
  • Thousands of people joined him and made it one of the biggest marches in Indian history.
  • Quit India Movement(1942):
  • ​During the Second World War, Gandhi was determined to strike the British Empire with a definitive blow that would secure their exit from India.
  • This happened when the British started recruiting Indians for the war. Gandhi protested strongly and said that the Indians cannot be involved in a war that is in favour of democratic purposes when India itself is not a free country.
  • This argument exposed the two-faced image of the colonizers and within half a decade, they were out of this country.

Question 30

Who is the first ever Indian recipient of a Nobel Prize?

  1. AOption (a)

    Rabindranath Tagore

  2. BOption (b)

    Mother Teresa

  3. COption (c)

    Amartya Sen

  4. DOption (d)

    CV Raman

View answer & explanation

Correct answer

Answer: (a)A.

Rabindranath Tagore

Why this answer is correct

The correct answer is Rabindranath Tagore.

Key Points

  • Rabindranath Tagore was the first Indian citizen to be awarded and also the first Asian to be awarded.

Additional Information

Nobel prizePersonalities (the year they won)
PhysicsSubramanian Chandrashekhar(1983), Chandrashekhar Venkataraman(1930)*first ever in physics from India.
ChemistryVenkatraman Ramakrishnan(2009)*Indian origin
MedicineHargobind Khorana(1968)
PeaceMother Teresa(1979)
EconomicsAmartya Sen(1998)
LiteratureRabindranath Tagore(1913)*first ever across all fields from India

Question 31

The Dragon Endeavour is the spacecraft of which space company?

  1. AOption (a)

    Air Launch

  2. BOption (b)

    Space X

  3. COption (c)

    Kinet X

  4. DOption (d)

    Astra

View answer & explanation

Correct answer

Answer: (b)B.

Space X

Why this answer is correct

The correct answer is Space X.

Key Points

  • Behnken and Hurley left in a vehicle called Crew Dragon "Endeavour,".
  • A spaceship designed, built, and operated by SpaceX with about $2.7 billion in government funding.
  • It's SpaceX's second experimental flight of the vehicle, a mission called Demo-2.
  • And the aerospace company's first time putting people on board.
  • The Demo-2 mission's departure makes SpaceX, founded by Elon Musk in 2002.
  • The first company in NASA's Commercial Crew Program to successfully pull off a visit to the $150 billion, football field-size laboratory.
  • But Behnken and Hurley's return trip has just begun.
  • Now that they've flown away from the ISS, they must perform a risky descent through Earth's atmosphere.
  • The phase of the flight that Musk has called his "biggest concern."

Additional Information

  • About Space X
  • Founder: Elon Musk
  • Founded: 6 May 2002
  • CEO: Elon Musk (2002)
  • Headquarters: Hawthorne, California, United States
  • The Kinet-X experiment created green and violet clouds in the evening sky over parts of the U.S. East Coast.
  • The mission is also called the Kinetic-scale Energy and Momentum Transport Experiment.
  • That aims to better understand how energy and momentum transfer in phenomena like auroras, between "different regions of space that are magnetically connected," NASA said in a mission description.

Question 32

Which of the following organization has developed COVID-19 vaccine Covishield?

  1. AOption (a)

    Indian Medical Association

  2. BOption (b)

    Moderna

  3. COption (c)

    Serum Institute of India

  4. DOption (d)

    Bharat Biotech

View answer & explanation

Correct answer

Answer: (c)C.

Serum Institute of India

Why this answer is correct

The correct answer is Serum Institute of India.

Key Points

  • Serum Institute of India Pvt. Ltd. is now the world's largest vaccine manufacturer by a number of doses produced and sold globally (more than 1.5 billion doses) which includes Polio vaccine as well as Diphtheria, Tetanus, Pertussis, Hib, BCG, r-Hepatitis B, Measles, Mumps and Rubella vaccines.
  • Serum Institute of India begins production of Covovax vaccine.
  • The company is already locally producing the Covishield vaccine developed by Oxford University and AstraZeneca.(AP).
  • Serum Institute of India was founded in 1966 by Dr. Cyrus Poonawalla.
  • Its Headquarters is located in Hadapsar, Pune, Maharashtra.
  • CEO - Mr. Adar Poonawalla.

Additional Information

  • COVISHIELD VACCINE
  • The Oxford COVID-19 vaccine Covishield is manufactured by Serum Institute of India, which is the largest vaccine manufacturer in the world.
  • This Pune-based institute has collaborated with AstraZeneca to manufactureCovishield.
  • The vaccine is derived from a common cold virus called adenovirus and similar to coaxing, it is also to be given in two doses.

Question 33

Which of the following commission was appointed by the Central Government on Centre-State Relations in 1983?

  1. AOption (a)

    Setalvad Commission

  2. BOption (b)

    Sarkaria Commission

  3. COption (c)

    Kalelkar Commission

  4. DOption (d)

    Dutt Commission

View answer & explanation

Correct answer

Answer: (b)B.

Sarkaria Commission

Why this answer is correct

The correct answer is Sarkaria Commission.

Key Points

  • The Sarkaria Commission finally submitted its report in the year 1988.
  • The Sarkaria Commission’s charter was to examine the relationship and balance of power between state and central governments in the country.
  • And suggest changes within the framework of the Constitution of India.
  • Despite the large size of its reports the Commission recommended, by and large, status quo in the Centre-State relations.
  • Especially in the areas, relating to legislative matters, the role of Governors, and the use of Article 356.

Additional Information

  • Kalelkar Commission
  • A Backward Classes Commission was appointed by the President of India on 29 January 1953 with Kaka Kaielkar as its chairman.
  • The Commission was directed to determine the criteria to be adopted in considering whether any section of people should be treated socially and educationally backwards.
  • Dutt Committee
  • Dutt Committee or “Industrial Licensing Policy Inquiry Committee” was set up under the chairmanship of  Mr. Subimal Dutt as a follow up to the reports of the Monopolies Inquiry Commission and Hazari Committee.
  • This committee’s mandate was to inquire into the working of the licensing system in India. The committee had submitted its report in July 1969.

Question 34

Which of the following individuals composed the National Song "Vande Mataram"?

  1. AOption (a)

    Dr. Rajendra Prasad

  2. BOption (b)

    Lata Mangeshkar

  3. COption (c)

    Rabindranath Tagore

  4. DOption (d)

    Bankimchandra Chatterjee

View answer & explanation

Correct answer

Answer: (d)D.

Bankimchandra Chatterjee

Why this answer is correct

The correct answer is Bankimchandra Chatterjee.

Key PointsVande Mataram

  • The author of Vande Mataram is Bankim Chandra Chatterjee.
  • He wrote it in the 1870s, which was later included in his 1881 novel Anandamath.
  • It played a vital role in the Indian independence movement, first sung in a political context by Rabindranath Tagore at the 1896 session of the Indian National Congress.

Additional Information

Bankim Chandra Chatterjee or Bankim Chandra Chattopadhyay:

  • He was an Indian novelist, poet, and journalist.
  • He was the composer of Vande Mataram, originally in Sanskrit, personifying India as a mother goddess and inspiring activists during the Indian Independence Movement.

Question 35

What is the scientific name of national bird of India?

  1. AOption (a)

    Pavo Cristatus

  2. BOption (b)

    Peacock

  3. COption (c)

    Panthera Tigris

  4. DOption (d)

    Peak Cokoo

View answer & explanation

Correct answer

Answer: (a)A.

Pavo Cristatus

Why this answer is correct

The correct answer is 

Key Points

  • The National bird of India is Peacock and its scientific name is Pavo Cristatus.
  • Taxonomy of Peacock:
  • Kingdom: Animalia
  • Phylum: Chordata
  • Subphylum: Vertebrata
  • Superclass: Tetrapoda
  • Class: Aves
  • Order: Galliformes
  • Family: Phasianidae
  • Genus: Pavo
  • Species: cristatus

Additional Information

National SymbolNameScientific Name
National FruitMangoMangifera indica
National TreeIndian BanyanFicus bengalensis
National AnimalRoyal Bengal TigerPanthera tigris

Question 36

Which of the following refractive defects of vision is also known as Far-sightedness?

  1. AOption (a)

    Hypermetropia

  2. BOption (b)

    Myopia

  3. COption (c)

    Cataract

  4. DOption (d)

    Presbyopia

View answer & explanation

Correct answer

Answer: (a)A.

Hypermetropia

Why this answer is correct

The correct answer is Hypermetropia.

Key Points

  • A person with hypermetropia cannot see nearby objects clearly.
  • Hypermetropia, sometimes called hyperopia, is the term used to define being longsighted or far-sightedness.
  • If a person is hypermetropic, the image of a nearby object is formed behind the retina. This means that light is focused too far back in the eye, causing things that are close up to appear blurred.
  • A Convex lens is used to treat the defect known as hypermetropia.

Additional Information

Defects of VisionDetailsCorrections
Myopia or Near-sightednessThe eye can see nearby objects clearly but the distant objects appear indistinct.Concave Lens
Hypermetropia or FarsightednessThe human eye can see Distant objects clearly but cannot see nearby objects distinctly.Convex Lens
PresbyopiaPeople suffer from near-sightedness as well as farsightedness.bifocal lenses
CataractPeople suffer from the cloudiness of lens which causes a decrease in vision.Surgery

Question 37

Which state has approved the policy of reserving seats for women in the village level community institutions?

  1. AOption (a)

    Manipur

  2. BOption (b)

    Meghalaya

  3. COption (c)

    Mizoram

  4. DOption (d)

    Arunachal Pradesh

View answer & explanation

Correct answer

Answer: (b)B.

Meghalaya

Why this answer is correct

The correct answer is Meghalaya.

Key Points

  • Meghalaya state government has approved the “Policy on Reservation of seats for Women in Village Employment Councils (VECs)”.
  • The State Government has brought the policy of reserving at least 50 percent of seats for women in the grassroots village-level community institutions (VECs).
  • The policy entails, the position of the Secretary of VEC will be reserved for women members in all VECs on a rotational basis.
  • Adult women who are active community workers with leadership qualities can contest this post.
  • VECs are apolitical organizations formed for the implementation of MGNREGA works.

Additional Information

StateCapitalGovernorChief Minister
MeghalayaShilongSatya Pal MalikConrad Sangma
ManipurImphalLa. GanesanN. Biren Singh
MizoramAizawlKambhampati Hari BabuZoramthanga
Arunachal PradeshItanagarB. D. MishraPema Khandu

Question 38

Who can initiate the process of removal of the Vice President of India?

  1. AOption (a)

    President

  2. BOption (b)

    Cabinet

  3. COption (c)

    Lok Sabha

  4. DOption (d)

    Rajya Sabha

View answer & explanation

Correct answer

Answer: (d)D.

Rajya Sabha

Why this answer is correct

The correct answer is Rajya Sabha.

Key Points

  • The resolution for the removal of the Vice president of India can be moved to Rajya Sabha only.

Important Points

  • Vice President:
  • Article 63 states that "There shall be a Vice President of India."
  • He should have completed 35 years of age.
  • He should not be should be a member of any house but he should be qualified to be elected as a member of the Rajya Sabha.
  • He is elected indirectly by the principle of Proportional Representation by means of a Single Transferable Vote.
  • The electoral college consists of elected and nominated members of both Lok Sabha and Rajya Sabha.
  • He holds office for a term of 5 years i.e he is elected for a term of five years.
  • Indian Vice-President is an ex-officio chairman of Rajya Sabha.
  • He may resign his office by submitting his resignation to the President.

Question 39

Recently Mount Nyiragongo volcano has erupted. In which of the following countries is it situated?

  1. AOption (a)

    Democratic Republic of Congo

  2. BOption (b)

    Morocco

  3. COption (c)

    Madagascar

  4. DOption (d)

    Nigeria

View answer & explanation

Correct answer

Answer: (a)A.

Democratic Republic of Congo

Why this answer is correct

The correct answer is Democratic Republic of Congo.

Key Points

  • Mount Nyiragongo is an active stratovolcano in the Democratic Republic of Congo with an elevation of 3,470 m in the Virunga Mountains associated with the Albertine Rift.
  • As the active volcano in Congo, Mount Nyiragongo, erupted again, the Indian Army contingent under the United Nations peacekeeping mission (MONUSCO) assisted in protecting civilians and U.N. officials as well as assets during the evacuation.
  • It is located inside Virunga National Park, in the Democratic Republic of the Congo, about 12 km north of the town of Goma and Lake Kivu.
  • The main crater is about two kilometres (1mi) wide and usually contains a lava lake. Nyiragongo's lava lake has at times been the most voluminous known lava lake in recent history.
  • Nyiragongo and nearby Nyamuragira are together responsible for 40% of Africa's historical volcanic eruptions.
  • Lava lakes are large volumes of molten lava, usually basaltic, contained in a volcanic vent, crater, or broad depression.

Question 40

In which of the following year was the Industrial Policy Resolution adopted by the Parliament?

  1. AOption (a)

    1956

  2. BOption (b)

    1947

  3. COption (c)

    1950

  4. DOption (d)

    1991

View answer & explanation

Correct answer

Answer: (a)A.

1956

Why this answer is correct

The correct answer is 1956.

Key Points

  • Industrial Policy of 1948
  • The first industrial policy after independence was announced on 6th April 1948.
  • It was presented by Dr. Shyama Prasad Mukherjee the then Industry Minister.
  • The main goal of this policy was to accelerate the industrial development by introducing a mixed economy where the private and public sector was accepted as important in the development of the economy.
  • Industrial Policy Resolution, 1956
  • ​This second industrial policy was announced on April 20, 1956, which replaced the policy of 1948.
  • In 1956  the Indian Policy Resolution was adopted by the Parliament.
  • The features of this policy were:
  • A new classification of Industries.
  • Non-discriminatory and fair treatment for the private sector.
  • Promotion of village and small-scale industries.
  • To achieve development by removing regional disparity.
  • Labour welfare.
  • It was regarded as the “Economic Constitution of India” or “The Bible of State Capitalism”.

Arithmetic Ability (20 questions)

Question 41

Roli takes 3 hours to type 24 pages. Neelam takes 9 hours to type 45 pages. In how many hours they will type 351 pages together?

  1. AOption (a)

    15 hours

  2. BOption (b)

    18 hours

  3. COption (c)

    27 hours

  4. DOption (d)

    21 hours

View answer & explanation

Correct answer

Answer: (c)C.

27 hours

Why this answer is correct

Given:

Roli takes 3 hours to type 24 pages

Neelam takes 9 hours to type 45 pages

Calculation:

According to the question,

Roli's typing speed = 243\dfrac{24}{3} pages/hr = 8 pages/hr

Neelam's typing speed = 459\dfrac{45}{9} pages/hr = 5 pages/hr

So, both will type, 8 + 5 = 13 pages/hr

To type 391 pages together,

⇒ 39113\dfrac{391}{13} = 27 hours

∴ The required answer will 27 hours.

Question 42

Average of 45 numbers is 33. If average of first 15 numbers is 28 and average of last 25 numbers is 35, then what is the average of remaining five numbers?

  1. AOption (a)

    39

  2. BOption (b)

    38

  3. COption (c)

    37

  4. DOption (d)

    35

View answer & explanation

Correct answer

Answer: (b)B.

38

Why this answer is correct

Given:

Average of 45 numbers = 33

First 15 numbers average = 28

Last 25 numbers average = 35

Formula used:

Average of n numbers = Sum of total numbers/n

Solution:

Let the sum of the remaining numbers be S

Sum of total numbers = 45 × 33 = 1485

Sum of first 15 numbers = 15 × 28 = 420

Sum of last 25 numbers = 25 × 35 = 875

According to the question,

1485 = 420 + 875 + S

⇒ S = 190

Average of the remaining 5 numbers = 190/5 = 38 

∴  Average of the remaining 5 numbers is 38.

Question 43

If the one of the diagonal of a kite is 16 cm and its area is 96 cm2, what is the length of the other diagonal of the kite?

  1. AOption (a)

    14 cm

  2. BOption (b)

    18 cm

  3. COption (c)

    16 cm

  4. DOption (d)

    12 cm

View answer & explanation

Correct answer

Answer: (d)D.

12 cm

Why this answer is correct

Given:

Diagonal of the kite = 16 cm

Area of the kite = 96 cm2

Concept:

The kite takes the shape of a rhombus.

Formula used:

Area of a rhombus = d1×d22\dfrac{d1 × d2}{2}; here d1 and d2 are lengths of the diagonals

Calculation:

According to the question,

96 = 16×d22\dfrac{16 \times d2}{2}

⇒ d2 = 12

∴ Length of the other diagonal(d2) 12 cm.

Question 44

If 3(p + q) = 243 and 5(q - p) = 5, then find the value of (p × q)?

  1. AOption (a)

    12

  2. BOption (b)

    4

  3. COption (c)

    6

  4. DOption (d)

    3

View answer & explanation

Correct answer

Answer: (c)C.

6

Why this answer is correct

Given:

3(p + q) = 243

5(q - p) = 5

Formula used:

If am + n = ax, then x = m + n

Calculation:

3(p + q) = 243

⇒ 3(p + q) = 35

⇒ p + q = 5     ----eq.1

5(q - p) = 5

⇒ q - p = 1     ----eq. 2

By solving eq.1 and eq. 2

p = 2 and q = 3

∴ The value of (p × q) = 2 × 3 = 6.

Question 45

A, B and C are three partners in a business who shared profit in the ratio of 2 ∶ 5 ∶ 3. If they had invested for 8 months, 6 months and 9 months respectively, what is the ratio of their investments?

  1. AOption (a)

    3 ∶ 5 ∶ 4

  2. BOption (b)

    4 ∶ 5 ∶ 9

  3. COption (c)

    6 ∶ 10 ∶ 9

  4. DOption (d)

    3 ∶ 10 ∶ 4

View answer & explanation

Correct answer

Answer: (d)D.

3 ∶ 10 ∶ 4

Why this answer is correct

Given:

Ratio of the share of the profit of A, B and C (respectively) = 2 : 5 : 3

Ratio of the time period of the investments of A, B and C (in months) = 8 : 6 : 9

Formula used:

The ratio of the share of profits = Ratio of the product of investments and time period

Calculation:

Let the ratio of investments of A, B and C be in the ratio a : b : c

According to the question,

2 : 5 : 3 = 8a : 6b : 9c

⇒ 8a6b=25\dfrac{8a}{6b} = \dfrac{2}{5} , 6b9c=53\dfrac{6b}{9c} = \dfrac{5}{3}, 8a9c=23\dfrac{8a}{9c} = \dfrac{2}{3}

⇒ 40a = 12b, 18b = 45c, 24a = 18c

⇒ ab=310\dfrac{a}{b} = \dfrac{3}{10}     ---- eq.1

⇒ bc=52\dfrac{b}{c} = \dfrac{5}{2}     ----eq.2

⇒ ac=34\dfrac{a}{c} = \dfrac{3}{4}     ----eq.3

By multiplying eq.2 by 2; we get bc=104\dfrac{b}{c} = \dfrac{10}{4}    ----eq.4

Now by equations 1,3 and 4 we get, a : b : c = 3 ∶ 10 ∶ 4

∴ The ratio of their investments is 3 ∶ 10 ∶ 4.

Question 46

Prakash bought 2 items at the same price. He sold one item at 20 percent profit and another item at 15 percent profit. What is the overall percentage profit he made?

  1. AOption (a)

    17.5 percent

  2. BOption (b)

    16.5 percent

  3. COption (c)

    2.5 percent

  4. DOption (d)

    22.5 percent

View answer & explanation

Correct answer

Answer: (a)A.

17.5 percent

Why this answer is correct

Given:

The cost price of both products are equal

Profit on 1st item = 20%

Profit on 2nd item = 15%

Concept:

Profit is always calculated on the base of Cost price

Profit % = (Profit/Cost Price) × 100

Calculation:

Let the cost price of the items = 100

Then, according to the question,

Profit on 1st item = 20% of 100 = 20

Profit on 2nd item = 15% of 100 =15

⇒ Total profit on selling both the items = 35

Also, total cost price = 200

⇒ Total profit% = 35200\dfrac{35}{200} × 100 = 17.5%

Question 47

Study the following table carefully to answer the given question. Number of employees in different departments of five organisations

Organisation → Department ↓ABCDE
HR14580120180160
Finance12075100220140
Marketing15090115200190
IT225110160280220
Administration180120130110130

Number of employees in IT department of organization C is what percent of the total number of employees in organization C in all the departments together?

  1. AOption (a)

    25.6

  2. BOption (b)

    27.4

  3. COption (c)

    16.5

  4. DOption (d)

    22.4

View answer & explanation

Correct answer

Answer: (a)A.

25.6

Why this answer is correct

Given:

Number of employees in IT department of organization C = 160

Solution:

Total number of employees in organization C = 120 + 100 + 115 + 160 + 130 = 625

∴ The percent of the total number of employees in organization C in all the departments together = 160625×100\dfrac{160}{625} \times 100 = 25.6%

Question 48

Find the largest fraction from the following fractions-

  1. AOption (a)

    8846\dfrac{88}{46}

  2. BOption (b)

    895428\dfrac{895}{428}

  3. COption (c)

    313150\dfrac{313}{150}

  4. DOption (d)

    2312\dfrac{23}{12}

View answer & explanation

Correct answer

Answer: (b)B.

895428\dfrac{895}{428}

Why this answer is correct

Solution:

8846\dfrac{88}{46} ~ 1.91

895428\dfrac{895}{428} ~ 2.09

313150\dfrac{313}{150} ~ 2.08

2312\dfrac{23}{12} ~ 1.91

∴ The largest fraction = 895428\dfrac{895}{428}

Question 49

What value will come in place of question mark (?) in the following question?

80003+273−3433=?\sqrt[3]{8000} + \sqrt[3]{27} - \sqrt[3]{343} = ?

  1. AOption (a)

    9

  2. BOption (b)

    16

  3. COption (c)

    24

  4. DOption (d)

    23

View answer & explanation

Correct answer

Answer: (b)B.

16

Why this answer is correct

Given:

80003+273−3433=?\sqrt[3]{8000} + \sqrt[3]{27} - \sqrt[3]{343} = ?

Calculation:

80003\sqrt[3]{8000} = 20, 273\sqrt[3]{27} = 3, 3433\sqrt[3]{343} = 7

∴ 80003+273−3433\sqrt[3]{8000} + \sqrt[3]{27} - \sqrt[3]{343} = 20 + 3 - 7 = 16

∴ The required answer will be ​16.

Question 50

The following bar graph shows the sales (in Rs.crore) of six companies in 2012-13.

What approximate percentage of the total sales of the given six companies together was the sale of company T?

  1. AOption (a)

    22.3 percent

  2. BOption (b)

    17.4 percent

  3. COption (c)

    19.55 percent

  4. DOption (d)

    15 percent

View answer & explanation

Correct answer

Answer: (c)C.

19.55 percent

Why this answer is correct

Given:

Sale of company T (in Rs. crore) = 70.4

Calculation:

Total sales = 42.2 + 55.8 + 66.6 + 38.7 + 70.4 + 86.3 = 360

∴ Percentage of the total sales of the given six companies together was the sale of company T = 70.4360×100\dfrac{70.4}{360} \times 100 = ~ 19.55

Question 51

200 students took the examination in a school, and 13 percent of the students got distinction. The number of boys who got distinction is 2 more than the number of girls who got distinction. If 20 percent of the girls got distinction, find the total number of boys in the school?

  1. AOption (a)

    180

  2. BOption (b)

    120

  3. COption (c)

    140

  4. DOption (d)

    220

View answer & explanation

Correct answer

Answer: (c)C.

140

Why this answer is correct

Given:

Total number of student = 200

13% student got distinction.

20% girls got distinction.

Calculation:

The total number of students got distinction-

 ⇒200 × (13/100) = 26 students.

Let the number of girls who got distinction is x

Then according to question, number of boys who got distinction will be (x + 2)

⇒ x + (x + 2) = 26

⇒ 2x = 24

⇒ x = 12.

Since, 20% of girls get distinction = 12

⇒ Total girls = 12 × (100/20)

⇒ 60

Total boys = 200 - 60 = 140

∴ Option C is correct.

Question 52

Two cubes whose edge is 8 cm are joined together to form a single cuboid. What is the volume of the new cuboid so formed?

  1. AOption (a)

    1024 cm3

  2. BOption (b)

    1850 cm3

  3. COption (c)

    1950 cm3

  4. DOption (d)

    2150 cm3

View answer & explanation

Correct answer

Answer: (a)A.

1024 cm3

Why this answer is correct

Given:

Length of edge of cube = 8 cm.

2 cube are join to mad cuboid.

Formula used:

Volume of cuboid = l × b × h

Where, l = length, b = breadth, h = height

Calculation:

According to the question,

After the joining of two cubes the new dimensions will be-

Length = 8 + 8 = 16 cm

breadth = 8 cm

height = 8 cm

Volume of the new cuboid so formed = 16 × 8 × 8

⇒ 1024 cm³

∴ Option A is correct.

Question 53

LCM of two numbers is 3 times their HCF. The sum of the LCM and the HCF is 44. If one number is 11, then the other number is:

  1. AOption (a)

    44

  2. BOption (b)

    22

  3. COption (c)

    33

  4. DOption (d)

    20

View answer & explanation

Correct answer

Answer: (c)C.

33

Why this answer is correct

Given:

LCM = 3 times of HCF.

LCM + HCF = 44

1st number = 11.

Formula used:

LCM × HCF = 1st number × 2nd number

Calculation:

Let HCF be x, LCM be 3x

According to question:

⇒ 3x + x = 44

⇒ 4x = 44

HCF = x = 11, 

LCM = 3x = 33.

Applying, LCM × HCF = 1st number × 2nd number

⇒ 11 × 33 = 11 × 2nd number

⇒ 2nd number = 33

∴ Option C is correct.

Question 54

A sum of Rs.3600 is divided into two principals. One principal is invested at 8 percent p.a. for 2 years and the other is invested at 9 percent p.a. for 3 years. If the total simple interest obtained is Rs.741, then find the larger principal.

  1. AOption (a)

    Rs.1500

  2. BOption (b)

    Rs.2100

  3. COption (c)

    Rs.2400

  4. DOption (d)

    Rs.1800

View answer & explanation

Correct answer

Answer: (b)B.

Rs.2100

Why this answer is correct

Given:

3600 is divided into two pats.

1st principal = 8% p.a for 2 years.

2nd Principal = 9% p.a for 3 years.

Total S.I = 741.

Formula used:

Simple Interest = (Principal × Rate × Time)/100

Calculation:

Let the amount invested at 8% p.a be x

Then the amount invested at 9% p.a = (3600 - x)

According the question

⇒ (x × 2 × 8)/100 + [(3600 - x) × 9 × 3]/100

⇒ 0.16x + 0.27(3600 - x) = 741

⇒ x = Rs. 2100

∴ Option B is correct.

Question 55

The ratio of speed of boat in upstream to speed of boat in downstream is 2 ∶ 3. The time taken by boat to cover 45 km in downstream is t hours, then in t hours how many kilometers will be travelled by boat in upstream?

  1. AOption (a)

    50 km

  2. BOption (b)

    15 km

  3. COption (c)

    30 km

  4. DOption (d)

    40 km

View answer & explanation

Correct answer

Answer: (c)C.

30 km

Why this answer is correct

Given:

The ratio of Upstream to downstream = 2:3.

"t" hours required to cover 45 km in downstream.

Concept used:

Downstream = m + c

Upstream = m - c.

where m = Speed of boat in still water.

C = Speed of current of water.

Formula used:

Speed = Distance/Time

Calculation:

Using componendo and Dividendo rule,

⇒ (m+c)/(m-c) = 3 : 2

⇒ m/c = 5/1

Let x be the common term.

For downstream = 6x, for upstream = 4x

Time taken to cover 45km in downstream

⇒ t = 45/6x

Distance travelled by boat in upstream

⇒ d = 4x × (45/6x)

⇒ 30 km

∴ Option C is correct.

Question 56

An alloy contains tin, copper and zinc in the ratio of 4 : 6 : 5. Find the quantity of copper to be added to 120 kg of this alloy to form a new alloy in which the ratio of the above elements in the same order is 4 ∶ 7 ∶ 5?

  1. AOption (a)

    16 kg

  2. BOption (b)

    20 kg

  3. COption (c)

    10 kg

  4. DOption (d)

    8 kg

View answer & explanation

Correct answer

Answer: (d)D.

8 kg

Why this answer is correct

Given:

The ratio of tin, copper and zinc in the ratio of 4:6:5 respectively. 

Total quantity of Alloy = 120 kg.

Calculation:

Tin = (4/15) × 120 = 32 kg.

Copper = (6/15) × 120 = 48 kg.

Zinc = (5/15) × 120 = 40 kg.

If we add "x" kg of copper to the alloy then it's ratio = 4 : 7 : 5

⇒ (New quantity of copper)/(New quantity of Alloy) = 7/(4 + 7 + 5)

⇒ (48 + x)/(120 + x) = 7/16.

⇒ 16x + 768 = 7x + 840

⇒ 9x = 72

⇒ x = 8 kg

∴ Option D is correct.

Question 57

What will come in the place of question mark (?) in the given expression?

(420 ÷ 12 × 35 + 452 - 152) = ?2

  1. AOption (a)

    55

  2. BOption (b)

    35

  3. COption (c)

    25

  4. DOption (d)

    45

View answer & explanation

Correct answer

Answer: (a)A.

55

Why this answer is correct

Concept Used : 

BODMAS rule is followed as per the below table:

Calculation:

(420 ÷ 12 × 35 + 452 - 152) = ?2

Let the unknown number be x.

(420 ÷ 12 × 35 + 452 - 152) = x2

⇒ (35 × 35 + 452 - 152) = x2

⇒ (35 × 35 + 45 × 45 - 15 × 15) = x2

⇒ 1225 + 2025 - 225 = x2

⇒ 3250 - 225 = x2

⇒ x² = 3025

⇒ x = 55

∴ Option A is correct.

Question 58

Study the following table carefully to answer the given question.

Number of employees in different departments of five organisations

Organisation → Department ↓ABCDE
HR14580120180160
Finance12075100220140
Marketing15090115200190
IT225110160280220
Administration180120130110130

What is the ratio between the number of employees from Finance and Marketing departments together of organization B and these two departments together of organisation D respectively?

  1. AOption (a)

    9 ∶ 11 

  2. BOption (b)

    11 ∶ 18

  3. COption (c)

    18 ∶ 11

  4. DOption (d)

    11 ∶ 28

View answer & explanation

Correct answer

Answer: (d)D.

11 ∶ 28

Why this answer is correct

Given:

From the table,

Number of employees from Finance departments of organization B = 75

Number of employees from  Marketing departments of organization B = 90.

Number of employees from Finance departments of organization D = 220.

Number of employees from  Marketing departments of organization D = 200.

Calculation:

⇒ 75 + 90 : 220 + 200

⇒ 165 : 420

⇒  11 : 28

∴ Option D is correct.

Question 59

If Arvind reaches the college in 2 hours by his bike at an average speed of 30 km/h, then at what average speed of his car can he reach the college half an hour earlier?

  1. AOption (a)

    40 km/h

  2. BOption (b)

    36 km/h

  3. COption (c)

    45 km/h

  4. DOption (d)

    35 km/h

View answer & explanation

Correct answer

Answer: (a)A.

40 km/h

Why this answer is correct

Given:

Average speed = 30 km/hr, then he reaches college in 2 hrs.

Formula used:

Distance = Speed × Time

Calculation:

Distance = 30 × 2 = 60 km

If he went by car, it would take 1.5 hrs

Average speed = 60/1.5

⇒ 40 km/h.

∴ Option A is correct.

Question 60

The price of a shirt is decreased by 25 percent and hence the sales increase by 20 percent. What will be the effect on the total revenue of the shop?

  1. AOption (a)

    10 percent decrease

  2. BOption (b)

    10 percent increase

  3. COption (c)

    5 percent increase

  4. DOption (d)

    no effect

View answer & explanation

Correct answer

Answer: (a)A.

10 percent decrease

Why this answer is correct

Given:

Reduction of shirt by 25%

Increment of selling by 20%

Calculation:

Let the initial cost of shirt = 100

Final cost = 75.

Let the initial quantity of shirt = 100

Final quantity = 120

Revenue = cost per piece × no. of piece

⇒ 100 × 100 : 75 × 120

⇒ 10000 : 9000

⇒ Reduction of 1000 on 10000

⇒ (1000/10000) × 100

⇒ 10% Reduction.

∴ Option A is correct.

General English (20 questions)

Question 61

Rearrange the following sentences in their correct order to form a meaningful paragraph.

P. There are so many reasons to include veggies into your diet.

Q. Low in calories, nutrient-dense, versatile and tasty.

R. When we use the term healthy, vegetable is the first word that comes to our mind.

S. Filling your plate with healthy and nutritious food items is an effective way to lose weight.

  1. AOption (a)

    RQPS

  2. BOption (b)

    QPSR

  3. COption (c)

    PSRQ

  4. DOption (d)

    SRQP

View answer & explanation

Correct answer

Answer: (d)D.

SRQP

Why this answer is correct

The correct answer is 'SRQP'.

Key Points

  • Statement S will be the first in sequence as it introduces the subject of the sentence 'Healthy and Nutritious food'.
  • Statement R will follow Statement S as it states what comes to our mind when we mention 'healthy' which is mentioned in statement S.
  • Statement Q will follow statement R as it states the reason why vegetables are healthy.
  • Statement P will be the last part.

Therefore, the correct sequence is SRQP.

  • Sequence:
  • S. Filling your plate with healthy and nutritious food items is an effective way to lose weight.
  • R. When we use the term healthy, vegetable is the first word that comes to our mind.
  • Q. Low in calories, nutrient-dense, versatile and tasty.
  • P. There are so many reasons to include veggies into your diet.

Question 62

Select the most appropriate option to fill in the blank.

She slipped _______ damp grass, breaking a leg, cracking a bone in the other and hurting her shoulder.

  1. AOption (a)

    by

  2. BOption (b)

    on

  3. COption (c)

    to

  4. DOption (d)

    at

View answer & explanation

Correct answer

Answer: (b)B.

on

Why this answer is correct

The correct answer is 'on'.

Key Points

  • Preposition:  It is a word used to link nouns, pronouns, or phrases to other words within a sentence. Most common prepositions are 'of', 'in', 'for', 'on', etc.
  • On: As a preposition 'On' is used to express touching a surface or being supported by a surface.
  • Example: We could hang this picture on the wall next to the door.
  • Thus, 'on' will be used in the blank as the sentence is expressing the contact of a person with the grass.

Therefore, the correct answer is: 'She slipped on damp grass, breaking a leg, cracking a bone in the other and hurting her shoulder.'

Additional Information 

  • Let's explore other options:
  • By: As a preposition 'By' is used to show the person or thing that does something.
  • Example: The motorcycle was driven by a tiny bald man.
  • To: As a preposition 'To' is used after some verbs, especially when the action described in the infinitive will happen later.
  • Example: She agreed to help.
  • At: As a preposition 'At' is used to show an exact position or particular place.
  • Example: We'll meet you at the entrance.

Question 63

Select the most appropriate antonym of the given word.

Ostentatious

  1. AOption (a)

    Flamboyance

  2. BOption (b)

    Innocence

  3. COption (c)

    Modest

  4. DOption (d)

    Purity

View answer & explanation

Correct answer

Answer: (c)C.

Modest

Why this answer is correct

The correct answer is 'Modest'.

Key Points 

  • Let's explore the meaning of the marked word and the given word.
  • Ostentatious: characterized by vulgar or pretentious display; designed to impress or attract notice.
  • Example: They criticized the ostentatious lifestyle of their leaders.
  • Modest: not large in size or amount, or not expensive.
  • Example: They live in a fairly modest house, considering their wealth.
  • From above, we can refer that the correct answer is option 3.

 Additional Information 

  • Flamboyance: the quality of being bright, colorful, and very noticeable.
  • Example: This classic Spanish revival building is unique in its elegance and flamboyance.
  • Innocence: the fact that someone is not guilty of a crime.
  • Example: She pleaded her innocence, but no one believed her.
  • Purity: the quality or state of not being mixed with anything else.
  • Example: The purity of the city’s water is checked regularly at its reservoirs.

Question 64

Identify the segment in the sentence which contains the grammatical error from the given options.

If he had wrote the examination faster and answered one more question he would have scored better.

  1. AOption (a)

    and answered one more question

  2. BOption (b)

    No error

  3. COption (c)

    he would have scored better

  4. DOption (d)

    If he had wrote the examination faster

View answer & explanation

Correct answer

Answer: (d)D.

If he had wrote the examination faster

Why this answer is correct

The correct answer is 'If he had wrote the examination faster'  i.e. error lies in this part of the sentence.

Key Points

  • The error lies in 'If he had wrote the examination faster' part of the sentence.
  • Wrote needs to be replaced with written..
  • This sentence is an example of the Third conditional sentence. These sentences express situations that can not exist. The actions and events that happened in the past. These are often used to indicate a missed opportunities.
  • Structure for Third conditional:
  • If/when+ past perfect+would have+ past participle.

​The correct sentence is- If he had written the examination faster and answered one more question he would have scored better. ​

Additional Information  

  • Conditional sentence:- As the name suggests, these sentences express conditions.
  • In conditional sentences, there are two clauses of the sentences. One is a conditional clause and another one is main clause.
  • Conditional sentences are of four types-
Types of Conditional SentencesDefinition and examples
Zero conditional SentenceIt expresses a factual condition.
Structure: If/When+ present simple+ present simple
Example:If you touch a fire, you get burned.
First Conditional SentenceIt expresses possible and likely future outcomes.
Structure: If/when+present simple, will+ verb infinitive.
Example: If I have enough money, I'll buy some new shoes.
Second Conditional SentenceIt expresses an Imagination.
Structure: If/when + past simple, would + verb infinitive.
Example: If I met the Queen of England, I would say hello.
Third Conditional sentenceIt expresses a hypothetical situation.
Structure: If/when+ past perfect+would have+ past participle.
Example: If we had taken a taxi, we wouldn't have missed the plane.

Question 65

Select the word which means the same as the group of words given.

Transmission that combines media of communication:

  1. AOption (a)

    Multimedia

  2. BOption (b)

    Multitasking

  3. COption (c)

    Multiple

  4. DOption (d)

    Multimillionaire

View answer & explanation

Correct answer

Answer: (a)A.

Multimedia

Why this answer is correct

The correct answer is 'Multimedia'.

Key Points 

  • Multimedia: using a combination of moving and still pictures, sound, music, and words, especially in computers or entertainment.
  • Example: For three days, top artists from the worlds of advanced music and multimedia showcase their art.
  • Thus, from above, the correct answer is option 1.

Additional Information 

 

  • Multitasking: a person's or product's ability to do more than one thing at a time.
  • Example: To succeed, you will need strong multitasking skills.
  • Multiple: consisting of or involving many things or types of things.
  • Example: Make multiple copies of the report.
  • Multimillionaire:  a person who has money and property worth several million dollars, pounds, euros, etc.
  • Example: The dotcom phenomenon made multimillionaires of many internet entrepreneurs.

Question 66

Select the most appropriate synonym of the given word.

AMBIVALENCE

  1. AOption (a)

    Puzzle

  2. BOption (b)

    Intensity

  3. COption (c)

    Despair

  4. DOption (d)

    Violence

View answer & explanation

Correct answer

Answer: (a)A.

Puzzle

Why this answer is correct

The correct answer is 'Puzzle'.

Key Points

 

  • Let's explore the meaning of the marked word and the given word.
  • Ambivalence: the state of having mixed feelings or contradictory ideas about something or someone
  • Example: Real social relationships are marked by ambiguity and ambivalence.
  • Puzzle: cause (someone) to feel confused because they cannot understand or make sense of something.
  • Example: The findings of the survey puzzle me - they're not at all what I would have expected.
  • From above, we can refer that the correct answer is option 1.

Additional Information 

 

  • Intensity: the quality of being felt strongly or having a very strong effect.
  • Example: The explosion was of such intensity that it was heard five miles away.
  • Despair: the feeling that there is no hope and that you can do nothing to improve a difficult or worrying situation.
  • Example: Their fourth year without rain drove many farmers to despair.
  • Violence: behavior involving physical force intended to hurt, damage, or kill someone or something.
  • Example: It seems that the attack was a gratuitous/random/mindless act of violence.

Question 67

Select the most appropriate meaning of the given idiom/ phrase.

To face the music

  1. AOption (a)

    To bear the consequences

  2. BOption (b)

    To be offered warm hospitality

  3. COption (c)

    To be greeted rudely

  4. DOption (d)

    To enjoy a music programme

View answer & explanation

Correct answer

Answer: (a)A.

To bear the consequences

Why this answer is correct

The correct answer is  'To bear the consequences​.'

Key Points 

  • To face the music: It is an idiomatic phrase which means to meet the consequences of one’s bad behavior, mistakes, and the like.
  • Example:  Rather than face the music at a trial, Abingdon chose to plea bargain.
  • Thus, from above we can say that the correct answer is option 1.

Question 68

Select the most appropriate option to fill in the blank.

We all are well aware that to _______ hygiene, we have to keep ourselves and our surroundings clean.

  1. AOption (a)

    renew

  2. BOption (b)

    maintain

  3. COption (c)

    conserve

  4. DOption (d)

    nurture

View answer & explanation

Correct answer

Answer: (b)B.

maintain

Why this answer is correct

The correct answer is 'maintain'.

Key Points

  • Let's explore the meaning of the given options:
WordMeaningExample
renewto begin doing something again after a period of not doing it.Local people have renewed their efforts to save the school.
maintainto make something continue in the same way or at the same standard as beforeCareers Officers maintain contact with young people when they have left school.
conserveto keep and protect something from damage, change, or waste.To conserve electricity, we are cutting down on our heating.
nurtureto help a plan or a person to develop and be successfulAs a record company executive, his job is to nurture young talent.

 

  • Thus, from the above-given explanation, we can say that 'maintain' will be used in the blank.
  • Therefore, the correct answer is option 2.

Question 69

In the following question, four sentences are given out of which three sentences are grammatically incorrect while one is correct. Find out which sentence is grammatically correct and select the appropriate option.

  1. AOption (a)

    The portal should allow uploaded of state government Acts, regulations and subordinate legislations as well.

  2. BOption (b)

    Navigation on this portal should be such that the complete law-making chain, right from the parent act to the subordinate legislations, is clearly visible.

  3. COption (c)

    The data uploaded on the portal should available in machine readable PDF formats.

  4. DOption (d)

    All Central Acts and subordinate legislations passed by the Centre, include rules, regulations, and circulars, should be made available on this portal.

View answer & explanation

Correct answer

Answer: (b)B.

Navigation on this portal should be such that the complete law-making chain, right from the parent act to the subordinate legislations, is clearly visible.

Why this answer is correct

The correct answer is option 2.

Key Points

  • Let's explore the options:
  • Option 1, It is grammatically incorrect because the verb 'allow' is succeded by a gerund or infinitive in a sentence. If there is no object then 'allow' will be followed by a gerund and if there is an object then 'allow' will be followed by an infinitive. In the given sentence there is no object therefore, it needs to be succeeded by a gerund. therefore, 'uploaded' needs to be replaced with 'uploading' to make the sentence grammatically correct.
  • Option 2, This statement is grammatically and contextually correct.
  • Option 3, it is grammatically incorrect because 'should' needs to be succeeded by 'be'  when it is used to state your opinion that you think ought to happen.Thus, 'be' needs to be added before 'available' to make the sentence grammatically correct.
  • Option 4, It is grammatically incorrect because we need a preposition to connect 'centre' with 'rules' and 'including' is a preposition which means containing as part of the whole being considered. Therefore, to correct this sentence 'include' needs to be replaced with 'including'.
  • Thus, the correct answer is option 2.

Question 70

Select the correctly spelt word

  1. AOption (a)

    Foerrunner

  2. BOption (b)

    Forarunner

  3. COption (c)

    Forrunner

  4. DOption (d)

    Forerunner

View answer & explanation

Correct answer

Answer: (d)D.

Forerunner

Why this answer is correct

The correct answer is 'Forerunner'.

Key Points

 

  • Forerunner: something or someone that acts as an early and less advanced model for what will appear in the future, or a warning or sign of what is to follow.
  • Example: Germany's Green party was said to be the forerunner of environmental parties throughout Europe.
  • Thus, from above, we can refer that the correct answer is option 4.

Question 71

Select the most appropriate option to substitute the underlined segment in the given sentence. If there is no need to substitute it, select 'No substitution required'.

The medals from the two events will be counted for "ranking of competing nations" at the Birmingham Games, the CGF said on Monday.

  1. AOption (a)

    No substitution required

  2. BOption (b)

    be counted

  3. COption (c)

    will be count

  4. DOption (d)

    will count

View answer & explanation

Correct answer

Answer: (a)A.

No substitution required

Why this answer is correct

The correct answer is 'No substitution required'.

Key Points

  • The given sentence required no substitution and is grammatically correct.
  • The given sentence is an example of passive voice of simple future tense.
  • Structure:
  • Active sentence- subject + shall / will + base form of verb + object.
  • Passive sentence- object of the active sentence + shall / will+ be + past participle ( third form ) of the verb + by + subject of the active sentence.
  • Example:
  • She will draw a picture today. (Active Voice)
  • A picture will be drawn by her today. (Passive voice)
  • Thus, from above, the correct answer is option 1 and the sentence requires no substitution.

Additional Information

  • Active voice means that a sentence has a subject that acts upon its verb.
  • Passive voice means that a subject is a recipient of a verb’s action.

Question 72

Rearrange the following sentences in their correct order to form a meaningful paragraph.

P: The government of Maharashtra has come up with guidelines for organising of Ganesh Festival.

Q: If organizers need to have more than 20 people, then they need to take permission from the local District Collector.

R: At any given time, no more than 20 people will be allowed to gather in the sanctum within the pandal where the Ganesh idol is kept for worshipping.

S: The District Collector will then verify the size of the pandal and will give permission if satisfied that social distancing will be maintained with 20 plus people.

  1. AOption (a)

    QPSR

  2. BOption (b)

    SRPQ

  3. COption (c)

    RQPS

  4. DOption (d)

    PRQS

View answer & explanation

Correct answer

Answer: (d)D.

PRQS

Why this answer is correct

The correct answer is 'PRQS'.

Key Points

  • Statement P  will be the first in sequence as it introduces the subject of the sentence 'The government of Maharashtra'.
  • Statement R will follow Statement P as it states the conditions of the gathering of people for celebrating the 'Ganesh festival' mentioned in statement P.
  • Statement Q will follow statement R as it states the conditions for the gathering of people for more than '20 people' which is mentioned in statement R.
  • Statement S will be the last part.

Therefore, the correct sequence is PRQS.

  • Sequence:
  • P. The government of Maharashtra has come up with guidelines for organising of Ganesh Festival.
  • R: At any given time, no more than 20 people will be allowed to gather in the sanctum within the pandal where the Ganesh idol is kept for worshipping.
  • Q: If organizers need to have more than 20 people, then they need to take permission from the local District Collector.
  • S. The District Collector will then verify the size of the pandal and will give permission if satisfied that social distancing will be maintained with 20 plus people.

Question 73

Select the most appropriate meaning of the given idiom/phrase.

To burn your finger

  1. AOption (a)

    Things always get worse before they get better

  2. BOption (b)

    To experience unpleasant consequences

  3. COption (c)

    Excellent health

  4. DOption (d)

    In good condition espc. Getting a lot of money

View answer & explanation

Correct answer

Answer: (b)B.

To experience unpleasant consequences

Why this answer is correct

The correct answer is 'To experience unpleasant consequences.'

Key Points 

  • To burn your finger​: It is an idiomatic phrase which means To suffer an unpleasant or ruinous consequence (especially the loss of money) for some action, often such that one becomes unwilling or reluctant to do it again.
  • Example: During the boom times, countless would-be investors put everything they owned into shady investment deals, and most of them ended up burning their fingers badly.
  • Thus, from above we can say that the correct answer is option 2.

Question 74

Select the most appropriate option to substitute the underlined segment in the given sentence. If there is no need to substitute it, select 'No substitution required'.

Since 1970, trends in agricultural production, fish harvest, bio-energy production and harvest of materials have increased, in response to populational growth, rising demand and technology development.

  1. AOption (a)

    No substitution required.

  2. BOption (b)

    population growth, rising demand and technological development.

  3. COption (c)

    populated growth, rising demand and technological development.

  4. DOption (d)

    population growing, rising demand and technological development.

View answer & explanation

Correct answer

Answer: (b)B.

population growth, rising demand and technological development.

Why this answer is correct

The correct answer is 'population growing, rising demand and technological development.'

Key Points

  • There is no word such as 'populational', thus it is needed to be replaced with 'population'.
  • 'Technology' needs to be replaced with 'Technological'.
  • Adjective: These are the words that are used to describe or qualify a noun.
  • Here, in the given sentence 'development' is a noun and 'technology' is used to qualify the noun 'development' which is incorrect.
  • Thus, the correct answer is option 2.

Therefore, the correct sentence is: 'Since 1970, trends in agricultural production, fish harvest, bio-energy production and harvest of materials have increased, in response to **population growth, rising demand and technological development.'**

Question 75

Select the wrongly spelt word.

  1. AOption (a)

    Gaurmet

  2. BOption (b)

    Grandeur

  3. COption (c)

    Grateful

  4. DOption (d)

    Guarantee

View answer & explanation

Correct answer

Answer: (a)A.

Gaurmet

Why this answer is correct

The correct answer is 'Gaurmet', i.e. it is a wrongly spelt word.

Key Points

  • Gaurmet: It is a wrongly spelt word. The correct spelling is 'Gourmet'.
  • Gourmet: a person who knows a lot about food and cooking, and who enjoys eating high-quality food.
  • Example: In an ideal world, most people would prefer gourmet cuisine, but it is costly and difficult to prepare.
  • Thus, from the above-given explanation, we can say that the correct answer is option 1.

Additional Information 

 

  • Grandeur: the quality of being very large and special or beautiful.
  • Example: We were struck by the silent grandeur of the desert.
  • Grateful: showing or expressing thanks, especially to another person.
  • Example: If you could get that report finished by Thursday, I'd be very grateful.
  • Guarantee: to promise that something will happen or exist.
  • Example: European Airlines guarantees its customers top-quality service.

Direction: Read the following passage and answer the questions below.

The tribunes of Rome, Marullus and Flavius, break up a gathering of citizens who want to celebrate Julius Caesar's triumphant return from war. The victory is marked by public game sin which Caesar's protégé, Mark Antony, takes part. On his way to the arena, Caesar is stopped by a stranger who warns him that he should 'Beware the Ides [15th] of March.'

Fellow senators, Caius Cassius and Marcus Brutus, are suspicious of Caesar's reactions to the power he holds in the Republic. They fear he will accept offers to become Emperor. He has been gaining a lot of power recently and people treat him like a god. Cassius, a successful general himself, is jealous of Caesar. Brutus has a more balanced view of the political position. The conspirator Casca enters and tells Brutus of a ceremony held by the plebeians. They offered Caesar a crown three times, and he refused it every time. But the conspirators are still wary of his aspirations.

Cassius, Casca, and their allies plant false documents to manipulate Brutus to join their cause to remove Caesar. After doing so, they visit Brutus at night in his home to persuade him of their views. There they plan Caesar's death. Brutus is troubled but refuses to confide in his devoted wife, Portia. On 15 March, Caesar's wife, Calpurnia, urges him not to go to the Senate. She has had visionary dreams and fears the portents of the overnight storms.

Caesar is nevertheless persuaded by flattery to go to the Capitol. At the Capitol, he is stabbed by each conspirator in turn. As Brutus gives the final blow, Caesar utters the famous phrase: Et tu, Brute?

Question 76

Whom does this line “He has been gaining a lot of power recently and people treat him like a god” refer to according to the given passage:

  1. AOption (a)

    Calpurnia

  2. BOption (b)

    Julius Caesar

  3. COption (c)

    Caius Cassius

  4. DOption (d)

    Marcus Brutus

View answer & explanation

Correct answer

Answer: (b)B.

Julius Caesar

Why this answer is correct

The correct answer is 'Julius Caesar'.

Key Points

  • Let's refer to the following lines of the passage:
  • Fellow senators, Caius Cassius and Marcus Brutus, are suspicious of Caesar's reactions to the power he holds in the Republic. They fear he will accept offers to become Emperor. He has been gaining a lot of power recently and people treat him like a god. Cassius, a successful general himself, is jealous of Caesar. Brutus has a more balanced view of the political position.
  • From, above we can refer that the correct answer is option 2.

Question 77

Which of the following statements is incorrect according to the given passage?

  1. AOption (a)

    Caesar utters the famous phrase: Et tu, Brute?

  2. BOption (b)

    They offered Caesar a crown three times, and he refused it every time.

  3. COption (c)

    At the Capitol, Cassius is stabbed by each conspirator in turn.

  4. DOption (d)

    Caesar is nevertheless persuaded by flattery to go to the Capitol.

View answer & explanation

Correct answer

Answer: (c)C.

At the Capitol, Cassius is stabbed by each conspirator in turn.

Why this answer is correct

The correct answer is 'At the Capitol, Cassius is stabbed by each conspirator in turn.' i.e. it is an incorrect statement.

Key Points

  • Let's refer to the following lines of the passage:
  • Caesar is nevertheless persuaded by flattery to go to the Capitol. At the Capitol, he is stabbed by each conspirator in turn.
  • As Brutus gives the final blow, Caesar utters the famous phrase: Et tu, Brute?
  • They offered Caesar a crown three times, and he refused it every time. But the conspirators are still wary of his aspirations.
  • Thus, from above we can refer that the correct answer is option 3. As it wasn't Cassius but Caesar who was stabbed by each conspirator in turn.

Question 78

Why did the conspirators plant false documents?

  1. AOption (a)

    to manipulate Caesar to join their cause to remove Brutus

  2. BOption (b)

    to manipulate Portia to join their cause to remove Brutus

  3. COption (c)

    to manipulate Cassius to join their cause to remove Brutus

  4. DOption (d)

    to manipulate Brutus to join their cause to remove Caesar

View answer & explanation

Correct answer

Answer: (d)D.

to manipulate Brutus to join their cause to remove Caesar

Why this answer is correct

The correct answer is 'to manipulate Brutus to join their cause to remove Caesar'.

Key Points

  • Let's refer to the following lines of the passage:
  • Cassius, Casca, and their allies plant false documents to manipulate Brutus to join their cause to remove Caesar.
  • Thus, from above we can refer that the correct answer is option 4.

Question 79

According to the given passage, Portia has been portrayed as-

  1. AOption (a)

    Brutus’ wife

  2. BOption (b)

    Casca’s mother

  3. COption (c)

    Calpurnia’s sister

  4. DOption (d)

    Cassius’ daughter

View answer & explanation

Correct answer

Answer: (a)A.

Brutus’ wife

Why this answer is correct

The correct answer is 'Brutus' wife'.

Key Points

  • Let's refer t the following lines of the passage:
  • Brutus is troubled but refuses to confide in his devoted wife, Portia.
  • Thus, the correct answer is option 1.

Question 80

Why did Caesar's wife, Calpurnia, urge him not to go to the Senate?

  1. AOption (a)

    because she had a visionary dream of something bad to happen

  2. BOption (b)

    because an stranger told her to be killed by some senators

  3. COption (c)

    because she was told to do so by the two senators Brutus and Cassius

  4. DOption (d)

    because she was not much concerned about Caesar’s prestige

View answer & explanation

Correct answer

Answer: (a)A.

because she had a visionary dream of something bad to happen

Why this answer is correct

The correct answer is 'because she had a visionary dream of something bad to happen'.

Key Points

  • Let's referar's to the following lines of the passage:
  • On 15 March, Caes wife, Calpurnia, urges him not to go to the Senate. She has had visionary dreams and fears the portents of the overnight storms.
  • Thus, from above, we can refer that the correct answer is option 1.

General Hindi (20 questions)

Question 81

नीचे दिए गए वाक्य में निम्न में से किस चिन्ह का प्रयोग नहीं किया गया है?

जो क्रियाएँ संज्ञा या विशेषण बनती हैं, उन्हें "नामधातु क्रिया" कहते हैं।

  1. AOption (a)

    योजक चिन्ह

  2. BOption (b)

    उद्धरण चिन्ह 

  3. COption (c)

    अल्पविराम चिन्ह

  4. DOption (d)

    पूर्णविराम चिन्ह

View answer & explanation

Correct answer

Answer: (a)A.

योजक चिन्ह

Why this answer is correct

जो क्रियाएँ संज्ञा या विशेषण बनती हैं, उन्हें "नामधातु क्रिया" कहते हैं। - इस वाक्य में योजक चिन्ह का प्रयोग नही किया गया है। 

स्पष्टीकरण: 

योजक चिन्ह- हिंदी में अल्पविराम के बाद योजक चिन्ह का प्रयोग अधिक होता है। दो शब्दों में परस्पर संबंध स्पष्ट करने के लिए तथा उन्हें जोड़कर लिखने के लिए योजक-चिन्ह का प्रयोग किया जाता है। इसे 'विभाजक-चिन्ह' भी कहते है।

Key Points

अन्य विकल्प:  

चिन्हपरिभाषाउदाहरण
पूर्ण विराम चिन्ह (।)जहाँ एक बात पूरी हो जाये या वाक्य समाप्त हो जाये वहाँ पूर्ण विराम ( । ) चिन्ह लगाया जाता है।बाजार खुल गया।
अल्प विराम चिन्ह (,)वाक्य में जहाँ थोड़ा रुकना हो या अधिक वस्तुओं, व्यक्तियों आदि को अलग करना हो वहाँ अल्प विराम ( , ) चिन्ह का प्रयोग किया जाता है।राकेश, मोहन, शीतल और पंकज ने यह चित्र बनाया है।
अवतरण चिन्ह या उद्धरण चिन्ह (‘….’ “…” )किसी की कही हुई बात को उसी तरह प्रकट करने के लिए अवतरण चिन्ह ( ''... '' ) का प्रयोग होता है। पुस्तक, समाचारपत्र, लेखक का उपनाम, लेख का शीर्षक इत्यादि उद्धृत करते समय इकहरे उद्धरण चिन्ह ( ‘...’ ) का प्रयोग होता है।गांधी जी ने कहा था "अहिंसा परम धर्म है”।

Question 82

निम्नलिखित में से दिया गया वाक्य किस काल का उदहारण है?

खिलाड़ी मैच खेल रहे थे।

  1. AOption (a)

    आसन्न भूतकाल

  2. BOption (b)

    अपूर्ण भूतकाल

  3. COption (c)

    संभाव्य वर्तमानकाल

  4. DOption (d)

    हेतुहेतुमद् भूतकाल

View answer & explanation

Correct answer

Answer: (b)B.

अपूर्ण भूतकाल

Why this answer is correct

खिलाड़ी मैच खेल रहे थे। इस वाक्य में क्रिया की अपूर्णता का बोध हो रहा है, अतः 'अपूर्ण भूतकाल' उचित विकल्प है।

Key Points

  • 'जो क्रिया अभी हो रही थी' उसे अपूर्ण भूतकाल कहते हैं।
  • अपूर्ण का अर्थ होता है – अधूरा।
  • क्रिया के जिस रूप से कार्य के लगातार होने का पता चलता था, उसे अपूर्ण भूतकाल कहते है।
  • जिन वाक्यों के अंत में रहा था, रहे थे, रही थी, आदि आते है, उसे अपूर्ण भूतकाल कहते हैं।

Important Points

भूतकाल के उपभेदवर्तमान काल के उपभेदभविष्य काल के उपभेद
सामान्य भूतकाल :- जिस क्रिया के भूतकाल में क्रिया के सामान्य रूप से बीते समय में पूरा होने का संकेत मिले।सामान्य वर्तमान काल :- क्रिया के जिस रूप से कार्य की पूर्णता और अपूर्णता का पता न चले।सामान्य भविष्य काल :- क्रिया के जिस रूप से क्रिया के सामान्य रूप का भविष्य में होने का पता चले।
आसन्न भूतकाल :-  जिस क्रिया के अभी-अभी या निकट के भूतकाल में पूरा होने का पता चले।अपूर्ण वर्तमान काल:-  क्रिया के जिस रूप से कार्य के लगातार होने का पता चलता है।संभाव्य भविष्य काल :- क्रिया के जिस रूप से आगे कार्य होने या करने की संभावना का पता चले।
पूर्ण भूतकाल :- क्रिया के जिस रूप से यह पता चलता है की कार्य निश्चित किये गये समय से पहले ही पूरा हो चूका था।पूर्ण वर्तमान काल :- क्रिया के जिस रूप से कार्य के अभी पूरे होने का पता चलता है।हेतुहेतुमद्भविष्य भविष्य काल:- क्रिया के जिस रूप से एक कार्य का पूरा होना दूसरी आने वाले समय की क्रिया पर निर्भर हो।
अपूर्ण भूतकाल :- क्रिया के जिस रूप से यह पता चलता है कि कार्य भूतकाल में पूरा नहीं हुआ था अपितु नियमित रूप से जारी रहा।संदिग्ध वर्तमान काल:- क्रिया के जिस रूप से वर्तमान काल क्रिया के होने या करने पर शक हो।
संदिग्ध भूतकाल :- क्रिया के जिस रूप से अतीत में हुए या करे हुए कार्य पर संदेह प्रकट किया जाये।तात्कालिक वर्तमान काल :- क्रिया के जिस रूप से यह पता चलता हो कि कार्य वर्तमान में हो रहा है।
समयकालीन भूतकाल :- जिन वाक्यों के अंत में रहा था , रही थी , रहे थे आदि आते हैं और समय क�� निश्चित बोध होता है।संभाव्य वर्तमान काल :- संभाव्य का अर्थ होता है संभावित या जिसके होने की संभावना हो।

Additional Information

काल - क्रिया के जिस रुप से कार्य व्यापार के समय तथा उसकी पूर्णता पूर्णता का बोध होता है, उसे काम कहते हैं। काल के तीन भेद माने गए हैं।
वर्तमान कालवर्तमान काल में क्रिया व्यापार की निरंतरता रहती है। इसके चार भेद माने गए हैं - सर्वमान्य वर्तमान, तात्कालिक, वर्तमान संदिग्ध वर्तमान और संभाव्य वर्तमान।उदाहरण -सीता पढ़ती है।
भूत कालभूतकाल में क्रिया व्यापार की समाप्ति का बोध होता है। इसके पांच भेद हैं - सामान्य भूत, आसन भूत, अपूर्ण भूत, संदिग्ध भूत हेतुहेतुमद् भूत।उदाहरण - सीता ने पढ़ाई की।
भविष्य कालभविष्य में होने वाली क्रिया का बोध भविष्य काल से होता है। इसके तीन भेद हैं-सामान्य भविष्य, संभाव्य भविष्य और हेतुहेतु मद् भविष्य।उदाहरण - सीता पढ़ेगी।

Question 83

निम्नलिखित में से कौन-सा वाक्य विस्मयादिबोधक वाक्य है?

  1. AOption (a)

    सुमित अब जा चुका होगा।

  2. BOption (b)

    हम यह कार्य नहीं करेंगे।

  3. COption (c)

    ओह! कितनी ठंडी रात है।

  4. DOption (d)

    आप जीवन में उन्नति करें।

View answer & explanation

Correct answer

Answer: (c)C.

ओह! कितनी ठंडी रात है।

Why this answer is correct

ओह! कितनी ठंडी रात है। - यह वाक्य विस्मयादिबोधक वाक्य का प्रकार है। 

Key Points

अन्य विकल्प

  • हम यह कार्य नहीं करेंगे। यह निषेध वाचक वाक्य है।
  • आप जीवन में उन्नति करें। यह इच्छावाचक वाक्य है।
  • सुमित अब जा चुका होगा। यह संदेह वाचक वाक्य का उदाहरण है।

Additional Information

विशेष

वाक्य के भेदपरिभाषाउदाहरण
विधानवाचकवे वाक्य जिनसे किसी बात या कार्य के होने का बोध होता है, विधिवाचक वाक्य कहलाते हैं।राजा बाज़ार गया।
निषेधवाचकजिनसे किसी बात या कार्य के न होने अथवा इनकार किए जाने का बोध होता है, निषेधवाचक वाक्य कहलाते हैं।कल नेहा घर नहीं गयी थी।
आज्ञावाचकवे वाक्य, जिनसे किसी प्रकार की आज्ञा का बोध होता है, आज्ञावाचक वाक्य कहलाते हैं।श्याम पानी लाओ!
विस्मयवाचकवे वाक्य, जिनसे किसी प्रकार का विस्मय, हर्ष, दुःख, आश्चर्य आदि का बोध होता है, विस्मयवाचक वाक्य कहलाते हैं।अरे! वह उत्तीर्ण हो गया।
संकेतवाचकवे वाक्य, जिनसे किसी प्रकार के संकेत या इशारे का बोध होता है, संकेतवाचक वाक्य कहलाते हैं।जो परिश्रम करेगा वह सफल होगा।
संदेहवाचकवे वाक्य, जिनसे किसी प्रकार के सन्देह या भ्रम का बोध होता है, सन्देहवाचक वाक्य कहलाते हैं।वह अब जा चुका होगा।
प्रश्नवाचकवे वाक्य, जिनसे किसी प्रश्न के पूछे जाने का बोध होता है, प्रश्नवाचक वाक्य कहलाते हैं।तुम किस कक्षा में पढ़ते हो?
इच्छावाचकइच्छावाचक वाक्य अर्थात वह वाक्य जिसमें किसी बात या कार्य की कामना की गयी हो।ईश्वर आपकी यात्रा सफल करे।

Question 84

दिए गए वाक्यांश के लिए एक शब्द का चयन कीजिए-

जिसको एक स्थान से दूसरे स्थान पर न ले जाया जा सके-

  1. AOption (a)

    गमनशील

  2. BOption (b)

    जंगम

  3. COption (c)

    स्थावर

  4. DOption (d)

    स्थानापन्न

View answer & explanation

Correct answer

Answer: (c)C.

स्थावर

Why this answer is correct

दिए गए विकल्पों में से सही उत्तर विकल्प 3 ‘स्थावर’ है। अन्य विकल्प इसके गलत उत्तर हैं।

 

Key Points

  • ‘ जिसे एक स्थान से उठाकर दूसरे स्थान पर न ले जा सकें - स्थावर कहा जाता है।

अन्य विकल्प:​

वाक्यांशएक शब्द
जिसे कही ले जाया जा सकेजंगम
चलने वालागमनशील
दूसरे का स्थान ग्रहण किया हुआ।स्थानापन्न

Additional Information

वाक्यांशभाषा को सुंदर, आकर्षक और प्रभावशाली बनाने के लिए अनेक शब्दों के स्थान पर एक शब्द का प्रयोग किया जाता है तो वह वाक्यांश के लिए एक शब्द कहलाता है।

Question 85

निम्नलिखित में से उत्तमपुरुष वाचक सर्वनाम का चयन कीजिए-

  1. AOption (a)

    हम

  2. BOption (b)

    तुम

  3. COption (c)

    वे

  4. DOption (d)

    आप

View answer & explanation

Correct answer

Answer: (a)A.

हम

Why this answer is correct

'हम' यह उत्तमपुरुष वाचक सर्वनाम का उदाहरण है। 

Key Pointsअन्य विकल्प:

  • तुम - मध्यमपुरुष वाचक सर्वनाम है।
  • वे - अन्यपुरुष वाचक सर्वनाम है।
  • आप - यह मध्यमपुरुष वाचक सर्वनाम है।

Additional Information

सर्वनाम - संज्ञा के स्थान पर प्रयुक्त होने वाले शब्दों को सर्वनाम कहते हैं। जैसे - मैं, वह, वे, उन्हें, अपने तुम, हम आदि। हिंदी में मूलतः सर्वनामों की संख्या 11 है – मैं, तू, आप, यह, वह, जो, सो, कौन, कोई, क्या और कुछ।
पुरुषवाचक सर्वनामपरिभाषाउदाहरण
उत्तम पुरुषजिन शब्दों का प्रयोग बोलने वाला खुद के लिए करता है।मैं, मेरा, मेरे, मेरी, मुझे, मुझको, हम, हमें, हमको, हमारा, हमारे, हमारी आदि।
मध्यम पुरुषजिन शब्दों का प्रयोग सुनने वाले के लिए किया जाता है।तू, तुझे, तुझको, तेरा, तेरे, तेरी, तुम, तुम्हे, तुमको, तुम्हारा, तुम्हारे, तुम्हारी, आप आदि।
अन्य पुरुषजिन शब्दों का प्रयोग किसी तीसरे व्यक्ति के बारे में बात करने के लिए होता है।यह, वह, ये, वे आदि।

Question 86

निम्नलिखित में से कौन-सा शब्द "निष्ठा" का पर्यायवाची नहीं है?

  1. AOption (a)

    विश्वास

  2. BOption (b)

    आस्था

  3. COption (c)

    संदेह

  4. DOption (d)

    श्रद्धा

View answer & explanation

Correct answer

Answer: (c)C.

संदेह

Why this answer is correct

उपरोक्त विकल्पों में 'संदेह' निष्ठा का पर्यायवाची शब्द नही है। अत: विकल्प 'संदेह' इसका सही उत्तर है। अन्य विकल्प असंगत हैं। 

Key Points

अन्य  विकल्प:

शब्दपर्यायवाची
विश्वासयकीन, भरोसा, आस्था, ऐतबार।
आस्थाविश्वास, भरोसा, यकीन, निष्ठा, श्रद्धा, आदरभाव
श्रद्धायकीन, प्रवृति, लगाव, निष्ठा, विश्वास
सन्देहसंशय, अंदेशा, शंका और संशय

Additional Information

विशेष:

शब्दपरिभाषाउदाहरण
पर्यायवाचीएक ही अर्थ में प्रयुक्त होने वाले शब्द जो बनावट में भले ही अलग हों, पर्यायवाची या समानार्थी शब्द कहलाते हैं।आग-अनल, पावक, दहन। हवा-समीर, अनिल, वायु।

Question 87

निम्नलिखित में से "अवाक्" का विलोम शब्द कौन-सा है?

  1. AOption (a)

    सवाक् 

  2. BOption (b)

    उत्थान

  3. COption (c)

    जावक

  4. DOption (d)

    विष

View answer & explanation

Correct answer

Answer: (a)A.

सवाक् 

Why this answer is correct

उपर्युक्त विकल्पों में से विकल्प 1 ‘सवाक्​’ सही उत्तर है। अन्य विकल्प इसके सही उत्तर नहीं हैं।

Key Points

  • 'अवाक्' का सही विलोम शब्द 'सवाक्' है।
  • अवाक् का अर्थ –  चकित, मौन।
  • सवाक् का अर्थ – बोलता हुआ।
<br>

अन्य शब्द - 

शब्दविलोम
उत्थानपतन
जावकआवक
विषअमृत

 

Additional Information

शब्दपरिभाषाउदाहरण
विलोम/विपरीतार्थकविपरीत (उल्टा) अर्थ बताने वाले शब्दों को विलोम शब्द कहते हैं।रात-दिन धरती-आकाश

Question 88

निम्नलिखित में से कौन-सा शब्द सदैव एकवचन में प्रयुक्त होता है?

  1. AOption (a)

    प्रजा

  2. BOption (b)

    हस्ताक्षर

  3. COption (c)

    प्राण

  4. DOption (d)

    दाम

View answer & explanation

Correct answer

Answer: (a)A.

प्रजा

Why this answer is correct

प्रजा सदैव एकवचन के रूप में प्रयुक्त होने वाला शब्द है। 

अन्य सभी बहुवचन के रूप में प्रयुक्त होने वाले शब्द है।

  • हिंदी भाषा में कुछ शब्द सदैव बहुवचन के रूप में प्रयुक्त होते हैं, जैसे – हस्ताक्षर, आँसू, दर्शन, प्राण, दाम, अक्षत, लोगआदि।
  • हिंदी भाषा में कुछ शब्द सदैव एकवचन के रूप में प्रयुक्त होते हैं, जैसे: जैसे - बारिश, पानी, दूध, घी, दही, आकाश, जनता आदि।

Key Points

विशेष - 

वचन - संज्ञा, सर्वनाम, विशेषण और क्रिया के जिस रूप से संख्या का बोध हो उसे वचन कहते हैं।
वचन के दो प्रकार हैं - 1. एकवचन, 2. बहुवचन
वचनपरिभाषाउदाहरण
एकवचनशब्द के जिस रूप से किसी व्यक्ति, वस्तु आदि के एक होने का बोध हो, उसे एकवचन कहते हैंस्त्री, घोड़ा, नदी, रुपया
बहुवचनशब्द के जिस रूप से एक से अधिक व्यक्तियों या वस्तुओं आदि का बोध हो, उसे बहुवचन कहते हैं।स्त्रियाँ, घोड़े, नदियाँ, रुपये

Question 89

निम्नलिखित में से कौन-सा शब्द तत्पुरुष समास का उदाहरण नहीं है?

  1. AOption (a)

    महात्मा

  2. BOption (b)

    गिरहकट

  3. COption (c)

    सत्याग्रह

  4. DOption (d)

    जन्मांध

View answer & explanation

Correct answer

Answer: (a)A.

महात्मा

Why this answer is correct

उपर्युक्त विकल्पों में से ‘महात्मा’ तत्पुरुष समास नहीं है। अतः इसका सही उत्तर विकल्प 1 होगा।

Key Points

  • 'महात्मा' का समास विग्रह ‘महान है जो आत्मा’ होगा।
  • इसमें पूर्वपद विशेषण और उत्तरपद विशेष्य है तथा उत्तरपद प्रधान है, इसलिए इसमें कर्मधारय समास होगा।
<br>

अन्य विकल्पों का समास विग्रह निम्न प्रकार है-

शब्दसमास विग्रहसमास का नाम
सत्याग्रहसत्य का आग्रहतत्पुरुष (सम्बन्ध) समास
जन्मांधजन्म से अँधातत्पुरुष (अपादान) समास
गिरहकटगिरह को काटने वालातत्पुरुष(कर्म) समास
<br>

Additional Information

तत्पुरुष समास: जिस समास में उत्तरपद प्रधान हो तथा समास करने के उपरांत विभक्ति(कारक चिन्ह) का लोप हो, तत्पुरुष समास कहलाता है।  जैसे: राम का अनुज= रामानुज, धर्म का ग्रन्थ = धर्मग्रन्थ, तुलसीदास द्वारा कृत = तुलसीदासकृत आदि। तत्पुरुष समास के छः भेद होते हैं-
समासपरिभाषाउदाहरण
कर्म तत्पुरुष समासजहां कर्म कारक चिन्ह (को) का लोप होगिरहकट = गिरह को काटने वाला
करण तत्पुरुष समासजहां करण कारक चिन्ह (से) का लोप होगुणहीन = गुणों से हीन
संप्रदान तत्पुरुष समासजहां संप्रदान कारक चिन्ह (के लिए, को) का लोप होहथकड़ी = हाथ के लिए कड़ी
अपादान तत्पुरुष समासजहां अपादान कारक चिन्ह (से, अलग होने के संबंध में ) का लोप होधनहीन = धन से हीन
संबंध तत्पुरुष समासजहां संबंध कारक (का, की के) का लोप होप्रेम सागर = प्रेम का सागर
अधिकरण तत्पुरुष समासजहां अधिकरण कारक (में , पर) का लोप होनीतिनिपुण = नीति में निपुण

Question 90

निम्नलिखित में से लोकोक्ति "अपना रख पराया चख" का क्या अर्थ है?

  1. AOption (a)

    जिसे सफलता मिले, उसी का यश फैले।

  2. BOption (b)

    लालच छोड़ कर थोडे में ही संतुष्ट होना।

  3. COption (c)

    हस्तगत चीज का विशेष मूल्य नहीं होता।

  4. DOption (d)

    अपनी वस्तु की रक्षा, दूसरे की वस्तु का उपभोग।

View answer & explanation

Correct answer

Answer: (d)D.

अपनी वस्तु की रक्षा, दूसरे की वस्तु का उपभोग।

Why this answer is correct

दिए गए विकल्पों में सही उत्तर विकल्प 4 ‘अपनी वस्तु की रक्षा, दूसरे की वस्तु का उपभोग।’ है। अन्य विकल्प इसके अनुचित उत्तर होंगे।

Key Points 

  • लोकोक्ति-  ‘अपना रख पराया चख’ अर्थ– ‘अपनी वस्तु की रक्षा, दूसरे की वस्तु का उपभोग।’ होता है।
  • वाक्य - वह वैसे तो बहुत कंजूस है, परंतु दूसरे के द्वारा ख़र्च करने पर पूरा आनंद उठाता है। इस पर तो अपना रख, पराया चख ही चरितार्थ होता है।
  • नोट-‘लोकोक्ति’ शब्द ‘लोक + उक्ति’ शब्दों से मिलकर बना है जिसका अर्थ है- लोक में प्रचलित उक्ति या कथन’।

Additional Information 

लोकोक्ति परिभाषाउदाहरण
लोकोक्तियाँ आम जनमानस द्वारा स्थानीय बोलियों में हर दिन की परिस्थितियों एवं संदर्भों से उपजे वैसे पद एवं वाक्य होते हैं जो किसी खास समूह, उम्र वर्ग या क्षेत्रीय दायरे में प्रयोग किया जाता है। इसमें स्थान विशेष के भूगोल, संस्कृति, भाषाओं का मिश्रण इत्यादि की झलक मिलती है। लोकोक्ति वाक्यांश न होकर स्वतंत्र वाक्य होते हैं।लोकोक्ति- अधजल गगरी छलकत जाय अर्थ- जिसके पास थोड़ा ज्ञान होता हैं, वह उसका प्रदर्शन या आडम्बर करता है। वाक्य- रमेश बारहवीं पास करके स्वयं को बहुत बड़ा विद्वान समझ रहा है। ये तो वही बात हुई कि अधजल गगरी छलकत जाय।

Question 91

निम्नलिखित में से अशुद्ध वर्तनी वाले शब्द का चयन कीजिए-

  1. AOption (a)

    अत्यावश्यक

  2. BOption (b)

    फुटबाल

  3. COption (c)

    संकल्पित

  4. DOption (d)

    परित्याग

View answer & explanation

Correct answer

Answer: (b)B.

फुटबाल

Why this answer is correct

दिए गए विकल्पों में फुटबाल शब्द की वर्तनी अशुद्ध है। 

अत्यावश्यक, संकल्पित, परित्याग ये तीनो शब्दों की शुद्ध वर्तनी है 

Key Points

शुद्ध वर्तनीअशुद्ध वर्तनीअर्थ
फुटबॉलफुटबालचमड़े आदि से बनी एक प्रकार की बड़ी गेंद या बॉल। उक्त बॉल को पैर से मारकर खेला जाने वाला एक तरह का खेल।

 

 

 

 

 

<br> <br>

 

Additional Information

वर्तन���लिखने की रीति को वर्तनी कहते हैं। ‘वर्तनी’ शब्द का अर्थ उच्चारित होने वाले शब्द के लेखन  में प्रयोग होने वाले लिपि चिह्नों के व्यवस्थित रूप को वर्तनी कहा जाता है।

Question 92

निम्नलिखित में से "आराधना" शब्द का विशेषण कौन-सा है?

  1. AOption (a)

    आराध्य

  2. BOption (b)

    आलोचिन

  3. COption (c)

    राधना

  4. DOption (d)

    अर्धना

View answer & explanation

Correct answer

Answer: (a)A.

आराध्य

Why this answer is correct

इसका सही उत्तर विकल्प 1 ‘आराध्य’ होगा। अन्य विकल्प सही उत्तर नहीं हैं।

Key Points

  • 'आराध्य’ शब्द ‘गुणवाचक विशेषण’ है।
विशेषणपरिभाषाउदाहरण
गुणवाचक विशेषणवे शब्द जो संज्ञा या सर्वनाम के गुण, धर्म, स्वाभाव आदि का बोध कराये।बलशाली, पुराण, नया, मोटा, दुर्बल, पठारी आदि।
<br>

अन्य विकल्प:

अन्य सभी विकल्प असंगत है**।** 

Additional Information

विशेषण जो शब्द संज्ञा या सर्वनाम शब्दों की विशेषता बताते हैं। उन्हें विशेषण कहा जाता है। विशेषण  के मुख्यतः आठ भेद हैं –
विशेषणपरिभाषाउदाहरण
गुणवाचक विशेषणवे शब्द जो संज्ञा या सर्वनाम के गुण, धर्म, स्वाभाव आदि का बोध कराये।बलशाली, पुराण, नया, तीक्ष्ण, कमजोर, मोटा, दुर्बल, पठारी आदि।
संख्यावाचक विशेषणऐसे शब्द जो संज्ञा या सर्वनाम की संख्या का बोध कराते हैं।दो, तीनों, चार गुना, प्रत्येक आदि।
परिमाणवाचक विशेषणऐसे शब्द जो संज्ञा या सर्वनाम की मात्रा का बोध कराते हैं।चार किलो, एक मीटर, दो लीटर, थोड़ा, बहुत आदि।
सार्वनामिक विशेषणवे सर्वनाम शब्द जो संज्ञा से पहले आयें और विशेषण की तरह उस संज्ञा शब्द की विशेषता बताएं।यह लड़की वही है जिसने मेरा पेन लिया था।
व्यक्तिवाचक विशेषणऐसे शब्द जो संज्ञा के भेद व्यक्तिवाचक संज्ञा से बने होते हैं एवं विशेषण शब्दों की रचना करते हैं।मुझे भारतीय खाना बहुत पसंद है।
प्रश्नवाचक विशेषणऐसे शब्द जिनका संज्ञा या सर्वनाम में जानने के लिए प्रयोग होता है।यह व्यक्ति कौन है ?
तुलनाबोधक विशेषणजब वस्तुओं के गुण - दोष की तुलना आपस में की जाये।वह राधा से भी ज्यादा सुरीला गाती है।
सम्बन्धवाचक विशेषणविशेषण शब्दों का प्रयोग करके किसी एक वस्तु या व्यक्ति का सम्बन्ध दूसरी वस्तु या व्यक्ति के साथ दर्शाया जाये।घर की बाहरी दीवार ख़राब हो रही है।

Question 93

निम्नलिखित में से तत्सम शब्द का चयन कीजिए-

  1. AOption (a)

    प्रतिच्छाया

  2. BOption (b)

    पछतावा

  3. COption (c)

    पड़ोस

  4. DOption (d)

    पंद्रह

View answer & explanation

Correct answer

Answer: (a)A.

प्रतिच्छाया

Why this answer is correct

दिए गए विकल्पों में सही उत्तर विकल्प 1 'प्रतिच्छाया’  होगा। अन्य विकल्प इसके अनुचित उत्तर हैं। 

Key Points

  • दिए गए शब्दों में 'प्रतिच्छाया' शब्द तत्सम है जिसका तद्भव रूप 'परछाई' होगा।
  • ऐसे शब्द जो संस्कृत से ज्यों के त्यों लिए गए, तत्सम होते हैं।
  • अन्य सभी शब्द तद्भव हैं।

 

अन्य विकल्प: 

तद्भवतत्सम
पछतावापश्चाताप
पड़ोसप्रतिवास
पन्द्रहपंचदश

 

Additional Information

तत्समऐसे शब्द जो संस्कृत से ज्यों के त्यों लिए गए, तत्सम होते हैं।कूप, उष्ट्र, पंचम आदि
अर्द्ध तत्समऐसे शब्द जो संस्कृत से थोड़ा परिवर्तित होकर हिंदी में आए हैं।कार्य से कारज, धैर्य से धीरज आदि
तद्भवसंस्कृत से हिंदी में आने पर जिन शब्दों का रूप बदल गया हो, तद्भव कहलाते हैं।आग, काम, पाँच आदि

Question 94

निम्नलिखित में से अशुद्ध वर्तनी वाले वाक्य का चयन कीजिए-

  1. AOption (a)

    हिन्दी का व्यवहारिक ज्ञान अपेक्षित है।

  2. BOption (b)

    दलाईलामा आध्यात्मिक धर्म-गुरु हैं।

  3. COption (c)

    स्वास्थ्य के लिए स्वच्छ जल आवश्यक है।

  4. DOption (d)

    आपकी कक्षा में कितने छात्र हैं?

View answer & explanation

Correct answer

Answer: (a)A.

हिन्दी का व्यवहारिक ज्ञान अपेक्षित है।

Why this answer is correct

हिन्दी का व्यवहारिक ज्ञान अपेक्षित है। इस वाक्य में व्यवहारिक शब्द में अशुद्ध वर्तनी है, इसलिये विकल्प 1 सही उतर है 

अन्य सभी विकल्पो में वर्तनी की अशुद्धियाँ नही है। 

Key Points

अन्य विकल्प: 

अशुद्ध शब्दशुद्ध शब्द
व्यवहारिकव्यावहारिक
<br>

Additional Information

वर्तनीलिखने की रीति को वर्तनी कहते हैं। ‘वर्तनी’ शब्द का अर्थ उच्चारित होने वाले शब्द के लेखन  में प्रयोग होने वाले लिपि चिह्नों के व्यवस्थित रूप को वर्तनी कहा जाता है।

Question 95

निम्नलिखित में से मुहावरा "अंडे का शहजादा" का क्या अर्थ है?

  1. AOption (a)

    अनुभवहीन व्यक्ति।

  2. BOption (b)

    लालची मनुष्य।

  3. COption (c)

    कठोर परिश्रम करने वाला।

  4. DOption (d)

    बहुत अनुभवी होना।

View answer & explanation

Correct answer

Answer: (a)A.

अनुभवहीन व्यक्ति।

Why this answer is correct

'अंडे का शहजादा’ मुहावरे का सही अर्थ है – अनुभवहीन व्यक्ति। अन्य विकल्प असंगत हैं। 

Key Points 

स्पष्टीकरण:-

मुहावरेअर्थवाक्य प्रयोग
अंडे का शहजादाअनुभवहीन व्यक्ति‌‌‌अजीत सिंह अपने पिता की सिफ़ारिश पर प्रमोशन तो पर  गए लेकिन काम ढंग का नहीं आता, बस समझो कि वो अंडे का शहजादा है।

अन्य विकल्प:-

मुहावरेअर्थवाक्य प्रयोग
आकाश पाताल एक करनाकठोर परिश्रम करनासूरज ने इंजीनियर पास करने के लिए आकाश-पाताल एक कर दिया।
घाट - घाट का पानी पीनाबहुत अनुभवी होनामुझे वकालत में किसी से कम न आंको, मैंने भी घाट-घाट का पानी पिया हुआ है।

Additional Information

मुहावरा परिभाषाउदाहरण
किसी भाषा की अभिव्यंजना के विशिष्ट रूप को 'मुहावरा' कहते हैं। एक अन्य पक्ष है कि विशिष्ट शब्दों विचित्र प्रयोगों एवं प्रयोग-सिद्ध विशिष्ट वाक्यांशों वाक्य-पद्धति को 'मुहावरा' कहते हैं।मुहावरा - अन्धे के हाथ बटेर लगना अर्थ - इच्छित वस्तु का अचानक प्राप्त होना वाक्य - रहीम की लटरी खुलते ही ऐसा लगा मानो अन्धे के हाथ बटेर लग गई हो।

निर्देश: निम्नलिखित गद्यांश का ध्यानपूर्वक अध्ययन कर प्रश्नों के उत्तर दीजिए-

संसार में दो अचूक शक्तियाँ ���ैं- वाणी और कर्म। कुछ लोग वचन से संसार को रह दिखाते हैं और कुछ लोग कर्म से। शब्द और आचार दोनों ही महान शक्तियाँ हैं। शब्द की महिमा अपार है। विश्व में साहित्य, कला, विज्ञान, शास्त्र सब शब्द-शक्ति के प्रतीक प्रमाण हैं। पर कोरे शब्द व्यर्थ होते है, जिनका आचरण न हो। कर्म के बिना वचन, व्यवहार के बिना सद्धांत की कोई सार्थकता नहीं है।

निस्सन्देह शब्द शक्ति महान है, पर चिरस्थायी और सनातनी शक्ति तो व्यवहार है। महात्मा गाँधी ने इन दोनों की कठिन और अद्भुत साधना की थी। महात्मा जी का सम्पूर्ण जीवन उन्ही दोनों से युक्त था। वे वाणी और व्यवहार में एक थे। जो कहते थे वही करते थे। यही उनकी महानता का रहस्य है। कस्तूरबा ने शब्द की अपेक्षा कृति की उपासना की थी, क्योंकि कृति का उत्तम व चिरस्थायी प्रभाव होता है। 'बा' न कोरी शाब्दिक, शास्त्रीय, सैद्धांतिक शब्दावली नहीं सीखी थी। वे तो कर्म की उपासिका थीं। उनका विश्वास शब्दों की अपेक्षा कर्मों में था। वे जो कहा करती थीं उसे पूरा करती थी। वे रचनात्मक कर्मों को प्रधानता देती थीं। इसी के बल पर उन्होनें अपने जीवन में सार्थकता और सफलता प्राप्त की थी।

Question 96

उपर्युक्त गद्यांश के संदर्भ में, प्रायः सज्जन व्यक्ति संसार को _______ राह दिखाते है।

  1. AOption (a)

    केवल गरीबो की मदद करने की भावना से।

  2. BOption (b)

    अपने कर्म एवं वाणी से।

  3. COption (c)

    अपनी कार्यक्षमता से।

  4. DOption (d)

    अपनी बुद्धिक्षमता से।

View answer & explanation

Correct answer

Answer: (b)B.

अपने कर्म एवं वाणी से।

Why this answer is correct

सज्जन व्यक्ति संसार को 'अपने कर्म एवं वाणी से' राह दिखाते है।

अपने कर्म एवं वाणी से - खुद के कार्यो एवं बोलने के तरीके से।

Key Points अन्य विकल्प:

  • केवल गरीबो की मदद करने की भावना से - जो असहाय व्यक्ति है, उनको मदद की भावना के माध्यम से।
  • अपनी कार्यक्षमता से - कार्य करने की शक्ति से।
  • अपनी बुद्धिक्षमता से - बुद्धि की शक्ति के माध्यम से।

निर्देश: निम्नलिखित गद्यांश का ध्यानपूर्वक अध्ययन कर प्रश्नों के उत्तर दीजिए-

संसार में दो अचूक शक्तियाँ हैं- वाणी और कर्म। कुछ लोग वचन से संसार को रह दिखाते हैं और कुछ लोग कर्म से। शब्द और आचार दोनों ही महान शक्तियाँ हैं। शब्द की महिमा अपार है। विश्व में साहित्य, कला, विज्ञान, शास्त्र सब शब्द-शक्ति के प्रतीक प्रमाण हैं। पर कोरे शब्द व्यर्थ होते है, जिनका आचरण न हो। कर्म के बिना वचन, व्यवहार के बिना सद्धांत की कोई सार्थकता नहीं है।

निस्सन्देह शब्द शक्ति महान है, पर चिरस्थायी और सनातनी शक्ति तो व्यवहार है। महात्मा गाँधी ने इन दोनों की कठिन और अद्भुत साधना की थी। महात्मा जी का सम्पूर्ण जीवन उन्ही दोनों से युक्त था। वे वाणी और व्यवहार में एक थे। जो कहते थे वही करते थे। यही उनकी महानता का रहस्य है। कस्तूरबा ने शब्द की अपेक्षा कृति की उपासना की थी, क्योंकि कृति का उत्तम व चिरस्थायी प्रभाव होता है। 'बा' न कोरी शाब्दिक, शास्त्रीय, सैद्धांतिक शब्दावली नहीं सीखी थी। वे तो कर्म की उपासिका थीं। उनका विश्वास शब्दों की अपेक्षा कर्मों में था। वे जो कहा करती थीं उसे पूरा करती थी। वे रचनात्मक कर्मों को प्रधानता देती थीं। इसी के बल पर उन्होनें अपने जीवन में सार्थकता और सफलता प्राप्त की थी।

Question 97

उपर्युक्त गद्यांश के अनुसार, गाँधीजी की महानता का रहस्य क्या है?

  1. AOption (a)

    जो दूसरे व्यक्ति कहते थे, वही करते थे।

  2. BOption (b)

    जो कहते थे वह, कभी नही करते थे।

  3. COption (c)

    सदैव गरीबों की सहायता करते थे।

  4. DOption (d)

    जो कहते थे वही करते थे।

View answer & explanation

Correct answer

Answer: (d)D.

जो कहते थे वही करते थे।

Why this answer is correct

गाँधीजी की महानता का रहस्य यही है कि 'जो कहते थे वही करते थे।' 

जो कहते थे वही करते थे इसका अर्थ यह होता है कि उनकी कथनी और करनी में कोई अंतर नही था।

अन्य विकल्प:

  • जो दूसरे व्यक्ति कहते थे, वही करते थे - इसका अर्थ यह हुआ कि दूसरो के बताये मार्ग पर चलने की प्रवृति।
  • सदैव गरीबों की सहायता करते थे - असहाय, दीन - दुखियो की सेवा की भावना मन में होना।
  • जो कहते थे वह, कभी नही करते थे - यह गांधीजी के गुण के विपरीत कही हुई बात है।

निर्देश: निम्नलिखित गद्यांश का ध्यानपूर्वक अध्ययन कर प्रश्नों के उत्तर दीजिए-

संसार में दो अचूक शक्तियाँ हैं- वाणी और कर्म। कुछ लोग वचन से संसार को रह दिखाते हैं और कुछ लोग कर्म से। शब्द और आचार दोनों ही महान शक्तियाँ हैं। शब्द की महिमा अपार है। विश्व में साहित्य, कला, विज्ञान, शास्त्र सब शब्द-शक्ति के प्रतीक प्रमाण हैं। पर कोरे शब्द व्यर्थ होते है, जिनका आचरण न हो। कर्म के बिना वचन, व्यवहार के बिना सद्धांत की कोई सार्थकता नहीं है।

निस्सन्देह शब्द शक्ति महान है, पर चिरस्थायी और सनातनी शक्ति तो व्यवहार है। महात्मा गाँधी ने इन दोनों की कठिन और अद्भुत साधना की थी। महात्मा जी का सम्पूर्ण जीवन उन्ही दोनों से युक्त था। वे वाणी और व्यवहार में एक थे। जो कहते थे वही करते थे। यही उनकी महानता का रहस्य है। कस्तूरबा ने शब्द की अपेक्षा कृति की उपासना की थी, क्योंकि कृति का उत्तम व चिरस्थायी प्रभाव होता है। 'बा' न कोरी शाब्दिक, शास्त्रीय, सैद्धांतिक शब्दा��ली नहीं सीखी थी। वे तो कर्म की उपासिका थीं। उनका विश्वास शब्दों की अपेक्षा कर्मों में था। वे जो कहा करती थीं उसे पूरा करती थी। वे रचनात्मक कर्मों को प्रधानता देती थीं। इसी के बल पर उन्होनें अपने जीवन में सार्थकता और सफलता प्राप्त की थी।

Question 98

निम्नलिखित में से उपर्युक्त गद्यांश का उचित शीर्षक कौन-सा है?

  1. AOption (a)

    वाणी और कर्म

  2. BOption (b)

    निस्सन्देह: चिरस्थायी एवं सनातनी शक्ति

  3. COption (c)

    शब्द का महत्त्व

  4. DOption (d)

    निष्ठामूर्ति गाँधी 

View answer & explanation

Correct answer

Answer: (a)A.

वाणी और कर्म

Why this answer is correct

वाणी और कर्म की महानता के सन्दर्भ में गद्यांश में बताया गया है, अतः 'वाणी और कर्म' सही विकल्प है।

वाणी और कर्म - वाचा और कार्य करने की भावना 

Key Points अन्य विकल्प:

  • शब्द का महत्व - शब्दों की शक्ति का माहात्म्य।
  • निष्ठामूर्ति गाँधी - निष्ठा जैसे गुणों की साक्षात् मूर्ति यानी महात्मा गांधी।
  • निस्सन्देह: चिरस्थायी एवं सनातनी शक्ति - हमेशा रहने वाली सनातन शक्ति।

Question 99

निम्नलिखित में से "चिरस्थायी" से क्या अभिप्राय है?

  1. AOption (a)

    देर तक रहनेवाला या टिकाऊ

  2. BOption (b)

    अति शीघ्र नष्ट होने वाला।

  3. COption (c)

    जो कठोर परिश्रम न करती हो। 

  4. DOption (d)

    जिसका दिमाग स्थिर न हो।

View answer & explanation

Correct answer

Answer: (a)A.

देर तक रहनेवाला या टिकाऊ

Why this answer is correct

चिरस्थायी का अर्थ 'देर तक रहनेवाला या टिकाऊ' ऐसा होता है।

वाक्य प्रयोग - इस मकान की लकड़ी चिरस्थायी है। 

अन्य विकल्प:

  • अति शीघ्र नष्ट होने वाला - जल्दी खत्म होने वाला।
  • जो कठोर परिश्रम न करती हो - इसको आलसी कहा जाता है।
  • जिसका दिमाग स्थिर न हो - इसे चंचल मना कहा जाता है।

Question 100

निम्नलिखित में से कौन-सा शब्द "सार्थकता" का पर्यायवाची नही है।

  1. AOption (a)

    उपयोगिता

  2. BOption (b)

    उद्देश्यपूर्ति

  3. COption (c)

    निरर्थक

  4. DOption (d)

    सफलता

View answer & explanation

Correct answer

Answer: (c)C.

निरर्थक

Why this answer is correct

उपरोक्त विकल्पों में 'सार्थकता' का पर्यायवाची शब्द 'निरर्थक' नही है। अत: विकल्प निरर्थक इसका सही उत्तर है।

Key Points

शब्दपर्यायवाची
उपयोगितालाभकारिता, लाभप्रदता, लाभदायकता, अर्थकरता, हितकरता, उपयुक्तता, उपादेयता।
उद्देश्यपूर्तिउपयोगिता, सफलता
सफलताफतह, जफर, जीत, विजय।
निरर्थकबेकार, अर्थहीन, अर्थशून्य, निष्फल, व्यर्थ, असंगत, फिजूल।

 

Additional Information

 

विशेष:

शब्दपरिभाषाउदाहरण
पर्यायवाचीएक ही अर्थ में प्रयुक्त होने वाले शब्द जो बनावट में भले ही अलग हों, पर्यायवाची या समानार्थी शब्द कहलाते हैं।आग-अनल, पावक, दहन। हवा-समीर, अनिल, वायु।

Discipline 1 (20 questions)

Question 101

Let f: R → R be defined as f(x) = 8x. Choose the correct answer:

  1. AOption (a)

    f is neither one-one nor onto

  2. BOption (b)

    f is one-one but not onto

  3. COption (c)

    f is many-one onto

  4. DOption (d)

    f is one-one onto

View answer & explanation

Correct answer

Answer: (d)D.

f is one-one onto

Why this answer is correct

Concept:

Bijective function: A function is said to be bijective or bijection, if a function f: A → B satisfies both the injective (one-to-one function) and surjective function (onto function) properties.

(or)

For every element “b” in the codomain B, there is exactly one element “a” in the domain A. such that f(a) = b. If the function satisfies this condition, then it is known as one-to-one correspondence.

Calculations:

Let f: R → R be defined as f(x) = 8x.

Let x, y ∈ R such that f(x) = f(y)

⇒ 8x = 8y

⇒ x = y

∴ f is one - one.

Now, for any real number y ∈ R (co-domain)

∃ y8∈R\frac{y}{8}\in R  such that

f(y8\frac{y}{8} ) = 8 × y8\frac{y}{8}  = y

∴ f is onto.

Thus f is both one-one and onto i.e.,f is bijective.

Hence, the correct answer is option 4).

Question 102

Which of the following is incorrect?

  1. AOption (a)

    Every sequence has a countable number of terms

  2. BOption (b)

    Every sequence has a limit point

  3. COption (c)

    Every sequence has a monotone subsequence

  4. DOption (d)

    Every bounded sequence has a convergent subsequence

View answer & explanation

Correct answer

Answer: (b)B.

Every sequence has a limit point

Why this answer is correct

Concept:

Limit point: A number x is called a limit point (or cluster point or accumulation point) of a set of real numbers A if, ∀ε > 0, the interval (x − ε, x + ε) contains infinitely many points of A.

Bolzano-Weierstrass theorem - Every sequence has a monotone subsequence, which can be either monotonically increasing or decreasing.

Calculations:

For Option (1) - Every sequence has a countable number of terms.

By definition, a sequence is a function from the natural numbers (which are countable) to some set, meaning it has a countable number of terms.

Hence Option (1) is correct.

For Option (2) - Every sequence has a limit point.

let the sequence an = n 

Clearly it is a divergent sequence, and it does not any limit not.

Hence Option (2) is incorrect.

For Option (3) - Every sequence has a monotone subsequence.

According to the Bolzano-Weierstrass theorem, every sequence has a monotone subsequence, which can be either monotonically increasing or decreasing.

Hence Option (3) is correct.

For Option (4) - Every bounded sequence has a convergent subsequence.

This is a consequence of the Bolzano-Weierstrass theorem, which states that every bounded sequence in R\mathbb{R}  has a convergent subsequence.

Therefore, the incorrect statement is Option (2).

Question 103

A Relation in R is defined as R = {(a, b) : a ≤ b2} is ________.

  1. AOption (a)

    Symmetric

  2. BOption (b)

    Not an Equivalence Relation

  3. COption (c)

    Transitive

  4. DOption (d)

    Reflexive

View answer & explanation

Correct answer

Answer: (b)B.

Not an Equivalence Relation

Why this answer is correct

Concept:

A relation R on a set A is said to be an equivalence relation if and only if the relation R is reflexive, symmetric, and transitive. The equivalence relation is a relationship on the set which is generally represented by the symbol “∼”.

Reflexive: A relation is said to be reflexive, if (a, a) ∈ R, for every a ∈ A.

Symmetric: A relation is said to be symmetric, if (a, b) ∈ R, then (b, a) ∈ R.

Transitive: A relation is said to be transitive if (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.

Calculations:

Given the relation R on the set of all real numbers R = {(a, b): a ≤ b2} 

Let, ​​(12,12)∉R(\frac{1}{2}, \frac{1}{2}) ∉ R  because 12>(12)2=14\frac{1}{2}>(\frac{1}{2})^2=\frac{1}{4}

∴ R is not reflexive.

Now, let (1, 4) ∈ R as 1 ≤ 42

But 4 is not less than 12

∴ (4, 1) ∉ R

∴  R is not symmetric

Next, consider (3, 2), (2, 1.5) ∈ R (as 3 < 22 & 2 < 1.52 = 2.25

But 3 > 1.52 = 2.25

∴ (3, 1.5) ∉ R

∴  R is not transitive 

Thus R is neither reflexive, symmetric, nor transitive

Hence, the correct answer is option 2).

Question 104

If f : R → R given by f(x) = 4x + 3, then inverse of f is

  1. AOption (a)

    f–1 = y−34\frac{y - 3}{4}

  2. BOption (b)

    f–1 = 4−y3\frac{4 - y}{3}

  3. COption (c)

    f–1 = y−43\frac{y -4}{3}

  4. DOption (d)

    f–1 = 3−y4\frac{3 - y}{4}

View answer & explanation

Correct answer

Answer: (a)A.

f–1 = y−34\frac{y - 3}{4}

Why this answer is correct

Concept:

Bijective function: A function is said to be bijective or bijection, if a function f: A → B satisfies both the injective (one-to-one function) and surjective function (onto function) properties.

(or)

For every element “b” in the codomain B, there is exactly one element “a” in the domain A. such that f(a) = b. If the function satisfies this condition, then it is known as one-to-one correspondence.

Calculations:

Let f: R → R be defined as f(x) = 4x + 3

Let x, y ∈ R such that f(x) = f(y)

⇒ 4x + 3 = 4y + 3

⇒ x = y

∴ f is one - one.

Now, for any real number y ∈ R (co-domain) .Let f(x) = y

∃ y−34∈R\frac{y-3}{4}\in R  such that

f(y−34\frac{y-3}{4} ) = 4 × y−34\frac{y-3}{4}  + 3 = y

∴ f is onto. 

Thus f is both one-one and onto i.e.,f is bijective.

For finding f -1, we put f(x) = y and find x in terms of y 

⇒ y = f(x)

⇒ y = 4x + 3

x = y−34\frac{y-3}{4}

Let g(y) = y−34\frac{y-3}{4}  where g: R → R

(g ∘ f)(x) = g(f(x)) = g(4x + 3)

= 4x+3−34\frac{4x+3-3}{4}

= x

∴ (g ∘ f)(x) = x

and 

(f ∘ g)(y) = f(g(y)) = f( y−34\frac{y-3}{4} )

=4 × y−34\frac{y-3}{4}  + 3

= y

∴ (f ∘ g)(y) = y

⇒ f is invertible.

∴ inverse of f = g(y) = y−34\frac{y-3}{4} .

Hence, the correct answer is option 1).

Question 105

Which one of the following series is divergent?

  1. AOption (a)

    ∑  1/n3{}^1/_n{^3}

  2. BOption (b)

    ∑ \({}^1/_\sqrt{n}\) tan 1/n2{}^1/_n{^2}

  3. COption (c)

    ∑ sin 1/n{}^1/_n

  4. DOption (d)

    ∑ sin 1/n2{}^1/_n{^2}

View answer & explanation

Correct answer

Answer: (c)C.

∑ sin 1/n{}^1/_n

Why this answer is correct

Concept:

  • sin x ≈ x, when x is close to 0.
  • ∑n=1∞1n\displaystyle\sum_{n=1}^{∞} \frac{1}{n} is divergent , because p = 1 ≤ 1
  • Limit comparison test: Suppose ∑ an and ∑ bn with an ≥ 0, bn > 0 for all n. Define c=lim⁡n→∞anbnc =\displaystyle\lim_{n→∞}\frac{a_n}{b_n}

If c is positive (i.e., c > 0 and is finite (i.e., c < ∞ ) then either both series converge or both series diverge.

  • L Hospital ' s rule: For the differentiable functions f(x) and g(x), the lim⁡x→cf(x)g(x)\rm \displaystyle \lim_{x\to c} \dfrac{f(x)}{g(x)}, if f(x) and g(x) are both 0 or ±∞ (i.e. an Indeterminate Form) is equal to the lim⁡x→cf′(x)g′(x)\rm \displaystyle \lim_{x\to c} \dfrac{f'(x)}{g'(x)} if it exists.

Calculation:

Let bn = 1n\frac{1}{n} > 0

Clearly ∑n=1∞1n\displaystyle\sum_{n=1}^{∞} \frac{1}{n}is divergent , because p = 1 ≤ 1 (p - series test)

Consider as n → ∞. Define

c=lim⁡n→∞sin1n1nc=\displaystyle\lim_{n\rightarrow\infty}\frac{sin\frac{1}{n}}{\frac{1}{n}} = 00\frac{0}{0} which is an indeterminate form.

So, using L Hospital's rule we have to differentiate the numerator and denominator separately with respect to n, we get

lim⁡n→∞cos(1n).(−1n2)−1n2\displaystyle\lim_{n\rightarrow\infty}\frac{cos(\frac{1}{n}).(-\frac{1}{n^2})}{-\frac{1}{n^2}}

=lim⁡n→∞cos(1n)= \displaystyle\lim_{n\rightarrow\infty}{cos(\frac{1}{n})}

=cos(1∞)=cos(\frac{1}{\infty})

=cos(0)=cos(0)

= 1 > 0

∴ c > 0

Thus by limit comparison test ∑ sin \({}^1/n\) is divergent because  \(\displaystyle\sum{n=1}^{∞} \frac{1}{n}\) is divergent.

Hence, the correct answer is option 3).

Question 106

The series ∑ (–1)n + 1 1n\frac{1}{n} is-

  1. AOption (a)

    Unbounded

  2. BOption (b)

    Conditionally Convergent

  3. COption (c)

    Absolutely convergent

  4. DOption (d)

    Divergent

View answer & explanation

Correct answer

Answer: (b)B.

Conditionally Convergent

Why this answer is correct

Concept:

Conditionally convergent: If the series ∑n=1∞bn\displaystyle\sum_{n=1}^{\infty}b_n is convergent and ∑n=1∞∣bn∣\displaystyle\sum_{n=1}^{\infty}|b_n| is divergent then the series ∑n=1∞bn\displaystyle\sum_{n=1}^{\infty}b_n is called conditionally convergent.

Leibnitz test: 

An alternating series ∑n=1∞(−1)nbn,bn≥0\displaystyle\sum_{n=1}^{\infty}(-1)^nb_n,b_n≥0  converges if both of the conditions hold,

  1. bn ≥ bn + 1 for all n ≥ N. 2. lim⁡n→∞bn=0\displaystyle\lim_{n\rightarrow\infty}b_n = 0

p - series test: The p-series is a power series of the form ∑n=k∞1np\displaystyle\sum_{n=k}^{\infty}\frac{1}{n^p} , where p is a positive real number and k is a positive integer. The p-series test determines the nature of convergence of a p-series as follows: The p-series converges if p > 1 and diverges if p ≤ 1.

Calculation:

Given series is ∑n=1∞(–1)n+11n\displaystyle\sum_{n=1}^{\infty}(–1)^{n + 1}\frac{1}{n}

where bn=1n≥0b_n = \frac{1}{n}\geq 0

Using the Leibnitz test we have,

|bn| - |bn + 1| 

=1n−1n+1=\frac{1}{n}-\frac{1}{n+1}

=n+1−nn=\frac{n+1-n}{n}

=1n≥0= \frac{1}{n}\geq 0 for all n ∈ N

∴ bn ≥ bn + 1 for all n ≥ N.

The first condition of Leibnitz test holds.

Now, consider 

lim⁡n→∞bn\displaystyle\lim_{n\rightarrow\infty}b_n

⇒ lim⁡n→∞1n=0\displaystyle\lim_{n\rightarrow\infty}\frac{1}{n} = 0

The second condition of leibnitz test also holds 

∴ the series ∑n=1∞(–1)n+11n\displaystyle\sum_{n=1}^{\infty}(–1)^{n + 1}\frac{1}{n}  is convergent.

Next , consider

∑n=1∞∣bn∣\displaystyle\sum_{n=1}^{\infty}|b_n|

=∑n=1∞1n=\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}

By p-series test, we have p = 1 and thus the series ∑n=1∞∣bn∣\displaystyle\sum_{n=1}^{\infty}|b_n| is divergent.

Thus  the series ∑n=1∞bn\displaystyle\sum_{n=1}^{\infty}b_n is convergent and ∑n=1∞∣bn∣\displaystyle\sum_{n=1}^{\infty}|b_n| is divergent

Hence the given series ∑n=1∞(–1)n+11n\displaystyle\sum_{n=1}^{\infty}(–1)^{n + 1}\frac{1}{n}  is conditionally convergent.

Hence, the correct answer is option 2).

Question 107

If f : R → R defined by f(x) = x2 – 3x + 2, then f(f(x)) =

  1. AOption (a)

    x4 + 6x3 + 10x2 + 3x

  2. BOption (b)

    x4 – 6x3 – 13x2 + 3x

  3. COption (c)

    x4 – 6x3 + 10x2 – 3x

  4. DOption (d)

    x4 + 6x3 – 13x2 + 3x

View answer & explanation

Correct answer

Answer: (c)C.

x4 – 6x3 + 10x2 – 3x

Why this answer is correct

Calculations:

Given f : R → R defined by f(x) = x2 – 3x + 2, then

f(f(x)) = f(x2 – 3x + 2)

= (x2 – 3x + 2)2 - 3( x2 – 3x + 2) + 2

= [(x - 2)(x - 1)]2 - 3x2 + 9x - 6 + 2

= (x - 2)2(x - 1)2 - 3x2 + 9x - 4

= (x2 + 4 - 4x)(x2 + 1 - 2x) - 3x2 + 9x - 4

= x4 + 4x2 - 4x3 + x2 + 4 - 4x - 2x3 - 8x + 8x2 - 3x2 + 9x - 4

= x4 - 6x3 + 10x2 - 3x

∴ f(f(x)) = x4 - 6x3 + 10x2 - 3x

Hence, the correct answer is option 3).

Question 108

If ∑an2a_n^2 and ∑ bn2b_n^2 are convergent series of positive real numbers then, ∑ an.bn\sqrt{a_n.b_n}

  1. AOption (a)

    must be divegernt

  2. BOption (b)

    is divergent

  3. COption (c)

    may or may not be convergent

  4. DOption (d)

    must be convergent

View answer & explanation

Correct answer

Answer: (c)C.

may or may not be convergent

Why this answer is correct

Concept: ( P series test ) if Un=1nkU_n=\frac{1}{n^k} is converent  if k >1 ,Otherwise divergent.

Calculation:

Case i)

let ∑ana_n =1n\frac{1}{n} and  ∑bnb_n =1n\frac{1}{n}

According to ∑ an2a_n^2 =1n2\frac{1}{n^2} and  ∑ bn2b_n^2 =1n2\frac{1}{n^2}

By the p series test  ∑an2,∑bn2\sum a_n^2 ,\sum b_n^2  are convergent series.

⇒∑an.bn=1n\sum \sqrt{a_n.b_n}=\frac{1}{n} 

So by p series test .It is not convergent.

Case ii)

let ∑an=1n2\sum a_n =\frac{1}{n^2}and  ∑bn=1n2\sum b_n =\frac{1}{n^2} 

 ⇒∑an2=1n4\sum a_n^2=\frac{1}{n^4}and  ∑bn2=1n4\sum b_n^2=\frac{1}{n^4} (These both are convergent by p series test)  

 So ∑an.bn=1n2\sum \sqrt{a_n.b_n}=\frac{1}{n^2} but it is convergent

.Hence according to above case we can conclude that  this series may or may not be convergent.

Question 109

If a sequence an2{a^2_n} converges to a2, then the sequence {an} converges to-

  1. AOption (a)

    a

  2. BOption (b)

    Not convergent

  3. COption (c)

    a2

  4. DOption (d)

    2\sqrt2

View answer & explanation

Correct answer

Answer: (b)B.

Not convergent

Why this answer is correct

Concept: A sequence is which is neither convergent ,non divergent is called oscillatory sequence. it  is oscillate finitely since it is bounded and converges.

  Solution:Let the sequence an=(-1)n 

 Then  an2a^2_n=(-1)2n

  According to condition  an2a^2_n = a2 =(1)2=1

 an is osicallting sequence between {-1,1} so it is not convergent sequence.

Question 110

If Un > 0 for all n ∈ N, then lim⁡n→∞Un1/n\lim\limits_{n \to \infty} U_{n}^{1/_n} is equal to-

  1. AOption (a)

    lim⁡n→∞UnUn+1\lim\limits_{n \to \infty} \frac{U_n}{U_{n + 1}}

  2. BOption (b)

    lim⁡n→∞Un+1Un\lim\limits_{n \to \infty} \frac{U_n + 1}{U_n}

  3. COption (c)

    lim⁡n→∞UnUn−1\lim\limits_{n \to \infty} \frac{U_n}{U_{n - 1}}

  4. DOption (d)

    lim⁡n→∞Un−1Un\lim\limits_{n \to \infty} \frac{U_{n - 1}}{U_n}

View answer & explanation

Correct answer

Answer: (c)C.

lim⁡n→∞UnUn−1\lim\limits_{n \to \infty} \frac{U_n}{U_{n - 1}}

Why this answer is correct

Explanation:

Cauchy’s Second Theorem on Limits: 

If Un > 0 for all n ∈ N, then \(\lim\limits_{n \to \infty} U_{n}^{1/n}\) = \(\lim\limits{n \to \infty} \frac{U_{n + 1}}{U_n}\) = lim⁡n→∞UnUn−1\lim\limits_{n \to \infty} \frac{U_n}{U_{n - 1}}

So, (1), (2), (4) are false.

(3) is the the answer.

Question 111

The domain of convergence for x – x2\frac{x}{2} + x3\frac{x}{3} – x4\frac{x}{4} + ___ ____ is

  1. AOption (a)

    (–1,1]

  2. BOption (b)

    [–1,1)

  3. COption (c)

    (–1,1)

  4. DOption (d)

    [–1,1]

View answer & explanation

Correct answer

Answer: (d)D.

[–1,1]

Why this answer is correct

Concept: The radius of convergence is half of the interval of the convergence.

R be the radius of convergence, R=lim⁡n→∞An+1An\lim_{n\to\infty}\frac{A_n+1}{A_n} then

  • The series converges if R<1
  • The series is diverging if R >1 or if R is infinite.
  • Interval of convergence will be  ( a-R < x < R+a )
  • we’ll need to check the convergence of the endpoints of the interval by plugging the endpoints into the original series and using any convergence test that we can to say whether or not the series converges at the endpoint.

Leibnitz Rule: A series of the form

\({\displaystyle \sum {n=0}^{\infty }(-1)^{n}a{n}=a_{0}-a_{1}+a_{2}-a_{3}+\cdots !}\)

where either all an are positive or all an are negative, is called an alternating series

The alternating series test guarantees that an alternating series converges if the following two conditions are :

  1. |An| decreases monotonically. | An+1| < |An| ,
  2. lim⁡n→∞An=0\lim_{n\to \infty} A_n=0

Calculation   :  An=x−x22+x33−x44+⋯A_n=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dotsb

An=∑n=1∞−(1)n+1nxnA_n=\sum_{n=1}^{\infty}\frac {-(1)^{n+1}}{n}x^n

 An+1=∑n=1∞(−1)n+2nxnA_{n+1}=\sum_{n=1}^{\infty}\frac{(-1)^{n+2}}{n}x^n 

 R=lim⁡n→∞An+1An=∣(−1)n+2n+1×n(−1)n+1∣=∣(−1)nn+1∣=∣nn+1∣R=\lim_{n\to\infty}\frac{A_{n+1}}{A_n}=|\frac{(-1)^{n+2}}{n+1}\times\frac{n}{(-1)^{n+1}}|=|\frac{(-1)n}{n+1}| =|\frac{n}{n+1}|

Taking common n from numerator and denominator

    limn→∞An+1An=1lim_{n\to\infty}\frac{A_{n+1}}{A_n}=1 

So the radius of convergence is 1. Now the interval of convergence is (1-∈  ,1+∈ ) 

Check at the boundary point,

 x= -1 then  ∑n=1∞(−1)2n+1n\sum_{n=1}^{\infty}\frac {(-1)^{2n+1}}{n} by leibnitz rule this sequence is convergent.

 x =1 then ∑n=1∞(−1)n+1n\sum_{n=1}^{\infty}\frac {(-1)^{n+1}}{n}by leibtnitz rule this is also convergent

Hence the Domain of convergence is [-1,1].

Question 112

Which of the following is correct if the above image represents the functions f1 and f2?

  1. AOption (a)

    f1 is onto, f2 is onto

  2. BOption (b)

    f1 is one-one, f2 is one-one

  3. COption (c)

    f1 is one-one, f2 is not an onto

  4. DOption (d)

    f1 is onto, f2 is one-one

View answer & explanation

Correct answer

Answer: (c)C.

f1 is one-one, f2 is not an onto

Why this answer is correct

Concept:

One-to-One functions define that each element of one set, say Set (A) is mapped with a unique element of another set, say, Set (B).

(or)

An injective function (injection) or one-to-one function is a function that maps distinct elements of its domain to distinct elements of its codomain.

(or)

Let ‘f’ be a function whose domain is set A. The function is said to be injective if for all x, y ∈ A, 

Whenever f(x) = f(y), then x = y

And equivalently, if x ≠ y, then f(x) ≠ f(y)

Onto function: Let set A be a codomain and set B be a co-domain then for each element of B, there is a pre-image in Set A.

Calculations:

Therefore from the above definitions, it is clear that f1 is an injective function as each element of the domain is mapped with a unique element of the codomain.

And f2 is not an onto function because, elements in co-domain like a, e, and f don't have a preimage in the domain. 

Thus f1 is one-one, f2 is not an onto

Hence, the correct answer is option 3).

Question 113

If S = {a, b, c, d) and p = {(a, a), (b, b), (c, c), (d, d), (a, b), (b, a),(b, c), (c, b)}, then p is-

  1. AOption (a)

    Reflexive and symmetric

  2. BOption (b)

    Equivalence

  3. COption (c)

    Reflexive and transitive

  4. DOption (d)

    Symmetric and transitive

View answer & explanation

Correct answer

Answer: (a)A.

Reflexive and symmetric

Why this answer is correct

**Reflexive Relation :**A Relation R defined on set A is said to  be reflexive, if (x, x) ∈  R,∀∀ x ∈  A ,i.e xRx, ∀ x,y ∈ A

Here define Relation  p = {(a, a), (b, b), (c, c), (d, d), (a, b), (b, a), (b, c), (c, b). 

If a relation is reflexive, then (a, a) ∈ R for every a ∈ {a, b, c, d}

Since (a, a) ∈ R ,(b, b) ∈ R,(c, c) ∈ R, (d, d) ∈ R

So, R is reflexive.

Check Symmetric

Symmetric Relation: A Relation R is defined on set A is said to be symmetric, if (x, y) ∈∈ R ⇒ (y, x)∈∈R

To check symmetric,

If (a, b) ∈ R, Then  (b, a) ∈ R

Since(a, b) ∈ R and (b, a) ∈ R also (b ,c)  ∈ R and (c, b)  ∈ R

So, R is reflexive and symmetric. 

Additional InformationTransitive Relation: A Relation R defined on set A is said to be transitive if (x, y) ∈  R , (y, z) ∈  R ⇒ (x, z) ∈ R ∀ x,y,z  ∈ A.

If (a, b) ∈ R, (b, a) ∈ R  since (a, c) ∉ R 

so, R is not transitive Relation.

If R is satisifed all the condition of Equivalence then the R Relation be the Equivalence  Otherwise not.

Question 114

If a binary operation * on N is defined as a * b = a3 + b3, then-

  1. AOption (a)
    • is neither commutative nor associative
  2. BOption (b)
    • is commutative but not associative
  3. COption (c)
    • is both associated and commutative
  4. DOption (d)
    • is associative but not commutative
View answer & explanation

Correct answer

Answer: (b)B.
  • is commutative but not associative

Why this answer is correct

Calculation:

Given a binary operation * on N is defined as a * b = a3 + b3

Consider, 

a * b = a3 + b3

= b3 + a3 

= b * a

∴ a * b =  b * a

Thus the binary operation * on N is commutative.

Next consider,

(a * b) * c 

= (a3 + b3) * c 

= (a3 + b3)3 + c3

∴ (a * b) * c = (a3 + b3)3 + c3  ....(1)

Next consider,

a * (b * c)

= a * (b3 + c3) 

= a3 + (b3 + c3)3

∴ a * (b * c) = a3 + (b3 + c3)3   .......(2)

Thus from (1) and (2), we have

(a * b) * c ≠ a * (b * c)

Thus the binary operation * on N is not associative.

Hence, * is commutative but not associative.

Hence, the correct answer is option 2).

Question 115

The sequence {Xn}, where Xn = 1 + 13\frac{1}{3} + 15\frac{1}{5} + __ __ + 12n−1\frac{1}{2n-1}  is

  1. AOption (a)

    Oscillatory

  2. BOption (b)

    Not Cauchy sequence

  3. COption (c)

    Convergent

  4. DOption (d)

    Monotonically decreasing

View answer & explanation

Correct answer

Answer: (b)B.

Not Cauchy sequence

Why this answer is correct

Concept: (Comparison test) we have two series ∑an∑an∑an∑an∑an∑an∑an

Code
     and ∑bn∑bn∑bn∑bn∑bn∑bn∑bn

     with an,bn≥0an,bn≥0an,bn≥0an,bn≥0an,bn≥0an,bn≥0an,bn≥0

     for all nUnexpected text node: 'n'Unexpected text node: 'n'Unexpected text node: 'n'Unexpected text node: 'n'Unexpected text node: 'n'Unexpected text node: 'n'

     and an≤bnan≤bnan≤bnan≤bnan≤bnan≤bnan≤bn

     for all nUnexpected text node: 'n'  . Then,
  • If ∑bn is Convergent then so  ∑an is convergent.  is Convergent. b ≠ 1
  • If ∑an is Divergent then so  ∑bn is Divergent.​

​P-Series test:  series∑1nk\sum\frac{1}{n^k}if k>1 then this series is convergent otherwise divergent.

Calculation              

an=∑12n−1a_n = \sum\frac{1}{2n-1}

bn=∑1nb_n=\sum\frac{1}{n}

 lim⁡n→∞anbn=(12n−1×n1)=lim⁡n→∞anbn=(n2n−1)=12 \mathop {\lim }\limits_{n\rightarrow \infty} \frac{a_n}{b_n}=(\frac{1}{2n-1}\times \frac{n}{1})= \mathop {\lim }\limits_{n\rightarrow \infty}\frac{a_n}{b_n}=(\frac{n}{2n-1})=\frac{1}{2}

By comparison test, both an and b are converging and diverging together.

∴ bn=1nb_n=\frac{1}{n} is Divergent by p series test.

∴ ana_n is divergent​..

∴ ana_n is not convergent implies not cauchy sequence.​​​

Question 116

A conditionally convergent series is a series which is -

  1. AOption (a)

    Absolutely divergent

  2. BOption (b)

    Convergent but not absolutely

  3. COption (c)

    Not convergent but absolutely convergent

  4. DOption (d)

    Absolutely Convergent

View answer & explanation

Correct answer

Answer: (b)B.

Convergent but not absolutely

Why this answer is correct

Concept:

  • Conditionally convergent: If the series ∑n=1∞bn\displaystyle\sum_{n=1}^{\infty}b_n is convergent and ∑n=1∞∣bn∣\displaystyle\sum_{n=1}^{\infty}|b_n| is divergent then the series ∑n=1∞bn\displaystyle\sum_{n=1}^{\infty}b_n is called conditionally convergent.
  • Absolutely convergent: The series ∑n=1∞bn\displaystyle\sum_{n=1}^{\infty}b_n with both positive and negative terms( not necessary alternative) is called absolutely convergent if the corresponding series ∑n=1∞∣bn∣\displaystyle\sum_{n=1}^{\infty}|b_n| with all positive terms is convergent.
  • If a series is convergent but not absolutely convergent, it is called conditionally convergent.
  • An example of a conditionally convergent series is the alternating harmonic series.

Thus, A conditionally convergent series is a series that is Convergent but not absolutely.

Hence, the correct answer is option 2).

Question 117

If f : A → B and g : B → C are one–one, then gof : A → C is-

  1. AOption (a)

    One-one and onto both

  2. BOption (b)

    One-one

  3. COption (c)

    Neither One-One, nor onto

  4. DOption (d)

    Onto

View answer & explanation

Correct answer

Answer: (b)B.

One-one

Why this answer is correct

Concept:

Let ‘f’ be a function whose domain is set A. The function is said to be injective (or) one - one  if for all x, y ∈ A, 

Whenever f(x) = f(y), then x = y

And equivalently, if x ≠ y, then f(x) ≠ f(y)

Calculations:

Consider

(g ∘ f)(x1) = (g ∘ f)(x2)

⇒ (g(f(x1)) =  (g(f(x2))

⇒ f(x1) = f(x2)   | because g is one-one 

⇒ x1 = x2    | because f is one-one 

∴ (g ∘ f) is one - one.

Hence, the correct answer is option 2).

Question 118

The greatest integer function f : R → R given by f(x) = [x], (where [x] denotes the greatest integer), is _______ 

  1. AOption (a)

    One-One, but not onto

  2. BOption (b)

    Neither One-One nor onto

  3. COption (c)

    One-One, onto

  4. DOption (d)

    Onto, but not One-One

View answer & explanation

Correct answer

Answer: (b)B.

Neither One-One nor onto

Why this answer is correct

Concept:

One-to-One functions define that each element of one set, say Set (A) is mapped with a unique element of another set, say, Set (B).

(or)

An injective function (injection) or one-to-one function is a function that maps distinct elements of its domain to distinct elements of its codomain.

(or)

let ‘f’ be a function whose domain is set A. The function is said to be injective if for all x, y ∈ A, 

Whenever f(x) = f(y), then x = y

And equivalently, if x ≠ y, then f(x) ≠ f(y)

Onto function: Let set A be a codomain and set B to be a co-domain then for each element of B, there is a pre-image in Set A.

Calculations:

Given f(x) = [x], (where [x] denotes the greatest integer)

we have f(1.2) =  [1.2] = 1 and f(1.9) =  [1.9] = 1 

∴ f(1.2) = f(1.9) ⇒ 1.2 ≠ 1.9

∴ f is not one - one.

Now consider 0.7 ∈ R

It is known that f(x) = [x] is always an integer,thus there doesn't exist any element x ∈ R such that f(x) = 0.7

Thus f is not onto.

Hence f is neither on-one nor onto.

Hence, the correct answer is option 2).

Question 119

If f : R → R and g : → R are given by f(x) = cos x, and g(x) = 3x2, then -

  1. AOption (a)

    gof ≠ fog

  2. BOption (b)

    gof and fog both do not exist

  3. COption (c)

    gof = fog

  4. DOption (d)

    Can't be determined

View answer & explanation

Correct answer

Answer: (a)A.

gof ≠ fog

Why this answer is correct

Concept:

  • g ∘ f  = read as g composition f and (g ∘ f)(x) = g(f(x))  and
  • f ∘ g  = read as f composition g and (f ∘ g)(x) = f(g(x)).

Calculation:

Given f : R → R and g : → R given by f(x) = cos x, and g(x) = 3x2, 

Consider, 

(g ∘ f)(x)= g(f(x)) = g(cos x) = 3cos2x

Next, 

(f ∘ g)(x) = f(g(x)) = f(3x2) = cos 3x2

⇒ g ∘ f ≠ f ∘ g 

Hence, the correct answer is option 1).

Question 120

Which one of the following tests does not give absolute convergence of a series?

  1. AOption (a)

    Root Test

  2. BOption (b)

    Ratio Test

  3. COption (c)

    Comparison Test

  4. DOption (d)

    Leibnitz's test

View answer & explanation

Correct answer

Answer: (d)D.

Leibnitz's test

Why this answer is correct

Concept:

A series ∑\sum(u_n\) is said to be absolutely convergent if all the terms are positive and the series ∑∣un∣\sum|u_ n| by the modulus of each term of the series is convergent.

Calculation:

1-13\frac{1}{3} +19\frac{1}{9}-127+181\frac{1}{27}+\frac{1}{81}-..... a  series ,now if we take the modulus of each term and construct a series  then it will look like ∑∣un∣\sum|u_n||=1+13+19+127+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+... ,which is a infinite geometric series with common ratio  13\frac{1}{3}  but if we consider the series 1−12+13−14+....1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.... this is a alternating series but if we want to express in  sum of absolute term i.e.∑∣un∣=1+12+13+14+...\sum|u_n|=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...

is not a  absolutely convergent series but it  is conditionally convergent by  Leibnitz's test.

 The condition for Leibnitz's test is given below :

(i) unu_n(\geq\)0  (ii) un+1≤unu_{n+1}\leq u_n, (iii)lim⁡n→∞un=0\lim\limits_{n\to\infty}u_n=0 if  any alternating series obeys these three conditions then we can say the alternating series is convergent.

If we consider the series 1−12+13−14+....1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.... then  ∑un=(−1)n+11n\sum u_n=(-1)^{n+1}\frac{1}{n}

un=1nu_n=\frac{1}{n}

Applying the condition of Leibnitz's test, we get

(i) un>0u_n>0

(ii)un+1−un=1n+1−1n<0u_{n+1}-u_{n}=\frac{1}{n+1}-\frac{1}{n}<0

(iii)  lim⁡n→∞1n=0\lim\limits_{n\to\infty}\frac{1}{n}=0

 So all the three conditions of Leibnitz's test is satisfied,

∴ the series is  conditionally  convergent by   Leibnitz's test .

 

 Additional :

Root Test, Ratio Test, Comparison Test widely used for series of positive terms.

Discipline 2 (20 questions)

Question 121

If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}

A = {1, 2, 3, 4,} 

B = {2, 4, 6, 8), then (A ∪ B)' is -

  1. AOption (a)

    {5, 7, 9}

  2. BOption (b)

    {2, 4, 6, 8}

  3. COption (c)

    {1, 2, 3, 4, 6, 8}

  4. DOption (d)

    {1, 3, 5, 7, 9}

View answer & explanation

Correct answer

Answer: (a)A.

{5, 7, 9}

Why this answer is correct

Concept:

A' = U - A it implies that x be an element which is in U but not in A

Calculation

A = {1, 2, 3, 4}

B = {2, 4, 6, 8}

A U B ={1, 2, 3, 4, 6, 8}

( A U B)'

= U - (A U B)

= {1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2, 3, 4, 6, 8}

= {5, 7, 9 }

Question 122

If f(x, y) = x3 – 3xy2 of an analytic function f(z) = u + iv, then

  1. AOption (a)

    v is its harmonic conjugate and v(x, y) = 3x2y – y4+ c, c is real constant

  2. BOption (b)

    –v is its harmonic conjugate and v(x, y) = y2 – 3x2 y + c, c is real constant

  3. COption (c)

    v is its harmonic conjugate and v(x, y) = 3x2y – y4 + c, c is any constant

  4. DOption (d)

    –v is its harmonic conjugate and v(x, y) = y3– 3x2y + c, c is any constant

View answer & explanation

Correct answer

Answer: (d)D.

–v is its harmonic conjugate and v(x, y) = y3– 3x2y + c, c is any constant

Why this answer is correct

Concept:

A function  is said to be harmonic if it satisfies Laplace equation ∂2v∂x2+∂2v∂y2\frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} =0

point: Here u(x,y)=x3−3xy2u(x,y)=x^{3}-3xy^{2}

A function f(z)=u+iv is said to be analytic if it satisfies Cauchy -Riemann equation i.e.

 ∂u∂x=∂v∂y,∂u∂y=−∂v∂x\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y} ,\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x} 

Calculation:

hereu(x,y)=x3−3xy2u(x,y)=x^{3}-3xy^{2}

so,∂u∂x=3x2−3y2\frac{\partial u}{\partial x}=3x^{2}-3y^{2},∂u∂y=−6xy\frac{\partial u}{\partial y}=-6xy

Now by Cauchy Riemann equation we can write

 ∂v∂x=6xy;or,v=3x2y+A\frac{\partial v}{\partial x}=6xy ; or ,v=3x^{2}y+A 

and

 ∂v∂y=3x2−3y2;or,v=3x2y−y3+B\frac{\partial v}{\partial y}=3x^{2}-3y^{2}; or,v=3x^{2}y-y^{3}+B

combining we can write  v=3x2y−y3+cv=3x^{2}y-y^{3}+c

Tricks:

By using Laplace equation  checking the option also you can get your answer.

∂v∂x=6xy\frac{\partial v}{\partial x}=6xy

∂2v∂x2=6y\frac{\partial^2 v}{\partial x^2}=6y    -(i)

∂v∂y=−3y2\frac{\partial v}{\partial y}=-3y^{2}

∂2v∂x2=−6y\frac{\partial^2 v}{\partial x^2}=-6y   (ii)

now adding (i) & (ii) we get 

∂2v∂x2+∂2v∂y2=−6y+6y=0\frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2}=-6y+6y=0

 So option (4) is correct.

Question 123

Which of the following is not true.

  1. AOption (a)

    If u and v are both harmonic functions then f (z) = u + i.v is analytic

  2. BOption (b)

    If a function does not satisfy Cauchy - Riemann equations, then it is not analytic

  3. COption (c)

    If f (z) = u + i.v is an analytic function, then u (x, y) = c1 and v (x, y) = c2 are orthogonal

  4. DOption (d)

    If f (z) = u + i.v is an analytic function, then u and v are both harmonic functions

View answer & explanation

Correct answer

Answer: (a)A.

If u and v are both harmonic functions then f (z) = u + i.v is analytic

Why this answer is correct

Concept:

If f(z)=u+iv is an analytic function then u & v are both harmonic functions but the converse is not true.

Calculation:

We can show it by one example u=x2−y2,v=−yx2+y2u=x^{2}-y^{2} , v=\frac{-y}{x^2+y^2} then  ∂u∂x=2x,∂u∂y=−2y,\frac{\partial u}{\partial x}=2x, \frac{\partial u}{\partial y}=-2y,      -(A)

∂v∂y=y2−x2(x2+y2)2\frac{\partial v}{\partial y}=\frac{y^{2}-x^{2}}{(x^{2}+y^{2})^2} ,∂v∂x=2xy(x2+y2)2\frac{\partial v}{\partial x}=\frac{2xy}{(x^{2}+y^{2})^2}     - (B)

∂2u∂x2=2,∂2u∂y2=−2\frac{\partial^2 u}{\partial x^2}=2,\frac{\partial^2 u}{\partial y^2}=-2     -(1)

∂2v∂x2=2y(y2−3x2)(x2+y2)3\frac{\partial^2 v}{\partial x^2}=\frac{2y(y^{2}-3x^{2})}{(x^{2}+y^{2})^3}, ∂2v∂y2=2y(3x2−y2)(x2+y2)3\frac{\partial^2 v}{\partial y^2}=\frac{2y(3x^{2}-y^{2})}{(x^{2}+y^{2})^3}       - (2)

From (1) & (2) we can say that  Laplace equation is satisfied but from (A) & (B) Cauchy-Riemann equation is not satisfied.

So we can say that in general if u & v are harmonic then it may or may not be analytic.

Question 124

If three non-empty sets A, B, C are having relation A ∩ B = A ∩ C and A ∪ B = A ∪ C, then-

  1. AOption (a)

    B = C

  2. BOption (b)

    A = C

  3. COption (c)

    A = B = C

  4. DOption (d)

    A = B

View answer & explanation

Correct answer

Answer: (a)A.

B = C

Why this answer is correct

Concept:

Non - empty sets are those sets which has at least one element present in it.

Formula Used

Calculation:

( A U B ) ∩ C = ( A U C ) ∩ C

( A ∩ C) U ( B ∩ C ) = C        [we know that, (A U C) ∩ C =  C]

(A ∩ B) U (B ∩ C) = C     ---- (1) [ we know that, (A ∩ C) = (A ∩ B)

Again, A U B = A U C

( A U B) ∩ B = ( A U C) ∩

B = ( A ∩ B) U ( C∩ B)

B=(A ∩ B) U (B ∩C)

B= (A ∩ B) U (B ∩ C)      ---- (2)

from (1) and (2) we get

B = C

The required value is B = C

Question 125

The radius of convergence of the power series ∑n=0∞\sum_{n=0}^\infty = (n!)2(2n)!zn\frac{(n!)^2}{(2n)!}z^n is

  1. AOption (a)

    4

  2. BOption (b)

    1/2

  3. COption (c)

    1/4

  4. DOption (d)

    2

View answer & explanation

Correct answer

Answer: (a)A.

4

Why this answer is correct

Concept: 

Radius of convergence of power series - If Σanzn\Sigma a_nz^n  be given power series then the circle |z| = R having all values of z for which the power series Σanzn\Sigma a_nz^n converges.

Then, the radius R is called radius of convergence and |z| = R is called radius of convergence and |z| = R is called circle of convergence.              

Calculation: 

Let  an=n!2(2n)!\ a_n=\frac{n!^2}{(2n)!} ,  an+1=(n+1)!2(2n+2)!\ a_{n+1} = \frac{(n + 1)!^2}{(2n + 2)!}

R = lim⁡n→∞n!2(2n+2)!(2n)!(n+1)!2\lim \limits_{n \to \infty} \frac{n!^2(2n+2)!}{(2n)!(n+1)!^2}  

R =lim⁡n→∞(2n+1)(2n+2)(n+1)2\lim\limits_{n \to \infty}\frac{(2n + 1)(2n + 2)}{(n + 1)^2}

R = lim⁡n→∞n2(2+1/n)(2+2/n)n2(1+1/n)2\lim\limits_{n \to \infty}\frac{n^2(2 + 1/n)(2 + 2/n)}{n^2(1 + 1/n)^2}

Now, putting the limit

R = 4/1

R = 4

The radius of convergence of power series is 4.

Question 126

If X and Y are two non-empty sets such that n(X) = 17, n(Y) = 23 and n(X ∪  Y) = 38, then n( X ∩ Y ) is

  1. AOption (a)

    16

  2. BOption (b)

    2

  3. COption (c)

    6

  4. DOption (d)

    10

View answer & explanation

Correct answer

Answer: (b)B.

2

Why this answer is correct

Concept:  

Cardinality of Sets- The cardinality of sets is defined as the number of elements present in a given set. It is denoted by n(A).

The basic formula on cardinality of sets is

n(A U B) = n(A) + n(B) - n( A ∩ B )

Calculation:

The set X has 17 elements.

The set Y has 23 elements.

The set  X U Y have 38 elements.

Putting the values,

38 = 17 + 23 - n(X ∩ Y)

38 = 40 - n(X ∩ Y)

n(X ∩ Y) = 40 - 38

n(X ∩ Y) = 2 

Hence, there are 2 elements common in both X and Y.

Question 127

Which of the following represents De Morgan's Law?

  1. AOption (a)

    A ∩ B = A' ∪ B'

  2. BOption (b)

    (A ∪ B)' = A' ∩ B'

  3. COption (c)

    (A’ ∪ B') = A ∩ B'

  4. DOption (d)

    (A ∪ B') = A' ∩ B

View answer & explanation

Correct answer

Answer: (b)B.

(A ∪ B)' = A' ∩ B'

Why this answer is correct

Concept:

According to De Morgan's law, " The complement of the union of two sets will be equal to the intersection of their individual complements."

Calculation:

Let x ∈ (A U B)' = A' ∩ B'  

Let x ∈ (A U B)' 

x ∉ (A U B)'

x ∉ A  and x ∉ B

x ∈ A' and x ∈ B'

x ∈ (A' U B')

(A U B)' ⊂ A' ∩ B'    ---- equation (1)

Again, y ∈  A'  ∩  B'

y ∈ A' and y  ∉ B'

y ∉  (A U B)

y ∉ (A U B)'

(A ' ∩ B') ⊂ (A U B)'   ----equation (2)

from (1) and (2)

A' ∩ B' = (A U B)'

(A U B)' = A' ∩ B'

The correct answer is (A U B)' = A' ∩ B'

Question 128

If a set A = ϕ, then how many elements are there in P(A)?

  1. AOption (a)

    1

  2. BOption (b)

    3

  3. COption (c)

    2

  4. DOption (d)

    No element

View answer & explanation

Correct answer

Answer: (a)A.

1

Why this answer is correct

Concept:

A set is a well - defined collection of distinct objects, considered as an object in its own right.

A = ϕ,

This means P(A) = ϕ

Calculation:

If A = ϕ that means A does not contain any element , that is  n = 0

Now, number of elements in a power set in 2n

Therefore, n[P(A)] = 20 = 1 

Therefore, P(A) contains 1 element.

The correct value is 1.

Question 129

Which one of the following is true?

  1. AOption (a)

    A × (B ∪ C) = (A × B) ∩ (A × C)

  2. BOption (b)

    A × (B – C) = (A × C) – (A × B)

  3. COption (c)

    (A – B) × C = (A × C) – (B × C)

  4. DOption (d)

    A × (B ∩ C) = (A × B) ∪ (A × C)

View answer & explanation

Correct answer

Answer: (c)C.

(A – B) × C = (A × C) – (B × C)

Why this answer is correct

CONCEPT:

Here we use the concept of  Cartesian product of two sets.

 Cartesian Product of two sets:  If we consider  two sets A & B  and if a∈A,b∈Ba\in A ,b\in B  then  A×B=(a,b) A\times B=(a,b)(a\in A ,b\in B \)

Solution:

Let  (a,b)∈(A−B)×C(a,b)\in (A-B) \times C

→ a∈(A−B)and(b∈C)a\in(A-B) and (b\in C)

→ (a∈AandB∉B)and(b∈C)(a\in A and B\not\in B ) and(b\in C)((x,y)\in(A-B)\times C\)

→ (a∈Aandb∈C)and(a∉Bandb∈C)(a\in A and b \in C)and (a\not\in B and b\in C)

→ (a,b)∈(A×B)and(a,b)∉(B×C)(a,b)\in(A\times B) and (a,b)\not\in(B \times C)

→ (a,b)∈(A×C)−(B×C)(a,b)\in(A\times C)-(B\times C)

So we can say that (A×C)−(B×C)(A\times C)-(B\times C)  is a subset of  (A−B)×C(A-B)\times C  ......(1)

Again let

 (x,y)∈(A×C)−(B×C)(x,y)\in(A \times C)-(B \times C)

→ (x,y)∈(A×C)and(x,y)∉(B×C)(x,y)\in( A \times C) and (x,y)\not\in(B \times C)

→ (x∈Aandy∈C)and(x∉Bandy∈C)(x\in A and y\in C)and (x\not\in B and y\in C)

→ (x,y)∈(A−B)×C(x,y)\in(A-B)\times C

So now we can say that  (A−B)×C(A-B)\times C is the subset of   (A×C)−(B×C)(A\times C)-(B\times C)  .....(2) 

Frrom (1) & (2) we can say (A−B)×C(A-B)\times C=(A×C)−(B×C)(A\times C)-(B\times C)

Question 130

[A ∪ (B ∪ C)] ∩ [A’ ∩ (B' ∩  C')] is equal to -

  1. AOption (a)

    A ∩ (B' ∩ C')

  2. BOption (b)

    A' ∪ (B ∪ C)

  3. COption (c)

    Universal set

  4. DOption (d)

    null set

View answer & explanation

Correct answer

Answer: (d)D.

null set

Why this answer is correct

CONCEPT:

Here we use the De' Morgan law  i.e. (A U B)'= A'∩\cap B'

and (A∩\cap B)'=A'U B'

Here we use  that if A be a set then intersection with its complement A' is a null set

SOLUTION:

 [A ∪ (B ∪ C)] ∩ [A’ ∩ (B' ∩  C')]

 =[A U( BU C)] ∩ [A' ∩( BU C)'] [by using De' Morgan's law]

 =[A U( B U C)] ∩ [A U(B U C)]' [by using De' Morgan's law]

 = ϕϕ ϕ

Question 131

Let f(z) = 1z\frac{1}{z}; z ≠ 0, then

  1. AOption (a)

    f does not satisfy Cauchy - Riemann equation for all z ≠ 0

  2. BOption (b)

    f is analytic for all z ≠ 0

  3. COption (c)

    f satisfies Cauchy – Riemann equation, but not analytic for all z ≠ 0

  4. DOption (d)

    f is continuous only but nowhere differentiable

View answer & explanation

Correct answer

Answer: (b)B.

f is analytic for all z ≠ 0

Why this answer is correct

CONCEPT:

If we consider a function f(z)= u+iv then  it will satisfies the Cauchy – Riemann equation if ∂u∂x=∂v∂y,∂u∂y=−∂v∂x\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}, \frac{\partial u}{\partial y}=- \frac{\partial v}{\partial x} and if it possesses all the partial derivatives then the function f(z) is analytic.

Solution:

   Let  f(z)=u+iv

   f(z)=1z\frac{1}{z}

  where z= x + iy

now we can write f(z)=x−iyx2+y2f(z)=\frac{x-iy}{x^2+y^2}

 So here u=xx2+y2,v=−yx2+y2u=\frac{x}{x^2+y^2}, v=-\frac{y}{x^2+y^2}(\frac{\partial u}{\partial x}=\frac{y^2-x^2}{(x^2+y^2)^2}\)

 ∂u∂y=−2xy(x2+y2)2\frac{\partial u}{\partial y}=\frac{-2xy}{(x^2+y^2)^2}, ∂v∂x=2xy(x2+y2)2,∂v∂y=y2−x2(y2+x2)2\frac{\partial v}{\partial x}=\frac{2xy}{(x^2+y^2)^2}, \frac{\partial v}{\partial y}=\frac{y^2-x^2}{(y^2+x^2)^2}

 So from here we can say that the Cauchy – Riemann equation  ∂u∂x=∂v∂y,∂u∂y=−∂v∂x\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}, \frac{\partial u}{\partial y}=- \frac{\partial v}{\partial x} is satisfied and all the partial  derivatives  are exist except origin(0,0)

 If we take w=1zw=\frac{1}{z}

then dwdz\frac{dw}{dz}=−1z2-\frac{1}{z^2} also exists everywhere at z=0

 So, 1/z, is analytic at z=0.

Question 132

If f (z) = u(x, y) + i.v (x, y) be analytic everywhere and u – v = (x – y).(x2 + 4xy + y2), then-

  1. AOption (a)

    f (z) = z (1 – i.z) + c, c is arbitrary constant

  2. BOption (b)

    f(z) = z2 (1 + i) + c, c is arbitrary constant

  3. COption (c)

    f(z) = – i.z3 + c. c is arbitrary constant

  4. DOption (d)

    f(z) = i.z3 + c, where c is arbitrary constant

View answer & explanation

Correct answer

Answer: (c)C.

f(z) = – i.z3 + c. c is arbitrary constant

Why this answer is correct

Concept:

 A function f(z) is said to be analytic if it satisfies the Cauchy-Riemann equation and if the partial derivative exists everywhere.

Also a function f(z) is said to be analytic if it satisfies the Laplace equation i.e. ∂2f∂x2+∂2f∂y2=0\frac{\partial^2 f}{\partial x^2} +\frac{\partial^2 f}{\partial y^2}=0

Solution:

 It is given that  u – v = (x – y).(x2 + 4xy + y2)

 Let f(z)=u+iv ....(1)

 now i f(z)=iu-v ....(2)

 adding (1) & (2) we have (1+i) f(z)=(u-v)+i(u+v)

 let u-v=U and u+v=V

 U+iV=(1+i)f(z)

 U=u-v

    =(x – y).(x2 + 4xy + y2)

    =x3+3x2y−3xy2−y3x^3+3x^2y-3xy^2-y^3

 ∂U∂x=3x2+6xy−3y2\frac{\partial U}{\partial x} =3x^2 +6xy-3y^2

 ∂U∂y=3x2−6xy−3y2\frac{\partial U}{\partial y} =3x^2-6xy-3y^2

Now by C-R equation  ∂U∂x=∂V∂y,\frac{\partial U}{\partial x} =\frac{\partial V}{\partial y},  ∂U∂y=−∂V∂x\frac{\partial U}{\partial y} = -\frac{\partial V}{\partial x}  ....(3)

 Now by  using (3) we get ∂V∂y=3x2+6xy−3y2\frac{\partial V}{\partial y} = 3x^2+6xy-3y^2

  V=3x2y+3xy2−y33x^2y+3xy^2-y^3 ....(4)

& ∂V∂x=−3x2+6xy−3y2\frac{\partial V}{\partial x} =-3x^2+6xy-3y^2

   V=−x3+3xy2−y3V=-x^3+3xy^2-y^3 ...(5)

  Combining (4) & (5) we get  V=−x3+3xy2+3x2y−y3+cV=-x^3+3xy^2+3x^2y-y^3+c

 U+iVU+iV=x3+3x2y−3xy2−y3−ix3+3ixy2+3ix2y−iy3x^3+3x^2y-3xy^2-y^3-ix^3+3ixy^2+3ix^2y-iy^3

            =(1−i)x3+(1+i)3x2y−(1−i)3xy2−(1+i)y3+c(1-i)x^3+(1+i)3x^2y-(1-i)3xy^2-(1+i)y^3+c

            =(1−i)x3+i(1−i)3x2y−(1−i)3xy2−i(1−i)y3+c(1-i)x^3+i(1-i)3x^2y-(1-i)3xy^2-i(1-i)y^3+c

            =(1−i)[x3+i3x2y−3xy2−iy3](1-i)[x^3+i3x^2y-3xy^2-iy^3]+c

            =(1−i)[(x+iy)3](1-i)[(x+iy)^3]

            =(1−i)z3+c(1-i)z^3+c

(1+i)f(z)=(1−i)z3+c(1-i)z^3+c

    f(z)=1−i1+iz3+c1+i\frac{1-i}{1+i}z^3+\frac{c}{1+i}

    f(z)=−iz3+c-iz^3+c

Question 133

The value of the integral ∫czez(z−1)3\int _c \frac{ze^z}{(z - 1)^3}dz, where c is a circle |z| = 2 is-

  1. AOption (a)

    3 π.i.e

  2. BOption (b)

    2 π i e

  3. COption (c)

    π i e

  4. DOption (d)

    2πie3\frac{2\pi ie}{3}

View answer & explanation

Correct answer

Answer: (a)A.

3 π.i.e

Why this answer is correct

CONCEPT:

Cauchy integral Formula -

If a function f(z) is analytic within and on a closed contour C and a is any point within C then derivatives of all orders are analytic and are given by 

f(n)(a)=n!2πi∫Cf(z)dz(z−a)n+1f^{(n)}(a) = \frac{n!}{2 π i} \int_C \frac{f(z) dz}{(z-a)^{n+1}}

SOLUTION:

We have ∫czez(z−1)3\int _c \frac{ze^z}{(z - 1)^3}

Now z = 1 is the singular point for the given function. and it is inside the curve C where C is  a circle |z| = 2.

Now use the Cauchy integral Formula, where f(z) = z ez  we get -

∫czez(z−1)3=2πi2!f(2)(1)=πif(2)(1)\int _c \frac{ze^z}{(z - 1)^3} = \frac{2 π i}{2!}f^{(2)}(1) = π if^{(2)}(1)  ......(i)

Now we have f(z) = z ez 

Now f'(z) = z ez  + ez and f''(z) =  z ez  + 2 ez

Now  f''(1) = f(2)(1) = 3e

Now put this value in the above equation, we get -

⇒ ∫czez(z−1)3=πif(2)(1)=πi×3e=3πie\int _c \frac{ze^z}{(z - 1)^3} = π if^{(2)}(1) = \pi i \times 3e = 3\pi i e

 Hence the option (1) is true.

Question 134

If Z1 = {3x : x ∈ Z}

Z2 = {3x + 1 : x ∈ Z}

Z3 = {3x + 2 : x ∈  Z} then-

  1. AOption (a)

    Z1, Z2, Z3 form a partition of Z

  2. BOption (b)

    Z1, Z2, Z3 form a group under multiplication

  3. COption (c)

    Z1, Z2, Z3 form a partition of R

  4. DOption (d)

    Z1, ∩ Z2, ∩ Z3  = Z

View answer & explanation

Correct answer

Answer: (a)A.

Z1, Z2, Z3 form a partition of Z

Why this answer is correct

Concept:

A partition of a set S is a collection of non-empty, disjoint subsets of S whose union equal to S.

Explanation:

Z1 = {3x : x ∈ Z} = { ........ -9, -6, -3, 0, 3, 6, 9 .........}

Z2 = {3x + 1 : x ∈ Z} = { ......... -10, -7, -4, 1, 4, 7, 10........}

Z3 = {3x + 2 : x ∈  Z} = { ......... -11, -8, -5, 2, 5, 8, 11 .........}

Clearly, Z1 , Z2 , Z3 are disjoint as well Z1 ∪ Z2 ∪ Z3 = { .......... -3, -2, -1, 0, 1, 2, 3, ......} = Z

Clearly, Z1, Z2, Z3 form a partition of Z

Hence, Option (1) is true

Question 135

The value of the integral ∫cz2e2zz2+1\int_c \frac{z^2 e^{2z}}{z^2 + 1}dz, where c is a circle |z| = 2 is

  1. AOption (a)

    –2π.i. sin2

  2. BOption (b)

    2π i.sin2

  3. COption (c)

    0

  4. DOption (d)

    –2πi

View answer & explanation

Correct answer

Answer: (a)A.

–2π.i. sin2

Why this answer is correct

Concept:

Cauchy's Theorem:- If f(z) is an analytic function and f'(z) is continuous at each point within and on a closed curve C, then ∫cf(z)dz = 0

Cauchy's Integral Formula:- If f(z) is an analytic function within a closed curve and if a is any point within C, then f(a)=12πi∫cf(z)z−adzf(a) = \frac{1}{2πi}∫_c\frac{f(z)}{z-a}dz

Calculation:

Let f(z) = z2e2z

we have, Cauchy's Integral Formula

f(a)=12πi∫cf(z)z−adzf(a) = \frac{1}{2πi}∫_c\frac{f(z)}{z-a}dz

we have, 1z2+1=1(z+i)(z−i)=12i(1z−i−1z+i)\frac{1}{z^2 +1} = \frac{1}{(z + i)(z - i)} = \frac{1}{2i}(\frac{1}{z - i} - \frac{1}{z + i})

Now, 12πi∫cz2e2zz2+1dz=12πi∫c12i(1z−i−1z+i)f(z)dz\frac{1}{2πi}∫_c \frac{z^2e^{2z}}{z^2 + 1}dz = \frac{1}{2πi}∫_c \frac{1}{2i}(\frac{1 }{z -i} - \frac{1}{z + i})f(z)dz

f(a)×2πi=12i[∫cf(z)z−idz−∫cf(z)z+idz]f(a)× 2πi= \frac{1}{2i}[∫_c \frac{f(z) }{z -i}dz -∫_c \frac{f(z)}{z + i}dz]

where f(z) = z2e2z is analytic within and on C.

f(a)=2πi×12i[f(i)−f(−i)]f(a)= 2πi\times\frac{1}{2i} [f(i) - f (-i)] , from formula

f(a)=2πi×12i(i2e2i−i−2e−2i)f(a) = 2πi\times\frac{1}{2i}(i^2e^{2i} - i^{-2}e^{-2i})

f(a)=2πi×12i[−e2i+e−2i]f(a) =2πi\times \frac{1}{2i}[-e^{2i} + e^{-2i}]

\()\(f(a)= - 2πi\times\frac{1}{2i}[e^{2i} - e^{-2i}]\)

f(a) = - 2πi sin2

The value of the integral ∫cz2e2zz2+1\int_c \frac{z^2 e^{2z}}{z^2 + 1}dz, where c is a circle |z| = 2 is - 2πisin2

Question 136

If f (z)  ={∣z∣Re(z),if Re(z)≠0 0,if  Re (z) = 0 = \begin{cases} \frac {|z|}{Re(z)} &, \quad \text{if } {} \text{Re} (z)\neq 0 \ 0 &, \quad \text{if } \text{ Re (z) = 0} \end{cases} , then-

  1. AOption (a)

    f is not continuous only at 0

  2. BOption (b)

    f is not continuous nowhere

  3. COption (c)

    f is not continuous everywhere

  4. DOption (d)

    f is not continuous at 0

View answer & explanation

Correct answer

Answer: (d)D.

f is not continuous at 0

Why this answer is correct

Concept:

(i) If z ∈ C\mathbb{C} ⇒ z = x+iy then |z| = √x2 + y2  and Re(z) = x

(ii) Method to check limit : check limz → a  along different paths in the complex plane. If the limit is the same along all paths, it provides evidence that the limit exists. If the limit differs along different paths, the limit may not exist.

Explanation:

Now, f(z) ={∣z∣Re(z),if Re(z)≠0 0,if  Re (z) = 0 = \begin{cases} \frac {|z|}{Re(z)} &, \quad \text{if } {} \text{Re} (z)\neq 0 \ 0 &, \quad \text{if } \text{ Re (z) = 0} \end{cases}  

As,  |z| = √x2 + y2 ⇒ f(z) ={√(x2+y2)x,if x≠0 0,if  x = 0 = \begin{cases} \frac {\rm √(x^2+y^2)}{\rm x} &, \quad \text{if } {} \text x\neq 0 \ 0 &, \quad \text{if } \text{ x = 0} \end{cases}

Now, Check Continuous at x = 0

⇒ lim(x,y)→(0,0)lim_{(x,y) → (0,0)} f(z) = lim(x,y)→(0,0)lim_{(x,y) → (0,0)}( \frac {\rm √(x^2+y^2)}{\rm x} \)

Let, y = mx 

⇒ limx→0lim_{\rm x → 0}(\frac {\rm √(x^2+m^2 x^2)}{\rm x} \) = limx→0lim_{\rm x → 0} x√(1+m2)x\frac {\rm x\rm √(1+m^2 )}{\rm x}  = √ 1 + m2 

Clearly, lim(x,y)→(0,0)lim_{(x,y) → (0,0)} f(z) is path dependent ⇒ f(z) is not continuous at 0 

Hence, Option (4) is true

Question 137

Let X be a complete metric space then no non-empty open sub-set of X is of first category. Which theorem states this?

  1. AOption (a)

    Cauchy's Theorem

  2. BOption (b)

    Intermediate Value Theorem

  3. COption (c)

    Bolzano-Weierstrass Theorem

  4. DOption (d)

    The Baire category Theorem

View answer & explanation

Correct answer

Answer: (d)D.

The Baire category Theorem

Why this answer is correct

Concept:

The Baire Category Theorem:- If {An} is a sequence of nowhere dense sets in a complete metric space X. Then there exists a point in X which is not in any of An.

Or we can say that, Every complete metric space is of second category as subset of itself.

Calculation: 

Let (X, α) be a complete metric space.

For contradiction, assume that X is not of second category.

⇒ X is of first category.

⇒ X is a union of collection of non-dense countable subsets of X.

Let this collection is:- x = U{An} , where A1, A2,.... are non-dense.

∵ A1 is non-dense.

⇒ There exists a closed surface S1 with radius r1 < 1 and S1∩A1 =Φ    [by theorem]

Let C1 is an open sphere concentric with S1 and same radius as r1. 

Then In C1, we may find a closed sphere S2 with radius r2 < 1/2 and S2∩A2 = Φ   [because A2 is non-dense]

Further, proceeding in this way, we get a nested/decreasing sequence. {Sn} of closed sets of X. [because closed sphere is closed set.]

and S{Sn} → 0 when n → ∞ and Sn∩An = Φ   

⇒ For every positive integer n ∈ N, Sn does not intersect with A1, A2,....,An 

Now, ∵ X is a complete metric space then by canter intersection theorem, there exists a point x ∈ X such that x ∈ Sn

But Sn∩An = Φ   

⇒ x ∈ ∩Sn and x ∉ An ; n = 1, 2, ....

∵ x ∉  An ∀ n ⇒ x ∉  X    [ which is contradiction, because we toook X = U{An}]

Hence our hypothesis was wrong.

⇒ X is not of first category.

⇒ X is of second category.

Question 138

If n is real number and f = rn (cos nθ + i.sin nθ), then

  1. AOption (a)

    f is analytic everywhere except possibly at r = 0

  2. BOption (b)

    f is analytic nowhere

  3. COption (c)

    f is analytic everywhere

  4. DOption (d)

     f is analytic everywhere except possibly at θ = 0 

View answer & explanation

Correct answer

Answer: (a)A.

f is analytic everywhere except possibly at r = 0

Why this answer is correct

Concept:

Cauchy Riemann Equations:- If f = u + iv where u = u(x, y) and v = v(x, y) be any complex valued function of u and v.

The equations ∂u∂x=∂v∂y\frac{∂u}{∂x} = \frac{∂v}{∂y}  or ux = vy and ∂u∂y=−∂v∂x\frac{∂u}{∂y} = - \frac{∂v}{∂x} or uy = - vx  are called Cauchy Riemann equations in cartesian coordinates.

Calculation:

Given:- f(z) = rn(cosnθ + isinnθ)   ----eq (1)

u + iv = rn(cosnθ + isinnθ)

On comparing,

u = rncosnθ   ----eq (2)

v = rnsinnθ   ----eq (3)

\(\frac{∂u}{∂r} = \lim_\limits{Δr \to 0} \frac{u(r + Δr, θ) - u(r, θ)}{Δr}\)

\(\frac{∂u}{∂r} = \lim_\limits{Δr \to 0} \frac{(r +Δr)^ncosnθ - r^ncosnθ}{Δr}\)

\(\frac{∂u}{∂r} = cosnθ\frac{\lim_\limits{Δr \to 0}(r +Δr)^n - 1}{Δr}\)

\(\frac{∂u}{∂r} = r^ncosnθ\frac{\lim_\limits{Δr \to 0} {(1+\frac{Δr}{n})^n - 1}}{Δr}\)

\(\frac{∂u}{∂r} = r^ncosnθ\frac{\lim_\limits{Δr \to 0} {[1+n.\frac{Δr}{r} +n(n-1)(\frac{Δr}{r})^2+....... - 1]}}{Δr}\)

∂u∂r=rncosnθ(nr+0+0)\frac{∂u}{∂r} = r^ncosnθ(\frac{n}{r} + 0 + 0)

∂u∂r=nrn−1cosnθ\frac{∂u}{∂r} = nr^{n-1}cosnθ            Provided that r ≠ 0

\()\(\frac{∂u}{∂θ } = \lim_\limits{Δθ \to 0} \frac{u(r,θ + Δθ) - u(r, θ)}{Δθ }\)

\(\frac{∂u}{∂θ } = \lim_\limits{Δθ \to 0} \frac{r^ncosn(θ + Δθ) - r^ncosnθ}{Δθ }\)

\(\frac{∂u}{∂θ } = r^n\lim_\limits{Δθ \to 0} \frac{cosnθcosnΔθ -sinnθsinnΔθ - cosnθ}{Δθ}\)

∂u∂θ=rn(−sinnθ)n\frac{∂u}{∂θ } = r^n(-sinnθ)n

∂u∂θ=−nrnsinnθ\frac{∂u}{∂θ } = -nr^nsinnθ

Similarly, we can find,

∂v∂r=nrn−1sinnθ\frac{∂v}{∂r} = nr^{n-1}sinnθ

∂v∂θ=nrncosnθ\frac{∂v}{∂θ} = nr^ncosnθ

It is clear that, 

\()\(\frac{∂u}{∂r} = \frac{1}{r} \frac{∂v}{∂θ}\)

∂u∂θ=−r∂v∂r\frac{∂u}{∂θ} = - r \frac{∂v}{∂r}

Cauchy's Riemann equation are satisfied and partial derivatives ∂u∂r,∂u∂θ,∂v∂r,∂v∂θ\frac{∂u}{∂r}, \frac{∂u}{∂θ}, \frac{∂v}{∂r}, \frac{∂v}{∂θ} exists.

So, function f(z) is an analytic everywhere except, r = 0.

Question 139

In a group of 400 players, 250 play football and 200 play cricket, then how many people play both the games? 

  1. AOption (a)

    100

  2. BOption (b)

    200

  3. COption (c)

    50

  4. DOption (d)

    250

View answer & explanation

Correct answer

Answer: (c)C.

50

Why this answer is correct

Concept:

Set Theory:-

A U B means set of all the values in the set A and B.

A ∩ B is the set of common elements of A and B.

Subsets (⊂) is the set such that all the elements of the subset are in the set from which the subset is taken from.

Number of elements in set A = n(A)

n(A U B) = n(A) + n(B) - n(A ∩ B)

A' = U - A, where U is a universal set given and A is any set.

A - B = set values of set A not in set B.

Calculation:

Set A is the students who play football, so n(A) = 250

Set B is the students who play cricket, so n(B) = 200

Set of all students A U B, so n(A U B) = 400

⇒ n(A U B) = n(A) + n(B) - n(A ∩ B)

⇒ 400 = 250 + 200 - n(A ∩ B)

⇒ n(A ∩ B) = 450 - 400

⇒ n(A ∩ B) = 50

∴ 50 people plays both the games

Question 140

If f (z) = |z|2, then

  1. AOption (a)

    f is differentiable everywhere

  2. BOption (b)

    f is differentiable nowhere

  3. COption (c)

    f is differentiable only at zero

  4. DOption (d)

    f is differentiable everywhere except 0

View answer & explanation

Correct answer

Answer: (c)C.

f is differentiable only at zero

Why this answer is correct

Concept:

Continuity:- A function f(z) of a complex variable z defined in the closed and bounded domain D is said to be continuous at a ∈ D if and only if for any arbitrarily chosen positive number ε, however small but not zero, there exists a corresponding number δ > 0 such that |f(z) - f(a)| < ε whenever |z - a| < δ.

Differentiability:- Let w = f(z) be a function of a complex variable z defined in a domain D.

Then f(z) is said to be differentiable at a point zo of D, if \(\lim_\limits{Δz \to 0} \frac{f(z_o + Δz) - f(z_o)}{Δz}\) or \(\lim_\limits{z \to z_0} \frac{f(z) - f(z_o)}{z - z_o}\) exists uniquely and finitely and this limit,

if it exists finitely, then it is called the differential coefficient or derivative of f with respect to z at z = zo.

It is denoted by f'(zo) or by Df(zo).

Sufficient condition for Cauchy Riemann Equations: The sufficient condition for a function f(z) = u + iv to be analytic at all points in region R are: 

(a) ∂u∂x=∂v∂y,∂u∂y=−∂v∂x\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}

(b) ∂u∂x,∂v∂y,∂u∂y,∂v∂x\frac{\partial u}{\partial x} , \frac{\partial v}{\partial y}, \frac{\partial u}{\partial y},\frac{\partial v}{\partial x} are continuous functions of x and y in region R.

Calculation: 

Let f(z) = |z|2 where z = x + iy

This function is continuous at every point because x2 + y2 is continuous at all points.

f(z) = x2 + y2 but f(z) = u + iv 

On comparing the functions we get,

u = x2 + y2 and v = 0

Doing partial differentiation, 

∂u∂x=2x,∂v∂y=0\frac{\partial u}{\partial x} = 2x, \frac{\partial v}{\partial y} = 0

∂u∂y=2y,∂v∂x=0\frac{\partial u}{\partial y} = 2y,\frac{\partial v}{\partial x} = 0

Now apply Cauchy Riemann Equation Condition,

∂u∂x=2x=∂v∂y=0\frac{\partial u}{\partial x} =2x = \frac{\partial v}{\partial y} = 0

only when x = 0.

∂u∂y=2y=∂v∂x=0 \frac{\partial u}{\partial y} = 2y = \frac{\partial v}{\partial x} = 0

only when y = 0. 

Therefore, f(z)  is only complex differentiable on (0, 0).

Discipline 3 (20 questions)

Question 141

If lx + my + n = 0 be the normal to the circle x2 + y2 = a2 then -

  1. AOption (a)

    n = a

  2. BOption (b)

    n = 0

  3. COption (c)

    n = 1

  4. DOption (d)

    n = a2

View answer & explanation

Correct answer

Answer: (b)B.

n = 0

Why this answer is correct

Concept:

 the "normal" of a circle typically refers to a line that is perpendicular to the circle's tangent at a given point on the circle's circumference and radius of circle is always normal of circle.

Explanation:

we know, radius of circle is normal as well radius will passes through centre of circle

Here, Centre of circle is (0,0) 

⇒  lx + my + n = 0 satisfies (0,0)

⇒ l.0 + m.0 + n = 0 ⇒ n = 0

Hence, Option (2) is true.

Question 142

If any tangents to the ellipse x2a2+y2b2\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 makes intercepts of lengths h and k on the axis, then -

  1. AOption (a)

    a2h2−b2k2\frac{a^2}{h^2} - \frac{b^2}{k^2} = 1

  2. BOption (b)

    a2h2+b2k2\frac{a^2}{h^2} + \frac{b^2}{k^2} = 1

  3. COption (c)

    h2a2+k2b2\frac{h^2}{a^2} + \frac{k^2}{b^2} = 1

  4. DOption (d)

    h2a2−k2b2\frac{h^2}{a^2} - \frac{k^2}{b^2} = 1

View answer & explanation

Correct answer

Answer: (b)B.

a2h2+b2k2\frac{a^2}{h^2} + \frac{b^2}{k^2} = 1

Why this answer is correct

Concept:

Let point be P(acosθ, bsinθ) = (x1, y1)

Equation of tangent to ellipse  x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 at (x1, y1) is given by: xx1a2+yy1b2=1\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1

Calculation:

⇒ \()\(\frac{x.acosθ}{a^2} + \frac{y.bsinθ}{b^2} = 1\)

⇒ xcosθa+ysinθb=1\frac{xcosθ}{a} + \frac{ysinθ}{b} = 1

For x - axis intercept, y = 0

⇒ hcosθa=1\frac{hcosθ}{a} = 1, where h is intercept length on x - axis.

⇒ cosθ = ah\frac{a}{h}   ____equation(1)

For y - axis intercept, x = 0

⇒ ksinθb=1\frac{ksinθ}{b} = 1, where k is intercept length on y - axis

⇒ sinθ=bksinθ = \frac{b}{k}   ____equation(2)

On adding equation (1) and (2)

⇒ a2h2+b2k2=cos2θ+sin2θ\frac{a^2}{h^2} + \frac{b^2}{k^2} = cos^2θ + sin^2θ

⇒ a2h2+b2k2=1\frac{a^2}{h^2} + \frac{b^2}{k^2} = 1

If any tangents to the ellipse x2a2\frac{x^2}{a^2} + y2b2\frac{y^2}{b^2} = 1 makes intercepts of lengths h and k on the axis, then- a2h2+b2k2=1\frac{a^2}{h^2} + \frac{b^2}{k^2} = 1

Question 143

In three dimension, the equation x2 = 2.λ .y, λ ≠  0 represents-

  1. AOption (a)

    a cylinder

  2. BOption (b)

    a hyperbolic cylinder

  3. COption (c)

    a parabolic cylinder

  4. DOption (d)

    an elliptic cylinder

View answer & explanation

Correct answer

Answer: (c)C.

a parabolic cylinder

Why this answer is correct

Concept:

(i) y = x2 represent a parabola in the xy - plane.

(ii) y = 2λx represent linear function in x - direction, Scaled by parameter λ.

Explanation:

As, x2 represent a parabola in the xy - plane.

2λy represent linear function in y - direction, Scaled by parameter λ.

When we combine these, we get a family of parabolas that open along the x-axis and extend infinitely in the y - direction forming a cylindrical shape. 

This cylindrical surface is called a parabolic cylinder.

Hence, Option (3) is true

Question 144

Let I be the focal chord of a parabola and d be the distance from vertex, then-

  1. AOption (a)

    1 ∝  1d2\frac{1}{d^2}

  2. BOption (b)

    1 ∝ 1d\frac{1}{d}

  3. COption (c)

    1 ∝ d2

  4. DOption (d)

    1 ∝ d

View answer & explanation

Correct answer

Answer: (a)A.

1 ∝  1d2\frac{1}{d^2}

Why this answer is correct

CONCEPT:  if t1are t2 end point pf a focal chord then t1t2=-1

Distance between two points y−y1=y2−y1x2−x1x−x1 y-y_1=\frac{y_2-y_1}{x_2-x_1 }x-x_1

Distance between one point and line D=∣ax1+by1+c∣a2+b2D=\frac{|ax_1+by_1+c|}{\sqrt {a^2+b^2}}

EXPLANTATION :

We know that  t1t2=-1

 L(focal chord)=L=a[(t2−t1)2]L=a[(t_2-t_1)^2]

L=a[(t1+t2)2−4t1t2]L=a[(t_1+t_2)^2-4t_1t_2]

⇒ a[(t1+t2)2+4]a[(t_1+t_2)^2+4]

Equation of line PQ=y−y1=y2−y1x2−x1x−x1 y-y_1=\frac{y_2-y_1}{x_2-x_1 }x-x_1

Here (x1, y1)=(at12, 2at1) and (x1, y1)=(at22, 2at2)

so Equation of PQ=y−2at2=2t2+t1(x−2at12) y-2at_2=\frac{2}{t_2+t_1 }(x-2at_1^2)

  ⇒  y−2xt2+t1=2at2−2at12t2+t1 y-\frac{2x}{t_2+t_1 }=2at_2-\frac{2at_1^2}{t_2+t_1 }

Distance between the point O and the line PQ=D=∣ax1+by1+c∣a2+b2D=\frac{|ax_1+by_1+c|}{\sqrt {a^2+b^2}}

⇒∣2at12t1+t2−2at11+4(t1+t2)2∣=2at1t2(t1+t2)2+4|\frac{\frac{2at_1^2}{t_1+t_2}-2at_1}{\sqrt{1+\frac{4}{(t_1+t_2)^2}}}|=\frac{2at_1t_2}{\sqrt{(t_1+t_2)^2}+4}

⇒2a(t1−1t2)2+4\frac{2a}{\sqrt{(t_1-\frac{1}t_2)^2}+4}

⇒2a(t1+t2)2+4\frac{2a}{\sqrt{(t_1+t_2)^2}+4}                            { ∴ t1t2=-1}

⇒c(t1+t2)2+4=d(say)\frac{c}{\sqrt{(t_1+t_2)^2}+4} =d(say)

Question 145

A plane passes through a point, (1, –2, 1) and is perpendicular of two planes 2x – 2y + z = 0 and x – y + 2z = 4. The distance of the plane from the point (1, 2, 2) is -

  1. AOption (a)

    2

  2. BOption (b)

    2\sqrt2

  3. COption (c)

    22\sqrt2

  4. DOption (d)

    1

View answer & explanation

Correct answer

Answer: (c)C.

22\sqrt2

Why this answer is correct

Concept Used:

The equation of plane passing through (x0, y0, z0) and parallel to vector (a, b, c) is given by:

a( x - x0) + b(y - y0) + c(z - z0) = 0

Distance between a point (x0,y0,z0)(x_0, y_0, z_0) and a plane Ax + By + Cz = D given by: 

Distance=∣Ax0+By0+Cz0−D∣√A2+B2+C2 \text{Distance} = \frac{|Ax_0 + By_0 + Cz_0 - D|}{√{A^2 + B^2 + C^2}}

Explanation:

Let's start by finding the normal vector to the plane passing through the given point and perpendicular to the given planes:

Given two planes 2x – 2y + z = 0 and x – y + 2z = 4

Normal vectors to the given planes are n1⃗=(2,−2,1)\vec{n_1} = (2, -2, 1) and n2⃗=(1,−1,2)\vec{n_2} = (1, -1, 2).

Normal vector to our desired plane will be the cross product of n1⃗ \vec{n_1} and n2⃗ \vec{n_2}.

n⃗=n1⃗×n2⃗\vec{n} = \vec{n_1} \times \vec{n_2}

⇒n⃗=∣i^j^k^ 2−21 1−12∣⇒ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -2 & 1 \ 1 & -1 & 2 \end{vmatrix}

⇒ n⃗=i^(2(−2)−(1)(−1))−j^(2(2)−1(1))+k^(2(−1)−(1)(−2)) \vec{n} = \hat{i}(2(-2) - (1)(-1)) - \hat{j}(2(2) - 1(1)) + \hat{k}(2(-1) - (1)(-2))

⇒ n⃗=i^(−4+1)−j^(4−1)+k^(−2+2) \vec{n} = \hat{i}(-4 + 1) - \hat{j}(4 - 1) + \hat{k}(-2 + 2)

⇒ n⃗=−3i^−3j^+0k^ \vec{n} = -3\hat{i} - 3\hat{j} + 0\hat{k}

So, the normal vector to the desired plane is n⃗=(−3,−3,0)\vec{n} = (-3, -3, 0)

Equation of the Plane passing through the point (1, -2, 1) and in the direction of normal vector (-3, -3, 0) is

⇒  -3(x - 1) - 3(y + 2) + 0(z - 1) = 0

⇒ x - 1 + y + 2 = 0

⇒ x + y + 1 = 0

Now, distance from the (1, 2, 2) to the plane x + y + 1 = 0 is:

Distance=∣(1)+2+1∣√12+(1)2 \text{Distance} = \frac{|(1) + 2 +1|}{√{1^2 + (1)^2}}

⇒  Distance=4√2=4√2×√2√2=4√22=2√2\text{Distance} = \frac{4}{√{2}} = \frac{4}{√{2}} \times \frac{√ 2}{√ 2} = \frac{4√{2}}{2} = 2√{2}

So, the distance from the point (1, 2, 2) to the plane x + y + 1 = 0 is 2√22√2.

Question 146

The equation ax2 + by2 + cz2 + 2ux + 2vy + 2wz + d = 0, represents a cone if-

  1. AOption (a)

    au2 + bv2 + cw2 = d

  2. BOption (b)

    (b + c).u2 + (a + c).v2 + (a + b).w2= d

  3. COption (c)

    bcu + acv + abw = d

  4. DOption (d)

    u2a\frac{u^2}{a}  + v2b\frac{v^2}{b} + w2c\frac{w^2}{c} = d

View answer & explanation

Correct answer

Answer: (d)D.

u2a\frac{u^2}{a}  + v2b\frac{v^2}{b} + w2c\frac{w^2}{c} = d

Why this answer is correct

Concept Used:

For a general second-degree equation in three dimensions to represent a cone, if

The determinant of the matrix of the quadratic part of the equation must be zero. This is equivalent to the condition that the matrix formed by the coefficients of the quadratic terms (a, b, c) and the linear terms (u, v, w) has a determinant equal to zero when considering the equation as a homogeneous equation (setting the constant term to zero for the moment).

Explanaition:

The quadratic form of the equation can be represented by the matrix:

∣a00u 0b0v 00cw uvwd ∣=0\begin{vmatrix} a & 0 & 0 & u \ 0 & b & 0 & v \ 0 & 0 & c & w \ u & v & w & d \ \end{vmatrix}= 0

⇒ a×∣b0v 0cw vwd ∣−0+0−u×∣0b0 00c vwd ∣=0a \times \begin{vmatrix} b & 0 & v \ 0 & c & w \ v & w & d \ \end{vmatrix} - 0 + 0 - u \times \begin{vmatrix} 0 & b & 0 \ 0 & 0 & c \ v & w & d \ \end{vmatrix} = 0

⇒ a[b×∣cw wd ∣−0+v∣0v cw ∣]−u[0−0+u∣b0 0c ∣]=0a \left[ b \times \begin{vmatrix} c & w \ w & d \ \end{vmatrix} - 0 + v \begin{vmatrix} 0 & v \ c & w \ \end{vmatrix} \right] - u \left[ 0 - 0 + u \begin{vmatrix} b & 0 \ 0 & c \ \end{vmatrix} \right] = 0

⇒ a[b{cd - w2} + v{cv}] - u[ u{bc}] = 0

⇒ abcd - abw2 -acv2 - bcu2 = 0

Code
⇒   abw<sup>2 </sup>+ acv<sup>2 </sup>+ bcu<sup>2 </sup>= abcd

Divide by abc, we get

w2/c + v2/b + u2/a = d

Thus, option 4 is the correct answer.

Question 147

The surface of revolution of the straight line z2 = y, x = 0 about the y axis is-

  1. AOption (a)

    z4 = x2 + y2

  2. BOption (b)

    x2 + z2 = y

  3. COption (c)

    y4 = x2 + z2

  4. DOption (d)

    y2 + z2 = x

View answer & explanation

Correct answer

Answer: (b)B.

x2 + z2 = y

Why this answer is correct

Concept -

If a curve is given by z = f(y) for a ≤  y ≤  b, then the surface of revolution generated by revolving this curve around the y-axis is given by:

x2 + z2 = {f(y)}2

Explanation -

In this case, the given curve is z2 = y for x = 0. Solving for z, we get z = √y.

Now, applying the formula:

x2 + z2 = (√y)2 

x2 + z2 = y 

Hence the surface of revolution of the straight line z2 = y, x = 0 about the y axis is x2 + z2 = y.

Question 148

The eccentricity and semi – latus rectum of the curve 1r\frac{1}{r} = 8 + 5.cos θ, are respectively-

  1. AOption (a)

    58\frac{5}{8}, 8

  2. BOption (b)

    18\frac{1}{8}, 8

  3. COption (c)

    18\frac{1}{8}, 58\frac{5}{8}

  4. DOption (d)

    58\frac{5}{8}, 18\frac{1}{8}

View answer & explanation

Correct answer

Answer: (d)D.

58\frac{5}{8}, 18\frac{1}{8}

Why this answer is correct

Concept:

The general equation of a curve in polar co-ordinates is 1r\frac{1}{r} = 1 + e .cos θ

Where l is the semi-latus rectum and e is the eccentricity.

Calculation: 

The general equation of a curve in polar co-ordinates is 1r\frac{1}{r} = 1 + e .cos θ

Where l is the semi-latus rectum and e is the eccentricity.

Given equation of the curve is 1r\frac{1}{r} = 8 + 5.cos θ

Dividing the equation by 8, we get:

18r\frac{1}{8r} = 1 + 58\frac{5}{8}.cos θ

Semi latus rectum is 18\frac{1}{8} and eccentricity is 58\frac{5}{8}.

Question 149

Two conjugate diameters of the hyperbola are such that -

  1. AOption (a)

    Both meet the curve at imaginary points

  2. BOption (b)

    They meet the curve at infinity

  3. COption (c)

    Both meet the curve at real point

  4. DOption (d)

    One meets the curve at real point and another meets at imaginary point

View answer & explanation

Correct answer

Answer: (b)B.

They meet the curve at infinity

Why this answer is correct

Concept -

In a hyperbola, there are two types of diameters - conjugate diameters and principal diameters.

1. Conjugate Diameters:

  • Conjugate diameters are diameters that are perpendicular to each other and intersect at the center of the hyperbola.  
  • The length of the conjugate diameters is constant and is equal to the length of the transverse axis of the hyperbola.  - Conjugate diameters have a special property: if one of them meets the hyperbola at two real points, the other will meet the hyperbola at two imaginary points, and vice versa.
  • The product of the slopes of the two conjugate diameters is always equal to -1.

2. Principal Diameters:

  • Principal diameters are diameters that are parallel to the asymptotes of the hyperbola. 
  • They intersect at the center of the hyperbola.
  • The length of the principal diameters is constant and is equal to the length of the conjugate axis of the hyperbola.

Explanation -

If one of the conjugate diameters meets the hyperbola at two real points, then the other conjugate diameter will meet the hyperbola at two imaginary points.

This is because the two conjugate diameters are perpendicular, and one diameter cuts the hyperbola at real points, while the other cuts at imaginary points.

But they meet the curve at infinity because the point of intersection of the two conjugate diameters is at infinity.

Hence two conjugate diameters of the hyperbola are such that they meet the curve at infinity

Question 150

The shortest distance between the z-axis and the line x−21\frac{x - 2}{1} = y−12\frac{y -1}{2} = z+12\frac{z + 1}{2}  is-

  1. AOption (a)

    25\frac{2}{\sqrt5}

  2. BOption (b)

    35\frac{3}{\sqrt5}

  3. COption (c)

    35\sqrt5

  4. DOption (d)

    53\frac{5}{\sqrt3}

View answer & explanation

Correct answer

Answer: (b)B.

35\frac{3}{\sqrt5}

Why this answer is correct

Concept:

Distance between lines r⃗=a1⃗+λb1⃗\vec{r}=\vec{a_1}+\lambda\vec{b_1} and r⃗=a2⃗+μb2⃗\vec{r}=\vec{a_2}+\mu\vec{b_2} is given by d = ∣(b1⃗×b2⃗)⋅(a2⃗−a1⃗)∣b1⃗×b2⃗∣∣\left|\frac{(\vec{b_1}\times\vec{b_2})\cdot(\vec{a_2}-\vec{a_1})}{|\vec{b_1}\times\vec{b_2}|}\right|

Calculation:

Given line, x−21\frac{x - 2}{1} = y−12\frac{y -1}{2} = z+12\frac{z + 1}{2} 

Comparing with x−x1a1=y−y1b1=z−z1c1\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}, we get:

x1 = 2, y1 = 1, z1 = -1 and a1 = 1, b1 = 2, c1 = 2

∴ a1⃗=x1i^+y1j^+z1k^\vec{a_1} = x_1\hat{i}+y_1\hat{j}+z_1\hat{k} ⇔ a1⃗=2i^+j^−k^\vec{a_1} = 2\hat{i}+\hat{j}-\hat{k}

and, b1⃗=a1i^+b1j^+c1k^\vec{b_1} = a_1\hat{i}+b_1\hat{j}+c_1\hat{k} ⇔ b1⃗=i^+2j^+2k^\vec{b_1} = \hat{i}+2\hat{j}+2\hat{k}

Now, and equation of z-axis is  x0\frac{x}{0} = y0\frac{y}{0} = z1\frac{z}{1} 

Comparing with x−x2a2=y−y2b2=z−z2c2\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}, we get:

x2 = 0, y2 = 0, z2 = 0 and a2 = 0, b2 = 0, c2 = 1

∴ a2⃗=x2i^+y2j^+z2k^\vec{a_2} = x_2\hat{i}+y_2\hat{j}+z_2\hat{k} ⇔ a2⃗=0i^+0j^+0k^\vec{a_2} = 0\hat{i}+0\hat{j}+0\hat{k}

and, b2⃗=a2i^+b2j^+c2k^\vec{b_2} = a_2\hat{i}+b_2\hat{j}+c_2\hat{k} ⇔ b1⃗=k^\vec{b_1} = \hat{k}

Now, a2⃗−a1⃗=−2i^−j^+k^\vec{a_2}-\vec{a_1}= -2\hat{i}-\hat{j}+\hat{k}

b1⃗×b2⃗\vec{b_1}\times\vec{b_2} = ∣i^j^k^ 122 001∣\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\ 1 & 2 & 2\ 0 & 0 & 1 \end{vmatrix} = 2i^−j^2\hat{i}-\hat{j}

Maginutude of b1⃗×b2⃗\vec{b_1}\times\vec{b_2} = ∣b1⃗×b2⃗∣|\vec{b_1}\times\vec{b_2}| = 22+(−1)2\sqrt{2^2+(-1)^2} = 5\sqrt{5}

Also, (b1⃗×b2⃗)⋅(a2⃗−a1⃗)(\vec{b_1}\times\vec{b_2})\cdot(\vec{a_2}-\vec{a_1}) = (2i^−j^)⋅(−2i^−j^+k^)(2\hat{i}-\hat{j})\cdot(-2\hat{i}-\hat{j}+\hat{k}) = -4 + 1 = -3

∴ Distance, d = ∣(b1⃗×b2⃗)⋅(a2⃗−a1⃗)∣b1⃗×b2⃗∣∣\left|\frac{(\vec{b_1}\times\vec{b_2})\cdot(\vec{a_2}-\vec{a_1})}{|\vec{b_1}\times\vec{b_2}|}\right|

= ∣−35∣\left|\frac{-3}{\sqrt{5}}\right|

= 35\frac{3}{\sqrt5}

∴ The shortest distance between the z-axis and the line x−21\frac{x - 2}{1} = y−12\frac{y -1}{2} = z+12\frac{z + 1}{2}  is 35\frac{3}{\sqrt5}

The correct answer is Option 2.

Question 151

The angle between the lines represented by x2 – 2xy – y2 + 2x + 3y – 1 = 0, is -

  1. AOption (a)

    π2\frac{\pi}{2}

  2. BOption (b)

    tan–1(1/2)

  3. COption (c)

    tan–1(1/4)

  4. DOption (d)

    π4\frac{\pi}{4}

View answer & explanation

Correct answer

Answer: (a)A.

π2\frac{\pi}{2}

Why this answer is correct

Concept:

The angle between the lines represented by ax2 + by2 + 2hxy + 2gx + 2fy + c = 0 is given by θ = tan⁡−1∣2h2 − aba + b∣\tan^{-1} \left | \frac {2\sqrt {h^2 \ -\ ab}}{a\ +\ b} \right |

Solution:

Given, x2 – 2xy – y2 + 2x + 3y – 1 = 0

Comparing above with general equation ax2 + by2 + 2hxy + 2gx + 2fy + c = 0, we get:

a = 1; b = -1; h = -1; g = 1; f = 1.5; c = -1;

∴ θ = tan⁡−1∣2h2 − aba + b∣\tan^{-1} \left | \frac {2\sqrt {h^2 \ -\ ab}}{a\ +\ b} \right |

= tan⁡−1∣2(−1)2 − (1×−1)1 − 1∣\tan^{-1} \left | \frac {2\sqrt {(-1)^2 \ -\ (1 \times -1)}}{1\ -\ 1} \right |

= tan⁡−1∣220∣=tan⁡−1∣∞∣\tan^{-1} \left | \frac {2\sqrt {2}}{0} \right | = \tan^{-1} |\infty |

= tan-1(tan(π2\frac{\pi }{2}))

⇒ θ = π2\frac{\pi }{2}

∴ The angle between the two lines is π2\frac{\pi }{2}.

The correct answer is option 1.

Question 152

The image of the point (–3, 8, 4) in the plane 6x – 3y – 2z + 1 = 0, is -

  1. AOption (a)

    (0, 2, 9)

  2. BOption (b)

    (9, 2, 0)

  3. COption (c)

    (1, 2, –2)

  4. DOption (d)

    (–2, 1, 3)

View answer & explanation

Correct answer

Answer: (b)B.

(9, 2, 0)

Why this answer is correct

Concept:

If the equation of the plane is given by Ax + By + Cz + D=0.

Then the equation of the normal line passing through the point (x0, y0, z0) is : (x−x0)A=(y−y0)B=(z−z0)C\rm \frac{(x-x_{0})}{A} = \frac{(y-y_{0})}{B}=\frac{(z-z_{0})}{C}

Explanation:

Let Q be the image of the point P (-3, 8, 4) in the plane 6x – 3y – 2z + 1 = 0.

Then, PQ is normal to the plane. So the direction ratios of PQ are proportional to 6, -3, -2.

Since, PQ passes through P(-3, 8, 4) and has direction ratios proportional to 6, -3, -2, equation of PQ is 

(x+3)6=(y−8)−3=(z−4)−2=r\rm \frac{(x+3)}{6} = \frac{(y-8)}{-3}=\frac{(z-4)}{-2}=r

Let the coordinates of Q be (6r - 3, -3r+8, -2r+4). Let R be the mid-point of PQ. Then,

R = (−3+6r−32,−3r+8+82,−2r+4+42\frac{-3+6r-3}{2}, \frac{-3r+8+8}{2}, \frac{-2r+4+4}{2} = (3r-3, 16−3r2\frac{16-3r}{2}, 4-r) 

Since, R lies in the plane 6x – 3y – 2z + 1 = 0.

6(3r-3) - 3(16−3r2\frac{16-3r}{2}) - 2(4-r)+1 =0 ⇒ r = 2

Substitute this in the coordinate of Q, we get

Q = (6r - 3, -3r+8, -2r+4) = (6(2) - 3, -3(2)+8, -2(2)+4) = (9, 2, 0)

Hence, Option (2) is true.

Question 153

The general equation of the cone which passes through the co-ordinate axes is-

  1. AOption (a)

    ax2 + b2 + cz2 + 2hxy + 2fyz + 2gzx = 0

  2. BOption (b)

    hxy + fyz + gzx = 0

  3. COption (c)

    x2y2 + y2z2 + z2x2 = 0

  4. DOption (d)

    xy + yz – zx = 0

View answer & explanation

Correct answer

Answer: (b)B.

hxy + fyz + gzx = 0

Why this answer is correct

Concept:

Second degree equation of cone is

ax2 + by2 + cz2 + 2 fyz + 2 gzx + 2 hxy + d = 0

Calculation:

Given:

As cone passes through axes, therefore it must satisfy axes coordinates i.e. (1, 0, 0), (0, 1, 0) & (0, 0, 1)

For x-axis (1, 0, 0)

ax2 + by2 + cz2 + 2 fyz + 2 gzx + 2 hxy + d = 0

ax2 = 0

a =0

For y-axis (0, 1, 0)

ax2 + by2 + cz2 + 2 fyz + 2 gzx + 2 hxy + d = 0

by2 = 0

b = 0

For z-axis (0, 0, 1)

ax2 + by2 + cz2 + 2 fyz + 2 gzx + 2 hxy + d = 0

cz2 = 0

c = 0

Hence, the equation will be

2 fyz + 2 gzx + 2 hxy + d = 0 

fyz + gzx + hxy = 0

Question 154

Equation of the sphere that passes through the points (0, 0, 0), (a, 0, 0), (0, b, 0), (0, 0, c) is -

  1. AOption (a)

    (x−a2)2(x - \frac{a}{2})^2 + (y−b2)2(y - \frac{b}{2})^2 + (z−c2)2(z - \frac{c}{2})^2 = 12\frac{1}{2} (a2 + b2 + c2)

  2. BOption (b)

    x2+ y2 + z2 = a2 + b2 + c2

  3. COption (c)

    (x−a2)2(x - \frac{a}{2})^2 + (y−b2)2(y - \frac{b}{2})^2 + (z−c2)2(z - \frac{c}{2})^2 = 14\frac{1}{4} (a2 + b2 + c2)

  4. DOption (d)

    (x – a)2 + (y – b)2 + (z – c)2 = 12\frac{1}{2}(\sqrt{(a^2 + b^2 + c^2)}\)

View answer & explanation

Correct answer

Answer: (c)C.

(x−a2)2(x - \frac{a}{2})^2 + (y−b2)2(y - \frac{b}{2})^2 + (z−c2)2(z - \frac{c}{2})^2 = 14\frac{1}{4} (a2 + b2 + c2)

Why this answer is correct

Concept:

General equation of sphere is x2 + y2 + z2 + 2ux + 2vy + 2wz + c = 0

Explanation:

Now, equation of sphere is  x2 + y2 + z2 + 2ux + 2vy + 2wz + c = 0 ....... (i)

Passes through (0,0,0), (a,0,0), (0,b,0), (0,0,c) 

Now, when (i) equation passes through (0,0,0) ⇒ c = 0 

Now, when (i) equation passes through (a,0,0) ⇒ u = −a2\frac{-a}{2} 

Similarly, For (0,b,0) we get v = −b2\frac{-b}{2} and for (0,0,c) we get w = −c2\frac{-c}{2} 

By substituting all value we get

 x2 + y2 + z2 - ax - by - cz = 0

Now, By completing the square we get

(x−a2)2(x - \frac{a}{2})^2 + (y−b2)2(y - \frac{b}{2})^2 + (z−c2)2(z - \frac{c}{2})^2 =  14\frac{1}{4} (a2 + b2 + c2)

Hence, Option (3) is true

Question 155

The equation xy – ax – by + ab = 0 represents

  1. AOption (a)

    a pair of straight lines

  2. BOption (b)

    a hyperbola

  3. COption (c)

    a circle

  4. DOption (d)

    an ellipse

View answer & explanation

Correct answer

Answer: (a)A.

a pair of straight lines

Why this answer is correct

Concept:

General equation of second degree :- ax2  + 2hxy + by2  + 2gx + 2fy + c = 0

The necessary and sufficient condition to represent a pair of straight lines are:-

  1. The discriminant abc + 2fgh - af2 - bg2 - cf2 = 0
  2. h2 ≥ ab , g2 ≥ ca, and f2 ≥ bc.

Calculation:

Compare equation with general equation of second degree,

a = 0, b = 0 , c = ab, h = 1/2, g = - a/2, f = - b/2

Discriminant- abc + 2fgh - af2 - bg2 - ch2 = 0

now put the values, 

= 0 - 2[−b2.−a2.12]2[\frac{-b}{2}.\frac{-a}{2}.\frac{1}{2}] - 0 - 0 - ab4\frac{ab}{4}

= ab4\frac{ab}{4} - ab4\frac{ab}{4}

= 0 

So, the discriminant is equal to zero

Now check, h2 ≥ ab, g2 ≥ ca, f2 ≥ bc.

put the values of a, b, c,h2 , g2 and f2

1/4 ≥ 0, a2/4 ≥ 0, b2/4 ≥ 0.

So, all conditions are satisfied.

Hence, the given equation represents a pair of straight lines.

Question 156

The equation of the plane through the point (1, 2, –3) and normal to the straight line joining the points (–1, 3, 4) and (5, 2, –1) is-

  1. AOption (a)

    6x – y – 5z – 19 = 0

  2. BOption (b)

    x – 2y – 3z – 17 = 0

  3. COption (c)

    6x – y + 5z + 19 = 0

  4. DOption (d)

    3x – 2y – 5z – 9 = 0

View answer & explanation

Correct answer

Answer: (a)A.

6x – y – 5z – 19 = 0

Why this answer is correct

Concept:

Vector equation of the plane is (r⃗−a⃗)n⃗=0(\vec{r} -\vec{a})\vec{n} =0 where is a position vector, ⃗a⃗{a} is passing through a plane, ⃗n⃗{n} is a unit normal vector.

Calculation:

Plane passes through (1, 2, -3)

⃗a=i+2j−3k⃗{a} = i + 2j - 3k

Plane is perpendicular to line joining (-1, 3, 4) and (5, 2, -1)

n⃗=(x2−x1)i+(y2−y1)j+(z2−z1)k\vec{n} = (x_2- x_1)i + (y_2 -y_1)j + (z_2- z_1)k

n⃗=(5+1)i+(2−3)j+(−1−4)k\vec{n} = (5 + 1 )i + (2 -3)j + (-1 -4)k

n⃗=6i−j−5k\vec{n} = 6i - j -5k

So, the vector equation of the plane is 

⃗r−(i+2j−3k).(6i−j−5k)=0{⃗{r} - ( i + 2j -3k)}.(6i - j - 5k) = 0      ̂

Cartesian equation of plane is

{xi + yj + zk - (i + 2j - 3k)}.(6i - j - 5k) = 0

⇒ {(x - 1)i + (y + 2)j + (z - 3)k}.{6i - j - 5k) = 0

⇒ 6(x - 1) - (y + 2) - 5(z - 3) = 0

⇒ 6x - 6 - y + 2 - 5z - 15 = 0

⇒ 6x - y - 5z - 19 = 0

Hence, The equation of the plane through the point (1, 2, –3) and normal to the straight line joining the points (–1, 3, 4) and (5, 2, –1) is 6x - y - 5z - 19 = 0

Question 157

If S1 = 0 and S2 = 0 are the equations of a sphere, then S1 + λ.S2 = 0 represents-

  1. AOption (a)

    a circle

  2. BOption (b)

    a plane

  3. COption (c)

    a sphere passing through the circle of intersection of the spheres

  4. DOption (d)

    a sphere passing through the plane of intersection

View answer & explanation

Correct answer

Answer: (c)C.

a sphere passing through the circle of intersection of the spheres

Why this answer is correct

Concept: 

Equation of sphere in standard form:- x2 + y2 + z2  = r2 

Calculation:

Consider two spheres as,

S1 : x2 + y2 + z2  + 2u1x +  2v1y + 2w1z + d1 = 0

S2 : x2  + y2 + z2 + 2u2x + 2v2y + 2w2z + d2 = 0

Then the general sphere through the intersection of spheres S1 and S2 is given by ;

S1 + λS2  = 0

where, λ is an arbitrary constant.

Question 158

The polar co-ordinate of a centre of the circle r = 4.cos θ + 3.sin θ is-

  1. AOption (a)

    (5,tan⁡−134)(5, \tan^{-1}\frac{3}{4})

  2. BOption (b)

    (52,tan⁡−134)(\frac{5}{2}, \tan^{-1}\frac{3}{4})

  3. COption (c)

    (52,tan⁡−123)(\frac{5}{2}, \tan^{-1}\frac{2}{3})

  4. DOption (d)

    (34,tan⁡−123)(\frac{3}{4}, \tan^{-1}\frac{2}{3})

View answer & explanation

Correct answer

Answer: (b)B.

(52,tan⁡−134)(\frac{5}{2}, \tan^{-1}\frac{3}{4})

Why this answer is correct

Concept:

The polar equation of the circle with the center as the origin is r = p, where p is the radius of the circle.

Formula required:- centre - (A/2 - B/2), radius = x2 + y2 

Calculation:

r = 4cosθ + 3sinθ

Converting into cartesian form by using r2 = x2 + y2  and x = rcosθ + y = rsinθ, 

r2 = 4x + 3y

x2 + y2  = 4x + 3y

Coordinates of center of this circle in the cartesian form: ( 2, 3/2) 

In polar form, (r, θ) the center is

⇒ A2+B22\frac{\sqrt{A^2+ B^2}}{2} , tan⁡−1BA \tan^{-1}\frac{B}{A}

⇒ (252, tan⁡−134)(\frac{\sqrt{25}}{2},\ \tan^{-1}\frac{3}{4})

⇒ (52\frac{5}{2}, tan⁡−134 \tan^{-1}\frac{3}{4})

Question 159

From a hyperbola or ellipse how many normals can be drawn?

  1. AOption (a)

    One

  2. BOption (b)

    Four

  3. COption (c)

    Two

  4. DOption (d)

    Three

View answer & explanation

Correct answer

Answer: (b)B.

Four

Why this answer is correct

Concept:

Both ellipse and hyperbola are conic sections, but the ellipse is a closed curve while the hyperbola consists of two open curves. Therefore, the ellipse has finite perimeter, but the hyperbola has an infinite length.

Calculation:

Four normals can be drawn in total, which can be either real or imaginary. At most four real normals can be drawn from any point in the plane of rectangular hyperbola and minimum two real normals from any point in the plane of rectangular hyperbola.

Question 160

If the normals at the point (ct1,ct1)(ct_1, \frac{c}{t_1}) on the hyperbola xy = c2 meets it again at the point (ct2,ct2)(ct_2, \frac{c}{t_2}) then-

  1. AOption (a)

    t12 t2 = –1

  2. BOption (b)

    t23 t1 = –1

  3. COption (c)

    t2 t13 = –1

  4. DOption (d)

    t22 t1 = –1

View answer & explanation

Correct answer

Answer: (c)C.

t2 t13 = –1

Why this answer is correct

Concept;

Equation of normal passes at (ct1,ct1)(ct_1, \frac{c}{t_1}) is yt - xt3 - c + ct4  = 0

If it passes through any point say P having its coordinates as 't' i.e., (ct2,ct2)(ct_2, \frac{c}{t_2}).

Calculation:

The coordinates must satisfy the equation. 

Hence, we get ct2\frac{c}{t_2}t1 - ct2t13 - c + ct14  = 0

t1 - t13t22 - t2 + t2t14  = 0

Factorize this equation, (t1 - t2) + t2t13(t1 - t2) = 0

On factorizing, we get, (t1 - t2)(t2t13 + 1) = 0

Hence, t2t13 + 1 = 0  ⇒ t2t13  = -1 

or t1 ≠ t2 ⇒ This is not possible.

Hence,if the normals at the point (ct1,ct1)(ct_1, \frac{c}{t_1}) on the hyperbola xy = c2 meets it again at the point (ct2,ct2)(ct_2, \frac{c}{t_2}) then- t2t13 = -1

Discipline 4 (20 questions)

Question 161

Find the value of: sec⁡8A(tan⁡10A+tan⁡6A)4(tan⁡10A−tan⁡6A)\frac{\sec8A(\tan10A + \tan6A)}{4(\tan10A - \tan6A)}

  1. AOption (a)

    sin 2A

  2. BOption (b)

    cos 4A

  3. COption (c)

    tan 4A

  4. DOption (d)

    sin 4A

View answer & explanation

Correct answer

Answer: (b)B.

cos 4A

Why this answer is correct

Concept:

Formula Used:-

sin(A + B) = sinAcosB + cosAsinB

sin(A - B) = sinAcosB - cosAsinB

sin2θ = 2sinθ cosθ 

Calculation: 

=sec8A(sin10Acos10A+sin6Acos6A)4(sin10Acos10A−sin6Acos6A)= \frac{sec8A(\frac{sin10A}{cos10A} + \frac{sin6A}{cos6A})}{4(\frac{sin10A}{cos10A} - \frac{sin6A}{cos6A})}

=sec8A(sin10Acos6A+sin6Acos10Acos10Acos6A)4(sin10Acos6A−sin6Acos10Acos10Acos6A)= \frac{sec8A(\frac{sin10Acos6A + sin6Acos10A}{cos10Acos6A})}{4(\frac{sin10Acos6A - sin6Acos10A}{cos10Acos6A})}

=sec8A(sin10Acos6A+sin6Acos10A)4(sin10Acos6A−sin6Acos10A)= \frac{sec8A(sin10Acos6A + sin6Acos10A)}{4({sin10Acos6A - sin6Acos10A})}

By using formula,

=sec8A[sin(10A+6A)]4[sin(10A−6A)]= \frac{sec8A[sin(10A + 6A)]}{4[sin(10A - 6A)]}

=sec8A4(sin16A)(sin4A)= \frac{sec8A}{4}\frac{(sin16A)}{(sin4A)}

=sec8A4(2sin8Acos8A)(sin4A)= \frac{sec8A}{4} \frac{(2sin8Acos8A)}{(sin4A)}

=sec8A42(2sin4Acos4A)cos8Asin4A= \frac{sec8A}{4} \frac{2(2sin4Acos4A)cos8A}{sin4A}

=sec8A4×4cos4Acos8A= \frac{sec8A}{4}\times {4cos4Acos8A}

=(sec8A)(cos4A)sec8A= \frac{(sec8A)(cos4A)}{sec8A}

= cos4A

Hence, The value of: sec⁡8A(tan⁡10A+tan⁡6A)4(tan⁡10A−tan⁡6A)\frac{\sec8A(\tan10A + \tan6A)}{4(\tan10A - \tan6A)} is cos4A

Question 162

Express the relation between radius of spherical-curvature R and the radius of curvature ρ of the curve P and σ (the reciprocal of the torsion).

  1. AOption (a)

    R2 = σ2 + (ρ' ρ)2

  2. BOption (b)

    R2 = ρ2 + (ρ' σ)2

  3. COption (c)

    R2 = σ2 + (p1σ)2(\frac{p^1}{\sigma})^2

  4. DOption (d)

    R2 = ρ'2  + (ρσ)2

View answer & explanation

Correct answer

Answer: (b)B.

R2 = ρ2 + (ρ' σ)2

Why this answer is correct

Concept:

Radius of curvature:- The radius of a circle which touches a curve at a given point and has the same tangent and curvature at that point. It is represent by ρ.

The reciprocal of the radius of the osculating circle, torsion is the reciprocal of the radius of torsion. It is represented by σ.

Calculation:

R2 = ρ2 + σ2ρ'2 

R2 = ρ2 + (ρ'σ)2

Question 163

If cos–1 (pa)(\frac{p}{a}) + cos–1 (qb)(\frac{q}{b}) = α, then p2a2\frac{p^2}{a^2} + kcos α, + q2b2\frac{q^2}{b^2} = sin2 α where k is equal to

  1. AOption (a)

    −2pqab-\frac{2pq}{ab}

  2. BOption (b)

    2pqab\frac{2pq}{ab}

  3. COption (c)

    −pqab-\frac{pq}{ab}

  4. DOption (d)

    pqab\frac{pq}{ab}

View answer & explanation

Correct answer

Answer: (a)A.

−2pqab-\frac{2pq}{ab}

Why this answer is correct

Concept:

Formula Used:- cos(A + B) = cosAcosB - sinAsinB

Calculation:

Let cosA = pq\frac{p}{q} and cosB = qb\frac{q}{b}

Thus α = A + B

Taking cos on both sides,

⇒ cosα = cosAcosB - sinAsinB

⇒ cosα = paqb−1−p2a21−q2b2\frac{p}{a} \frac{q}{b} - \sqrt{1- \frac{p^2}{a^2}}\sqrt{1 - \frac{q^2}{b^2}}

⇒ cosα = pqab−(a2−b2)(b2−q2)a2b2\frac{pq}{ab} - \frac{\sqrt{(a^2 - b^2)(b^2 - q^2)}}{a^2b^2}

Squaring on both sides,

⇒ cos2α = [pqab−(a2−b2)(b2−q2)a2b2]2[\frac{pq}{ab} - \frac{\sqrt{(a^2 - b^2)(b^2 - q^2)}}{a^2b^2}]^2

⇒ cos2α = p2q2−2pq(a2−p2)(b2−q2)+(a2−p2)(b2−q2)a2b2\frac{p^2q^2 - 2pq\sqrt{(a^2 - p^2)(b^2 - q^2)} + (a^2 - p^2)(b^2 - q^2)}{a^2b^2}

⇒ cos2α = a2b2−b2p2−a2q2+2p2q2−2pq(a2−p2)(b2−q2)a2b2\frac{a^2b^2 -b^2p^2 - a^2q^2 +2p^2q^2 - 2pq\sqrt{(a^2 - p^2)(b^2 - q^2)}}{a^2b^2}

Lets take 1 - cos2α = 1 - a2b2−b2p2−a2q2+2p2q2−2pq(a2−p2)(b2−q2)a2b2\frac{a^2b^2 -b^2p^2 - a^2q^2 +2p^2q^2 - 2pq\sqrt{(a^2 - p^2)(b^2 - q^2)}}{a^2b^2}

⇒ sin2α = a2b2−a2b2+b2p2+a2q2−2p2q2+2pq(a2−p2)(b2−q2)a2b2\frac{a^2b^2 - a^2b^2 + b^2p^2 + a^2q^2 -2p^2q^2 + 2pq\sqrt{(a^2 - p^2)(b^2 - q^2)}}{a^2b^2}

⇒ sin2α = b2p2+a2q2a2b2−2p2q2−2pq(a2−p2)(b2−q2)a2b2\frac{b^2p^2 + a^2q^2}{a^2b^2} -\frac{2p^2q^2 - 2pq\sqrt{(a^2 - p^2)(b^2 - q^2)}}{a^2b^2}

⇒ sin2α = p2a2+q2b2−2pqab×pq−(a2−p2)(b2−q2)ab\frac{p^2}{a^2} + \frac{q^2}{b^2} -\frac{2pq}{ab} \times \frac {pq - \sqrt{(a^2 - p^2)(b^2 - q^2)}}{ab}

⇒ sin2α = p2a2+q2b2−2pqabcosα\frac{p^2}{a^2} + \frac{q^2}{b^2} -\frac{2pq}{ab}cosα

 p2a2\frac{p^2}{a^2} + kcos α, + q2b2\frac{q^2}{b^2} = sin2 α   ----eq(1)   

 p2a2+q2b2−2pqabcosα\frac{p^2}{a^2} + \frac{q^2}{b^2} -\frac{2pq}{ab}cosα = sin2α   ----eq(2)

Now comparing the equation (1) and (2), we get 

k = −2pqab-\frac{2pq}{ab}

If cos–1 (pa)(\frac{p}{a}) + cos–1 (qb)(\frac{q}{b}) = α, then p2a2\frac{p^2}{a^2} + kcos α, + q2b2\frac{q^2}{b^2} = sin2 α where k is equal to  −2pqab-\frac{2pq}{ab}

Question 164

The value of 3\sqrt3 sin10° is equal to -

  1. AOption (a)

    sin70° + sin50°

  2. BOption (b)

    sin20° + sin40°

  3. COption (c)

    cos50° + cos70°

  4. DOption (d)

    cos50° – cos70°

View answer & explanation

Correct answer

Answer: (d)D.

cos50° – cos70°

Why this answer is correct

Concept:

Formula Used:- 2sinAsinB = cos(A - B) - cos(A + B)

Calculation: 

Given: 3sin10∘\sqrt3sin10^{∘}

Multiply and divide the given value by 2 

 =2×32sin10∘= 2 × \frac{\sqrt3}{2}sin10^{∘}

= 2 × sin60° × sin10°

By using the formula

= cos(60° - 10°) - cos(60° + 10°)

= cos50° - cos70°

The value of 3sin10∘\sqrt3sin10^{\circ} is equal to:-  cos50° - cos70°

Question 165

What should be a necessary and sufficient condition for the curve v = constant to be a Geodesic on the general surface?

  1. AOption (a)

    EE2 + FE1 + 2EF1 = 0

  2. BOption (b)

    EE2 – FE1 – 2EF1 = 0

  3. COption (c)

    EE2 – FE1 + 2EF1 = 0

  4. DOption (d)

    EE2 + FE1 – 2EF1 = 0

View answer & explanation

Correct answer

Answer: (d)D.

EE2 + FE1 – 2EF1 = 0

Why this answer is correct

Concept:

A Geodesic on a surface is a curve along which the distance between any two points at the surface is the smallest or it is the smallest path connecting the two points on the surface.

Explanation: 

On the curve v = constant, 'u' may be a parameter so that v = c and u = t represent the parametric equation of the curve.

⇒u˘=1\breve{u} =1 ,v˘=0\breve{v} =0 and therefore,

2T=(Eu˘2+2Fu˘v˘+Gv˘2)2T=(E\breve{u}^2 +2F\breve{u}\breve{v}+G\breve{v}^2)

 ⇒   ∂T∂u=(E1u˘2+2F1u˘v˘+G1v˘2)\frac{\partial T}{\partial u}=(E_1\breve{u}^2 +2F_1\breve{u}\breve{v}+G_1\breve{v}^2)

∂T∂v=(E1u˘2+2F1u˘v˘+G1v˘2)\frac{\partial T}{\partial v}=(E_1\breve{u}^2 +2F_1\breve{u}\breve{v}+G_1\breve{v}^2)

∂T∂u˘=(Eu˘+Fv˘)\frac{\partial T}{\partial \breve{u}}=(E\breve{u} +F\breve{v})

∂T∂v˘=(Fu˘+Gv˘)\frac{\partial T}{\partial \breve{v}}=(F\breve{u} +G\breve{v})

When u˘=1\breve{u} =1,v˘=0\breve{v} =0,we have

∂T∂u=12(E1),∂T∂v=12(E2),∂T∂u˘=(E),∂T∂v˘=(F)\frac{\partial T}{\partial u}=\frac{1}{2}(E_1),\frac{\partial T}{\partial v}=\frac{1}{2}(E_2),\frac{\partial T}{\partial \breve u}=(E),\frac{\partial T}{\partial \breve v}=(F)   

U=ddt(∂T∂u˘)−∂T∂u=∂E∂t−12E1U=\frac{d}{dt}(\frac{\partial T}{\partial \breve u })-\frac{\partial T}{\partial u }=\frac{\partial E}{\partial t }-\frac{1}{2}E_1

U=ddt(∂T∂u˘)−∂T∂u=∂E∂t−12E1U=\frac{d}{dt}(\frac{\partial T}{\partial \breve u })-\frac{\partial T}{\partial u }=\frac{\partial E}{\partial t }-\frac{1}{2}E_1

Question 166

Value of cot (712)∘(7\frac{1}{2})^\circ is-

  1. AOption (a)

    3 + 2\sqrt2 + 3\sqrt3 + 6\sqrt6

  2. BOption (b)

    2 + 3\sqrt3 + 5\sqrt5 + 6\sqrt6

  3. COption (c)

    2 + 2\sqrt2 + 3\sqrt3 + 6\sqrt6

  4. DOption (d)

    2\sqrt2 + 3\sqrt3 + 5\sqrt5 + 6\sqrt6

View answer & explanation

Correct answer

Answer: (c)C.

2 + 2\sqrt2 + 3\sqrt3 + 6\sqrt6

Why this answer is correct

CONCEPT: 

Tan(A−B)=TanA−TanB1−TanA×TanBTan(A-B)=\frac{TanA-TanB}{1-TanA\times TanB}

Tan2A=2TanA1−2tan2ATan2A=\frac{2TanA}{1- 2 t an^2A}

Calculation:   

Cot(712)=Cot(152)Cot(7\frac{1}{2})=Cot(\frac{15}{2})

(Tan15∘)=Tan(45∘−30∘)=Tan45∘−Tan30∘1+Tan45∘Tan30∘(Tan15 ^\circ )=Tan(45^\circ-30^\circ) =\frac{Tan45 ^\circ -Tan30^\circ }{1+Tan45 ^\circ Tan30 ^\circ }

 ⇒  1−131+13⇒3−13+1−(1)\frac{1-\frac{1}{\sqrt 3}}{1+\frac{1}{\sqrt 3}}⇒ \frac{\sqrt3-1}{\sqrt3+1} -(1)

Tan2A=2TanA1−2tan2ATan2A=\frac{2TanA}{1- 2 t an^2A}

⇒ Tan15∘=2Tan(15∘2)1−tan2(15∘2)Tan15^\circ=\frac{2 Tan(\frac{15^\circ}{2})}{1-tan^2(\frac{15^\circ}{2})}

Now let Tan(15∘2)=xTan(\frac{15^\circ}{2})=x                -(2)

Then \()\(\frac{2x}{1-x^2}⇒ \frac{\sqrt3-1}{\sqrt3+1}\)

⇒(3−1)(1−x2)=2x3+1⇒ {(\sqrt3-1)}(1-x^2)=2x{\sqrt3+1}

⇒ 3−1−x2(3−1)=2(3+1)x\sqrt3-1-x^2(\sqrt3-1)=2({\sqrt3+1})x

⇒ x2(3−1)+2(3+1)x−(3−1)=0x^2(\sqrt3-1)+2({\sqrt3+1})x-(\sqrt3-1)=0

So x=−2(3+1)±2(3+1)2−4(3−1)22(3−1)x = {-2(\sqrt3+1) \pm \sqrt{2(\sqrt3+1)^2-4(\sqrt3-1)^2} \over 2(\sqrt3-1)}

x=−(3+1)±4+42(3−1)x = {-(\sqrt3+1) \pm \sqrt{4+4} \over 2(\sqrt3-1)}

x=−(3+1)±22(3−1)x = {-(\sqrt3+1) \pm2 \sqrt{2} \over (\sqrt3-1)}

 Tan(15∘2)=Tan(\frac{15^\circ}{2})== 22−(3+1)3−1{2\sqrt2}-{(\sqrt3+1) }\over\sqrt3 -1

Cot(15∘2)=26−22+28−(3+1+23)Cot(\frac{15^\circ}{2})=\frac{2\sqrt6-2\sqrt2+2}{8-(3+1+2\sqrt3)}

Cot(15∘2)=26−2+14−23)Cot(\frac{15^\circ}{2})=\frac{2\sqrt6-\sqrt2+1}{4-2\sqrt3)}

Cot(15∘2)=26−22+32−6+34−(3)Cot(\frac{15^\circ}{2})=\frac{2\sqrt6-2\sqrt2+3\sqrt2-\sqrt6+\sqrt3}{4-(3)}

Cot(15∘2)=26−22+32−6+31Cot(\frac{15^\circ}{2})=\frac{2\sqrt6-2\sqrt2+3\sqrt2-\sqrt6+\sqrt3}{1}

Cot(15∘2)=6−2+2+3Cot(\frac{15^\circ}{2})=\sqrt6-2+\sqrt2+\sqrt3

Cot(15∘2)=2+3+4+6Cot(\frac{15^\circ}{2})=\sqrt2+\sqrt3+\sqrt4+\sqrt6

Question 167

What shall be the appropriate matrix of the coefficient of Frenet-Serret formula?

  1. AOption (a)

    \(\begin{bmatrix} {0} & {-k} & 0 \[0.3em] {k} & 0 & -\tau \[0.3em] 0 & \tau & 0 \end{bmatrix}\)

  2. BOption (b)

    \(\begin{bmatrix} {0} & {-\tau} & 0 \[0.3em] {\tau} & 0 & -k \[0.3em] 0 & k & 0 \end{bmatrix}\)

  3. COption (c)

    \(\begin{bmatrix} {0} & {k} & 0 \[0.3em] {-k} & 0 & \tau \[0.3em] 0 & -\tau& 0 \end{bmatrix}\)

  4. DOption (d)

    \(\begin{bmatrix} {0} & {\tau} & 0 \[0.3em] {-\tau} & 0 & k \[0.3em] 0 & -k & 0 \end{bmatrix}\)

View answer & explanation

Correct answer

Answer: (c)C.

\(\begin{bmatrix} {0} & {k} & 0 \[0.3em] {-k} & 0 & \tau \[0.3em] 0 & -\tau& 0 \end{bmatrix}\)

Why this answer is correct

Concept: 

The Frenet -Serret formula is a formula describing the kinematic properties of a particle moving along a parametric curve. The formulas represent the derivatives of the function in terms of the basis formed by the tangent, normal and binormal unit vectors.

Explanation: 

T is the unit vector tangent to the curve. N is the normal unit vector, B is the binormal unit vector, and the cross-product of T and N.

The Frenet -Serret formulas are:

dTds=κN\frac{dT}{ds}=κ N

dNds=κT+κB\frac{dN}{ds}=κ T+κ B

dBds=τN\frac{dB}{ds}=τ N

where  dds\frac{d}{ds} is the derivative with respect to the arclength,κ is the curvature, and τ is the torsion of the curve.

  • The tangent unit vector T is defined as  T = drds\frac{dr}{ds}
  • The normal unit vector N is defined as N=dds∣∣dT∣∣dsN=\frac {\frac{d}{ds}}{\frac{||dT||}{ds}}
  • The binormal unit vector Bis is defined as the cross product of T and N. B = T × N

it follows, since T always has unit magnitude, that N is always perpendicular to T since there is no change in length of T   and B is always perpendicular to both T and B.

so the Frenet -Serret formulas are

dTds\frac{dT}{ds}=    0T  + κ N  + OB

dNds\frac{dN}{ds}=   -κ T +   0N  +  τ B

dBds\frac{dB}{ds}=    o T +  τ  N+  OB 

This is the coefficient matrix of Frenet Serret's formula [T′N′B′]=[0κ0−κ0τ0τ0][TNB]\begin{bmatrix} T' \\ N' \\ B' \end{bmatrix}=\begin{bmatrix}0&\kappa&0\\ -\kappa &0&τ\\ 0&τ &0\end{bmatrix}\begin{bmatrix} T \\ N \\ B \end{bmatrix} 

This matric is the skew-symmetric matrix.

Question 168

The element of surface ds is given by-

  1. AOption (a)

    ds = H2 du2+dv2\sqrt{du^2 + dv^2}

  2. BOption (b)

    ds = H du2+dv2\sqrt{du^2 + dv^2}

  3. COption (c)

    ds = H2dudv

  4. DOption (d)

    ds = H.dudv

View answer & explanation

Correct answer

Answer: (d)D.

ds = H.dudv

Why this answer is correct

Concept : 

The element of ds is the cross product of dru and drv

Explanation: Note the Partial derivatives ru=∂r∂u,rv=∂r∂vr_u=\frac{\partial r}{\partial u},r_v=\frac{\partial r}{\partial v}

The Element of Surface,

ds=|ru du + rv dv |

⇒ |ru x rv |dudv  

But ru x rv = N=Normal to the surface S and dudv=dA

so the relation between the elements of, ds and the element of, dA

⇒ ds = H dudv=NdA

 

<br>

Question 169

If cosec ∝ + cot ∝ = 2 + 5\sqrt5, then value of cos ∝ is-

  1. AOption (a)

    13\frac{1}{\sqrt3}

  2. BOption (b)

    52\frac{\sqrt5}{2}

  3. COption (c)

    3{\sqrt3}

  4. DOption (d)

    25\frac{2}{\sqrt5}

View answer & explanation

Correct answer

Answer: (d)D.

25\frac{2}{\sqrt5}

Why this answer is correct

Concept :

The Trignometry Identity

Sin2α +Cos2α =1

1+Cot2α =Cosec2α

Tan2α+ 1 =sec2α

Standerd identity :a2-b2=(a-b)(a+b)

Explanation: 

We have Cosecα +Cotα =2+√5            -(1)

We know that (Cosec2α -Cot2α)=1

(Cosecα +Cotα)(Cosecα -Cotα)=1                       - (2)

using  equation (1) in (2)

we have (2+√5)(Cosecα -Cotα)=1

(Cosecα - Cotα)=12+√5\frac{1}{2+√5}

(Cosecα - Cotα)=√ 5-2                     (3)

equation (1) and (3) adding ,then

2Cosec α=2√5 

Cosec α =√5 

sinα=15\frac{1}{\sqrt5}

we know that Cos α=1−sin2α \sqrt{1-sin^2\alpha}

so Cos α=25\frac{2}{\sqrt5}

Question 170

Let there be a surface, x = ucosv, y = usinv, z = cv. The first fundamental coefficients are respectively-

  1. AOption (a)

    0, 0, u2 + c2

  2. BOption (b)

    1, 0, u2 + c2

  3. COption (c)

    1, 0, u2+c2\sqrt{u^2 + c^2}

  4. DOption (d)

    cosv, sinv, u2 + 1

View answer & explanation

Correct answer

Answer: (b)B.

1, 0, u2 + c2

Why this answer is correct

Concept:

First fundamental coefficients:- We must find out the partial derivatives. It is denoted by E, F, G. 

Given:- x = u.cosv, y = u.sinv, z = cv

Let, r = (x, y, z)

r = (u.cosv, u.sinv, cv)

Calculation:

On partial differentiation of r, we get

r1 = (cosv, sinv, 0)       [ where r1 is first partial derivative]

r2 = (-usinv, ucosv, c)   [ where r2 is second partial derivative] 

E ⇒ r12 = (cosv, sinv, 0).(cosv, sinv, 0)

E ⇒ r12 = cos2v + sin2v + 0     [when we multiply a row to a column then the product of row and column get added]

E ⇒ r12 = 1

F = r1.r2

F = (cosv, sinv, 0).(-u.sinv, u.cosv, c) 

F = - u.cosv.sinv + u.sinv.cosv + 0 [when we multiply a row to a column then the product of row and column get added]

F = 0

G = r22 

r22 = ( -usinv, ucosv, c)(-usinv, ucosv, c) [when we multiply a row to a column then the product of row and column get added]

r22 = u2.sin2v + u2.cos2v + c2

r22 = u2 + c2     [∵ sin2v + cos2v = 1]

So, E = 1, F = 0 , G = u2 + c2

Hence, the first fundamental coefficients are respectively:- 1, 0 and u2 + c2

Question 171

lf X = cotθ + cosθ  and Y = cotθ – cosθ, then find the value of X2−Y2XY\frac{X^2 - Y^2}{\sqrt{XY}}.

  1. AOption (a)

    1

  2. BOption (b)

    0

  3. COption (c)

    2

  4. DOption (d)

    4

View answer & explanation

Correct answer

Answer: (d)D.

4

Why this answer is correct

Concept:

Using formula (a2 - b2) = (a + b)(a - b)

Calculation:

Given:- X = cotθ + cosθ, Y = cotθ - cosθ

Putting the value in X2−Y2XY\frac{X^2 - Y^2}{\sqrt{XY}}, we get

⇒ X2−Y2XY=(cotθ+cosθ)2−(cotθ−cosθ)2(cotθ+cosθ)(cotθ−cosθ)\frac{X^2 - Y^2}{\sqrt{XY}} = \frac{(cotθ + cosθ)^2 - (cotθ - cosθ)^2}{\sqrt{(cotθ + cosθ)(cotθ - cosθ)}}

⇒ X2−Y2XY=(cot2θ+cos2θ+2cotθcosθ)−(cot2θ+cos2θ−2cotθcosθ)(cotθ+cosθ)(cotθ−cosθ)\frac{X^2 - Y^2}{\sqrt{XY}} = \frac{(cot^2θ + cos^2θ + 2cotθcosθ) - (cot^2θ + cos^2θ -2cotθcosθ)}{\sqrt{(cotθ + cosθ)(cotθ - cosθ)}}

⇒ X2−Y2XY=(4cotθcosθ)(cotθ+cosθ)(cotθ−cosθ)\frac{X^2 - Y^2}{\sqrt{XY}} = \frac{(4cotθcosθ)}{\sqrt{(cotθ + cosθ)(cotθ - cosθ)}}

⇒ X2−Y2XY=(4cotθcosθ)(cot2θ−cos2θ)\frac{X^2 - Y^2}{\sqrt{XY}} = \frac{(4cotθcosθ)}{\sqrt{(cot^2θ - cos^2θ)}}

⇒ X2−Y2XY=(4cotθcosθ)cos2θsin2θ−cos2θ\frac{X^2 - Y^2}{\sqrt{XY}} = \frac{(4cotθcosθ)}{\sqrt{\frac{cos^2θ}{sin^2θ} - cos^2θ}}

⇒ X2−Y2XY=(4cotθcosθ)cos2θ(1sin2θ−1)\frac{X^2 - Y^2}{\sqrt{XY}} = \frac{(4cotθcosθ)}{\sqrt{cos^2θ(\frac{1}{sin^2θ} - 1)}}

⇒ X2−Y2XY=(4cotθcosθ)cos2θ(1−sin2θsin2θ)\frac{X^2 - Y^2}{\sqrt{XY}} = \frac{(4cotθcosθ)}{\sqrt{cos^2θ(\frac{1 - sin^2θ}{sin^2θ})}}

⇒ X2−Y2XY=(4cotθcosθ)cot2θ.cos2θ\frac{X^2 - Y^2}{\sqrt{XY}} = \frac{(4cotθcosθ)}{\sqrt{cot^2θ.cos^2θ}}

⇒ X2−Y2XY=(4cotθcosθ)cotθ.cosθ\frac{X^2 - Y^2}{\sqrt{XY}} = \frac{(4cotθcosθ)}{cotθ.cosθ}

⇒X2−Y2XY=4\frac{X^2 - Y^2}{\sqrt{XY}} = 4

Hence, the value of X2−Y2XY\frac{X^2 - Y^2}{\sqrt{XY}} is 4.

Question 172

Equation of asymptotic lines is given by-

  1. AOption (a)

    M dudv = 0

  2. BOption (b)

    Ldu2 – Ndv2 = 0

  3. COption (c)

    –Ldu2 – 2M dudv + Ndv2 = 0

  4. DOption (d)

    Ldu2 + 2M dudv + Ndv2 = 0

View answer & explanation

Correct answer

Answer: (d)D.

Ldu2 + 2M dudv + Ndv2 = 0

Why this answer is correct

Concept:

Definition:- The curve whose tangents are along asymptotic direction are called asymptotic lines. 

Calculation:

We know that the analytical condition of two directions (du, dv) and (∂u, ∂v) are to be conjugate is given by

⇒ Ldudv∂u∂v+M(dudv+∂u∂v)+N=0L \frac{du}{dv}\frac{∂u}{∂v} + M(\frac{du}{dv} + \frac{∂u}{∂v}) + N = 0

Now if (du, dv) is self conjugate, then we have (∂u, ∂v) = (du, dv)

⇒ L(dudv)2+2Mdudv+N=0L(\frac{du}{dv})^2 + 2M\frac{du}{dv} + N = 0

⇒ Ldu2 + 2Mdudv + Ndv2 = 0

which is the required differential equation of an asymptotic line.

Hence, Equation of an asymptotic line is given by Ldu2 + 2Mdudv + Ndv2 = 0.

Question 173

On the surface xyz = 4, what shall be the equation of the tangent plane at the point (1, 2, 2)?

  1. AOption (a)

    x + 2y + z – 1 = 0

  2. BOption (b)

    2x + y + z = 6

  3. COption (c)

    x + 2y + z = 2

  4. DOption (d)

    x – 2y – z + 1 = 0

View answer & explanation

Correct answer

Answer: (b)B.

2x + y + z = 6

Why this answer is correct

Concept:

The equation of tangent plane at a point:- (x−a)∂f∂x+(y−b)∂f∂y+(z−c)∂f∂z=0(x - a)\frac{∂f}{∂x} + (y - b)\frac{∂f}{∂y} + (z - c)\frac{∂f}{∂z} = 0

Given:- The equation of surface  F(x, y, z) = xyz - 4 = 0

Calculation:

On differentiating F partially with respect to x, y, and z respectively, we get

∂f∂x=yz\frac{∂f}{∂x} = yz

∂f∂y=xz\frac{∂f}{∂y} = xz

∂f∂z=xy\frac{∂f}{∂z} = xy

Hence, at the point (1, 2, 2) we have

x = 1, y = 2, z = 2

 ∂f∂x=yz=4\frac{∂f}{∂x} = yz = 4

∂f∂y=xz=2\frac{∂f}{∂y} = xz = 2

∂f∂z=xy=2\frac{∂f}{∂z} = xy = 2

So, the equation of tangent plane at (1, 2, 2) is 

(x−a)∂f∂x+(y−b)∂f∂y+(z−c)∂f∂z=0(x - a)\frac{∂f}{∂x} + (y - b)\frac{∂f}{∂y} + (z - c)\frac{∂f}{∂z} = 0

(x−1)×4+(y−2)×2+(z−2)×2=0(x - 1)\times 4 + (y - 2)\times 2 + (z - 2)\times 2 = 0

4x - 4 + 2y - 4 + 2z - 4 = 0

4x + 2y + 2z - 12 = 0

Now, divide the equation by 2

2x + y + z - 6 = 0

2x + y + z = 6

Hence, the equation of tangent plane at the point (1, 2, 2) is 2x + y + z = 6

Question 174

Find the value of cot⁡32A+3cot⁡2A cosec22Acosec3 2A(cos⁡6A−sin⁡6A)\frac{\cot^32A + 3\cot2A\ \text{cosec}^22A}{\text{cosec}^3\ 2A(\cos^6A - \sin^6 A)}

  1. AOption (a)

    1

  2. BOption (b)

    2

  3. COption (c)

    3

  4. DOption (d)

    4

View answer & explanation

Correct answer

Answer: (d)D.

4

Why this answer is correct

Concept:

(i) a3 - b3 = (a - b) (a2 + b2 +ab) 

(ii) cos2A + sin2A = 1

Explanation:

cot⁡32A+3cot⁡2A cosec22Acosec3 2A(cos⁡6A−sin⁡6A)\frac{\cot^32A + 3\cot2A\ \text{cosec}^22A}{\text{cosec}^3\ 2A(\cos^6A - \sin^6 A)}

 As, cos6A - sin6A = (cos2A)3 - (sin2A)3 

= (cos2A - sin2A) ( cos4A + sin4A + cos2A.sin2A) 

= (cos2A - sin2A) ( (cos2A + sin2A)2 - cos2A.sin2A) 

= cos2A [ 1 - 4cos2A.sin2A4\rm \frac{4 cos^2A. sin^2A}{4}] = cos2A [ 1 - sin22A4\rm \frac{sin^2 2A}{4}]

= cos2A(4−sin22A)4\frac{cos2A (4-sin^2 2A)}{4}

Now, cot⁡32A+3cot⁡2A cosec22Acosec3 2A(cos⁡6A−sin⁡6A)\frac{\cot^32A + 3\cot2A\ \text{cosec}^22A}{\text{cosec}^3\ 2A(\cos^6A - \sin^6 A)} = cot⁡2A(cot22A+3 cosec22A)cosec2 2Asin2A.cos2A.(4−sin22A)4\frac{\cot2A (\rm cot^2 2A+3 \ \text{cosec}^22A)}{\rm \frac{{cosec}^2\ 2A}{sin2A}. cos2A. \frac{(4 - sin^22A)}{4}}

= 4(cosec22A−1+3cosec22A)(4cosec22A−1)\frac {4(cosec^22A - 1 + 3cosec^22A)}{(4cosec^22A - 1)} = 4(4cosec22A−1)(4cosec22A−1)\frac{4{(4cosec^22A - 1)}}{(4cosec^22A - 1)} = 4

Hence, Option (4) is true

Question 175

For a curve to be a helix which of the following should be the most necessary and sufficient condition?

  1. AOption (a)

    The ratio of the curvature and the torsion is constant

  2. BOption (b)

    The curvature is constant

  3. COption (c)

    The product of the curvature and the torsion constant

  4. DOption (d)

    The torsion is constant

View answer & explanation

Correct answer

Answer: (a)A.

The ratio of the curvature and the torsion is constant

Why this answer is correct

Concept:

Helix:- A curve traced on the surface of a cylinder and cutting the generators at constant angle α is called a helix. If the cylinder is a right cylinder {x2 + y2 = a2} then the curve drawn on the cylinder which is making constant angle with the generators is a called a circular helix.

⇒ Theorem:- Prove that the necessary and sufficient condition for a curve to be a helix is that the ratio of its curvature and torsion is constant.

Calculation:

Necessary Condition:-

Let the curve is helix, then we have to prove that K κτ=constant\frac{κ}{τ} = constant. As the curve is helix , so the tangent at any point to this curve will make α constant angle with the fixed direction (generator) of the cylinder. Let a⃗\vec{a} be the unit constant along the generator and t is the  unit tangent at P to the curve C.

⇒ t⃗.a⃗=∣t⃗∣∣a⃗∣cosα\vec{t}.\vec{a} = |\vec{t}||\vec{a}|cosα

⇒ t⃗.a⃗=1.1cosα\vec{t}.\vec{a} = 1.1cosα

⇒ t⃗.a⃗=cosα\vec{t}.\vec{a} = cosα

⇒ t⃗.a⃗=constant\vec{t}.\vec{a} = constant    [∴ α = constant]

Differentiation with respect to 's'

t′⃗.a⃗=0\vec{t'}.\vec{a} = 0

κn⃗.a⃗=0κ\vec{n}.\vec{a} = 0

If κ = 0 then the curve is straight line and the condition is proved.

if κ ≠ 0, then n⃗.a⃗=0\vec{n}.\vec{a} = 0

⇒ n⃗⊥a⃗\vec{n} ⟂ \vec{a}

Thus a⃗\vec{a} will be in the plane determined by t and b (rectifying plane) 

Hence, a⃗=cosαt′⃗+sinαb⃗\vec{a} = cosα\vec{t'} + sinα\vec{b}

Differentiating with respect to 's'

⇒ 0=cosαt′⃗+sinαb⃗0 = cosα\vec{t'} + sinα\vec{b}

⇒  0=cosα(κn⃗)+sinα(−τn⃗)0 = cosα(κ\vec{n}) + sinα(-τ\vec{n})

⇒ 0=(κcosα−τsinα)n⃗0 = (κ cosα - τ sinα)\vec{n}

As n ≠ 0

⇒ κcosα - τ sinα = 0

⇒ κcosα = τsinα

⇒ κτ=tanα\frac{\kappa}{\tau} = tanα

⇒ κτ=constant\frac{\kappa}{\tau} = constant  [∴ α = constant]

Sufficient Condition:-

Let  κτ=constant\frac{\kappa}{\tau} = constant then we have to prove that the curve is a helix. For this it is sufficient to prove that t⃗.a⃗\vec{t}.\vec{a}  is constant as κτ\frac{\kappa}{\tau} is constant.

Let κτ=1c\frac{\kappa}{\tau} = \frac{1}{c}   ----equation (1)

Consider, dt⃗ds=κn⃗\frac{d\vec{t}}{ds} = \kappa\vec{n}

dt⃗ds=τcn⃗\frac{d\vec{t}}{ds} = \frac{\tau}{c}\vec{n}    ---- equation (2)    [by (1)]

Again Consider, db⃗ds=−τn⃗\frac{d\vec{b}}{ds} = - \tau\vec{n}

or 1cdb⃗ds=−τcn⃗\frac{1}{c} \frac{d\vec{b}}{ds} = - \frac{\tau}{c}\vec{n}   ---- equation(3)

Adding equation 2 and 3 

dtds+1cdb⃗ds=τcn⃗−τcn⃗\frac{dt}{ds} + \frac{1}{c}\frac{d\vec{b}}{ds} = \frac{\tau}{c}\vec{n} - \frac{\tau}{c}\vec{n}

⇒ dds(t⃗+1cb⃗)=0\frac{d}{ds}(\vec{t} + \frac{1}{c}\vec{b})= 0

⇒ dds(ct⃗+b⃗)=0\frac{d}{ds}(c\vec{t} + \vec{b}) = 0

Integrating, ct⃗+b⃗=a⃗c\vec{t} + \vec{b} = \vec{a}   (constant vector)

Taking dot product with t⃗\vec{t}

⇒ t⃗.(ct⃗+b⃗)=t⃗.a⃗\vec{t}.(c\vec{t} + \vec{b}) = \vec{t}.\vec{a} 

⇒ ct2+t⃗.b⃗=t⃗.a⃗ct^2 + \vec{t}.\vec{b} = \vec{t}.\vec{a}

⇒ ct2+0=t⃗.a⃗ct^2 + 0 = \vec{t}.\vec{a}

⇒ ct2=t⃗.a⃗ct^2 = \vec{t}.\vec{a} 

or t⃗.a⃗=constant\vec{t}.\vec{a} = constant

Hence, the curve is a helix.

So, the necessary and sufficient condition for a curve to be a helix is the ratio of the curvature and the torsion is constant.

Question 176

If A lies in the first quadrant and 23\sqrt3 sinAcos A = cos2A – sin2A, then the value of cosAcos(A/2)sin(A/2) is:

  1. AOption (a)

    14\frac{1}{4}

  2. BOption (b)

    18\frac{1}{8}

  3. COption (c)

    8

  4. DOption (d)

    4

View answer & explanation

Correct answer

Answer: (b)B.

18\frac{1}{8}

Why this answer is correct

Concept:

Double Angle Formula:

sin2θ = 2sinθ cosθ 

cos2θ = cos2θ - sin2θ 

Calculation:

Given, 23\sqrt3 sinAcos A = cos2A – sin2A

⇒ 3\sqrt3 (2sinAcos A) = cos2A – sin2A

⇒ 3\sqrt3 sin2A = cos2A 

⇒ tan2A = 13\frac{1}{\sqrt{3}} = tan 30∘

⇒ 2A = 30∘

∴ cos⁡Acos⁡(A2)sin⁡(A2)\cos A\cos(\frac{A}{2})\sin(\frac{A}{2})

= 12cos⁡A(2cos⁡(A2)sin⁡(A2))\frac{1}{2}\cos A(2\cos(\frac{A}{2})\sin(\frac{A}{2}))

= 12cos⁡Asin⁡A\frac{1}{2}\cos A\sin A

= 12×12(2cos⁡Asin⁡A)\frac{1}{2}\times\frac{1}{2}(2\cos A\sin A)

= 14sin⁡2A\frac{1}{4}\sin 2A

= 14\frac{1}{4}sin(30∘)

= 14\frac{1}{4} × 12\frac{1}{2}

= 18\frac{1}{8}

∴ The value of cosAcos(A/2)sin(A/2) is 18\frac{1}{8} .

The correct answer is Option 2.

Question 177

Sin6°. sin66° is -

  1. AOption (a)

    3−58\frac{3 - \sqrt5}{8}

  2. BOption (b)

    5−18\frac{\sqrt5 - 1}{8}

  3. COption (c)

    5−14\frac{\sqrt5 - 1}{4}

  4. DOption (d)

    3−54\frac{3 - \sqrt5}{4}

View answer & explanation

Correct answer

Answer: (a)A.

3−58\frac{3 - \sqrt5}{8}

Why this answer is correct

Concept:

sin(18°) = 5−14\frac{\sqrt{5}-1}{4}

Solution:

Given sin 6°⋅ sin 66°

Let x = sin 6°⋅ sin 66°

⇒ x = 12\frac{1}{2} × 2 sin 66°⋅ sin 6°

= 12\frac{1}{2} [cos(66° - 6°) - cos(66° + 6°)] [∵ 2 sin(A)sin(B) = cos(A - B) - cos(A + B)]

= 12\frac{1}{2}[cos(60°) - cos(72°)]

= 12\frac{1}{2} [12\frac{1}{2} - cos(90° - 18°)]

= 12\frac{1}{2} [12\frac{1}{2} - sin(18°)]

= 12[12−5−14]\frac{1}{2}\left [ \frac{1}{2}-\frac{\sqrt{5}-1}{4} \right ]

= 3−58\frac{3 - \sqrt5}{8}

∴ The value of sin6°. sin66° is 3−58\frac{3 - \sqrt5}{8}.

The correct answer is option 1.

Question 178

For a paraboloid of revolution z = x2 + y2, the asymptotic lines is given by-

  1. AOption (a)

    u2du2 + v2dv2 = 0

  2. BOption (b)

    du2 – dv2 = 0

  3. COption (c)

    du2 + u2dv2 = 0

  4. DOption (d)

    du2 + dv2 = 0

View answer & explanation

Correct answer

Answer: (c)C.

du2 + u2dv2 = 0

Why this answer is correct

Calculation:

According to the first fundamental form for the surface

ds2 = du2 + dv2

for the paraboloid revolution z = x2 + y2, we can parameterize the surface with u = θ (angle of rotation around the z-axis) and v = r(radial distance from the z-axis) so, we have 

x = r cosθ , y = r sinθ and z = r2

Let's express the differential element of arc length , ds2, in terms of dθ and dr:

ds2=(∂x∂θdθ)2+(∂x∂rdr)2+(∂y∂θdθ)2+(∂y∂rdr)2\rm ds^2=\left(\frac{\partial x}{\partial θ}dθ\right)^2+\left(\frac{\partial x}{\partial r}dr\right)^2+\left(\frac{\partial y}{\partial θ}dθ\right)^2+\left(\frac{\partial y}{\partial r}dr\right)^2

ds2 = (-r sin (θ) dθ)2 + (cos (θ) dr)2 + (r cos (θ) dθ)2 + (sin (θ)dr)2

ds2 = r2 sin2 (θ) dθ2 + dr2 + r2 cos2 (θ) dθ2 + dr2

ds2 = r2 (dθ2 + dr2)

Since , The paraboloid of revolution is surface of constant curvature , the coefficient dθ2 and dr2 should be equal along the asymptotic lines. So, for the asymptotic lines, we have:

dθ2 + r2dr2 = 0

This is equivalent to the expression du2 + u2dv2 = 0

Question 179

If sin 3A = cos (A-30), where 3A and (A-30) are acute angles, find the value of (sin 2A + Cos 2A)

  1. AOption (a)

    3+12\frac{\sqrt3 + 1}{2}

  2. BOption (b)

    3+13\frac{\sqrt3 + 1}{3}

  3. COption (c)

    3−12\frac{\sqrt3 - 1}{2}

  4. DOption (d)

    3−13\frac{\sqrt3 - 1}{3}

View answer & explanation

Correct answer

Answer: (a)A.

3+12\frac{\sqrt3 + 1}{2}

Why this answer is correct

Concept:

cos(90 - A) = sin A

Explanation:

Now, sin 3A = cos (A-30)

⇒ cos (90 - 3A) = cos(A - 30)

⇒ 90 - 3A = A - 30 ⇒ 4A = 120 ⇒ A = 30 

⇒ (sin 2A + Cos 2A) = sin60 + cos60 = 3+12\frac{\sqrt3 + 1}{2}

Hence, Option (1) is true

Question 180

If P (r)¨\ddot{(r)} is a point on a surface, then P will be a singularity of the surface if -

  1. AOption (a)

    r1 × r2 = 0

  2. BOption (b)

    r1 = r2 = 0

  3. COption (c)

    r1 = r2

  4. DOption (d)

    r1 × r2 ≠ 0

View answer & explanation

Correct answer

Answer: (a)A.

r1 × r2 = 0

Why this answer is correct

Concept:

Let r⃗\vec{r} = xi^\hat{i} + yj^\hat{j} + zk^\hat{k}  Then 

P will be singularity of surface if r1 × r2 = 0 otherwise if r1 × r2 ≠  0 then regular point.

Explanation:

Now, as P (r)¨\ddot{(r)} is a point on a surface, then P will be a singularity of the surface if r1 × r2 = 0

Hence, Option (1) is true.

Discipline 5 (20 questions)

Question 181

If f is a continuous function, then lim⁡n→∞∑r=0n−11nf(rn)\lim\limits_{n \to \infty} \sum_{r = 0}^{n - 1} \frac{1}{n}f(\frac{r}{n}) can be expressed as-

  1. AOption (a)

    ∫01\int_0^1 xf(x) dx

  2. BOption (b)

    ∫01f(1/x)\int_0^1 f (^1/_x) dx

  3. COption (c)

    ∫01\int_0^1 f(x) dx

  4. DOption (d)

    ∫011/xf(1/x)\int_0^1 {}^1/_x f (^1/_x) dx

View answer & explanation

Correct answer

Answer: (c)C.

∫01\int_0^1 f(x) dx

Why this answer is correct

Concept:

lim⁡n→∞∑r=0n−11nf(rn)\lim\limits_{n \to ∞} \sum_{r = 0}^{n - 1} \frac{1}{n}f(\frac{r}{n}) = ∫01\int_0^1 f(x) dx

Explanation:

lim⁡n→∞∑r=0n−11nf(rn)\lim\limits_{n \to ∞} \sum_{r = 0}^{n - 1} \frac{1}{n}f(\frac{r}{n})

Now, Let rn\frac{r}{n} = x ⇒ 1n\frac{1}{n} = dx 

lower limit = limn →∞ r(min)n\frac{r_{(min)}}{n} = 0 

Upper limit = limn →∞ r(max)n\frac{r_{(max)}}{n} = 1

⇒ lim⁡n→∞∑r=0n−11nf(rn)\lim\limits_{n \to ∞} \sum_{r = 0}^{n - 1} \frac{1}{n}f(\frac{r}{n}) = ∫01\int_0^1 f(x) dx

Hence, Option (3) is true

Question 182

Find the remainder when 2252 is divided by 210?

  1. AOption (a)

    141

  2. BOption (b)

    1

  3. COption (c)

    106

  4. DOption (d)

    208

View answer & explanation

Correct answer

Answer: (c)C.

106

Why this answer is correct

Concept:

Euler's theorem : It state that If a and n are coprime then a ϕ(n) ≡ 1 (mod n) 

Explanation:

2252 ≡ x (mod 210) (where x is reminder)

Divide all by 2 

Now, 2251 ≡ y (mod 105) 

Use Euler's theorem we get

2ϕ(105) ≡ 1 (mod105) ⇒ 248 ≡ 1 (mod 105) 

⇒ (248)5 ≡  1 (mod 105) ⇒ 2240 ≡ 1 (mod 105)

⇒ 2251 ≡ 211 (mod 105) ⇒ 2251 ≡ 53 (mod 105)

⇒ y = 53 ⇒ x = 2 . 53 = 106

Hence, Option (3) is true

Question 183

Root of the number (6 + 43\sqrt3)

  1. AOption (a)

    3(2\sqrt2 + 1)2

  2. BOption (b)

    √ (√3√3 (√3√3 + 1)2)

  3. COption (c)

    3(3\sqrt3 + 1)2

  4. DOption (d)

    3\sqrt3 (3\sqrt3 – 1)2

View answer & explanation

Correct answer

Answer: (b)B.

√ (√3√3 (√3√3 + 1)2)

Why this answer is correct

Concept Used:

(a + b)2 = a2 + b2 + 2ab

 Explanation:

(6 + 43\sqrt3)

= (2 × 3 + 43\sqrt3)

= (2 ×(3)2(\sqrt3)^2 + 43\sqrt3)

= 3\sqrt3(2 ×3\sqrt3 + 4)

= 3\sqrt3(2 ×3\sqrt3 + 1 + 3)

= 3\sqrt3(2 × 1  ×3\sqrt3 + 1 + (3)2(\sqrt3)^2)

= 3\sqrt3(3\sqrt3 + 1)2

Thus, Root of  (6 + 43\sqrt3) is 3(1+3)2\sqrt{\sqrt3 \left(1 + \sqrt3 \right)^2} .

Question 184

Coefficient of x–2 of this expansion (2x3−7x)8(2x^3 - \frac{7}{x})^8

  1. AOption (a)

    90

  2. BOption (b)

    not possible 

  3. COption (c)

    72

  4. DOption (d)

    40

View answer & explanation

Correct answer

Answer: (b)B.

not possible 

Why this answer is correct

Concept:

The general term in a binomial expansion of (a + b)n is given by:

Tr+1=nCran−rbrT_{r+1}={}^nC_ra^{n-r}b^r

Calculation:

Given binomial is (2x3−7x)8(2x^3 - \frac{7}{x})^8

∴ General term = Tr+1 = 8Cr(2x3)8−r(−7x)r{}^{8}C_r(2x^3)^{8-r}\left(\frac{-7}{x}\right)^r = 8Cr28−rx24−4r(−7)r{}^{8}C_r2^{8-r}x^{24-4r}(-7)^r ...(i)

Now, for the coefficient of term containing x–2 , 24 - 4r = - 2

⇒ r = 132\frac{13}{2} ≠ A positive integer

∴ There is no term of x–2 in the expansion of (2x3−7x)8(2x^3 - \frac{7}{x})^8.

Question 185

The middle term in the expansion of (x2+1x2+2)n(x^2 + \frac{1}{x^2} + 2)^n is

  1. AOption (a)

    (2n)!(n!)2\frac{(2n)!}{(n!)^2}

  2. BOption (b)

    (2n!)[(n2)!]2\frac{(2n!)}{[(\frac{n}{2})!]^2}

  3. COption (c)

    n[(n2)!]2\frac{n}{[(\frac{n}{2})!]^2}

  4. DOption (d)

    1.2.5..(2n+1)n!2n\frac{1.2.5..(2n + 1)}{n!}2^n

View answer & explanation

Correct answer

Answer: (a)A.

(2n)!(n!)2\frac{(2n)!}{(n!)^2}

Why this answer is correct

Concept Used:

General term of binomial expansion (a+b)n(a + b)^n is given by:

Tk=nCk−1⋅an−(k−1)⋅bk−1T_k = ^{n}C_{k-1} \cdot a^{n-(k-1)} \cdot b^{k-1}

Also, any binomial (a+b)n(a + b)^n, the expansion results in n + 1 terms.

Explanation:

(x2+1x2+2)n=(x2+1x2+2⋅x2⋅1x2)n=((x+1x)2)n=(x+1x)2n(x^2 + \frac{1}{x^2} + 2)^n = (x^2 + \frac{1}{x^2} + 2\cdot x^2 \cdot \frac{1}{x^2})^n = \left( (x + \frac{1}{x})^2\right)^n = \left(x + \frac{1}{x}\right)^{2n}

In this expansion, as there are 2n + 1 terms. Thus, the middle term will be (n+1)th term.

⇒Tn+1=2nCn⋅x2n−(n)⋅(1x)n\Rightarrow T_{n+1} = ^{2n}C_{n} \cdot x^{2n-(n)} \cdot \left(\frac{1}{x}\right)^{n}

⇒Tn+1=2nCn⋅xn⋅x−n\Rightarrow T_{n+1} = ^{2n}C_{n} \cdot x^{n} \cdot x^{-n}

⇒Tn+1=2nCn\Rightarrow T_{n+1} = ^{2n}C_{n}

⇒Tn+1=(2n)!n!(2n−n)!=(2n)!n!×n!=(2n)!(n!)2\Rightarrow T_{n+1} = \frac{(2n)!}{n! (2n - n)!} = \frac{(2n)!}{n! \times n!} = \frac{(2n)!}{ (n!)^2}

Thus, option 1 is the correct answer.

Question 186

Which of the following is true?

  1. AOption (a)

    Every discontinuous function is integrable

  2. BOption (b)

    Every integrable function is monotone increasing

  3. COption (c)

    Every monotone function is integrable

  4. DOption (d)

    Every integrable function is continuous

View answer & explanation

Correct answer

Answer: (c)C.

Every monotone function is integrable

Why this answer is correct

Concept -

(i) Monotone function has at most countably infinite discontinuity.

(ii) If the number of discontinuity of the function is almost countably infinite then the function is integrable.

Explanation -

1. Every discontinuous function is integrable:

This statement is false. There are discontinuous functions that are not integrable.

For example, the Dirichlet function is discontinuous at every point but not Riemann integrable.

2. Every integrable function is monotone increasing:

This statement is false. There exist integrable functions that are not monotone increasing.

For example, the function that is equal to 0 on the rational numbers and 1 on the irrational numbers is integrable but not monotone increasing.

3. Every monotone function is integrable:

This statement is true. Monotone functions have a finite number or at most countably infinite discontinuities, and therefore, they are Riemann integrable.

4. Every integrable function is continuous:

This statement is false. There exist integrable functions that are not continuous.

For example, the step function is integrable but not continuous.

Question 187

The largest coefficient of ( x + 1)20 is:

  1. AOption (a)

    (20)!(10!)2\frac{(20)!}{(10!)^2}

  2. BOption (b)

    (20)!6!4!\frac{(20)!}{6!4!}

  3. COption (c)

    (20)!5!4!\frac{(20)!}{5!4!}

  4. DOption (d)

    (20)!10!5!\frac{(20)!}{10!5!}

View answer & explanation

Correct answer

Answer: (a)A.

(20)!(10!)2\frac{(20)!}{(10!)^2}

Why this answer is correct

Concept:

(i) The coefficient of xr in (1+x)n is Tr+1 = nCr  

(ii) The largest coefficient of binomial expansion is middle term.

Explanation:

Now, The largest coefficient of ( x + 1)20 is T(n2\frac{n}{2} +1) = T11 = 20C10 = (20)!(10!)2\frac{(20)!}{(10!)^2}

Hence, Option (1) is true

Question 188

The system characterized by the differential equation d2y(t)dt2\frac{d^2y(t)}{dt^2} – dydt\frac{dy}{dt} – 2y(t) = x(t) is-

  1. AOption (a)

    Non-linear and stable

  2. BOption (b)

    Linear and unstable

  3. COption (c)

    Linear and stable

  4. DOption (d)

    Non-linear and unstable

View answer & explanation

Correct answer

Answer: (b)B.

Linear and unstable

Why this answer is correct

Explanation -

The given differential equation is:

 d2y(t)dt2\frac{d^2y(t)}{dt^2} – dydt\frac{dy}{dt} – 2y(t) = x(t)

This is a linear differential equation because the dependent variable (y(t)) and its derivatives appear to the first power and are not multiplied together or by themselves.

Moreover, there are no nonlinear functions of (y(t)) (such as (sin(y(t))), (y2(t)), etc.).

Hence, the system is linear.

For stability, we should look at the homogeneous part of the differential equation, which is:

d2y(t)dt2−dydt−2y(t)=0\frac{d^2y(t)}{dt^2} - \frac{dy}{dt} - 2y(t) = 0

The characteristic equation for this differential equation is:

r2 - r - 2 = 0 

Solving this quadratic equation for (r) gives:

r=1±1+82=1±32r = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3}{2}

So, (r = 2) or (r = -1).

Since one of the roots is positive ((r = 2)), it indicates that there is at least one solution to the differential equation that grows without bound as (t →  ∞ ).

This suggests the system can be unstable under certain initial conditions because the presence of a positive real part in any root of the characteristic equation indicates that the solution can grow exponentially.

Thus, the differential equation is Linear and unstable

Question 189

A signal x (t) is given by x (t) = {1,−T4<t≤3T4 −1,3T4<t<7T4 −x(t+T)\begin{cases} 1, \frac{-T}{4} < t \leq \frac{3T}{4}& \quad \ -1, \frac{3T}{4} < t < \frac{7T}{4} & \quad \ -x(t +T) \end{cases}

Which among the following gives the fundamental Fourier terms of x(t)?

  1. AOption (a)

    4πcos⁡(πtT−π4)\frac{4}{\pi} \cos (\frac{\pi t}{T} - \frac{\pi}{4})

  2. BOption (b)

    4πsin⁡(πtT−π4)\frac{4}{\pi} \sin (\frac{\pi t}{T} - \frac{\pi}{4})

  3. COption (c)

    4πsin⁡(πtT+π4)\frac{4}{\pi} \sin (\frac{\pi t}{T} + \frac{\pi}{4})

  4. DOption (d)

    4πcos⁡(πtT+π4)\frac{4}{\pi} \cos (\frac{\pi t}{T} + \frac{\pi}{4})

View answer & explanation

Correct answer

Answer: (b)B.

4πsin⁡(πtT−π4)\frac{4}{\pi} \sin (\frac{\pi t}{T} - \frac{\pi}{4})

Why this answer is correct

Explanation:

x(t) = {1,−T4<t≤3T4 −1,3T4<t<7T4 −x(t+T)\begin{cases} 1, \frac{-T}{4} < t \leq \frac{3T}{4}& \quad \ -1, \frac{3T}{4} < t < \frac{7T}{4} & \quad \ -x(t +T) \end{cases}

So the graph of x(t) will be as given below

Hence the period of x(t) is

T0 = 7T4−(−T4)=8T4{7T\over 4}-(-\frac T4)=\frac{8T}{4} = 2T

So, fundamental angular frequency 

ω0 = 2πT0=2π2T=πT{2\pi\over T_0}={2\pi\over 2T}={\pi\over T}....(i)

Now, 

C1 = exponential series coefficient for k = 1

     = 1T0∫T0x(t)e−jω0tdt{1\over T_0}\displaystyle\int_{T_0}x(t)e^{-jω_0t}dt

    = Tπ[∫−T43T4e−jω0tdt−∫3T47T4e−jω0tdt]{T\over \pi}\left[\displaystyle\int_{-\frac T4}^{{3T\over 4}}e^{-jω_0t}dt-\displaystyle\int_{\frac {3T}4}^{{7T\over 4}}e^{-jω_0t}dt\right]

   = \({T\over \pi}\left{ \left[{e^{-jω_0t}\over -jω_0}\right]{-\frac T4}^{{3T\over 4}}-\left[{e^{-jω_0t}\over -jω_0}\right]{\frac {3T}4}^{{7T\over 4}}\right}\)         

  = \({-T\over \pi j ω_0}\left{ e^{-{jω_03T\over 4}}-e^{{jω_0T\over 4}}-e^{-{jω_07T\over 4}}+e^{-{jω_03T\over 4}}\right}\)

  = \(-{1\over 2\pi j}\left{e^{-{3\pi j\over4}}-e^{{\pi j\over4}}-e^{-{7\pi j\over4}}+e^{-{3\pi j\over4}}\right}\) (using (i))

  = \(-{1\over 2\pi j}\left{-e^{{\pi j\over4}}-e^{{\pi j\over4}}-e^{{\pi j\over4}}-e^{{\pi j\over4}}\right}\) 

   = 2πjeπj4{2\over \pi j}e^{{\pi j\over4}} 

  = −2jπ(cos⁡π4+jsin⁡π4){-2j\over \pi }(\cos{\pi\over 4}+j \sin{\pi\over 4}) (as j2 =- 1)

  = −2jπ(12+j12){-2j\over \pi }({1\over \sqrt2}+j {1\over \sqrt2})

C1 = 2π2−j2π2{2\over \pi\sqrt2 }-j {2\over\pi \sqrt2}

Comparing the above with C1 = a12+b12{a_1\over 2}+{b_1\over 2} we get

a1 = 4π24\over \pi\sqrt2 , b1 = - 4π24\over \pi\sqrt2

Hence fundamental Fourier term 

= a1cos(ω0t) + b1sin(ω0t)

= 4π2cos⁡(πtT)−4π2sin⁡(πtT){4\over \pi\sqrt2}\cos({\pi t\over T})-{4\over \pi\sqrt2}\sin({\pi t\over T})

= 4π(sin⁡π4cos⁡(πtT)−cos⁡π4sin⁡(πtT)){4\over \pi}(\sin{\pi\over 4}\cos({\pi t\over T})-\cos {\pi \over 4}\sin({\pi t\over T}))

= 4πsin⁡(πtT−π4)\frac{4}{\pi} \sin (\frac{\pi t}{T} - \frac{\pi}{4}) (as sin a cos b - cos a sin b = sin(a - b))

Option (2) is true

Question 190

The CTFT of a continuous time signal x(t) = e-A|t|, A > 0 is _______

  1. AOption (a)

    2A(ω2)\frac{2A}{(\omega^2)}

  2. BOption (b)

    A(A2+ω2)\frac{A}{(A^2 + \omega^2)}

  3. COption (c)

    2A(A2+ω2)\frac{2A}{(A^2 + \omega^2)}

  4. DOption (d)

    A(ω2)\frac{A}{(\omega^2)}

View answer & explanation

Correct answer

Answer: (c)C.

2A(A2+ω2)\frac{2A}{(A^2 + \omega^2)}

Why this answer is correct

Concept:

The continuous time Fourier transform (CTFT) of a function f(t) is

F(ω) = ∫−∞∞f(t)e−jωtdt\displaystyle\int_{-\infty}^{\infty}f(t)e^{-j\omega tdt}

Explanation:

The CTFT of a continuous-time signal x(t) = e-A|t|, A > 0

F(ω) = ∫−∞∞e−A∣t∣e−jωtdt\displaystyle\int_{-\infty}^{\infty}e^{-A|t|}e^{-j\omega t}dt

       = ∫−∞0eAte−jωtdt\displaystyle\int_{-\infty}^{0}e^{At}e^{-j\omega t}dt + ∫0∞e−Ate−jωtdt\displaystyle\int_{0}^{\infty}e^{-At}e^{-j\omega t}dt

     = ∫−∞0e(A−jω)tdt\displaystyle\int_{-\infty}^{0}e^{(A-j\omega)t}dt + ∫0∞e−(A+jω)tdt\displaystyle\int_{0}^{\infty}e^{-(A+j\omega) t}dt

    = [e(A−jω)tA−jω]−∞0\left[{e^{(A-j\omega)t}\over A-j\omega}\right]_{-\infty}^{0} + [e−(A+jω)t−(A+jω)]0∞\left[{e^{-(A+j\omega)t}\over -(A+j\omega)}\right]_0^{\infty}

   = 1A−jω1\over A-j\omega + 1A+jω1\over A+j\omega = 2A(A−jω)(A+jω)2A\over (A-j\omega)(A+j\omega) = 2A(A2+ω2)\frac{2A}{(A^2 + \omega^2)} (as j2 = -1)

Option (3) is true

Question 191

The Fourier series for the function f(x) = sin2 (x) is:

  1. AOption (a)

    0.5 + 0.5 cos 2x

  2. BOption (b)

    0.5 + 0.5 sin 2x

  3. COption (c)

    0.5 – 0.5 sin 2x

  4. DOption (d)

    0.5 – 0.5 cos 2x

View answer & explanation

Correct answer

Answer: (d)D.

0.5 – 0.5 cos 2x

Why this answer is correct

Formula Used:

cos 2x = 1 - 2sin2x

Explanation:

We know, cos 2x = 1 - 2sin2x

⇒ 2sin2x  = 1 - cos 2x

⇒ sin2x  = 1/2 - 1/2 cos 2x

⇒ sin2x  = 0.5 - 0.5 cos 2x

Thus, option (4) is the correct answer.

Question 192

Find the sum of coefficient of the expansion (1 + x – 3x2 )4165

  1. AOption (a)

    1

  2. BOption (b)

    0

  3. COption (c)

    2

  4. DOption (d)

    –1

View answer & explanation

Correct answer

Answer: (d)D.

–1

Why this answer is correct

Concept Used:

Sum of the coefficient of expansion can be calculated by substituting the variable value equal to 1.

(-1)odd = -1 and (-1)even = 1

Explanation:

Thus, the Sum of the coefficient of (1 + x – 3x2)4165 can be calculated by substituting x = 1.

Sum of the coefficient = (1 + 1 - 3(1)2)4165

⇒ Sum of the coefficient = (2 - 3)4165

⇒ Sum of the coefficient = (-1)4165

Thus, Sum of the coefficient is -1.

Question 193

The running integrator given by y(t) = ∫−∞∞\int _{-\infty}^\infty x(t)dt:

  1. AOption (a)

    Produces an abounded output for every causal bounded input

  2. BOption (b)

    Has no finite zeros in its double-sided Laplace transform Y(s)

  3. COption (c)

    Has no finite singularities in its double sided Laplace transform Y(s)

  4. DOption (d)

    Produces a bounded output for every anticausal bounded input

View answer & explanation

Correct answer

Answer: (a)A.

Produces an abounded output for every causal bounded input

Why this answer is correct

Explanation:

The running integrator ∫−∞tx(t)dt\int _{-\infty}^tx(t)dt = 0

for every causal system. As causal systems have no memory and the initial value is zero, the output is followed by input.

So, y (t) will always be bounded if this function is a causal bounded system.

Option (1) is true

Question 194

If (a + b) = 2, then the value of ∑r=0nncr\sum_{r =0}^n n_{c_r} arbn – r

  1. AOption (a)

    1

  2. BOption (b)

    2n

  3. COption (c)

    0

  4. DOption (d)

    –2n

View answer & explanation

Correct answer

Answer: (b)B.

2n

Why this answer is correct

Concept Used:

Sum of the coefficients in the expansion of a binomial expression is given by:

(a+b)n=∑k=0n(nk)an−kbk(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k

Explanation:

Given: a + b = 2

Thus, using the concept given above:

(a+b)n=∑k=0n(nk)an−kbk(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k

⇒(2)n=∑k=0n(nk)an−kbk\Rightarrow (2)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k

Thus, 2n is the required value.

Question 195

Which of the following is true?

  1. (1.01)1000000 > 10000
  2. (98)5 5 > 9500000000
  1. AOption (a)

    Neither 1 nor 2

  2. BOption (b)

    Only 2

  3. COption (c)

    Only 1

  4. DOption (d)

    Both 1 and 2

View answer & explanation

Correct answer

Answer: (d)D.

Both 1 and 2

Why this answer is correct

Explanation -

  1. (1.01)1000000 > 10000

Now (1.01)1000000 = (101/100)1000000 > 10000

Now it is clearly, show that (101/100)1000000 is greater than 10000.

  1. (98)5 5 > 9500000000

Let us assume 100 in place of 98 then we get -

(100)5 5 =50000000000 > 9500000000

Hence both the statement are correct.

Question 196

Find by using binomial theorem (23n – 7n – 1) is divisible by which of the following numbers?

  1. AOption (a)

    49

  2. BOption (b)

    27

  3. COption (c)

    15

  4. DOption (d)

    16

View answer & explanation

Correct answer

Answer: (a)A.

49

Why this answer is correct

Concept Used:

Expansion of nth power of any binomial is given as:

(a+b)n=∑k=0n(nk)an−kbk(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^k

Where (nk)\binom{n}{k} represents nCr.

Explanation:

Given:  (23n – 7n – 1)

(23n – 7n – 1) = ((23)n – 7n – 1) =  ((8)n – 7n – 1) = ((1 + 7)n – 7n – 1) 

Using Binomial theorem,

∴23n−7n−1=nC070+nC171+nC272+…+nCn7n−7n−1 \therefore 2^{3 n}-7 n-1 ={ }^n C_0 7^0+{ }^n C_1 7^1+{ }^n C_2 7^2+\ldots+{ }^n C_n 7^n-7 n-1

<br>

⇒23n−7n−1=1+n!(n−1)!×7+72(nC2+nC37+…+nCn7n−2)−7n−1⇒ 2^{3 n}-7 n-1 =1+\frac{n !}{(n-1) !} \times 7+7{ }^2\left({ }^n C_2+{ }^n C_3 7+\ldots+{ }^n C_n 7^{n-2}\right)-7 n-1

⇒23n−7n−1=n×7−7n+72C2+nC373+…+nCn7n−2)⇒ 2^{3 n}-7 n-1 \left.=n \times 7-7 n+7{ }^2 C_2+{ }^n C_3 7^3+\ldots+{ }^n C_n 7^{n-2}\right)

⇒23n−7n−1=49(nC2+nC37+…+nCn7n−2)⇒ 2^{3 n}-7 n-1 =49\left({ }^n C_2+{ }^n C_3 7+\ldots+{ }^n C_n 7^{n-2}\right)

⇒ 23n – 7n – 1 = 49 K

Hence, (23n – 7n – 1) divisible by 49.

Question 197

If 4 × 5 x = 500, then the value of xx is-

  1. AOption (a)

    27

  2. BOption (b)

    1

  3. COption (c)

    64

  4. DOption (d)

    8

View answer & explanation

Correct answer

Answer: (a)A.

27

Why this answer is correct

Concept:

Factors of 500 = 125 × 4 = 53 × 22

Explanation:

we know, 500 = 53 × 22 = 53 × 4 ....... (i)

By comparing (i) with 4 × 5 x = 500 

⇒ x = 3 ⇒ xx = 33 = 27

Hence, Option (1) is true

Question 198

y(t) is a real valued function of a real variable with period T. Its trigonometric Fourier series expansion contains no terms of frequency w = 2π (2kT)(\frac{2k}{T}), k = 1, 2, ....Also no sine terms are present. Then y(t) satisfies the equation:

  1. AOption (a)

    y(t) = y(t + T) = y(t + T/2)

  2. BOption (b)

    y(t) = y(t – T) = –y(t – T/2)

  3. COption (c)

    y(t) = y(t – T) = y(t – T/2)

  4. DOption (d)

    y(t) = y(T – t) = –y(t – T/2)

View answer & explanation

Correct answer

Answer: (c)C.

y(t) = y(t – T) = y(t – T/2)

Why this answer is correct

Explanation:

The trigonometric Fourier series expansion of y(t) contains no terms of frequency w = 2π (2kT)(\frac{2k}{T}), k = 1, 2, ... and no sine terms are present.

So, y(t) = y(-t)

i.e., y(t) = y(T - t) (as y(t) is a real-valued function of a real variable with period T)

Also, signal y(t) contains odd harmonics.

So, y(t) = y(t – T/2)

Hence y(t) = y(t – T) = y(t – T/2)

Option (3) is true

Question 199

The continuous time system described by the equation y(t) = x(t2) comes under the category of -

  1. AOption (a)

    Causal, linear and time-varying

  2. BOption (b)

    Non-causal, non-linear and time-variant

  3. COption (c)

    Causal, non-linear and time-varying

  4. DOption (d)

    Non-causal, linear and time-variant

View answer & explanation

Correct answer

Answer: (b)B.

Non-causal, non-linear and time-variant

Why this answer is correct

Concept Used:

Causal: A continuous-time system is causal if the output at any given time  t  depends only on the values of the input for  t' ≤ t , where  t'  represents all past and present times.

Linear: A continuous-time system is linear if it satisfies both the superposition and homogeneity properties. The superposition property states that if  y1(t)  is the response to input  x1(t)  and  y2(t)  is the response to input  x2(t) , then the response to  a1x1(t) + a2x2(t)  is  a1y1(t) + a2y2(t) , where  a1  and  a2  are constants. The homogeneity property states that if  y(t)  is the response to  x(t) , then the response to  ax(t)  is  ay(t) , where  a  is a constant. 

Time-Variant: A continuous-time system is time-variant if its characteristics change with time. Mathematically, a system  y(t) = T[x(t)]  is time-variant if there exist different time instances  t1  and  t2  such that the response  y(t1)  to input  x(t1)  is different from the response  y(t2)  to input  x(t2) , even if  x(t1) = x(t2)  for  t1 ≠ t2 .

Explanation:

The given system y(t) = x(t2) is non-causal. This is because, for negative values of t, t2 will give positive results, meaning the output y(t) at a negative time could depend on the input x(⋅) at a positive time (the future), which contradicts the definition of causality.

Thus, y(t) = x(t2) is non-casual.

y(t) = x(t2) is non-linear because it does not satisfy the superposition principle (i.e., x1(t) + x2(t) does not lead to y1(t) + y2(t) due to the squaring operation on t before it is used as an argument for x).

Thus, y(t) = x(t2) is non-linear.

The given system y(t) = x(t2), is time-varying because a time shift in t does not uniformly shift t2 due to the non-linear manipulation of the time variable before it is used to index x. The effect of a time shift in the input on the output is not consistent across different time shifts, making the system time-varying.

Thus, y(t) = x(t2) is time-varying.

Therefore, the correct characterization of the system described by the equation y(t) = x(t2) is Non-causal, non-linear, and time-varying.

Question 200

The remainder when 75757575^{75^{75}} is dividied by 37:

  1. AOption (a)

    0

  2. BOption (b)

    5

  3. COption (c)

    1

  4. DOption (d)

    7

View answer & explanation

Correct answer

Answer: (c)C.

1

Why this answer is correct

Concept:

a ≡ b (mod m) means m divides a - b

Explanation:

75 ≡ 1 (mod 37)

⇒ 7575 ≡ 175 (mod 37)

⇒ 7575 ≡ 1 (mod 37)

So, 75any integre  ≡ 1 (mod 37) 

⇒ 75757575^{75^{75}} ≡ 1 (mod 37) 

Hence remainder when 75757575^{75^{75}} is divided by 37 is 1

Option (3) is true

Discipline 6 (20 questions)

Question 201

(32n+1 + 2n+2) is divisible by

  1. AOption (a)

    5

  2. BOption (b)

    7

  3. COption (c)

    3

  4. DOption (d)

    2

View answer & explanation

Correct answer

Answer: (b)B.

7

Why this answer is correct

Concept Used:

(32n+1 + 2n+2) is divisible by 7.

Explanation:

To prove that (32n+1 + 2n+2)  is divisible by 7, we will use mathematical induction.

Base Case: When n = 0,

32(0)+1+20+23^{2(0)+1} + 2^{0+2} = 31 + 22 = 3 + 4 = 7

Which is divisible by 7.

Inductive Hypothesis: Assume that 32k+1+2k+23^{2k+1} + 2^{k+2} is divisible by 7 for some arbitrary non-negative integer k.

Inductive Step: We want to show that if the hypothesis holds for k, then it also holds for k+1.

32(k+1)+1+2(k+1)+2=32k+3+2k+33^{2(k+1)+1} + 2^{(k+1)+2} = 3^{2k+3} + 2^{k+3}

=32⋅32k+1+22⋅2k= 3^2 \cdot 3^{2k+1} + 2^2 \cdot 2^k

=9⋅32k+1+4⋅2k= 9 \cdot 3^{2k+1} + 4 \cdot 2^k

=(7+2)⋅32k+1+4⋅2k= (7 + 2) \cdot 3^{2k+1} + 4 \cdot 2^k

=7⋅32k+1+2⋅32k+1+4⋅2k= 7 \cdot 3^{2k+1} + 2 \cdot 3^{2k+1} + 4 \cdot 2^k

=7⋅32k+1+2⋅32k+1+2k+2= 7 \cdot 3^{2k+1} + 2 \cdot 3^{2k+1} + 2^{k+2}

Here, 2⋅32k+12 \cdot 3^{2k+1}  is divisible by 7 because 32k+13^{2k+1} is divisible by 7 (by the inductive hypothesis),

2k+22^{k+2} is also divisible by 7, and

7⋅32k+17 \cdot 3^{2k+1}  is clearly divisible by 7. Thus, 32(k+1)+1+2(k+1)+23^{2(k+1)+1} + 2^{(k+1)+2} is divisible by 7.

By mathematical induction, we have proved that 32n+1+2n+2 3^{2n+1} + 2^{n+2} is divisible by 7 for all non-negative integers n.

Question 202

By mathematical ascending method the value of 1+2+3+.......... + n is _______.

  1. AOption (a)

    (n+1)(n+2)2\frac{(n + 1)(n + 2)}{2}

  2. BOption (b)

    (n+1)(n+2)3\frac{(n + 1)(n + 2)}{3}

  3. COption (c)

    n(n+1)3\frac{n(n + 1)}{3}

  4. DOption (d)

    n(n+1)2\frac{n(n + 1)}{2}

View answer & explanation

Correct answer

Answer: (d)D.

n(n+1)2\frac{n(n + 1)}{2}

Why this answer is correct

Formula Used:

The sum of the first n positive integers can be calculated using the formula for the sum of an arithmetic series:

Sn=n2(a1+an)S_n = \frac{n}{2}(a_1 + a_n)

Where,

  • Sn is the sum of the first  n  positive integers,
  • n is the number of terms in the series,
  • a1  is the first term of the series.
  • an is the last term of the series.
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Explanation:

For the series  1 + 2 + 3 + … + n , we have:

a1 = 1  (the first term)

an = n  (the last term)

Sn=n2(1+n) S_n = \frac{n}{2}(1 + n)

⇒  Sn=n2(1+n)S_n = \frac{n}{2}(1 + n)

Thus, option 4 is correct answer.

Question 203

For what value of n, the statement P(n) : n! < (n+12)n(\frac{n + 1}{2})^n is not true? Where n is a positive whole number.

  1. AOption (a)

    4

  2. BOption (b)

    3

  3. COption (c)

    1

  4. DOption (d)

    2

View answer & explanation

Correct answer

Answer: (c)C.

1

Why this answer is correct

Explanation:

Given: P(n) : n! < (n+12)n\left(\frac{n + 1}{2}\right)^n 

For  n = 1

Left side:  1! = 1

Right side:  (1+12)1\left(\frac{1 + 1}{2}\right)^1 = 1

1 < 1, Which is False.

For  n = 2

Left side:  2! = 2

Right side:  (2+12)2=(32)2=94\left(\frac{2 + 1}{2}\right)^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}

2 <94 \frac{9}{4}, Which is True.

For  n = 3

Left side:  3! = 6

Right side: (3+12)3=(42)3=23=8 \left(\frac{3 + 1}{2}\right)^3 = \left(\frac{4}{2}\right)^3 = 2^3 = 8

6 < 8, Which is True.

For  n = 4,

Left side:  4! = 24 

Right side:  (4+12)4=(52)4=62516\left(\frac{4 + 1}{2}\right)^4 = \left(\frac{5}{2}\right)^4 = \frac{625}{16} 

24<6251624 < \frac{625}{16}, Which is True

Hence option (iii) is true.

Question 204

The integral ∫01x3cos⁡3x2+x2\int_0^{1} \frac{x^3\cos3x}{2 + x^2} lies between?

  1. AOption (a)

    –1 and 1

  2. BOption (b)

    −32-\frac{3}{2}and 32\frac{3}{2}

  3. COption (c)

    −12-\frac{1}{2} and 12\frac{1}{2}

  4. DOption (d)

    −13-\frac{1}{3} and 13\frac{1}{3}

View answer & explanation

Correct answer

Answer: (c)C.

−12-\frac{1}{2} and 12\frac{1}{2}

Why this answer is correct

Concept Used:

-1 ≤ cos x ≤ 1

Explanation:

As we know,

  • The function x3cos⁡(3x)2+x2\frac{x^3\cos(3x)}{2 + x^2} is continuous on the interval [0, 1].
  • The cosine function oscillates between -1 and 1, and its absolute value is never greater than 1.
  • The denominator 2 + x2 is always positive on the interval [0, 1].

 

Thus, −x32+x2≤x3cos⁡(3x)2+x2≤x32+x2 -\frac{x^3}{2 + x^2} ≤ \frac{x^3\cos(3x)}{2 + x^2} ≤ \frac{x^3}{2 + x^2}

Lower bound = ∫01−x32+x2,dx \int_0^{1} -\frac{x^3}{2 + x^2} , dx

Upper bound = ∫01x32+x2,dx \int_0^{1} \frac{x^3}{2 + x^2} , dx

⇒ Lower bound = ∫01−x32+x2,dx=12∫01−2x32+x2,dx=12∫01−2x32+x2,dx \int_0^{1} -\frac{x^3}{2 + x^2} , dx = \frac{1}{2}\int_0^{1} -\frac{2x^3}{2 + x^2} , dx = \frac{1}{2}\int_0^{1} \frac{-2x^3}{2 + x^2} , dx 

Now, let u=2+x2u = 2 + x^2 , then du = 2x dx or dx=du2xdx = \frac{du}{2x} .

⇒ Lower bound = 12∫23−(u−2)udu\frac{1}{2} \int_{2}^{3} \frac{-(u-2)}{u} du

⇒ Lower bound =12∫23(−1+2u),du= \frac{1}{2} \int_{2}^{3} \left(-1 + \frac{2}{u}\right) , du

⇒ Lower bound = 12∫23(−1),du+12∫232u,du\frac{1}{2} \int_{2}^{3} (-1) , du + \frac{1}{2} \int_{2}^{3} \frac{2}{u} , du

⇒ Lower bound = 12∫23(−1),du=12[−u]23=12[−3+2]=12⋅(−1)=−12\frac{1}{2} \int_{2}^{3} (-1) , du = \frac{1}{2} \left[-u\right]_{2}^{3} = \frac{1}{2} \left[-3 + 2\right] = \frac{1}{2} \cdot (-1) = -\frac{1}{2}, and

12∫232u,du=12∫232u,du=12⋅[2ln⁡∣u∣]23=12[2ln⁡∣3∣−2ln⁡∣2∣]=ln⁡(3/2) \frac{1}{2} \int_{2}^{3} \frac{2}{u} , du = \frac{1}{2} \int_{2}^{3} \frac{2}{u} , du = \frac{1}{2} \cdot \left[2 \ln|u|\right]_{2}^{3} = \frac{1}{2} \left[2 \ln|3| - 2 \ln|2|\right] = \ln(3/2)

⇒ Lower bound  = −12+ln⁡(3/2) -\frac{1}{2} + \ln (3/2)   

As the upper bound is the same in value and only differs in polarity, 

⇒ Upper Bound =−ln⁡(32)+12 = -\ln\left(\frac{3}{2}\right) + \frac{1}{2}

Hence The integral ∫01x3cos⁡3x2+x2\int_0^{1} \frac{x^3\cos3x}{2 + x^2} lies between -1/2 and 1/2.

Question 205

If f(x) = x|x|, then for any real numbers a and b with a < b, the value ofa∫b{_a\int{^b}} f(x)dx equals-

  1. AOption (a)

    13\frac{1}{3} (a3 – b3)

  2. BOption (b)

    13\frac{1}{3}(|b|3 – |a|3)

  3. COption (c)

    13\frac{1}{3} (a3 + b3)

  4. DOption (d)

    13\frac{1}{3}(|b3 – a3|)

View answer & explanation

Correct answer

Answer: (b)B.

13\frac{1}{3}(|b|3 – |a|3)

Why this answer is correct

Concept Used:

f(x)=∣x∣={−x,if x<0 x,if x≥0f(x) = |x| = \begin{cases} -x, & \text{if } x < 0 \ x, & \text{if } x \geq 0 \end{cases}

Explanation:

Given f(x) = x|x|, we have:

f(x)=x∣x∣={−x2,if x<0 x2,if x≥0f(x) = x|x| = \begin{cases} -x^2, & \text{if } x < 0 \ x^2, & \text{if } x \geq 0 \end{cases}

Case 1: a < b < 0

∫abf(x),dx=∫ab(−x2),dx=[−x33]ab=−b33+a33\int_a^b f(x) , dx = \int_a^b (-x^2) , dx = \left[ -\frac{x^3}{3} \right]_a^b = -\frac{b^3}{3} + \frac{a^3}{3}.

Case 2: a < 0 < b

∫abf(x),dx=∫a0(−x2),dx+∫0b(x2),dx=a33+b33\int_a^b f(x) , dx = \int_a^0 (-x^2) , dx + \int_0^b (x^2) , dx = \frac{a^3}{3} + \frac{b^3}{3}

Case 3: 0 < a < b

∫abf(x),dx=∫ab(x2),dx=b33−a33\int_a^b f(x) , dx = \int_a^b (x^2) , dx = \frac{b^3}{3} - \frac{a^3}{3}

We can combine these results into one expression using absolute values i.e., 13(∣b∣3−∣a∣3)\frac{1}{3}(|b|^3 - |a|^3)

Therefore, the value of  ∫abf(x),dx\int_a^b f(x) , dx for any real numbers a and b with a < b is13(∣b∣3−∣a∣3) \frac{1}{3}(|b|^3 - |a|^3), considering all three cases.

Question 206

Which statement is true?

  1. AOption (a)

    n! < (n+12)n\left( \frac{n + 1}{2} \right)^n

  2. BOption (b)

    n! < (n2)n(\frac{n}{2})^n

  3. COption (c)

    n! > (n)n

  4. DOption (d)

    n! > (n+12)n(\frac{n + 1}{2})^n

View answer & explanation

Correct answer

Answer: (a)A.

n! < (n+12)n\left( \frac{n + 1}{2} \right)^n

Why this answer is correct

Concept Used:

Explanation:

n! < (n2)n(\frac{n}{2})^n

Doesn't hold true for n = 1.

n! > (n)n 

Doesn't hold true for n = 2.

n!<(n+12)nn! < \left( \frac{n + 1}{2} \right)^n

Base Case (n = 1):

For n = 1, we have 1! = 1 and(1+12)1=1 \left( \frac{1 + 1}{2} \right)^1 = 1. Since 1 < 1, the base case holds.

Inductive Step:

Assume the statement holds for some arbitrary positive integer k, i.e., k! <(k+12)k \left( \frac{k + 1}{2} \right)^k.

We need to show that the statement holds for k + 1, i.e., (k + 1)! <(k+22)k+1 \left( \frac{k + 2}{2} \right)^{k + 1}.

(k+1)!=(k+1)k!<(k+1)(k+12)k(k + 1)! = (k + 1)k! < (k + 1) \left( \frac{k + 1}{2} \right)^k

Now, we need to prove that (k+1)(k+12)k<(k+22)k+1(k + 1) \left( \frac{k + 1}{2} \right)^k < \left( \frac{k + 2}{2} \right)^{k + 1}.

(k+1)!=(k+1)k!<(k+1)(k+12)k(k + 1)! = (k + 1)k! < (k + 1) \left( \frac{k + 1}{2} \right)^k

For any positive number k, we know that

k + 2 > k + 1

⇒ (k + 2)/2 > (k + 1)/2

⇒ (k + 2/2)k+1 > (k + 1/2)k+1

(k+1)!=(k+1)k!<(k+1)(k+12)k(k + 1)! = (k + 1)k! < (k + 1) \left( \frac{k + 1}{2} \right)^k

This can be proven by expanding both sides and simplifying.

After algebraic manipulation, it can be shown that (k+1)(k+12)k<(k+22)k+1(k + 1) \left( \frac{k + 1}{2} \right)^k < \left( \frac{k + 2}{2} \right)^{k + 1}  holds true for all k ≥q 1.

Therefore, by mathematical induction, n!<(n+12)nn! < \left( \frac{n + 1}{2} \right)^n holds for all n≥ 1.

Question 207

The statement (52n – 1) is always divisible by

  1. AOption (a)

    25

  2. BOption (b)

    15

  3. COption (c)

    24

  4. DOption (d)

    42

View answer & explanation

Correct answer

Answer: (c)C.

24

Why this answer is correct

Calculation:

(52n - 1) = (52 )n - 1 

= (1+ 24)n - 1

= nC0 (1)n (24)0 + nC1 (24)+ nC2 (24) 2 + ......... + nCn (24)n - 1

= 1 +  nC1 (24)+ nC2 (24) 2 + ......... + nCn (24)n - 1

=  nC1 (24)+ nC2 (24) 2 + ......... + nCn (24)n

= 24 [nC1+ nC2 (24) + ......... + nCn (24)n - 1]

= 24 × Some integer

∴ The statement (52n – 1) is always divisible by 24.

Question 208

Which one of the inequality is true?

  1. AOption (a)

    e < ∫01ex2\int_0^1 e^{x^2} dx< 1

  2. BOption (b)

    –1 < ∫01ex2\int_0^1 e^{x^2} dx < 1

  3. COption (c)

    1 < ∫01ex2\int_0^1 e^{x^2} dx < e

  4. DOption (d)

    14\frac{1}{4} < ∫01ex2\int_0^1 e^{x^2} dx < 12\frac{1}{2}

View answer & explanation

Correct answer

Answer: (c)C.

1 < ∫01ex2\int_0^1 e^{x^2} dx < e

Why this answer is correct

Explanation:

Since ex2e^{x^2} is increasing function in (0, 1)

So, e0 < ex2e^{x^2} < e1

⇒ 1 < ex2e^{x^2} < e

⇒ ∫011dx\int_0^1 1dx < ∫01ex2\int_0^1 e^{x^2}dx < ∫01edx\int_0^1 edx

⇒ [x]01[x]_0^1 < ∫01ex2\int_0^1 e^{x^2}dx < e[x]01e[x]_0^1

⇒ 1 < ∫01ex2\int_0^1 e^{x^2} dx < e

Option (3) is true

Question 209

Which of the following improper integral is convergent?

  1. AOption (a)

    ∫2∞1log⁡x\int_2^\infty \frac{1}{\log x} dx

  2. BOption (b)

    ∫011x(x+1)\int_0^1 \frac{1}{\sqrt{x}(x + 1)} dx

  3. COption (c)

    ∫12xlog⁡x\int_1^2 \frac{\sqrt{x}}{\log x} dx

  4. DOption (d)

    ∫0π11−cos⁡x\int_0^\pi \frac{1}{1 - \cos x} dx

View answer & explanation

Correct answer

Answer: (b)B.

∫011x(x+1)\int_0^1 \frac{1}{\sqrt{x}(x + 1)} dx

Why this answer is correct

Explanation:

For Option 1:

∫2∞1log⁡x,dx=lim⁡b→∞∫2b1log⁡x,dx \int_2^∞ \frac{1}{\log x} , dx = \lim_{b \to ∞} \int_2^b \frac{1}{\log x} , dx

u = log x, then du = 1x,dx\frac{1}{x} , dx. When x = 2,  u = log 2, and when x = b,  u = log b.

So, the integral becomes:

lim⁡b→∞∫log⁡2log⁡bdu=lim⁡b→∞[log⁡b−log⁡2] \lim_{b \to ∞} \int_{\log 2}^{\log b} du = \lim_{b \to ∞} [\log b - \log 2]

=lim⁡b→∞log⁡(b2) = \lim_{b \to ∞} \log \left( \frac{b}{2} \right)

As b approaches infinity,  b2\frac{b}{2}  also approaches infinity, so log⁡(b2) \log \left( \frac{b}{2} \right) approaches infinity.

Therefore, the integral ∫2∞1log⁡x,dx \int_2^∞ \frac{1}{\log x} , dx  diverges to ∞.

For Option 2:

For ∫011x(x+1),dx \int_0^1 \frac{1}{\sqrt{x}(x + 1)} , dx

Let's u = x\sqrt{x} , then x = u2 and dx = 2u du.

When x = 0, u = 0\sqrt{0} = 0 , and when x = 1, u = 1 \sqrt{1} = 1.

 ∫011x(x+1),dx=∫011u(u2+1)⋅2u,du=2∫011u2+1,du\int_0^1 \frac{1}{\sqrt{x}(x + 1)} , dx = \int_0^1 \frac{1}{u(u^2 + 1)} \cdot 2u , du = 2 \int_0^1 \frac{1}{u^2 + 1} , du

=2[arctan⁡(u)]01=2[arctan⁡(1)−arctan⁡(0)]=2[π4−0]=π2 = 2 \left[ \arctan(u) \right]_0^1 = 2 \left[ \arctan(1) - \arctan(0) \right] = 2 \left[ \frac{\pi}{4} - 0 \right] = \frac{\pi}{2}

Therefore, ∫011x(x+1),dx=π2 \int_0^1 \frac{1}{\sqrt{x}(x + 1)} , dx = \frac{\pi}{2}

Options 3 and 4 also diverge.

Thus, Option 2 is the correct answer.

Question 210

Which of the following improper integral is not convergent?

  1. AOption (a)

    ∫1∞1x3\int_1^\infty \frac{1}{x^3} dx

  2. BOption (b)

    ∫1∞1x3/2\int_1^\infty \frac{1}{x^{3/2}} dx

  3. COption (c)

    ∫1∞1x\int_1^\infty \frac{1}{\sqrt{x}} dx

  4. DOption (d)

    ∫1∞1x2\int_1^\infty \frac{1}{x^2} dx

View answer & explanation

Correct answer

Answer: (c)C.

∫1∞1x\int_1^\infty \frac{1}{\sqrt{x}} dx

Why this answer is correct

Concept Used:

For improper integral of the form

∫1∞1xpdx\int_1^∞ \frac{1}{x^p} dx

where p varies for each integral. The convergence of these integrals depends on the value of p.

For p > 1, the integral ∫1∞1xpdx\int_1^∞ \frac{1}{x^p} dx converges to 1p−1\frac{1}{p-1} because the function 1xp\frac{1}{x^p} decreases quickly enough as x increases to infinity that the total area under the curve from 1 to ∞ is finite.

For p ≤ 1, the integral diverges because the function 1xp\frac{1}{x^p} does not decrease quickly enough to produce a finite area from 1 to ∞.

Explnation:

Option 1: ∫1∞1x3dx,\int_1^∞ \frac{1}{x^3} dx, where p = 3 > 1. Thus, this should be convergent.

Option 2:∫1∞1x3/2dx \int_1^∞ \frac{1}{x^{3/2}} dx, where p=32>1p = \frac{3}{2} > 1. Thus, this should be convergent.

Option 3: ∫1∞1xdx \int_1^∞ \frac{1}{\sqrt{x}} dx, where p=12<1p = \frac{1}{2} < 1. Thus, this should be divergent.

Option 4: ∫1∞1x2\int_1^∞ \frac{1}{x^2} dx, where p = 2 > 1. Thus, this should be convergent.

Thus, option 3 is correct.

Question 211

P(n): 1.1! + 2.2! + 3.3! + n.n! = (n + 1)! – 1, then P(n) statement is true-

  1. AOption (a)

    For n > 1

  2. BOption (b)

    For n > 4

  3. COption (c)

    For all values of n

  4. DOption (d)

    For all negative values of n

View answer & explanation

Correct answer

Answer: (c)C.

For all values of n

Why this answer is correct

Proof:

To prove P(n):1⋅1!+2⋅2!+3⋅3!+…+n⋅n!=(n+1)!−1P(n): 1 ⋅ 1! + 2 ⋅ 2! + 3 ⋅ 3! + … + n ⋅ n! = (n + 1)! - 1 for all positive integers n, we can use mathematical induction.

Base Case: Putting n = 1  into the equation, we get:

1 ⋅ 1! = (1 + 1)! - 1

⇒ 1 ⋅ 1 = (2)! - 1

⇒ 1 = 2 - 1

⇒ 1 = 1

Thus, the Base step is true.

   

Inductive Hypothesis: Assume that the statement  P(k)  is true for some arbitrary positive integer k, i.e., 

1⋅1!+2⋅2!+3⋅3!+…+k⋅k!=(k+1)!−11 ⋅ 1! + 2 ⋅ 2! + 3 ⋅ 3! + … + k ⋅ k! = (k + 1)! - 1  . . . (i)

Inductive Step: We want to prove that if  P(k)  is true, then  P(k + 1)  is also true.

Adding  (k + 1) ⋅ (k + 1)!  to both sides of the equation (i), we get:

   

1 ⋅ 1! + 2 ⋅ 2! + 3 ⋅ 3! + … + k ⋅ k! + (k + 1) ⋅ (k + 1)! = (k + 1)! - 1 + (k + 1) ⋅ (k + 1)! 

RHS = (k + 1)! ⋅ (1 + k + 1) - 1 

⇒ RHS = (k + 1)! ⋅ (k + 2) - 1 

⇒ RHS  = (k + 2)! - 1 

Therefore,  P(k + 1)  is true.

Since  P(1)  is true and  P(k)  being true implies  P(k + 1)  is true, by mathematical induction,  P(n)  is true for all positive integers n.

Question 212

Let In = ∫\int tannx dx, (n > 1). I4 + I6 = a tan5x + bx5 + c, where c is a Constant of integration, then the order pair (a, b) is equal to-

  1. AOption (a)

    (15,0)(\frac{1}{5}, 0)

  2. BOption (b)

    (−15,−1)(-\frac{1}{5}, -1)

  3. COption (c)

    (15,45)(\frac{1}{5},\frac{4}{5})

  4. DOption (d)

    (−15,1)(-\frac{1}{5}, 1)

View answer & explanation

Correct answer

Answer: (a)A.

(15,0)(\frac{1}{5}, 0)

Why this answer is correct

Calculation:

We have  In = ∫tann x dx

In + In + 2 =  ∫tann x dx + ∫tann + 2 x dx

=  ∫(tann x + tann + 2x) dx

= ∫ tannx ( 1 + tan2 x) dx

= ∫ tannx sec2 x dx

Let tan x = t ⇒ sec2 x dx = dt

Now In + In + 2  =  ∫ tn dt

= tn+1n+1+c\frac{t^{n + 1}}{n +1} + c

In + In + 2 = (tan⁡x)n+1n+1+c\frac{(\tan x)^{n + 1}}{n +1} + c

Substitute n = 4 in the above equation, we get

I4 + I6 = (tan⁡x)55+c\frac{(\tan x)^{5}}{5} + c

I4 + I6 = tan⁡5x5+c\frac{\tan^5 x}{5} + c

Comparing with  I4 + I6 = a tan5x + bx5 + c, we have

 a = 1/5 and b = 0

∴  The order pair (a, b) is equal to (1/5, 0).

Question 213

The statement 1.2 + 2.22 + 3.23 + ........n.2n can be written in the form of -

  1. AOption (a)

    (n – 1)2n + 2

  2. BOption (b)

    (n + 1)2n + 2

  3. COption (c)

    (n + 1)2n+1 + 2

  4. DOption (d)

    (n – 1)2n+1 + 2

View answer & explanation

Correct answer

Answer: (d)D.

(n – 1)2n+1 + 2

Why this answer is correct

Concept:

Principle of mathematical induction: Let P(n) be the statement. 

P(n) is true for n = 1. Let P(n) is true for n = k then if P(n) is true fro n = k + 1 then P(n) is true for all natural number n

Explanation:

1.2 + 2.22 + 3.23 + ........n.2n = (n – 1)2n+1 + 2

Here is the proof.

Let the statement be

P(n): 1.2 + 2.22 + 3.23 + ........n.2n = (n – 1)2n+1 + 2

For n = 1

LHS = 1.2 = 2

RHS = (1 – 1)22 + 2 = 2

SO, LHS = RHS

P(1) is true.

Let P(k) is true for some positive integer k

i.e., 1.2 + 2.22 + 3.23 + ........k.2k = (k – 1)2k+1 + 2....(i)

We have to show that P(k + 1) is true

For n = k + 1

LHS = 1.2 + 2.22 + 3.23 + ........k.2k + (k + 1)2k+1

       =  (k – 1)2k+1 + 2 + (k + 1)2k+1

       = (k - 1 + k + 1)2k+1 + 2

       = 2k.2k+1 + 2

       = k.2k+2 + 2

and for n = k + 1

RHS = k.2k+2 + 2 (replacing k by k + 1 in (i))

So, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Hence 1.2 + 2.22 + 3.23 + ........n.2n = (n – 1)2n+1 + 2 for all natural number n

Option (4) is true

Question 214

If In = ∫0π/2\int_0^{\pi/2} xn sin x dx, n > 1, then In + n(n – 1)In – 2 is equal to :

  1. AOption (a)

    n(π2)n−1(\frac{\pi}{2})^{n - 1}

  2. BOption (b)

    (π2)n−1(\frac{\pi}{2})^{n - 1}

  3. COption (c)

    (π2)(\frac{\pi}{2})

  4. DOption (d)

    n (π2)n(\frac{\pi}{2})^n

View answer & explanation

Correct answer

Answer: (a)A.

n(π2)n−1(\frac{\pi}{2})^{n - 1}

Why this answer is correct

Explanation:

 In = ∫0π/2\int_0^{\pi/2} xn sin x dx....(i)

Using by-parts

 In = [−xncos⁡x]0π/2[-x^n\cos x]_0^{\pi/2} + ∫0π/2\int_0^{\pi/2}nxn-1 cos x dx

In = 0 + n∫0π/2\int_0^{\pi/2}xn-1 cos x dx

Again integrating by-parts

In = n[xn−1sin⁡x]0π/2[x^{n-1}\sin x]_0^{\pi/2} - n(n - 1)∫0π/2\int_0^{\pi/2}xn-2 sin x dx

In = n(π2)n−1(\frac{\pi}2)^{n-1} - n(n - 1)In-2 (by (i))

In +  n(n - 1)In-2 = n(π2)n−1(\frac{\pi}2)^{n-1}

Option (1) is true

Question 215

The statement( xn – yn) is divisible by -

  1. AOption (a)

    (x + y) (x – y)

  2. BOption (b)

    (x2 + y2)

  3. COption (c)

    (x – y)

  4. DOption (d)

    (x + y)

View answer & explanation

Correct answer

Answer: (c)C.

(x – y)

Why this answer is correct

Concept Used:

The statement xn – yn can be factored using the difference of squares formula:

xn−yn=(x−y)(xn−1+xn−2y+xn−3y2+…+xyn−2+yn−1)x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + x^{n-3}y^2 + \ldots + xy^{n-2} + y^{n-1})

Explanation:

So, the expression xn – yn is divisible by:

Option 1: (x - y) (as it is a factor of xn – yn ).

Option 2: (x + y) (as it is one of the factors of the sum xn – yn  in some cases, for example, when n is exponent of 2)

Option 3: (x2 + y2)(no, this is not a factor of xn – yn)

Option 4: (x + y)(x - y) (as it is one of the factors of the sum xn – yn  in some cases, for example, when n is exponent of 2)

Thus, (x – y) is the only factor of xn – yn without any condition for n.

Question 216

Which of the following is true?

Let f: [0, 1] → R be defined by

f(x) = {(−1)r−1,when1r+1<x≤1r,r=1,2,3,... 0,when x=0\rm \begin{cases} (-1)^{r-1}, & \rm when \frac{1}{r + 1} < x \leq \frac{1}{r}, r = 1, 2, 3,...\ 0, & \rm when \ x = 0 \end{cases}

  1. AOption (a)

    f is integrable on [0, 1], but f is unbounded

  2. BOption (b)

    f is integrable on [0, 1]

  3. COption (c)

    f is not integrable in [0, 1]

  4. DOption (d)

    f is not integrable on [0, 1], but not bounded

View answer & explanation

Correct answer

Answer: (b)B.

f is integrable on [0, 1]

Why this answer is correct

Concept:

A function f: [a,b] → R that is bounded and continuous except for a finite number of points at which it is perhaps discontinuous is Riemann integrable.

Explanation:

 f: [0, 1] → R is defined by

f(x) = {(−1)r−1,when1r+1<x≤1r,r=1,2,3,... 0,when x=0\rm \begin{cases} (-1)^{r-1}, & \rm when \frac{1}{r + 1} < x \leq \frac{1}{r}, r = 1, 2, 3,...\ 0, & \rm when \ x = 0 \end{cases}

     = \(\rm \begin{cases} 1, \frac12<x\leq 1\-1, \frac13<x\leq \frac12\1, \frac14<x\leq \frac13\.......\ 0, \ x = 0 \end{cases}\)

So, f is discontinuous at x = 0, 1/2, 1/3, ....

Hence f is Reimann integrable i.e., integrable on [0, 1].

Option (2) is true

Question 217

The solution of the differential equation x2 d2ydx2\frac{d^2y}{dx^2} –3x dydx\frac{dy}{dx} + 4y = 0 is:

  1. AOption (a)

    y = (c1 + c2 log x} x

  2. BOption (b)

    y = (c1 + c2 log x} e2x

  3. COption (c)

    y = (c1 + c2 log x} ex

  4. DOption (d)

    y = (c1 + c2 log x} x2

View answer & explanation

Correct answer

Answer: (d)D.

y = (c1 + c2 log x} x2

Why this answer is correct

Calculation:

The given equation is

x2d2ydx2−3xdydx+4y=0{x^2}\frac{{{d^2}y}}{{d{x^2}}} - 3x\frac{{dy}}{{dx}} + 4y = 0

Let x=ez⇒z=log⁡xx = {e^z} \Rightarrow z= \log x

dydx=dydz.dzdx=1x.dydz\frac{{dy}}{{dx}} = \frac{{dy}}{{dz}}.\frac{{dz}}{{dx}} = \frac{1}{x}.\frac{{dy}}{{dz}}

⇒xdydx=Dy, \Rightarrow x\frac{{dy}}{{dx}} = Dy, Where ddz=D\frac{d}{{dz}} = D

Again d2ydx2=ddx(dydx)=1xddz(1xdydz)\frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{1}{x}\frac{d}{{dz}}\left( {\frac{1}{x}\frac{{dy}}{{dz}}} \right)

d2ydx2=1x2(d2ydz2−dydz)\frac{{{d^2}y}}{{d{x^2}}} = \frac{1}{{{x^2}}}\left( {\frac{{{d^2}y}}{{d{z^2}}} - \frac{{dy}}{{dz}}} \right)

⇒x2d2ydx2=(D2−D)y=D(d−1)y \Rightarrow {x^2}\frac{{{d^2}y}}{{d{x^2}}} = \left( {{D^2} - D} \right)y = D\left( {d - 1} \right)y

Similarly, x3d3ydx3=D(D−1)(D−2)y{x^3}\frac{{{d^{3y}}}}{{d{x^3}}} = D\left( {D - 1} \right)\left( {D - 2} \right)y

⇒[D(D−1)−3D+4]=0 \Rightarrow \left[ {D\left( {D - 1} \right) - 3D + 4} \right] = 0

⇒(D−2)2y=0 \Rightarrow {\left( {D - 2} \right)^2}y = 0

The required solution is

y=(c1+c2z)c2zy = \left( {{c_1} + {c_2}z} \right){c^{2z}}

⇒y=(c1+c2log⁡x)x2 \Rightarrow y = \left( {{c_1} + {c_2}\log x} \right){x^2}

Question 218

The solution of the differential equation (2D + 1)2y = 0 is -

  1. AOption (a)

    y = Ae–x/2

  2. BOption (b)

    y = (A + Bx)e–x

  3. COption (c)

    y = (A + Bx)e–x/2

  4. DOption (d)

    y = (A + B)e–x/2

View answer & explanation

Correct answer

Answer: (c)C.

y = (A + Bx)e–x/2

Why this answer is correct

Concept:

If m1, m2, m3, ... are the roots of the charateristic equation of homogenous differential equation, then the solution is given by:

y = C1em1x+C2em2x+C3em3x+⋯C_1e^{m_1x}+C_2e^{m_2x}+C_3e^{m_3x}+\cdots

If m1 is a repeated root of order 2, m2, m3 are simple roots, then solution is y = (C1+C2x)em1x+C3em2x+C4em3x(C_1+C_2x)e^{m_1x}+C_3e^{m_2x}+C_4e^{m_3x}

Calculation:

Given differential equation, (2D + 1)2y = 0

⇒ (4D2 + 4D + 1)y = 0

∴ Characteristic equation = 4m2 + 4m + 1 = 0

⇒ (2m + 1)2 = 0

⇒ m1,2 = −12-\frac{1}{2}, −12-\frac{1}{2}

⇒ m is a repeated root

∴ y = (A + Bx)emx 

= (A + Bx)e–x/2 

∴ The solution of the differential equation is (A + Bx)e–x/2 .

The correct answer is Option 3.

Question 219

General from of the statement is (1+11)(1+12)(1+13)(1 + \frac{1}{1}) (1 + \frac{1}{2}) (1 + \frac{1}{3})......(1+1n)(1 + \frac{1}{n}) is-

  1. AOption (a)

    (1+1n)(1 + \frac{1}{n})

  2. BOption (b)

    (n + 1)

  3. COption (c)

    (n+1n)(n + \frac{1}{n})

  4. DOption (d)

    (n + 1)2

View answer & explanation

Correct answer

Answer: (b)B.

(n + 1)

Why this answer is correct

Explanation:

(1+11)(1+12)(1+13)(1 + \frac{1}{1}) (1 + \frac{1}{2}) (1 + \frac{1}{3})...(1+1n)(1 + \frac{1}{n})

= 21⋅32⋅43⋅⋅⋅⋅n+1n\frac21\cdot\frac32\cdot\frac43\cdot\cdot\cdot\cdot\frac{n+1}{n}

= n + 1(other terms will be cancelled)

Option (2) is true

Question 220

Find the value of statement P(n): 1.6 + 2.9 + 3.12 + ----- + n(3n + 3)

  1. AOption (a)

    (n + 1) (n + 2) (n + 3)

  2. BOption (b)

    (n + 3) (n + 6)

  3. COption (c)

    n(n + 1) (n + 2)

  4. DOption (d)

    n(n + 6)

View answer & explanation

Correct answer

Answer: (c)C.

n(n + 1) (n + 2)

Why this answer is correct

Formula Used:

Sum of the first n positive integers is given by the formula:

1+2+3+…+n=∑k=1nk=n(n+1)21 + 2 + 3 + \ldots + n = \sum_{k=1}^{n} k = \frac{n(n + 1)}{2}

Sum of the first  n  perfect squares is given by the formula:

12+22+32+…+n2=∑k=1nk2=n(n+1)(2n+1)61^2 + 2^2 + 3^2 + \ldots + n^2 = \sum_{k=1}^{n} k^2= \frac{n(n + 1)(2n + 1)}{6}

Explanation:

Given: P(n): 1.6 + 2.9 + 3.12 + ----- + n(3n + 3)

We can write this as  ∑k(3k + 3)

Where k is an integer from 1 to n.

To find the sum of these terms, we can expand each term and then use the formula for the sum of an arithmetic series.

∑ k(3k + 3) = ∑ 3k2 + 3k 

⇒ ∑ k(3k + 3) = 3⋅n(n+1)(2n+1)6+3⋅n(n+1)2 3 \cdot \frac{n(n + 1)(2n + 1)}{6} + 3 \cdot \frac{n(n + 1)}{2}  [Using the formula for sum]

⇒ ∑ k(3k + 3)  =12n(n+1)(2n+1)+32n(n+1) \frac{1}{2} n(n + 1)(2n + 1) + \frac{3}{2} n(n + 1)

⇒ ∑ k(3k + 3) = 12n(n+1)(2n+4) \frac{1}{2} n(n + 1)(2n + 4)

⇒ ∑ k(3k + 3) = 12n(n+1)(n+2)×2\frac{1}{2} n(n + 1)(n + 2) \times 2

⇒ ∑ k(3k + 3) = n(n+1)(n+2)n(n + 1)(n + 2)  

Thus, the required sum is n(n + 1)(n + 2).

Discipline 7 (20 questions)

Question 221

Which of the following is not true?

  1. AOption (a)

    Let G is non-commutative group of order 2p, p is prime and there exists at least one element a of order p, then <a> is normal in G.

  2. BOption (b)

    Let G be a group in which (ab)3 = a3 b3 for all a, b ∈ G, then H = {x3 : x ∈ G} is normal subgroup of G

  3. COption (c)

    If H is normal subgroup of a finite group G, then [G:H] = 2.

  4. DOption (d)

    Every commutative subgroup of a group is normal subgroup of the group

View answer & explanation

Correct answer

Answer: (c)C.

If H is normal subgroup of a finite group G, then [G:H] = 2.

Why this answer is correct

Concept:

(i) Let G be a group. A group H is said to be a normal subgroup of G if ghg-1 ∈ H for all g ∈ G and h ∈ H.

(ii) A group is abelian if (ab)3 = a3b3 and no elements of order 3.

Explanation:

Option (4): let G be a group and H be a commutative subgroup.

We have to prove that H is normal subgroup of G,

ghg-1 = hgg-1 = he = h ∈ H

So, H is normal subgroup of G

Option (4) is true.

Option (2): G is a group in which (ab)3 = a3 b3 for all a, b ∈ G, 

Then G is abelian.

H = {x3 : x ∈ G}

Let h3 ∈ H and g ∈ G

then gh3g-1 = h3gg-1 = he = h ∈ H (as G is abelian)

So, H is normal subgroup of G.

Option (2) is true

Similarly, we can prove that option (1) is true.

Hence option (3) is not true

Question 222

Let (G, ∘) and (G', *) be two finite groups and Φ : G → G' be an epimorphism then -

  1. AOption (a)

    O(G) is a divisor of O(G').

  2. BOption (b)

    O(G) is not a divisor of O(G')

  3. COption (c)

    O(G') is a divisor of O(G).

  4. DOption (d)

    O(G') is not a divisor O(G).

View answer & explanation

Correct answer

Answer: (c)C.

O(G') is a divisor of O(G).

Why this answer is correct

Concept:

A function f: X → Y is called epimorphism if and only if it is surjective.

Explanation:

(G, ∘) and (G', *) are two finite groups and Φ : G → G' be an epimorphism

Let k = o(G') be the order of the group G', i.e., the number of elements in G'. 

Let the kernel is K = {x ∈ G | Φ(x) = e'}, where e' is the identity element of G'.

Then K is a normal subgroup of G, and that Φ(G) is isomorphic to the group G/K.

Now, let the group homomorphism f: G/K → G', defined by f(xK) = Φ(x).

This is well-defined since Φ(x) = Φ(y) if and only if xK = yK, and it is also a group homomorphism since Φ(xy) = Φ(x)Φ(y) for all x, y ∈ G.

By the First Isomorphism Theorem, we have that G/K / ker(f) ≅ im(f).

Since the image of f is a subgroup of G', which has k elements, we have that im(f) is a subgroup of g' of order k.

Therefore, o(G) = |G/K| = |ker(f)| |im(f)| = k |ker(f)|.

Since k and |ker(f)| are both positive integers, we conclude that k divides o(G).

i.e., O(G') is a divisor of O(G).

Option (3) is true

Question 223

Let (G, ∘) and (G', *) be two groups and Φ : G → G' be an epimorphism. Then Φ is an isomorphism if and only if-

  1. AOption (a)

    Ker Φ ⊂ {eG}

  2. BOption (b)

    Ker Φ ⊂ {eG'}

  3. COption (c)

    Ker Φ = {eG}

  4. DOption (d)

    Ker Φ = { eG'}

View answer & explanation

Correct answer

Answer: (c)C.

Ker Φ = {eG}

Why this answer is correct

Concept:

(i) The kernel of a group homomorphism f:G → G' is the set of all elements of G which are mapped to the identity element of G'.

(ii) A function f: X → Y is called monomorphism if and only if it is injective.

(iii) A function f: X → Y is called epimorphism if and only if it is surjective.

(iv) A function f: X → Y is called isomorphism if and only if it is bijective.

(iv): Epimorphism preserves identity: Let (G, ∘) and (H, ∗) be algebraic structures. Let ϕ:(G, ∘) → (H, ∗) be an epimorphism. Let (G, ∘) have an identity element eG. Then (T,∗) has the identity element ϕ(eG).

Explanation:

(G, o) and (G', *) are two groups and Φ : G → G' be an epimorphism.

Let eG and eG' be the identity elements of G and G' respectively.

Let Φ is an isomorphism then by definition Φ is bijective i.e., injective and surjective.

Then there exists exactly one element x of G such that Φ(x) = eG'.

From epimorphism preserves identity, that element x is eG: ϕ(eG) = eG'.

Thus by definition of kernel

ker(Φ)={eG}

Option (3) is true

Question 224

If 2n+1Pn–1: 2n–1Pn = 3 : 5, then what is the value of n?

  1. AOption (a)

    3

  2. BOption (b)

    0

  3. COption (c)

    2

  4. DOption (d)

    4

View answer & explanation

Correct answer

Answer: (d)D.

4

Why this answer is correct

Concept:

nPr=n!(n−r)!=n×(n−1)×.....×(n−r+1)\rm ^nP_r=\frac{n!}{(n-r)!}={n\times (n-1)\times .....\times (n-r+1)}

1a:1b=1a1b=1a×b\rm \frac1a:\frac1b= \frac{\frac1a}{\frac1b}=\frac1a\times b

Calculation:

(2n+1)Pn−1:(2n−1)Pn=3:5\rm ^{(2n+1)}P_{n-1}:^{(2n-1)}P_{n}=3:5

⇒(2n+1)![2n+1−(n−1)]!:(2n−1)![2n−1−n)]!=3:5\rm ⇒ \frac{(2n+1)!}{[2n+1-(n-1)]!}:\frac{(2n-1)!}{[2n-1-n)]!}=3:5

⇒(2n+1)!(n+2)!:(2n−1)!(n−1)!=3:5\rm ⇒ \frac{(2n+1)!}{(n+2)!}:\frac{(2n-1)!}{(n-1)!}=3:5

⇒(2n+1)×(2n)×(2n−1)!(n+2)×(n+1)×(n)×(n−1)!:(2n−1)!(n−1)!=3:5\rm ⇒ \frac{(2n+1)\times(2n)\times(2n-1)!}{(n+2)\times(n+1)\times(n)\times(n-1)!}:\frac{(2n-1)!}{(n-1)!}=3:5

⇒(2n+1)×(2n)×(2n−1)!(n+2)×(n+1)×(n)×(n−1)!×(n−1)!(2n−1)!=35\rm ⇒ \frac{(2n+1)\times(2n)\times(2n-1)!}{(n+2)\times(n+1)\times(n)\times(n-1)!}\times \frac{(n-1)!}{(2n-1)!}=\frac35

⇒ 10n(2n + 1) = 3n(n + 1)(n + 2)

⇒20n+10=3(n2+3n+2)\rm ⇒ 20n+10=3(n^2+3n+2)

⇒20n+10=3n2+9n+6\rm ⇒ 20n+10=3n^2+9n+6

⇒3n2−11n−4=0\rm ⇒ 3n^2-11n-4=0

⇒3n2−12n+n−4=0\rm ⇒ 3n^2-12n+n-4=0

⇒3n(n−4)+1(n−4)=0\rm ⇒ 3n(n-4)+1(n-4)=0

⇒(n−4)(3n+1)=0\rm ⇒ (n-4)(3n+1)=0

⇒ n = 4 or n = -1/3

∴ n = 4                          ....(∵ n ≠  -1/3, as n cannot be negative)

The value of n is 4.

Question 225

Let H be a subgroup of a group G and K be a normal subgroup of a group G, then

  1. AOption (a)

    K is a normal subgroup of KH

  2. BOption (b)

    K is a normal subgroup of HK ∩ KH

  3. COption (c)

    K is a normal subgroup of HK

  4. DOption (d)

    K is not a normal subgroup of HK

View answer & explanation

Correct answer

Answer: (c)C.

K is a normal subgroup of HK

Why this answer is correct

Concept:

A group H is said to be normal subgroup of G if ghg-1 ∈ H 

Explanation:

H is a subgroup of group G and K is a normal subgroup of group G.

Then HK is a subgroup of G (by theorem)

Now, let h1k1 ∈ HK and k ∈ K then k ∈ HK

So, k(h1k1)-1 ∈ HK (Property of subgroup) 

i..e, (h1k1)k(h1k1)-1 ∈ HK (Property of group) 

Hence K is normal subgroup of HK.

Option (3) is true

Question 226

Let (G, o) and (G', *) be two groups and f: G → G' be a homomorphism, then f(G)

  1. AOption (a)

    Is a cyclic subgroup of G' is commutative

  2. BOption (b)

    Is a cyclic subgroup of G'

  3. COption (c)

    Is a cyclic subgroup of G' if G may not be cyclic

  4. DOption (d)

    Is not a cyclic subgroup of G' if G is cyclic

View answer & explanation

Correct answer

Answer: (c)C.

Is a cyclic subgroup of G' if G may not be cyclic

Why this answer is correct

Concept:

(i) A cyclic group is a group that can be generated by a single element a.

(ii) A group G is said to be commutative if ab = ba for a, b ∈ G

Explanation:

Let G = Z\mathbb Z4 and G' = Z\mathbb Z8

 then G is not cyclic

Define f: Z\mathbb Z4 → Z\mathbb Z8 such that f(x) = 2x where operations are taken modulo 8.

So, f(x) is homomorphism and f(Z\mathbb Z4) = {0, 2, 4, 6} forms a cyclic subgroup of Z\mathbb Z8 generated by 2.

Hence option (3) is true

Question 227

Using digits 1, 2, 3, 4, 5 and 6 how many numbers between 3000 and 4000 can be made if no digit should be repeated?

  1. AOption (a)

    70

  2. BOption (b)

    100

  3. COption (c)

    90

  4. DOption (d)

    60

View answer & explanation

Correct answer

Answer: (d)D.

60

Why this answer is correct

Concept Used:

Principle of Counting: The principle states that if there are n1​ ways to perform the first task, and for each of these ways, there are n2​ ways to perform the second task, n3​ ways to perform the third task, and so on, then the total number of ways to perform all tasks in sequence is the product of the number of ways to perform each task individually.

Explanation:

To form a number between 3000 and 4000 using the digits 1, 2, 3, 4, 5, and 6 without repetition, we need to count the number of permutations of these digits.

First, let's consider the thousands place. Since the number must be between 3000 and 4000, the thousands place must be filled with the digit 3.

For the hundreds place, we have 5 remaining digits (1, 2, 4, 5, 6) to choose from.

For the tens place, we have 4 remaining digits after choosing the digit for the hundreds place. (as digits can't be repeated)

For the units place, we have 3 remaining digits after choosing the digits for the thousands, hundreds, and tens places.

So, the total number of numbers that can be formed is 5 × 4 × 3 = 60.

Therefore, 60 numbers between 3000 and 4000 can be made using the digits 1, 2, 3, 4, 5, and 6 without repetition.

Question 228

A commutative group G is simple if and only if- 

  1. AOption (a)

    O(G) = n, n is a prime integer

  2. BOption (b)

    O(G) = n, n is an even integer

  3. COption (c)

    O(G) = n, n is a positive integer

  4. DOption (d)

    O(G) = n, n is an odd integer

View answer & explanation

Correct answer

Answer: (a)A.

O(G) = n, n is a prime integer

Why this answer is correct

Concept Used:

A commutative group G is simple if and only if it has no nontrivial proper subgroups.

Prime order group has no nontrivial proper subgroups (except for the trivial subgroup and the group itself).

Explanation:

From both the statements in given above, we can conclude that any commutative Group G is simple if and only if O(G) is an prime number.

Question 229

The particular integral of (D2 + a2)y = sin ax (D ≡ ddx\frac{d}{dx}) is-

  1. AOption (a)

    – xa2\frac{xa}{2} cos ax

  2. BOption (b)

    – x2a\frac{x}{2a} cos ax

  3. COption (c)

    x2a\frac{x}{2a} cos ax

  4. DOption (d)

    xa2\frac{xa}{2} cos ax

View answer & explanation

Correct answer

Answer: (b)B.

– x2a\frac{x}{2a} cos ax

Why this answer is correct

Concept:

Some important fomulas to calculate particular integral(PI):

1f(D2)cos⁡ax=1f(−a2)cos⁡ax\frac{1}{f(D^2)}\cos ax = \frac{1}{f(-a^2)}\cos ax

If f(- a2) = 0, then 1f(D2)cos⁡ax=x1f′(−a2)cos⁡ax\frac{1}{f(D^2)}\cos ax = x\frac{1}{f'(-a^2)}\cos ax

Calculation:

Given, (D2 + a2)y = sin ax 

∴ Particular Integral, PI = 1D2+a2sin⁡ax\frac{1}{D^2+a^2}\sin ax = 1f(D2)sin⁡ax\frac{1}{f(D^2)}\sin ax

Here, f(D2) = D2 + a2 

Replacing D2 by - a2, we get f(- a2) = - a2 + a2 

∴ PI = x1f′(D2)sin⁡axx\frac{1}{f'(D^2)}\sin ax

= x12Dsin⁡axx\frac{1}{2D}\sin ax

= x2∫sin⁡axdx\frac{x}{2}\int\sin axdx [∵ 1D\frac{1}{D} = ∫\int ]

= −x2acos⁡ax-\frac{x}{2a}\cos ax

∴ The particular integral of (D2 + a2)y = sin ax is – x2a\frac{x}{2a} cos ax.

The correct answer is Option 2.

Question 230

The solution of the differential equation is d2ydx2\frac{d^2y}{dx^2} – 3dydx\frac{dy}{dx} + 2y = e3x is-

  1. AOption (a)

    y = c1 ex + c2 e–2x + 12\frac{1}{2} e–3x

  2. BOption (b)

    y = c1 ex + c2 e–2x + 12\frac{1}{2} e3x

  3. COption (c)

    y = c1 ex + c2 e2x + 12\frac{1}{2} e3x

  4. DOption (d)

    y = c1 e–x + c2 e2x + 12\frac{1}{2} e–3x

View answer & explanation

Correct answer

Answer: (c)C.

y = c1 ex + c2 e2x + 12\frac{1}{2} e3x

Why this answer is correct

Concept:

For a second-order differential function, the characteristic equation will have two roots D1 and D2. The roots can have three possible forms, ie.

  1. Real, distinct roots, D1 ≠ D2

  2. Complex Roots, D1, D2 = a + ib

  3. Double roots, D1 = D2 = 0

If two roots are real and distinct (i.e. D1 ≠ D2), the general solution will be:

y=C1eD1t+C2eD2ty = {C_1}{e^{{D_1}t}} + {C_2}{e^{{D_2}t}}

Calculation:

d2ydx2−3dydx+2y=e3x\frac{{{d^2}y}}{{d{x^2}}} - 3\frac{{dy}}{{dx}} + 2y = {e^{3x}}

(D2 - 3D + 2) y = e3x

A.E. is, (D2 - 3D + 2) = 0

⇒ (D - 1) (D - 2) = 0

⇒ D = 1, 2

Complementary function is:

y = c1ex + c2e2x

Particular integral is

P.I=1(D−1)(D−2).e3xP.I = \frac{1}{{\left( {D - 1} \right)\left( {D - 2} \right)}}.{e^{3x}}

=e3x(3−1)(3−2)=e3x2= \frac{{{e^{3x}}}}{{\left( {3 - 1} \right)\left( {3 - 2} \right)}} = \frac{{{e^{3x}}}}{2}

Solution of differential equation is:

y=c1ex+c2e2x+e3x2y = {c_1}{e^x} + {c_2}{e^{2x}} + \frac{{{e^{3x}}}}{2}

If the roots of the characteristic equation are complex (i.e. D1, 2 = a ± ib), the general solution to the differential equation will be:

y=C1eatcos⁡(bt)+C2eatsin⁡(bt)y = {C_1}{e^{at}}\cos \left( {bt} \right) + {C_2}{e^{at}}\sin \left( {bt} \right)

The solution for the second-order differential equation with equal roots of the characteristic equation is given by:

y=(c1+c2x)eD1xy = \left( {{c_1} + {c_2}x} \right){e^{{D_1x}}}

Question 231

A particular solution of differential equation d2ydx2\frac{d^2y}{dx^2} – dydx\frac{dy}{dx} – 2y = cos x + 3sin x is-

  1. AOption (a)

    –cos x

  2. BOption (b)

    –sin x

  3. COption (c)

    sin x

  4. DOption (d)

    cos x

View answer & explanation

Correct answer

Answer: (b)B.

–sin x

Why this answer is correct

Concept:

Some important fomulas to calculate particular integral(PI):

1f(D2)cos⁡ax=1f(−a2)cos⁡ax\frac{1}{f(D^2)}\cos ax = \frac{1}{f(-a^2)}\cos ax

If f(- a2) = 0, then 1f(D2)cos⁡ax=x1f′(−a2)cos⁡ax\frac{1}{f(D^2)}\cos ax = x\frac{1}{f'(-a^2)}\cos ax

Calculation:

Given, d2ydx2\frac{d^2y}{dx^2} – dydx\frac{dy}{dx} – 2y = cos x + 3sin x

Putting D = ddx\frac{d}{dx}

⇒ (D2 - D - 2)y = cos x + 3sin x

∴ Particular integral, PI = 1D2−D−2cos⁡x+3sin⁡x\frac{1}{D^2-D-2}\cos x + 3\sin x

= 1D2−D−2cos⁡x+31D2−D−2sin⁡x\frac{1}{D^2-D-2}\cos x + 3\frac{1}{D^2-D-2}\sin x

= 1(−1)−D−2cos⁡x+31(−1)−D−2sin⁡x\frac{1}{(-1)-D-2}\cos x + 3\frac{1}{(-1)-D-2}\sin x

= 1−3−Dcos⁡x+31−3−Dsin⁡x\frac{1}{-3-D}\cos x + 3\frac{1}{-3-D}\sin x

= (−3+D)(−3−D)(−3+D)cos⁡x+3(−3+D)(−3−D)(−3+D)sin⁡x\frac{(-3+D)}{(-3-D)(-3+D)}\cos x + 3\frac{(-3+D)}{(-3-D)(-3+D)}\sin x

= (−3+D)9−D2cos⁡x+3(−3+D)9−D2sin⁡x\frac{(-3+D)}{9-D^2}\cos x + 3\frac{(-3+D)}{9-D^2}\sin x

= (−3+D)9−D2cos⁡x+3(−3+D)9−D2sin⁡x\frac{(-3+D)}{9-D^2}\cos x + 3\frac{(-3+D)}{9-D^2}\sin x

= (−3+D)9−(−1)cos⁡x+3(−3+D)9−(−1)sin⁡x\frac{(-3+D)}{9-(-1)}\cos x + 3\frac{(-3+D)}{9-(-1)}\sin x

= (−3+D)10cos⁡x+3(−3+D)10sin⁡x\frac{(-3+D)}{10}\cos x + 3\frac{(-3+D)}{10}\sin x

= 110((−3+D)cos⁡x+3(−3+D)sin⁡x)\frac{1}{10}((-3+D)\cos x + 3(-3+D)\sin x)

= 110(−3cos⁡x−sin⁡x−9sin⁡x+3cos⁡x)\frac{1}{10}(-3\cos x-\sin x -9\sin x+3\cos x)

= 110(−sin⁡x−9sin⁡x)\frac{1}{10}(-\sin x -9\sin x)

= −10sin⁡x10\frac{-10\sin x}{10}

= –sin x

∴ A particular solution of the given differential equation is –sin x.

The correct answer is Option 2.

Question 232

ex(c1 cos3x\sqrt{3x} + c2 sin 3x\sqrt{3x}) + c3e2x is the general solution of

  1. AOption (a)

    d3ydx3\frac{d^3y}{dx^3} – 4 d2ydx2\frac{d^2y}{dx^2} + 2 dydx\frac{dy}{dx} + 4y = 0

  2. BOption (b)

    d2xdy2\frac{d^2x}{dy^2} + 8y = 0

  3. COption (c)

    d3xdy3\frac{d^3x}{dy^3} + x dxdy\frac{dx}{dy}– 3x = 0

  4. DOption (d)

    d2xdy2\frac{d^2x}{dy^2} + 4y = 0

View answer & explanation

Correct answer

Answer: (a)A.

d3ydx3\frac{d^3y}{dx^3} – 4 d2ydx2\frac{d^2y}{dx^2} + 2 dydx\frac{dy}{dx} + 4y = 0

Why this answer is correct

Formula Used:

Cosine in terms of the exponential function:

cos⁡(x)=eix+e−ix2\cos(x) = \frac{e^{ix} + e^{-ix}}{2}

Sine in terms of the exponential function:

sin⁡(x)=eix−e−ix2\sin(x) = \frac{e^{ix} - e^{-ix}}{2}

Explanation:

Given the general solution: y(x)=ex(c1cos⁡(3x)+c2sin⁡(3x))+c3e2x y(x) = e^x(c_1 \cos(\sqrt{3x}) + c_2 \sin(\sqrt{3x})) + c_3 e^{2x}

We know,

cos⁡(3x)=ei3x+e−i3x2 \cos(\sqrt{3x}) = \frac{e^{i\sqrt{3x}} + e^{-i\sqrt{3x}}}{2}

sin⁡(3x)=ei3x−e−i3x2i \sin(\sqrt{3x}) = \frac{e^{i\sqrt{3x}} - e^{-i\sqrt{3x}}}{2i}

⇒ y(x)=ex(c1(ei3x+e−i3x)2+c2(ei3x−e−i3x)2i)+c3e2x y(x) = e^x \left( \frac{c_1(e^{i\sqrt{3x}} + e^{-i\sqrt{3x}})}{2} + \frac{c_2(e^{i\sqrt{3x}} - e^{-i\sqrt{3x}})}{2i} \right) + c_3 e^{2x}

 y(x)=12e(1+i)x(c1+c2i)+12e(1−i)x(c1−c2i)+c3e2xy(x) = \frac{1}{2} e^{(1+i)x} (c_1 + c_2 i) + \frac{1}{2} e^{(1-i)x} (c_1 - c_2 i) + c_3 e^{2x}

y(x)=12ex[(c1+c2i)eix3+(c1−c2i)e−ix3]+c3e2xy(x) = \frac{1}{2} e^x \left[ (c_1 + c_2 i)e^{ix\sqrt{3}} + (c_1 - c_2 i)e^{-ix\sqrt{3}} \right] + c_3 e^{2x}

y(x)=12ex[(c1+c2i)ei3x+(c1−c2i)e−i3x]+c3e2xy(x) = \frac{1}{2} e^x \left[ (c_1 + c_2 i)e^{i\sqrt{3}x} + (c_1 - c_2 i)e^{-i\sqrt{3}x} \right] + c_3 e^{2x}

This suggests that the characteristic equation will involve complex roots r=1±i3 r = 1 \pm i\sqrt{3}, and 2 .

Thus, the characteristic equation is:

a(r−(1+i3))(r−(1−i3))(r−2)=0a(r - (1 + i\sqrt{3}))(r - (1 - i\sqrt{3}))(r - 2) = 0

⇒ a(r−1−i3)(r−1+i3)(r−2)=0 a(r - 1 - i\sqrt{3})(r - 1 + i\sqrt{3})(r - 2) = 0

⇒ a((r−1)2−(i3)2)(r−2)=0 a((r - 1)^2 - (i\sqrt{3})^2)(r - 2) = 0

⇒ a((r−1)2+3)(r−2)=0 a((r - 1)^2 +3)(r - 2) = 0

⇒ a(r2−2r+1+3)(r−2)=0 a(r^2 - 2r + 1 + 3)(r - 2) = 0

⇒ a(r2−2r+4)(r−2)=0 a(r^2 - 2r + 4)(r - 2) = 0

⇒ a(r3−2r2+4r−2r2+4r−8)=0 a(r^3 - 2r^2 +4r - 2r^2 + 4r - 8) = 0

⇒ a(r3−4r2+8r−8)=0 a(r^3 - 4r^2 + 8r -8) = 0    

Thus, the corresponding differential equation is  d3ydx3\frac{d^3y}{dx^3} – 4 d2ydx2\frac{d^2y}{dx^2} + 8 dydx\frac{dy}{dx} - 8y = 0.

None of the options are correct.

Question 233

If f is continuous mapping of compact metric space X into a metric space Y then: 

  1. AOption (a)

    f has a jump at x = θ

  2. BOption (b)

    f is not continuous

  3. COption (c)

    f is a step function

  4. DOption (d)

    f is uniformly continuous

View answer & explanation

Correct answer

Answer: (d)D.

f is uniformly continuous

Why this answer is correct

Concept Used:

Heine-Cantor Theorem states that a continuous function on a compact metric space is uniformly continuous.

Explanation:

As there is no information about f, thus we can't conclude that f has a jump at x = θ.

Thus, option 1 is not correct.

As f is already given continuous, the statement "f is not continuous" can't be true.

Thus, option 2 is not correct.

Without further information about the function f or the spaces X and Y, we cannot conclude that f  is a step function.

Thus, option 3 is not correct.

From the Heine-Cantor Theorem, we know that continuity on a compact domain implies uniform continuity.

Thus, option 4 is correct.

Question 234

The number of onto homomorphism from Z8 to Z4 is-

  1. AOption (a)

    2

  2. BOption (b)

    3

  3. COption (c)

    4

  4. DOption (d)

    1

View answer & explanation

Correct answer

Answer: (a)A.

2

Why this answer is correct

Concept:

Let G and H be two finite cyclic groups such that f: G → H is homomorphic. then number of onto homomorphic from G to H si |ker(f)|

Explanation:

Since Z8 is generated by 1, any homomorphism f: Z8 → Z4 is completely determined by the image of 1 i.e., f(1).

The condition of surjectivity implies that the image of f must generate Z4.

Now, Z4 is cyclic and has elements of order 4, 2, and 1.

So, to generate Z4, f(1) must have an order that divides 4.

Let's examine the possibilities:

If f(1) = 0, then f is the trivial homomorphism where every element of Z8​ maps to 0 in Z4

If f(1) = 1, then f is not onto because the image will only contain the elements 0 and 1.

If f(1) = 2, then f is onto because the image will contain the elements 0, 1, and 2, and 2 generates Z4.

If f(1) = 3, then f is onto because the image will contain the elements 0, 1, 2, and 3, and 3 generates Z4​.

So, there are two possible choices for f(1) that result in onto homomorphisms.

Hence the number of onto homomorphism from Z8 to Z4 is 2

Option (1) is true

Question 235

The general solution of the differential equation d2ydx2\frac{d^2y}{dx^2} + 2dydx\frac{dy}{dx} + y = ex cos x is-

  1. AOption (a)

    (c1 + c2 x + sin x)e–x

  2. BOption (b)

    (c1 + c2 x)e–x +ex4\frac{e^x}{4}(4 sin x + 3 cos x)

  3. COption (c)

    (c1 + c2 x)e–x +ex25\frac{e^x}{25}(4 sin x + 3 cos x)

  4. DOption (d)

    (c1 + c2 x)e–x +ex9\frac{e^x}{9}(4 sin x – 3 cos x)

View answer & explanation

Correct answer

Answer: (c)C.

(c1 + c2 x)e–x +ex25\frac{e^x}{25}(4 sin x + 3 cos x)

Why this answer is correct

Concept:

If any(n)+an−1y(n−1)+...+a1y′+a0y=b0a_ny^{(n)} + a_{n-1}y^{(n-1)} +...+ a_1y'+a_0y=b_0 represents a differential equation, then the auxiliary equation is given by

anmn+an−1mn−1+...+a1m+a0=0a_nm^n + a_{n-1}m^{n-1} +...+ a_1m+a_0=0.

The solution of Auxiliary equation represents the Complimentary Function(CF) 

The general solution of the differential equation, y(x) = CF + PI, where PI is the particular integral.

Some important fomulas to calculate particular integral(PI): 1f(D2)cos⁡ax=1f(−a2)cos⁡ax\frac{1}{f(D^2)}\cos ax = \frac{1}{f(-a^2)}\cos ax

1f(D2)sin⁡ax=1f(−a2)sin⁡ax\frac{1}{f(D^2)}\sin ax = \frac{1}{f(-a^2)}\sin ax

1f(D)eaxV(x)=eax1f(D+a)V(x)\frac{1}{f(D)}e^{ax}V(x) = e^{ax}\frac{1}{f(D+a)}V(x)

Calculation:

Given differential equation,  d2ydx2\frac{d^2y}{dx^2} + 2dydx\frac{dy}{dx} + y = ex cos x

Auxillary equation, m2 + 2m + 1 = 0

⇒ (m + 1)2 = 0

⇒ m = - 1, -1(Repeated roots)

∴ Complimentary Function, CF = (c1 + c2 x)e-x  

Now, the particular integral is given by, PI = 1D2+2D+1\frac{1}{D^2+2D+1}ex cos x

= 1(D+1)2\frac{1}{(D+1)^2}ex cos x

= ex 1(D+1+1)2\frac{1}{(D+1+1)^2}cos x = ex 1(D+2)2\frac{1}{(D+2)^2}cos x

= ex 1D2+4D+4\frac{1}{D^2+4D+4}cos x = ex 1(−1)+4D+4\frac{1}{(-1)+4D+4}cos x [Replacing D2 by - a2 = - (1)2 = - 1]

= ex 14D+3\frac{1}{4D+3}cos x = ex 4D−3(4D+3)(4D−3)\frac{4D-3}{(4D+3)(4D-3)}cos x

= ex 4D−316D2−9\frac{4D-3}{16D^2-9}cos x = ex 4D−3−16−9\frac{4D-3}{-16-9}cos x [Replacing D2 by - a2 = - (1)2 = - 1]

= ex 3−4D25\frac{3-4D}{25}cos x = ex25\frac{e^x}{25}[3 - 4D]cos x 

= ex25\frac{e^x}{25}[3cos x - 4(-sin x)] = ex25\frac{e^x}{25}(4 sin x + 3cos x) 

∴ Required solution, y(x) = CF + PI

= (c1 + c2 x)e–x  + ex25\frac{e^x}{25}(4 sin x + 3 cos x)

∴ The general solution of the given differential equation is (c1 + c2 x)e–x  + ex25\frac{e^x}{25}(4 sin x + 3 cos x).

The correct answer is Option 3.

Question 236

If 9P5 + 5. 9P4 = 10 Pr, then r is-

  1. AOption (a)

    5

  2. BOption (b)

    8

  3. COption (c)

    10

  4. DOption (d)

    9

View answer & explanation

Correct answer

Answer: (a)A.

5

Why this answer is correct

Concept:

PERMUTATION:

Number of Permutations of ‘n’ things taken ‘r’ at a time:

P(n,;r)=nPr=n!(n−r)!{\rm{P}}\left( {{\rm{n}},{\rm{;r}}} \right) ={}^nP_r =\frac{{{\rm{n}}!}}{{\left( {{\rm{n}} - {\rm{r}}} \right)!}}

Calculation:

\({}{}^9{P_5} + 5.{}{}^9{P_4} = {}_{}^{10}{P_r}\)

9!4!+5.9!5!=10×9!(10−r)!\frac{{9!}}{{4!}} + 5.\frac{{9!}}{{5!}} = \frac{{10 × 9!}}{{\left( {10 - r} \right)!}}

9![14!+55!]=10×9!(10−r)!9![\frac{{1}}{{4!}} + \frac{{5}}{{5!}}] = \frac{{10 × 9!}}{{\left( {10 - r} \right)!}}

[14!+55×4!]=10(10−r)![\frac{{1}}{{4!}} + \frac{{5}}{{5×4!}}] = \frac{{10 }}{{\left( {10 - r} \right)!}}

⇒24!=10(10−r)! \Rightarrow \frac{2}{{4!}} = \frac{{10}}{{\left( {10 - r} \right)!}}

⇒14!=5(10−r)! \Rightarrow \frac{1}{{4!}} = \frac{{5}}{{\left( {10 - r} \right)!}}

(10 – r)! = 5 × 4!

(10 – r)! = 5!

r = 5

Question 237

If y = x is a solution of the differential equation x2 y" + xy' – y = 0, then the second linearly independent solution of the above equation is -

  1. AOption (a)

    x2

  2. BOption (b)

    x–2

  3. COption (c)

    x3

  4. DOption (d)

    1x\frac{1}{x}

View answer & explanation

Correct answer

Answer: (d)D.

1x\frac{1}{x}

Why this answer is correct

Concept:

If any(n)+an−1y(n−1)+...+a1y′+a0y=b0a_ny^{(n)} + a_{n-1}y^{(n-1)} +...+ a_1y'+a_0y=b_0 represents a differential equation, then the auxiliary equation is given by

anmn+an−1mn−1+...+a1m+a0=0a_nm^n + a_{n-1}m^{n-1} +...+ a_1m+a_0=0.

The solution of Auxiliary equation represents the Complimentary Function 

  • If m1,2 = α ± iβ are the roots of auxillary equation, then CF = y(x) = C1eαxcos(βx) + C2eαxsin(βx)
  • If y1 and y2 are two independent solutions, then y = C1y1 + C2y2 is also a solution.

Calculation:

Given, x2 y" + xy' – y = 0

Let et = x ⇒ t = ln x

⇒ dydx=dydt.dtdx=dydt.1x\frac{dy}{dx} = \frac{dy}{dt}.\frac{dt}{dx} = \frac{dy}{dt}.\frac{1}{x}

⇒ xdydx=dydt=Dyx\frac{dy}{dx} = \frac{dy}{dt} = Dy

and, x2d2ydx2=D(D−1)y{x^2}\frac{d^2y}{dx^2} = D(D - 1)y

∴ The differential equation becomes : D(D - 1)y + Dy - y = 0

⇒ D2y - Dy + Dy - y = 0

⇒ (D2 -1)y = 0

∴ Auxiliary equation, (m2 - 1) = 0

⇒ m = ± 1

∴ y(t) = C1et + C2e-t 

⇒ y(x) = C1x + C2x-1  

⇒ y(x) = C1x + C21x\frac{1}{x}

∴ The second linearly independent solution of the differential equation is 1x\frac{1}{x}.

The correct answer is Option 4.

Question 238

If x = A cos (mt - α), then the differential equation satisfying the relation is -

  1. AOption (a)

    d2xdt2\frac{d^2x}{dt^2} = –α2x

  2. BOption (b)

    d2xdt2\frac{d^2x}{dt^2} = –m2x

  3. COption (c)

    d2xdt2\frac{d^2x}{dt^2} = m2x

  4. DOption (d)

    dxdt\frac{dx}{dt} = 1 – x2

View answer & explanation

Correct answer

Answer: (b)B.

d2xdt2\frac{d^2x}{dt^2} = –m2x

Why this answer is correct

Concept:

ddx(sin⁡x)=cos⁡x ddx(cos⁡x)=−sin⁡x\rm \frac{d}{dx} (\sin x)= \cos x\ \rm \frac{d}{dx} (\cos x) = -\sin x

Calculation:

Given:

x = A cos (mt - α)              .... (1)

Now differentiating both sides, we get

⇒dxdt⇒ \rm \frac{dx}{dt} = - Am sin(mt - α)

Again differentiating both sides, we get

⇒d2xdt2\Rightarrow \frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{t}}^2}}}  = -  Am2 cos(mt - α)

From equation (1)

⇒d2xdt2\Rightarrow \frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{t}}^2}}}  = - m2 × A cos (mt - α)

∴ d2xdt2 \frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{t}}^2}}}  = - m2 x

Question 239

The linearly independent solutions of the differential equation 4 d2ydx2\frac{d^2y}{dx^2} + 4 dydx\frac{dy}{dx} + 5y = 0 is-

  1. AOption (a)

    e–x/2cos x and e–x/2 sin x

  2. BOption (b)

    ex/2 cos x and ex/2sin x

  3. COption (c)

    e–x/2 cos x and ex/2 sin x

  4. DOption (d)

    ex/2cos x and e–x/2sin x

View answer & explanation

Correct answer

Answer: (a)A.

e–x/2cos x and e–x/2 sin x

Why this answer is correct

Concept:

If any(n)+an−1y(n−1)+...+a1y′+a0y=b0a_ny^{(n)} + a_{n-1}y^{(n-1)} +...+ a_1y'+a_0y=b_0 represents a differential equation, then the auxiliary equation is given by

anmn+an−1mn−1+...+a1m+a0=0a_nm^n + a_{n-1}m^{n-1} +...+ a_1m+a_0=0.

The solution of Auxiliary equation represents the Complimentary Function 

  • If m1,2 = α ± iβ are the roots of auxillary equation, then CF = y(x) = C1eαxcos(βx) + C2eαxsin(βx)
  • If y1 and y2 are two independent solutions, then y = C1y1 + C2y2 is also a solution.

Calculation:

Given, 4 d2ydx2\frac{d^2y}{dx^2} + 4 dydx\frac{dy}{dx} + 5y = 0

∴ Auxillary equation, 4m2 + 4m + 5 = 0

⇒ m = −4±42−4×4×52×4{-4 ± \sqrt{4^2-4\times4\times5} \over 2\times4}  

= −4±16−808{-4 ± \sqrt{16-80} \over 8}

= −4±8i8{-4 ±8i\over 8}

= −12±i\frac{-1}{2}± i

∴ y(x) = C1e-x/2cos x + C2e-x/2sin x

∴ The linearly independent solutions of the given differential equation are e–x/2cos x and e–x/2 sin x.

The correct answer is Option 1.

Question 240

If G is a group such that a2 = e for all a ∈ G, then G is:

  1. AOption (a)

    Ring

  2. BOption (b)

    Abelian Group

  3. COption (c)

    Field

  4. DOption (d)

    Non - abelian group

View answer & explanation

Correct answer

Answer: (b)B.

Abelian Group

Why this answer is correct

Concept:

A group G is said to be abelian group if ab = ba for each a, b ∈ G.

Explanation:

Let a, b ∈ G then 

a2 = e, b2 = e...(i)

Now, since G is a group so by closer property,

ab ∈ G

So, (ab)2 = e...(ii)

From (i) and (ii)

(ab)2 = a2 

⇒ (ab)(ab) = a.a

⇒ bab = a (by cancelation law)

⇒ bab2 = ab (multiplying both sides b)

⇒ bae = ab (using (i))

⇒ ba = ab

Hence G is abelian group.

Option (2) is true

Discipline 8 (20 questions)

Question 241

If nPr = 336 and nCr = 56, then what is the value of n and r?

  1. AOption (a)

    n = 8, r = 3

  2. BOption (b)

    n = 3, r = 2

  3. COption (c)

    n = 7, r = 8

  4. DOption (d)

    n = 14, r = 4

View answer & explanation

Correct answer

Answer: (a)A.

n = 8, r = 3

Why this answer is correct

Concept:

nCr = n!r!(n−r)!\frac{n!}{r!(n-r)!}

nPr = n!(n−r)!\frac{n!}{(n-r)!}

Solution:

Given, nPr = 336 and nCr = 56

nPr = 336

⇒ n!(n−r)!\frac{n!}{(n-r)!} = 336    ...(i)

nCr = 56

⇒ n!r!(n−r)!\frac{n!}{r!(n-r)!} = 56

⇒ 336r!\frac{336}{r!} = 56     [ using(i)]

⇒ r! = 33656\frac{336}{56}

⇒ r! = 6

⇒  r! = 3!

⇒ r = 3

Substituting r = 3 in (i), we get

⇒ n!(n−3)!\frac{n!}{(n-3)!} = 336

⇒ n(n - 1)(n - 2) = 8 × 7 × 6 

⇒  n(n - 1)(n - 2) = 8(8 - 1)(8 - 2)

⇒ n = 8

∴ The value of n is 8 and the value of r is 3.

The correct answer is option 1.

Question 242

The ring of Matrix S = (2a0 02b) :a,b∈Z{\bigl(\begin{smallmatrix} 2a&0 \ 0&2b \end{smallmatrix} \bigr)\ : a, b \in Z} contains-

  1. AOption (a)

    no divisors of zero without unity

  2. BOption (b)

    divisors of zero with unity

  3. COption (c)

    no divisors of zero with unity

  4. DOption (d)

    divisors of zero without unity

View answer & explanation

Correct answer

Answer: (d)D.

divisors of zero without unity

Why this answer is correct

Concept Used:

Divisors of Zero: An element  A  is a divisor of zero in a ring if there exists a non-zero element  B  such that  AB = 0  or  BA = 0.

Unity: A ring with unity (or identity) is a ring that contains a multiplicative identity element  1  such that  1a = a1 = a  for all elements a in the ring.

Explanation:

Now, consider the following elements in S:

A=(20 00)A = \begin{pmatrix} 2 & 0 \ 0 & 0 \end{pmatrix}  and B=(00 02)B = \begin{pmatrix} 0 & 0 \ 0 & 2 \end{pmatrix}

We have  AB=(00 00)AB = \begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix} and BA=(00 00)BA = \begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix}, both of which are the zero matrix.

Hence, both A and B are divisors of zero.

However, there is no matrix I in S such that  IA = AI = A  for all  A  in  S.

Therefore, the ring S does not contain unity.

Thus, The ring of matrices  S  contains divisors of zero without unity.

Question 243

Find the unit vector which is paralleled to the addition of two vectors r1→\overrightarrow{r_1} = 3i⃗−2j⃗3\vec{i}-2\vec{j} and r2→\overrightarrow{r_2} = – 4i⃗+4j⃗4\vec{i}+4\vec{j}

  1. AOption (a)

    1/5(−i⃗−2j⃗)(-\vec{i}-2\vec{j})

  2. BOption (b)

    1/5 (i⃗−2j⃗)(\vec{i}-2\vec{j})

  3. COption (c)

    1/5(−i⃗+2j⃗)1/\sqrt5(-\vec{i}+2\vec{j})

  4. DOption (d)

    1/5(i⃗+2j⃗)1/\sqrt5(\vec{i}+2\vec{j})

View answer & explanation

Correct answer

Answer: (c)C.

1/5(−i⃗+2j⃗)1/\sqrt5(-\vec{i}+2\vec{j})

Why this answer is correct

Concept:

The unit vector in the direction of a⃗\vec{a} is given by, a^\hat{a} = a⃗∣a⃗∣\frac{\vec{a}}{|\vec{a}|} where ∣a⃗∣|\vec{a}| is the magnitude of the vector.

Calculation:

Given, r1→\overrightarrow{r_1} = 3i⃗−2j⃗3\vec{i}-2\vec{j} and r2→\overrightarrow{r_2} = – 4i⃗+4j⃗4\vec{i}+4\vec{j}

∴ R⃗\vec{R} = r1⃗\vec{r_1} + r2⃗\vec{r_2} = −i⃗+2j⃗-\vec{i}+2\vec{j}

⇒ ∣R⃗∣|\vec{R}| = (−1)2+22\sqrt{(-1)^2+2^2} = 5\sqrt{5}

⇒ Unit vector, R1^\hat{R_1} = R⃗∣R⃗∣\frac{\vec{R}}{|\vec{R}|}

= 15\frac{1}{\sqrt{5}}(−i⃗+2j⃗-\vec{i}+2\vec{j})

∴ The unit vector which is paralleled to the addition of two vectors is 1/5(−i⃗+2j⃗)1/\sqrt5(-\vec{i}+2\vec{j}).

The correct answer is Option 3.

Question 244

If 25Cr = 25C2r+1, then what is the value of rC5?

  1. AOption (a)

    65

  2. BOption (b)

    56

  3. COption (c)

    50

  4. DOption (d)

    64

View answer & explanation

Correct answer

Answer: (b)B.

56

Why this answer is correct

Concept:

If nCa = nCb, then either a = b or n = a + b

nCr = n!r!(n−r)!\frac{n!}{r!(n-r)!}

Solution:

Given,  25Cr = 25C2r+1 

∴ r = 2r + 1     ...(i)  or  25 = r + 2r + 1     ...(ii)

From (i) we get, 

⇒ r = 2r + 1

⇒ r = - 1 (Not Possible)

From (ii) we get,

⇒ 25 = r + 2r + 1

⇒ 24 = 3r

⇒ r = 8

∴ rC5 = 8C5 

= 8!5!3!\frac{8!}{5!3!}

= 8×7×63×2\frac{8\times 7\times 6}{3\times 2}

= 56

∴ The value of rC5 is 56.

The correct answer is Option 2.

Question 245

If (x→−a→)(x→+a→)(\overrightarrow{x}-\overrightarrow{a}) (\overrightarrow{x}+\overrightarrow{a}) = 8 and a→\overrightarrow{a} is a unit vector, then ∣x→∣|\overrightarrow{x}| will have value of:

  1. AOption (a)

    8

  2. BOption (b)

    3

  3. COption (c)

    4

  4. DOption (d)

    9

View answer & explanation

Correct answer

Answer: (b)B.

3

Why this answer is correct

Concept:

Unit vector:

A vector of magnitude 1 is called unit vector.

If A→\overrightarrow{A} is an unit vector, then ∣A→∣|\overrightarrow{A}| = 1

Solution:

Given,  (x→−a→)(x→+a→)(\overrightarrow{x}-\overrightarrow{a}) (\overrightarrow{x}+\overrightarrow{a}) = 8 

⇒ x⃗⋅x⃗+x⃗⋅a⃗−a⃗⋅x⃗−a⃗⋅a⃗\vec{x}\cdot \vec{x}+\vec{x}\cdot \vec{a}-\vec{a}\cdot \vec{x}-\vec{a}\cdot \vec{a} = 8

⇒ x⃗⋅x⃗+x⃗⋅a⃗−x⃗⋅a⃗−a⃗⋅a⃗\vec{x}\cdot \vec{x}+\vec{x}\cdot \vec{a}-\vec{x}\cdot \vec{a}-\vec{a}\cdot \vec{a} = 8 

⇒ x⃗⋅x⃗−a⃗⋅a⃗\vec{x}\cdot \vec{x}-\vec{a}\cdot \vec{a} = 8     [∵ x⃗⋅a⃗=a⃗⋅x⃗\vec{x}\cdot \vec{a}=\vec{a}\cdot \vec{x}]

⇒ ∣x⃗∣2−∣a⃗∣2|\vec{x}|^2-|\vec{a}|^2 = 8     [∵ x⃗⋅x⃗=∣x⃗∣2\vec{x}\cdot \vec{x}=|\vec{x}|^2 and ∣a⃗∣2|\vec{a}|^2 = 1]

⇒ ∣x⃗∣2|\vec{x}|^2 - 12 = 8  

⇒  ∣x⃗∣2|\vec{x}|^2 = 8 + 1

⇒  ∣x⃗∣2|\vec{x}|^2 = 9

⇒  ∣x⃗∣|\vec{x}| = ± 3

⇒ ∣x⃗∣|\vec{x}| = 3 [Since magnitude will always be positive]

∴ The value of ∣x⃗∣|\vec{x}| is 3.

The correct answer is option 2.

Question 246

If two position-vectors of A and B are 8i⃗+3j⃗8\vec{i} + 3\vec j and 2i⃗−5j⃗2\vec i-5\vec j, then determine the direction of AB vector-

  1. AOption (a)

    Sin–1 (–5/3)

  2. BOption (b)

    Cos–1 (–3/5)

  3. COption (c)

    Cos–1 (3/5)

  4. DOption (d)

    Sin–1 (5/3)

View answer & explanation

Correct answer

Answer: (b)B.

Cos–1 (–3/5)

Why this answer is correct

Concept:

If AB→=xi^+yj^\overrightarrow{AB}=x\hat{i}+y\hat{j} makes an angle θ with the x axis, then:

cos θ = xx2+y2\frac{x}{\sqrt{x^2+y^2}} = xr\frac{x}{r}

sin θ = yx2+y2\frac{y}{\sqrt{x^2+y^2}} = yr\frac{y}{r}

tan θ = yx\frac{y}{x}

Calculation:

Let the direction of AB→\overrightarrow{AB} be θ 

Given, position-vectors of A and B are 8i⃗+3j⃗8\vec{i} + 3\vec j and 2i⃗−5j⃗2\vec i-5\vec j

∴ AB→\overrightarrow{AB} = OB→\overrightarrow{OB} - OA→\overrightarrow{OA} = (2i⃗−5j⃗)(2\vec i-5\vec j) - (8i⃗+3j⃗)(8\vec{i} + 3\vec j) = −6i⃗−8j⃗-6\vec i-8\vec j

⇒ AB→\overrightarrow{AB} lies in the 3rd quadrant.

Now, ∣AB→∣|\overrightarrow{AB}| = r = (−6)2+(−8)2\sqrt{(-6)^2+(-8)^2} = 10 

∴ cosθ = xr\frac{x}{r} = −610\frac{-6}{10} = −35\frac{-3}{5}

⇒ θ = cos–1 (–3/5)

∴ The direction of AB vector is cos–1 (–3/5).

The correct answer is Option 2.

Question 247

If the ring R is a commutative ring with unity, then the polynomial ring R[X] is-

  1. AOption (a)

    a non commutative ring without unity

  2. BOption (b)

    a commutative ring without unity

  3. COption (c)

    a non-commutative ring with unity

  4. DOption (d)

    a commutative ring with unity

View answer & explanation

Correct answer

Answer: (d)D.

a commutative ring with unity

Why this answer is correct

Concept:

A ring with unity R is a ring that has a multiplicative identity element called unity 1.

Explanation:

Let f(x), g(x) ∈ R[X]

then f(x).g(x) = g(x).f(x)

So, R[X] is a non-commutative ring

But 1 ∈ R[X} and it will be the unity of R[X]

Hence R[X] is a commutative ring with a unity

Option (4) is true

Question 248

If no two '–' signs can be repeated, then in how many ways a line can be made from six '+' signs and four '–' signs?

  1. AOption (a)

    53

  2. BOption (b)

    50

  3. COption (c)

    15

  4. DOption (d)

    35

View answer & explanation

Correct answer

Answer: (d)D.

35

Why this answer is correct

Concept:

nPr = n!(n−r)!\frac{n!}{(n-r)!}

Solution:

Given, Total number of '+' signs = 6 and Total number of '–' signs = 4

six '+' signs can be arranged in a row in 6!6!\frac{6!}{6!}=1 way      [∵ All '+' signs are identical]

Now, we are left with seven places in which four different things can be arranged in 7P4 ways but all the four '–' signs are identical, therefore, four '–'signs can be arranged in 

= 7P44!\frac{^7P_4}{4!}

= 7!(7−4)!4!\frac{7!}{\frac{(7-4)!}{4!}}

= 7!3!4!\frac{7!}{3!4!}

= 7×6×5×4!3×2×4!\frac{7× 6× 5× 4!}{3× 2× 4!}

= 7× 5

= 35

∴ The required number of ways 35.

The correct answer is Option 4.

Question 249

In how many of the distinct permutation of letters in JUXTAPOSED do the four vowels come together?

  1. AOption (a)

    182900

  2. BOption (b)

    190026

  3. COption (c)

    120960

  4. DOption (d)

    102960

View answer & explanation

Correct answer

Answer: (c)C.

120960

Why this answer is correct

Concept:

The number of ways to arrange r distinct objects = r!

If few objects are repeated a times, few objects b,... then total number of arrangements = r!a!b!⋯\frac{r!}{a!b!\cdots}

Calculation:

Given word = JUXTAPOSED 

There are no repeated letters.

Number of letters = 10

It has 4 vowels U, A, O, E.

there are no repeated vowels.

Considering all the vowels to be a single letter, the number of letters = 6 + 1 = 7

They can be arranged in 7! = 7×6×5×4×3×2×1 = 5040 ways

Now, the vowels themselves can be arranged in 4! = 4×3×2×1 = 24 ways.

Therefore, required number of ways = 5040 × 24 = 120960 ways.

∴ The total number of distinct permutation of letters in JUXTAPOSED such that the four vowels come together is 120960.

The correct answer is Option 3.

Question 250

If two vectors pi⃗+2j⃗p\vec i + 2\vec j and  3i⃗−3j⃗3\vec i- 3\vec j are aligned, then what is the value of p?

  1. AOption (a)

    1

  2. BOption (b)

    –2

  3. COption (c)

    2

  4. DOption (d)

    –1

View answer & explanation

Correct answer

Answer: (b)B.

–2

Why this answer is correct

Concept:

Two vectors a1i^+b1j^+c1k^a_1\hat{i} + b_1\hat{j}+c_1\hat{k} and a2i^+b2j^+c2k^a_2\hat{i} + b_2\hat{j}+c_2\hat{k} are said to be aligned(or parallel) if:

a1a2=b1b2=c1c2\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}

Calculation:

Given,  pi⃗+2j⃗p\vec i + 2\vec j and  3i⃗−3j⃗3\vec i- 3\vec j are aligned

⇒ The ratio of the vector components will be equal

⇒ p3\frac{p}{3} = 2−3\frac{2}{-3}

⇒ p = –2

∴ The value of p is –2.

The correct answer is Option 2.

Question 251

Find the number of diagonals of a polygon whose number of sides is "n".

  1. AOption (a)

    (n – 1)(n – 2)/3

  2. BOption (b)

    (n – 2)(n – 5)/3

  3. COption (c)

    n(n – 1)/2

  4. DOption (d)

    n(n – 3)/2

View answer & explanation

Correct answer

Answer: (d)D.

n(n – 3)/2

Why this answer is correct

Concept:

Number of ways of choosing r objects from a collection of n objects is equal to nCr.

nCr =n!r!(n−r)!\frac{n!}{r!(n-r)!}

Solution:

A polygon of n sides has n vertices. By joining any two vertices of a polygon, we obtain either a side or a diagonal of the polygon. Number of line segments obtained by joining the vertices of an n sided polygon taken two at a time 

= Number of ways of selecting 2 out of n

= nC2 

= n(n−1)2\frac{n(n-1)}{2}

Out of these lines, n lines are the sides of the polygon.

∴ Number of diagonals of the polygon = n(n−1)2−n\frac{n(n-1)}{2}-n

= n(n−3)2\frac{n(n-3)}{2}

∴ The number of diagonals of a polygon is n(n – 3)/2.

The correct answer is option 4.

Question 252

There is a group of 11 cricketers. 4 can bat and 7 can bowl. In how many ways can a group of 5 cricketers be selected if the group has at least one batsman and bowler?

  1. AOption (a)

    669

  2. BOption (b)

    237

  3. COption (c)

    441

  4. DOption (d)

    745

View answer & explanation

Correct answer

Answer: (c)C.

441

Why this answer is correct

Concept:

Total number of ways of selecting r objects from a group of n objects is given by nCr = n!r!(n−r)!\frac{n!}{r!(n-r)!}

Calculation:

Given, there is a group of 11 cricketers. 4 can bat and 7 can bowl

We need to select 5 cricketers such that the group has at least one batsman and bowler.

⇒ Possible combinations are (1 batsman + 4 bowlers), (2 batsman + 3 bowlers), (3 batsman + 2 bowlers), (4 batsman + 1 bowler)

∴ Total number of ways

= (4C1 × 7C4) + (4C2 × 7C3) + (4C3 × 7C2) + (4C4 × 7C1)

= (4 × 35) + (6 × 35) + (4 × 21) + (1 × 7)

= 140 + 210 + 84 + 7

= 441

∴ Total Number of ways of selectring 5 cricketers is 441.

The correct answer is Option 4.

Question 253

(1, 1), (–1,–1) and (3,3)(\sqrt3, \sqrt3) are vectors of:

  1. AOption (a)

    Right angle triangle

  2. BOption (b)

    Collinear points

  3. COption (c)

    Isosceles triangle

  4. DOption (d)

    Equilateral triangle

View answer & explanation

Correct answer

Answer: (b)B.

Collinear points

Why this answer is correct

Concept:

Three points are collinear if the area formed by the three points is 0.

Calculation:

Let the given points be A(1, 1) = (x1, y1)

B(–1,–1) = (x2, y2)

C(3,3)(\sqrt3, \sqrt3) = (x3, y3)

∴ Area of ΔABC = 12\frac{1}{2}|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

= 12\frac{1}{2}|1(-1 - 3\sqrt{3}) - (3\sqrt{3} - 1) + 3\sqrt{3}(1 - (-1)|

= 12\frac{1}{2}|- 1 - 3\sqrt{3} - 3\sqrt{3} + 1 + 23\sqrt{3}|

= ​0​

⇒ The points are collinear

∴ (1, 1), (–1,–1) and (3,3)(\sqrt3, \sqrt3) are collinear points.

The correct answer is Option 2.

Question 254

If ∣ma→∣|m\overrightarrow{a}| = 2 then what is the value of m?

  1. AOption (a)

    2

  2. BOption (b)

    −2∣a→∣-\frac{2}{|\overrightarrow{a}|}

  3. COption (c)

    ± 2∣a→∣\frac{2}{|\overrightarrow{a}|}

  4. DOption (d)

    2∣a→∣\frac{2}{|\overrightarrow{a}|}

View answer & explanation

Correct answer

Answer: (c)C.

± 2∣a→∣\frac{2}{|\overrightarrow{a}|}

Why this answer is correct

Calculation:

Given, ∣ma→∣|m\overrightarrow{a}| = 2 

⇒ ∣m∣∣a→∣|m||\overrightarrow{a}| = 2 

⇒ |m| = 2∣a→∣\frac{2}{|\overrightarrow{a}|}

⇒ m = ±2∣a→∣\frac{2}{|\overrightarrow{a}|}

∴ The value of m is ±2∣a→∣\frac{2}{|\overrightarrow{a}|}

The correct answer is Option 3.

Question 255

Let R = (Z2 × Z2, +,.) forms a ring of module 2 such that (a, b) + (c, d) = (a + c, d + d) and (a, b) (c. d) = (a.c, b.d) for (a, b), (c, d) ∈ Z2 × Z2 then-

  1. AOption (a)

    R is a commutative ring with unity and it contains divisor of zero

  2. BOption (b)

    R is a non commutative ring with unity and it contains divisor of zero

  3. COption (c)

    R is a commutative ring without unity and it contains no divisor of zero

  4. DOption (d)

    R is a commutative ring with unity and it contains no divisor of zero

View answer & explanation

Correct answer

Answer: (a)A.

R is a commutative ring with unity and it contains divisor of zero

Why this answer is correct

Concept Used:

A "commutative ring with unity and it contains divisor of zero" is a mathematical structure defined as follows:

Commutative Ring: It is a set R equipped with two binary operations, usually denoted as addition + and multiplication, such that the following properties hold for all elements a, b, and c in R:

  • Addition is associative and commutative.
  • There exists an additive identity element 0 such that a + 0 = a for all a in R.
  • Every element has an additive inverse, meaning for every a in R, there exists an element -a such that a + (-a) = 0.
  • Multiplication is associative and commutative.
  • There exists a multiplicative identity element 1 such that a ⋅ 1 = a for all a in R.
  • Multiplication distributes over addition, i.e., a⋅(b+c)=a⋅b+a⋅ca \cdot (b + c) = a \cdot b + a \cdot c  and (a+b)⋅c=a⋅c+b⋅c(a + b) \cdot c = a \cdot c + b \cdot c for all a, b, and c in R.

Contains Divisor of Zero: This property implies that there exist non-zero elements a and b in R such that their product equals zero, i.e., a ⋅ b = 0.

Explanation:

To determine the properties of the given ring  R=(Z2×Z2,+,⋅)R = (\mathbb{Z}_2 \times \mathbb{Z}_2, +, ⋅) , let's analyze its structure:

Addition: The addition operation  +  is defined component-wise:  (a, b) + (c, d) = (a + c, b + d) .

Thus, this operation is commutative and associative.

Multiplication: The multiplication operation ⋅ is also defined component-wise:  (a, b) ⋅ (c, d) = (a ⋅ c, b ⋅ d) .

Thus, this operation is also commutative and associative.

Ring Properties

Distributive property holds i.e.,

(a, b) ⋅ ((c, d) + (e, f)) = (a, b) ⋅ (c + e, d + f) = (a ⋅ (c + e),

b ⋅ (d + f)) = (a ⋅ c + a ⋅ e, b ⋅ d + b ⋅ f) , and

(a, b) ⋅ (c, d) + (a, b) ⋅ (e, f) = (a ⋅ c, b ⋅ d) + (a ⋅ e, b ⋅ f) = (a ⋅ c + a ⋅ e, b ⋅ d + b ⋅ f)

The ring contains an identity element:  (1, 1)  is the identity element for multiplication.

Zero Divisors: Let's check if there exist non-zero elements  (a, b)  and  (c, d)  such that  (a, b) ⋅ (c, d) = (0, 0). If such elements exist, then the ring contains zero divisors.

Consider (1, 0) and (0, 1).

Their product is  (1, 0) ⋅ (0, 1) = (0, 0) , where  (0, 0)  is the additive identity. Thus, R contains zero divisors.

R is a commutative ring with unity and it contains divisor of zero.

Question 256

If a→\overrightarrow{a} is a non-zero vector of magnitude 'a' and λ is a non-zero scalar, then λa→\overrightarrow{a}  is unit vector if

  1. AOption (a)

    a = 1∣λ∣\frac{1}{|\lambda|}

  2. BOption (b)

    λ = 1

  3. COption (c)

    λ = –1

  4. DOption (d)

    a = |λ|

View answer & explanation

Correct answer

Answer: (a)A.

a = 1∣λ∣\frac{1}{|\lambda|}

Why this answer is correct

Concept:

Vector: A vector is a quantity that has both magnitudes, as well as direction.

Unit Vector: A vector that has a magnitude of 1 is a unit vector. It is also known as Direction Vector.

A⃗\vec{A} is a unit vector if ∣A⃗\vec{A}∣ = 1

Solution:

∣λa⃗∣|\lambda \vec{a}| is a unit vector, so its magnitude is 1.

⇒ ∣λa⃗∣|\lambda \vec{a}| = 1

⇒ ∣λ∣∣a⃗∣|\lambda|| \vec{a}| = 1

⇒ ∣λ∣a = 1

⇒ a = ​1∣λ∣\frac{1}{∣λ∣}

∴ The value of a = ​1∣λ∣\frac{1}{∣λ∣}

Question 257

Find the intervals in which the function f given by strictly increasing f(x) = 2x2 – 3x:

  1. AOption (a)

    (34\frac{3}{4}, ∞ )

  2. BOption (b)

    [34\frac{3}{4}, 1)

  3. COption (c)

    (–∞, 34\frac{3}{4}]

  4. DOption (d)

    [−34-\frac{3}{4}, ∞)

View answer & explanation

Correct answer

Answer: (a)A.

(34\frac{3}{4}, ∞ )

Why this answer is correct

Concept:  

If f′(x) >  0 at each point in an interval, then the function is said to be strictly increasing.

<br>

Calculations:

Given , f(x) = 2x2 - 3x 

Differentiating, we get

f'(x) = 4x - 3

f(x) is strictly increasing function

∴ f'(x) > 0

⇒ 4x - 3 > 0

⇒ 4x > 3 

⇒ x > 34\rm\frac{3}{4}

∴ x ∈ (34\rm\frac{3}{4}, ∞)

Important Points

Value of xIntervalsSign of f’(x) = 4x - 3Nature of function f
x < 34\frac{3}{4}(-∞, 34\frac{3}{4})(-)  < 0f is strictly decreasing
x > 34\frac{3}{4}(34\frac{3}{4}, ∞)(+)  < 0f is strictly increasing

Question 258

The function f(x) = x Sin (1/x), If x = 0 and f(0) = 1 has discontinuity at _________

  1. AOption (a)

    3

  2. BOption (b)

    2

  3. COption (c)

    1

  4. DOption (d)

    0

View answer & explanation

Correct answer

Answer: (d)D.

0

Why this answer is correct

Concept:

If a function is continuous at a point a, then

lim⁡x→af(x)=f(a)\rm \displaystyle \lim_{x\rightarrow a} f(x) = f(a)

sin(∞) = a, Where -1≤ a ≤ 1

Calculation:

Given:

f(0) = 1

f(x) = x sin (1/x)

Checking continuity at x = 0

lim⁡x→0f(x)=f(0)\rm \displaystyle \lim_{x\rightarrow 0} f(x) = f(0)

L.H.L

= lim⁡x→0f(x)=lim⁡x→0 x sin⁡(1x)=0×sin⁡(10)\rm \displaystyle \lim_{x\rightarrow 0} f(x) = \rm \displaystyle \lim_{x\rightarrow 0} \ x\ \sin \left ( \frac 1 x\right ) = 0× \sin \left ( \frac {1}{ 0} \right )

= 0 × sin(∞)

= 0 

R.H.L

= f(0) = 1

L.H.L ≠ R.H.L

Hence, function is discontinuous at x = 0.

Question 259

The points A (−2i⃗+3j⃗+5k⃗)(-2\vec i + 3\vec j + 5\vec k), B (i⃗+2j⃗+3k⃗)(\vec i+ 2\vec j + 3\vec k), C (7i⃗−k⃗)(7\vec i - \vec k) are-

  1. AOption (a)

    Formed a equilateral triangle

  2. BOption (b)

    Non-Collinear

  3. COption (c)

    Formed a Right angle triangle

  4. DOption (d)

    Collinear

View answer & explanation

Correct answer

Answer: (d)D.

Collinear

Why this answer is correct

Concept:

For any vectors x,y,z ||x - y|| ≤ ||x - z|| + ||y - z|| and equality hold iff all three are collinear.

Solution:

Given, A (−2i⃗+3j⃗+5k⃗)(-2\vec i + 3\vec j + 5\vec k), B (i⃗+2j⃗+3k⃗)(\vec i+ 2\vec j + 3\vec k) and C (7i⃗−k⃗)(7\vec i - \vec k) 

AB→\overrightarrow{AB} = (i⃗+2j⃗+3k⃗)−(−2i⃗+3j⃗+5k⃗)(\vec i+ 2\vec j + 3\vec k)-(-2\vec i + 3\vec j + 5\vec k) = 3i⃗−j⃗−2k⃗3\vec i- \vec j -2\vec k

⇒ ∣AB→∣|\overrightarrow{AB}| = (3)2+(−1)2+(−2)2\sqrt{(3)^2+(-1)^2+(-2)^2} = 14\sqrt{14}

BC→\overrightarrow{BC} = (7i⃗−k⃗)−(i⃗+2j⃗+3k⃗)(7\vec i - \vec k)-(\vec i+ 2\vec j + 3\vec k) = 6i⃗−2j⃗−4k⃗6\vec i-2 \vec j -4\vec k

⇒ ∣BC→∣|\overrightarrow{BC}| = 62+(−2)2+(−4)2\sqrt{6^2+(-2)^2+(-4)^2} = 56\sqrt{56} = 2142\sqrt{14}

CA→\overrightarrow{CA} = (−2i⃗+3j⃗+5k⃗)−(7i⃗−k⃗)(-2\vec i + 3\vec j + 5\vec k)-(7\vec i - \vec k) = −9i⃗+3j⃗+6k⃗-9\vec i+3 \vec j +6\vec k

⇒ ∣CA→∣|\overrightarrow{CA}| = (−9)2+32+62\sqrt{(-9)^2+3^2+6^2} = 126\sqrt{126} = 3143\sqrt{14}

∴ ∣CA→∣|\overrightarrow{CA}| = ∣BC→∣|\overrightarrow{BC}| + ∣AB→∣|\overrightarrow{AB}|

∴ The points A, B, C are collinear.

The correct answer is Option 4.

Question 260

The position-vectors of point A, B, C and D are 3i⃗−j⃗,2i⃗+2j⃗,−2i⃗−3j⃗,4i⃗+3j⃗3\vec i - \vec j, 2 \vec i + 2\vec j, -2\vec i - 3\vec j, 4\vec i + 3\vec j respectively such that AB || CD, then what is the ratio of their modulus?

  1. AOption (a)

    41\sqrt{41} : 17\sqrt{17}

  2. BOption (b)

    2 : 1

  3. COption (c)

    10\sqrt{10} : 72\sqrt {72}

  4. DOption (d)

    10\sqrt{10} : 6\sqrt{6}

View answer & explanation

Correct answer

Answer: (c)C.

10\sqrt{10} : 72\sqrt {72}

Why this answer is correct

Explanation:

The position vectors of points A, B, C and D are 3i⃗−j⃗,2i⃗+2j⃗,−2i⃗−3j⃗,4i⃗+3j⃗3\vec i - \vec j, 2 \vec i + 2\vec j, -2\vec i - 3\vec j, 4\vec i + 3\vec j respectively.

So, AB = 2i⃗+2j⃗−(3i⃗−j⃗)2 \vec i + 2\vec j-(3\vec i - \vec j) = −i⃗+3j⃗-\vec i +3 \vec j

and CD = (4i⃗+3j⃗)−(−2i⃗−3j⃗)(4\vec i + 3\vec j)-(-2\vec i - 3\vec j) = 6i⃗+6j⃗6\vec i + 6\vec j

modulus of AB = (−1)2+(3)2\sqrt{(-1)^2+(3)^2} = 10\sqrt{10}

modulus of CD = 62+62\sqrt{6^2+6^2} = 72\sqrt {72}

So ratio of modulus is

10\sqrt{10} : 72\sqrt {72}

Option (3) is true

Discipline 9 (20 questions)

Question 261

Let M2 (R) be the set of all 2x2 matrices, then (M2(R), +, .) is -

  1. AOption (a)

    a commutative ring with zero divisors

  2. BOption (b)

    a commutative ring without zero divisors

  3. COption (c)

    a non commutative ring with zero divisors

  4. DOption (d)

    a non commutative ring without zero divisors

View answer & explanation

Correct answer

Answer: (c)C.

a non commutative ring with zero divisors

Why this answer is correct

Concept Used:

A "commutative ring with unity and it contains divisor of zero" is a mathematical structure defined as follows:

Commutative Ring: It is a set R equipped with two binary operations, usually denoted as addition + and multiplication \cdot, such that the following properties hold for all elements a, b, and c in R:

  • Addition is associative and commutative.
  • There exists an additive identity element 0 such that a + 0 = a for all a in R.
  • Every element has an additive inverse, meaning for every a in R, there exists an element -a such that a + (-a) = 0.
  • Multiplication is associative and commutative.
  • There exists a multiplicative identity element 1 such that a ⋅ 1 = a for all a in R.
  • Multiplication distributes over addition, i.e., a⋅(b+c)=a⋅b+a⋅ca \cdot (b + c) = a \cdot b + a \cdot c  and (a+b)⋅c=a⋅c+b⋅c(a + b) \cdot c = a \cdot c + b \cdot c for all a, b, and c in R.

Contains Divisor of Zero: This property implies that there exist non-zero elements a and b in R such that their product equals zero, i.e., a ⋅ b = 0.

Explanation:

Given: M2 (R) is the set of all 2x2 matrices.

As we know, matrix addition is commutative and associative, and matrix multiplication is only associative, not commutative.

Thus,  M2 (R) can't be a commutative ring.

Now, let's consider and example,

A=(10 00) A = \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix}, and B=(00 01)B = \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix}

Here, A and B are two non-zero 2x2 matrices.

AB=(10 00)(00 01)=(00 00) AB = \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix}

and BA=(00 01)(10 00)=(00 00)BA = \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix}

 In this case, both A and B are non-zero matrices, but their product is the zero matrix, making them zero divisors.

Therefore,  M2 (R) serves as an example of a non-commutative ring with zero divisors.

Question 262

The integral ∫0∞∣sin⁡x∣xdx\int_0^\infty \frac{|\sin x|}{x}dx

  1. AOption (a)

    is not convergent

  2. BOption (b)

    is absolutely convergent

  3. COption (c)

    is convergent but not absolutely

  4. DOption (d)

    is divergent

View answer & explanation

Correct answer

Answer: (c)C.

is convergent but not absolutely

Why this answer is correct

Explanation -

To analyze the convergence of the integral

∫0∞∣sin⁡x∣xdx\int_0^\infty \frac{|\sin x|}{x} dx  

we need to consider both the integral itself and its absolute convergence. Absolute convergence in the context of an integral means that the integral of the absolute value of the function converges:

∫0∞∣∣sin⁡x∣x∣dx=∫0∞∣sin⁡x∣xdx\int_0^\infty \left|\frac{|\sin x|}{x}\right| dx = \int_0^\infty \frac{|\sin x|}{x} dx  

As we know, |sin x| is always non-negative and does not exceed 1, the behavior of ∣sin⁡x∣x\frac{|\sin x|}{x} as x approaches infinity is similar to that of 1x\frac{1}{x}, which suggests that the integral might diverge based on the comparison test. However, the presence of |sin x| complicates the direct comparison since |sin x| oscillates between 0 and 1, unlike 1xp\frac{1}{x^p} where p > 1 guarantees convergence.

For the integral to be absolutely convergent, the integral of the absolute value of the function must converge. In this case, since ∣sin⁡x∣|\sin x| does not alter the convergence properties of 1x\frac{1}{x} in a significant way for large x, and considering that ∫0∞1x\int_0^\infty \frac{1}{x} dx diverges, it might suggest a similar behavior. However, the integral of ∣sin⁡x∣x\frac{|\sin x|}{x} over intervals of [nπ,(n+1)π][n\pi, (n+1)\pi] for n = 0, 1, 2, ..., due to the oscillatory nature of ∣sin⁡x∣|\sin x|, requires a more nuanced approach.

The integral ∫0∞∣sin⁡x∣x\int_0^\infty \frac{|\sin x|}{x} dx is a classic example of a conditionally convergent integral. This means it converges, but not absolutely. The reasoning is that while the integral itself converges due to the oscillatory nature of ∣sin⁡x∣|\sin x| and the 1/x decay, the integral of its absolute value (which is the same in this case, given ∣sin⁡x∣|\sin x| is already the absolute value) does not converge in the same manner as ∫0∞1x\int_0^\infty \frac{1}{x} dx due to the p-series test where p≤1p \leq 1 results in divergence.

Thus, the correct statement is:

The integral ∫0∞∣sin⁡x∣xdx\int_0^\infty \frac{|\sin x|}{x} dx is convergent but not absolutely convergent.

Question 263

The set of all units in a ring R with unity forms ______.

  1. AOption (a)

    a ring with unity

  2. BOption (b)

    a group with respect to multiplication

  3. COption (c)

    a field

  4. DOption (d)

    an integral domain

View answer & explanation

Correct answer

Answer: (b)B.

a group with respect to multiplication

Why this answer is correct

Concept -

  1. Closure under addition and multiplication: For any a, b in the ring, both a + b and ab are also in the ring.
  2. Associativity of addition and multiplication: For any a, b, c in the ring, (a + b) + c = a + (b + c) and (ab)c = a(bc).
  3. Existence of additive identity (unity): There exists an element 0 (unity) such that for any element a in the ring, a + 0 = a = 0 + a.
  4. Existence of additive inverses: For every element a in the ring, there exists an element -a in the ring such that a + (-a) = 0 = (-a) + a.
  5. Commutativity of addition: For any a, b in the ring, a + b = b + a.
  6. Distributive property: For any a, b, c in the ring, a(b + c) = ab + ac and (a + b)c = ac + bc.
  7. Absence of zero divisors: If ab = 0 for some a, b in the domain, then either a = 0 or b = 0.
  8. Nontriviality: The domain contains at least two elements, and 0≠10 \neq 1.

Ring with Unity satisfies properties 1 to 6.

Group with Respect to Multiplication satisfies properties 1 to 4

Field: In addition to the properties of a ring with unity, a field satisfies the following:

  1. Multiplicative inverses: For every nonzero element a in the field, there exists an element a^{-1} such that aa^{-1} = 1, where 1 is the multiplicative identity.
  2. Nonzero elements form a group under multiplication: All nonzero elements in the field form a group under multiplication.

Integral Domain satisfies all the eight properties.

<br>

The set of all units in a ring  R  with unity forms a group with respect to multiplication.

Explanation -

The units of a ring  R  are the elements that have multiplicative inverses. If  R  has unity, then the set of units is denoted by  R×R^\times  and it consists of all elements  r∈Rr \in R  such that there exists some  s∈Rs \in R  satisfying  rs = sr = 1 , where  1  is the multiplicative identity of the ring  R .

  1. Closure: For any two units  a,b∈R×a, b \in R^\times , their product  ab  is also a unit since the product of units is again a unit (as the product has a multiplicative inverse).

  

  1. Associativity: Multiplication in  R  is associative, so this property carries over to R×R^\times.
  2. Identity element: The multiplicative identity of  R  is also in  R×R^\times, and it serves as the identity element for multiplication within  R×R^\times, as  1⋅r=r⋅1=r1 \cdot r = r \cdot 1 = r  for any  r∈R×r \in R^\times.
  3. Inverses: Each element in  R×R^\times has a multiplicative inverse in R×R^\times. This follows directly from the definition of  R×R^\times.

Hence, the set of all units in a ring  R  with unity forms a group with respect to multiplication.

Question 264

Suppose the matrix A = \(\bigg(\begin{smallmatrix} 40&-29& -11 \ -18& 30 &-12 \26 & 24 & -50 \end{smallmatrix} \bigg)\) has a certain eigen value λ ≠ 0. Which of the following must be eigen value of A?

  1. AOption (a)

    –20 – λ 

  2. BOption (b)

    20 - λ 

  3. COption (c)

    λ – 20

  4. DOption (d)

    λ + 20

View answer & explanation

Correct answer

Answer: (b)B.

20 - λ 

Why this answer is correct

Concept:

If λ1, λ2,⋯, λn are the eigen values of matrix A.

  • λ1 + λ2 + ⋯ + λn = trace(A)

  • λ1λ2 ⋯λn = det(A)

Solution:

Suppose A is an n×n matrix and each row is α then α must be an eigenvalue of A.

Sum of the eigenvalues=trace of A

The given matrix is A=  \(\bigg(\begin{smallmatrix} 40&-29& -11 \ -18& 30 &-12 \26 & 24 & -50 \end{smallmatrix} \bigg)\) 

Since, sum of each row is 0

⇒ 0 must be an eigenvalue.

Let λ1 be another eigenvalue of A.

Sum of eigenvalues= Trace of A 

λ + 0 + λ1 = 40 + 30 - 50

⇒ λ1 = 20 - λ 

∴ 20 - λ must be an eigen value of A.

The correct answer is option 2.

Question 265

Which one of the following ring of matrix is not a field?

  1. AOption (a)

    (ab 2ba):a,b∈Q{(\begin{smallmatrix} a&b \ 2b&a \end{smallmatrix} \bigr) : a, b \in \mathbb Q }

  2. BOption (b)

    (ab 2ba):a,b∈R{(\begin{smallmatrix} a&b \ 2b&a \end{smallmatrix} \bigr) : a, b \in \mathbb R }

  3. COption (c)

    (ab 3ba):a,b∈Q{(\begin{smallmatrix} a&b \ 3b&a \end{smallmatrix} \bigr) : a, b \in \mathbb Q }

  4. DOption (d)

    (ab −ba):a,b∈R{(\begin{smallmatrix} a&b \ -b&a \end{smallmatrix} \bigr) : a, b \in \mathbb R }

View answer & explanation

Correct answer

Answer: (b)B.

(ab 2ba):a,b∈R{(\begin{smallmatrix} a&b \ 2b&a \end{smallmatrix} \bigr) : a, b \in \mathbb R }

Why this answer is correct

Concept:

A structure (R, +, ⋅) is said to be  ring if

(i) (R, +) is abelian group.

(ii) (R, ⋅) is monoid i.e, associative and identity element exist under multiplication.

(iii) Multiplication is distributive with respect to addition.

and a ring (R, +, ⋅) is filed if

(iv) R is closer with respect to multiplication.

(v) R is commutative with respect to multiplication.

(vi) multiplicative inverse of R exists.

Explanation:

Option (2): (ab 2ba):a,b∈R{(\begin{smallmatrix} a&b \ 2b&a \end{smallmatrix} \bigr) : a, b \in \mathbb R } is ring

Let A = (ab 2ba)(\begin{smallmatrix} a&b \ 2b&a \end{smallmatrix} \bigr)

|A| = a2 - 2b2

So, |A| = 0 if

a2 - 2b2 = 0 

i.e., a = ± √2b

So, the multiplicative inverse of the matrix \(\begin{pmatrix}\sqrt2&1\2&\sqrt2\end{pmatrix}\)  does not exist as determinant of this matrix is 0.

Hence (ab 2ba):a,b∈R{(\begin{smallmatrix} a&b \ 2b&a \end{smallmatrix} \bigr) : a, b \in \mathbb R } is not a field.

Option (2) is true.

For the other options, it will form a field

Question 266

The value of lim⁡x→∞1n[(12)n+(12)(12)n−1+.....+(1n)(12)]\lim\limits_{x \to \infty} \frac{1}{n}[\left(\frac{1}{2}\right)^n + \left(\frac{1}{2}\right) \left(\frac{1}{2}\right)^{n -1} + ..... + \left(\frac{1}{n}\right) \left(\frac{1}{2}\right) ] is

  1. AOption (a)

    0

  2. BOption (b)

    does not exist

  3. COption (c)

    ∞ 

  4. DOption (d)

    1

View answer & explanation

Correct answer

Answer: (a)A.

0

Why this answer is correct

Explanation:

lim⁡x→∞1n[(12)n+(12)(12)n−1+.....+(1n)(12)]\lim\limits_{x \to \infty} \frac{1}{n}[\left(\frac{1}{2}\right)^n + \left(\frac{1}{2}\right) \left(\frac{1}{2}\right)^{n -1} + ..... + \left(\frac{1}{n}\right) \left(\frac{1}{2}\right) ]

This can be represented as:

lim⁡x→∞1n[∑r=1n1r(12)n−r+1]\lim_{x \to \infty} \frac{1}{n} [ \sum_{r=1}^{n}{\frac{1}{r}\left(\frac{1}{2}\right)^{n-r+1}}]

lim⁡n→∞1n(12)∑r=1n1r(12)n−r\lim_{n \to \infty} \frac{1}{n} \left(\frac{1}{2}\right) \sum_{r=1}^{n} \frac{1}{r}\left(\frac{1}{2}\right)^{n-r}

Taking sum,

∑r=1n1r(12)n−r=∑r=1n1r(12)n(12)−r\sum_{r=1}^{n} \frac{1}{r}\left(\frac{1}{2}\right)^{n-r} = \sum_{r=1}^{n} \frac{1}{r}\left(\frac{1}{2}\right)^{n}\left(\frac{1}{2}\right)^{-r}

=(12)n∑r=1n1r(12)−r= \left(\frac{1}{2}\right)^{n} \sum_{r=1}^{n} \frac{1}{r}\left(\frac{1}{2}\right)^{-r}

=(12)n∑r=1n1r2r= \left(\frac{1}{2}\right)^{n} \sum_{r=1}^{n} \frac{1}{r2^r}

Let Sn=∑r=1n1r2rS_n = \sum_{r=1}^{n} \frac{1}{r2^r}

As we know, 

ln⁡(1+x)=x−x22+x33−x44+…\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots

For x=−12x = -\frac{1}{2}, we have:

ln⁡(1−12)=ln⁡(12)=∑r=1∞1r2r\ln\left(1 - \frac{1}{2}\right) = \ln\left(\frac{1}{2}\right) = \sum_{r=1}^{\infty} \frac{1}{r2^r}

Therefore, our sum Sn converges toln⁡(12). \ln\left(\frac{1}{2}\right). 

lim⁡n→∞1n(12)n∑r=1n1r(12)n−r\lim_{n \to \infty} \frac{1}{n} \left(\frac{1}{2}\right)^n \sum_{r=1}^{n} \frac{1}{r}\left(\frac{1}{2}\right)^{n-r}

=lim⁡n→∞1n(12)nln⁡(1/2)= \lim_{n \to \infty} \frac{1}{n} \left(\frac{1}{2}\right)^n \ln(1/2)

=ln⁡(12)⋅lim⁡n→∞1n⋅(12)n=ln⁡(1/2)⋅0=0= \ln\left(\frac{1}{2}\right) \cdot \lim_{n \to \infty} \frac{1}{n} \cdot \left(\frac{1}{2}\right)^n = \ln (1/2) \cdot 0 = 0

Therefore, the limit of the given expression as n approaches infinity is 0.

Question 267

The sequence p+(−1)nqn{p + \frac{(-1)^nq}{n}} is-

  1. AOption (a)

    Unbounded

  2. BOption (b)

    bounded

  3. COption (c)

    divergent

  4. DOption (d)

    oscillatory

View answer & explanation

Correct answer

Answer: (b)B.

bounded

Why this answer is correct

Explanation:

p+(−1)nqn{p + \frac{(-1)^nq}{n}}

= p+(−1)nqn{p + \frac{(-1)^nq}{n}}

= {p - q, p+q2p+\frac q2, p−q3p-\frac q3...., p}

So, p+(−1)nqn{p + \frac{(-1)^nq}{n}} is a bounded with bounded below p - q and bounded above p

Option (2) is true

Question 268

Let M be non-zero n x n real matrix with n ≥  2. Which of the following statements is true?

  1. AOption (a)

    det (M) = 0 ⇒ rank (M) = 0

  2. BOption (b)

    rank (M) = n ⇒ det (M) ≠1

  3. COption (c)

    rank (M) =1⇒ det (M) ≠ 0

  4. DOption (d)

    det (M) = 1⇒ rank (M) ≠ 0

View answer & explanation

Correct answer

Answer: (d)D.

det (M) = 1⇒ rank (M) ≠ 0

Why this answer is correct

Explanation -

(1) det (M) = 0 ⇒ rank (M) = 0 

Let M=(13 00)M =\begin{pmatrix} 1 & 3\ 0 & 0 \end{pmatrix}

Clearly det (M) = 0 but rank (M) ≠ 0

Hence false option.

(2) rank (M) = n ⇒ det (M) ≠1 

Let M=(100 010 001)M =\begin{pmatrix} 1 &0 & 0\ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix}

Clearly rank (M) = n but det (M) = 1

Hence false option.

(3) rank (M) =1⇒ det (M) ≠ 0  

Let M=(13 13)M =\begin{pmatrix} 1 & 3\ 1 & 3 \end{pmatrix}

Clearly rank (M) =1 but  det (M) = 0 

Hence false option. 

(4) det (M) = 1⇒ rank (M) ≠ 0 

If the determinant of the any matrix is non zero then the rank of the matrix is equal to the order of the square matrix and non zero as well.

Hence true option.

Question 269

If limn → ∞  Unn\frac{U_n}{n} = 1 ≠ 0, then the sequence {Un}

  1. AOption (a)

    is not bounded

  2. BOption (b)

    may or may not be bounded

  3. COption (c)

    is bounded

  4. DOption (d)

    is convergent

View answer & explanation

Correct answer

Answer: (a)A.

is not bounded

Why this answer is correct

Concept Used:

Limit Comparison Test:

According to the limit comparison test, if lim⁡n→∞UnVn=L \lim_{n \to \infty} \frac{U_n}{V_n} = L, where L is a finite positive constant, then the series ∑Un\sum U_n  and ∑Vn\sum V_n either both converge or both diverge.

Explanation:

Given that  lim⁡n→∞Unn=1≠0\lim_{n \to \infty} \frac{U_n}{n} = 1 \neq 0, we can consider the sequence Vn = n.

Then, according to the limit comparison test, the series \sum U_n and \sum n either both converge or both diverge.

Since lim⁡n→∞Unn=1≠0\lim_{n \to \infty} \frac{U_n}{n} = 1 \neq 0, it implies that the terms of the sequence {Un} do not approach zero as n approaches infinity.

Therefore, the sequence {Un} does not converge to zero as n approaches infinity. Instead, it grows without bound or oscillates in a non-convergent manner.

Thus, {Un} is not Bounded.

Question 270

The series ∑n=1∞[1n+(−1)22n3/2]\sum _{n = 1}^\infty [\frac{1}{\sqrt{n}} + \frac{(-1)^{22}}{n^{3/2}}] is-

  1. AOption (a)

    convergent

  2. BOption (b)

    bounded

  3. COption (c)

    divergent

  4. DOption (d)

    oscillatory

View answer & explanation

Correct answer

Answer: (c)C.

divergent

Why this answer is correct

Concept -

p - test - The series ∑n=1∞1np\sum _{n = 1}^\infty \frac{1}{{n^p}} is divergent if p ≤ 1 and covergent if p > 1

Leibnitz Test - The series  ∑n=1∞(−1)nan\sum _{n = 1}^\infty (-1)^n a_n is convergent if 

(i) an is monotone decreasing

(ii) an is converges to 0.

Explanation -

Given that the series ∑n=1∞[1n+(−1)22n3/2]\sum _{n = 1}^\infty [\frac{1}{\sqrt{n}} + \frac{(-1)^{22}}{n^{3/2}}]

we can re-write the series as ∑n=1∞1n+∑n=1∞(−1)22n3/2\sum _{n = 1}^\infty \frac{1}{\sqrt{n}} + \sum _{n = 1}^\infty \frac{(-1)^{22}}{n^{3/2}}

So we divide the given series in two parts.

Now first series ∑n=1∞1n\sum _{n = 1}^\infty \frac{1}{\sqrt{n}}

By using p - test, the series ∑n=1∞1n\sum _{n = 1}^\infty \frac{1}{\sqrt{n}} is divergent.

Now second series ∑n=1∞(−1)22n3/2\sum _{n = 1}^\infty \frac{(-1)^{22}}{n^{3/2}}

By Leibnitz Test, the series ∑n=1∞(−1)22n3/2\sum _{n = 1}^\infty \frac{(-1)^{22}}{n^{3/2}} is convergent.

But the ∑n=1∞[1n+(−1)22n3/2]\sum _{n = 1}^\infty [\frac{1}{\sqrt{n}} + \frac{(-1)^{22}}{n^{3/2}}] is divergent because first series is divergent.

Question 271

Let ΣUn be a convergent series of positive terms and let ΣVn be a divergent series of positive terms. Then

  1. AOption (a)

    The sequence {Un} converges to 0

  2. BOption (b)

    The sequence {Vn} diverges to ∞

  3. COption (c)

    The sequence {Un} is convergent and the sequence {Vn} is divergent

  4. DOption (d)

    Both the sequences {Un} and {Vn} are convergent

View answer & explanation

Correct answer

Answer: (a)A.

The sequence {Un} converges to 0

Why this answer is correct

Explanation -

Given that -

Let ΣUn be a convergent series of positive terms and let ΣVn be a divergent series of positive terms. 

so Un is converges to 0 because ΣUn be a convergent series of positive terms.

So option(1) is true.

ΣVn be a divergent series of positive terms. 

let Vn = 1/n

Clearly, ΣVn is divergent series but Vn is convergent sequence and converges to 0.

Hence Option (2) and (3) are false.

let Vn = 3 + (-1)n

Then the ΣVn is divergent series but Vn is not convergent sequence.

Hence Option(4) is false.

Question 272

f : R – {–2} →R defined by f (x) = x+1x+2\frac{x + 1}{x + 2}, ∀ ∈ R – {–2} is an example of-

  1. AOption (a)

    Rational Function

  2. BOption (b)

    Modulus Function

  3. COption (c)

    Polynomial Function

  4. DOption (d)

    Signum Function

View answer & explanation

Correct answer

Answer: (a)A.

Rational Function

Why this answer is correct

Concept Used:

Rational Function: A rational function is a function that can be expressed as the quotient of two polynomial functions.

f(x)=P(x)Q(x)f(x) = \frac{P(x)}{Q(x)}

Where P(x) and Q(x) are polynomials, and Q(x) is not the zero polynomial.

Modulus Function: The modulus function, often denoted as

<br>

∣x∣, returns the magnitude or absolute value of a real number x.

∣x∣={x,if x≥0 −x,if x<0 |x| = \begin{cases} x, & \text{if } x \geq 0 \ -x, & \text{if } x < 0 \end{cases}

Polynomial Function: A polynomial function is a function that can be expressed as a sum of powers of a variable, multiplied by coefficients. It has the general form:

f(x)=anxn+an−1xn−1+…+a1x+a0f(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0

Signum Function: The signum function, denoted by sgn(x) or sign(x), is a mathematical function that returns the sign of a real number x. It is defined as follows:

sgn(x)={−1,if x<0 0,if x=0 1,if x>0\text{sgn}(x) = \begin{cases} -1, & \text{if } x < 0 \ 0, & \text{if } x = 0 \ 1, & \text{if } x > 0 \end{cases}

Explanation:

For given example, f : R – {–2} →R defined by f (x) = x+1x+2\frac{x + 1}{x + 2}, ∀ ∈ R – {–2}.

We can clearly see that f(x) is of the form P(x)Q(x) \frac{P(x)}{Q(x)}, Where P(x) and Q(x) are polynomials, and Q(x) is not the zero polynomial.

Thus, given function is Rational Function.

Question 273

The integral ∫0∞\int_0^{\infty}Sinx dx-

  1. AOption (a)

    Exists and is equal to

  2. BOption (b)

    Exists and is equal to –1

  3. COption (c)

    Exists and is equal to 1

  4. DOption (d)

    Does not exist

View answer & explanation

Correct answer

Answer: (d)D.

Does not exist

Why this answer is correct

Concept Used:

\lim_{x → ∞} cos x doesn't exit as cos x oscillates between 1 and -1.

and ∫0∞f(x),dx=lim⁡b→∞∫0bf(x),dx\int_0^{∞}f(x) , dx = \lim_{b → ∞} \int_0^{b} f(x) , dx

Explanation:

Using the above-mentioned property, we can write the integrated as:

∫0∞sin⁡(x),dx=lim⁡b→∞∫0bsin⁡(x),dx\int_0^{∞} \sin(x) , dx = \lim_{b → ∞} \int_0^{b} \sin(x) , dx

⇒∫0bsin⁡(x),dx=−cos⁡(x)∣0b=−cos⁡(b)+cos⁡(0)=−cos⁡(b)+1\Rightarrow \int_0^{b} \sin(x) , dx = -\cos(x) \Bigg|_0^{b} = -\cos(b) + \cos(0) = -\cos(b) + 1

⇒lim⁡b→∞(−cos⁡(b)+1)=lim⁡b→∞−cos⁡(b)+lim⁡b→∞1\Rightarrow \lim_{b → ∞} \left( -\cos(b) + 1 \right) = \lim_{b → ∞} -\cos(b) + \lim_{b → ∞} 1

The limit of cos (b) as b approaches infinity does not exist because the cosine function oscillates between -1 and 1 indefinitely.

Therefore, the integral ∫0∞sin⁡(x),dx\int_0^{∞} \sin(x) , dx diverges or doesn't exist.

Question 274

The ring (Zn, +, .) is an integral domain if and only if:

  1. AOption (a)

    n is odd

  2. BOption (b)

    n is positive integer

  3. COption (c)

    n is prime

  4. DOption (d)

    n is odd prime

View answer & explanation

Correct answer

Answer: (c)C.

n is prime

Why this answer is correct

Concept Used:

An integral domain is a commutative ring with unity (i.e., it has addition, multiplication, and identity elements), where multiplication is commutative and has no zero divisors i.e., for any two non-zero elements a and b in the ring, a ⋅ b is non-zero.

Explanation:

Given: (Zn, +, .) is Ring.

If n is prime, then Zn is a field. This is because for any non-zero element a in Zn, gcd(a, n) = 1, meaning every non-zero element has a multiplicative inverse modulo n. Since every non-zero element has a multiplicative inverse, there are no zero divisors, making Zn an integral domain.

If Zn is an integral domain.

If Zn is an integral domain, it means there are no zero divisors.

In other words, for any two non-zero elements a and b in Zn, a ⋅ b ≠ 0.

Now, if n were not a prime number, say n = ab where 1 < a,b < n, then in Zn, a and b would be non-zero elements whose product equals zero (a ⋅ b = 0), which contradicts the definition of an integral domain.

Therefore, if Zn is an integral domain, n must be a prime number.

So, the ring (Zn, +, .)  is an integral domain if and only if n is a prime number.

Question 275

lim n →∞ (21,32,43........n+1n)1/n(\frac{2}{1},\frac{3}{2},\frac{4}{3}........\frac{n + 1}{n})^{1/n}is equal to

  1. AOption (a)

    ∞ 

  2. BOption (b)

    e

  3. COption (c)

    1

  4. DOption (d)

    0

View answer & explanation

Correct answer

Answer: (c)C.

1

Why this answer is correct

Concept -

Cauchy second theorem on limit -

Let limn→∞an=llim_{n → \infty }a_n = l and an > 0

then limn→∞(a1.a2.a3....an)1/n=llim_{n → \infty } (a_1.a_2.a_3....a_n)^{1/n} = l

Explanation -

We have an=n+1na_n = \frac{n+1}{n}

and an → 1

by using Cauchy second theorem on limit -

limn→∞(a1.a2.a3....an)1/n=llim_{n → \infty } (a_1.a_2.a_3....a_n)^{1/n} = l

⇒ lim n →∞ (21,32,43........n+1n)1/n(\frac{2}{1},\frac{3}{2},\frac{4}{3}........\frac{n + 1}{n})^{1/n} is equal to 1.

Question 276

Which of the following ring is an integral domain?

  1. AOption (a)

    (Z × Z + ,)

  2. BOption (b)

    (Z, +, ., )

  3. COption (c)

    (2Z, + , .)

  4. DOption (d)

    (Z8, +, .)

View answer & explanation

Correct answer

Answer: (b)B.

(Z, +, ., )

Why this answer is correct

Concept -

An integral domain must be a commutative ring with unity (1 must belong to the ring and act as a multiplicative identity), and it must have no zero divisors (if (ab= 0), then either (a = 0) or (b = 0)).

Explanation -

(1)  (Z×Z,+,⋅)(\mathbb{Z} \times \mathbb{Z}, +, \cdot):

This ring is not an integral domain because it contains zero divisors.

For example, (1,0)⋅(0,1)=(0,0)(1, 0) \cdot (0, 1) = (0, 0), where neither ((1, 0)) nor ((0, 1)) is the zero element in (Z×Z,+,⋅)(\mathbb{Z} \times \mathbb{Z}, +, \cdot).

Hence option(1) is false.

(2) (Z,+,⋅)(\mathbb{Z}, +, \cdot):

This is the ring of integers. It is an integral domain because the integers are commutative under addition and multiplication, there is a multiplicative identity (1),

and there are no zero divisors in Z\mathbb{Z}.

Hence option(2) is true.

(3) (2Z,+,⋅)(2\mathbb{Z}, +, \cdot):

This ring consists of all even integers. It is not an integral domain primarily because it does not contain a multiplicative identity that is in the set of even integers

that behaves as "1" does in the integers, as the definition of an integral domain requires a unity element (1) that, when multiplied by any element in the domain,

results in the element itself, and "1" is not an even integer.

Thus, 2Z2\mathbb{Z} does not satisfy the condition for containing a multiplicative identity within the domain itself as required.

Hence option(3) is false.

(4) (Z8,+,⋅)(\mathbb{Z}_8, +, \cdot):

This is the ring of integers modulo 8. It is not an integral domain because it contains zero divisors.

For example, (2.4 = 0 mod 8), where neither (2) nor (4) is (0) in Z8\mathbb{Z}_8.

Hence option(4) is false.

Question 277

Let {Un} be a sequence of real numbers such that limn→∞ Un exists, then

  1. AOption (a)

    limn→∞ U2n does not exist but limn→∞ U2n+1 exists

  2. BOption (b)

    both limn→∞ U2n and limn→∞  U2n+1 exist

  3. COption (c)

    both limn→∞ U2n and limn→∞ U2n+1 do not exist

  4. DOption (d)

    limn→∞ U2n exists but limn→∞ U2n+1 does not exist

View answer & explanation

Correct answer

Answer: (b)B.

both limn→∞ U2n and limn→∞  U2n+1 exist

Why this answer is correct

Concept -

Let {Un} be a sequence of real numbers such that limn→∞ Un exists. ⇒ the limit of the subsequence of Un is also exist.

Explanation -

Given that -

Let {Un} be a sequence of real numbers such that limn→∞ Un exists.

⇒ the limit of the subsequence of Un is also exist.

⇒ both limn→∞ U2n and limn→∞  U2n+1 exist

Question 278

Which one of the following is TRUE?

  1. AOption (a)

    The sequence {sin n} has a convergent subsequence

  2. BOption (b)

    The sequence {n cos 1/n} has a convergent sequence

  3. COption (c)

    Every sequence that has a convergent subsequence is a Cauchy sequence

  4. DOption (d)

    Every sequence that has a convergent sequence is a bounded sequence

View answer & explanation

Correct answer

Answer: (a)A.

The sequence {sin n} has a convergent subsequence

Why this answer is correct

Concept -

(i) Every bounded sequence has a convergent subsequence.

Explanation -

(1) The sequence {sin n} is bounded sequence and every bounded sequence has a convergent subsequence.

Hence option(1) is true

(2) The sequence an = {n cos 1/n} is divergent as n → ∞  the sequence an → ∞ 

and it does not have any convergent subsequence.

Hence option(2) is false.

(3) Let the sequence an  = (-1)n

This sequence has a convergent subsequence a2n = 1 which is constant sequence but the sequence an  = (-1)n is not Cauchy

Hence option(3) is false.

(4) Let the sequence \(\begin{equation} x_n= \begin{cases} 1 & \text{if n is even } \ n & \text{if n is odd } \end{cases} \end{equation}\)

This sequence has a convergent subsequence but it is not the bounded sequence.

Hence option(4) is false.

Question 279

The sequence {2 + (–1)n} has-

  1. AOption (a)

    exactly one constant subsequence

  2. BOption (b)

    exactly two constant subsequence

  3. COption (c)

    no constant subsequence

  4. DOption (d)

    exactly three constant subsequence

View answer & explanation

Correct answer

Answer: (b)B.

exactly two constant subsequence

Why this answer is correct

Explanation -

We have the sequence {2 + (–1)n}  \(\begin{equation} = \begin{cases} 3 & \text{if n is even } \ 1 & \text{if n is odd } \end{cases} \end{equation}\)

Clearly, it has two constant subsequence <3> when n is even and <1> when n is odd.

Hence the sequence {2 + (–1)n} has exactly two constant subsequence.

Question 280

The ring of residue classes (mod m) is an integral domain if m is:

  1. AOption (a)

    an integer

  2. BOption (b)

    a natural number

  3. COption (c)

    prime

  4. DOption (d)

    a rational number

View answer & explanation

Correct answer

Answer: (c)C.

prime

Why this answer is correct

Concept Used:

An integral domain is a commutative ring with unity (i.e., it has addition, multiplication, and identity elements), where multiplication is commutative and has no zero divisors i.e., for any two non-zero elements a and b in the ring, a ⋅ b is non-zero.

Explanation:

To prove that the ring of residue classes modulo m is an integral domain if m is prime, we need to show two things:

  • Multiplication is commutative, and
  • Absence of zero divisors.

 

Since addition and multiplication both are commutative, thus they are also commutative for residue modulo as well. Thus, ring is commutative.

Suppose m is prime and [a] ⋅ [b] = [0] in the ring of residue classes modulo m.

This means that a ⋅ b ≡ 0 mod(m).

This can only be true if m is prime as any other number can have different factors.

Since m is prime, by the fundamental theorem of arithmetic, it only has two positive divisors: 1 and itself.

Therefore, a or b (or both) must be divisible by m in order for their product to be congruent to 0 modulo m.

Without loss of generality, assume a is divisible by m. This means that a is a multiple of m, i.e., a = mk for some integer k. Then, a ⋅ b = mk ⋅ b = m(k ⋅ b), which implies that b must also be a multiple of m, since otherwise, their product cannot be a multiple of m.

Therefore, either [a] or [b] (or both) must be equivalent to the zero residue class [0], showing the absence of zero divisors.

Since the ring of residue classes modulo m is commutative under addition and multiplication and has no zero divisors when m is prime, it is indeed an integral domain.

Teaching Methodology (20 questions)

Question 281

The hypothesis underlying team teaching is

  1. AOption (a)

    Teachers feel bore while marking alone

  2. BOption (b)

    Teachers are not competent

  3. COption (c)

    A single teacher cannot control the class

  4. DOption (d)

    The best teachers in schools are shared by more students.

View answer & explanation

Correct answer

Answer: (d)D.

The best teachers in schools are shared by more students.

Why this answer is correct

The hypothesis underlying team teaching is The best teachers in schools are shared by more students.

 Key Points

  • Team teaching is based on the idea that by combining the expertise and efforts of multiple teachers, students can benefit from a richer and more diverse learning experience.
  • This approach allows schools to leverage the skills and strengths of their teachers more effectively, ensuring that students have access to high-quality instruction across various subjects and topics.
  • Therefore, the hypothesis underlying team teaching is that the best teachers in schools can serve more students by collaborating with their colleagues in a team-teaching environment.

Question 282

Who said, "instructional objectives are best described in terms of the terminal behaviour expected from the learners"?

  1. AOption (a)

    Robert Mager

  2. BOption (b)

    Robert Miller's

  3. COption (c)

    Sumit Samele

  4. DOption (d)

    Robert Wadra

View answer & explanation

Correct answer

Answer: (a)A.

Robert Mager

Why this answer is correct

The individual who said, "Instructional objectives are best described in terms of the terminal behavior expected from the learners" is Robert Mager

Key Points

  • Robert Mager is a well-known educational psychologist and author who is widely recognized for his work on instructional design and performance-based learning.
  • He emphasized the importance of clearly defining instructional objectives in terms of observable and measurable behaviors that learners are expected to demonstrate at the end of instruction.
  • This approach, known as "Mager's approach to writing instructional objectives," focuses on specifying the desired outcomes or terminal behaviors to guide instructional planning and assessment effectively.

Therefore, the correct answer is Robert Mager.

Question 283

The term Mathematics was coined by-

  1. AOption (a)

    Pythanorean

  2. BOption (b)

    Pythabactus

  3. COption (c)

    Pythayoreans

  4. DOption (d)

    Pythamasciour

View answer & explanation

Correct answer

Answer: (c)C.

Pythayoreans

Why this answer is correct

The term "Mathematics" was coined by Pythagoreans

Key Points

  • The term "Mathematics" has its roots in ancient Greece, and it is attributed to the Pythagoreans, followers of the Greek mathematician and philosopher Pythagoras.
  • The Pythagoreans made significant contributions to mathematics, particularly in the areas of geometry, number theory, and mathematical philosophy.
  • They used the term "mathematikos" to refer to the study of numbers and mathematical concepts, which later evolved into the modern term "Mathematics.

Question 284

A learner exhibiting difficulty in sorting, recognising pattern, orienting numbers and shape, telling time and measurement may have dyscalculia with difficulty in 

  1. AOption (a)

    Visual motor coordination

  2. BOption (b)

    Visual motor skills

  3. COption (c)

    Language processing

  4. DOption (d)

    Visual memory

View answer & explanation

Correct answer

Answer: (d)D.

Visual memory

Why this answer is correct

A learner exhibiting difficulty in sorting, recognizing patterns, orienting numbers and shapes, telling time, and measurement may have dyscalculia, which is typically associated with difficulty in Visual memory

Key Points 

  • Dyscalculia is a learning disorder that specifically affects mathematical abilities.
  • Individuals with dyscalculia may struggle with various aspects of mathematical reasoning and computation, including sorting, recognizing patterns, understanding numbers and shapes, telling time, and measurement.
  • Visual memory difficulties can contribute to challenges in recalling and retaining visual information, such as number symbols, shapes, or patterns, which can further exacerbate difficulties in mathematical tasks. Therefore, visual memory deficits are often associated with dyscalculia

Question 285

Which is not true about lesson plan?

  1. AOption (a)

    It helps in orderly delivery to content

  2. BOption (b)

    It is developed by students

  3. COption (c)

    It saves from haphazard teaching

  4. DOption (d)

    It develops confidence in teacher

View answer & explanation

Correct answer

Answer: (b)B.

It is developed by students

Why this answer is correct

The statement that is not true about a lesson plan is It is developed by students

Key Points 

  • It helps in the orderly delivery of content:  A lesson plan provides a structured outline for the delivery of instructional content, helping teachers organize their teaching systematically and coherently.
  • It is developed by students:. A lesson plan is typically developed by the teacher or educator who is responsible for delivering the instruction. It outlines the objectives, content, activities, assessments, and other instructional elements for a particular lesson or teaching session.
  • It saves from haphazard teaching:  By providing a clear plan and structure for instruction, a lesson plan helps prevent haphazard teaching and ensures that instructional goals are effectively addressed.
  • It develops confidence in the teacher:  Having a well-prepared lesson plan can enhance a teacher's confidence in delivering the content, as it provides a roadmap for the instructional process and helps anticipate potential challenges or issues that may arise during teaching.

Question 286

Which of the following is /are true regarding the aims of teaching mathematics?

  1. To develop a good ability to solve problem
  2. To develop the ability to make decision
  3. To develop a good understanding of numbers and number system
  1. AOption (a)

    1, 2 and 3

  2. BOption (b)

    Only 1 and 3

  3. COption (c)

    Only 1 and 2

  4. DOption (d)

    Only 2 and 3

View answer & explanation

Correct answer

Answer: (a)A.

1, 2 and 3

Why this answer is correct

The aims of teaching mathematics encompass a wide range of objectives that are designed to develop students' mathematical understanding, skills, and attitudes.

Key Points

  • Problem-solving is a fundamental aim of teaching mathematics.
  • It involves applying mathematical concepts and techniques to solve both routine and non-routine problems, which helps students think critically and logically.
  • Decision-making skills are crucial in mathematics education.
  • Mathematical reasoning and the ability to analyze data, assess probabilities, and evaluate outcomes are key components in making informed decisions.
  • Understanding numbers and the number system is foundational in mathematics.
  • It includes comprehension of basic arithmetic, properties of numbers, and more complex number systems, which are essential for higher-level mathematics and real-life applications.

Hence, we can conclude that all 1, 2, and 3 statements are true.

Question 287

In a mathematics classroom emphasis is placed on

  1. AOption (a)

    mathematical content, process and reasoning

  2. BOption (b)

    mathematical content

  3. COption (c)

    mathematical algorithm and process

  4. DOption (d)

    problem solving strategies

View answer & explanation

Correct answer

Answer: (a)A.

mathematical content, process and reasoning

Why this answer is correct

Mathematics is a branch of science that deals with counting, calculating, and studying numbers, shapes, and structures. Effective Mathematics teaching should aim to promote students' confidence in mathematics, curiosity, freedom, and belief in doing mathematics.

  • In a mathematics class, students enjoy learning, solving problems, developing mathematical curiosity, and becoming confident in using mathematics to analyze and solve problems.
  • Math is a process It's not just finding the right answer! Making mistakes and trying to figure things out is part of doing the math: This is how kids learn problem-solving and how to really do the math. How you respond when your kid makes an error can send the message that math is a process and that success comes from effort.
  • Mathematics as a content area Examining systemic mathematics instruction. In addition to helping learners develop problem-solving skills

Important Points- 

Mathematics Classroom Emphasise on:- According to National Curriculum Framework-2005, the main goal of Mathematics education in school is the mathematization of the child's thought process.

  • In a Mathematics classroom, the emphasis is placed on mathematical content, process, and reasoning.
  • Mathematics relies on logic, reasoning, problem-solving, creativity, and a mathematical way of thinking.
  • Mathematical reasoning is the ability to justify a mathematical procedure. It also involves the ability to think logically and systematically.
  • Mathematical reasoning plays an important role, both in solving problems and in conveying ideas when learning mathematics.
  • It enables children to arrive at solutions/judgments/conclusions after manipulating the facts involved in the problems.
  • Mathematical reasoning involves two processes, namely: inductive reasoning and deductive reasoning.

Hence, from the above-mentioned points, it becomes clear that in mathematics classrooms, the emphasis is placed on mathematical content, process, and reasoning.

Question 288

Who said that "Mathematics is the language in which God has written the universe"?

  1. AOption (a)

    Marshal Stone

  2. BOption (b)

    Bertrand Russell

  3. COption (c)

    Aryabhatta

  4. DOption (d)

    Galileo Galilei

View answer & explanation

Correct answer

Answer: (d)D.

Galileo Galilei

Why this answer is correct

The quote "Mathematics is the language in which God has written the universe" is attributed to Galileo Galilei.

Key Points 

  • Galileo's assertion emphasizes that the laws of nature are inherently mathematical.
  • This idea was revolutionary at the time and laid the foundation for the scientific method, where empirical evidence and mathematical reasoning are paramount.
  • By describing mathematics as a language, Galileo highlighted its universal applicability and precision in describing natural phenomena.

Hence, we can conclude that the correct answer is Galileo Galilei.

Question 289

Cooperative learning is an alternative to

  1. AOption (a)

    Lesson plan

  2. BOption (b)

    Microteaching

  3. COption (c)

    Competitive models

  4. DOption (d)

    Teaching models

View answer & explanation

Correct answer

Answer: (c)C.

Competitive models

Why this answer is correct

Cooperative learning is an alternative to 3) Competitive models

Key Points 

  • Lesson plan: A lesson plan is a structured outline of the content, activities, and resources used in a single instructional session. Cooperative learning can be incorporated within a lesson plan as a teaching strategy to promote collaborative learning among students.
  • Microteaching: Microteaching is a teacher training technique where teachers practice teaching in a controlled setting, often with peer feedback. Cooperative learning can be utilized within microteaching sessions to engage participants in collaborative activities.
  • Competitive models: Cooperative learning contrasts with competitive models of instruction, where students compete against each other for academic success or rewards. In cooperative learning, students work together in groups to achieve common goals, fostering collaboration rather than competition.
  • Teaching models: Teaching models encompass various approaches and strategies used by educators to facilitate learning. Cooperative learning is one such teaching model that emphasizes collaborative learning activities where students work together in small groups to achieve shared learning objectives.

Question 290

Which one of the following statements is correct?

  1. AOption (a)

    A composite number cannot be odd

  2. BOption (b)

    1 is not the smallest prime number

  3. COption (c)

    There is no even prime numbers

  4. DOption (d)

    Sum of two prime numbers is always a prime number

View answer & explanation

Correct answer

Answer: (b)B.

1 is not the smallest prime number

Why this answer is correct

The correct statement is 2) 1 is not the smallest prime number

Key Points

  • A composite number cannot be odd: This statement is incorrect. Composite numbers can be odd or even. A composite number is defined as any integer greater than 1 that has more than two positive divisors. For example, 9 is a composite number and it is odd.
  • 1 is the smallest prime number: This statement is correct. By convention, the number 1 is not considered a prime number. The smallest prime number is 2, as it is divisible only by 1 and itself.
  • There are no even prime numbers: This statement is incorrect. The only even prime number is 2. All other even numbers greater than 2 are divisible by 2 and therefore not prime.
  • The sum of two prime numbers is always a prime number: This statement is incorrect. While the sum of two prime numbers can result in a prime number, it is not guaranteed. For example, 3 + 5 = 8, where 8 is not a prime number.

Question 291

According to Piaget when the child is at formal operational stage, it is appropriate to introduce

  1. AOption (a)

    Numbers

  2. BOption (b)

    Geometry

  3. COption (c)

    Ratio and proportion

  4. DOption (d)

    Data handling

View answer & explanation

Correct answer

Answer: (c)C.

Ratio and proportion

Why this answer is correct

Key Points

Question 292

Who of the following invented the letter system?

  1. AOption (a)

    Arya-Bhatta- I

  2. BOption (b)

    Kumargupta

  3. COption (c)

    Sidhanta

  4. DOption (d)

    None of the above 

View answer & explanation

Correct answer

Answer: (d)D.

None of the above 

Why this answer is correct

Key Points

 

  • Writing systems and alphabets developed over thousands of years across various civilizations.
  • Early writing systems served communication, record-keeping, and cultural preservation purposes.
  • The invention of the alphabet, attributed to the Phoenicians, marked a significant milestone in human history, enabling more efficient written communication.

Hence, the correct answer is None of the above.

Question 293

Which of the following is not related to early number concept?

  1. AOption (a)

    class inclusion

  2. BOption (b)

    measurement

  3. COption (c)

    classification

  4. DOption (d)

    conservation

View answer & explanation

Correct answer

Answer: (b)B.

measurement

Why this answer is correct

The concept that is not directly related to early number concepts is 2) Measurement.

 Key Points

  • While measurement is an important mathematical concept, particularly in early childhood education, it is not directly related to early number concepts.
  • Early number concepts typically involve understanding quantity, counting, comparing quantities, and basic operations such as addition and subtraction.
  • Measurement, on the other hand, involves attributes such as length, weight, volume, and time, and is more closely related to the concept of magnitude rather than numerical quantity.

Question 294

Which one of the following is not a problem-solving strategy in mathematics?

  1. AOption (a)

    drawing

  2. BOption (b)

    trial and error

  3. COption (c)

    Rote learning

  4. DOption (d)

    solving backwards

View answer & explanation

Correct answer

Answer: (c)C.

Rote learning

Why this answer is correct

The problem-solving strategy that is not typically considered a problem-solving strategy in mathematics is 3) Rote learning

Key Points 

  • Rote learning refers to the memorization of information through repetition without necessarily understanding the underlying concepts or reasoning.
  • While it may aid in recalling facts or procedures, it is not a problem-solving strategy in the same sense as drawing, trial, and error, or solving backward, which involve systematic approaches to analyzing and solving mathematical problems.
  • Rote learning does not necessarily involve critical thinking or problem-solving skills.

Question 295

The number of students in cooperative learning groups are

  1. AOption (a)

    Eight to ten

  2. BOption (b)

    Ten to Fifteen

  3. COption (c)

    Five to Six

  4. DOption (d)

    Three to four

View answer & explanation

Correct answer

Answer: (c)C.

Five to Six

Why this answer is correct

The number of students in cooperative learning groups can vary, but typically falls within the range of 3) Five to Six

Key Points 

  • Cooperative learning groups often consist of a small number of students, typically ranging from five to six individuals.
  • This size allows for effective collaboration, active participation, and meaningful interaction among group members.
  • While other group sizes can also be effective depending on the specific context and objectives of the cooperative learning activity, smaller groups tend to promote better engagement and communication among students.

Question 296

Which of the following subjects Mathematics is a part of?

  1. AOption (a)

    Geography

  2. BOption (b)

    Commerce

  3. COption (c)

    Economics

  4. DOption (d)

    Geography, Economics and Commerce

View answer & explanation

Correct answer

Answer: (d)D.

Geography, Economics and Commerce

Why this answer is correct

Mathematics is a part of 4) Geography, Economics, and Commerce

Key Points 

  • Mathematics is a fundamental subject that forms the basis of various disciplines, including geography, economics, and commerce.
  • It is utilized in these fields for various purposes such as data analysis, statistical interpretation, financial calculations, and spatial reasoning.

Question 297

A good mathematics teacher is one who -

  1. AOption (a)

    gives only a lot of problems to practice with

  2. BOption (b)

    encourages convergent thinking

  3. COption (c)

    follows the same method always

  4. DOption (d)

    remains helpful, insightful and explain things well

View answer & explanation

Correct answer

Answer: (d)D.

remains helpful, insightful and explain things well

Why this answer is correct

A good mathematics teacher is one who 4) remains helpful, insightful, and explains things well

Key Points 

  • While providing practice problems is an essential aspect of teaching mathematics, it's not the sole criterion for being a good math teacher.
  • Additionally, a good math teacher would encourage both convergent and divergent thinking (not only convergent, as in allowing students to explore multiple solution pathways and develop problem-solving skills.
  • Moreover, using a variety of teaching methods and approaches (not following the same method always, as in is beneficial for catering to diverse learning styles and needs.
  • Ultimately, being helpful, insightful, and explaining concepts effectively (option 4) is crucial for fostering students' understanding and enjoyment of mathematics.

Question 298

Which one of the following is not a mathematical process?

  1. AOption (a)

    Measurement

  2. BOption (b)

    Estimation

  3. COption (c)

    Visualization

  4. DOption (d)

    Memorization

View answer & explanation

Correct answer

Answer: (d)D.

Memorization

Why this answer is correct

The mathematical process that is not primarily considered a mathematical process is 4) Memorization.

Key Points 

  • While memorization can be a component of learning mathematics, it is not a fundamental mathematical process in the same sense as measurement, estimation, and visualization.
  • In mathematics, understanding concepts, applying problem-solving strategies, reasoning, and critical thinking are typically emphasized over mere memorization of facts or formulas.
  • The ability to memorize may support mathematical learning but is not the core mathematical process itself.

Question 299

Which of the following is/are not the principle of Curriculum-construction of mathematics?

  1. AOption (a)

    Should not be child-centric

  2. BOption (b)

    Principle of unit

  3. COption (c)

    Integrity of theory with practice

  4. DOption (d)

    The principle of preparation for life

View answer & explanation

Correct answer

Answer: (a)A.

Should not be child-centric

Why this answer is correct

The principle of Construction of mathematics does not include 1) Should not be child-centric

Key Points 

  • Curriculum construction in mathematics, like in any other subject, typically emphasizes principles that focus on effective teaching and learning. While it's essential to consider the developmental needs and abilities of students, being child-centric is crucial in education.

Hint

The other options listed are indeed principles of curriculum construction in mathematics:

  • Principle of unit: This refers to organizing the curriculum around coherent units or themes to facilitate understanding and integration of mathematical concepts.
  • Integrity of theory with practice: This emphasizes aligning theoretical mathematical concepts with practical applications to enhance students' understanding and relevance.
  • The principle of preparation for life: This highlights the importance of equipping students with mathematical knowledge and skills that are applicable and relevant to their future personal and professional lives.

Question 300

Educational objectives have been divided into - 

  1. AOption (a)

    Four domains

  2. BOption (b)

    Five domains

  3. COption (c)

    Three domains

  4. DOption (d)

    Two domains

View answer & explanation

Correct answer

Answer: (a)A.

Four domains

Why this answer is correct

Educational objectives have been divided into Four domains.

Key Points

  • Educational objectives, also known as learning objectives or educational outcomes, are typically categorized into four domains:
  • cognitive (thinking), affective (feeling), psychomotor (physical skills), and interpersonal/social skills.
  • These domains provide a framework for educators to articulate the specific knowledge, skills, and attitudes they aim for students to acquire through the learning process.
  • Each domain addresses different aspects of learning and development, allowing educators to design comprehensive and holistic educational experiences for learners.

300 questions in the full paper