Previous Year Paper

UPPSC LT Grade Assistant Teacher (Mathematics) Official Paper (Held On: 29 Jul, 2018) (Previous Year Paper)

150 questions · 120 minutes · with answers · free

Part 1 (30 questions)

1

Which of the following census years is known as the 'Year of Great Divide' in India? 

  1. ((a))

    1911

  2. ((b))

    1921

  3. ((c))

    1951

  4. ((d))

    1991

Show Answer
Answer: ((b))

1921

The correct answer is 1921.

Key Points

  • The year 1921 is referred to as the 'Year of Great Divide' in India due to a significant demographic shift.
  • It marked the first recorded decline in India's population growth rate, primarily due to high mortality caused by pandemics like the Spanish Flu and famines.
  • Before 1921, India's population growth was largely stagnant or fluctuating due to lack of medical advancements and high death rates.
  • Post-1921, with gradual improvements in healthcare, sanitation, and living conditions, India's population began to steadily grow.
  • The 'Great Divide' signifies the transition from an era of fluctuating population growth to one of consistent and rapid growth.

Additional Information

  • Spanish Flu Pandemic (1918-1920):
  • It was one of the deadliest pandemics in history, claiming millions of lives globally, including a significant impact on India.
  • India was one of the worst-hit countries, with an estimated death toll of over 12-18 million.
  • Pre-1921 Mortality Factors:
  • Lack of medical facilities, frequent famines, and high infant mortality rates contributed to stagnant population growth.
  • Diseases such as plague, cholera, and malaria were widespread and uncontrolled.
  • Post-1921 Demographic Trends:
  • Gradual improvements in healthcare, vaccination programs, and sanitation led to a decline in death rates.
  • India's population began to grow exponentially due to a combination of high birth rates and reduced mortality.
  • Census in India:
  • The first Census of India was conducted in 1872 under British rule, but it was incomplete.
  • Starting from 1881, censuses were conducted every 10 years systematically.
2

SRI method is related to

  1. ((a))

    wheat

  2. ((b))

    cotton

  3. ((c))

    mustard

  4. ((d))

    paddy

Show Answer
Answer: ((d))

paddy

The correct answer is paddy.

Key Points

  • The System of Rice Intensification (SRI) is a method of cultivating paddy (rice) aimed at increasing yield and resource efficiency.
  • SRI involves planting younger seedlings (8-12 days old) with wider spacing to allow better root and canopy growth.
  • This method emphasizes reduced water usage by keeping the soil moist but not continuously flooded, unlike traditional paddy cultivation.
  • SRI promotes the use of organic fertilizers and discourages excessive chemical inputs, making it eco-friendly and sustainable.
  • Farmers using SRI have reported significant increases in yield with reduced costs, especially in water-scarce regions.

Additional Information

  • Key Principles of SRI:
  • Planting single seedlings instead of clumps to reduce competition for nutrients.
  • Wide spacing of plants in a grid pattern enhances sunlight exposure and airflow.
  • Intermittent irrigation (wetting and drying) rather than continuous flooding reduces water wastage.
  • Benefits of SRI:
  • Increased yields (up to 30-50% in some cases) due to better root and plant health.
  • Reduced water usage (by 25-50%) compared to traditional methods.
  • Minimized input costs by encouraging organic inputs and reducing dependency on chemical fertilizers.
  • Environmental Impact: SRI helps in reducing greenhouse gas emissions by limiting the decomposition of organic matter in flooded conditions, which releases methane.
  • Global Adoption: SRI has been successfully adopted in many countries, including India, Indonesia, Vietnam, and Madagascar, benefiting smallholder farmers.
  • Challenges: Adoption of SRI requires skill development and labor-intensive practices, which can be barriers for large-scale farming or regions with limited labor availability.
3

Which of the following pairs is not correctly matched?

Crop         :     Insect-pest

  1. ((a))

    Groundnut : Pod borer

  2. ((b))

    Gram : Pod borer

  3. ((c))

    Paddy : Banka 

  4. ((d))

    Maize : Stem borer 

Show Answer
Answer: ((a))

Groundnut : Pod borer

The correct answer is Groundnut : Pod borer.

Key Points

  • The Pod borer (Helicoverpa armigera) is primarily associated with crops like pulses and cotton, not groundnut.
  • Groundnut crops are more commonly affected by pests like red hairy caterpillar (Amsacta albistriga) and aphids, rather than pod borers.
  • In contrast, pod borers are known pests of gram (chickpea) and other leguminous crops.
  • The other pairs listed in the options, such as gram with pod borer, paddy with banka, and maize with stem borer, are correctly matched based on pest-crop associations.
  • Thus, the pair Groundnut : Pod borer is incorrectly matched in the given options.

Additional Information

  • Pod borer (Helicoverpa armigera):
  • A major pest affecting leguminous crops like chickpea, pigeon pea, and cotton.
  • It damages the pods and seeds, leading to significant yield loss.
  • Stem borer:
  • A common pest for cereals like maize, rice, and sorghum.
  • It bores into the stem, affecting nutrient and water transport in the plant.
  • Banka (Rice gall midge):
  • A pest associated with paddy (rice), causing the formation of silver shoots or gall-like structures.
  • It stunts plant growth and reduces grain yield.
  • Groundnut pests:
  • Common pests include aphids, red hairy caterpillars, and thrips.
  • Effective pest management includes cultural practices, crop rotation, and biological control methods.
  • Integrated Pest Management (IPM):
  • An approach combining biological, cultural, mechanical, and chemical methods to manage pests sustainably.
  • IPM minimizes environmental impact and promotes long-term pest control.
4

The rotation intensity of Maize-Potato-Mung bean is

  1. ((a))

    100%

  2. ((b))

    200%

  3. ((c))

    250%

  4. ((d))

    300%

Show Answer
Answer: ((d))

300%

The correct answer is 300%.

Key Points

  • Crop rotation intensity refers to the number of times crops are planted and harvested on the same piece of land within a year.
  • This specific rotation provides an effective balance of nutrients, improves soil health, and minimizes pest and disease buildup.
  • The Maize-Potato-Mung bean rotation is commonly practiced in regions where conditions support all three crops in a single agricultural calendar year.
  • Cropping intensity is calculated as the ratio of the total cropped area to the total cultivable area, expressed as a percentage. It indicates how intensively the land is being used for crop production over a year.
  • The cropping intensity can be calculated by using the following formula:

Cropping Intensity=(Total Cropped AreaTotal Cultivable Area)×100\text{Cropping Intensity} = \left( \frac{\text{Total Cropped Area}}{\text{Total Cultivable Area}} \right) \times 100

In the case of Maize-Potato-Mung bean:

  • Maize, Potato, and Mung bean are grown sequentially within a single year on the same field.
  • This means the field is being used three times in a year.

So, for a given area of land, the total cropped area over the year would be:

Total Cropped Area=(1 crop of Maize+1 crop of Potato+1 crop of Mung bean)×Total Cultivable Area \text{Total Cropped Area} = (1 \text{ crop of Maize} + 1 \text{ crop of Potato} + 1 \text{ crop of Mung bean}) \times \text{Total Cultivable Area}

Total Cropped Area=3×Total Cultivable Area \text{Total Cropped Area} = 3 \times \text{Total Cultivable Area}

Therefore, the cropping intensity would be

Cropping Intensity=(3×Total Cultivable AreaTotal Cultivable Area)×100\text{Cropping Intensity} = \left( \frac{3 \times \text{Total Cultivable Area}}{\text{Total Cultivable Area}} \right) \times 100

Cropping Intensity = 300%

Additional Information

  • Crop Rotation:
  • It is the practice of growing different types of crops sequentially on the same field to improve soil fertility and health.
  • It helps in breaking pest and disease cycles and optimizing the use of soil nutrients.
  • Maize:
  • A cereal crop known for its high carbohydrate content and suitability for various soil types.
  • Usually sown in the summer (Kharif) season in India.
  • Potato:
  • A tuber crop rich in carbohydrates, typically grown in the winter (Rabi) season in India.
  • It requires well-drained, sandy-loam soil with moderate temperatures.
  • Mung Bean:
  • A legume crop rich in protein, commonly grown in the summer or post-monsoon season.
  • It improves soil fertility by fixing atmospheric nitrogen into the soil.
  • Benefits of Crop Rotation:
  • Enhances soil structure and organic matter content.
  • Reduces reliance on chemical fertilizers and pesticides.
  • Improves overall farm productivity and sustainability.
5

Which of the following pairs is not correctly matched?

 Crop               Variety

  1. ((a))

    Groundnut : Kaushal 

  2. ((b))

    Mustard : Vardan 

  3. ((c))

    Linseed : Chamatkar 

  4. ((d))

    Gram : Udai 

Show Answer
Answer: ((a))

Groundnut : Kaushal 

The correct answer is Groundnut : Kaushal.

Key Points

  • Kaushal is not a recognized or widely known variety of groundnut. It is incorrectly matched in the given list.
  • Recognized varieties of groundnut include TMV-2, JL-24, K-6, and TAG-24, which are commonly cultivated in India.
  • The other crop-variety pairs mentioned, such as Mustard : Vardan, Linseed : Chamatkar, and Gram : Udai, are correctly matched and are widely cultivated varieties.
  • Crop-variety matching is crucial for agricultural development as it ensures the selection of the right seeds for specific soil and climatic conditions.
  • Misidentification of crop varieties can lead to incorrect agricultural practices and lower productivity.

Additional Information

  • Groundnut:
  • Groundnut, also known as peanut, is a major oilseed crop grown in tropical and subtropical regions.
  • It is rich in protein and oil, making it an important part of the agricultural economy.
  • India is one of the largest producers of groundnut globally, with major cultivation in Gujarat, Andhra Pradesh, and Tamil Nadu.
  • Crop Varieties:
  • Crop varieties are developed through selective breeding and genetic improvement to enhance yield, disease resistance, and adaptability.
  • Popular mustard varieties include Pusa Bold, Vardan, and Rohini, which are known for their high oil content and resistance to pests.
  • Linseed varieties like Chamatkar are valued for their high oil and fiber content.
  • Importance of Correctly Matched Crop Varieties:
  • Ensures optimal yield and quality by matching crops with suitable soil types and climatic conditions.
  • Reduces risk of crop failure by using varieties resistant to local pests and diseases.
  • Promotes sustainable agricultural practices and contributes to food security.
  • Agricultural Research in India:
  • Organizations like the Indian Council of Agricultural Research (ICAR) play a key role in developing and disseminating improved crop varieties.
  • State Agricultural Universities and research institutes collaborate to ensure farmers have access to high-quality seeds.
6

Which of the following diseases cannot be cured by antibiotics? 

  1. ((a))

    Tuberculosis

  2. ((b))

    Tetanus

  3. ((c))

    Measles

  4. ((d))

    Cholera

Show Answer
Answer: ((c))

Measles

The correct answer is Measles.

Key Points

  • Measles is a viral disease caused by the Measles morbillivirus, which cannot be treated with antibiotics as they are effective only against bacterial infections.
  • Antibiotics target bacteria by disrupting their cell walls, protein synthesis, or DNA replication, but viruses like measles lack these structures, rendering antibiotics ineffective.
  • The best preventive measure against measles is vaccination, particularly the Measles-Mumps-Rubella (MMR) vaccine, which provides long-term immunity.
  • Management of measles focuses on supportive care, including hydration, antipyretics, and vitamin A supplementation to prevent complications like blindness and pneumonia.

Additional Information

  • Antibiotics vs. Viruses: Antibiotics are ineffective against viruses because viruses replicate inside host cells and lack the cellular machinery targeted by antibiotics.
  • Vaccination: Vaccines are the most effective way to prevent viral diseases. The MMR vaccine is highly effective against measles, with two doses providing around 97% immunity.
  • Complications of Measles: Severe complications include pneumonia, encephalitis, and subacute sclerosing panencephalitis (SSPE), a rare but fatal brain disorder.
  • Global Impact: Despite being preventable, measles remains a significant global health issue, causing over 140,000 deaths annually, primarily in unvaccinated children (WHO, 2022).
  • Other Diseases in the List: Tuberculosis, tetanus, and cholera are bacterial infections and can be treated with appropriate antibiotics, unlike measles.
7

Which of the following pairs is not correctly matched?

  1. ((a))

    Computer : Charles Babbage

  2. ((b))

    Radio : Karl Benz

  3. ((c))

    Barometer : E. Torricelli

  4. ((d))

    Dynamo : Michael Faraday

Show Answer
Answer: ((b))

Radio : Karl Benz

The correct answer is Radio : Karl Benz.

Key Points

  • Charles Babbage is regarded as the "Father of the Computer" for conceptualizing the first mechanical computer known as the Analytical Engine.
  • Radio was invented by Guglielmo Marconi, who successfully demonstrated wireless telegraphy and is often referred to as the "Father of Radio."
  • E. Torricelli invented the barometer in 1643, making significant contributions to the field of atmospheric pressure measurement.
  • Michael Faraday invented the dynamo, a device for converting mechanical energy into electrical energy, laying the foundation for modern electromagnetic technology.

Additional Information

  • Charles Babbage: His Analytical Engine was the precursor to modern computers, integrating concepts such as an arithmetic logic unit and control flow.
  • Guglielmo Marconi: He was awarded the Nobel Prize in Physics in 1909 for his pioneering work in wireless communication.
  • Evangelista Torricelli: An Italian physicist and mathematician, Torricelli's invention of the barometer was a breakthrough in studying atmospheric pressure.
  • Michael Faraday: Known for his discovery of electromagnetic induction, Faraday's work is fundamental to electrical engineering and physics.
  • Karl Benz: He was the inventor of the first automobile powered by an internal combustion engine, but he is not associated with the invention of the radio.
8

The communication satellites are invariably

  1. ((a))

    revolving at their own speed

  2. ((b))

    stationary

  3. ((c))

    geostationary

  4. ((d))

    changing their track and speed

Show Answer
Answer: ((c))

geostationary

The correct answer is geostationary.

Key Points

  • Communication satellites are placed in a geostationary orbit, which means they remain fixed relative to a specific location on Earth's surface.
  • A geostationary orbit is located at an altitude of approximately 35,786 kilometers (22,236 miles) above the equator.
  • These satellites revolve around the Earth at the same rotational speed as the Earth’s rotation, achieving a stationary position relative to the ground.
  • Geostationary satellites are essential for uninterrupted services such as telecommunications, television broadcasting, and weather forecasting.
  • Examples of communication satellites in geostationary orbit include the INSAT series (India) and Intelsat series (global).

Additional Information

  • Geostationary Orbit:
  • It is a type of geosynchronous orbit directly above the Earth’s equator.
  • Satellites in this orbit have an orbital period equal to the Earth’s rotational period (24 hours).
  • Advantages of Geostationary Satellites:
  • Continuous coverage of a specific area on Earth.
  • Ideal for real-time communication services, such as satellite TV and internet.
  • Other Satellite Orbits:
  • Low Earth Orbit (LEO): Used for Earth observation and remote sensing (e.g., the International Space Station).
  • Medium Earth Orbit (MEO): Primarily used for navigation systems like GPS.
  • Geostationary Satellite Launch:
  • These satellites are launched using powerful rockets to reach the required altitude.
  • Once in orbit, their position is maintained using onboard thrusters.
9

For which substance among the following, conductivity increases with temperature?

  1. ((a))

    Copper

  2. ((b))

    Germanium

  3. ((c))

    Silver

  4. ((d))

    Iron

Show Answer
Answer: ((b))

Germanium

The correct answer is Germanium.

Key Points

  • Germanium is a semiconductor material, and its electrical conductivity increases with temperature due to the generation of more charge carriers (electrons and holes).
  • In semiconductors like Germanium, the energy gap between the valence and conduction bands allows more electrons to jump to the conduction band at higher temperatures, enhancing conductivity.
  • Unlike metals, where conductivity decreases with temperature, semiconductors exhibit an inverse relationship because of their unique band structure.
  • Germanium is widely used in electronic devices such as transistors and diodes due to its temperature-dependent conductivity.
  • This property makes Germanium useful in temperature-sensitive applications, including infrared optics and thermistors.

Additional Information

  • Semiconductors:
  • These are materials with a conductivity level between that of conductors and insulators, such as Germanium and Silicon.
  • Their conductivity is highly dependent on temperature and doping (addition of impurities).
  • Temperature Effect on Metals vs Semiconductors:
  • In metals, conductivity decreases with temperature due to increased lattice vibrations causing electron scattering.
  • In semiconductors, conductivity increases with temperature as more electrons gain energy to cross the energy gap into the conduction band.
  • Energy Bands:
  • The valence band contains electrons involved in bonding, while the conduction band contains free electrons that contribute to conductivity.
  • The energy gap between these bands determines the material's electrical properties.
  • Applications of Germanium:
  • Germanium is used in electronics, infrared devices, fiber optics, and detectors due to its excellent semiconductor properties.
  • Its temperature-dependent conductivity also makes it suitable for thermistors and temperature sensors.
  • Other Semiconductors:
  • Silicon is another widely used semiconductor, but it has a higher bandgap compared to Germanium, making it more suitable for high-temperature applications.
10

The area of a regular hexagon of side 232\sqrt{3} cm is

  1. ((a))

    12312\sqrt{3} cm2

  2. ((b))

    18218\sqrt{2} cm2

  3. ((c))

    18 cm2

  4. ((d))

    18318\sqrt{3} cm2

Show Answer
Answer: ((d))

18318\sqrt{3} cm2

Given:

Side of hexagon (a) = 2√3 cm

Formula used:

Area of hexagon = (3√3/2) × a2

Calculation:

Area = (3√3/2) × (2√3)2

⇒ Area = (3√3/2) × (4 × 3)

⇒ Area = (3√3/2) × 12

⇒ Area = 18√3 cm2

∴ The correct answer is option (4).

11

If  2x+2x=3,2x+\frac{2}{x}=3, then the value of x3+1x3+2x^3+\frac{1}{x^3}+2

  1. ((a))

    38\frac{3}{8}

  2. ((b))

    198\frac{19}{8}

  3. ((c))

    218\frac{21}{8}

  4. ((d))

    78\frac{7}{8}

Show Answer
Answer: ((d))

78\frac{7}{8}

Given:

2x + (2/x) = 3

Formula Used:

To find the value of x3 + (1/x3) + 2, we use the following identity:

[x + (1/x)]3 = x3 + (1/x3) + 3[x + (1/x)]

Calculations:

2x + (2/x) = 3

x + (1/x) = 3/2

⇒ Use the identity:

[x + (1/x)]3 = x3 + (1/x3) + 3[x + (1/x)]

⇒ (3/2)3 = x3 + (1/x3) + 3 × (3/2)

⇒ 27/8 = x3 + (1/x3) + 9/2

⇒ x3 + (1/x3) = (27/8) - (9/2)

⇒ x3 + (1/x3) = -9/8

⇒ Add 2 to the result:

x3 + (1/x3) + 2 = (-9/8) + 2

⇒ x3 + (1/x3) + 2 = 7/8

∴ The correct answer is option (4).

12

If one of the roots of the quadratic equation 2x2+px+4=0 is 2, then the other root is

  1. ((a))

    -2

  2. ((b))

    -1

  3. ((c))

    +1

  4. ((d))

    +2

Show Answer
Answer: ((c))

+1

Given:

Quadratic equation: 2x2 + px + 4 = 0

One root (x₁) = 2

Formula used:

If the roots of a quadratic equation are x₁ and x₂, then:

Sum of roots = -b/a

Product of roots = c/a

Calculations:

Given that one root is x₁ = 2, let the other root be x₂.

Using the formula for product of roots:

⇒ x₁ × x₂ = c/a

⇒ 2 × x₂ = 4/2

⇒ x₂ = 2 / 2

⇒ x₂ = 1

∴ The other root is 1, and the correct answer is option (3).

13

In which State was the military exercise 'Vijay Prahar' held in May 2018?

  1. ((a))

    Maharashtra

  2. ((b))

    Gujarat

  3. ((c))

    Rajasthan

  4. ((d))

    Madhya Pradesh

Show Answer
Answer: ((c))

Rajasthan

The correct answer is Rajasthan.

Key Points

  • The military exercise 'Vijay Prahar' was conducted by the Indian Army in May 2018 in Rajasthan, specifically in the Mahajan Field Firing Range.
  • This exercise involved over 25,000 troops of the South Western Command, showcasing advanced combat strategies and readiness.
  • 'Vijay Prahar' focused on testing integrated firepower and maneuverability in a joint operational scenario involving infantry, artillery, armored units, and air support.
  • The exercise aimed to validate offensive strategies to ensure preparedness for future conflicts.
  • It also emphasized the use of modern warfare techniques and technologies such as precision targeting and network-centric operations.

Additional Information

  • South Western Command:
  • One of the seven operational commands of the Indian Army, headquartered in Jaipur, Rajasthan.
  • Responsible for securing India's western borders and conducting strategic military exercises.
  • Military Exercises:
  • India conducts various military exercises, both domestic and international, to enhance combat readiness and foster cooperation with other nations.
  • Examples include 'Operation Gagan Strike,' 'Yudh Abhyas,' and 'Indra' among others.
  • Precision Targeting:
  • A modern warfare technique that uses advanced technologies like drones, satellite imagery, and precision-guided munitions.
  • Helps in minimizing collateral damage and increasing operational efficiency.
  • Network-Centric Operations:
  • A military strategy that leverages information technology to link forces, enabling real-time data sharing and enhanced decision-making.
  • It plays a crucial role in modern warfare for achieving faster and more coordinated responses.
14

Who has won the Women Singles Title of Badminton in Commonwealth Games, 2018?

  1. ((a))

    Saina Nehwal

  2. ((b))

    P. V. Sindhu

  3. ((c))

    K. Gilmour

  4. ((d))

    Michelle Li

Show Answer
Answer: ((a))

Saina Nehwal

The correct answer is Saina Nehwal.

Key Points

  • Saina Nehwal won the Women Singles Title in Badminton at the Commonwealth Games 2018 held in Gold Coast, Australia.
  • She defeated fellow Indian shuttler P. V. Sindhu in the finals with a score of 21-18, 23-21.
  • It was a historic moment as it marked an all-Indian final in the Women’s Singles event of Badminton at the Commonwealth Games.
  • This victory added to Saina Nehwal's illustrious career, making her one of the most celebrated badminton players in India.
  • Saina had previously won the same title at the 2010 Commonwealth Games in New Delhi, making her a two-time Commonwealth Games gold medalist in this category.

Additional Information

  • Commonwealth Games 2018:
  • The 2018 Commonwealth Games were held in Gold Coast, Australia, from April 4 to April 15, 2018.
  • India secured an impressive tally of 66 medals, including 26 gold, 20 silver, and 20 bronze, finishing 3rd in the overall medal rankings.
  • Badminton contributed significantly to India’s medal count, with players like Saina Nehwal, P. V. Sindhu, and Kidambi Srikanth excelling.
  • Badminton in Commonwealth Games:
  • Badminton has been a part of the Commonwealth Games since 1966.
  • India has consistently been one of the top-performing nations in badminton at the Commonwealth Games.
  • Saina Nehwal's Career Highlights:
  • She is the first Indian shuttler to win an Olympic medal (bronze at the London 2012 Olympics).
  • She became the first Indian woman to achieve the World No. 1 ranking in badminton in 2015.
  • Saina has won numerous BWF titles, including the prestigious All England Open runner-up position in 2015.
  • Women's Singles Final (2018 CWG):
  • The final match between Saina Nehwal and P. V. Sindhu was closely contested, showcasing the high level of talent in Indian badminton.
  • Saina's experience and aggressive gameplay were key factors in her victory over Sindhu.
15

In the World Press Freedom Index, 2018, India is placed at

  1. ((a))

    135th

  2. ((b))

    136th

  3. ((c))

    138th

  4. ((d))

    137th

Show Answer
Answer: ((c))

138th

The correct answer is 138th

  • India’s rank in the 2018 World Press Freedom Index 138th.

Key Points

  • The Index is compiled and published by Reporters without borders.
  • Reporters without borders is a Paris based non-profit organisation.
  • India ranked 142nd in the 2020 World Press Freedom Index
  • Around 180 countries were ranked in the index.
  • The World Press Freedom Index 2018 was topped by Norway followed by Finland, Sweden, Netherlands and Denmark.

Additional Information

  • According to the 2020 World Press Freedom Index, Pakistan ranked 145 dropping three places, Bangladesh ranked 151, North Korea was at 180th
16

In which of the following texts, it is stated that those who could not speak Sanskrit language correctly were called 'Mlecchas'?

  1. ((a))

    Shvetashvatara Upanishad

  2. ((b))

    Gopaths Brahmana

  3. ((c))

    Brihadaranyaka Upanishad

  4. ((d))

    Shatapatha Brahmana

Show Answer
Answer: ((d))

Shatapatha Brahmana

The correct answer is Shatapatha Brahmana.

Key Points

  • The term 'Mleccha' was used in ancient Indian texts to describe people who could not speak Sanskrit properly or adhered to non-Vedic practices.
  • The Shatapatha Brahmana, a prose text associated with the Shukla Yajurveda, provides references to the term 'Mleccha' in the context of linguistic and cultural distinctions.
  • The Shatapatha Brahmana categorizes those outside the Vedic cultural framework, including those with different linguistic practices, as 'Mlecchas.'
  • This text highlights the importance of Sanskrit as the language of Vedic rituals and knowledge, often contrasting it with non-Sanskrit speakers.
  • The term 'Mleccha' was not necessarily derogatory but was used to mark linguistic and cultural differences.

Additional Information

  • Shatapatha Brahmana:
  • One of the most significant Brahmana texts associated with the Shukla Yajurveda.
  • It provides detailed explanations of Vedic rituals, ceremonies, and their symbolic meanings.
  • The text is divided into several chapters (Adhyayas) and sections (Kandas).
  • Mleccha:
  • A term used in ancient India to denote people outside the Vedic cultural sphere.
  • It was often associated with non-Sanskrit speakers or those who did not follow Vedic practices.
  • Sanskrit Language:
  • Considered the sacred language of the Vedas and Vedic rituals.
  • Proficiency in Sanskrit was seen as a marker of adherence to Vedic traditions.
  • Vedic Literature:
  • Comprises four Vedas (Rigveda, Yajurveda, Samaveda, Atharvaveda), Brahmanas, Aranyakas, and Upanishads.
  • Brahmanas like the Shatapatha Brahmana provide explanations and instructions for Vedic rituals.
  • Cultural Significance:
  • The distinction between 'Arya' (followers of Vedic culture) and 'Mleccha' reflects the cultural and linguistic diversity of ancient India.
  • These categorizations were used to understand and organize the complex social fabric of the time.
17

Match List-I with List-II and select the correct answer using the codes given below the Lists:

List-I (King)List-II (Spouse)
A. Chandragupta I1. Dutta Devi
B. Samudragupta2. Kuberanaga
C. Chandragupta II3. Kumara Devi
D. Kumaragupta I4. Ananta Devi

Codes:

  1. ((a))

    A-2, B-3, C-4, D-1

  2. ((b))

    A-3, B-2, C-4, D-1

  3. ((c))

    A-3, B-1, C-2, D-4

  4. ((d))

    A-4, B-3, C-1, D-2

Show Answer
Answer: ((c))

A-3, B-1, C-2, D-4

The correct answer is A-3, B-1, C-2, D-4.

Key Points

  • Chandragupta I was married to Kumara Devi, a princess of the Lichchhavi dynasty, which helped establish the Gupta Empire.
  • Samudragupta was married to Dutta Devi, who is mentioned in inscriptions as his queen.
  • Chandragupta II was married to Kuberanaga, a Naga princess, which strengthened alliances and expanded the empire.
  • Kumaragupta I was married to Ananta Devi, as per historical records and inscriptions from the Gupta period.

Additional Information

  • Gupta Empire: The Gupta dynasty is considered one of the golden ages of Indian history, marked by advancements in art, science, and culture.
  • Samudragupta: Known as the "Napoleon of India," Samudragupta expanded the empire significantly and was an accomplished poet and musician.
  • Chandragupta I's marriage: His marriage to Kumara Devi of the Lichchhavi dynasty played a crucial role in the Gupta Empire's rise to power.
  • Gupta inscriptions: Many details about Gupta rulers and their spouses are derived from inscriptions, such as the Allahabad Pillar Inscription.
  • Political alliances: Marriages in the Gupta era were often used to strengthen political alliances and consolidate power within and outside the empire.
18

With reference to the book Arthashastra, which of the following statements is/are correct?

  1. It is the oldest masterpiece on Indian State Policy.
  2. There is no description of Mauryan empire and administration in this book. Select the correct answer using the codes given below.

Codes:

  1. ((a))

    1 only

  2. ((b))

    2 only

  3. ((c))

    Both 1 and 2

  4. ((d))

    Neither 1 nor 2

Show Answer
Answer: ((a))

1 only

The correct answer is 1 only.

Key Points

  • The Arthashastra, authored by Kautilya (Chanakya), is considered the oldest known treatise on Indian State Policy, providing insights into governance, administration, and economics.
  • It was composed during the Mauryan period, around the 3rd century BCE, and is regarded as a masterpiece in ancient political and economic thought.
  • The book extensively describes the Mauryan empire's administrative structure, including the duties of the king, ministers, and other officials, as well as strategies for internal and external security.
  • Statement 2 is incorrect as the Arthashastra does provide a detailed description of the Mauryan empire and its administration, which aligns closely with historical accounts of the period.
  • Thus, only statement 1 is correct, making the correct answer 1 only.

Additional Information

  • Kautilya (Chanakya): A key figure in Indian history who served as an advisor and prime minister to Chandragupta Maurya. His ideas laid the foundation for the Mauryan Empire's administration.
  • Structure of Arthashastra: The book is divided into 15 sections (adhikaranas) and covers topics such as statecraft, military strategy, economic policy, and social welfare.
  • Core Concepts: The Arthashastra emphasizes concepts like dharma (moral duty), danda (punishment or enforcement), and niti (policy or strategy) as the pillars of governance.
  • Comparison with Other Texts: Unlike religious texts like the Vedas or epics like the Mahabharata, the Arthashastra is a secular and pragmatic guide to governance and administration.
  • Rediscovery: The text was lost for centuries and was rediscovered in 1905 by R. Shamasastry, who translated and published it, bringing it back to prominence in modern times.
19

Who among the following addressed Delhi as one of the greatest cities in the world?

  1. ((a))

    Ibn Batuta

  2. ((b))

    Alberuni

  3. ((c))

    Farishta

  4. ((d))

    Abul Fazl

Show Answer
Answer: ((a))

Ibn Batuta

The correct answer is Ibn Batuta.

Key Points

  • Ibn Batuta, a Moroccan traveler and scholar, visited India during the reign of Muhammad bin Tughlaq (14th century).
  • He served as the Qadi (judge) in the court of Muhammad bin Tughlaq and documented his travels in his work, "Rihla" (The Journey).
  • In his writings, Ibn Batuta described Delhi as one of the greatest and most magnificent cities of the world during that period.
  • He marveled at the city’s architectural brilliance, bustling markets, and the grandeur of the Sultan’s court.
  • Ibn Batuta's account remains a valuable source of information about the socio-economic and cultural life of medieval India.

Additional Information

  • Rihla (The Journey): This is the travelogue of Ibn Batuta, documenting his travels across the Islamic world, including North Africa, the Middle East, India, and China.
  • Delhi under Muhammad bin Tughlaq: The city was an important cultural and political hub during the Tughlaq Dynasty, known for its ambitious projects and administrative reforms.
  • Medieval Travelers: Along with Ibn Batuta, other notable travelers to India include Marco Polo, Al-Biruni, and Niccolò de' Conti, who documented various aspects of Indian life.
  • Qadi Role: A Qadi is a judge ruling in accordance with Islamic law (Sharia). Ibn Batuta’s appointment as Qadi signifies his deep understanding of Islamic jurisprudence.
  • Architectural Brilliance of Delhi: During the medieval period, Delhi was adorned with iconic structures like the Qutub Minar, Alai Darwaza, and tombs, showcasing Indo-Islamic architecture.
20

Who is known as the Father of India's Local Self-Government?

  1. ((a))

    Lord Lytton

  2. ((b))

    Lord Ripon

  3. ((c))

    Lord Curzon

  4. ((d))

    Lord Dalhousie

Show Answer
Answer: ((b))

Lord Ripon

The correct answer is Lord Ripon.

Key Points

  • Lord Ripon is known as the Father of Local Self-Government in India due to his significant reforms in the local governance system.
  • In 1882, Lord Ripon introduced the Local Self-Government Resolution, which laid the foundation for democratic decentralization in India.
  • His reforms aimed at empowering local bodies like municipalities and district boards, making them responsible for public services such as education, health, and sanitation.
  • He emphasized the importance of local participation in governance and entrusted local bodies with financial and administrative autonomy.
  • The resolution is considered a landmark in Indian administrative history, as it promoted the idea of grassroots-level governance.

Additional Information

  • Lord Ripon's Governance:
  • He served as the Viceroy of India from 1880 to 1884 under Queen Victoria's reign.
  • He is also known for repealing the Vernacular Press Act, which was seen as oppressive by Indians.
  • Local Self-Government:
  • It refers to the administration of local areas by locally elected representatives, ensuring decentralization of power.
  • This concept is crucial for grassroots democracy and promoting citizen participation in governance.
  • Historical Context:
  • Before Lord Ripon, local governance was largely controlled by colonial officials, with minimal participation from Indians.
  • Ripon's reforms were a step towards self-rule and inspired future movements for independence.
  • Impact of Ripon's Reforms:
  • The resolution laid the foundation for the current Panchayati Raj system in rural areas and municipal governance in urban areas.
  • It also emphasized the need for financial independence of local bodies, which remains a key aspect of local governance today.
21

At least how many days are required to give the prior notice for the impeachment of the President of India?

  1. ((a))

    7 days

  2. ((b))

    14 days

  3. ((c))

    21 days

  4. ((d))

    30 days

Show Answer
Answer: ((b))

14 days

The correct answer is Option 2 (14 days).

Key Points

  • The impeachment process of the President of India requires a prior notice of at least 14 days.
  • The notice must be signed by one-fourth of the total members of the house initiating the motion.
  • The impeachment motion is based on the grounds of violation of the Constitution.
  • After the notice is given, the motion must be supported by a two-thirds majority of the total membership of both houses of Parliament.
  • The process for impeachment is detailed under Article 61 of the Indian Constitution.

Additional Information

  • Impeachment: The President of India can be removed from office through impeachment for the violation of the Constitution.
  • Article 61: This article outlines the procedure for impeachment, which includes investigation and voting by Parliament.
  • Grounds for Impeachment: Violation of the Constitution is the only explicit ground for impeachment mentioned in the Indian Constitution.
  • Two-thirds Majority: The motion for impeachment must be passed by a two-thirds majority of the total membership of each house of Parliament.
  • Role of Judiciary: The charges against the President are investigated by a committee formed for this purpose before the voting process in Parliament takes place.
22

Who administers the oath of office and secrecy to the Governor of a State in India?

  1. ((a))

    The President of India

  2. ((b))

    The Vice President of India

  3. ((c))

    The Chief Justice of the High Court of the State

  4. ((d))

    The Speaker of the Legislative Assembly of the State

Show Answer
Answer: ((c))

The Chief Justice of the High Court of the State

The correct answer is The Chief Justice of the High Court of the State.

Key Points

  • The Governor of a State in India takes the oath of office and secrecy administered by the Chief Justice of the High Court of the State.
  • The oath is in accordance with the provisions of Article 159 of the Indian Constitution.
  • The oath includes allegiance to the Constitution of India, upholding sovereignty and integrity, and performing duties faithfully.
  • If the Chief Justice of the High Court is unavailable, the senior-most judge of the High Court administers the oath.
  • The process ensures the Governor's constitutional role and accountability as the nominal head of the state.

Additional Information

  • Article 159 of the Indian Constitution: It specifies the oath or affirmation taken by the Governor while entering office.
  • Governor's Role: The Governor serves as the constitutional head of a state and exercises powers as per the Constitution.
  • Appointment of Governor: The Governor is appointed by the President of India under Article 155.
  • Tenure: The Governor holds office for a term of five years but can be removed earlier by the President or resign voluntarily.
  • High Court Structure: Each state has a High Court, and its Chief Justice is the senior-most judicial authority in the state.
23

Which part of our Constitution envisages a three-tier system of Panchayati Raj?

  1. ((a))

    Part IX

  2. ((b))

    Part X

  3. ((c))

    Part XI

  4. ((d))

    Part XII

Show Answer
Answer: ((a))

Part IX

The correct answer is Part IX.

Key Points

  • Part IX of the Indian Constitution is titled "The Panchayats" and provides for a three-tier system of Panchayati Raj for rural self-governance.
  • It was added to the Constitution through the 73rd Constitutional Amendment Act, 1992, which came into effect on April 24, 1993.
  • The three tiers include Gram Panchayat (village level), Panchayat Samiti (block level), and Zila Parishad (district level).
  • The Panchayati Raj system is aimed at decentralization of power to ensure greater participation of people in governance at the grassroots level.
  • Articles 243 to 243-O under Part IX deal with the structure, powers, and functioning of Panchayati Raj institutions (PRIs).

Additional Information

  • 73rd Constitutional Amendment Act, 1992: This amendment provided a constitutional status to Panchayati Raj institutions and aimed at strengthening local self-governance in rural areas.
  • Eleventh Schedule: This schedule of the Constitution contains 29 subjects that are to be devolved to Panchayati Raj institutions, such as agriculture, rural housing, health, and education.
  • Gram Sabha: It is a body consisting of persons registered in the electoral rolls of a village within the area of a Gram Panchayat. It acts as the foundation of the Panchayati Raj system.
  • State Election Commission: The 73rd Amendment mandates the establishment of a State Election Commission to conduct free and fair elections to Panchayati Raj institutions.
  • Reservation of Seats: The Act provides for the reservation of seats for Scheduled Castes (SCs), Scheduled Tribes (STs), and women (not less than one-third of the total seats) in every Panchayat.
24

Which of the following States has no oil refinery?

  1. ((a))

    Gujarat

  2. ((b))

    Kerala

  3. ((c))

    Chhattisgarh

  4. ((d))

    West Bengal

Show Answer
Answer: ((c))

Chhattisgarh

The correct answer is Chhattisgarh.

Key Points

  • Chhattisgarh does not have any operational oil refinery as of now, while states like Gujarat, Kerala, and West Bengal have established refineries.
  • Gujarat hosts major refineries like the Reliance Jamnagar Refinery, which is the largest refinery in the world.
  • Kerala has the Kochi Refinery, a major petroleum refinery in southern India.
  • West Bengal has the Haldia Refinery, which is a key facility under the Indian Oil Corporation.

Additional Information

  • Oil Refinery: An industrial facility that processes crude oil into usable petroleum products like gasoline, diesel, jet fuel, and other petrochemicals.
  • Major Oil Refining States in India: States like Gujarat, Maharashtra, Tamil Nadu, and Assam are home to significant refinery operations.
  • Jamnagar Refinery: Located in Gujarat, it is the largest oil refinery complex in the world, operated by Reliance Industries.
  • Strategic Importance: Oil refineries are critical for energy security and economic growth as they cater to domestic and industrial fuel needs.
  • Indian Oil Corporation (IOC): A leading public sector company in India that operates many refineries across the country, including the Haldia Refinery in West Bengal.
25

Which of the following rivers does not flow in Australia?

  1. ((a))

    Hunter River

  2. ((b))

    Flinders River

  3. ((c))

    Orange River

  4. ((d))

    Gilbert River

Show Answer
Answer: ((c))

Orange River

The correct answer is Orange River.

Key Points

  • The Orange River is located in southern Africa and does not flow in Australia.
  • It is one of the longest rivers in Africa, originating in the Drakensberg Mountains of Lesotho and flowing westward into the Atlantic Ocean.
  • Rivers such as the Hunter River, Flinders River, and Gilbert River are located in Australia, making them incorrect options.
  • The Orange River is vital for water supply and irrigation in the region but is geographically limited to the African continent.
  • This exclusion of the Orange River highlights its distinct location compared to the rivers mentioned in the question.

Additional Information

  • Hunter River: Located in New South Wales, Australia, it is known for its significance in agriculture and coal mining.
  • Flinders River: The longest river in Queensland, Australia, it flows into the Gulf of Carpentaria and is critical for regional ecosystems.
  • Gilbert River: Also located in Queensland, Australia, it contributes to the Gulf Country’s water systems and supports cattle grazing industries.
  • Orange River Basin: Covers regions in South Africa, Namibia, and Botswana, providing water resources for mining, agriculture, and energy production.
  • River Classification: Rivers are classified based on geography, ecosystem roles, and their contribution to human activities such as irrigation, transportation, and energy generation.
26

Which of the following States recorded decrease in its population in 2011 Census?

  1. ((a))

    Kerala

  2. ((b))

    Sikkim

  3. ((c))

    Nagaland

  4. ((d))

    Manipur

Show Answer
Answer: ((c))

Nagaland

The correct answer is Nagaland.

Key Points

  • Nagaland was the only Indian state to record a decrease in its population according to the 2011 Census data.
  • The population of Nagaland declined from 1,988,636 in 2001 to 1,978,502 in 2011, marking a negative growth rate of -0.58%.
  • The decline was attributed to migration, discrepancies in earlier census data, and other socio-economic factors.
  • Population changes in Nagaland raised concerns about the accuracy of enumeration processes in the region.
  • This negative population growth is a unique case among Indian states during the 2011 Census.

Additional Information

  • Census of India:
  • The Census of India is conducted every 10 years and is the most comprehensive source of demographic data in the country.
  • The 2011 Census was the 15th National Census since 1872 and covered all states and union territories.
  • It provides data on population size, density, literacy rate, sex ratio, and other socio-economic indicators.
  • Population Growth Rate:
  • Population growth rate refers to the percentage change in the population over a specific time period.
  • A negative growth rate indicates a decline in the population size during the period.
  • Factors Affecting Population Decline:
  • Migration: Movement of people to other regions for better opportunities can lead to population decline.
  • Errors in census enumeration can also affect population data accuracy.
  • Socio-political factors, such as conflicts, may influence demographic trends.
  • Significance of Census Data:
  • Census data is crucial for policy-making, development planning, and resource allocation.
  • It helps identify trends in population dynamics and socio-economic changes over time.
27

Which of the following is the most effective measure of population control according to Malthus?

  1. ((a))

    War

  2. ((b))

    Disaster

  3. ((c))

    Birth control

  4. ((d))

    Social evils

Show Answer
Answer: ((a))

War

The correct answer is War.

Key Points

  • According to Thomas Malthus, war is one of the key "positive checks" that limits population growth by reducing the population through mortality.
  • Malthus categorized population control measures into two types: preventive checks (e.g., moral restraint, delayed marriage) and positive checks (e.g., famine, disease, war).
  • War was emphasized as a natural and inevitable mechanism to bring population levels back in alignment with the available resources.
  • Malthus argued that without such "positive checks," human population growth would outpace food production, leading to widespread scarcity and suffering.
  • His theory is based on the principle that population grows geometrically while food supply grows arithmetically, creating a gap that needs to be corrected by such checks.

Additional Information

  • Thomas Malthus and his Theory:
  • Thomas Robert Malthus was an 18th-century British economist and demographer.
  • In his famous work, An Essay on the Principle of Population (1798), he proposed that population growth, if unchecked, would lead to resource depletion.
  • He introduced the concepts of "preventive checks" and "positive checks" to manage population growth.
  • Preventive Checks:
  • These are measures to reduce birth rates, such as moral restraint, delayed marriage, and birth control.
  • Malthus supported moral restraint as the most ethical preventive measure.
  • Positive Checks:
  • These are natural and man-made factors that increase death rates, including famine, disease, and war.
  • Malthus believed these checks were necessary to prevent overpopulation and maintain the balance between population and resources.
  • Criticism of Malthusian Theory:
  • Modern economists argue that technological advancements in agriculture and resource management have disproved Malthus's dire predictions of widespread famine and resource scarcity.
  • Critics also highlight that social, economic, and political factors greatly influence population dynamics beyond Malthus's simplistic model.
28

Which of the following is not a biome?

  1. ((a))

    Desert

  2. ((b))

    Grassland

  3. ((c))

    Ecosystem

  4. ((d))

    Tundra

Show Answer
Answer: ((c))

Ecosystem

The correct answer is Ecosystem.

Key Points

  • A biome is a large ecological area on the Earth's surface, classified based on its climate, flora, and fauna. Examples include desert, tundra, and grassland.
  • An ecosystem, on the other hand, refers to a community of living organisms interacting with their physical environment, such as a pond or forest.
  • Ecosystem is not a biome; it is a smaller functional unit within a biome where biotic (living) and abiotic (non-living) components interact.
  • Biomes are broader classifications, encompassing multiple ecosystems. For example, the desert biome includes various ecosystems such as sand dunes, oases, and rocky deserts.

Additional Information

  • Definition of Biome: A biome is characterized by specific climate conditions and distinct types of plants and animals adapted to it. Common examples include forests, grasslands, deserts, and tundras.
  • Definition of Ecosystem: An ecosystem includes all living organisms and their interactions with the non-living environment within a specific location. It can vary in size, from a small pond to a large forest.
  • Difference Between Biome and Ecosystem: Biomes are larger geographical areas with similar climatic conditions, while ecosystems are smaller units within biomes where organisms interact with their environment.
  • Major Biomes: Earth’s major biomes include tropical rainforests, savannas, deserts, grasslands, tundras, and aquatic biomes like freshwater and marine ecosystems.
  • Examples of Ecosystems: Examples include coral reefs, mangroves, wetlands, and urban ecosystems, each functioning uniquely within a biome.
29

Dudhwa National Park is situated in which of the following States?

  1. ((a))

    Assam

  2. ((b))

    Uttarakhand

  3. ((c))

    Rajasthan

  4. ((d))

    Uttar Pradesh

Show Answer
Answer: ((d))

Uttar Pradesh

The correct answer is Uttar Pradesh.

Key Points

  • Dudhwa National Park is located in the Lakhimpur Kheri district of Uttar Pradesh, India.
  • It was established in 1977 and later designated as a tiger reserve in 1987 under Project Tiger.
  • The park spans an area of approximately 490 square kilometers and is part of the Terai ecosystem.
  • Dudhwa is renowned for its population of Bengal tigers, swamp deer (Barasingha), and Indian rhinoceros.
  • The park is also a haven for birdwatchers, with over 400 species of resident and migratory birds recorded.

Additional Information

  • Terai Ecosystem: The park is part of the Terai belt, a marshy grassland ecosystem found in the foothills of the Himalayas, known for its rich biodiversity.
  • Fauna: Key species include the Bengal tiger, Indian rhinoceros, Asian elephant, Barasingha, and Gangetic dolphin (in nearby rivers).
  • Flora: The park consists of dense sal forests, grasslands, and wetlands, providing a diverse habitat for wildlife.
  • Conservation Challenges: The park faces threats such as habitat destruction, human-wildlife conflict, and poaching, prompting ongoing conservation efforts.
  • Tourism: Dudhwa National Park attracts eco-tourists and wildlife enthusiasts, with safaris and guided tours offered to explore its unique biodiversity.
30

According to the Fourth Round of National Family Health Survey, the current TFR (Total Fertility Rate-children per woman) is

  1. ((a))

    2.2

  2. ((b))

    3.2

  3. ((c))

    4.2

  4. ((d))

    4.5

Show Answer
Answer: ((a))

2.2

The correct answer is 2.2.

Key Points

  • The Total Fertility Rate (TFR) in India according to the Fourth Round of the National Family Health Survey (NFHS-4) was reported to be 2.2 children per woman.
  • A TFR of 2.2 indicates that India is moving towards achieving replacement-level fertility (2.1 children per woman).
  • The decline in TFR is attributed to increased awareness of family planning, better access to contraception, and socio-economic improvements.
  • The TFR varies significantly across states, with some states like Kerala and Tamil Nadu having a TFR below 2, while others like Bihar and Uttar Pradesh report higher rates.
  • The reduction in TFR is a critical factor in stabilizing population growth and achieving sustainable development goals (SDGs).

Additional Information

  • Total Fertility Rate (TFR):
  • TFR is the average number of children that would be born to a woman over her lifetime, based on current birth rates.
  • A TFR of 2.1 is considered the replacement level, where the population remains stable without migration.
  • National Family Health Survey (NFHS):
  • NFHS is a large-scale, multi-round survey conducted in a representative sample of households across India.
  • It provides essential data on health, nutrition, fertility, family planning, and mortality.
  • Factors Influencing TFR Decline:
  • Improved access to education, particularly for women.
  • Increased availability and use of contraceptives.
  • Rising urbanization and economic development.
  • Government initiatives promoting family planning and awareness campaigns.
  • Implications of Low TFR:
  • Lower TFR leads to slower population growth, reducing the strain on resources and public services.
  • However, prolonged low TFR can result in an aging population, leading to a higher dependency ratio in the future.

Part 2 (120 questions)

31

If θ\theta  is real, then

  1. ((a))

    cos(iθ\theta )=i coshθ\theta

  2. ((b))

    sin(iθ\theta )=i sinhθ\theta

  3. ((c))

    tan(iθ\theta)=tanhθ\theta

  4. ((d))

    cot(iθ\theta )=icothθ\theta

Show Answer
Answer: ((b))

sin(iθ\theta )=i sinhθ\theta

Calculation:

If θR\theta \in \mathbb{R} then trigonometric functions with imaginary arguments relate to hyperbolic functions using the identities:

cos(iθ)=cosh(θ) \cos(i\theta) = \cosh(\theta)

sin(iθ)=isinh(θ) \sin(i\theta) = i\sinh(\theta)

tan(iθ)=itanh(θ) \tan(i\theta) = i\tanh(\theta)

cot(iθ)=icoth(θ) \cot(i\theta) = -i\coth(\theta)

Now, analyzing the options given:

  1. cos(iθ)=icosh(θ) \cos(i\theta) = i\cosh(\theta) Incorrect, correct form is cos(iθ)=cosh(θ) \cos(i\theta) = \cosh(\theta)

  2. sin(iθ)=isinh(θ) \sin(i\theta) = i\sinh(\theta) Correct

  3. tan(iθ)=tanh(θ) \tan(i\theta) = \tanh(\theta) Incorrect, correct form is tan(iθ)=itanh(θ) \tan(i\theta) = i\tanh(\theta)

  4. cot(iθ)=icoth(θ) \cot(i\theta) = i\coth(\theta) Incorrect, correct form is cot(iθ)=icoth(θ) \cot(i\theta) = -i\coth(\theta)

Hence, the correct answer is Option 2.

Additional Information

Important Hyperbolic Identities:

cosh2xsinh2x=1 \cosh^2 x - \sinh^2 x = 1

Even/Odd Properties of Hyperbolic Functions:

sinh(x)=sinh(x)(odd function) \sinh(-x) = -\sinh(x) \quad \text{(odd function)}

cosh(x)=cosh(x)(even function) \cosh(-x) = \cosh(x) \quad \text{(even function)}

tanh(x)=tanh(x)(odd function) \tanh(-x) = -\tanh(x) \quad \text{(odd function)}

coth(x)=coth(x)(odd function) \coth(-x) = -\coth(x) \quad \text{(odd function)}

Inverse Hyperbolic Functions:

sinh1x=ln(x+x2+1) \sinh^{-1}x = \ln(x + \sqrt{x^2 + 1})

cosh1x=ln(x+x21) \cosh^{-1}x = \ln(x + \sqrt{x^2 - 1})

tanh1x=12ln(1+x1x) \tanh^{-1}x = \frac{1}{2} \ln\left(\frac{1 + x}{1 - x}\right)

32

Ifz=x+iy,z = x+iy, where i=1,i=\sqrt{-1},   then z3z+3=2\left|\frac{z-3}{z+3}\right|=2 represents a circle, whose centre and radius, respectively, are

  1. ((a))

    (5,0),5

  2. ((b))

    (−5,0),2

  3. ((c))

    (−5,0),3

  4. ((d))

    (−5,0),4

Show Answer
Answer: ((d))

(−5,0),4

Calculation:

z3z+3=2 \left| \frac{z - 3}{z + 3} \right| = 2 , where z = x + iy 

z3z+3=2(x3)2+y2(x+3)2+y2=2 \Rightarrow \frac{|z - 3|}{|z + 3|} = 2 \Rightarrow \frac{√{(x - 3)^2 + y^2}}{√{(x + 3)^2 + y^2}} = 2

(x3)2+y2(x+3)2+y2=4 \Rightarrow \frac{(x - 3)^2 + y^2}{(x + 3)^2 + y^2} = 4

x26x+9+y2=4(x2+6x+9+y2) \Rightarrow x^2 - 6x + 9 + y^2 = 4(x^2 + 6x + 9 + y^2)

x26x+9+y2=4x2+24x+36+4y2 \Rightarrow x^2 - 6x + 9 + y^2 = 4x^2 + 24x + 36 + 4y^2

3x230x273y2=0 \Rightarrow -3x^2 - 30x - 27 - 3y^2 = 0

x2+10x+y2+9=0 \Rightarrow x^2 + 10x + y^2 + 9 = 0

x2+10x+25+y2=16 \Rightarrow x^2 + 10x + 25 + y^2 = 16 (completing the square)

(x+5)2+y2=16 \Rightarrow (x + 5)^2 + y^2 = 16

Comparing (x+5)2+y2=16(x + 5)^2 + y^2 = 16  with the standard form, we have:

Centre = ( -5, 0)

Radius = √16 = 4

Hence, the correct answer is Option 4.

33

If ω(1)\omega (\ne1) is a cube root of unity, then the value of (1ω+ω2)5+(1+ωω2)532)(1-\omega+\omega^2)^5+(1+\omega-\omega^2)^5 - 32) is

  1. ((a))

    0

  2. ((b))

    -32

  3. ((c))

    32

  4. ((d))

    -64

Show Answer
Answer: ((a))

0

Calculation:

Given expression:

(1ω+ω2)5+(1+ωω2)532\left(1 - \omega + \omega^2\right)^5 + \left(1 + \omega - \omega^2\right)^5 - 32

Using the identity for cube roots of unity:

⇒ 1+ω+ω2=0ω2+ω=11 + \omega + \omega^2 = 0 \Rightarrow \omega^2 + \omega = -1

Simplify each term:

⇒ 1ω+ω2=2ω1 - \omega + \omega^2 = -2\omega

⇒ 1+ωω2=2ω21 + \omega - \omega^2 = -2\omega^2

Substitute into the original expression:

⇒ (2ω)5+(2ω2)532(-2\omega)^5 + (-2\omega^2)^5 - 32

Calculate powers:

=(2)5ω5+(2)5(ω2)532=32ω532ω1032= (-2)^5 \omega^5 + (-2)^5 (\omega^2)^5 - 32 = -32 \omega^5 - 32 \omega^{10} - 32

Use ω3=1 \omega^3 = 1 to reduce powers:

⇒ ω5=ω2,ω10=ω\omega^5 = \omega^2, \quad \omega^{10} = \omega

So expression becomes:

⇒ 32ω232ω32=32(ω2+ω+1)-32 \omega^2 - 32 \omega - 32 = -32 (\omega^2 + \omega + 1)

Since ω2+ω+1=0\omega^2 + \omega + 1 = 0 we have:

⇒ 32×0=0 -32 \times 0 = 0

Hence, the correct answer is Option 1.

34

The value of 34i\sqrt{3-4i} is

  1. ((a))

    2+i

  2. ((b))

    1+i

  3. ((c))

    1−i

  4. ((d))

    2−i

Show Answer
Answer: ((d))

2−i

Calculation:

z=34iRe(z)=3,Im(z)=4 z = 3 - 4i \Rightarrow \text{Re}(z) = 3, \text{Im}(z) = -4

z=32+(4)2=9+16=25=5 |z| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5

z=±(z+Re(z)2+isgn(Im(z))zRe(z)2) \sqrt{z} = \pm \left( \sqrt{ \frac{|z| + \text{Re}(z)}{2} } + i \cdot \operatorname{sgn}(\text{Im}(z)) \cdot \sqrt{ \frac{|z| - \text{Re}(z)}{2} } \right)

34i=±(5+32i532) \Rightarrow \sqrt{3 - 4i} = \pm \left( \sqrt{ \frac{5 + 3}{2} } - i \cdot \sqrt{ \frac{5 - 3}{2} } \right)

=±(4i1)=±(2i) = \pm \left( \sqrt{4} - i \cdot \sqrt{1} \right) = \pm (2 - i)

Hence, the correct answer is Option 4.

35

If cos(x+iy)=cosα+isinαcos(x+iy)=cosα+isinα , then the value of  (cosh2y+cos2x)(cosh2y+cos2x) is

  1. ((a))

    1

  2. ((b))

    2

  3. ((c))

    -2

  4. ((d))

    2\sqrt{2}

Show Answer
Answer: ((b))

2

Calculation:

cos(x+iy)=cosα+isinα \cos(x + iy) = \cos \alpha + i \sin \alpha

cos(x+iy)=cosxcoshyisinxsinhy \Rightarrow \cos(x + iy) = \cos x \cosh y - i \sin x \sinh y

cosxcoshy=cosα,sinxsinhy=sinα \Rightarrow \cos x \cosh y = \cos \alpha, \quad -\sin x \sinh y = \sin \alpha

(cosxcoshy)2+(sinxsinhy)2=cos2α+sin2α=1 \Rightarrow (\cos x \cosh y)^2 + (\sin x \sinh y)^2 = \cos^2 \alpha + \sin^2 \alpha = 1

cos2xcosh2y+sin2xsinh2y=1 \Rightarrow \cos^2 x \cosh^2 y + \sin^2 x \sinh^2 y = 1

cos2x(1+sinh2y)+sin2xsinh2y=1 \Rightarrow \cos^2 x (1 + \sinh^2 y) + \sin^2 x \sinh^2 y = 1

cos2x+sinh2y(cos2x+sin2x)=1 \Rightarrow \cos^2 x + \sinh^2 y (\cos^2 x + \sin^2 x) = 1

cos2x+sinh2y=1sinh2y=1cos2x \Rightarrow \cos^2 x + \sinh^2 y = 1 \Rightarrow \sinh^2 y = 1 - \cos^2 x

cosh2y=2sinh2y+1,cos2x=2cos2x1 \Rightarrow \cosh 2y = 2 \sinh^2 y + 1, \quad \cos 2x = 2 \cos^2 x - 1

cosh2y+cos2x=2(1cos2x)+1+2cos2x1=2 \Rightarrow \cosh 2y + \cos 2x = 2(1 - \cos^2 x) + 1 + 2 \cos^2 x - 1 = 2

Hence, the correct answer is Option 2.

36

The three cube roots of z=8iz=−8i are

  1. ((a))

    2i,3i,3i2i,− \sqrt{3}−i,\sqrt{3}−i

  2. ((b))

    2i,3i,3i-2i,− \sqrt{3} −i, \sqrt{3} −i

  3. ((c))

    2i,3i,3+i2i,− \sqrt{3} −i, \sqrt{3} +i

  4. ((d))

    2i,3i,3+i2i, \sqrt{3} −i, -\sqrt{3} +i

Show Answer
Answer: ((a))

2i,3i,3i2i,− \sqrt{3}−i,\sqrt{3}−i

Calculation:

z=8i=8ei3π2 z = -8i = 8e^{i\frac{3\pi}{2}}

Let z=reiθ, then the cube roots are given by z3=r3ei(θ+2πk3) for k=0,1,2 \text{Let } z = re^{i\theta}, \text{ then the cube roots are given by } \sqrt[3]{z} = \sqrt[3]{r} \cdot e^{i\left(\frac{\theta + 2\pi k}{3}\right)} \text{ for } k = 0, 1, 2

8ei3π23=2ei(3π/2+2πk3) \Rightarrow \sqrt[3]{8e^{i\frac{3\pi}{2}}} = 2e^{i\left(\frac{3\pi/2 + 2\pi k}{3}\right)}

For k = 0:

z0=2eiπ2=2i z_0 = 2e^{i\frac{\pi}{2}} = 2i

For k = 1:

z1=2ei7π6=2(cos7π6+isin7π6)=2(3212i)=3i z_1 = 2e^{i\frac{7\pi}{6}} = 2\left(\cos\frac{7\pi}{6} + i\sin\frac{7\pi}{6}\right) = 2\left(-\frac{\sqrt{3}}{2} - \frac{1}{2}i\right) = -\sqrt{3} - i

For k = 2:

z2=2ei11π6=2(cos11π6+isin11π6)=2(3212i)=3i z_2 = 2e^{i\frac{11\pi}{6}} = 2\left(\cos\frac{11\pi}{6} + i\sin\frac{11\pi}{6}\right) = 2\left(\frac{\sqrt{3}}{2} - \frac{1}{2}i\right) = \sqrt{3} - i

∴ the cube roots of z=8i z = -8i are 2i, 3i, 3i 2i,\ -\sqrt{3} - i,\ \sqrt{3} - i .

Hence, the correct answer is Option 1.

37

If  Im(z12z+1)=4\text{Im}\left(\frac{z-1}{2z+1}\right)=-4    then the locus of z is

  1. ((a))

    an ellipse

  2. ((b))

    a parabola

  3. ((c))

    a straight line

  4. ((d))

    a circle

Show Answer
Answer: ((d))

a circle

Calculation:

z=x+iyz1=x1+iy,2z+1=2x+1+2iy z = x + iy \Rightarrow z - 1 = x - 1 + iy,\quad 2z + 1 = 2x + 1 + 2iy

⇒ z12z+1=(x1)+iy(2x+1)+2iy \frac{z - 1}{2z + 1} = \frac{(x - 1) + iy}{(2x + 1) + 2iy}

⇒ Im(a+ibc+id)=bcadc2+d2 \text{Im} \left( \frac{a + ib}{c + id} \right) = \frac{bc - ad}{c^2 + d^2}

Let a=x1,b=y,c=2x+1,d=2y \text{Let } a = x - 1, \quad b = y, \quad c = 2x + 1, \quad d = 2y

Im=y(2x+1)(x1)(2y)(2x+1)2+4y2=2xy+y2xy+2y(2x+1)2+4y2=3y(2x+1)2+4y2 \Rightarrow \text{Im} = \frac{y(2x + 1) - (x - 1)(2y)}{(2x + 1)^2 + 4y^2} = \frac{2xy + y - 2xy + 2y}{(2x + 1)^2 + 4y^2} = \frac{3y}{(2x + 1)^2 + 4y^2}

Given: 3y(2x+1)2+4y2=4 \text{Given: } \frac{3y}{(2x + 1)^2 + 4y^2} = -4

3y=4((2x+1)2+4y2) \Rightarrow 3y = -4\left((2x + 1)^2 + 4y^2\right)

3y=4(4x2+4x+1+4y2) \Rightarrow 3y = -4(4x^2 + 4x + 1 + 4y^2)

3y=16x216x416y2 \Rightarrow 3y = -16x^2 - 16x - 4 - 16y^2

16x2+16x+16y2+3y+4=0 \Rightarrow 16x^2 + 16x + 16y^2 + 3y + 4 = 0

This is a general equation of a circle since  x2 and y2 have equal coefficients and the same signs

Hence, the correct answer is Option 4.

38

If  f(z)=(x2+ay2)+ibxyf(z)=(x ^2 +ay ^2 ) +ibxy   is a complex analytic function of z=x+iy,z=x+iy, then the value of a+ba+b is

  1. ((a))

    0

  2. ((b))

    1

  3. ((c))

    -1

  4. ((d))

    2

Show Answer
Answer: ((b))

1

Calculation:

f(z)=(x2+ay2)+ibxy f(z) = (x^2 + a y^2) + i b x y is analytic where z=x+iy z = x + i y .

Let u=x2+ay2 u = x^2 + a y^2 and v=bxy v = b x y be the real and imaginary parts of f(z) f(z) .

By Cauchy-Riemann equations,

ux=vyanduy=vx \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \text{and} \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}

Calculate partial derivatives:

ux=2x,uy=2ay \frac{\partial u}{\partial x} = 2x, \quad \frac{\partial u}{\partial y} = 2 a y

vx=by,vy=bx \frac{\partial v}{\partial x} = b y, \quad \frac{\partial v}{\partial y} = b x

From Cauchy-Riemann equations,

2x=bx    b=2 2x = b x \implies b = 2

2ay=by    2a=b    2a=2    a=1 2 a y = - b y \implies 2 a = -b \implies 2 a = -2 \implies a = -1

Therefore,

a+b=1+2=1 a + b = -1 + 2 = 1

Hence, the correct answer is Option 2.

39

Which one of the following is false?

  1. ((a))

    f(z)=zˉf(z) = \bar{z}   is nowhere analytic.

  2. ((b))

    f(z)=z2f(z) = z^2   is everywhere analytic.

  3. ((c))

    f(z)=z2f(z) = |z|^2   is analytic at z=0.

  4. ((d))

    f(z)=ezf(z) = e^z   is analytic everywhere.

Show Answer
Answer: ((c))

f(z)=z2f(z) = |z|^2   is analytic at z=0.

Calculation:

(a) f(z)=zˉ (a)\ f(z) = \bar{z}

zˉ\bar{z} is the complex conjugate of z . It does not satisfy the Cauchy-Riemann equations anywhere in C.\mathbb{C} .

Therefore, it is nowhere analytic.

(b) f(z)=z2 (b)\ f(z) = z^2

This is a polynomial function. All polynomials in z are entire functions, i.e., analytic everywhere on C.\mathbb{C}.

Therefore, it is analytic everywhere.

(c) f(z)=z2=zzˉ (c)\ f(z) = |z|^2 = z\bar{z}

We analyze this at z = 0 . Let z = x + iy , then:

f(z)=z2=x2+y2 f(z) = |z|^2 = x^2 + y^2

This is a real-valued function and depends on both z and zˉ\bar{z} , so it is not holomorphic. Let's apply the Cauchy-Riemann equations:

Let u(x,y)=x2+y2u(x, y) = x^2 + y^2 ,v(x,y)=0v(x, y) = 0

ux=2x,uy=2y,vx=0,vy=0 u_x = 2x,\quad u_y = 2y,\quad v_x = 0,\quad v_y = 0

At (0,0): ux=0u_x = 0, uy = 0 , so Cauchy-Riemann equations hold at the point, but the function is still not complex differentiable in any neighborhood of 0. Hence, it is not analytic at z = 0 

(d) f(z)=ez (d)\ f(z) = e^z

This is the complex exponential function, which is entire (analytic everywhere in C \mathbb{C}

Therefore, it is analytic everywhere.

Hence, the correct answer is Option (c).

40

For  zCz \in C  the inequality  z+i>zi|z+i| > |z-i|   is

  1. ((a))

    always true

  2. ((b))

    never true

  3. ((c))

    true for Re z>0\text{Re }z > 0

  4. ((d))

    true for Im z>0\text{Im }z > 0

Show Answer
Answer: ((d))

true for Im z>0\text{Im }z > 0

Calculation:

z=x+iy z = x + iy

⇒ z+i=x2+(y+1)2 |z + i| = \sqrt{x^2 + (y+1)^2}

⇒ zi=x2+(y1)2 |z - i| = \sqrt{x^2 + (y-1)^2}

⇒ z+i>zix2+(y+1)2>x2+(y1)2 |z + i| > |z - i| \Rightarrow \sqrt{x^2 + (y+1)^2} > \sqrt{x^2 + (y-1)^2}

x2+(y+1)2>x2+(y1)2 \Rightarrow x^2 + (y+1)^2 > x^2 + (y-1)^2

(y+1)2>(y1)2 \Rightarrow (y+1)^2 > (y-1)^2

y2+2y+1>y22y+1 \Rightarrow y^2 + 2y + 1 > y^2 - 2y + 1

2y>2y4y>0y>0 \Rightarrow 2y > -2y \Rightarrow 4y > 0 \Rightarrow y > 0

∴  the inequality  |z + i| > |z - i|  holds true if and only if  Imz>0\ \operatorname{Im} z > 0

Hence, the correct answer is Option 4.

41

The value of 33x21+3xdx\int_{-3}^{3} \frac{x^2}{1+3x} dx  is

  1. ((a))

    13\frac{1}{3}

  2. ((b))

    19\frac{1}{9}

  3. ((c))

    3

  4. ((d))

    9

Show Answer
Answer: ((d))

9

Calculation:

We are asked to evaluate the definite integral:

33x21+3x,dx \int_{-3}^{3} \frac{x^2}{1 + 3^x} , dx

Let f(x)=x21+3x f(x) = \frac{x^2}{1 + 3^x} . We analyze the symmetry of the function:

f(x)=(x)21+3x=x21+13x=x23x3x+1 f(-x) = \frac{(-x)^2}{1 + 3^{-x}} = \frac{x^2}{1 + \frac{1}{3^x}} = \frac{x^2 \cdot 3^x}{3^x + 1}

So,

f(x)+f(x)=x21+3x+x23x1+3x=x2 f(x) + f(-x) = \frac{x^2}{1 + 3^x} + \frac{x^2 \cdot 3^x}{1 + 3^x} = x^2

Therefore,

33f(x),dx=33x21+3x,dx=33f(x)+f(x)2,dx=33x22,dx \int_{-3}^{3} f(x) , dx = \int_{-3}^{3} \frac{x^2}{1 + 3^x} , dx = \int_{-3}^{3} \frac{f(x) + f(-x)}{2} , dx = \int_{-3}^{3} \frac{x^2}{2} , dx

Since x2 x^2 is even, we simplify:

33x22,dx=203x22,dx=03x2,dx \int_{-3}^{3} \frac{x^2}{2} , dx = 2 \cdot \int_{0}^{3} \frac{x^2}{2} , dx = \int_{0}^{3} x^2 , dx

=[x33]03=273=9 = \left[ \frac{x^3}{3} \right]_0^3 = \frac{27}{3} = 9

Hence, the correct answer is Option (d): 9.

42

The area bounded by the curves  y=sinx,y=cosxy=sinx, y=cosx  and  y-axis is

  1. ((a))

    2+1\sqrt{2}+1

  2. ((b))

    21\sqrt{2}-1

  3. ((c))

    2(21)2(\sqrt{2}-1)

  4. ((d))

    2+12\frac{\sqrt{2}+1}{2}

Show Answer
Answer: ((b))

21\sqrt{2}-1

Calculation:

We are given then curves  y=sinx,y=cosxy=sinx, y=cosx  and  y-axis is

First, find the point of intersection sinx=cosxtanx=1x=π4 \sin x = \cos x \Rightarrow \tan x = 1 \Rightarrow x = \frac{\pi}{4}

We need to evaluate the area between the curves from x=0 to x=π4. x = 0 \text{ to } x = \frac{\pi}{4}.

A=0π4(cosxsinx),dx A = \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) , dx

=[sinx+cosx]0π4 = \left[ \sin x + \cos x \right]_0^{\frac{\pi}{4}}

=(sinπ4+cosπ4)(sin0+cos0) = \left( \sin \frac{\pi}{4} + \cos \frac{\pi}{4} \right) - \left( \sin 0 + \cos 0 \right)

=(22+22)(0+1) = \left( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \right) - (0 + 1)

=21 = \sqrt{2} - 1

Hence, the correct answer is Option (b).

43

The limit  limx2ax+b3x2\lim_{x \to 2} \frac{\sqrt{ax+b}-3}{x-2}12\frac{1}{2} . then the value of a, b is

  1. ((a))

    a=3,b=3a=3,b=3

  2. ((b))

    aba \ne b

  3. ((c))

    a=0,b=4a=0,b=4

  4. ((d))

    a=2,b=1a=2,b=1

Show Answer
Answer: ((a))

a=3,b=3a=3,b=3

Calculation:

limx2ax+b3x2=12 \lim_{x \to 2} \frac{\sqrt{ax + b} - 3}{x - 2} = \frac{1}{2}

To form an indeterminate 00, 2a+b=32a+b=9(1) \frac{0}{0},\ \sqrt{2a + b} = 3 \Rightarrow 2a + b = 9 \quad \text{(1)}

Multiply numerator and denominator by the conjugate ax+b3x2ax+b+3ax+b+3 \frac{\sqrt{ax + b} - 3}{x - 2} \cdot \frac{\sqrt{ax + b} + 3}{\sqrt{ax + b} + 3}

=(ax+b)9(x2)(ax+b+3) = \frac{(ax + b) - 9}{(x - 2)(\sqrt{ax + b} + 3)}

Using (1), we write: ax+b9=a(x2) \text{Using (1), we write: } ax + b - 9 = a(x - 2)

a(x2)(x2)(ax+b+3)=aax+b+3 \Rightarrow \frac{a(x - 2)}{(x - 2)(\sqrt{ax + b} + 3)} = \frac{a}{\sqrt{ax + b} + 3}

limx2aax+b+3=a6=12 \lim_{x \to 2} \frac{a}{\sqrt{ax + b} + 3} = \frac{a}{6} {=} \frac{1}{2}

a=3 \Rightarrow a = 3

Substitute in (1): 2(3)+b=9b=3 \text{Substitute in (1): } 2(3) + b = 9 \Rightarrow b = 3

Hence, the correct answer is Option 1:

44

Consider the following statements:

I. y=xy = |x| is differentiable at x=0x=0

II. y=xxy = x|x| is differentiable everywhere.

Which of the above statements is/are true?

  1. ((a))

    Only I

  2. ((b))

    Only II

  3. ((c))

    Both I and II

  4. ((d))

    Neither I nor II

Show Answer
Answer: ((b))

Only II

Calculation:

Check differentiability of y=x y = |x| at x=0 x=0 :

⇒ y={x,x0 x,x<0 y = \begin{cases} x, & x \geq 0 \ -x, & x < 0 \end{cases}

Left-hand derivative at 0:

⇒ limh00+h0h=limh0h0h=1 \lim_{h \to 0^-} \frac{|0 + h| - |0|}{h} = \lim_{h \to 0^-} \frac{-h - 0}{h} = -1

Right-hand derivative at 0:

⇒ limh0+0+h0h=limh0+h0h=1 \lim_{h \to 0^+} \frac{|0 + h| - |0|}{h} = \lim_{h \to 0^+} \frac{h - 0}{h} = 1

Since 11 -1 \neq 1 , y=x y = |x| is not differentiable at x=0 x=0 .

Now check differentiability of y=xx y = x|x| :

⇒ y={x2,x0 x2,x<0 y = \begin{cases} x^2, & x \geq 0 \ -x^2, & x < 0 \end{cases}

Derivative for x>0 x > 0 :

⇒ y=2x y' = 2x

Derivative for x<0 x < 0 :

⇒ y=2x y' = -2x

Left-hand derivative at 0:

⇒ limh0h20h=limh0h=0 \lim_{h \to 0^-} \frac{-h^2 - 0}{h} = \lim_{h \to 0^-} -h = 0

Right-hand derivative at 0:

⇒ limh0+h20h=limh0+h=0 \lim_{h \to 0^+} \frac{h^2 - 0}{h} = \lim_{h \to 0^+} h = 0

Since left and right derivatives are equal and differentiable elsewhere, y=xx y = x|x| is differentiable everywhere.

∴ only Statement II is true.

Hence, the correct answer is Option 2.

45

If 1u=x2+y2+z2\frac{1}{u} = \sqrt{x^2+y^2+z^2} then  xux+yuy+zuzx\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} + z\frac{\partial u}{\partial z}  is equal to

  1. ((a))

    0

  2. ((b))

    2u2u

  3. ((c))

    u-u

  4. ((d))

    u2u^2

Show Answer
Answer: ((c))

u-u

Calculation:

Given,

1u=x2+y2+z2 \frac{1}{u} = \sqrt{x^2 + y^2 + z^2}

We need to evaluate:

xux+yuy+zuz x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z}

Let r=x2+y2+z2 r = \sqrt{x^2 + y^2 + z^2} , then

u=1r u = \frac{1}{r}

Now, compute the partial derivatives:

ux=ddx(1r)=1r2rx=xr3 \frac{\partial u}{\partial x} = \frac{d}{dx} \left( \frac{1}{r} \right) = -\frac{1}{r^2} \cdot \frac{\partial r}{\partial x} = -\frac{x}{r^3}

uy=yr3,uz=zr3 \frac{\partial u}{\partial y} = -\frac{y}{r^3}, \quad \frac{\partial u}{\partial z} = -\frac{z}{r^3}

Now substitute in the given expression:

x(xr3)+y(yr3)+z(zr3) x \cdot \left( -\frac{x}{r^3} \right) + y \cdot \left( -\frac{y}{r^3} \right) + z \cdot \left( -\frac{z}{r^3} \right)

=x2+y2+z2r3=r2r3=1r=u = -\frac{x^2 + y^2 + z^2}{r^3} = -\frac{r^2}{r^3} = -\frac{1}{r} = -u

∴ the expression is equal to u -u .

Hence, the correct answer is Option (c).

46

The differential equation of the straight lines at a fixed distance p from the origin is

  1. ((a))

    (xyy)2=p2(1+y2)(xy'-y)^2 = p^2(1+y'^2)

  2. ((b))

    (xy+y)2=p2(1+y2)(xy'+y)^2 = p^2(1+y'^2)

  3. ((c))

    (xy)2=p2(1+y2)(x-y')^2 = p^2(1+y'^2)

  4. ((d))

    (x+y)2=p2(1+y2)(x+y')^2 = p^2(1+y'^2)

Show Answer
Answer: ((a))

(xyy)2=p2(1+y2)(xy'-y)^2 = p^2(1+y'^2)

Calculation:

Given: y=mx+c,m=y \text{Given: } y = mx + c, \quad m = y'

Perpendicular distance from origin=p=c1+m2c=p1+m2 \text{Perpendicular distance from origin} = p = \frac{|c|}{\sqrt{1 + m^2}} \Rightarrow |c| = p \sqrt{1 + m^2}

c=±p1+(y)2 \Rightarrow c = \pm p \sqrt{1 + (y')^2}

Equation of the line: y=yx+cyyx=c \text{Equation of the line: } y = y' x + c \Rightarrow y - y' x = c

yyx=±p1+(y)2 \Rightarrow y - y' x = \pm p \sqrt{1 + (y')^2}

(yyx)2=p2(1+(y)2) \Rightarrow (y - y' x)^2 = p^2 (1 + (y')^2)

(xyy)2=p2(1+(y)2) \Rightarrow (x y' - y)^2 = p^2 (1 + (y')^2)

Therefore, the differential equation of straight lines at fixed distance p from the origin is:

(xyy)2=p2(1+(y)2) (x y' - y)^2 = p^2 (1 + (y')^2)

Hence, the correct answer is Option 1.

47

The solution of the differential equation  yxdydx=a(y2+dydx)y - x \frac{dy}{dx} =a( y^2+ \frac{dy}{dx}) is

  1. ((a))

    (x+a)(1ay)=cy(x+a)(1-ay) = cy

  2. ((b))

    (x+a)(1+ay)=cy(x+a)(1+ay) = cy

  3. ((c))

    (x+a)(1+ay)=cx(x+a)(1+ay) = cx

  4. ((d))

    (y+a)(1+ax)=cy(y+a)(1+ax) = cy

Show Answer
Answer: ((a))

(x+a)(1ay)=cy(x+a)(1-ay) = cy

Calculation:

Given: yxdydx=a(y2+dydx)y - x \frac{dy}{dx} = a\left(y^2 + \frac{dy}{dx} \right)

yay2=dydx(x+a)\Rightarrow y - a y^2 = \frac{dy}{dx}(x + a)

dydx=yay2x+a\Rightarrow \frac{dy}{dx} = \frac{y - a y^2}{x + a}

dyy(1ay)=dxx+a\Rightarrow \frac{dy}{y(1 - a y)} = \frac{dx}{x + a}

(1y+a1ay)dy=dxx+a\Rightarrow \int \left( \frac{1}{y} + \frac{a}{1 - a y} \right) dy = \int \frac{dx}{x + a}

lnyln1ay=lnx+a+lnC\Rightarrow \ln|y| - \ln|1 - a y| = \ln|x + a| + \ln C

ln(y1ay)=ln(C(x+a)\Rightarrow \ln \left( \frac{y}{1 - a y} \right) = \ln(C(x + a)

y1ay=C(x+a)\Rightarrow \frac{y}{1 - a y} = C(x + a)

y=C(x+a)(1ay)\Rightarrow y = C(x + a)(1 - a y)

y=C(x+a)aC(x+a)y\Rightarrow y = C(x + a) - a C(x + a) y

y+aC(x+a)y=C(x+a)\Rightarrow y + a C(x + a) y = C(x + a)

y(1+aC(x+a))=C(x+a)\Rightarrow y(1 + a C(x + a)) = C(x + a)

y=C(x+a)1+aC(x+a)\Rightarrow y = \frac{C(x + a)}{1 + a C(x + a)}

(x+a)(1ay)=cy\Rightarrow (x + a)(1 - a y) = c y

Hence, the correct answer is Option 1.

48

The value of c in in Lagrange's mean value theorem for f(x)=x(x1)f(x)=x(x−1) in [1,2] is

  1. ((a))

    54\frac{5}{4}

  2. ((b))

    32\frac{3}{2}

  3. ((c))

    74\frac{7}{4}

  4. ((d))

    95\frac{9}{5}

Show Answer
Answer: ((b))

32\frac{3}{2}

Calculation:

f(x)=x(x1) f(x) = x(x - 1)

a=1,b=2 a = 1, \quad b = 2

f(1)=1(11)=0 f(1) = 1(1 - 1) = 0

f(2)=2(21)=2 f(2) = 2(2 - 1) = 2

f(c)=f(2)f(1)21=201=2 \Rightarrow f'(c) = \frac{f(2) - f(1)}{2 - 1} = \frac{2 - 0}{1} = 2

f(x)=x2xf(x)=2x1 f(x) = x^2 - x \Rightarrow f'(x) = 2x - 1

2c1=2c=32 \Rightarrow 2c - 1 = 2 \Rightarrow c = \frac{3}{2}

Hence, the correct answer is Option 2

49

If x=a(cost+tsint)andy=a(sinttcost),x=a(cost+tsint) and y=a(sint−tcost), then the value of d2ydx2\frac{d^2y}{dx^2}

  1. ((a))

    tasec3t\frac{t}{a} \text{sec}^3 t

  2. ((b))

    atsec3tat \text{sec}^3 t

  3. ((c))

    1asec3tt\frac{1}{a} \frac{\text{sec}^3 t}{t}

  4. ((d))

    asec3tt\frac{{a}\text{sec}^3 t}{t}

Show Answer
Answer: ((c))

1asec3tt\frac{1}{a} \frac{\text{sec}^3 t}{t}

Calculation:

x=a(cost+tsint),y=a(sinttcost) x = a(\cos t + t \sin t), \quad y = a(\sin t - t \cos t)

dxdt=a(sint+sint+tcost)=atcost \frac{dx}{dt} = a(-\sin t + \sin t + t \cos t) = a t \cos t

dydt=a(cost(costtsint))=atsint \frac{dy}{dt} = a(\cos t - (\cos t - t \sin t)) = a t \sin t

dydx=dydtdxdt=atsintatcost=tant \Rightarrow \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{a t \sin t}{a t \cos t} = \tan t

d2ydx2=ddt(tant)÷dxdt=sec2tatcost \Rightarrow \frac{d^2y}{dx^2} = \frac{d}{dt}(\tan t) \div \frac{dx}{dt} = \frac{\sec^2 t}{a t \cos t}

=1atcostsec2t=sec3tat = \frac{1}{a t \cos t} \cdot \sec^2 t = \frac{\sec^3 t}{a t}

Hence, the correct answer is Option 3.

50

If limx0xsin(1x)=A \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) =A  and limxxsin(1x)=B \lim_{x \to \infty} x \sin\left(\frac{1}{x}\right) = B  then which of the following is true?

  1. ((a))

    A=B=0A=B=0

  2. ((b))

    A=0 and B=A=0 \text{ and } B=\infty

  3. ((c))

    A=1 and B=A=1 \text{ and } B=\infty

  4. ((d))

    A=0 and B=1A=0 \text{ and } B=1

Show Answer
Answer: ((d))

A=0 and B=1A=0 \text{ and } B=1

Calculation:

limx0xsin(1x) \lim_{x \to 0} x \sin\left(\frac{1}{x}\right)

xxsin(1x)x \Rightarrow -x \leq x \sin\left(\frac{1}{x}\right) \leq x

limx0xsin(1x)=0 \Rightarrow \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) = 0

A=0 \therefore A = 0

limxxsin(1x) \lim_{x \to \infty} x \sin\left(\frac{1}{x}\right)

sin(1x)1xas x \Rightarrow \sin\left(\frac{1}{x}\right) \approx \frac{1}{x} \quad \text{as } x \to \infty

xsin(1x)x1x=1 \Rightarrow x \sin\left(\frac{1}{x}\right) \approx x \cdot \frac{1}{x} = 1

limxxsin(1x)=1 \Rightarrow \lim_{x \to \infty} x \sin\left(\frac{1}{x}\right) = 1

B=1 \therefore B = 1

Hence, the correct answer is Option (d)

51

The solution of the differential equation  (x+2y3)dydx=y, y(0)=1(x+2y^3)\frac{dy}{dx}=y\textbf{, } y(0)=1 is

  1. ((a))

    x+yy3=0x+y-y^3=0

  2. ((b))

    xy+y3=0x-y+y^3=0

  3. ((c))

    x+2y2y3=0-x+2y-2y^3=0

  4. ((d))

    x+2y2y3=0x+2y-2y^3=0

Show Answer
Answer: ((a))

x+yy3=0x+y-y^3=0

Calculation:

Given, (x+2y3)dydx=y,y(0)=1 (x + 2y^3) \frac{dy}{dx} = y, \quad y(0) = 1

Rewrite the equation as: dydx=yx+2y3 \frac{dy}{dx} = \frac{y}{x + 2y^3}

Invert both sides to get:

⇒ dxdy=x+2y3y=xy+2y2 \frac{dx}{dy} = \frac{x + 2y^3}{y} = \frac{x}{y} + 2y^2

Substitute x=uy x = uy , so

⇒ dxdy=u+ydudy \frac{dx}{dy} = u + y \frac{du}{dy}

Plug into the equation:

⇒ u+ydudy=u+2y2 u + y \frac{du}{dy} = u + 2y^2

Cancel u u on both sides:

⇒ ydudy=2y2 y \frac{du}{dy} = 2y^2

Divide both sides by y y dudy=2y \frac{du}{dy} = 2y

Integrate both sides:

⇒ du=2y,dyu=y2+C \int du = \int 2y , dy \Rightarrow u = y^2 + C

Recall substitution x=uy x = uy :

⇒ x=y(y2+C)=y3+Cy x = y(y^2 + C) = y^3 + Cy

Rewrite as: xy3=Cy x - y^3 = Cy

Apply initial condition y(0)=1 y(0) = 1 :

⇒ 01=C×1C=1 0 - 1 = C \times 1 \Rightarrow C = -1

Substitute back C=1 C = -1 :

 xy3=yx+yy3=0 x - y^3 = -y \Rightarrow x + y - y^3 = 0

∴ the solution of the differential equation is x+yy3=0 x + y - y^3 = 0 .

Hence, the correct answer is Option (a).

52

limx0+e1x1e1x+1\lim_{x \to 0^+} \frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}  is equal to

  1. ((a))

    -1

  2. ((b))

    1

  3. ((c))

    0

  4. ((d))

    2

Show Answer
Answer: ((b))

1

Calculation:

Given, limx0+e1/x1e1/x+1 \lim_{x \to 0^+} \frac{e^{1/x} - 1}{e^{1/x} + 1}

As x0+ x \to 0^+ , we have:

1x+e1/x \frac{1}{x} \to +\infty \Rightarrow e^{1/x} \to \infty

So the expression becomes:

e1/x1e1/x+11+1= \frac{e^{1/x} - 1}{e^{1/x} + 1} \approx \frac{\infty - 1}{\infty + 1} = \frac{\infty}{\infty}

Now divide numerator and denominator by e1/x e^{1/x} :

11e1/x1+1e1/x \Rightarrow \frac{1 - \frac{1}{e^{1/x}}}{1 + \frac{1}{e^{1/x}}}

As e1/x1e1/x0 e^{1/x} \to \infty \Rightarrow \frac{1}{e^{1/x}} \to 0 , so:

101+0=11=1 \Rightarrow \frac{1 - 0}{1 + 0} = \frac{1}{1} = 1

Hence, the correct answer is Option (b)

53

The function  ϕ(x)=(xa)m(xb)n \phi(x) = (x - a)^m (x - b)^n on the interval [a,b] satisfied the conditions of  Rolle's Theorem, when

  1. ((a))

    m,n are positive integers

  2. ((b))

    m,n are positive integers and a<b

  3. ((c))

    a<b

  4. ((d))

    m>n

Show Answer
Answer: ((b))

m,n are positive integers and a<b

Calculation:

Given, ϕ(x)=(xa)m(xb)n \phi(x) = (x - a)^m (x - b)^n

We need to determine when this function satisfies the conditions of Rolle's Theorem.

Rolle's Theorem states: A function f(x) f(x) satisfies the theorem on the interval [a,b][a, b] if:

The function is continuous on [a,b][a, b]

The function is differentiable on (a,b)(a, b)

f(a)=f(b) f(a) = f(b)

Let’s verify these for ϕ(x) \phi(x) :

  1. ϕ(x) \phi(x) is a polynomial function ⟶ continuous and differentiable everywhere.
  2. ϕ(a)=(aa)m(ab)n=0 \phi(a) = (a - a)^m (a - b)^n = 0
  3. ϕ(b)=(ba)m(bb)n=0 \phi(b) = (b - a)^m (b - b)^n = 0

So, ϕ(a)=ϕ(b)=0 \phi(a) = \phi(b) = 0

Hence, all conditions of Rolle’s Theorem are satisfied if:

m,n m, n are positive integers (to define powers properly)

a<b a < b to define a valid interval [a,b][a, b]

Hence, the correct answer is Option (b).

54

Let  f:RRf: R \to R  be a differentiable function such that  f(x2)=4x21f'(x^2)=4x^2-1  for x>0x>0  and  f(1)=1f(1)=1 . Then f(4)f(4)  is

  1. ((a))

    64

  2. ((b))

    30

  3. ((c))

    42

  4. ((d))

    28

Show Answer
Answer: ((d))

28

Calculation:

f(x2)=4x21 f'(x^2) = 4x^2 - 1

Let t=x2x=t t = x^2 \Rightarrow x = \sqrt{t}

Then, f(t)=4t1 f'(t) = 4t - 1

Integrating both sides:

f(t)=(4t1)dt=2t2t+C f(t) = \int (4t - 1) dt = 2t^2 - t + C

Using the initial condition f(1)=1 f(1) = 1 :

1=2(1)21+C1=21+CC=0 1 = 2(1)^2 - 1 + C \Rightarrow 1 = 2 - 1 + C \Rightarrow C = 0

So, f(t)=2t2t f(t) = 2t^2 - t

Now, find f(4) f(4) :

f(4)=2(4)24=2164=324=28 f(4) = 2(4)^2 - 4 = 2 \cdot 16 - 4 = 32 - 4 = 28

Hence, the correct answer is Option 4.

55

If y=xx...to infinity, then xdydxy = x^{x^{...^{\text{to infinity}}}}\textbf{, then }x \frac{dy}{dx} is equal to

  1. ((a))

    y2yxlogex\frac{y^2}{y-x\log_e x}

  2. ((b))

    y2xylogex\frac{y^2}{x-y\log_e x}

  3. ((c))

    y21ylogex\frac{y^2}{1-y\log_e x}

  4. ((d))

    y2ylogex1\frac{y^2}{y\log_e x-1}

Show Answer
Answer: ((c))

y21ylogex\frac{y^2}{1-y\log_e x}

Calculation:

y=xxx y = x^{x^{x^{\cdots}}}

y=xy \Rightarrow y = x^y

lny=ylnx \Rightarrow \ln y = y \ln x

1ydydx=dydxlnx+yx \Rightarrow \frac{1}{y} \cdot \frac{dy}{dx} = \frac{dy}{dx} \ln x + \frac{y}{x}

dydx=ydydxlnx+y2x \Rightarrow \frac{dy}{dx} = y \cdot \frac{dy}{dx} \ln x + \frac{y^2}{x}

dydx(1ylnx)=y2x \Rightarrow \frac{dy}{dx}(1 - y \ln x) = \frac{y^2}{x}

xdydx=y21ylnx \Rightarrow x \cdot \frac{dy}{dx} = \frac{y^2}{1 - y \ln x}

Hence, the correct answer is Option (c).

56

If x=t,y=loge(cost),x=t, y=\log_e(\cos t), t[0,π4]t \in \left[0, \frac{\pi}{4}\right] , then the value of 0π4(dxdt)2+(dydt)2dt\int_0^{\frac{\pi}{4}} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt  is

  1. ((a))

    loge(2+1)\log_e(\sqrt{2}+1)

  2. ((b))

    loge(21)\log_e(\sqrt{2}-1)

  3. ((c))

    2loge(2+1)\sqrt{2}\log_e(\sqrt{2}+1)

  4. ((d))

    2loge(21)\sqrt{2}\log_e(\sqrt{2}-1)

Show Answer
Answer: ((a))

loge(2+1)\log_e(\sqrt{2}+1)

Calculation:

Given: x=t,y=loge(cost),t[0,π4] x = t,\quad y = \log_e(\cos t),\quad t \in \left[ 0, \frac{\pi}{4} \right]

We are to evaluate:

0π4(dxdt)2+(dydt)2,dt \int_0^{\frac{\pi}{4}} \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } , dt

Since x=tdxdt=1 \text{Since } x = t \Rightarrow \frac{dx}{dt} = 1

⇒ y=loge(cost)dydt=ddt[log(cost)]=sintcost=tant y = \log_e(\cos t) \Rightarrow \frac{dy}{dt} = \frac{d}{dt}[\log(\cos t)] = \frac{-\sin t}{\cos t} = -\tan t

Substituting in the integral:

⇒ 0π41+tan2t,dt \int_0^{\frac{\pi}{4}} \sqrt{1 + \tan^2 t} , dt

Using identity: 1+tan2t=sec2t1+tan2t=sect 1 + \tan^2 t = \sec^2 t \Rightarrow \sqrt{1 + \tan^2 t} = \sec t

Thus, the integral becomes:

⇒ 0π4sect,dt \int_0^{\frac{\pi}{4}} \sec t , dt

We know: sect,dt=logesect+tant+C \int \sec t , dt = \log_e|\sec t + \tan t| + C

0π4sect,dt=[loge(sect+tant)]0π4 \Rightarrow \int_0^{\frac{\pi}{4}} \sec t , dt = \left[ \log_e(\sec t + \tan t) \right]_0^{\frac{\pi}{4}}

At t=π4t = \frac{\pi}{4}  sect=2, tant=1loge(2+1)\sec t = \sqrt{2},\ \tan t = 1 \Rightarrow \log_e(\sqrt{2} + 1)

At t = 0 sect=1, tant=0loge(1)=0\sec t = 1,\ \tan t = 0 \Rightarrow \log_e(1) = 0

Final Answer=loge(2+1) \therefore \text{Final Answer} = \log_e(\sqrt{2} + 1)

Hence, the correct answer is Option (a):

57

The value of  limn[1n+1+1n+2++16n]\lim_{n \to \infty} \left[\frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{6n}\right]

  1. ((a))

    0

  2. ((b))

    loge2\log_e 2

  3. ((c))

    loge3\log_e 3

  4. ((d))

    loge6\log_e 6

Show Answer
Answer: ((d))

loge6\log_e 6

Calculation:

We are given: limn(1n+1+1n+2++16n) \lim_{n \to \infty} \left( \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{6n} \right)

Rewrite the sum as: r=15n1n+r \sum_{r = 1}^{5n} \frac{1}{n + r}

Divide numerator and denominator by n n :

=r=15n1/n1+rn = \sum_{r = 1}^{5n} \frac{1/n}{1 + \frac{r}{n}}

Now, let rn=x \frac{r}{n} = x . As r r goes from 1 1 to 5n 5n , x x goes from 1n \frac{1}{n} to 5 5 .

Hence, the expression becomes a Riemann sum:

limnr=15n1/n1+rn=0511+x,dx \lim_{n \to \infty} \sum_{r=1}^{5n} \frac{1/n}{1 + \frac{r}{n}} = \int_0^5 \frac{1}{1 + x} , dx

Now compute the integral:

0511+x,dx=[loge(1+x)]05=loge(6)loge(1)=loge(6) \int_0^5 \frac{1}{1 + x} , dx = \left[ \log_e(1 + x) \right]_0^5 = \log_e(6) - \log_e(1) = \log_e(6)

Hence, the correct answer is Option (d).

58

limn(1+sinan)n\lim_{n \to \infty} \left(1+\sin\frac{a}{n}\right)^n is equal to

  1. ((a))

    ee

  2. ((b))

    eae^a

  3. ((c))

    e2ae^{2a}

  4. ((d))

    0

Show Answer
Answer: ((b))

eae^a

Calculation:

We are given: limn(1+sinan)n \lim_{n \to \infty} \left( 1 + \sin \frac{a}{n} \right)^n

As n n \to \infty an0 \frac{a}{n} \to 0 so we use the approximation:

sinanan \sin \frac{a}{n} \approx \frac{a}{n}

Thus, the expression becomes:

limn(1+an)n \lim_{n \to \infty} \left( 1 + \frac{a}{n} \right)^n

This is a standard exponential limit form:

limn(1+kn)n=ek \lim_{n \to \infty} \left( 1 + \frac{k}{n} \right)^n = e^k

Here, k=aLimit=ea k = a \Rightarrow \text{Limit} = e^a

Hence, the correct answer is Option (b)

59

The value of 01000ex[x]dx\int_0^{1000} e^{x - [x]} dx

  1. ((a))

    e10001e^{1000}-1

  2. ((b))

    e10001e1\frac{e^{1000}-1}{e-1}

  3. ((c))

    1000(e1)1000(e-1)

  4. ((d))

    e11000\frac{e-1}{1000}

Show Answer
Answer: ((c))

1000(e1)1000(e-1)

Calculation:

Given, 01000exx,dx \int_0^{1000} e^{x - \lfloor x \rfloor} , dx

Let f(x)=exx f(x) = e^{x - \lfloor x \rfloor} . This function is periodic because the fractional part xx x - \lfloor x \rfloor repeats every 1 unit.

So, f(x) f(x) has period T=1 T = 1 .

Using the periodicity property of definite integrals:

⇒ 0nTf(x),dx=n0Tf(x),dx \int_0^{nT} f(x) , dx = n \int_0^T f(x) , dx

Here, n=1000 n = 1000 and T=1 T = 1

01000exx,dx=100001exx,dx \int_0^{1000} e^{x - \lfloor x \rfloor} , dx = 1000 \int_0^1 e^{x - \lfloor x \rfloor} , dx

On the interval x[0,1) x \in [0, 1) , we know x=0 \lfloor x \rfloor = 0

exx=ex e^{x - \lfloor x \rfloor} = e^x

So,

⇒ 01exx,dx=01ex,dx=ex01=e1 \int_0^1 e^{x - \lfloor x \rfloor} , dx = \int_0^1 e^x , dx = e^x \big|_0^1 = e - 1

⇒ 01000exx,dx=1000(e1) \int_0^{1000} e^{x - \lfloor x \rfloor} , dx = 1000(e - 1)

Hence, the correct answer is Option (c).

60

The value of x2exdx\int x^2 e^x dx is

  1. ((a))

    2ex+c2e^x+c

  2. ((b))

    (x2+2)ex+c(x^2+2)e^x+c

  3. ((c))

    (x2+2x+2)ex+c(x^2+2x+2)e^x+c

  4. ((d))

    (x22x+2)ex+c(x^2-2x+2)e^x+c

Show Answer
Answer: ((d))

(x22x+2)ex+c(x^2-2x+2)e^x+c

Calculation:

Given, x2ex,dx \int x^2 e^x , dx

Use integration by parts: u,dv=uvv,du \int u , dv = uv - \int v , du

Let u=x2du=2x,dx u = x^2 \Rightarrow du = 2x , dx and

dv=exdxv=ex dv = e^x dx \Rightarrow v = e^x

⇒ x2ex,dx=x2ex2xex,dx \int x^2 e^x , dx = x^2 e^x - \int 2x e^x , dx

Now, apply integration by parts again on 2xex,dx \int 2x e^x , dx

Let u=2xdu=2,dx u = 2x \Rightarrow du = 2 , dx , dv=exdxv=ex dv = e^x dx \Rightarrow v = e^x

⇒ 2xex,dx=2xex2exdx=2xex2ex \int 2x e^x , dx = 2x e^x - \int 2 e^x dx = 2x e^x - 2e^x

Substitute back:

⇒ x2ex,dx=x2ex(2xex2ex) \int x^2 e^x , dx = x^2 e^x - (2x e^x - 2e^x)

⇒ =x2ex2xex+2ex = x^2 e^x - 2x e^x + 2e^x

⇒ =(x22x+2)ex+C = (x^2 - 2x + 2) e^x + C

Hence, the correct answer is Option (d).

61

The value of 0xdx(1+x)(1+x2)\int_0^\infty \frac{xdx}{(1+x)(1+x^2)} is

  1. ((a))

    π2\frac{\pi}{2}

  2. ((b))

    π4\frac{\pi}{4}

  3. ((c))

    π3\frac{\pi}{3}

  4. ((d))

    π8\frac{\pi}{8}

Show Answer
Answer: ((b))

π4\frac{\pi}{4}

Calculation:

Given, 0x,dx(1+x)(1+x2) \int_0^\infty \frac{x , dx}{(1 + x)(1 + x^2)}

Let this integral be I I .

Use substitution: x=1tdx=1t2dt x = \frac{1}{t} \Rightarrow dx = -\frac{1}{t^2} dt

Change the limits accordingly: as x0t x \to 0 \Rightarrow t \to \infty and xt0 x \to \infty \Rightarrow t \to 0

So, I=0x(1+x)(1+x2),dx=01t(1+1t)(1+1t2)(1t2)dt I = \int_0^\infty \frac{x}{(1 + x)(1 + x^2)} , dx = \int_\infty^0 \frac{\frac{1}{t}}{\left(1 + \frac{1}{t}\right)\left(1 + \frac{1}{t^2}\right)} \cdot \left(-\frac{1}{t^2}\right) dt

Simplify numerator: 1t3 -\frac{1}{t^3}

Simplify denominator: (1+1t)(1+1t2)=(t+1)(t2+1)t3 \left(1 + \frac{1}{t}\right)\left(1 + \frac{1}{t^2}\right) = \frac{(t + 1)(t^2 + 1)}{t^3}

Therefore, the integrand becomes: 1t3t3(t+1)(t2+1)=1(t+1)(t2+1) \frac{-1}{t^3} \cdot \frac{t^3}{(t + 1)(t^2 + 1)} = \frac{-1}{(t + 1)(t^2 + 1)}

Flip the limits to remove the negative:

⇒ I=01(x+1)(x2+1)dx I = \int_0^\infty \frac{1}{(x + 1)(x^2 + 1)} dx

Add the original and transformed integrals:

⇒ 2I=0(x(1+x)(1+x2)+1(x+1)(x2+1))dx 2I = \int_0^\infty \left( \frac{x}{(1 + x)(1 + x^2)} + \frac{1}{(x + 1)(x^2 + 1)} \right) dx

Take LCM and simplify the integrand:

⇒ x+1(1+x)(1+x2)=11+x2 \frac{x + 1}{(1 + x)(1 + x^2)} = \frac{1}{1 + x^2}

⇒ 2I=011+x2dx=[tan1x]0=π2 2I = \int_0^\infty \frac{1}{1 + x^2} dx = \left[ \tan^{-1} x \right]_0^\infty = \frac{\pi}{2}

I=π4 \Rightarrow I = \frac{\pi}{4}

Hence, the correct answer is Option (b).

62

If u=(x2+y2)12 u=(x^2+y^2)^{\frac{1}{2}} and  x3+y3+3axy=5a2x^3+y^3+3axy=5a^2 then the value of  dudx\frac{du}{dx}  at  (a.a)  is

  1. ((a))

    aa

  2. ((b))

    a2a^2

  3. ((c))

    3a23a^2

  4. ((d))

    None of these

Show Answer
Answer: ((d))

None of these

Given:

u=x2+y2=(x2+y2)12 u = \sqrt{x^2 + y^2} = (x^2 + y^2)^{\frac{1}{2}}

x3+y3+3axy=5a2 x^3 + y^3 + 3 a x y = 5 a^2

⇒ dudx=x+ydydxx2+y2=x+ydydxu \frac{du}{dx} = \frac{x + y \frac{dy}{dx}}{\sqrt{x^2 + y^2}} = \frac{x + y \frac{dy}{dx}}{u}

⇒ 3x2+3y2dydx+3a(y+xdydx)=0 3x^2 + 3 y^2 \frac{dy}{dx} + 3 a \left( y + x \frac{dy}{dx} \right) = 0

3y2dydx+3axdydx=3x23ay \Rightarrow 3 y^2 \frac{dy}{dx} + 3 a x \frac{dy}{dx} = -3 x^2 - 3 a y

dydx(3y2+3ax)=3x23ay \Rightarrow \frac{dy}{dx} (3 y^2 + 3 a x) = -3 x^2 - 3 a y

dydx=(x2+ay)y2+ax \Rightarrow \frac{dy}{dx} = \frac{-(x^2 + a y)}{y^2 + a x}

⇒ dydx=(a2+aa)a2+aa=2a22a2=1 \frac{dy}{dx} = \frac{-(a^2 + a \cdot a)}{a^2 + a \cdot a} = \frac{-2 a^2}{2 a^2} = -1

Evaluate u  at (a, a)

u=a2+a2=a2 u = \sqrt{a^2 + a^2} = a \sqrt{2}

Calculate  dudx\frac{du}{dx}  at (a, a) 

dudx=a+a(1)a2=0a2=0 \frac{du}{dx} = \frac{a + a (-1)}{a \sqrt{2}} = \frac{0}{a \sqrt{2}} = 0

Hence, the value of dudx \frac{du}{dx} at (a, a) is 0

Hence, the correct answer is Option d.

63

The solution of the differential equation 1x2dy+1y2dx=0 \sqrt{1-x^2} dy + \sqrt{1-y^2} dx = 0  (x<1,y<1)(|x|<1, |y|<1) is

  1. ((a))

    x1y2+y1x2=cx\sqrt{1-y^2}+y\sqrt{1-x^2}=c

  2. ((b))

    sin1y+sin1x=c\sin^{-1} y + \sin^{-1} x = c

  3. ((c))

    x21x2+y21y2=c\frac{x^2}{\sqrt{1-x^2}} + \frac{y^2}{\sqrt{1-y^2}} = c

  4. ((d))

    1x2+1y2=c\sqrt{1-x^2}+\sqrt{1-y^2}=c

Show Answer
Answer: ((b))

sin1y+sin1x=c\sin^{-1} y + \sin^{-1} x = c

Calculation:

Given the differential equation:

1x2,dy+1y2,dx=0 \sqrt{1 - x^2} , dy + \sqrt{1 - y^2} , dx = 0

Rewriting the equation:

1x2,dy=1y2,dx \sqrt{1 - x^2} , dy = -\sqrt{1 - y^2} , dx

Separate the variables:

dy1y2=dx1x2 \frac{dy}{\sqrt{1 - y^2}} = -\frac{dx}{\sqrt{1 - x^2}}

Integrating both sides:

dy1y2=dx1x2 \int \frac{dy}{\sqrt{1 - y^2}} = -\int \frac{dx}{\sqrt{1 - x^2}}

Using the standard integral:

sin1y=sin1x+C \sin^{-1} y = -\sin^{-1} x + C

Rewriting:

sin1x+sin1y=C \sin^{-1} x + \sin^{-1} y = C

Hence, the correct solution option 2.

64

If  u=logex3+y3x+yu=\log_e\frac{x^3+y^3}{x+y}  , then the value of xux+yuyx\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y} is

  1. ((a))

    uu

  2. ((b))

    2

  3. ((c))

    0

  4. ((d))

    u+1u+1

Show Answer
Answer: ((b))

2

Calculation:

Let f(x,y)=x3+y3x+y f(x, y) = \frac{x^3 + y^3}{x + y} , so that u=logf(x,y) u = \log f(x, y) .

 

Note that:

Numerator x3+y3 x^3 + y^3 is homogeneous of degree 3

Denominator x+y x + y is homogeneous of degree 1

Therefore, f(x,y) f(x, y) is homogeneous of degree 2.

By Euler’s Theorem:

xfx+yfy=2f x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = 2f

Now using chain rule on u=logf u = \log f :

xux+yuy=1f(xfx+yfy)=1f2f=2 x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{1}{f}(x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y}) = \frac{1}{f} \cdot 2f = 2

Hence, the correct answer is Option (b)

65

If x+2y=8x + 2y = 8 then the maximum value of xy is

  1. ((a))

    20

  2. ((b))

    16

  3. ((c))

    24

  4. ((d))

    8

Show Answer
Answer: ((d))

8

Calculation:

Given, x+2y=8 x + 2y = 8

We are required to find the maximum value of xy xy .

Using the AM ≥ GM inequality:

⇒ x+2y22xy \frac{x + 2y}{2} \geq \sqrt{2xy}

Substitute the given constraint x+2y=8 x + 2y = 8 :

⇒ 822xy42xy \frac{8}{2} \geq \sqrt{2xy} \Rightarrow 4 \geq \sqrt{2xy}

Square both sides:

⇒ 162xyxy8 16 \geq 2xy \Rightarrow xy \leq 8

So, the maximum value of xy xy is 8.

This value is achieved when equality holds in AM ≥ GM, i.e., when:

x=2y x = 2y

Substitute into the constraint:

x+2y=82y+2y=8y=2, x=4 x + 2y = 8 \Rightarrow 2y + 2y = 8 \Rightarrow y = 2, \ x = 4

Then xy=4×2=8 xy = 4 \times 2 = 8

Hence, the correct answer is Optin 4.

66

The equation of the tangent at θ=π2\theta=\frac{\pi}{2} to the curve x=a(θ+sinθ)x=a(\theta+\sin\theta)y=a(1+cosθ)y=a(1+\cos\theta) is

  1. ((a))

    xy=a(π2+2)x−y=a(\frac{π}{2}+2)

  2. ((b))

    xy=aπ2x−y=\frac{aπ}{2}

  3. ((c))

    x+y=a(π2+2)x+y=a(\frac{π}{2}+2)

  4. ((d))

    x+y=aπ2x+y=\frac{aπ}{2}

Show Answer
Answer: ((c))

x+y=a(π2+2)x+y=a(\frac{π}{2}+2)

Calculation:

Given,

x=a(θ+sinθ),y=a(1+cosθ) x = a(\theta + \sin\theta), \quad y = a(1 + \cos\theta)

We need to find the equation of the tangent at θ=π2 \theta = \frac{\pi}{2} .

Differentiate both parametric equations with respect to θ \theta :

⇒ dxdθ=a(1+cosθ),dydθ=a(sinθ) \frac{dx}{d\theta} = a(1 + \cos\theta), \quad \frac{dy}{d\theta} = a(-\sin\theta)

Now compute dydx \frac{dy}{dx} using the chain rule:

⇒ dydx=dydθdxdθ=asinθa(1+cosθ)=sinθ1+cosθ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-a\sin\theta}{a(1 + \cos\theta)} = \frac{-\sin\theta}{1 + \cos\theta}

Substitute θ=π2 \theta = \frac{\pi}{2} :

⇒ sinπ2=1,cosπ2=0 \sin\frac{\pi}{2} = 1, \quad \cos\frac{\pi}{2} = 0

dydx=11+0=1 \Rightarrow \frac{dy}{dx} = \frac{-1}{1 + 0} = -1

Now find the point of tangency:

⇒ x=a(π2+sinπ2)=a(π2+1) x = a\left(\frac{\pi}{2} + \sin\frac{\pi}{2}\right) = a\left(\frac{\pi}{2} + 1\right)

⇒ y=a(1+cosπ2)=a(1+0)=a y = a(1 + \cos\frac{\pi}{2}) = a(1 + 0) = a

Now use the point-slope form of the line:

⇒ ya=1(xa(π2+1)) y - a = -1 \cdot \left(x - a\left(\frac{\pi}{2} + 1\right)\right)

Simplify:

⇒ ya=x+a(π2+1) y - a = -x + a\left(\frac{\pi}{2} + 1\right)

x+y=a(π2+2) \Rightarrow x + y = a\left(\frac{\pi}{2} + 2\right)

Hence, the correct answer is Option (c)

67

The area bounded by the curves y=x1y = |x|-1 and y=x+1y =- |x|+1

  1. ((a))

    1

  2. ((b))

    2

  3. ((c))

    2\sqrt{2}

  4. ((d))

    4

Show Answer
Answer: ((b))

2

Calculation:

Given,

y=x1 y = |x| - 1

y=x+1 y = -|x| + 1

These two V-shaped curves intersect at:

x1=x+12x=2x=1x=±1 |x| - 1 = -|x| + 1 \Rightarrow 2|x| = 2 \Rightarrow |x| = 1 \Rightarrow x = \pm1

So the region of interest lies between x=1 x = -1 and x=1 x = 1 .

From symmetry, we calculate the area in the first quadrant and double it:

Area=201[(x+1)(x1)]dx \text{Area} = 2 \int_0^1 \left[(-x + 1) - (x - 1)\right] dx

=201(2x+2)dx = 2 \int_0^1 (-2x + 2) dx

Integrating:

=2[x2+2x]01=2(1+2)=2 = 2 \left[ -x^2 + 2x \right]_0^1 = 2(-1 + 2) = 2

Hence, the correct answer is Option (2)

68

The slope of the tangent at the point P(x,y) on a curve is  y+3x+2-\frac{y+3}{x+2} .If the curve passes through the origin, then the equation of the curve is

  1. ((a))

    xy+2y+3x=0xy+2y+3x=0

  2. ((b))

    x2y2+2x3y=0x ^2 −y ^2 +2x−3y=0

  3. ((c))

    xy+6x=0xy+6x=0

  4. ((d))

    xy2y+3x=0 xy−2y+3x=0

Show Answer
Answer: ((a))

xy+2y+3x=0xy+2y+3x=0

Calculation:

Given,

The slope of the tangent to the curve at point (x,y) (x, y) is:

⇒ dydx=y+3x+2 \frac{dy}{dx} = -\frac{y + 3}{x + 2}

We solve this differential equation by separating the variables:

⇒ dyy+3=dxx+2 \frac{dy}{y + 3} = -\frac{dx}{x + 2}

Integrating both sides:

⇒ dyy+3=dxx+2 \int \frac{dy}{y + 3} = -\int \frac{dx}{x + 2}

⇒ logy+3=logx+2+logC \log|y + 3| = -\log|x + 2| + \log C

Using log properties:

⇒ logy+3=log(Cx+2) \log|y + 3| = \log \left( \frac{C}{x + 2} \right)

Exponentiating both sides:

⇒ y+3=Cx+2 |y + 3| = \frac{C}{|x + 2|}

Removing modulus by absorbing sign into the constant:

⇒ (y+3)(x+2)=C (y + 3)(x + 2) = C

The curve passes through the origin, so at (0,0) (0, 0) :

⇒ (0+3)(0+2)=C6=C (0 + 3)(0 + 2) = C \Rightarrow 6 = C

Substitute back:

⇒ (y+3)(x+2)=6 (y + 3)(x + 2) = 6

Expanding the equation:

⇒ xy+2y+3x+6=6xy+2y+3x=0 xy + 2y + 3x + 6 = 6 \Rightarrow xy + 2y + 3x = 0

Hence, the correct answer is Option (a).

69

If y(x)  is a solution of the differential equation dydx+2xy=x\frac{dy}{dx} + 2xy = x , y(0)=0y(0)=0 then limxy(x)\lim_{x \to \infty} y(x) is

  1. ((a))

    12-\frac{1}{2}

  2. ((b))

    -1

  3. ((c))

    12\frac{1}{2}

  4. ((d))

    1

Show Answer
Answer: ((c))

12\frac{1}{2}

Calculation:

The differential equation is:

dydx+2xy=x \frac{dy}{dx} + 2xy = x with initial condition y(0)=0 y(0) = 0

This is a first-order linear differential equation of the form:

dydx+P(x)y=Q(x) \frac{dy}{dx} + P(x)y = Q(x)

Where: P(x)=2x P(x) = 2x and Q(x)=x Q(x) = x

The integrating factor (IF) is:

⇒ IF=e2x,dx=ex2 IF = e^{\int 2x,dx} = e^{x^2}

Multiplying the entire equation by the integrating factor:

⇒ ex2dydx+2xex2y=xex2 e^{x^2} \cdot \frac{dy}{dx} + 2x e^{x^2} y = x e^{x^2}

The left-hand side becomes:

⇒ ddx(yex2)=xex2 \frac{d}{dx}(y \cdot e^{x^2}) = x e^{x^2}

Integrating both sides:

⇒ ddx(yex2),dx=xex2dx \int \frac{d}{dx}(y \cdot e^{x^2}),dx = \int x e^{x^2} dx

Let u=x2du=2xdxxdx=12du u = x^2 \Rightarrow du = 2x dx \Rightarrow x dx = \frac{1}{2} du

⇒ xex2dx=12eudu=12ex2 \int x e^{x^2} dx = \frac{1}{2} \int e^u du = \frac{1}{2} e^{x^2}

So, yex2=12ex2+C y \cdot e^{x^2} = \frac{1}{2} e^{x^2} + C

Divide both sides by ex2 e^{x^2} :

⇒ y=12+Cex2 y = \frac{1}{2} + C e^{-x^2}

Apply the initial condition y(0)=0 y(0) = 0 :

⇒ 0=12+CC=12 0 = \frac{1}{2} + C \Rightarrow C = -\frac{1}{2}

So the solution becomes:

⇒ y(x)=1212ex2 y(x) = \frac{1}{2} - \frac{1}{2} e^{-x^2}

Now take the limit as x x \to \infty :

⇒ limxy(x)=12120=12 \lim_{x \to \infty} y(x) = \frac{1}{2} - \frac{1}{2} \cdot 0 = \frac{1}{2}

Hence, the correct answer is Option (c).

70

If (2+sinx)y+1dydx=cosx \frac{(2 + \sin x)}{y + 1} \cdot \frac{dy}{dx} = -\cos x  and  y(0)=1y(0)=1  then y(π2)y\left(\frac{\pi}{2}\right)  is equal to

  1. ((a))

    1

  2. ((b))

    23\frac{2}{3}

  3. ((c))

    13-\frac{1}{3}

  4. ((d))

    13\frac{1}{3}

Show Answer
Answer: ((d))

13\frac{1}{3}

Calculation:

Given, (2+sinx)y+1dydx=cosx \frac{(2 + \sin x)}{y + 1} \cdot \frac{dy}{dx} = -\cos x with y(0)=1 y(0) = 1

Multiply both sides by y+1 y + 1 :

⇒ (2+sinx)dydx=(y+1)cosx (2 + \sin x) \cdot \frac{dy}{dx} = - (y + 1) \cos x

Separate the variables:

⇒ dyy+1=cosx2+sinxdx \frac{dy}{y + 1} = -\frac{\cos x}{2 + \sin x} dx

Integrate both sides:

⇒ dyy+1=cosx2+sinxdx \int \frac{dy}{y + 1} = \int -\frac{\cos x}{2 + \sin x} dx

Let u=2+sinxdu=cosx,dx u = 2 + \sin x \Rightarrow du = \cos x , dx

Then:

⇒ cosx2+sinxdx=1udu=ln2+sinx \int -\frac{\cos x}{2 + \sin x} dx = -\int \frac{1}{u} du = -\ln|2 + \sin x|

Left-hand side:

dyy+1=lny+1 \int \frac{dy}{y + 1} = \ln|y + 1|

So, lny+1=ln2+sinx+C \ln|y + 1| = -\ln|2 + \sin x| + C

Apply log properties:

⇒ lny+1+ln2+sinx=Cln(y+1)(2+sinx)=C \ln|y + 1| + \ln|2 + \sin x| = C \Rightarrow \ln|(y + 1)(2 + \sin x)| = C

Exponentiate both sides:

⇒ (y+1)(2+sinx)=eC=K (y + 1)(2 + \sin x) = e^C = K

Apply the initial condition y(0)=1 y(0) = 1 :

⇒ (1+1)(2+sin0)=K22=4K=4 (1 + 1)(2 + \sin 0) = K \Rightarrow 2 \cdot 2 = 4 \Rightarrow K = 4

So the equation becomes:

⇒ (y+1)(2+sinx)=4 (y + 1)(2 + \sin x) = 4

Now compute y(π2) y\left( \frac{\pi}{2} \right) :

⇒ (y+1)(2+sin(π2))=4(y+1)(3)=4 (y + 1)(2 + \sin\left(\frac{\pi}{2}\right)) = 4 \Rightarrow (y + 1)(3) = 4

⇒ y+1=43y=13 y + 1 = \frac{4}{3} \Rightarrow y = \frac{1}{3}

Hence, the correct answer is Option (d).

71

The mean weight of 9 items is 15 kg. If one more item is added, the mean weight becomes 16 kg. Then the weight of the 10th item is

  1. ((a))

    35 kg

  2. ((b))

    30 kg

  3. ((c))

    25 kg

  4. ((d))

    20 kg

Show Answer
Answer: ((c))

25 kg

Calculation:

Given, The mean weight of 9 items is 15,kg 15 , \text{kg} .

Total weight of 9 items = 9×15=135,kg 9 \times 15 = 135 , \text{kg}

After adding one more item, the mean becomes 16,kg 16 , \text{kg} .

Total weight of 10 items = 10×16=160,kg 10 \times 16 = 160 , \text{kg}

So, weight of the 10th item = 160135=25,kg 160 - 135 = 25 , \text{kg}

Hence, the correct answer is Option (c).

72

If  P(A)=715, P(B)=815P(A) = \frac{7}{15}\textbf{, } P(B) = \frac{8}{15}  and P(A UB)=1115P(A \ U B) = \frac{11}{15} then P(A/B)P(A/B)

  1. ((a))

    38\frac{3}{8}

  2. ((b))

    12\frac{1}{2}

  3. ((c))

    78\frac{7}{8}

  4. ((d))

    58\frac{5}{8}

Show Answer
Answer: ((b))

12\frac{1}{2}

Formula used:

P(AB)=P(AB)P(B) P(A | B) = \frac{P(A ∩ B)}{P(B)}

P(A ∩ B) = P(A) + P(B) - P(A U B)

Calculation:

P(A ∩ B) = 7/15 + 8/15 - 11/15 = (7 + 8 - 11)/15 = 4/15

So,

P(A|B) = P(A ∩ B) / P(B) = (4/15) / (8/15) = 4/8 = 1/2

Hence, the correct answer is Option (b).

73

A coin is thrown 6 times. The probability of getting exactly four heads is

  1. ((a))

    14\frac{1}{4}

  2. ((b))

    34\frac{3}{4}

  3. ((c))

    516\frac{5}{16}

  4. ((d))

    1564\frac{15}{64}

Show Answer
Answer: ((d))

1564\frac{15}{64}

Calculation:

A coin is thrown 6 times and the probability of getting exactly 4 heads is to be calculated.

Using the Binomial Probability Formula:

P(X=r)=(nr)prqnr P(X = r) = \binom{n}{r} p^r q^{n-r}

Where,

n=6 n = 6 (number of trials)

r=4 r = 4 (number of heads)

p=12 p = \frac{1}{2} (probability of head)

q=1p=12 q = 1 - p = \frac{1}{2}

Substitute the values:

P(X=4)=(64)(12)4(12)2=(64)(12)6 P(X=4) = \binom{6}{4} \left(\frac{1}{2}\right)^4 \left(\frac{1}{2}\right)^2 = \binom{6}{4} \left(\frac{1}{2}\right)^6

Calculate the combination:

(64)=(62)=6×52×1=15 \binom{6}{4} = \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15

Therefore,

P(X=4)=15×164=1564 P(X=4) = 15 \times \frac{1}{64} = \frac{15}{64}

Hence, the correct answer is Option (d).

74

A bag contains 8 red and 5 white balls. Three balls are drawn at random. The probability that one ball is red and two balls are white, is

  1. ((a))

    40143\frac{40}{143}

  2. ((b))

    80146\frac{80}{146}

  3. ((c))

    10296\frac{10}{296}

  4. ((d))

    5296\frac{5}{296}

Show Answer
Answer: ((a))

40143\frac{40}{143}

Calculation:

Given, A bag contains 8 red and 5 white balls. Three balls are drawn at random.

We need to find the probability that exactly one ball is red and two balls are white.

Total number of balls = 8 + 5 = 13

Calculate the total number of ways to draw 3 balls:

(133)=13×12×113×2×1=286 \binom{13}{3} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286

Calculate the favorable outcomes:

Number of ways to choose 1 red ball from 8 red balls: (81)=8 \binom{8}{1} = 8

Number of ways to choose 2 white balls from 5 white balls: (52)=5×42×1=10 \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10

Total favorable outcomes = 8×10=80 8 \times 10 = 80

Therefore, the required probability is:

P=80286=40143 P = \frac{80}{286} = \frac{40}{143}

Hence, the correct answer is Option (a).

75

The mean of 1, 3, 4, 5, 7, 4 is n. The numbers 3, 2, 2, 4, 3, p, 3 have mean n−1 and median q. Then p+q is

  1. ((a))

    6

  2. ((b))

    4

  3. ((c))

    7

  4. ((d))

    5

Show Answer
Answer: ((c))

7

Calculation:

Calculate mean n:

 

n=1+3+4+5+7+46=246=4 n = \frac{1 + 3 + 4 + 5 + 7 + 4}{6} = \frac{24}{6} = 4

Calculate mean of the second set:

3+2+2+4+3+p+37=n1=41=3 \frac{3 + 2 + 2 + 4 + 3 + p + 3}{7} = n - 1 = 4 - 1 = 3

Simplify:

17+p7=3    17+p=21    p=4 \frac{17 + p}{7} = 3 \implies 17 + p = 21 \implies p = 4

Arrange the second set in ascending order to find median q:

2,2,3,3,3,4,4 2, 2, 3, 3, 3, 4, 4

The median q is the 4th value:

q=3 q = 3

Calculate p + q

p+q=4+3=7 p + q = 4 + 3 = 7

Hence, the correct answer is Option (c).

76

If a hyperbola, whose parametric equations are x=ct,y=ctx=ct, y=\frac{c}{t} meets any circle with centre at (0,0) in four points, determined by the parametric values t1,t2,t3t_1,t_2,t_3 and t4t_4​, then the value oft1t2t3t4t_1 t_2 t_3 t_4​ is

  1. ((a))

    c2c^2

  2. ((b))

    c2-c^2

  3. ((c))

    -1

  4. ((d))

    1

Show Answer
Answer: ((d))

1

Calculation:

Given, A hyperbola with parametric equations:

x=ct x = ct , y=ct y = \frac{c}{t}

It meets a circle centered at the origin in four points with parameter values t1,t2,t3,t4 t_1, t_2, t_3, t_4

We need to find the value of t1t2t3t4 t_1 t_2 t_3 t_4 .

Equation of the circle with radius r is:

x2+y2=r2 x^2 + y^2 = r^2

Substitute the parametric equations:

(ct)2+(ct)2=r2 (ct)^2 + \left(\frac{c}{t}\right)^2 = r^2

Simplify:

c2t2+c2t2=r2 c^2 t^2 + \frac{c^2}{t^2} = r^2

Multiply both sides by t2 t^2

c2t4r2t2+c2=0 c^2 t^4 - r^2 t^2 + c^2 = 0

Let z=t2 z = t^2 , so:

c2z2r2z+c2=0 c^2 z^2 - r^2 z + c^2 = 0

The roots of this quadratic arez1=t12 z_1 = t_1^2  and and  z2=t22z_2 = t_2^2

The product of the roots t1t2t3t4t_1 t_2 t_3 t_4 is:

(z1)(z1)(z2)(z2)=z1z2=c2c2=1 (\sqrt{z_1})(-\sqrt{z_1})(\sqrt{z_2})(-\sqrt{z_2}) = z_1 \cdot z_2 = \frac{c^2}{c^2} = 1

Hence, the correct answer is Option d.

77

The product of the perpendiculars drawn from the foci of an ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 on any tangent to it, is

  1. ((a))

    a2a^2

  2. ((b))

    b2b^2

  3. ((c))

    -1

  4. ((d))

    2

Show Answer
Answer: ((b))

b2b^2

Calculation:

Given,The ellipse equation: x2a2+y2b2=1 \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

The foci are at points:

(±c,0) ( \pm c, 0 ) , where c=a2b2 c = \sqrt{a^2 - b^2}

Let the equation of the tangent be:

⇒ y=mx+c1 y = mx + c_1

The perpendicular distance from a point (x0,y0) (x_0, y_0) to line Ax+By+C=0 Ax + By + C = 0 is:

⇒ d=Ax0+By0+CA2+B2 d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}

Distances from the foci (c,0) (c,0) and (c,0) (-c,0) to the tangent are:

⇒ d1=mc+c11+m2,d2=mc+c11+m2 d_1 = \frac{|mc + c_1|}{\sqrt{1 + m^2}}, \quad d_2 = \frac{|-mc + c_1|}{\sqrt{1 + m^2}}

Product of perpendiculars is:

d1d2=(mc+c1)(mc+c1)1+m2=c12m2c21+m2 d_1 d_2 = \frac{|(mc + c_1)(-mc + c_1)|}{1 + m^2} = \frac{|c_1^2 - m^2 c^2|}{1 + m^2}

Using the tangent condition of ellipse:

⇒ c12=a2m2+b2 c_1^2 = a^2 m^2 + b^2

Substituting c2=a2b2 c^2 = a^2 - b^2 , we get:

⇒ d1d2=a2m2+b2m2(a2b2)1+m2=b2(1+m2)1+m2=b2 d_1 d_2 = \frac{|a^2 m^2 + b^2 - m^2(a^2 - b^2)|}{1 + m^2} = \frac{b^2 (1 + m^2)}{1 + m^2} = b^2

Hence, the product of the perpendiculars drawn from the foci of the ellipse to any tangent is b2 b^2 .

78

Let y=mx+cy=mx+c be the equation of normal to the parabola y2=4ax y ^2 =4ax at  (am2,2am)(am^2,−2am). Then c is equal to

  1. ((a))

    am3am^3

  2. ((b))

    2am+am3−2am+am^3

  3. ((c))

    2am+am32am+am^3

  4. ((d))

    2amam3-2am-am^3

Show Answer
Answer: ((d))

2amam3-2am-am^3

Calculation:

Given, y2=4ax y^2 = 4ax

Equation of normal at point (am2,2am) (am^2, -2am) is y=mx+c y = mx + c

The slope of the tangent at (am2,2am) (am^2, -2am) is

⇒ dydx=2ay=2a2am=1m \frac{dy}{dx} = \frac{2a}{y} = \frac{2a}{-2am} = -\frac{1}{m}

The slope of the normal is the negative reciprocal of the tangent slope,

slope of normal=m \text{slope of normal} = m

Equation of normal passing through (am2,2am) (am^2, -2am) :

⇒ y(2am)=m(xam2) y - (-2am) = m(x - am^2)

Simplify the equation:

⇒ y+2am=mxam3 y + 2am = mx - am^3

⇒ y=mxam32am y = mx - am^3 - 2am

Comparing with y=mx+c y = mx + c , we get

⇒ c=am32am c = -am^3 - 2am

Hence, the correct answer is Option 4.

79

If the sum of the slopes of the lines x2−2λxy−7y2=0 is four times their product, then the value of λ is

  1. ((a))

    =1

  2. ((b))

    2

  3. ((c))

    -2

  4. ((d))

    1

Show Answer
Answer: ((b))

2

Calculation:

Given,

ax2+2hxy+by2=0 ax^2 + 2hxy + by^2 = 0

a=1,2h=2λh=λ,b=7 a = 1, \quad 2h = -2\lambda \Rightarrow h = -\lambda, \quad b = -7

The slopes m1 and m2 of the lines satisfy:

⇒ m1+m2=2hb m_1 + m_2 = -\frac{2h}{b}

⇒ m1m2=ab m_1 m_2 = \frac{a}{b}

Substituting the values:

⇒ m1+m2=2(λ)7=2λ7 m_1 + m_2 = -\frac{2(-\lambda)}{-7} = -\frac{2\lambda}{7}

⇒ m1m2=17=17 m_1 m_2 = \frac{1}{-7} = -\frac{1}{7}

Given condition:

⇒ m1+m2=4m1m2 m_1 + m_2 = 4 m_1 m_2

Substitute sum and product of slopes:

⇒ 2λ7=4×(17) -\frac{2\lambda}{7} = 4 \times \left(-\frac{1}{7}\right)

Simplify:

⇒ 2λ7=47 -\frac{2\lambda}{7} = -\frac{4}{7}

Multiply both sides by 7:

⇒ 2λ=4 -2\lambda = -4

Divide by -2:

⇒ λ=2 \lambda = 2

Hence, the correct answer is Option b.

80

The distance between the foci of a hyperbola is 16 units and its eccentricity is 2\sqrt{2} , Its equation is

  1. ((a))

    x2y2=32x^2−y^2=32

  2. ((b))

    2x2y2=322x ^2 −y ^2 =32

  3. ((c))

    x22y2=32x ^2 −2y ^2 =32

  4. ((d))

    3x23y2=323x ^2 −3y ^2 =32

Show Answer
Answer: ((a))

x2y2=32x^2−y^2=32

Calculation:

Distance between the foci, 2c=16 2c = 16 units

Eccentricity, e=2 e = \sqrt{2}

Calculate c c :

c=162=8 c = \frac{16}{2} = 8

Using the formula for eccentricity of hyperbola:

e=caa=ce=82=42 e = \frac{c}{a} \Rightarrow a = \frac{c}{e} = \frac{8}{\sqrt{2}} = 4\sqrt{2}

For a hyperbola, the relation between a a , b b , and c c is:

c2=a2+b2 c^2 = a^2 + b^2

Substitute values:

82=(42)2+b264=32+b2b2=32 8^2 = (4\sqrt{2})^2 + b^2 \Rightarrow 64 = 32 + b^2 \Rightarrow b^2 = 32

Equation of hyperbola is:

x2a2y2b2=1 \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1

Substitute a2=32 a^2 = 32 and b2=32 b^2 = 32 :

x232y232=1x2y2=32 \frac{x^2}{32} - \frac{y^2}{32} = 1 \Rightarrow x^2 - y^2 = 32

Hence, the correct answer is Option (a).

81

For what values of k, the line y=kx+2 will be tangent to the conic 4x2−9y2=36?

  1. ((a))

    ± 23\frac{2}{3}

  2. ((b))

    ± 223\frac{2\sqrt2}3

  3. ((c))

    ± 89\frac{8}9

  4. ((d))

    ± 423\frac{4\sqrt2}3

Show Answer
Answer: ((b))

± 223\frac{2\sqrt2}3

Calculation:

Line equation: y=kx+2 y = kx + 2

Conic equation: 4x29y2=36 4x^2 - 9y^2 = 36

We need to find values of k k for which the line is tangent to the conic.

Substitute y=kx+2 y = kx + 2 into conic equation:

⇒ 4x29(kx+2)2=36 4x^2 - 9(kx + 2)^2 = 36

⇒ 4x29(k2x2+4kx+4)=36 4x^2 - 9(k^2 x^2 + 4kx + 4) = 36

⇒ 4x29k2x236kx36=36 4x^2 - 9k^2 x^2 - 36kx - 36 = 36

⇒ (49k2)x236kx72=0 (4 - 9k^2) x^2 - 36kx - 72 = 0

For tangency, the quadratic in x x must have exactly one solution, so its discriminant D=0 D = 0 :

⇒ D=(36k)24(49k2)(72)=0 D = (-36k)^2 - 4 (4 - 9k^2)(-72) = 0

Calculate discriminant:

⇒ 1296k2+288(49k2)=0 1296 k^2 + 288 (4 - 9k^2) = 0

⇒ 1296k2+11522592k2=0 1296 k^2 + 1152 - 2592 k^2 = 0

⇒ 1296k2+1152=0 -1296 k^2 + 1152 = 0

⇒ 1296k2=1152k2=11521296=89 1296 k^2 = 1152 \Rightarrow k^2 = \frac{1152}{1296} = \frac{8}{9}

k=±223 k = \pm \frac{2 \sqrt{2}}{3}

Hence, the correct answer is Option (b).

82

The locus of the centres of circles, that passes through the origin and cuts off a length 6 from the line y=4, is

  1. ((a))

    x2−8y+25=0

  2. ((b))

    x2−8y−25=0

  3. ((c))

    x2+8y−25=0

  4. ((d))

    None of the above

Show Answer
Answer: ((c))

x2+8y−25=0

Calculation:

Given: Line: y=4 y = 4

The locus of the centers of circles passing through the origin and cutting off a length 6 from the line y = 4  is to be found.

Let the center of such a circle be (h, k) and radius  r .

The circle passes through the origin, so:h2+k2=r2 h^2 + k^2 = r^2

The circle cuts a chord of length 6 from the line y = 4 . The distance from the center to the line y = 4 is: k4 |k - 4|

Length of chord l = 6 , so the perpendicular distance from center to chord satisfies:

l=2r2(k4)26=2r2(k4)2 l = 2 \sqrt{r^2 - (k - 4)^2} \Rightarrow 6 = 2 \sqrt{r^2 - (k - 4)^2}

Simplify:

3=r2(k4)29=r2(k4)2 3 = \sqrt{r^2 - (k - 4)^2} \Rightarrow 9 = r^2 - (k - 4)^2

Substitute r2=h2+k2 r^2 = h^2 + k^2 :

9=h2+k2(k4)2 9 = h^2 + k^2 - (k - 4)^2

⇒ 9=h2+k2(k28k+16) 9 = h^2 + k^2 - (k^2 - 8k + 16)

⇒ 9=h2+k2k2+8k169=h2+8k16 9 = h^2 + k^2 - k^2 + 8k - 16 \Rightarrow 9 = h^2 + 8k - 16

⇒ h2+8k=25 h^2 + 8k = 25

Replace h  by x  and k  by y , the locus is:

x2+8y25=0 x^2 + 8y - 25 = 0

Hence, the correct answer is Option (c).

83

The image of the point (3, 5, 7) in the plane 2x+ y+z=6 is 

  1. ((a))

    (5, 1, 3) 

  2. ((b))

    (5, -1, 3)

  3. ((c))

    (5, 1, -3)

  4. ((d))

    (-5, 1, 3) 

Show Answer
Answer: ((d))

(-5, 1, 3) 

Calculation:

Given,Point P(3, 5, 7) 

Plane equation: 2x+y+z=6 2x + y + z = 6

Rewrite the plane equation in standard form:

2x+y+z6=0 2x + y + z - 6 = 0

So, A=2,B=1,C=1,D=6 A = 2, \quad B = 1, \quad C = 1, \quad D = -6

Calculate numerator:

Ax1+By1+Cz1+D=2(3)+1(5)+1(7)6=12 Ax_1 + By_1 + Cz_1 + D = 2(3) + 1(5) + 1(7) - 6 = 12

Calculate denominator:

A2+B2+C2=22+12+12=6 A^2 + B^2 + C^2 = 2^2 + 1^2 + 1^2 = 6

Using formula for image of point in plane,

x=x12A(Ax1+By1+Cz1+D)A2+B2+C2=32×2×126=5 x' = x_1 - \frac{2A(Ax_1 + By_1 + Cz_1 + D)}{A^2 + B^2 + C^2} = 3 - \frac{2 \times 2 \times 12}{6} = -5

y=y12B(Ax1+By1+Cz1+D)A2+B2+C2=52×1×126=1 y' = y_1 - \frac{2B(Ax_1 + By_1 + Cz_1 + D)}{A^2 + B^2 + C^2} = 5 - \frac{2 \times 1 \times 12}{6} = 1

z=z12C(Ax1+By1+Cz1+D)A2+B2+C2=72×1×126=3 z' = z_1 - \frac{2C(Ax_1 + By_1 + Cz_1 + D)}{A^2 + B^2 + C^2} = 7 - \frac{2 \times 1 \times 12}{6} = 3

Hence, the image of the point is (5,1,3) (-5, 1, 3) .

Hence, the correct answer is Option d.

84

The direction cosines of a line segment. whose projections on the coordinate axes axes are (-6, 3, 2) are

  1. ((a))

    67,37,27-\frac{6}7, \frac{3}7, \frac{2}7

  2. ((b))

    67,37,27\frac{6}7, \frac{3}7, \frac{2}7

  3. ((c))

    67,37,27\frac{6}7, -\frac{3}7, \frac{2}7

  4. ((d))

    None of the above

Show Answer
Answer: ((a))

67,37,27-\frac{6}7, \frac{3}7, \frac{2}7

Calculation:

Given,

The projections of the line segment on the coordinate axes are:

6,3,2 -6, \quad 3, \quad 2

Find the length of the line segment:

L=(6)2+32+22=36+9+4=49=7 L = \sqrt{(-6)^2 + 3^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7

Calculate the direction cosines (projections divided by length):

l=67,m=37,n=27 l = \frac{-6}{7}, \quad m = \frac{3}{7}, \quad n = \frac{2}{7}

The direction cosines are (67,37,27) \left( -\frac{6}{7}, \frac{3}{7}, \frac{2}{7} \right)

Hence, the correct option is (a).

85

If the line x23=y34=z45 and x1a=y23=z34\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5} \text{ and } \frac{x-1}{a}=\frac{y-2}{3}=\frac{z-3}{4} are coplanar, then a is equal to

  1. ((a))

    1

  2. ((b))

    2

  3. ((c))

    3

  4. ((d))

    4

Show Answer
Answer: ((b))

2

Calculation:

Given the lines:x23=y34=z45\frac{x-2}{3} = \frac{y-3}{4} = \frac{z-4}{5}

and x1a=y23=z34\frac{x-1}{a} = \frac{y-2}{3} = \frac{z-3}{4}

These two lines are coplanar if the scalar triple product of direction vectors and the vector connecting points on both lines is zero.

Direction vector of first line, d1=(3,4,5)\vec{d_1} = (3, 4, 5)

Direction vector of second line, d2=(a,3,4)\vec{d_2} = (a, 3, 4)

Vector connecting points on the two lines: p=(21,32,43)=(1,1,1)\vec{p} = (2-1, 3-2, 4-3) = (1, 1, 1)

Calculate scalar triple product:

⇒ p(d1×d2)=0\vec{p} \cdot (\vec{d_1} \times \vec{d_2}) = 0

First find cross product d1×d2\vec{d_1} \times \vec{d_2}

⇒ ijk 345 a34=i(4×45×3)j(3×45×a)+k(3×34×a)\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 3 & 4 & 5 \ a & 3 & 4 \end{vmatrix} = \mathbf{i}(4 \times 4 - 5 \times 3) - \mathbf{j}(3 \times 4 - 5 \times a) + \mathbf{k}(3 \times 3 - 4 \times a)

=i(1615)j(125a)+k(94a)=i(1)j(125a)+k(94a)= \mathbf{i}(16 - 15) - \mathbf{j}(12 - 5a) + \mathbf{k}(9 - 4a) = \mathbf{i}(1) - \mathbf{j}(12 - 5a) + \mathbf{k}(9 - 4a)

Now scalar triple product:

⇒ p(d1×d2)=(1)(1)+(1)((125a))+(1)(94a)\vec{p} \cdot (\vec{d_1} \times \vec{d_2}) = (1)(1) + (1)(-(12 - 5a)) + (1)(9 - 4a)

=112+5a+94a=(112+9)+(5a4a)=2+a=0 = 1 - 12 + 5a + 9 - 4a = (1 - 12 + 9) + (5a - 4a) = -2 + a = 0

⇒ a=2a = 2

Hence, the correct answer is Option b.

86

The length of perpendicular from (1, 2, 3) to the line x63=y72=z72\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2} is

  1. ((a))

    3

  2. ((b))

    17\sqrt{17}

  3. ((c))

    7

  4. ((d))

    20\sqrt{20}

Show Answer
Answer: ((c))

7

Calculation:

Given, The Point P (1, 2, 3) 

The line in symmetric form x63=y72=z72 \frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{-2}

Identify a point on the line and the direction vector.

Point on line A=(6,7,7) A = (6, 7, 7)

Direction vecto d=(3,2,2) \vec{d} = (3, 2, -2)

Vector from point A to P is

⇒ AP=(16,27,37)=(5,5,4) \vec{AP} = (1 - 6, 2 - 7, 3 - 7) = (-5, -5, -4)

Calculate the cross product AP×d\vec{AP} \times \vec{d}

⇒ AP×d=ijk 554 322=18i22j+5k\vec{AP} \times \vec{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -5 & -5 & -4 \ 3 & 2 & -2 \end{vmatrix} = 18 \mathbf{i} - 22 \mathbf{j} + 5 \mathbf{k}

Magnitude of AP×d\vec{AP} \times \vec{d}

⇒ AP×d=182+(22)2+52=324+484+25=833 |\vec{AP} \times \vec{d}| = \sqrt{18^2 + (-22)^2 + 5^2} = \sqrt{324 + 484 + 25} = \sqrt{833}

Magnitude of direction vectord \vec{d}

⇒ d=32+22+(2)2=9+4+4=17 |\vec{d}| = \sqrt{3^2 + 2^2 + (-2)^2} = \sqrt{9 + 4 + 4} = \sqrt{17}

Length of perpendicular from P  to the line is given by:

⇒ d=AP×dd=83317=83317=49=7 d = \frac{|\vec{AP} \times \vec{d}|}{|\vec{d}|} = \frac{\sqrt{833}}{\sqrt{17}} = \sqrt{\frac{833}{17}} = \sqrt{49} = 7

Hence, the correct answer is Option 3.

87

If cosα\alpha, cosβ\beta, cosγ\gamma are the direction cosines of a straight line, then (sin2 α\alpha + sin2 β\beta  + sin2 γ\gamma) is equal to.

  1. ((a))

    0

  2. ((b))

    1

  3. ((c))

    3

  4. ((d))

    2

Show Answer
Answer: ((d))

2

Calculation:

Given, Direction cosines of a straight line are cosα,cosβ,cosγ \cos\alpha, \cos\beta, \cos\gamma

We know,cos2α+cos2β+cos2γ=1 \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1

Also, sin2θ=1cos2θ \sin^2\theta = 1 - \cos^2\theta

Therefore,

⇒ sin2α+sin2β+sin2γ=(1cos2α)+(1cos2β)+(1cos2γ) \sin^2\alpha + \sin^2\beta + \sin^2\gamma = (1 - \cos^2\alpha) + (1 - \cos^2\beta) + (1 - \cos^2\gamma)

=3(cos2α+cos2β+cos2γ) = 3 - (\cos^2\alpha + \cos^2\beta + \cos^2\gamma)

=31=2 = 3 - 1 = 2

Hence, the correct answer is Option (d)

88

The radius of the spherex2+y2+z2xyz=0x ^2 +y ^2 +z ^2 −x−y−z=0 is

  1. ((a))

    32\frac{3}{2}

  2. ((b))

    32\frac{\sqrt{3}}{2}

  3. ((c))

    12\frac{1}{\sqrt{2}}

  4. ((d))

    3\sqrt{3}

Show Answer
Answer: ((b))

32\frac{\sqrt{3}}{2}

Calculation:

Given, the equation of the sphere is:

x2+y2+z2xyz=0 x^2 + y^2 + z^2 - x - y - z = 0

Compare this with the general form of the sphere:

x2+y2+z2+2ux+2vy+2wz+d=0 x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0

We get:

2u=1u=12 2u = -1 \Rightarrow u = -\frac{1}{2}

2v=1v=12 2v = -1 \Rightarrow v = -\frac{1}{2}

2w=1w=12 2w = -1 \Rightarrow w = -\frac{1}{2}

The formula for the radius of the sphere is:

r=u2+v2+w2d r = \sqrt{u^2 + v^2 + w^2 - d}

Substitute the values:

r=(12)2+(12)2+(12)20 r = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 - 0}

r=14+14+14=34=32 r = \sqrt{\frac{1}{4} + \frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}

Hence, the correct answer is Option (b).

89

The conic 5x26xy+5y2+26x22y+29=05x ^2 −6xy+5y ^2 +26x−22y+29=0 represents

  1. ((a))

    a circle

  2. ((b))

    a parabola

  3. ((c))

    a hyperbola

  4. ((d))

    an ellipse

Show Answer
Answer: ((d))

an ellipse

Calculation:

Given the general conic equation: 5x26xy+5y2+26x22y+29=0 5x^2 - 6xy + 5y^2 + 26x - 22y + 29 = 0

Compare this with the general second-degree equation:

Ax2+Bxy+Cy2+Dx+Ey+F=0 Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0

Here, A=5, B=6, C=5 A = 5, \ B = -6, \ C = 5

We calculate the discriminant:

Δ=B24AC \Delta = B^2 - 4AC

Δ=(6)24(5)(5)=36100=64 \Delta = (-6)^2 - 4(5)(5) = 36 - 100 = -64

Since Δ<0 \Delta < 0 and A=C A = C but B0 B \neq 0 , the conic is an ellipse.

Hence, the correct answer is Option (d).

90

The coordinates of the point, where the line   x21=y+31=z16\frac{x-2}{-1}=\frac{y+3}{1}=\frac{z-1}{6} intersects the plane 2x+y+z=7,2x+y+z=7, are

  1. ((a))

    (2,1,−7)

  2. ((b))

    (7,−1,2)

  3. ((c))

    (1,−2,7)

  4. ((d))

    (2,−7,1)

Show Answer
Answer: ((c))

(1,−2,7)

Calculation:

Given the line: x21=y+31=z16=t \frac{x - 2}{-1} = \frac{y + 3}{1} = \frac{z - 1}{6} = t

Parametric equations:

x=2t, y=3+t, z=1+6t x = 2 - t, \ y = -3 + t, \ z = 1 + 6t

Plane equation: 2x+y+z=7 2x + y + z = 7

Substitute parametric values into plane:

2(2t)+(3+t)+(1+6t)=7 2(2 - t) + (-3 + t) + (1 + 6t) = 7

42t3+t+1+6t=72+5t=7t=1 4 - 2t - 3 + t + 1 + 6t = 7 \Rightarrow 2 + 5t = 7 \Rightarrow t = 1

Now substitute t=1 t = 1 :

x=1, y=2, z=7 x = 1, \ y = -2, \ z = 7

Hence, the correct answer is Option (c)

91

If A is a 3×3 non-singular matrix, then det(adj A) is equal to

  1. ((a))

    2detA

  2. ((b))

    3detA

  3. ((c))

    (detA)2

  4. ((d))

    (detA)3

Show Answer
Answer: ((c))

(detA)2

Calculation:

Given,

A is a 3×3 non-singular matrix.

We need to find det(adj(A))\det(\text{adj}(A)).

Concept:

If A is an n×nn \times n non-singular matrix, then:

det(adj(A))=(detA)n1 \det(\text{adj}(A)) = (\det A)^{n-1}

Here, n = 3 , so:

det(adj(A))=(detA)2 \det(\text{adj}(A)) = (\det A)^2

Hence, the correct answer is Option c.

92

The composite mapping fg f∘g of the maps 

f:RR,f(x)=sinxf:R→R,f(x)=sinx

g:RR,g(x)=x2g:R→R,g(x)=x^ 2

is

  1. ((a))

    sinx+x2sinx+x^ 2

  2. ((b))

    sin(x2)sin(x^ 2 )

  3. ((c))

    (sinx)2(sinx) ^ 2

  4. ((d))

    sinxx2\frac{sin x}{x^2}

Show Answer
Answer: ((b))

sin(x2)sin(x^ 2 )

Calculation:

Given: f(x) = sin x and 

g(x) =  x2

By definition of composition:

fg)(x)=f(g(x))f \circ g)(x) = f(g(x))

Substitute g(x)=x2g(x) = x^2 :

f(g(x))=f(x2)=sin(x2)f(g(x)) = f(x^2) = \sin(x^2)

Hence, the correct answer is Option (b).

93

A square matrix P satisfies P2=IPP^2=I-P If Pn=5I8PP ^ n = 5I - 8P  then n is equal to

  1. ((a))

    4

  2. ((b))

    5

  3. ((c))

    6

  4. ((d))

    7

Show Answer
Answer: ((c))

6

Calculation:

Given:

P2=IP P^2 = I - P

Pn=5I8P P^n = 5I - 8P

We compute higher powers of P recursively:

P2=IP P^2 = I - P

⇒ P3=PP2=PP2=P(IP)=2PI P^3 = P \cdot P^2 = P - P^2 = P - (I - P) = 2P - I

⇒ P4=PP3=P(2PI)=2P2P=2(IP)P=2I3P P^4 = P \cdot P^3 = P(2P - I) = 2P^2 - P = 2(I - P) - P = 2I - 3P

⇒ P5=PP4=P(2I3P)=2P3P2=2P3(IP)=5P3I P^5 = P \cdot P^4 = P(2I - 3P) = 2P - 3P^2 = 2P - 3(I - P) = 5P - 3I

⇒ P6=PP5=P(5P3I)=5P23P=5(IP)3P=5I8P P^6 = P \cdot P^5 = P(5P - 3I) = 5P^2 - 3P = 5(I - P) - 3P = 5I - 8P

This matches the given equation Pn=5I8P P^n = 5I - 8P .

Hence, the correct answer is Option c.

94

The number of solutions of log4(x1)=log2(x3)\log_4(x-1)=\log_2(x-3)is

  1. ((a))

    2

  2. ((b))

    3

  3. ((c))

    1

  4. ((d))

    0

Show Answer
Answer: ((c))

1

Calculation:

Given, log4(x1)=log2(x3) \log_4 (x-1) = \log_2 (x-3)

Rewrite the base 4 logarithm as base 2:

⇒ log4(x1)=log2(x1)log24=log2(x1)2 \log_4 (x-1) = \frac{\log_2 (x-1)}{\log_2 4} = \frac{\log_2 (x-1)}{2}

So the equation becomes:

⇒ log2(x1)2=log2(x3) \frac{\log_2 (x-1)}{2} = \log_2 (x-3)

Multiply both sides by 2:

⇒ log2(x1)=2log2(x3) \log_2 (x-1) = 2 \log_2 (x-3)

Use logarithm power rule on the right:

⇒ log2(x1)=log2((x3)2) \log_2 (x-1) = \log_2 \left((x-3)^2\right)

Since the logs are equal, their arguments must be equal:

⇒ x1=(x3)2 x - 1 = (x - 3)^2

Expand the right side:

⇒ x1=x26x+9 x - 1 = x^2 - 6x + 9

Bring all terms to one side:

⇒ 0=x26x+9x+1 0 = x^2 - 6x + 9 - x + 1

⇒ 0=x27x+10 0 = x^2 - 7x + 10

⇒ (x5)(x2)=0x - 5)(x - 2) = 0

x=5orx=2 x = 5 \quad \text{or} \quad x = 2

Check domain restrictions:

x1>0x>1 x - 1 > 0 \Rightarrow x > 1

x3>0x>3 x - 3 > 0 \Rightarrow x > 3

Check the solutions:

x=2 x=2 : fails x > 3, so discard

x=5 x=5 : satisfies both domain restrictions

 the number of solutions is 1 (i.e., x=5

Hence, the correct answer is Option C.

95

The eigenvalues of the matrix A=[ahg 0b0 0cc]A=\begin{bmatrix} a & h & g \ 0 & b & 0 \ 0 & c & c \end{bmatrix}   are

  1. ((a))

    a, h, g

  2. ((b))

    a, g, c

  3. ((c))

    a, h, c

  4. ((d))

    a, b, c

Show Answer
Answer: ((d))

a, b, c

Calculation: 

The matrix A A is an upper triangular matrix of the form:

A=[ahg 0b0 00c] A = \begin{bmatrix} a & h & g \ 0 & b & 0 \ 0 & 0 & c \end{bmatrix}

For any triangular matrix (upper or lower), the eigenvalues are the entries on the main diagonal.

Therefore, the eigenvalues of matrix A A are:

a,b,c a, b, c

Hence, the correct answer is Option (d) .

96

A cyclic group having only one generator can have at most

  1. ((a))

    1 element

  2. ((b))

    2 elements

  3. ((c))

    3 elements

  4. ((d))

    4 elements

Show Answer
Answer: ((b))

2 elements

Calculation:

Given, a cyclic group having only one generator can have at most:

The number of generators of a cyclic group of order n is given by Euler's totient function φ(n) \varphi(n) .

So, if a cyclic group has only one generator, then:

⇒ φ(n)=1 \varphi(n) = 1

Euler's totient function φ(n)=1 \varphi(n) = 1 only when n=1 n = 1 or n=2 n = 2 .

Check the possible orders:

⇒ n=1orn=2 n = 1 \quad \text{or} \quad n = 2

If n=1 n = 1 , the group has 1 element.

If n=2 n = 2 , the group has 2 elements.

Therefore, a cyclic group having only one generator can have at most 2 elements.

∴ the answer is Option (b) 2 elements.

97

Every diagonal element of a skew-symmetric matrix is

  1. ((a))

    zero

  2. ((b))

    unity

  3. ((c))

    non-zero

  4. ((d))

    purely imaginary

Show Answer
Answer: ((a))

zero

Calculation:

A skew-symmetric matrix A A is a square matrix that satisfies the condition:

AT=A A^T = -A

where AT A^T is the transpose of A A .

For the diagonal elements of a skew-symmetric matrix:

aii=aii a_{ii} = -a_{ii}

which implies

2aii=0    aii=0 2a_{ii} = 0 \implies a_{ii} = 0

So, every diagonal element of a skew-symmetric matrix is zero.

Hence, the Correct answer is Option a.

98

The number of real solutions of the equation x2+5x+4=0|x|^2+5|x|+4=0 is

  1. ((a))

    4

  2. ((b))

    2

  3. ((c))

    1

  4. ((d))

    0

Show Answer
Answer: ((d))

0

Calculation:

Given,

x2+5x+4=0 |x|^2 + 5|x| + 4 = 0

Let  t = |x| . Since x0|x| \geq 0 , we rewrite the equation as:

t2+5t+4=0 t^2 + 5t + 4 = 0

Factorizing the quadratic:

(t+4)(t+1)=0 (t + 4)(t + 1) = 0

So,

t=4ort=1 t = -4 \quad \text{or} \quad t = -1

But both values are negative and since t=x0t = |x| \geq 0 , there are no valid solutions.

∴ The number of real solutions is 0.

Hence, the correct answe is Option d.

99

The sum of the infinite series 112.12+12.34.12212.34.56.123+........1 - \frac{1}{2}. \frac{1}{2}+ \frac{1}{2}. \frac{3}{4}.\frac{1}{2^2}- \frac{1}{2}. \frac{3}{4}. \frac{5}{6}.\frac{1}{2^3}+........\infty is

  1. ((a))

    23\sqrt{\frac{2}{3}}

  2. ((b))

    13\sqrt{\frac{1}3}

  3. ((c))

    3\sqrt{3}

  4. ((d))

    32\sqrt{\frac{3}{2}}

Show Answer
Answer: ((a))

23\sqrt{\frac{2}{3}}

Calculation:

Given, the infinite series: 11212+1234122123456123+ 1 - \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{1}{2^2} - \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{1}{2^3} + \cdots

Express the general term as:

⇒ Tn=(1)n1135(2n3)2462(n1)12n1 T_n = (-1)^{n-1} \frac{1 \cdot 3 \cdot 5 \cdots (2n - 3)}{2 \cdot 4 \cdot 6 \cdots 2(n-1)} \cdot \frac{1}{2^{n-1}}

Using double factorial notation:

⇒ Tn=(1)n1(2n3)!!(2n2)!!12n1 T_n = (-1)^{n-1} \frac{(2n-3)!!}{(2n-2)!!} \cdot \frac{1}{2^{n-1}}

Simplify the factorials using:

⇒ (2n)!!=2nn!and(2n1)!!=(2n)!2nn! (2n)!! = 2^n n! \quad \text{and} \quad (2n-1)!! = \frac{(2n)!}{2^n n!}

Rewrite the fraction:

⇒ (2n3)!!(2n2)!!=(2(n1)n1)22n2 \frac{(2n-3)!!}{(2n-2)!!} = \frac{\binom{2(n-1)}{n-1}}{2^{2n - 2}}

Thus, the term becomes:

⇒ Tn=(1)n1(2(n1)n1)123n3 T_n = (-1)^{n-1} \binom{2(n-1)}{n-1} \cdot \frac{1}{2^{3n - 3}}

Sum the series starting at n=1 :

⇒ S=n=1(1)n1(2(n1)n1)123n3 S = \sum_{n=1}^\infty (-1)^{n-1} \binom{2(n-1)}{n-1} \frac{1}{2^{3n - 3}}

Letting m = n-1 , rewrite the sum:

⇒ S=m=0(1)m(2mm)(18)m S = \sum_{m=0}^\infty (-1)^m \binom{2m}{m} \left(\frac{1}{8}\right)^m

Using the generating function for central binomial coefficients with alternating signs:

⇒ m=0(1)m(2mm)xm=11+4x \sum_{m=0}^\infty (-1)^m \binom{2m}{m} x^m = \frac{1}{\sqrt{1 + 4x}}

Substitute x=18x = \frac{1}{8}

⇒ S=11+12=132=23 S = \frac{1}{\sqrt{1 + \frac{1}{2}}} = \frac{1}{\sqrt{\frac{3}{2}}} = \sqrt{\frac{2}{3}}

Hence, the sum of the infinite series is: 23 \sqrt{\frac{2}{3}}

100

The sum of three numbers in arithmetic progression is 51 and the product of first and third terms is 273. The common difference of this progression is

  1. ((a))

    5

  2. ((b))

    4

  3. ((c))

    3

  4. ((d))

    6

Show Answer
Answer: ((b))

4

Calculation:

The sum of three numbers in arithmetic progression is 51, and the product of the first and third terms is 273.

Let the three numbers in A.P. be:

ad, a, a+d a - d,\ a,\ a + d

Use the sum condition

⇒ (ad)+a+(a+d)=3a=51a=513=17 (a - d) + a + (a + d) = 3a = 51 \Rightarrow a = \frac{51}{3} = 17

Use the product condition

⇒ (ad)(a+d)=a2d2=273 (a - d)(a + d) = a^2 - d^2 = 273

⇒ 172d2=273289d2=273 17^2 - d^2 = 273 \Rightarrow 289 - d^2 = 273

⇒ d2=289273=16d=16=4 d^2 = 289 - 273 = 16 \Rightarrow d = \sqrt{16} = 4

Hence, the correct answer is Option (b)

101

The harmonic mean of two numbers is 4. If their arithmetic mean A and geometric mean G satisfy the equation 2A+G2=272A+G^2=27, then the numbers are

  1. ((a))

    1, 3

  2. ((b))

    1, 4

  3. ((c))

    3, 6

  4. ((d))

    None of the above

Show Answer
Answer: ((c))

3, 6

Calculation:

Given, Harmonic Mean (H) = 4

And the condition: 2A+G2=27 2A + G^2 = 27

Let the two numbers be x x and y y

Use the harmonic mean formula:

⇒ H=2xyx+y=4 H = \frac{2xy}{x + y} = 4

Arithmetic mean and geometric mean:

⇒ A=x+y2 A = \frac{x + y}{2} ,    G=xyG2=xy G = \sqrt{xy} \Rightarrow G^2 = xy

Plug into the equation:

⇒ 2A+G2=27(x+y)+xy=27 2A + G^2 = 27 \Rightarrow (x + y) + xy = 27   ... (i)

From harmonic mean:

⇒ 2xyx+y=42xy=4(x+y)xy=2(x+y) \frac{2xy}{x + y} = 4 \Rightarrow 2xy = 4(x + y) \Rightarrow xy = 2(x + y)   ... (ii)

Substitute (ii) in (i):

⇒ (x+y)+2(x+y)=273(x+y)=27x+y=9 (x + y) + 2(x + y) = 27 \Rightarrow 3(x + y) = 27 \Rightarrow x + y = 9

Then from (ii): xy=2(9)=18 xy = 2(9) = 18

Solve quadratic equation:

⇒ t29t+18=0t=9±81722=9±32t=3, 6 t^2 - 9t + 18 = 0 \Rightarrow t = \frac{9 \pm \sqrt{81 - 72}}{2} = \frac{9 \pm 3}{2} \Rightarrow t = 3,\ 6

Hence, the correct answer is Option (c)

102

Let A be a 3×3 matrix with eigenvalues 1,−1,0. Then the value of I+A100|I+A^{100}|  is

  1. ((a))

    6

  2. ((b))

    4

  3. ((c))

    27

  4. ((d))

    100

Show Answer
Answer: ((b))

4

Calculation:

Note: There was a discrepancy in the original options of the Question, and we have modified the options accordingly. 

Given: Let A A be a 3×33 \times 3 matrix with eigenvalues 1,1,01, -1, 0.

We are required to find I+A100 \left| I + A^{100} \right| .

Eigenvalues of A A are 1,1,0 1, -1, 0 .

Eigenvalues of A100 A^{100} will be 1100,(1)100,0100 1^{100}, (-1)^{100}, 0^{100} , i.e., 1,1,0 1, 1, 0 .

\Rightarrow Eigenvalues of A100 A^{100} are 1,1,0 1, 1, 0 .

Now, eigenvalues of I+A100 I + A^{100} will be 1+λ 1 + \lambda , where λ \lambda is an eigenvalue of A100 A^{100} .

\Rightarrow Eigenvalues of I+A100 I + A^{100} are:

1+1=21 + 1 = 2,

1+1=21 + 1 = 2,

1+0=11 + 0 = 1.

Therefore, I+A100=2×2×1=4 \left| I + A^{100} \right| = 2 \times 2 \times 1 = 4 .

Hence, the correct value is Option 2.

103

Let G be a group with identity element e. Let a,b∈G be such that a5=e and aba−1=b2. Then o(b) is

  1. ((a))

    17

  2. ((b))

    23

  3. ((c))

    29

  4. ((d))

    31

Show Answer
Answer: ((d))

31

Calculation:

Given: G G is a group with identity element e e . Let a,bG a, b \in G such that a5=e a^5 = e and aba1=b2 aba^{-1} = b^2 .

We need to find the order of b b , i.e., the smallest positive integer n n such that bn=e b^n = e .

Now, apply conjugation repeatedly:

aba1=b2 aba^{-1} = b^2

a2ba2=a(b2)a1=(aba1)2=(b2)2=b4 a^2 b a^{-2} = a (b^2) a^{-1} = (aba^{-1})^2 = (b^2)^2 = b^4

a3ba3=a(b4)a1=(aba1)4=(b2)4=b8 a^3 b a^{-3} = a (b^4) a^{-1} = (aba^{-1})^4 = (b^2)^4 = b^8

a4ba4=a(b8)a1=(b2)8=b16 a^4 b a^{-4} = a (b^8) a^{-1} = (b^2)^8 = b^{16}

a5ba5=a(b16)a1=(b2)16=b32 a^5 b a^{-5} = a (b^{16}) a^{-1} = (b^2)^{16} = b^{32}

But since a5=e a^5 = e , we know that conjugation by a5 a^5 must give the original element:

a5ba5=bb32=bb31=e a^5 b a^{-5} = b \Rightarrow b^{32} = b \Rightarrow b^{31} = e

∴  the order of b b is 31 .

Hence, the correct answer is Option d.

104

Every square matrix can be expressed as

  1. ((a))

    a Hermitian matrix

  2. ((b))

    a skew-symmetric matrix

  3. ((c))

    sum of symmetric and skew-symmetric matrices

  4. ((d))

    None of the above

Show Answer
Answer: ((c))

sum of symmetric and skew-symmetric matrices

Calculation:

Let A A be any square matrix. Then,

A=12(A+AT)+12(AAT) A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T)

Now,

12(A+AT) \frac{1}{2}(A + A^T) is symmetric, because:

(12(A+AT))T=12(AT+A)=12(A+AT) \left(\frac{1}{2}(A + A^T)\right)^T = \frac{1}{2}(A^T + A) = \frac{1}{2}(A + A^T)

and

12(AAT) \frac{1}{2}(A - A^T) is skew-symmetric, because:

(12(AAT))T=12(ATA)=12(AAT) \left(\frac{1}{2}(A - A^T)\right)^T = \frac{1}{2}(A^T - A) = -\frac{1}{2}(A - A^T)

Hence, any square matrix is the sum of a symmetric matrix and a skew-symmetric matrix.

Hence, the correct answer is Option c.

105

The sum of the infinite series  12!+1+23!+1+2+34!+1+2+3+45!+\frac{1}{2!}+\frac{1+2}{3!}+\frac{1+2+3}{4!}+\frac{1+2+3+4}{5!}+\dots\infty   is

  1. ((a))

    2e

  2. ((b))

    3e

  3. ((c))

    3e2\frac{3e}{2}

  4. ((d))

    e2\frac{e}{2}

Show Answer
Answer: ((d))

e2\frac{e}{2}

Calculation:

The infinite series is 12!+1+23!+1+2+34!+1+2+3+45!+\frac{1}{2!} + \frac{1+2}{3!} + \frac{1+2+3}{4!} + \frac{1+2+3+4}{5!} + \cdots

Note that 1+2++(n1)=(n1)n21 + 2 + \cdots + (n - 1) = \frac{(n - 1)n}{2}

So, the general term is: 1+2++(n1)n!=(n1)n2n!=n12(n1)!\frac{1 + 2 + \cdots + (n - 1)}{n!} = \frac{(n - 1)n}{2n!} = \frac{n - 1}{2(n - 1)!}

Therefore, the series becomes:

n=2n12(n1)!\sum_{n=2}^{\infty} \frac{n - 1}{2(n - 1)!}

Change the index by letting k=n1k = n - 1, so when n=2n = 2, k=1k = 1. Hence,

k=1k2k!=12k=1kk!\sum_{k=1}^{\infty} \frac{k}{2k!} = \frac{1}{2} \sum_{k=1}^{\infty} \frac{k}{k!}

We know: k=1kk!=e\sum_{k=1}^{\infty} \frac{k}{k!} = e

So, the final result is:

12e=e2\frac{1}{2} \cdot e = \frac{e}{2}

Hence, the correct answer is Option (d).

106

The characteristic roots of the matrix  A=[54 12]A = \begin{bmatrix} 5 & 4 \ 1 & 2 \end{bmatrix}    are

  1. ((a))

    1, 6

  2. ((b))

     -1, 6

  3. ((c))

    -1, -6

  4. ((d))

    1, -6

Show Answer
Answer: ((a))

1, 6

Calculation:

The matrixA=[54 12] A = \begin{bmatrix} 5 & 4 \ 1 & 2 \end{bmatrix}

To find the characteristic roots (eigenvalues), solve the characteristic equation:

det(AλI)=0 \det(A - \lambda I) = 0

5λ4 12λ=0 \Rightarrow \begin{vmatrix} 5 - \lambda & 4 \ 1 & 2 - \lambda \end{vmatrix} = 0

Calculate the determinant:

(5λ)(2λ)4=0 (5 - \lambda)(2 - \lambda) - 4 = 0

(105λ2λ+λ2)4=0 \Rightarrow (10 - 5\lambda - 2\lambda + \lambda^2) - 4 = 0

λ27λ+6=0 \Rightarrow \lambda^2 - 7\lambda + 6 = 0

Solve the quadratic equation:

(λ1)(λ6)=0 (\lambda - 1)(\lambda - 6) = 0

λ=1,6 \Rightarrow \lambda = 1, 6

Hence, the correct answer is Option a

107

For square matrices A and B, which of the following is true?

  1. ((a))

    (AB)=AB(AB)'=A'B'

  2. ((b))

    (A+B)=A+B(A+B)'=A'+B'

  3. ((c))

    (AB)1=A1B1(AB)^{-1}=A^{-1}B^{-1}

  4. ((d))

    (A+B)1=A1+B1(A+B)^{-1}=A^{-1}+B^{-1}

Show Answer
Answer: ((b))

(A+B)=A+B(A+B)'=A'+B'

Calculation:

We’re given four statements involving operations on square matrices A and B. Let’s analyze each one:

(a) (AB)=AB(AB)' = A'B'

This is false because the transpose of a product reverses the order:

(AB)=BA(AB)' = B'A'

(b) (A+B)=A+B(A + B)' = A' + B'

This is true. Transposition distributes over matrix addition.

(c) (AB)1=A1B1(AB)^{-1} = A^{-1}B^{-1}

This is false. The inverse of a product also reverses the order:

(AB)1=B1A1(AB)^{-1} = B^{-1}A^{-1}

(d) (A+B)1=A1+B1(A + B)^{-1} = A^{-1} + B^{-1}

This is false. Inverse of a sum doesn't simplify like this in general.

Hence, the correct answer is Option (b).

108

The characteristic roots of a Hermitian matrix are

  1. ((a))

    real

  2. ((b))

    purely imaginary

  3. ((c))

    complex numbers

  4. ((d))

    None of the above

Show Answer
Answer: ((a))

real

Calculation:

Let A A be a Hermitian matrix, and let λ \lambda be an eigenvalue with eigenvector x0 \mathbf{x} \neq \mathbf{0} .

Then by definition: Ax=λx A \mathbf{x} = \lambda \mathbf{x}

Taking the Hermitian transpose (conjugate transpose) of both sides: xHA=λxH \mathbf{x}^H A = \lambda^* \mathbf{x}^H

Now multiply the original equation on the left by xH \mathbf{x}^H :

xHAx=λxHx(1) \mathbf{x}^H A \mathbf{x} = \lambda \mathbf{x}^H \mathbf{x} \quad \text{(1)}

From the conjugate transpose, we also get:

xHAx=λxHx(2) \mathbf{x}^H A \mathbf{x} = \lambda^* \mathbf{x}^H \mathbf{x} \quad \text{(2)}

Comparing (1) and (2):

λxHx=λxHx \lambda \mathbf{x}^H \mathbf{x} = \lambda^* \mathbf{x}^H \mathbf{x}

Since xHx \mathbf{x}^H \mathbf{x} is real and positive (it's the squared norm), we can divide both sides:

λ=λ \lambda = \lambda^*

This implies that λ \lambda is real.

Hence, the characteristic roots (eigenvalues) of a Hermitian matrix are real.

The correct answer is (a) real.

109

The generator/generators of the cyclic group a,a2,a3,a4=ea, a^2, a^3, a^4=e is/are

  1. ((a))

    a4a ^ 4

  2. ((b))

    a2a ^ 2

  3. ((c))

    a4,a2a^4,a^2

  4. ((d))

    a,a3a,a^3

Show Answer
Answer: ((d))

a,a3a,a^3

Calculation:

Given: The cyclic group a,a2,a3,a4=e { a, a^2, a^3, a^4 = e }

The generators of a cyclic group Cn are elements aka^k such that gcd(k,n)=1 \gcd(k, n) = 1 .

For n = 4, calculate:

gcd(1,4)=1 \gcd(1, 4) = 1  a1=a→ a^1 = a  is a generator

gcd(2,4)=2 \gcd(2, 4) = 2  a2→ a^2 is not a generator

gcd(3,4)=1 \gcd(3, 4) = 1 a3 → a^3 is a Generatore

gcd(4,4)=4 \gcd(4, 4) = 4  a4=e→ a^4 = e is not a generator

 

Hence, the correct answer is Option d.

110

The value of the determinant  4316 3574 1732\begin{vmatrix} 43 & 1 & 6 \ 35 & 7 & 4 \ 17 & 3 & 2 \end{vmatrix}  is

  1. ((a))

    0

  2. ((b))

    56

  3. ((c))

    756

  4. ((d))

    964

Show Answer
Answer: ((a))

0

Calculation:

Given:  4316 3574 1732 \begin{vmatrix} 43 & 1 & 6 \ 35 & 7 & 4 \ 17 & 3 & 2 \ \end{vmatrix}

Det=a(eifh)b(difg)+c(dheg)\text{Det} = a(ei - fh) - b(di - fg) + c(dh - eg)

a=43,b=1,c=6,d=35,e=7,f=4,g=17,h=3,i=2 a = 43,\quad b = 1,\quad c = 6,\quad d = 35,\quad e = 7,\quad f = 4,\quad g = 17,\quad h = 3,\quad i = 2

Det=43(7243)1(352417)+6(353717)\Rightarrow \text{Det} = 43(7 \cdot 2 - 4 \cdot 3) - 1(35 \cdot 2 - 4 \cdot 17) + 6(35 \cdot 3 - 7 \cdot 17)

Det=43(1412)1(7068)+6(105119)\Rightarrow \text{Det} = 43(14 - 12) - 1(70 - 68) + 6(105 - 119)

Det=86284\Rightarrow \text{Det} = 86 - 2 - 84 = 0 

Hence, the correct answer is Option (a).

111

If the maximum and minimum values of (5+6cosθ+2cos2θ) satisfy the quadratic equation x2−px+q=2, then p, q are respectively

  1. ((a))

    55/4, 47/4

  2. ((b))

    12, 13

  3. ((c))

    14, 13

  4. ((d))

    13, 14

Show Answer
Answer: ((a))

55/4, 47/4

Calculation:

Note: There was a discrepancy in the original options, and none of them matched the correct answer. Therefore, we have updated the options accordingly.

Given, the function f(θ)=5+6cosθ+2cos2θ f(\theta) = 5 + 6 \cos \theta + 2 \cos 2\theta

Rewrite using the double angle formula cos2θ=2cos2θ1 \cos 2\theta = 2 \cos^2 \theta - 1 :

⇒ f(θ)=5+6cosθ+2(2cos2θ1)=3+6cosθ+4cos2θ f(\theta) = 5 + 6 \cos \theta + 2 (2 \cos^2 \theta - 1) = 3 + 6 \cos \theta + 4 \cos^2 \theta

Let x=cosθ x = \cos \theta , then

⇒ f(x)=3+6x+4x2 f(x) = 3 + 6x + 4x^2 , where x[1,1] x \in [-1, 1]

Since f(x) f(x) is quadratic with positive leading coefficient, the minimum is at vertex:

Vertex at x=b2a=62×4=34 x = -\frac{b}{2a} = -\frac{6}{2 \times 4} = -\frac{3}{4}

Evaluate f(x) f(x) at vertex:

⇒ f(34)=3+6×(34)+4×(34)2=34 f\left(-\frac{3}{4}\right) = 3 + 6 \times \left(-\frac{3}{4}\right) + 4 \times \left(-\frac{3}{4}\right)^2 = \frac{3}{4}

Evaluate f(x) f(x) at boundaries:

⇒ f(1)=36+4=1 f(-1) = 3 - 6 + 4 = 1

⇒ f(1)=3+6+4=13 f(1) = 3 + 6 + 4 = 13

So the minimum value of f(θ) f(\theta) is 34 \frac{3}{4} , and maximum value is 13 13

These values satisfy the quadratic equation:

⇒ x2px+q=2 x^2 - p x + q = 2

Rewrite as: ⇒ x2px+(q2)=0 x^2 - p x + (q - 2) = 0

Sum of roots = maximum + minimum = 13+34=554 13 + \frac{3}{4} = \frac{55}{4}

Product of roots = maximum × minimum = 13×34=394 13 \times \frac{3}{4} = \frac{39}{4}

From the quadratic formula, sum of roots = p p and product of roots = q2 q - 2

∴  p=554 p = \frac{55}{4}

⇒ q2=394=9.75    q=394+2=474 q - 2 = \frac{39}{4} = 9.75 \implies q = \frac{39}{4} + 2 = \frac{47}{4}

∴  the values are p=554 p = \frac{55}{4} and 474\frac{47}{4}.

Hence, the correct answer is Option a.

112

The sum of the series 72+70+68+⋯+40 is

  1. ((a))

    950

  2. ((b))

    952

  3. ((c))

    954

  4. ((d))

    956

Show Answer
Answer: ((b))

952

Calculation:

Given: The arithmetic series

72+70+68++40 72 + 70 + 68 + \ldots + 40

 

Use the formula l=a+(n1)d l = a + (n - 1)d to find number of terms:

⇒ 40=72+(n1)(2) 40 = 72 + (n - 1)(-2)

⇒ 40=722n+2 40 = 72 - 2n + 2

⇒ 2n=7440=34n=17 2n = 74 - 40 = 34 \Rightarrow n = 17

Now use the sum formula for A.P.:

⇒ Sn=n2(a+l) S_n = \frac{n}{2}(a + l)

⇒ S17=172(72+40)=172×112=19042=952 S_{17} = \frac{17}{2}(72 + 40) = \frac{17}{2} \times 112 = \frac{1904}{2} = 952

Hence, the correct answer option b.

113

Given that the set Z of integers forms a group under the binary operation ∗, defined by a∗b=a+b+1, a,b∈Z. The inverse of -2 in the group is

  1. ((a))

    2

  2. ((b))

    4

  3. ((c))

    -2

  4. ((d))

    0

Show Answer
Answer: ((d))

0

Calculation:

A binary operation ab=a+b+1 a * b = a + b + 1 on the set of integers Z \mathbb{Z} .

We are asked to find the inverse of 2 -2 under this operation.

Find the identity element e e such that:

ae=a a * e = a for all aZ a \in \mathbb{Z}

Using the operation:

⇒ ae=a+e+1=ae+1=0e=1 a * e = a + e + 1 = a \Rightarrow e + 1 = 0 \Rightarrow e = -1

So, the identity element is 1 -1 .

Find the inverse of 2 -2 .

We want an element x x such that:

2x=1 -2 * x = -1 (identity element)

Apply the operation:

⇒ 2x=2+x+1=x1 -2 * x = -2 + x + 1 = x - 1

Set it equal to the identity:

⇒ x1=1x=0 x - 1 = -1 \Rightarrow x = 0

Hence, the correct answer is Option d.

114

The sum of first ten terms of the series 121+177+1165+\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+\dots is

  1. ((a))

    10129\frac{10}{129}

  2. ((b))

    20129\frac{20}{129}

  3. ((c))

    30129\frac{30}{129}

  4. ((d))

    40129\frac{40}{129}

Show Answer
Answer: ((a))

10129\frac{10}{129}

Calculation:

Given: The sum of the first ten terms of the series:

121+177+1165+\frac{1}{21} + \frac{1}{77} + \frac{1}{165} + \cdots

Observe the denominators: 21 =3×7 3 \times 7, 77 = 7×117 \times 11, 165=11×15165 = 11 \times 15, etc.

This suggests the general term is: Tn=1(4n1)(4n+3)T_n = \frac{1}{(4n - 1)(4n + 3)}

Use partial fractions:

⇒ 1(4n1)(4n+3)=A4n1+B4n+3\frac{1}{(4n - 1)(4n + 3)} = \frac{A}{4n - 1} + \frac{B}{4n + 3}

Solving gives: A=14, B=14A = \frac{1}{4},\ B = -\frac{1}{4}

So,

⇒ 1(4n1)(4n+3)=14(4n1)14(4n+3)\frac{1}{(4n - 1)(4n + 3)} = \frac{1}{4(4n - 1)} - \frac{1}{4(4n + 3)}

Sum first 10 terms:

⇒ n=110(14(4n1)14(4n+3))\sum_{n=1}^{10} \left( \frac{1}{4(4n - 1)} - \frac{1}{4(4n + 3)} \right)

This is a telescoping series. Most terms cancel out:

1431443=1121172\Rightarrow \frac{1}{4 \cdot 3} - \frac{1}{4 \cdot 43} = \frac{1}{12} - \frac{1}{172}

Convert to common denominator:

⇒ 112=43516, 1172=3516\frac{1}{12} = \frac{43}{516},\ \frac{1}{172} = \frac{3}{516}

433516=40516=10129\Rightarrow \frac{43 - 3}{516} = \frac{40}{516} = \frac{10}{129}

∴ The sum of the first 10 terms is 10129\frac{10}{129}.

Hence, the correct answe is Option a.

115

The condition that the equations ax2+bx+c=0,ax2+bx+c=0ax^2+bx+c=0, a'x^2+b'x+c'=0   have a common root is

  1. ((a))

    (bcbc)2=(caca)(abab)(bc'-b'c)^2 = (ca'-c'a)(ab'-a'b)

  2. ((b))

    (abab)2=(caca)(bcbc)(ab'-a'b)^2 = (ca'-c'a)(bc'-b'c)

  3. ((c))

    (caca)2=(bcbc)(abab)(ca'-c'a)^2 = (bc'-b'c)(ab'-a'b)

  4. ((d))

    None of the above

Show Answer
Answer: ((c))

(caca)2=(bcbc)(abab)(ca'-c'a)^2 = (bc'-b'c)(ab'-a'b)

Calculation:

Given: The equations ax2+bx+c=0 ax^2 + bx + c = 0 and ax2+bx+c=0 a'x^2 + b'x + c' = 0 have a common root.

let the common root be x x .

Then, x=acacabab x = \frac{a'c - ac'}{ab' - a'b}

Substitute back and simplify:

⇒ (acacabab)2=bcbcabab\left( \frac{a'c - ac'}{ab' - a'b} \right)^2 = \frac{bc' - b'c}{ab' - a'b}

Multiplying both sides by (abab)2(ab' - a'b)^2:

⇒ (acac)2=(bcbc)(abab)(a'c - ac')^2 = (bc' - b'c)(ab' - a'b)

Hence, the correct condition is:

⇒ (acac)2=(bcbc)(abab)(a'c - ac')^2 = (bc' - b'c)(ab' - a'b)

Hence, the correct answer is Option c.

116

The value of p for which the sum of the squares of the roots of the equation x2(p2)xp+1=0x^2−(p−2)x−p+1=0 is minimum, will be

  1. ((a))

    0

  2. ((b))

    1

  3. ((c))

    2

  4. ((d))

    3

Show Answer
Answer: ((b))

1

Calculation:

Given: x2(p2)xp+1=0 x^2 - (p - 2)x - p + 1 = 0

Let the roots be α,β \alpha, \beta

α+β=p2 \Rightarrow \alpha + \beta = p - 2

αβ=p+1 \Rightarrow \alpha \beta = -p + 1

α2+β2=(α+β)22αβ \Rightarrow \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta

=(p2)22(p+1) = (p - 2)^2 - 2(-p + 1)

=p24p+4+2p2 = p^2 - 4p + 4 + 2p - 2

=p22p+2 = p^2 - 2p + 2

⇒The expression is minimized at p=(2)2×1=1p = \frac{-(-2)}{2 \times 1} = 1

Hence, the correct answer is Option (b)

117

The domain of the function  f(x)=log2(x+3)x2+3x+2f(x)=\frac{\log_2(x+3)}{x^2+3x+2}  is

  1. ((a))

    R1,2R-{-1,-2}

  2. ((b))

    (2,)(-2,\infty)

  3. ((c))

    R1,2,3R-{-1,-2,-3}

  4. ((d))

    (3,)1,2(-3,\infty)-{-1,-2}

Show Answer
Answer: ((d))

(3,)1,2(-3,\infty)-{-1,-2}

Calculation:

Given: f(x)=log2(x+3)x2+3x+2 f(x) = \frac{\log_2 (x + 3)}{x^2 + 3x + 2}

To find the domain of f(x) , consider the following conditions:

log2(x+3) \log_2 (x + 3) is defined if x+3>0x>3 x + 3 > 0 \Rightarrow x > -3

Denominator x2+3x+20 x^2 + 3x + 2 \neq 0

Factor the denominator:

x2+3x+2=(x+1)(x+2) x^2 + 3x + 2 = (x + 1)(x + 2)

So denominator is zero at x=1 x = -1 and x=2 x = -2

Therefore, domain is:(3,)1,2(-3,\infty)-{-1,-2}

Hence, the correct answer is Option (d).

118

Let ∗ be a binary operation defined on the set of positive rational numbers Q+ by the rule a∗b=ab3\frac{ab}{3}, ∀a,b∈Q+. Then the inverse of 4∗6 is

  1. ((a))

    98\frac{9}{8}

  2. ((b))

    23\frac{2}{3}

  3. ((c))

    38\frac{3}{8}

  4. ((d))

    32\frac{3}{2}

Show Answer
Answer: ((a))

98\frac{9}{8}

Calculation:

Given: ab=ab3,a,bQ+a * b = \frac{ab}{3}, \quad \forall a,b \in Q^+

Find the inverse of 464 * 6 under the operation * .

46=4×63=8\Rightarrow 4 * 6 = \frac{4 \times 6}{3} = 8

Let the inverse of 8 be xx such that 8x=e8 * x = e, where ee is the identity element.

Find identity ee:

ae=aae3=ae=3\Rightarrow a * e = a \Rightarrow \frac{a e}{3} = a \Rightarrow e = 3

Now find xx:

8x=38x3=38x=9x=98\Rightarrow 8 * x = 3 \Rightarrow \frac{8 x}{3} = 3 \Rightarrow 8x = 9 \Rightarrow x = \frac{9}{8}

∴ the inverse of 4 * 6 is 98\frac{9}{8}

Therefore, the correct answer is Option (a).

119

The Least order of non- Abelian group is 

  1. ((a))

    4

  2. ((b))

    5

  3. ((c))

    6

  4. ((d))

    8

Show Answer
Answer: ((c))

6

Calculation:

We are asked to find the least order of a non-Abelian group.

Let us consider the symmetric group Sn S_n , which represents all permutations of n n elements.

The order of the symmetric group is given by:

O(Sn)=n! O(S_n) = n!

Now, consider S3 S_3 :

O(S3)=3!=3×2×1=6 O(S_3) = 3! = 3 \times 2 \times 1 = 6

The group S3 S_3 is non-Abelian because for some permutations a a and b b , we have abba ab \ne ba .

Hence, the least order of a non-Abelian group is 6.

120

if the function f:RRf : R → R is defined by f(x)=x2+xf(x) = x^2 + x then the function f is

  1. ((a))

    one-one but not onto

  2. ((b))

    onto but not one-one

  3. ((c))

    both one-one and onto

  4. ((d))

    neither one-one nor onto

Show Answer
Answer: ((d))

neither one-one nor onto

Calculation:

Given function: f(x)=x2+x f(x) = x^2 + x , with domain and codomain RR \mathbb{R} \to \mathbb{R} .

Check One-One

Let f(0)=02+0=0 f(0) = 0^2 + 0 = 0 and f(1)=(1)2+(1)=11=0 f(-1) = (-1)^2 + (-1) = 1 - 1 = 0

So, f(0)=f(1) f(0) = f(-1) but 01f 0 \ne -1 \Rightarrow f is not one-one.

Check Onto

To check surjectivity, solve f(x)=yx2+x=yx2+xy=0 f(x) = y \Rightarrow x^2 + x = y \Rightarrow x^2 + x - y = 0

This quadratic has real solutions only if discriminant D0D=1+4y0y14 D \geq 0 \Rightarrow D = 1 + 4y \geq 0 \Rightarrow y \geq -\frac{1}{4}

Hence, range of f f is [14,)R \left[ -\frac{1}{4}, \infty \right) \ne \mathbb{R} \Rightarrow not onto.

Hence, the function is neither one-one nor onto.

Hence, the correct answer is Option 4.

121

Consider the following statements:

I. If A  is a skew-symmetric matrix, then A2 is symmetric.

II. Trace of a skew-symmetric matrix of an odd order is always zero.

Which of the above statements is/are true?

  1. ((a))

    Only I

  2. ((b))

    Only II

  3. ((c))

    Both I and II

  4. ((d))

    Neither I nor II

Show Answer
Answer: ((c))

Both I and II

Calculation:

Statement I: If A A is a skew-symmetric matrix, then AT=A A^T = -A .

Now, (A2)T=(AA)T=ATAT=(A)(A)=A2A2 (A^2)^T = (AA)^T = A^T A^T = (-A)(-A) = A^2 \Rightarrow A^2 is symmetric.

So, Statement I is true.

Statement II: In a skew-symmetric matrix, diagonal entries are always zero because aii=aiiaii=0 a_{ii} = -a_{ii} \Rightarrow a_{ii} = 0 .

Hence, trace = sum of diagonal elements = 0.

This holds true for all orders, especially for odd order.

So, Statement II is also true.

Hence, both Statement I and Statement II are true.

Hence, the correct answer is Option 3.

122

The system of equations

x+2y+3z=1x + 2y + 3z = 1

2x+y+3z=22x+y+3z=2

x+y+2z=3x + y + 2z = 3

has

  1. ((a))

    no solution

  2. ((b))

    unique solution

  3. ((c))

    infinite solutions

  4. ((d))

    None of the above

Show Answer
Answer: ((a))

no solution

Calculation:

Given system of equations:

 x+2y+3z=1 x + 2y + 3z = 1 ......(1)

 2x+y+3z=2 2x + y + 3z = 2 .......(2)

 x+y+2z=3 x + y + 2z = 3 ........(3)

Write the coefficient matrix A A :

A=123 213 112 A = \begin{vmatrix} 1 & 2 & 3 \ 2 & 1 & 3 \ 1 & 1 & 2 \end{vmatrix}

A=1×(1×23×1)2×(2×23×1)+3×(2×11×1)|A| = 1 \times (1 \times 2 - 3 \times 1) - 2 \times (2 \times 2 - 3 \times 1) + 3 \times (2 \times 1 - 1 \times 1)

=1×(23)2×(43)+3×(21)=12+3=0= 1 \times (2 - 3) - 2 \times (4 - 3) + 3 \times (2 - 1) = -1 - 2 + 3 = 0

Since A=0 |A| = 0 , the system may have either no solution or infinitely many solutions.

Calculate Ax |A_x| by replacing the first column with constants:

Ax=123 213 312 A_x = \begin{vmatrix} 1 & 2 & 3 \ 2 & 1 & 3 \ 3 & 1 & 2 \end{vmatrix}

Calculate Ax |A_x| :

Ax=1×(1×23×1)2×(2×23×3)+3×(2×11×3)|A_x| = 1 \times (1 \times 2 - 3 \times 1) - 2 \times (2 \times 2 - 3 \times 3) + 3 \times (2 \times 1 - 1 \times 3)

=1×(23)2×(49)+3×(23)=1+103=6= 1 \times (2 - 3) - 2 \times (4 - 9) + 3 \times (2 - 3) = -1 + 10 - 3 = 6

Calculate Ay |A_y| by replacing the second column with constants:

Ay=113 223 132 A_y = \begin{vmatrix} 1 & 1 & 3 \ 2 & 2 & 3 \ 1 & 3 & 2 \end{vmatrix}

Calculate Ay |A_y| :

Ay=1×(2×23×3)1×(2×23×1)+3×(2×32×1)|A_y| = 1 \times (2 \times 2 - 3 \times 3) - 1 \times (2 \times 2 - 3 \times 1) + 3 \times (2 \times 3 - 2 \times 1)

=1×(49)1×(43)+3×(62)=51+12=6= 1 \times (4 - 9) - 1 \times (4 - 3) + 3 \times (6 - 2) = -5 - 1 + 12 = 6

Calculate Az |A_z| by replacing the third column with constants:

Az=121 212 113 A_z = \begin{vmatrix} 1 & 2 & 1 \ 2 & 1 & 2 \ 1 & 1 & 3 \end{vmatrix}

Calculate Az |A_z| :

Az=1×(1×32×1)2×(2×32×1)+1×(2×11×1)|A_z| = 1 \times (1 \times 3 - 2 \times 1) - 2 \times (2 \times 3 - 2 \times 1) + 1 \times (2 \times 1 - 1 \times 1)

=1×(32)2×(62)+1×(21)=18+1=6= 1 \times (3 - 2) - 2 \times (6 - 2) + 1 \times (2 - 1) = 1 - 8 + 1 = -6

Since A=0 |A| = 0 but at least one of Ax,Ay,Az |A_x|, |A_y|, |A_z| is non-zero, the system has no solution (it is inconsistent).

Hence, the correct answer is Option 1

123

if A is a 2 ×2 matrix such that trace(A) = 6 , |A| = 12 hen trace(A−1) is

  1. ((a))

    1/2

  2. ((b))

    1/3

  3. ((c))

    1/6

  4. ((d))

    1

Show Answer
Answer: ((a))

1/2

Calculation:

Let A=[ab cd] A = \begin{bmatrix} a & b \ c & d \end{bmatrix}

Given: trace(A)=a+d=6 \text{trace}(A) = a + d = 6 and A=adbc=12 |A| = ad - bc = 12

The inverse of a 2×2 matrix is:

A1=1adbc[db ca] A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}

So, trace(A1)=a+dadbc \text{trace}(A^{-1}) = \frac{a + d}{ad - bc}

trace(A1)=612=12 \Rightarrow \text{trace}(A^{-1}) = \frac{6}{12} = \frac{1}{2}

Hence, the correct answer is Option 1

Alternate Method

Trace (A-1) = Tr(A)A\frac{Tr(A)}{|A|}

=612=12= \frac{6}{12} = \frac{1}{2}

124

If f(x1x)=x31x3f (x - \frac{1}{x}) = x^3 - \frac{1}{x^3} then the value of f(1) is

  1. ((a))

    -2

  2. ((b))

    -1

  3. ((c))

    0

  4. ((d))

    4

Show Answer
Answer: ((d))

4

Calculation:

Given: f(x1x)=x31x3 f\left(x - \frac{1}{x}\right) = x^3 - \frac{1}{x^3}

We are asked to find: f(1) f(1)

Let t=x1xf(t)=x31x3 t = x - \frac{1}{x} \Rightarrow f(t) = x^3 - \frac{1}{x^3}

Now, to find f(1) f(1) , solve x1x=1 x - \frac{1}{x} = 1

Multiply both sides by x x :

x21=xx2x1=0 x^2 - 1 = x \Rightarrow x^2 - x - 1 = 0

Now use the identity:

x31x3=(x1x)3+3(x1x) x^3 - \frac{1}{x^3} = \left(x - \frac{1}{x} \right)^3 + 3\left(x - \frac{1}{x} \right)

Since x1x=1 x - \frac{1}{x} = 1 , we substitute:

f(1)=13+3(1)=1+3=4 f(1) = 1^3 + 3(1) = 1 + 3 = 4

Hence, the correct answer is Option 4.

125

For the equation |x2| +|x| - 6 =0

  1. ((a))

    there is only one root

  2. ((b))

    the sum of roots is -1

  3. ((c))

    the product of roots is -4

  4. ((d))

    there are four roots

Show Answer
Answer: ((c))

the product of roots is -4

Calculation:

Given equation: x2+x6=0 |x|^2 + |x| - 6 = 0

Let y=x y = |x| , then the equation becomes: y2+y6=0 y^2 + y - 6 = 0

Solving the quadratic

y=2 \Rightarrow y = 2 or y=3 y = -3

Since y=x0 y = |x| \geq 0 , discard y=3 y = -3 .

So, x=2x=±2 |x| = 2 \Rightarrow x = \pm 2

Thus, the roots are: x=2,2 x = 2, -2

Sum of roots: 2+(2)=0 2 + (-2) = 0

Product of roots: 2(2)=4 2 \cdot (-2) = -4

Hence, the correct answer is Option 3.

126

If the roots of the equation (ab)x2+(ca)x+(bc)=0(a−b)x^2+(c−a)x+(b−c)=0 are equal, then a, b and c are in

  1. ((a))

    arithmetic progression

  2. ((b))

    geometric progression

  3. ((c))

    harmonic progression

  4. ((d))

    none of the above

Show Answer
Answer: ((a))

arithmetic progression

Calculation:

Given equation: (ab)x2+(ca)x+(bc)=0 (a - b)x^2 + (c - a)x + (b - c) = 0

Given that the roots are equal, we know that: α=β \alpha = \beta

Let α=β=1 \alpha = \beta = 1

From sum of roots formula: α+β=BA=caab \alpha + \beta = -\frac{B}{A} = -\frac{c - a}{a - b}

Since α+β=1+1=2 \alpha + \beta = 1 + 1 = 2 ,

we have: 2=caab 2 = -\frac{c - a}{a - b}

Multiply both sides: 2(ab)=ac 2(a - b) = a - c

2a2b=ac2aa+c=2ba+c=2b 2a - 2b = a - c \Rightarrow 2a - a + c = 2b \Rightarrow a + c = 2b

This is the condition for Arithmetic Progression: 2b=a+c 2b = a + c

Hence, the correct answer is (a) Arithmetic Progression.

127

If f(x)=cosxandg(x)=sinx,f(x)=cos∣x∣ and g(x)=sin∣x∣, then

  1. ((a))

    both f and g are even functions

  2. ((b))

    both f and g are odd functions

  3. ((c))

    f is an even function and g is an odd function

  4. ((d))

    f is an odd function and g is an even function

Show Answer
Answer: ((a))

both f and g are even functions

Calculation:

Given functions: f(x)=cosxf(x) = \cos |x| and g(x)=sinxg(x) = \sin |x|

Check whether (f) and (g) are even or odd functions.

Recall: A function h(x) is even if h(x)=h(x) and odd if h(x)=h(x).\text{Recall: A function } h(x) \text{ is even if } h(-x) = h(x) \text{ and odd if } h(-x) = -h(x).

Evaluate f(-x):

f(x)=cosx=cosx=f(x)\Rightarrow f(-x) = \cos|-x| = \cos|x| = f(x)

So, (f) is an even function.

Evaluate g(-x)

g(x)=sinx=sinx=g(x)\Rightarrow g(-x) = \sin|-x| = \sin|x| = g(x)

So, g is also an even function.

∴ both (f) and (g) are even functions.

Hence, the correct answer is Option (a).

128

If f(x)=1xx2 xx21 x21xf(x)=\begin{vmatrix} 1 & x & x^2 \ x & x^2 & 1 \ x^2 & 1 & x \end{vmatrix}​​ then the value of \()f(\(\sqrt[3]{3}\))  is

  1. ((a))

    -6

  2. ((b))

    6

  3. ((c))

    4

  4. ((d))

    -4

Show Answer
Answer: ((d))

-4

Calculation:

Given: f(x)=1xx2 xx21 x21xf(x) = \begin{vmatrix} 1 & x & x^2 \ x & x^2 & 1 \ x^2 & 1 & x \end{vmatrix}

f(x)=1x21 1xxx1 x2x+x2xx2 x21\Rightarrow f(x) = 1 \cdot \begin{vmatrix} x^2 & 1 \ 1 & x \end{vmatrix} - x \cdot \begin{vmatrix} x & 1 \ x^2 & x \end{vmatrix} + x^2 \cdot \begin{vmatrix} x & x^2 \ x^2 & 1 \end{vmatrix}

f(x)=1(x31)x(0)+x2(xx4)\Rightarrow f(x) = 1(x^3 - 1) - x(0) + x^2(x - x^4)

f(x)=x31+x3x6\Rightarrow f(x) = x^3 - 1 + x^3 - x^6

f(x)=2x3x61\Rightarrow f(x) = 2x^3 - x^6 - 1

Now evaluate at: x=33x = \sqrt[3]{3}

x3=3, x6=9\Rightarrow x^3 = 3,\ x^6 = 9

f(33)=2(3)91=691=4\Rightarrow f(\sqrt[3]{3}) = 2(3) - 9 - 1 = 6 - 9 - 1 = -4

Hence, the correct answer is Option (d)

129

Let R be a relation on a set A and let IA​ denote the identity relation on A. Then R is antisymmetric if and only if

  1. ((a))

    R=R1R=R^{-1}

  2. ((b))

    RR1IAR \cup R^{-1} \subseteq I_A

  3. ((c))

    RR1IAR \cap R^{-1} \subseteq I_A

  4. ((d))

    None of the above

Show Answer
Answer: ((c))

RR1IAR \cap R^{-1} \subseteq I_A

Calculation:

Given, a relation R R on a set A A , and IA I_A denotes the identity relation on A A . We are asked to find the correct condition for which R is antisymmetric.

Definition: A relation R R is antisymmetric if:

⇒ (a,b)R and (b,a)Ra=b (a, b) \in R \text{ and } (b, a) \in R \Rightarrow a = b

This means that if both (a,b) (a, b) and (b,a) (b, a) are in R R , then they must be the same element (i.e., a=b a = b ), implying that the pair lies in the identity relation.

Mathematical Condition:

⇒ RR1IA R \cap R^{-1} \subseteq I_A

This condition states that the intersection of the relation R R and its inverse R1 R^{-1} must only contain identity elements — satisfying the definition of antisymmetry.

Now let's evaluate the options:

Option 1: R=R1 R = R^{-1} → This is the condition for a symmetric relation, not antisymmetric. 

Option 2: RR1IA R \cup R^{-1} \subseteq I_A → Too strict and not a valid characterization. 

Option 3: RR1IA R \cap R^{-1} \subseteq I_A → This is exactly the definition of antisymmetry. 

Option 4: None of the above → Incorrect, since Option 3 is valid. 

 the correct answer is Option 3: .

130

If x is the first term of a geometric progression and the sum of its infinite terms is 13\frac{1}{3}​, then x lies in the interval

  1. ((a))

    0<x<120 < x < \frac{1}{2}

  2. ((b))

    1<x<14-1 < x < \frac{1}{4}

  3. ((c))

    12<x<12-\frac{1}{2} < x < \frac{1}{2}

  4. ((d))

    0<x<230 < x < \frac{2}{3}

Show Answer
Answer: ((d))

0<x<230 < x < \frac{2}{3}

Calculation:

Given: x x is the first term of a geometric progression, and the sum of its infinite terms is 13 \frac{1}{3}

The formula for the sum of an infinite geometric progression is: S=a1r S = \frac{a}{1 - r} , where r<1 |r| < 1

Let the first term a=x a = x and the common ratio be r r . Then:

⇒ x1r=13 \frac{x}{1 - r} = \frac{1}{3}

x=1r3 \Rightarrow x = \frac{1 - r}{3}

Now, solving for r r in terms of x x :

⇒ r=13x r = 1 - 3x

For the sum to exist, the common ratio must satisfy: r<113x<1 |r| < 1 \Rightarrow |1 - 3x| < 1

Now solving: 1<13x<1 -1 < 1 - 3x < 1

2<3x<0 \Rightarrow -2 < -3x < 0

0<x<23 \Rightarrow 0 < x < \frac{2}{3}

Hence, the correct answer is Option (d

131

If  n=0rn=s,r<1,\sum_{n=0}^{\infty} r^n=s, |r|<1,  then  n=0r2n\sum_{n=0}^{\infty} r^{2n}    is equal to:

  1. ((a))

    s22s+1\frac{s^2}{2s+1}

  2. ((b))

    s22s1\frac{s^2}{2s-1}

  3. ((c))

    2ss21\frac{2s}{s^2-1}

  4. ((d))

    s2s^2

Show Answer
Answer: ((b))

s22s1\frac{s^2}{2s-1}

Calculation:

Given: n=0rn=s\sum_{n=0}^\infty r^n = s

11r=s\Rightarrow \frac{1}{1 - r} = s

1r=1s\Rightarrow 1 - r = \frac{1}{s}

r=11s\Rightarrow r = 1 - \frac{1}{s}

We need to find:

n=0r2n=11r2\sum_{n=0}^{\infty} r^{2n} = \frac{1}{1 - r^2}

Compute r2r^2:

⇒ r2=(11s)2=12s+1s2r^2 = \left(1 - \frac{1}{s}\right)^2 = 1 - \frac{2}{s} + \frac{1}{s^2}

So:

⇒ 1r2=1(12s+1s2)=2s1s21 - r^2 = 1 - \left(1 - \frac{2}{s} + \frac{1}{s^2}\right) = \frac{2}{s} - \frac{1}{s^2}

Taking the reciprocal:

⇒ n=0r2n=11r2=12s1s2=s22s1\sum_{n=0}^{\infty} r^{2n} = \frac{1}{1 - r^2} = \frac{1}{\frac{2}{s} - \frac{1}{s^2}} = \frac{s^2}{2s - 1}

Hence, the correct answer is Option (b).

132

The infinite series 11P+12P+13P+14P+\frac{1}{1^P}+\frac{1}{2^P}+\frac{1}{3^P}+\frac{1}{4^P}+\dots\infty is convergent, if:

  1. ((a))

    p=0p=0

  2. ((b))

    p<1p<1

  3. ((c))

    p=1p=1

  4. ((d))

    p>1p>1

Show Answer
Answer: ((d))

p>1p>1

Calculation:

Given infinite series: n=11np \sum_{n=1}^{\infty} \frac{1}{n^p}

This is a p-series, and its convergence depends on the value of p .

The rule for convergence of a p-series is:

If p1diverges, and if p>1converges\text{If } p \leq 1 \Rightarrow \text{diverges, and if } p > 1 \Rightarrow \text{converges}

This result is based on the integral test applied to the function f(n)=1npf(n) = \frac{1}{n^p}

Therefore, the series:

11p+12p+13p+\frac{1}{1^p} + \frac{1}{2^p} + \frac{1}{3^p} + \cdots

is convergent if and only if p>1 p > 1 .

Hence, the correct answer is Option (d.

133

Which one of the following sequences is not convergent?

  1. ((a))

    (1+(1)n)(1+(-1)^n)

  2. ((b))

    (nn+1)\left(\frac{n}{n+1}\right)

  3. ((c))

    (1+(1)nn)\left(1+\frac{(-1)^n}{n}\right)

  4. ((d))

    None of the above

Show Answer
Answer: ((a))

(1+(1)n)(1+(-1)^n)

Calculation: 

For (a), when n is even: an=1+1=2 \text{For (a), when } n \text{ is even: } a_n = 1 + 1 = 2

For (a), when n is odd: an=11=0 \text{For (a), when } n \text{ is odd: } a_n = 1 - 1 = 0

Sequence alternates between 0 and 2, so it does not converge. \Rightarrow \text{Sequence alternates between 0 and 2, so it does not converge.}

For (b), as n,an=nn+1=11+1n1 \text{For (b), as } n \to \infty, a_n = \frac{n}{n+1} = \frac{1}{1 + \frac{1}{n}} \to 1

Sequence converges to 1. \Rightarrow \text{Sequence converges to 1.}

For (c), as n,(1)nn0, so an1 \text{For (c), as } n \to \infty, \frac{(-1)^n}{n} \to 0, \text{ so } a_n \to 1

Sequence converges to 1. \Rightarrow \text{Sequence converges to 1.}

Hence, only sequence (a) is not convergent.

134

If (1x+x2)n=a0+a1x+a2x2++a2nx2n(1-x+x^2)^n = a_0+a_1x+a_2x^2+\dots+a_{2n}x^{2n}  ,then  (a0+a2+a4++a2n)(a_0+a_2+a_4+\dots+a_{2n})  is equal to:

  1. ((a))

    3n12\frac{3^n-1}{2}

  2. ((b))

    3n+12\frac{3^n+1}{2}

  3. ((c))

    3n+22\frac{3^n+2}{2}

  4. ((d))

    3n22\frac{3^n-2}{2}

Show Answer
Answer: ((b))

3n+12\frac{3^n+1}{2}

Calculation:

Given: (1x+x2)n=a0+a1x+a2x2++a2nx2n (1 - x + x^2)^n = a_0 + a_1 x + a_2 x^2 + \cdots + a_{2n} x^{2n}

We need to find: a0+a2+a4++a2n a_0 + a_2 + a_4 + \cdots + a_{2n}

Using the property of sums of coefficients for even powers, we write:

⇒ f(1)=k=02nak=a0+a1+a2++a2n f(1) = \sum_{k=0}^{2n} a_k = a_0 + a_1 + a_2 + \cdots + a_{2n}

⇒ f(1)=k=02nak(1)k=a0a1+a2a3++(1)2na2n f(-1) = \sum_{k=0}^{2n} a_k (-1)^k = a_0 - a_1 + a_2 - a_3 + \cdots + (-1)^{2n} a_{2n}

Add both equations:

⇒ f(1)+f(1)=2(a0+a2+a4++a2n) f(1) + f(-1) = 2 (a_0 + a_2 + a_4 + \cdots + a_{2n})

Therefore,

⇒ a0+a2+a4++a2n=f(1)+f(1)2 a_0 + a_2 + a_4 + \cdots + a_{2n} = \frac{f(1) + f(-1)}{2}

Calculate f(1)  and f(-1):

⇒ f(1)=(11+1)n=1n=1 f(1) = (1 - 1 + 1)^n = 1^n = 1

⇒ f(1)=(1(1)+(1)2)n=(1+1+1)n=3n f(-1) = (1 - (-1) + (-1)^2)^n = (1 + 1 + 1)^n = 3^n

∴  a0+a2+a4++a2n=1+3n2 a_0 + a_2 + a_4 + \cdots + a_{2n} = \frac{1 + 3^n}{2}

Hence, the correct answer is option b.

135

Every subgroup of an Abelian group is not

  1. ((a))

    cyclic

  2. ((b))

    Abelian

  3. ((c))

    normal

  4. ((d))

    None of the above

Show Answer
Answer: ((a))

cyclic

Solution:

In group theory, an Abelian group \text{Abelian group} is a group where the group operation is commutative, i.e., for any elements a,ba, b, ab=baab = ba.

Every subgroup of an Abelian group is Abelian because the subgroup inherits the commutative property from the group.

Every subgroup of an Abelian group is normal because in Abelian groups, all subgroups are normal (the conjugate of any subgroup element is the element itself since ghg1=hg h g^{-1} = h).

However, not every subgroup of an Abelian group is cyclic.

A cyclic group is generated by a single element. An Abelian group may have subgroups that are not cyclic (for example, the group Z2×Z2 \mathbb{Z}_2 \times \mathbb{Z}_2 is Abelian but not cyclic).

  Every subgroup of an Abelian group is not necessarily cyclic.

Hence, the correct answer is option a.

136

If a×b2+ab2=144∣ a × b ∣ ^2 +∣ a ⋅ b ∣ ^2 =144 and a=4∣a∣=4, then b∣b∣ is equal to

  1. ((a))

    12

  2. ((b))

    8

  3. ((c))

    4

  4. ((d))

    3

Show Answer
Answer: ((d))

3

Calculation:

Given: a×b2+ab2=144 |\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 = 144 and a=4 |\vec{a}| = 4

We know the identity: a×b2+ab2=a2b2 |\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2

Substituting the given values: 144=(4)2×b2 144 = (4)^2 \times |\vec{b}|^2

144=16×b2 144 = 16 \times |\vec{b}|^2

Dividing both sides by 16: b2=14416=9 |\vec{b}|^2 = \frac{144}{16} = 9

Taking the square root: b=3 |\vec{b}| = 3

Hence, the correct answer is Option (d).

137

If F=x2yi^+xj^+2yzk^\vec{F} = x^2y\hat{i} + x\hat{j} + 2yz\hat{k} , then the value of div curlF\vec{F} is

  1. ((a))

    0

  2. ((b))

    1

  3. ((c))

    2

  4. ((d))

    3

Show Answer
Answer: ((a))

0

Calculation:

Given: F=x2yi^+xzj^+2yzk^\vec{F} = x^2 y \hat{i} + x z \hat{j} + 2 y z \hat{k}

We need to find: div(curl F)\text{div}(\text{curl} \ \vec{F})

Recall the vector calculus identity:

div(curl F)=0\text{div}(\text{curl} \ \vec{F}) = 0

This identity states that the divergence of the curl of any vector field is always zero.

∴ , the value of div(curl F)\text{div}(\text{curl} \ \vec{F}) is 00

Therefore, the correct answer is Option (a).

138

If a\vec a and b\vec b are constant vectors, then ([r,a,b])\nabla ([\vec r, \vec a, \vec b]) is equal to

  1. ((a))

    0^\hat{0}

  2. ((b))

    (ab)r(\vec{a} \cdot \vec{b})\vec{r}

  3. ((c))

    a×b\vec{a} \times \vec{b}

  4. ((d))

    (a×b)r(\vec{a} \times \vec{b})|\vec{r}|

Show Answer
Answer: ((c))

a×b\vec{a} \times \vec{b}

Calculation:

Given: ([r,a,b]) \nabla([\mathbf{r}, \mathbf{a}, \mathbf{b}])

[r,a,b]=r(a×b)[\mathbf{r}, \mathbf{a}, \mathbf{b}] = \mathbf{r} \cdot (\mathbf{a} \times \mathbf{b})

(r(a×b))\Rightarrow \nabla(\mathbf{r} \cdot (\mathbf{a} \times \mathbf{b}))

(rc) where c=a×b\Rightarrow \nabla(\mathbf{r} \cdot \mathbf{c}) \text{ where } \mathbf{c} = \mathbf{a} \times \mathbf{b}

c since (rc)=c (gradient of dot product with constant vector)\Rightarrow \mathbf{c} \text{ since } \nabla(\mathbf{r} \cdot \mathbf{c}) = \mathbf{c} \text{ (gradient of dot product with constant vector)}

a×b\Rightarrow \mathbf{a} \times \mathbf{b}

Hence, the correct answer is Option b.

139

The value of (c×a)×(a×b)(\vec c × \vec a)×(\vec a×\vec b) is

  1. ((a))

    0^\hat{0}

  2. ((b))

    [bca]b[\vec{b}\vec{c}\vec{a}]\vec{b}

  3. ((c))

    [cab]a[\vec{c}\vec{a}\vec{b}]\vec{a}

  4. ((d))

    [abc]a[\vec{a}\vec{b}\vec{c}]\vec{a}

Show Answer
Answer: ((d))

[abc]a[\vec{a}\vec{b}\vec{c}]\vec{a}

Calculation:

Given: (c×a)×(a×b)(\vec{c} \times \vec{a}) \times (\vec{a} \times \vec{b})

Using vector triple product identity: p×(q×r)=q(pr)r(pq)\vec{p} \times (\vec{q} \times \vec{r}) = \vec{q}(\vec{p} \cdot \vec{r}) - \vec{r}(\vec{p} \cdot \vec{q})

Let: p=c×a,q=a,r=b\vec{p} = \vec{c} \times \vec{a}, \quad \vec{q} = \vec{a}, \quad \vec{r} = \vec{b}

(c×a)×(a×b)=a[(c×a)b]b[(c×a)a]\Rightarrow (\vec{c} \times \vec{a}) \times (\vec{a} \times \vec{b}) = \vec{a} \big[(\vec{c} \times \vec{a}) \cdot \vec{b}\big] - \vec{b} \big[(\vec{c} \times \vec{a}) \cdot \vec{a}\big]

(c×a)a=0(\vec{c} \times \vec{a}) \cdot \vec{a} = 0  ,  

(c×a)×(a×b)=a[(c×a)b]\Rightarrow (\vec{c} \times \vec{a}) \times (\vec{a} \times \vec{b}) = \vec{a} \big[(\vec{c} \times \vec{a}) \cdot \vec{b}\big]

Using scalar triple product identity: (c×a)b=a(b×c)(\vec{c} \times \vec{a}) \cdot \vec{b} = \vec{a} \cdot (\vec{b} \times \vec{c})

(c×a)×(a×b)=a[a(b×c)]\Rightarrow (\vec{c} \times \vec{a}) \times (\vec{a} \times \vec{b}) = \vec{a} [\vec{a} \cdot (\vec{b} \times \vec{c})]

Hence, the value is [a(b×c)]a[\vec{a} \cdot (\vec{b} \times \vec{c})] \vec{a}, which corresponds to Option (d).

140

div (r×a \vec r \times \vec a ), where vec a\vec {a} is a constant vector,is equal to

  1. ((a))

    0

  2. ((b))

    a∣ \vec{a} ∣(|\vec{a}|\)

  3. ((c))

    r∣ \vec{r} ∣(|\vec{r}|\)

  4. ((d))

    ar\vec{a} ⋅ \vec{r} (\vec{a}. \vec{r}\)

Show Answer
Answer: ((a))

0

Calculation:

Given: div(r×a),a=constant vector \text{div} (\vec{r} \times \vec{a}), \quad \vec{a} = \text{constant vector}

Write r×a \vec{r} \times \vec{a} as a determinant:

r×a=i^j^k^ xyz a1a2a3=i^(ya3za2)j^(xa3za1)+k^(xa2ya1) \vec{r} \times \vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ x & y & z \ a_1 & a_2 & a_3 \end{vmatrix} = \hat{i}(y a_3 - z a_2) - \hat{j}(x a_3 - z a_1) + \hat{k}(x a_2 - y a_1)

Now find the divergence:

div(r×a)=x(ya3za2)+y[(xa3za1)]+z(xa2ya1) \text{div} (\vec{r} \times \vec{a}) = \frac{\partial}{\partial x}(y a_3 - z a_2) + \frac{\partial}{\partial y}[-(x a_3 - z a_1)] + \frac{\partial}{\partial z}(x a_2 - y a_1)

Since a \vec{a} is constant, partial derivatives of components with respect to unrelated variables are zero:

x(ya3za2)=0,y[(xa3za1)]=0,z(xa2ya1)=0 \Rightarrow \frac{\partial}{\partial x}(y a_3 - z a_2) = 0, \quad \frac{\partial}{\partial y}[-(x a_3 - z a_1)] = 0, \quad \frac{\partial}{\partial z}(x a_2 - y a_1) = 0

Therefore,

div(r×a)=0 \text{div} (\vec{r} \times \vec{a}) = 0

Hence, the correct answer is Option (a):

141

If vectors A\vec{A} and B\vec{B} are irrotational, then

  1. ((a))

    A×B\vec{A} × \vec{B}  is irrotational

  2. ((b))

    A×B\vec{A} × \vec{B}  is solenoidal

  3. ((c))

    AB\vec{A} - \vec{B}  is rotational

  4. ((d))

    None of the above

Show Answer
Answer: ((b))

A×B\vec{A} × \vec{B}  is solenoidal

Calculation:

Given: If vectors A and B are irrotational, then \text{If vectors } \vec{A} \text{ and } \vec{B} \text{ are irrotational, then}

×A=0,×B=0 \Rightarrow \nabla × \vec{A} = \vec{0}, \quad \nabla × \vec{B} = \vec{0}

Using the vector identity

×(A×B)=A(B)B(A)+(B)A(A)B \nabla × (\vec{A} × \vec{B}) = \vec{A} (\nabla \cdot \vec{B}) - \vec{B} (\nabla \cdot \vec{A}) + (\vec{B} \cdot \nabla) \vec{A} - (\vec{A} \cdot \nabla) \vec{B}

Since A\vec{A} and B\vec{B} are irrotational, their curls are zero but divergence may not be zero

Hence, the curl of A\vec{A} × B\vec{B} is not necessarily zero, but

(A×B)=0\nabla \cdot (\vec{A} \times \vec{B}) = 0

∴ A×B\vec{A} \times \vec{B} is solenoidal

Hence, the correct answer is Option (b).

142

The vector rr3\frac{\vec{r}}{|\vec{r}|^3}​, where r=xi^+yj^+zk^\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}, is

  1. ((a))

    only solenoidal

  2. ((b))

    only irrotational

  3. ((c))

    both solenoidal and irrotational

  4. ((d))

    neither solenoidal nor irrotational

Show Answer
Answer: ((b))

only irrotational

Calculation:

Given vector field: F=rr3 \vec{F} = \frac{\vec{r}}{|\vec{r}|^3} , where r=xi^+yj^+zk^ \vec{r} = x\hat{i} + y\hat{j} + z\hat{k}

Check if the field is irrotational (curl-free)

Note that F \vec{F} is the gradient of the scalar potential ϕ=1r \phi = -\frac{1}{|\vec{r}|}

⇒ F=(1r) \vec{F} = \nabla \left( -\frac{1}{|\vec{r}|} \right)

And the curl of a gradient is always zero:

⇒ ×F=0 \nabla \times \vec{F} = 0

So, F \vec{F} is irrotational everywhere (except possibly at the origin).

Check if the field is solenoidal (divergence-free)

We use the known identity from vector calculus:

⇒ (rr3)=4πδ(r) \nabla \cdot \left( \frac{\vec{r}}{|\vec{r}|^3} \right) = 4\pi \delta(\vec{r})

 

Where δ(r) \delta(\vec{r}) is the Dirac delta function centered at the origin.

This implies:

For all r0 \vec{r} \neq 0 , F=0 \nabla \cdot \vec{F} = 0

At the origin, the divergence is not zero; it's a singularity

Hence, the field is not divergence-free at the origin ⇒ not strictly solenoidal.

Conclusion:

F \vec{F} is irrotational (curl-free everywhere).

But F \vec{F} is not solenoidal (not divergence-free at the origin).

 the vector field is only irrotational

Hence, the correct answer is Option (b).

143

If A×B=C×D\vec{A} \times \vec{B}=\vec{C} \times \vec{D} and A×C=B×D\vec{A} \times \vec{C}=\vec{B} \times \vec{D}, then vectors AD\vec{A}-\vec{D} and BC\vec{B}-\vec{C} are

  1. ((a))

    equal

  2. ((b))

    parallel

  3. ((c))

    perpendicular

  4. ((d))

     inclined at an angle of 60°

Show Answer
Answer: ((b))

parallel

Calculation:

Given: 

Rearranging the first equation: 

Using vector identity: 

This implies vectors and are parallel.

Hence, the vectors and are parallel, so the correct answer is Option (b).

144

If  a,b,c\vec{a}, \vec{b}, \vec{c}  are non-coplanar unit vectors such that a×(b×c)=b+c2\vec{a} \times (\vec{b} \times \vec{c})=\frac{\vec{b}+\vec{c}}{\sqrt{2}}  , then the angle between a\vec{a} and b\vec{b} is

  1. ((a))

    3π4\frac{3π}{4}

  2. ((b))

    π4\frac{π}{4}

  3. ((c))

    π2\frac{π}{2}

  4. ((d))

    ππ

Show Answer
Answer: ((a))

3π4\frac{3π}{4}

Calculation:

Given: a,b,c \vec{a}, \vec{b}, \vec{c} are non-coplanar unit vectors such that

⇒ a×(b×c)=b+c2 \vec{a} \times (\vec{b} \times \vec{c}) = \frac{\vec{b} + \vec{c}}{\sqrt{2}}

We use the vector triple product identity:

 ⇒ a×(b×c)=(ac)b(ab)c \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}

So comparing both sides:

(ac)b(ab)c=b+c2 (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = \frac{\vec{b} + \vec{c}}{\sqrt{2}}

Matching coefficients:

⇒ ac=12 \vec{a} \cdot \vec{c} = \frac{1}{\sqrt{2}} , and ab=12ab=12 -\vec{a} \cdot \vec{b} = \frac{1}{\sqrt{2}} \Rightarrow \vec{a} \cdot \vec{b} = -\frac{1}{\sqrt{2}}

Now, since ab=abcosθ \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta and both are unit vectors:

⇒ cosθ=12θ=cos1(12) \cos\theta = -\frac{1}{\sqrt{2}} \Rightarrow \theta = \cos^{-1} \left(-\frac{1}{\sqrt{2}}\right)

θ=3π4 \Rightarrow \theta = \frac{3\pi}{4}

Hence, the correct answer is Option (a).

145

If v1,v2,v3 \vec{v_1}, \vec{v_2}, \vec{v_3} are three non-zero vectors such that v1×v2=v3,v2×v3=v1 \vec{v_1} \times \vec{v_2}=\vec{v_3}, \vec{v_2} \times \vec{v_3}=\vec{v_1} then

  1. ((a))

    v1=v2|\vec{v_1}|=|\vec{v_2}|

  2. ((b))

    v2=v3|\vec{v_2}|=|\vec{v_3}|

  3. ((c))

    v1=v3|\vec{v_1}|=|\vec{v_3}|

  4. ((d))

    v2=v1×v3\vec{v_2}=\vec{v_1} \times \vec{v_3}

Show Answer
Answer: ((c))

v1=v3|\vec{v_1}|=|\vec{v_3}|

Calculation:

Given: V1×V2=V3,V2×V3=V1\vec{V}_1 \times \vec{V}_2 = \vec{V}_3,\quad \vec{V}_2 \times \vec{V}_3 = \vec{V}_1

V2×(V1×V2)=V1\Rightarrow \vec{V}_2 \times (\vec{V}_1 \times \vec{V}_2) = \vec{V}_1

Using the vector triple product identity:

a×(b×c)=(ac)b(ab)c\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}

V2×(V1×V2)=(V2V2)V1(V2V1)V2\Rightarrow \vec{V}_2 \times (\vec{V}_1 \times \vec{V}_2) = (\vec{V}_2 \cdot \vec{V}_2)\vec{V}_1 - (\vec{V}_2 \cdot \vec{V}_1)\vec{V}_2

V22V1(V1V2)V2=V1\Rightarrow |\vec{V}_2|^2 \vec{V}_1 - (\vec{V}_1 \cdot \vec{V}_2)\vec{V}_2 = \vec{V}_1

(V221)V1=(V1V2)V2\Rightarrow (|\vec{V}_2|^2 - 1)\vec{V}_1 = (\vec{V}_1 \cdot \vec{V}_2)\vec{V}_2

To satisfy this condition, let:

V2=1|\vec{V}_2| = 1

V1V2=0\vec{V}_1 \cdot \vec{V}_2 = 0

Since V3=V1×V2\vec{V}_3 = \vec{V}_1 \times \vec{V}_2 and this vector is orthogonal to both, we conclude:

V1=V3|\vec{V}_1| = |\vec{V}_3|

Hence, the correct answer is Option (c).

146

The equationz3z+3=2\left|\frac{z-3}{z+3}\right|=2 represents

  1. ((a))

    a parabola

  2. ((b))

    a hyperbola

  3. ((c))

    a circle

  4. ((d))

    an ellipse

Show Answer
Answer: ((c))

a circle

Calculation:

Given: z3z+3=2\left| \frac{z - 3}{z + 3} \right| = 2  where  z = x + i y  is a complex number.

z3z+3=2 \Rightarrow \frac{|z - 3|}{|z + 3|} = 2

z3=2z+3 \Rightarrow |z - 3| = 2 |z + 3|

(x3)2+y2=2(x+3)2+y2 \Rightarrow \sqrt{(x - 3)^2 + y^2} = 2 \sqrt{(x + 3)^2 + y^2}

(x3)2+y2=4((x+3)2+y2) \Rightarrow (x - 3)^2 + y^2 = 4 \left( (x + 3)^2 + y^2 \right)

x26x+9+y2=4(x2+6x+9+y2) \Rightarrow x^2 - 6x + 9 + y^2 = 4(x^2 + 6x + 9 + y^2)

x26x+9+y2=4x2+24x+36+4y2 \Rightarrow x^2 - 6x + 9 + y^2 = 4x^2 + 24x + 36 + 4y^2

x26x+9+y24x224x364y2=0 \Rightarrow x^2 - 6x + 9 + y^2 - 4x^2 - 24x - 36 - 4y^2 = 0

3x230x273y2=0 \Rightarrow -3x^2 - 30x - 27 - 3y^2 = 0

x2+10x+9+y2=0(dividing by 3) \Rightarrow x^2 + 10x + 9 + y^2 = 0 \quad \text{(dividing by } -3 \text{)}

(x2+10x+25)+y2=9+25(completing the square) \Rightarrow (x^2 + 10x + 25) + y^2 = -9 + 25 \quad \text{(completing the square)}

(x+5)2+y2=16 \Rightarrow (x + 5)^2 + y^2 = 16

This represents a circle with center (-5, 0) and radius 4.

Hence, the correct answer is Option (c)

147

If xn=cos(π2n)+isin(π2n),nNx_n = \cos\left(\frac{\pi}{2^n}\right) + i\sin\left(\frac{\pi}{2^n}\right), n \in \mathbb{N} then limn(x1x2x3xn)\lim_{n\to\infty} (x_1 \cdot x_2 \cdot x_3 \cdots x_n) is

  1. ((a))

    0

  2. ((b))

    -1

  3. ((c))

    1

  4. ((d))

    2

Show Answer
Answer: ((b))

-1

Calculation:

Given: xn=cos(π2n)+isin(π2n) x_n = \cos\left(\frac{\pi}{2^n}\right) + i \sin\left(\frac{\pi}{2^n}\right)

We are asked to evaluate: limnx1x2x3xn \lim_{n \to \infty} x_1 x_2 x_3 \cdots x_n

x1x2x3xn=cos(π2+π22+π23+)+isin(π2+π22+π23+) \Rightarrow x_1 x_2 x_3 \cdots x_n = \cos\left(\frac{\pi}{2} + \frac{\pi}{2^2} + \frac{\pi}{2^3} + \cdots \right) + i \sin\left(\frac{\pi}{2} + \frac{\pi}{2^2} + \frac{\pi}{2^3} + \cdots \right)

The series inside the angle is a geometric series:

n=1π2n=π2(1112)=π22=π \sum_{n=1}^\infty \frac{\pi}{2^n} = \frac{\pi}{2} \cdot \left( \frac{1}{1 - \frac{1}{2}} \right) = \frac{\pi}{2} \cdot 2 = \pi

x1x2x3=cos(π)+isin(π) \Rightarrow x_1 x_2 x_3 \cdots = \cos(\pi) + i \sin(\pi)

Now, using standard trigonometric values:

cos(π)=1,sin(π)=0 \cos(\pi) = -1, \quad \sin(\pi) = 0

x1x2x3=1+0i=1 \Rightarrow x_1 x_2 x_3 \cdots = -1 + 0i = -1

Hence, the correct answer is Option (b)

148

If f(z)={u(x,y)+iv(x,y),for z0 0,for z=0f(z) = \begin{cases} u(x, y) + iv(x, y), & \text{for } z \ne 0 \ 0, & \text{for } z = 0 \end{cases} where u(x,y)=x3y3x2+y2,u(x, y) = \frac{x^3 - y^3}{x^2 + y^2}, v(x,y)=x3+y3x2+y2v(x, y) = \frac{x^3 + y^3}{x^2 + y^2} then the value of  limz0f(z)f(0)z0\lim_{z \to 0} \frac{f(z) - f(0)}{z - 0} along y = x wiil be

  1. ((a))

    1 - i 

  2. ((b))

    1i2\frac{1 - i}{2}

  3. ((c))

    1 + i

  4. ((d))

    1+i2\frac{1 + i}{2}

Show Answer
Answer: ((d))

1+i2\frac{1 + i}{2}

Calculation:

We are given a function f(z)=u(x,y)+iv(x,y) f(z) = u(x, y) + iv(x, y) for z0 z \neq 0 , and f(0)=0 f(0) = 0 , where:

u(x,y)=x3y3x2+y2,v(x,y)=x3+y3x2+y2 u(x, y) = \frac{x^3 - y^3}{x^2 + y^2}, \quad v(x, y) = \frac{x^3 + y^3}{x^2 + y^2}

We are to find: limz0f(z)f(0)z0 \lim_{z \to 0} \frac{f(z) - f(0)}{z - 0} along the path y=x y = x

Substituting y=x y = x , we get:

u(x,x)=x3x32x2=0 u(x, x) = \frac{x^3 - x^3}{2x^2} = 0

v(x,x)=x3+x32x2=x v(x, x) = \frac{x^3 + x^3}{2x^2} = x

So, f(z)=ix f(z) = ix , and since z=x+ix=x(1+i) z = x + ix = x(1 + i) along this path, we now evaluate:

f(z)z=ixx(1+i)=i1+i \frac{f(z)}{z} = \frac{ix}{x(1+i)} = \frac{i}{1+i}

Multiplying numerator and denominator by the conjugate of the denominator:

i1+i1i1i=i(1i)(1+i)(1i)=ii21(1)=i+12 \frac{i}{1+i} \cdot \frac{1-i}{1-i} = \frac{i(1 - i)}{(1 + i)(1 - i)} = \frac{i - i^2}{1 - (-1)} = \frac{i + 1}{2}

Hence, the value of the limit is 1+i2 \frac{1+i}{2} .

The correct answer is (d).

149

If  a=cos4π3+isin4π3a = \cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3} then the value of (1+a2)3n\left(\frac{1+a}{2}\right)^{3n}   is

  1. ((a))

    (1)n(-1)^n

  2. ((b))

    123n\frac{1}{2^{3n}}

  3. ((c))

    (1)n23n\frac{(-1)^n}{2^{3n}}

  4. ((d))

    (1)n+1(-1)^n+1

Show Answer
Answer: ((c))

(1)n23n\frac{(-1)^n}{2^{3n}}

Calculation:

Given: a=cos(4π3)+isin(4π3) a = \cos\left(\frac{4\pi}{3}\right) + i \sin\left(\frac{4\pi}{3}\right)

a=ei4π3 \Rightarrow a = e^{i \frac{4\pi}{3}}

We need to compute: (1+a2)3n \left( \frac{1 + a}{2} \right)^{3n}

1+a=1+ei4π3=12i32 \Rightarrow 1 + a = 1 + e^{i \frac{4\pi}{3}} = \frac{1}{2} - i\frac{\sqrt{3}}{2}

1+a2=14i34 \Rightarrow \frac{1 + a}{2} = \frac{1}{4} - i\frac{\sqrt{3}}{4}

This is a complex number with magnitude:

14i34=141+3=12 \left| \frac{1}{4} - i\frac{\sqrt{3}}{4} \right| = \frac{1}{4} \sqrt{1 + 3} = \frac{1}{2}

And argument: θ=tan1(3/41/4)=π3 \theta = \tan^{-1}\left( \frac{-\sqrt{3}/4}{1/4} \right) = -\frac{\pi}{3}

So, 1+a2=12cis(π3) \frac{1 + a}{2} = \frac{1}{2} \text{cis}\left(-\frac{\pi}{3}\right)

(1+a2)3n=(12)3ncis(nπ) \Rightarrow \left( \frac{1 + a}{2} \right)^{3n} = \left( \frac{1}{2} \right)^{3n} \cdot \text{cis}(-n\pi)

=123n(cos(nπ)+isin(nπ))=(1)n23n = \frac{1}{2^{3n}} \cdot (\cos(n\pi) + i \sin(n\pi)) = \frac{(-1)^n}{2^{3n}}

Hence, the correct answer is Option (c).

150

If ω(1)\omega (\neq 1) is cube root of unity, then the value of  (1+ω2+2ω)3n(1+ω+2ω2)3n(1+\omega^2+2\omega)^{3n}-(1+\omega+2\omega^2)^{3n} is

  1. ((a))

     0

  2. ((b))

    1

  3. ((c))

    ω\omega

  4. ((d))

    ω2\omega^2

Show Answer
Answer: ((a))

 0

Calculation:

Given: If ω1 is a cube root of unity, then evaluate \text{If } \omega \neq 1 \text{ is a cube root of unity, then evaluate}

(1+ω2+2ω)3n(1+ω+2ω2)3n (1 + \omega^2 + 2\omega)^{3n} - (1 + \omega + 2\omega^2)^{3n}

Using the identity of cube roots of unity: 1+ω+ω2=0 \Rightarrow \text{Using the identity of cube roots of unity: } 1 + \omega + \omega^2 = 0

1+ω=ω2and1+ω2=ω \Rightarrow 1 + \omega = -\omega^2 \quad \text{and} \quad 1 + \omega^2 = -\omega

(1+ω2+2ω)=(1+ω2)+2ω=ω+2ω=ω \Rightarrow (1 + \omega^2 + 2\omega) = (1 + \omega^2) + 2\omega = -\omega + 2\omega = \omega

(1+ω+2ω2)=(1+ω)+2ω2=ω2+2ω2=ω2 \Rightarrow (1 + \omega + 2\omega^2) = (1 + \omega) + 2\omega^2 = -\omega^2 + 2\omega^2 = \omega^2

(1+ω2+2ω)3n(1+ω+2ω2)3n=ω3n(ω2)3n \Rightarrow \therefore (1 + \omega^2 + 2\omega)^{3n} - (1 + \omega + 2\omega^2)^{3n} = \omega^{3n} - (\omega^2)^{3n}

ω3=1ω3n=1and(ω2)3n=1 \Rightarrow \omega^3 = 1 \Rightarrow \omega^{3n} = 1 \quad \text{and} \quad (\omega^2)^{3n} = 1

Expression=11=0 \Rightarrow \text{Expression} = 1 - 1 = 0

Hence, the value of the expression is 0, and the correct answer is Option (a).

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