Official Paper

UPPSC LT Grade Assistant Teacher Maths Official Paper Held On: 06 Dec 2025 Shift 1 (Previous Year Paper)

150 questions · 120 minutes · with answers · free

Part 1 (30 questions)

1

In the Gross Domestic Product (GDP) of India, the largest share in the current years has come from:

  1. ((a))

    Large-scale industries

  2. ((b))

    Service sector

  3. ((c))

    Small and cottage industries

  4. ((d))

    Agricultural sector

Show Answer
Answer: ((b))

Service sector

The correct answer is Service sector.

Key Points

  • The Service sector has the largest contribution to the Gross Domestic Product (GDP) of India in recent years. It accounts for more than 50% of the GDP.
  • This sector includes industries like Information Technology (IT), telecommunications, banking, healthcare, education, hospitality, and real estate.
  • The growth of the IT and Software industry in cities such as Bengaluru, Hyderabad, and Pune has significantly boosted the contribution of the service sector.
  • India is a global leader in IT and Business Process Outsourcing (BPO) services, which has further propelled its service sector growth.
  • The service sector also significantly contributes to employment generation, especially in urban areas, and is a key driver of India's economic development.
  • It is supported by increasing urbanization, rising disposable incomes, and a growing middle-class population that demands more services.
  • The service sector has played a critical role in transforming India into a knowledge-based economy, with a focus on innovation and technology-driven solutions.

Additional Information

  • Large-scale industries
  • These industries include steel, automobile, textiles, and petrochemicals.
  • They play a significant role in the industrial and manufacturing growth of the country.
  • However, their contribution to the GDP is smaller compared to the service sector.
  • Small and cottage industries
  • These industries involve handicrafts, handlooms, and artisanal products.
  • They are vital for rural employment and preserving India's cultural heritage.
  • Despite their importance, their GDP contribution is relatively limited compared to the service sector.
  • Agricultural sector
  • The agricultural sector was historically the largest contributor to India's GDP but has now been surpassed by the service sector.
  • It still remains the primary source of livelihood for a large portion of the population, particularly in rural areas.
  • This sector includes activities like crop production, fishing, and forestry.
2

The frequency of ultrasonic waves is:

  1. ((a))

    More than 50,000 Hz 

  2. ((b))

    More than 20,000 Hz

  3. ((c))

    20 Hz to 20,000 Hz 

  4. ((d))

    Less than 20 Hz

Show Answer
Answer: ((b))

More than 20,000 Hz

The Correct answer is More than 20,000 Hz.

Key Points

  • Ultrasonic waves are defined as sound waves with a frequency higher than the upper limit of human hearing, which is approximately 20,000 Hz.
  • These waves are extensively used in medical imaging, such as ultrasound scans, and also in industrial applications like sonography and cleaning processes.
  • They are utilized in non-destructive testing to evaluate the integrity of materials without causing damage.
  • Ultrasonic waves are also employed in animal communication. For instance, bats use these high-frequency sound waves for echolocation.
  • The production of ultrasonic waves often involves the use of piezoelectric crystals, which vibrate rapidly when subjected to electrical energy.

Additional Information

  • 20 Hz to 20,000 Hz
  • This range represents the audible frequency range for humans, commonly referred to as the hearing range.
  • Humans can detect sounds within this range, but anything beyond this (like ultrasonic waves) is inaudible.
  • Less than 20 Hz
  • Sound waves with frequencies below 20 Hz are termed infrasonic waves.
  • These are used in seismology to study earthquakes and volcanic eruptions.
  • Some animals, like elephants, communicate using infrasonic frequencies.
  • More than 50,000 Hz
  • While this is a correct description of ultrasonic waves, the standard definition begins at 20,000 Hz, making "More than 20,000 Hz" the more precise answer.
  • Applications of frequencies above 50,000 Hz include advanced medical imaging and scientific research.
3

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

List I (Tribe)List II (Location)
A.Maasaii.Kalahari Desert
B.Semaiii.New Zealand
C.Bushmeniii.Malaysia
D.Maoriiv.East Africa

 

<br> <br>
  1. ((a))

    A-i, B-iv, C-iii, D-ii

  2. ((b))

    A-i, B-ii, C-iv, D-iii

  3. ((c))

    A-iv, B-ii, C-i, D-iii

  4. ((d))

    A-iv, B-iii, C-i, D-ii

Show Answer
Answer: ((d))

A-iv, B-iii, C-i, D-ii

The Correct answer is A-iv, B-iii, C-i, D-ii.

Key Points

  • Maasai are a tribal group found in East Africa, mainly in regions of Kenya and Tanzania.
  • The Semai are an indigenous tribe located in Malaysia, particularly in the central regions of the country.
  • The Bushmen, also known as San people, are indigenous hunter-gatherers living in the Kalahari Desert, which spans parts of Botswana, Namibia, and South Africa.
  • The Maori are the indigenous people of New Zealand, known for their rich cultural heritage and traditions.

Additional Information

  • Maasai
  • The Maasai tribe is known for its colorful attire and nomadic lifestyle.
  • They primarily rely on cattle herding as their source of livelihood.
  • Semai
  • The Semai tribe is recognized for its emphasis on peaceful coexistence and avoidance of conflict.
  • They practice shifting cultivation and are skilled in basket weaving.
  • Bushmen
  • The Bushmen are known for their profound knowledge of the desert ecosystem.
  • They use traditional methods for hunting and gathering food.
  • Maori
  • The Maori are famous for their distinctive tattoo art called ta moko.
  • Their traditional war dance, haka, is internationally recognized.
4

The famous “Dudhsagar Falls” is situated on which river?

  1. ((a))

    Mandovi river

  2. ((b))

    Bhadra river

  3. ((c))

    Sharavati river

  4. ((d))

    Ulhas river

Show Answer
Answer: ((a))

Mandovi river

The Correct answer is Mandovi River.

Key Points

  • The Dudhsagar Falls is a spectacular waterfall located in the Indian state of Goa.
  • It is situated on the Mandovi River, which originates in the Western Ghats and flows westward into the Arabian Sea.
  • The name "Dudhsagar" translates to "Sea of Milk" in English, owing to the waterfall's appearance of milky white water cascading down the rocks.
  • It is one of the tallest waterfalls in India, with a height of approximately 310 meters (1017 feet).
  • The waterfall is located in the Bhagwan Mahaveer Sanctuary and Molem National Park, which are rich in biodiversity.
  • The site is famous as a tourist attraction and is particularly popular among trekkers and nature enthusiasts.
  • The waterfall can be accessed via a railway line and is visible from the Chennai-Vasco railway route.
  • The Mandovi River is often referred to as the "Lifeline of Goa" because it serves as a crucial water source for the region.

Additional Information

  • Bhadra River
  • The Bhadra River originates in the Western Ghats in Karnataka.
  • It is a tributary of the Tungabhadra River, which eventually merges into the Krishna River.
  • The river is famous for the Bhadra Wildlife Sanctuary, which is home to a variety of flora and fauna.
  • Sharavati River
  • The Sharavati River is located in Karnataka and is best known for forming the Jog Falls, one of India's highest waterfalls.
  • This river is utilized for hydroelectric projects, contributing to the state's power supply.
  • Ulhas River
  • The Ulhas River is located in the state of Maharashtra.
  • It flows westward and empties into the Arabian Sea.
  • The river basin supports several towns and cities, including Kalyan, Thane, and Ulhasnagar.
5

Which crop is known as the "Camel of Crops"?

  1. ((a))

    Sorghum 

  2. ((b))

    Bajra

  3. ((c))

    Ragi

  4. ((d))

    Kodo

Show Answer
Answer: ((a))

Sorghum 

The Correct answer is Sorghum.

Key Points

  • Sorghum is widely known as the "Camel of Crops" due to its ability to survive in arid and semi-arid regions where other crops may fail.
  • It is a cereal crop that is highly adaptable to drought conditions, extreme heat, and poor-quality soil, making it a vital food and fodder crop in many parts of the world.
  • Sorghum is a staple food crop in many developing countries, especially in Africa and South Asia.
  • The crop is rich in nutritional content and provides essential nutrients such as protein, dietary fiber, and minerals like iron and phosphorus.
  • It is used for multiple purposes, including food (in the form of porridge, bread, and beverages), animal feed, and as a raw material for producing ethanol.
  • Sorghum’s ability to grow with minimal water requirements and its resilience to harsh climates make it a critical crop for ensuring food security in water-scarce regions.
  • Different varieties of sorghum are cultivated for grain production, syrup extraction, and biofuel production.

Additional Information

  • Bajra
  • Bajra, also known as Pearl Millet, is another drought-resistant crop commonly grown in arid and semi-arid regions.
  • It is rich in fiber, protein, and essential minerals like magnesium and calcium, making it a healthy food choice.
  • Bajra is primarily grown in India and Africa and is used to prepare flatbreads, porridge, and other traditional dishes.
  • Ragi
  • Ragi, also known as Finger Millet, is a staple food in many parts of India and Africa.
  • It is a nutrient-rich crop, particularly high in calcium, making it beneficial for bone health.
  • Ragi is widely used to make porridge, flatbreads, and malted beverages.
  • Kodo
  • Kodo Millet is a highly nutritious and drought-resistant millet grown in parts of India.
  • It is rich in dietary fiber, protein, and antioxidants, making it a healthy choice for those with diabetes or gluten intolerance.
  • Kodo is commonly used to prepare porridge, upma, and dosa.
6

Biodiversity Hotspots are defined on which of the following basis ? 

  1. ((a))

    Vegetation 

  2. ((b))

    All organisms

  3. ((c))

    Crop diversity

  4. ((d))

    Animals

Show Answer
Answer: ((a))

Vegetation 

The Correct answer is Vegetation.

Key Points

  • Biodiversity hotspots are areas that are exceptionally rich in species diversity, particularly plant species, and are under threat of destruction.
  • The concept of biodiversity hotspots was introduced by Norman Myers in 1988.
  • One of the primary criteria for defining a biodiversity hotspot is the presence of at least 1,500 species of vascular plants as endemics, which means they are not found anywhere else in the world.
  • Another important criterion is that the region must have lost at least 70% of its original vegetation, making the conservation of remaining vegetation crucial.
  • Vegetation acts as the primary producer in ecosystems, supporting the survival of other organisms like animals and microbes, which is why it is the basis for defining biodiversity hotspots.
  • Examples of biodiversity hotspots include the Himalayas, the Western Ghats, the Sundaland (including the Nicobar Islands), and the Indo-Burma region in India.
  • Conservation efforts in biodiversity hotspots focus on protecting their vegetation as it directly impacts the survival of other species and the overall ecosystem health.

Additional Information

  • All organisms
  • Although biodiversity includes all organisms, the concept of biodiversity hotspots specifically focuses on plant species and their endemism, as they form the base for all ecosystems.
  • Crop diversity
  • Crop diversity refers to the variety of crops and their genetic traits, which is essential for agriculture but is not a criterion for defining biodiversity hotspots.
  • Animals
  • While animal species are an integral part of biodiversity, biodiversity hotspots are primarily defined based on vegetation and plant endemism.
  • Animals depend on vegetation for food and habitat, making vegetation the most critical factor for biodiversity hotspots.
7

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

List I (Name of Scheme)List II (Main Objective)
A.MAAi.Benefits under PMMVY
B.LaQshyaii.Promotion of breast feeding
C.SUMANiii.Improvement in labour room and maternity OT
D.MCP Cardiv.Free healthcare to pregnant women and newborn child
  1. ((a))

    A-ii, B-iii, C-iv, D-i

  2. ((b))

    A-iv, B-iii, C-i, D-ii 

  3. ((c))

    A-ii, B-iv, C-iii, D-i

  4. ((d))

    A-iv, B-i, C-iii, D-ii

Show Answer
Answer: ((a))

A-ii, B-iii, C-iv, D-i

The correct answer is Option 1.

Key Points

  • MAA (Mothers’ Absolute Affection): The primary objective of this scheme is the promotion of breastfeeding. It aims to create awareness among mothers about the benefits of breastfeeding and to provide support to ensure proper breastfeeding practices.
  • LaQshya: This scheme focuses on the improvement in labour rooms and maternity operation theatres. It aims to enhance the quality of care during delivery and immediate postpartum care.
  • SUMAN (Surakshit Matritva Aashwasan): This initiative provides free healthcare to pregnant women and newborns. It ensures respectful and quality maternity care with zero expense to the beneficiary.
  • MCP Card (Mother and Child Protection Card): The MCP card is used for tracking the benefits under the Pradhan Mantri Matru Vandana Yojana (PMMVY). It records essential details of antenatal, postnatal care, and immunization for the mother and the child.
  • The correct mapping of the schemes and their objectives is as follows:
  • A - ii (MAA: Promotion of breastfeeding).
  • B - iii (LaQshya: Improvement in labour room and maternity OT).
  • C - iv (SUMAN: Free healthcare to pregnant women and newborn child).
  • D - i (MCP Card: Benefits under PMMVY).

Additional Information

  • MAA (Mothers’ Absolute Affection):
  • This scheme is a part of the National Health Mission (NHM) and was launched by the Ministry of Health and Family Welfare.
  • It emphasizes the importance of exclusive breastfeeding for the first six months of life and continued breastfeeding for up to two years or more.
  • It includes awareness campaigns, counseling sessions, and training programs for healthcare providers.
  • LaQshya:
  • Launched under the National Health Mission (NHM), it is aimed at reducing maternal and neonatal mortality.
  • This initiative ensures adherence to quality standards in labour rooms and maternity OTs.
  • It focuses on improving infrastructure, infection control, and patient satisfaction.
  • SUMAN (Surakshit Matritva Aashwasan):
  • Launched in 2019 by the Government of India, this scheme provides assured, dignified, and respectful maternity care at no cost.
  • It includes free tests, medicines, transport, and delivery services.
  • The initiative also aims to reduce preventable maternal and newborn deaths.
  • MCP Card (Mother and Child Protection Card):
  • The card is a joint initiative of the Ministry of Women and Child Development and the Ministry of Health and Family Welfare.
  • It serves as a tool for monitoring and ensuring the delivery of health and nutrition services to pregnant women and children.
  • It is also used for availing benefits under schemes like PMMVY, Janani Suraksha Yojana (JSY), etc.
8

When was "Sarva Shiksha Abhiyan" launched in India?

  1. ((a))

    2007

  2. ((b))

    2001 

  3. ((c))

    2014

  4. ((d))

    2020

Show Answer
Answer: ((b))

2001 

The Correct answer is 2001.

Key Points

  • Sarva Shiksha Abhiyan (SSA) was launched in the year 2001 by the Government of India.
  • It is a flagship program aimed at achieving the goal of universalization of elementary education in a time-bound manner.
  • The program focuses on providing free and compulsory education to children in the age group of 6-14 years as per the provisions of Article 21A of the Indian Constitution.
  • The SSA initiative is built to bridge the gaps in education by targeting disadvantaged groups, including girls, scheduled castes, scheduled tribes, and children with disabilities.
  • The program is financed by both the Central and State governments under a shared funding pattern.
  • SSA also emphasizes improving the quality of education, teacher training, and infrastructure development in schools.
  • Later, SSA was integrated into the Samagra Shiksha Abhiyan in 2018 to consolidate various education-related programs.
  • The initiative complements the objectives of the Right to Education (RTE) Act, 2009, ensuring every child receives access to quality education.

Additional Information

  • 2007
  • This option does not align with the launch of Sarva Shiksha Abhiyan. However, it is worth noting that during 2007, there was an emphasis on enhancing the SSA framework to improve access to education for marginalized communities.
  • 2014
  • In this year, the Government of India launched several new initiatives related to education, but it is not the year of the SSA launch. For instance, the Beti Bachao Beti Padhao scheme was introduced to promote girl child education.
  • 2020
  • In 2020, the National Education Policy (NEP 2020) was introduced, which brought transformative changes to the education sector. However, it is unrelated to the launch of SSA.
9

Which one of the following writs can be issued by High Court to secure the liberty of an individual?

  1. ((a))

    Quo warranto

  2. ((b))

    Mandamus

  3. ((c))

    Habeas Corpus

  4. ((d))

    Prohibition

Show Answer
Answer: ((c))

Habeas Corpus

The Correct answer is Habeas Corpus.

Key Points

  • The writ of Habeas Corpus is issued to secure the liberty of an individual.
  • It is a Latin term that means "you may have the body."
  • This writ is used to release a person who has been unlawfully detained or imprisoned.
  • Under this writ, the court directs the person or authority who has detained another person to bring the detainee before the court.
  • The purpose of this writ is to ensure that no individual is deprived of their liberty without legal justification.
  • The Article 32 and Article 226 of the Indian Constitution empower the Supreme Court and High Courts, respectively, to issue this writ.
  • It acts as a safeguard against arbitrary arrests and ensures protection of fundamental rights provided under the Constitution of India.

Additional Information

  • Quo warranto
  • The writ of Quo Warranto is issued to prevent a person from holding a public office without legal authority.
  • It ensures that only a legally qualified person occupies a public office.
  • The court can question the eligibility of the officeholder under this writ.
  • Mandamus
  • The writ of Mandamus is issued by a court to direct a public official or governmental body to perform a duty that they are obligated to perform.
  • It cannot be issued against private individuals or entities.
  • This writ ensures the proper functioning of public authorities.
  • Prohibition
  • The writ of Prohibition is issued by a higher court to a lower court or tribunal to prevent it from exceeding its jurisdiction or acting contrary to the law.
  • It is issued when a lower court proceeds with a case that it does not have the authority to hear.
  • This writ is preventive in nature.
10

Among the crops listed below, which one is not a coarse cereal crop (Millets)?

  1. ((a))

    Wheat

  2. ((b))

    Maize

  3. ((c))

    Jowar

  4. ((d))

    Baira

Show Answer
Answer: ((a))

Wheat

The Correct answer is Wheat.

Key Points

  • Wheat is not considered a coarse cereal crop (Millet).
  • It is a staple food grain and is classified as a cereal crop, but not as a millet.
  • Wheat is primarily grown in the temperate regions, requiring a cool climate during the growing season and bright sunshine at the time of maturity.
  • It is rich in carbohydrates and a good source of dietary fiber and proteins, making it a significant food crop globally.
  • Wheat is cultivated in several countries, with major producers being India, China, the USA, and Russia.
  • Unlike millets, wheat is not drought-resistant and requires more irrigation and fertile soil for optimal growth.

Additional Information

  • Maize
  • Maize is a cereal crop often categorized as a coarse cereal.
  • It is used both for human consumption and as livestock feed.
  • Maize is a versatile crop that is grown in tropical, subtropical, and temperate regions.
  • It is a major source of starch and is processed into various products like corn oil and corn syrup.
  • Jowar
  • Jowar, also called sorghum, is a major millet grown in India.
  • It is highly drought-resistant and thrives in semi-arid regions.
  • Jowar is a rich source of protein, fiber, and essential minerals.
  • It is widely consumed in India and Africa as part of daily diets.
  • Bajra
  • Bajra, also known as pearl millet, is another common coarse cereal crop.
  • It is well-adapted to arid and semi-arid regions due to its high resistance to drought.
  • Bajra is rich in energy, protein, fiber, and essential nutrients like iron and magnesium.
  • It is a staple food crop in Rajasthan, Gujarat, and Maharashtra.
11

Which country abolished its Two Child policy in June 2025 ?

  1. ((a))

    Japan

  2. ((b))

    India

  3. ((c))

    Australia

  4. ((d))

    Vietnam

Show Answer
Answer: ((d))

Vietnam

The correct answer is Vietnam.

Key Points

  • Vietnam abolished its Two-Child Policy in June 2025, marking a significant change in its population control measures.
  • The Two-Child Policy was initially implemented to address concerns related to overpopulation and its impact on economic resources.
  • In recent years, Vietnam faced challenges due to declining birth rates, leading to a shrinking workforce and an aging population.
  • The abolition of the Two-Child Policy is aimed at encouraging citizens to have more children and thereby balancing the population structure.
  • Vietnam's government has also been working on policies to address issues related to childcare, education, healthcare, and financial support for families.
  • This decision aligns with global trends where countries like China have relaxed similar policies to combat demographic challenges.
  • Vietnam has been implementing progressive measures to improve its socio-economic stability while adapting to changing demographic needs.

Additional Information

  • Japan
  • Japan has one of the world’s lowest birth rates and has implemented policies to encourage families to have more children, but it does not have a strict Two-Child Policy.
  • The country faces challenges related to its aging population and workforce shortages.
  • Japan is known for its advanced healthcare system and efforts to improve family welfare.
  • India
  • India does not have a nationwide Two-Child Policy, though some states promote policies to control population growth.
  • The government focuses on measures like family planning, education, and awareness campaigns to manage population issues.
  • India has a high population density and faces challenges related to resource allocation and infrastructure development.
  • Australia
  • Australia does not have a Two-Child Policy and encourages population growth through immigration and support for families.
  • The country has a low population density and focuses on maintaining a balanced demographic structure.
  • Australia is known for its progressive welfare system and support for families.
12

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

List I (Article)List II (Subject)
A.Article 51Ai.Abolition of titles
B.Article 14ii.Fundamental duties
C.Article 17iii.Abolition of untouchability
D.Article 18iv.Equality before law
  1. ((a))

    A-ii, B-iv, C-iii, D-i

  2. ((b))

    A-i, B-ii, C-iii, D-iv

  3. ((c))

    A-ii, B-iii, C-iv, D-i

  4. ((d))

    A-ii, B-iv, C-i, D-iii

Show Answer
Answer: ((a))

A-ii, B-iv, C-iii, D-i

The Correct answer is A-ii, B-iv, C-iii, D-i.

Key Points

  • Article 51A: This article in the Indian Constitution enumerates the Fundamental Duties of citizens. It highlights responsibilities such as protecting sovereignty, promoting harmony, safeguarding public property, and striving for excellence.
  • Article 14: It guarantees equality before law and equal protection of the laws to all persons within the territory of India. This is a cornerstone of the principle of non-discrimination.
  • Article 17: This article focuses on the Abolition of Untouchability. It prohibits the practice of untouchability in any form and declares such acts as offenses punishable by law.
  • Article 18: It deals with the Abolition of Titles, aiming to prevent the creation of titles that promote inequalities, except for military and academic distinctions.
  • The match between the articles and their subjects ensures a clear understanding of their respective significance in the Indian Constitution.

Additional Information

  • Article 51A
  • Introduced by the 42nd Amendment Act of 1976.
  • It emphasizes duties such as respecting national symbols, protecting the environment, and upholding the spirit of brotherhood.
  • There are 11 Fundamental Duties listed under this article.
  • Article 14
  • Forms the basis of the Right to Equality.
  • It applies to both citizens and non-citizens within Indian territory.
  • It ensures that no person is above the law and mandates fairness in legal proceedings.
  • Article 17
  • Prohibits the practice of untouchability in any form.
  • Ensures equality and dignity to individuals irrespective of their caste.
  • Legal provisions to enforce this article are implemented through the Protection of Civil Rights Act, 1955.
  • Article 18
  • Abolishes titles like Raja, Maharaja, Sir, etc., except for military and academic distinctions.
  • Aims to promote equality and discourage hereditary privileges.
  • Ensures that titles do not interfere with democratic ideals.
13

The Durand Cup is associated with which sport ?

  1. ((a))

    Cricket

  2. ((b))

    Basketball

  3. ((c))

    Football

  4. ((d))

    Hockey

Show Answer
Answer: ((c))

Football

The Correct answer is Football.

Key Points

  • The Durand Cup is one of the oldest football tournaments in the world and the oldest in India.
  • It was first held in the year 1888 and is named after Mortimer Durand, who was then the Foreign Secretary in British India.
  • This tournament is a symbol of India’s football heritage and has been a platform for showcasing football talent.
  • The Indian Army has been instrumental in organizing the Durand Cup for many years.
  • It features participation from some of India’s top football teams, including those from the Indian Super League (ISL), I-League, and other domestic leagues.
  • The tournament usually takes place annually and includes teams from both professional and amateur circuits.
  • The Durand Cup is often seen as a testing ground for young and emerging football talent in India.
  • The winner of the Durand Cup is awarded a set of three trophies: the President’s Cup, the Durand Cup, and the Shimla Trophy.

Additional Information

  • Cricket
  • Cricket is the most popular sport in India and is governed by the Board of Control for Cricket in India (BCCI).
  • Famous tournaments associated with cricket in India include the Indian Premier League (IPL), Ranji Trophy, and Vijay Hazare Trophy.
  • Cricket is played in different formats such as Test Matches, One-Day Internationals (ODIs), and T20s.
  • Basketball
  • Basketball is governed by the Basketball Federation of India (BFI).
  • The sport is gaining popularity in India, especially in schools and colleges.
  • India participates in international basketball tournaments, including the FIBA Asia Cup.
  • Hockey
  • Hockey is India’s national sport and has a rich history, with the Indian men’s hockey team winning multiple Olympic gold medals.
  • Major tournaments include the Hockey India League (HIL) and the Sultan Azlan Shah Cup.
  • The Indian women’s hockey team has also gained prominence in recent years, performing well in international events like the Olympics and World Cup.
14

Which building was built in Fatehpur Sikri by Emperor Akbar ?'

  1. ((a))

    Moti Mahal

  2. ((b))

    Heera Mahal

  3. ((c))

    Panch Mahal

  4. ((d))

    Rang Mahal

Show Answer
Answer: ((c))

Panch Mahal

The Correct answer is Panch Mahal.

Key Points

  • Panch Mahal is a five-storied architectural marvel built by Emperor Akbar in Fatehpur Sikri.
  • It was constructed in the 16th century and served as a place for the emperor to relax and enjoy the cool breeze.
  • The structure is also referred to as the "Badgir" or "wind catcher" tower due to its design that allows natural ventilation.
  • The building is built using red sandstone, a signature material in Mughal architecture.
  • The five stories of the Panch Mahal are in diminishing size as they ascend, giving it a pyramidal structure.
  • Each story is supported by pillars, with the ground floor having the maximum number of pillars (84).
  • The structure was primarily used as a pleasure palace and as a place for informal gatherings.
  • It is a reflection of the blend of Persian, Islamic, and Indian architectural styles that flourished during Akbar's reign.

Additional Information

  • Moti Mahal
  • Moti Mahal translates to the "Pearl Palace" and is a historical structure found in various locations in India, such as Rajasthan and Madhya Pradesh.
  • It is not located in Fatehpur Sikri nor was it built by Emperor Akbar.
  • Heera Mahal
  • The term "Heera Mahal" means "Diamond Palace," but it is not associated with Akbar's architectural works or Fatehpur Sikri.
  • Rang Mahal
  • Rang Mahal, meaning "Painted Palace," is a common name for palaces found in several parts of India, especially in Rajasthan.
  • It was not built by Akbar nor is it located in Fatehpur Sikri.
15

During a bill seeking the promotion and regulation of online gaming, it was disclosed by the Government that people lose close to ₹ 20,000 crore every year. When was the bill introduced in the Lok Sabha ? 

  1. ((a))

    August 30, 2025 

  2. ((b))

    August 20, 2025 

  3. ((c))

    July 17, 2025

  4. ((d))

    September 2, 2025

Show Answer
Answer: ((b))

August 20, 2025 

The correct answer is August 20, 2025.

Key Points

  • The bill for the promotion and regulation of online gaming was introduced in the Lok Sabha on August 20, 2025.
  • The Government disclosed that people lose approximately ₹20,000 crore every year due to online gaming activities, which highlights the significance of regulating this sector.
  • The bill aims to establish specific rules and regulations to ensure fair practices in online gaming and prevent financial losses or exploitation.
  • The introduction of this bill is a part of the Government's efforts to address the economic and social consequences of unregulated online gaming activities in India.
  • This initiative aligns with the broader goal of promoting safe and responsible gaming practices while preventing illegal activities such as fraud and gambling addiction.

Additional Information

  • August 30, 2025
  • This date does not align with the timeline of the bill's introduction in the Lok Sabha.
  • Thu Jul 17 2025 00:00:00 GMT+0530 (India Standard Time)
  • This date refers to a day in July 2025, which does not correspond to the actual date of the bill's introduction.
  • Tue Sep 02 2025 00:00:00 GMT+0530 (India Standard Time)
  • This date is in September 2025, whereas the correct date of introduction was in August 2025.
16

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

List I (Revolution)List II (Production)
A.Whitei.Oil seeds
B.Blueii.Food grains
C.Yellowiii.Milk
D.Greeniv.Fish
<br> <br>
  1. ((a))

    A-ii, B-iv, C-iii, D-i

  2. ((b))

    A-i, B-iii, C-iv, D-ii

  3. ((c))

    A-iii, B-ii, C-i, D-iv

  4. ((d))

    A-iii, B-iv, C-i, D-ii

Show Answer
Answer: ((d))

A-iii, B-iv, C-i, D-ii

The Correct answer is Option 4.

Key Points

  • White Revolution is associated with Milk production. This revolution was spearheaded by the visionary leader Dr. Verghese Kurien, also known as the "Father of the White Revolution" in India.
  • Blue Revolution refers to the growth and development of Fish production. It aimed to increase the production of aquatic organisms and ensure sustainable fisheries.
  • Yellow Revolution is linked to the production of Oil seeds. This revolution aimed to make India self-reliant in edible oils production.
  • Green Revolution is connected to the increased production of Food grains, particularly wheat and rice, through the use of high-yield variety seeds, irrigation, fertilizers, and pesticides. It was initiated in the 1960s under the leadership of M.S. Swaminathan.
  • The correct matching is:
  • A - iii (White - Milk)
  • B - iv (Blue - Fish)
  • C - i (Yellow - Oil seeds)
  • D - ii (Green - Food grains)

Additional Information

  • White Revolution
  • The White Revolution in India was launched in 1970 by the National Dairy Development Board (NDDB).
  • It significantly increased milk production, making India the largest producer of milk and dairy products globally.
  • The program associated with this revolution was called Operation Flood.
  • Blue Revolution
  • The Blue Revolution was aimed at increasing aquaculture and fish production.
  • It includes both marine and inland fisheries.
  • It promotes new technologies, sustainable fishing practices, and infrastructure development.
  • Yellow Revolution
  • The Yellow Revolution primarily focused on increasing the production of edible oils like mustard, sunflower, and groundnut oil.
  • The revolution gained momentum in the 1980s.
  • Green Revolution
  • The Green Revolution was initiated in the 1960s to achieve self-sufficiency in food grain production.
  • It introduced high-yield variety (HYV) seeds, improved irrigation facilities, and modern agricultural techniques.
  • The revolution initially focused on the states of Punjab, Haryana, and Uttar Pradesh.
17

Which one of the following weeds is commonly called "Gajar Ghas" or "Congress grass" ?

  1. ((a))

    Parthenium

  2. ((b))

    Bathua

  3. ((c))

    Gokharoo (Xanthium)

  4. ((d))

    Junglee Jowar

Show Answer
Answer: ((a))

Parthenium

The Correct answer is Parthenium.

Key Points

  • Parthenium is commonly known as "Gajar Ghas" or "Congress Grass" in India.
  • It is an invasive weed species introduced in India accidentally in the 1950s through imported wheat from the United States.
  • This weed is known for causing severe health issues such as skin allergies, respiratory problems, and asthma in humans.
  • It also has a detrimental impact on agriculture as it reduces soil fertility, affects crop yields, and competes with native vegetation.
  • Parthenium is classified as a noxious weed due to its rapid spread and adverse effects on ecosystems.
  • It is primarily found in open lands, roadsides, and agricultural fields in India.
  • The weed produces chemicals such as parthenin, which suppresses the growth of other plants nearby—a phenomenon known as allelopathy.
  • Efforts to control Parthenium include mechanical removal, biological control using insects like the Mexican beetle (Zygogramma bicolorata), and chemical herbicides.

Additional Information

  • Bathua
  • Bathua is an edible leafy vegetable commonly consumed in India.
  • It is rich in vitamins, minerals, and antioxidants, making it highly nutritious.
  • Bathua is often used in preparing dishes like parathas, soups, and curries.
  • Unlike Parthenium, Bathua is not harmful but rather beneficial for human consumption.
  • Gokharoo (Xanthium)
  • Gokharoo, also known as Xanthium, is a medicinal plant used in Ayurveda.
  • It is often used for treating urinary tract infections, kidney stones, and skin diseases.
  • Gokharoo produces spiny fruits that can stick to animal fur or clothing for seed dispersal.
  • Junglee Jowar
  • Junglee Jowar refers to wild sorghum, a type of grass.
  • It grows naturally in grasslands and agricultural areas and is known for its drought tolerance.
  • Unlike Parthenium, Junglee Jowar is not invasive or harmful; it has limited agricultural significance.
18

Which political leader forcefully raised the issue of reservation for Backward Castes in Bihar for the first time?

  1. ((a))

    Sharad Yadav

  2. ((b))

    Lalu Prasad Yadav

  3. ((c))

    Karpoori Thakur

  4. ((d))

    Mulayam Singh Yadav

Show Answer
Answer: ((c))

Karpoori Thakur

The Correct answer is Karpoori Thakur.

Key Points

  • Karpoori Thakur was a prominent Indian politician and a social reformer from Bihar who forcefully raised the issue of reservation for Backward Castes in the state for the first time.
  • He was the Chief Minister of Bihar during two different terms: 1970-71 and 1977-79.
  • Thakur is known for implementing the reservation policy for the Backward Classes in Bihar, which was a landmark decision in social justice.
  • He introduced a quota system for the Backward Classes in government jobs and educational institutions, ensuring their representation and opportunities.
  • His efforts were aimed at addressing the socio-economic inequalities faced by marginalized communities.
  • Karpoori Thakur was also a staunch supporter of socialism and worked extensively for the upliftment of the underprivileged sections of society.
  • He is remembered as a leader who advocated for the rights of the oppressed and worked tirelessly to bring about social and political reforms.

Additional Information

  • Sharad Yadav
  • Sharad Yadav was a senior Indian politician and member of the Lok Sabha.
  • He was associated with issues concerning social justice, but he is not credited with raising the reservation issue in Bihar for the first time.
  • Lalu Prasad Yadav
  • Lalu Prasad Yadav, a former Chief Minister of Bihar, is well-known for his support of social justice and empowerment of Backward Castes.
  • However, he addressed the reservation issue after Karpoori Thakur's initial efforts.
  • Mulayam Singh Yadav
  • Mulayam Singh Yadav was a prominent leader from Uttar Pradesh and former Chief Minister of the state.
  • His work largely focused on social justice for Backward Castes in Uttar Pradesh rather than Bihar.
19

Which leader was against the idea of strengthening (Autonomy) local government bodies?

  1. ((a))

    B.R. Ambedkar

  2. ((b))

    Vinoba Bhave

  3. ((c))

    Mahatma Gandhi 

  4. ((d))

    M.A. Ayyangar

Show Answer
Answer: ((a))

B.R. Ambedkar

The correct answer is B.R. Ambedkar.

Key Points

  • B.R. Ambedkar, the principal architect of the Indian Constitution, was skeptical of the idea of strengthening local government bodies, particularly during the formative years of India's independence.
  • He believed that local bodies lacked the capacity, organization, and infrastructure to handle governance effectively.
  • Ambedkar emphasized the need for a strong central government to ensure uniformity and equity in governance across the country.
  • He was particularly concerned about the caste-based inequalities prevalent in rural areas, which could perpetuate discrimination if too much power was given to local institutions.
  • Ambedkar advocated for the protection of marginalized groups, such as Scheduled Castes and Scheduled Tribes, which he believed could be best ensured through a centralized framework.
  • He also opined that strengthening local governance prematurely could lead to poor implementation of policies and hinder socio-economic reforms.

Additional Information

  • Vinoba Bhave
  • Vinoba Bhave was a disciple of Mahatma Gandhi and is known for his role in the Bhoodan Movement, aimed at redistributing land to the landless.
  • He supported decentralization and believed in empowering local communities through self-governance.
  • Mahatma Gandhi
  • Mahatma Gandhi was a staunch advocate of Panchayati Raj and believed in the importance of strengthening local self-government.
  • He envisioned a decentralized political structure where villages would function as self-reliant units.
  • M.A. Ayyangar
  • M.A. Ayyangar was the second Speaker of the Lok Sabha in India and played an active role in parliamentary procedures.
  • He was not directly associated with the debate on local governance but focused more on legislative and parliamentary governance.
20

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

List I (Buddhist Councils)List II (Places)
A.Firsti.Pataliputra
B.Secondii.Vaishali
C.Thirdiii.Rajgrih
D.Fourthiv.Kashmir
<br>
  1. ((a))

    A-iv, B-iii, C-ii, D-i

  2. ((b))

    A-ii, B-iii, C-iv, D-i

  3. ((c))

    A-iii, B-ii, C-i, D-iv

  4. ((d))

    A-ii, B-i, C-iii, D-iv

Show Answer
Answer: ((c))

A-iii, B-ii, C-i, D-iv

The Correct answer is A-iii, B-ii, C-i, D-iv.

Key Points

  • The First Buddhist Council was held at Rajgrih, shortly after the death of Gautama Buddha, around 483 BCE. It was presided over by Mahakassapa and aimed at compiling Buddha's teachings.
  • The Second Buddhist Council was convened at Vaishali, approximately 100 years after Buddha's death. It focused on resolving disputes regarding monastic practices and the interpretation of the Vinaya.
  • The Third Buddhist Council took place at Pataliputra under the patronage of Emperor Ashoka around 250 BCE. It aimed at purifying the Buddhist teachings and removing heretical views.
  • The Fourth Buddhist Council was held in Kashmir during the reign of King Kanishka, a Kushan ruler, around the 1st century CE. This council focused on the division of Buddhism into two sects: Mahayana and Theravada.
  • These councils were crucial in shaping the development of Buddhism and its spread across the Indian subcontinent and beyond.

Additional Information

  • Rajgrih
  • Rajgrih was the ancient capital of the Magadha kingdom and is located in present-day Bihar, India.
  • It is known for its association with both Buddhism and Jainism.
  • Rajgrih is also home to various ancient Buddhist sites such as the Griddhakuta Hill, where Buddha delivered many sermons.
  • Vaishali
  • Vaishali was an ancient city located in modern-day Bihar, India.
  • It holds significance in both Buddhist and Jain traditions.
  • Vaishali is considered the birthplace of Lord Mahavira, the 24th Tirthankara of Jainism.
  • Pataliputra
  • Pataliputra, present-day Patna, was a major city in ancient India and the capital of the Mauryan Empire under Emperor Ashoka.
  • This city was a hub of administrative, cultural, and religious activities during ancient times.
  • It played a pivotal role in the spread of Buddhism across India and beyond.
  • Kashmir
  • Kashmir served as a significant center for Buddhism during the reign of King Kanishka.
  • It is noted for its role in the establishment and growth of the Mahayana sect of Buddhism.
  • The Fourth Buddhist Council held here led to significant developments in Buddhist philosophy and literature.
21

The hump of the camel is made up of:

  1. ((a))

    Cartilage

  2. ((b))

    Lymph

  3. ((c))

    Fat tissue

  4. ((d))

    Collagen fibres

Show Answer
Answer: ((c))

Fat tissue

The correct answer is Fat tissue.

Key Points

  • The hump of a camel is primarily made up of fat tissue, which serves as an energy reserve.
  • It helps camels survive in desert climates where food and water are scarce.
  • When food is not available, the fat in the hump is metabolized for energy and water.
  • The hump may shrink and droop if the camel uses up the stored fat during extended periods of food scarcity.
  • Contrary to popular belief, the hump does not store water but contributes indirectly to a camel's survival in arid conditions.
  • The ability of camels to metabolize fat efficiently from their hump is an adaptation to their desert environment.
  • The hump also helps in thermoregulation by reducing heat exposure to the rest of the body.

Additional Information

  • Cartilage
  • Cartilage is a flexible connective tissue found in areas such as joints, the ear, and the nose.
  • It provides structural support and reduces friction in joints but is not present in the camel's hump.
  • Lymph
  • Lymph is a colorless fluid that circulates in the lymphatic system, playing a role in immune responses.
  • It is not related to the structure of a camel’s hump.
  • Collagen fibers
  • Collagen fibers are a major component of connective tissues and are responsible for providing strength and structure.
  • They are found in skin, tendons, and ligaments but not in the camel’s hump.
22

Who used the term "Biodiversity" for the first time?

  1. ((a))

    R.H. Whittaker

  2. ((b))

    R.D. Barnes

  3. ((c))

    Walter G. Rosen

  4. ((d))

    N. Myers

Show Answer
Answer: ((c))

Walter G. Rosen

The Correct answer is Walter G. Rosen.

Key Points

  • The term "Biodiversity" was first introduced by Walter G. Rosen in 1985.
  • He used this term during the planning for the "National Forum on BioDiversity" held in Washington D.C., USA.
  • The term "Biodiversity" is a short form of Biological Diversity, which refers to the variety of life forms on Earth, including plants, animals, microorganisms, and their ecosystems.
  • Rosen's contribution helped in emphasizing the importance of conserving the Earth’s vast variety of life forms and their interactions.
  • Later, the term gained global recognition and became widely used in the fields of ecology, environmental science, and conservation biology.
  • Today, biodiversity is a critical concept in tackling issues like climate change, habitat destruction, species extinction, and ecosystem degradation.

Additional Information

  • R.H. Whittaker
  • R.H. Whittaker was a prominent ecologist who proposed the Five Kingdom Classification in 1969.
  • The classification includes the kingdoms Monera, Protista, Fungi, Plantae, and Animalia.
  • He is known for his contributions to the study of plant ecology and the concept of biomes.
  • R.D. Barnes
  • R.D. Barnes was a zoologist and is widely recognized for his work on invertebrate zoology.
  • He authored the book "Invertebrate Zoology", which is a comprehensive study of invertebrates.
  • N. Myers
  • N. Myers is an environmentalist known for introducing the concept of "biodiversity hotspots" in 1988.
  • Biodiversity hotspots are regions with significant levels of species richness and are under threat due to human activities.
  • His work has been instrumental in focusing conservation efforts on ecologically critical areas.
23

The number of diagonals in a hexagon is:

  1. ((a))

    9

  2. ((b))

    18

  3. ((c))

    15

  4. ((d))

    12

Show Answer
Answer: ((a))

9

Given:

Number of sides of hexagon (n) = 6

Formula used:

Number of diagonals = n(n − 3) ÷ 2

Calculations:

⇒ Number of diagonals = 6(6 − 3) ÷ 2

⇒ = 6 × 3 ÷ 2

⇒ = 18 ÷ 2

⇒ = 9

∴ The number of diagonals in a hexagon is 9.

24

If 20% of a number is 50, then what is that number?

  1. ((a))

    300

  2. ((b))

    150

  3. ((c))

    250

  4. ((d))

    350

Show Answer
Answer: ((c))

250

Given:

20% of a number = 50

Formula used:

Percentage = (Part ÷ Whole) × 100

Calculations:

⇒ 20% = 50

⇒ 20 ÷ 100 × Number = 50

⇒ Number = 50 × 100 ÷ 20

⇒ Number = 250

∴ The required number is 250.

25

When did the All India Congress Committee meet in Bombay, where the famous "Quit India" resolution was passed?

  1. ((a))

    9th August, 1942

  2. ((b))

    7th August, 1942

  3. ((c))

    8th August, 1942

  4. ((d))

    6th August, 1942

Show Answer
Answer: ((c))

8th August, 1942

The Correct answer is 8th August, 1942.

Key Points

  • The famous Quit India resolution was passed on 8th August, 1942 during a meeting of the All India Congress Committee (AICC) held in Bombay.
  • This resolution marked the beginning of the Quit India Movement, a major campaign led by the Indian National Congress against British rule in India.
  • The movement was initiated during World War II, when Indians demanded an end to British colonial rule.
  • The resolution was introduced by Mahatma Gandhi, who called for a mass civil disobedience campaign that included non-cooperation and peaceful protests.
  • The primary slogan of the movement was “Do or Die”, urging Indians to fight for independence with determination.
  • Key leaders of the Indian National Congress, including Jawaharlal Nehru, Sardar Patel, and Maulana Azad, supported the resolution.
  • As a result of this movement, the British government arrested most of the Congress leaders, including Mahatma Gandhi, leading to widespread protests across India.
  • The Quit India Movement is considered one of the final steps toward India's independence, which was achieved on 15th August, 1947.

Additional Information

  • Key Characteristics of the Quit India Movement
  • The movement involved participation from various sections of Indian society, including farmers, workers, students, and women.
  • The British government responded with severe repression, including arrests, violence, and censorship.
  • Despite the challenges, the movement united Indians in their demand for independence and showcased the strength of nonviolent resistance.
  • It inspired similar movements in other colonies seeking independence from imperial rule.
  • Historical Context of 1942
  • The Quit India Movement was launched during World War II, when the British were facing challenges both internationally and domestically.
  • The movement highlighted the growing impatience among Indians for independence.
  • It also exposed the weakening grip of the British Empire on its colonies.
26

 If the LCM of two numbers is 2079 and their HCF is 27 and if one of the numbers is 189, then find the other number.

  1. ((a))

    297

  2. ((b))

    792

  3. ((c))

    279

  4. ((d))

    927

Show Answer
Answer: ((a))

297

Given:

LCM of two numbers = 2079

HCF of two numbers = 27

One number = 189

Formula used:

Product of two numbers = LCM × HCF

Calculations:

⇒ Product of two numbers = 2079 × 27

⇒ Product = 56133

⇒ Other number = Product ÷ Given number

⇒ Other number = 56133 ÷ 189

⇒ Other number = 297

∴ The other number is 297.

27

Which of these metals reacts with water at normal temperature?

  1. ((a))

    Sodium

  2. ((b))

    Mercury 

  3. ((c))

    Aluminium

  4. ((d))

    Magnesium

Show Answer
Answer: ((a))

Sodium

The Correct answer is Sodium.

Key Points

  • Sodium is a highly reactive metal that reacts vigorously with water at normal temperature.
  • The reaction of sodium with water produces sodium hydroxide (NaOH) and releases hydrogen gas, along with a significant amount of heat.
  • Sodium reacts so intensely that it can cause fire or explosions when exposed to water.
  • To prevent accidental reactions, sodium is usually stored in substances like kerosene oil or mineral oil, which keep it away from moisture.
  • This reaction is an example of the property of alkali metals, which are known for their high reactivity with water.
  • The vigorous reaction of sodium with water is often utilized in educational demonstrations to explain chemical reactivity.

Additional Information

  • Mercury
  • Mercury is a liquid metal at room temperature and does not react with water under normal conditions.
  • It has a wide range of industrial applications, including in thermometers, barometers, and electrical switches.
  • Mercury is toxic and requires careful handling due to its poisonous effects on humans and the environment.
  • Aluminium
  • Aluminium is a metal that does not react with water at room temperature due to the presence of a protective oxide layer on its surface.
  • It is widely used in applications like construction, aerospace, packaging, and electrical components.
  • Magnesium
  • Magnesium reacts with water, but this reaction occurs slowly at room temperature and is much faster in hot water.
  • The reaction produces magnesium hydroxide and releases hydrogen gas.
  • Magnesium is commonly used in alloys, fireworks, and lightweight materials.
28

The 2024 Lok Sabha elections in India were conducted in how many phases?

  1. ((a))

    6

  2. ((b))

    7

  3. ((c))

    9

  4. ((d))

    5

Show Answer
Answer: ((c))

9

The Correct answer is 9 phases.

Key Points

  • The 2024 Lok Sabha elections in India were conducted in 9 phases, as per the Election Commission of India.
  • Conducting the elections in multiple phases ensures a smooth and fair election process in the world's largest democracy.
  • Each phase covers different constituencies, allowing the Election Commission to efficiently manage resources like security forces and polling personnel.
  • Multi-phase elections are crucial in India due to its vast geographical diversity and population size.
  • The 9-phase election process allows for better monitoring, reducing the chances of malpractices and ensuring transparency.
  • In previous elections, such as the 2019 Lok Sabha elections, a similar multi-phase strategy was adopted.
  • Important factors like law and order, logistics, and political climate are considered while deciding the number of phases.
  • The phased approach also ensures that voters in each region have adequate time to prepare and participate in the electoral process.

Additional Information

  • Importance of Multi-Phase Elections
  • Multi-phase elections help maintain law and order in sensitive regions by deploying security forces effectively.
  • It provides enough time for electoral officials to prepare and ensures a more organized voting process.
  • This approach also allows for better voter turnout as people have sufficient time to plan their participation.
  • Single-Phase vs Multi-Phase Elections
  • Single-phase elections are quicker but are usually preferred in smaller countries or regions with fewer logistical challenges.
  • Multi-phase elections are ideal for large democracies like India, where organizing elections in a single day is not feasible.
29

Consider the following Acts and arrange them in the correct  chronological order:

I. Charter Act

II. Pitt's India Act

III. The Indian Councils Act

IV. The Regulating Act

Select the correct answer from the codes given below:

  1. ((a))

    IV, I, II, III

  2. ((b))

    IV, III, II, I

  3. ((c))

    IV, II, I, III

  4. ((d))

    I, II, III, IV

Show Answer
Answer: ((c))

IV, II, I, III

The Correct answer is IV, II, I, III.

Key Points

  • The chronological order of the Acts is as follows:
  • Regulating Act (1773): This was the first attempt by the British Parliament to regulate the affairs of the East India Company in India. It introduced the post of the Governor-General of Bengal and established a Supreme Court at Calcutta.
  • Pitt's India Act (1784): Named after William Pitt the Younger, this act distinguished between the commercial and political functions of the East India Company. It established a Board of Control to supervise civil, military, and revenue affairs.
  • Charter Act (1813): This act marked the beginning of the era of British government control over the East India Company. It ended the monopoly of the company over trade, allowing other British merchants to trade with India, except for tea and trade with China.
  • The Indian Councils Act (1861): This act introduced non-official Indian members in the legislative councils and laid the foundation for the parliamentary system in India.
  • The correct chronological order is: Regulating Act (1773), Pitt's India Act (1784), Charter Act (1813), The Indian Councils Act (1861).

Additional Information

  • Regulating Act (1773):
  • The act established the Office of Governor-General of Bengal, with Warren Hastings being the first to hold the position.
  • The Supreme Court of Judicature was set up in Calcutta, consisting of one chief justice and three judges.
  • It was the first act to assert the right of the British Parliament to regulate the East India Company.
  • Pitt's India Act (1784):
  • This act created the Board of Control to oversee the political affairs of the East India Company.
  • It allowed the British government to have direct control over Indian administration while leaving the company with administrative and commercial functions.
  • Charter Act (1813):
  • The act ended the trade monopoly of the East India Company in India, except for trade in tea and trade with China.
  • It allowed Christian missionaries to enter India for the purpose of religious and social reforms.
  • The Indian Councils Act (1861):
  • This act marked the beginning of the legislative councils in India and introduced the concept of representative institutions.
  • It provided for the inclusion of Indians in the legislative process, laying the foundation for later reforms.
30

Which one of the following states receives floods in the winter months?

  1. ((a))

    Gujarat

  2. ((b))

    Tamil Nadu

  3. ((c))

    Madhya Pradesh

  4. ((d))

    Maharashtra

Show Answer
Answer: ((b))

Tamil Nadu

The correct answer is Tamil Nadu.

Key Points

  • Tamil Nadu receives floods during the winter months, primarily due to the Northeast Monsoon, which occurs between the months of October and December.
  • The state is geographically located in such a way that it receives most of its rainfall during the retreating monsoon season, unlike other parts of India that primarily depend on the Southwest Monsoon.
  • Major rivers such as the Cauvery, Vaigai, and Thamirabarani often overflow during heavy rains, contributing to flooding in several parts of the state.
  • Several districts, including Chennai, Cuddalore, and Nagapattinam, are highly vulnerable to floods during this period due to their low-lying nature and proximity to the coast.
  • Urban areas like Chennai often face severe waterlogging because of inadequate drainage infrastructure and high-intensity rainfall.
  • Floods during this period significantly affect agriculture, infrastructure, and livelihoods in Tamil Nadu, making disaster preparedness crucial.
  • The Tamil Nadu government implements strategies such as early warning systems and flood mitigation projects to minimize the impact of floods during winter.

Additional Information

  • Gujarat
  • Gujarat primarily receives rainfall during the Southwest Monsoon from June to September. It is not prone to winter floods.
  • Occasionally, Gujarat experiences cyclonic storms from the Arabian Sea, which may lead to flooding, but this generally occurs outside the winter season.
  • Madhya Pradesh
  • Madhya Pradesh is a landlocked state and relies predominantly on the Southwest Monsoon for its rainfall.
  • Winter floods are uncommon in Madhya Pradesh as the state does not experience significant rainfall during this period.
  • Maharashtra
  • Maharashtra also depends on the Southwest Monsoon for its primary rainfall.
  • Floods in Maharashtra usually occur during the monsoon season, particularly in regions like Konkan, which includes Mumbai, but winter floods are not prevalent.

Part 2 (120 questions)

31

The sum of coefficients of the polynomial (1 + x - 3x2)15​ is:

  1. ((a))

    -1

  2. ((b))

    0

  3. ((c))

    1

  4. ((d))

    None of the above

Show Answer
Answer: ((a))

-1

Calculation:

The sum of the coefficients of a polynomial P(x) is found by setting the variable

x= 1

Let P(x)=anxn+an1xn1++a1x+a0 P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0

Then the sum of coefficients is P(1)=an+an1++a1+a0 P(1) = a_n + a_{n-1} + \dots + a_1 + a_0

Given the polynomial:

P(x)=(1+x3x2)15 P(x) = (1 + x - 3x^2)^{15}

Substitute x=1 x = 1 into the expression:

P(1)=(1+13(1)2)15 P(1) = (1 + 1 - 3(1)^2)^{15}

P(1)=(1+13)15 P(1) = (1 + 1 - 3)^{15}

P(1)=(23)15 P(1) = (2 - 3)^{15}

P(1)=(1)15 P(1) = (-1)^{15}

Since the power is odd:

P(1)=1 P(1) = -1

The sum of coefficients is -1

32

If |z1| < 1 and |z2| < 1, then value of z1z21z1z2\rm \left|\frac{z_{1}-z_{2}}{1-z_{1} z_{2}}\right| is:

  1. ((a))

    equal to 1

  2. ((b))

    less than 1

  3. ((c))

    greater than 1

  4. ((d))

    None of the above

Show Answer
Answer: ((b))

less than 1

Calculation:

Given complex numbers z1 and z2 such that z1<1 |z_1| < 1 and z2<1 |z_2| < 1

We consider the expression z1z21z1z2 \left|\frac{z_1 - z_2}{1 - \overline{z_1}z_2}\right|

For any two points inside the unit disk, the following inequality holds:

z1z21z1z2<1 \left|\frac{z_1 - z_2}{1 - \overline{z_1}z_2}\right| < 1

This is a standard result from complex analysis (Schwarz–Pick inequality)

∴  The value of z1z21z1z2 \left|\frac{z_1 - z_2}{1 - \overline{z_1}z_2}\right| is strictly less than 1

33

Consider the series:

I. n=1cos1n\sum_{n=1}^{\infty} \cos \frac{1}{n}

II. n=11nsin1n\sum_{n=1}^{\infty} \frac{1}{n} \sin \frac{1}{n}

Here,

  1. ((a))

    I converges but II does not

  2. ((b))

    Neither I nor II converges

  3. ((c))

    I and II both converge

  4. ((d))

    II converges but I does not

Show Answer
Answer: ((d))

II converges but I does not

Calculation:

We study the convergence of the following two series:

I. n=1cos(1n) \sum_{n=1}^{\infty} \cos\left(\frac{1}{n}\right)

II. n=11nsin(1n) \sum_{n=1}^{\infty} \frac{1}{n}\sin\left(\frac{1}{n}\right)

Series I:

As n n \to \infty , we have 1n0 \frac{1}{n} \to 0 and hence cos(1n)cos0=1 \cos\left(\frac{1}{n}\right) \to \cos 0 = 1

Since the general term does not tend to zero, the series cos(1n) \sum \cos\left(\frac{1}{n}\right) is divergent

Series II:

For small values of x, we know that sinxx \sin x \approx x

So for large n, sin(1n)1n \sin\left(\frac{1}{n}\right) \approx \frac{1}{n}

Hence, 1nsin(1n)1n2 \frac{1}{n}\sin\left(\frac{1}{n}\right) \approx \frac{1}{n^2}

Since 1n2 \sum \frac{1}{n^2} is a convergent series, 1nsin(1n) \sum \frac{1}{n}\sin\left(\frac{1}{n}\right) also converges by Limit Comparison Test.

Series I diverges and Series II converges.

∴ The correct answer is II converges but I does not

34

The area between parabola y2 = 4ax and its latus rectum is:

  1. ((a))

    83\frac{8}{3}πa

  2. ((b))

    8πa2

  3. ((c))

    πa2

  4. ((d))

    83\frac{8}{3}a2

Show Answer
Answer: ((d))

83\frac{8}{3}a2

Calculation:

Give equation of parabola  y2 = 4ax

⇒focus (a, 0)

So, the equation of the latus rectum is x = a

Since, the parabola is symmetric about the x-axis, the required area is twice the area bounded by the curve in the first quadrant from x = 0 to x = a

From the equation y2=4ax y^2 = 4ax , we get y=2ax y = 2\sqrt{a}\sqrt{x} for the upper half:

Required Area A = 20ay,dx 2 \int_{0}^{a} y , dx

A=20a2ax1/2,dx A = 2 \int_{0}^{a} 2\sqrt{a} x^{1/2} , dx

A=4a0ax1/2,dx A = 4\sqrt{a} \int_{0}^{a} x^{1/2} , dx

Using the power rule for integration:

A=4a[x3/23/2]0a A = 4\sqrt{a} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{a}

A=4a×23[a3/20] A = 4\sqrt{a} \times \frac{2}{3} \left[ a^{3/2} - 0 \right]

A=83aa3/2 A = \frac{8}{3}\sqrt{a} \cdot a^{3/2}

Since a=a1/2 \sqrt{a} = a^{1/2} :

A=83a(1/2+3/2) A = \frac{8}{3} a^{(1/2 + 3/2)}

A=83a2 A = \frac{8}{3} a^2

∴ The Correct Option is  83a2 \frac{8}{3} a^2

35

Consider the following two statements:

I. n=1un\sum_{n=1}^{\infty} u_{n} is convergent ⇒ n=1un\sum_{n=1}^{\infty}\left|u_{n}\right| is convergent.

II. n=1un\sum_{n=1}^{\infty}\left|u_{n}\right| is convergent ⇒ n=1un\sum_{n=1}^{\infty} u_{n} is convergent.

Which of the above statements is/are correct?

  1. ((a))

    Only II

  2. ((b))

    Only I

  3. ((c))

    Neither I nor II

  4. ((d))

    Both I and II

Show Answer
Answer: ((a))

Only II

Concept:

There are two types of convergence of a series:

  1. Absolute convergence: un \sum |u_n| is convergent.
  2. Conditional convergence: un \sum u_n is convergent but un \sum |u_n| is divergent.

Calculation:

Statement I:

un \sum u_n is convergent un \Rightarrow \sum |u_n| is convergent.

This statement is false

Counter Example: (1)n+1n \sum \frac{(-1)^{n+1}}{n } is convergent, but (1)n+1n=1n \sum |\frac{(-1)^{n+1}}{n} | = \sum \frac{1}{n} is divergent

So, convergence of un \sum u_n does not guarantee convergence of un \sum |u_n|

Statement II:

un \sum |u_n| is convergent un \Rightarrow \sum u_n is convergent.

This statement is true

If a series converges absolutely, then it always converges

The correct option is Only II

36

The pole of the plane Ix + my + nz = p with respect to the sphere x2 + y2 + z2 = a2 is:

  1. ((a))

    (lap,map,nap)\rm \left(\frac{l \mathrm{a}}{\mathrm{p}}, \frac{\mathrm{ma}}{\mathrm{p}}, \frac{\mathrm{na}}{\mathrm{p}}\right)

  2. ((b))

    (lpa2,mpa2,npa2)\rm \left(\frac{l}{p} a^{2}, \frac{m}{p} a^{2}, \frac{n}{p} a^{2}\right)

  3. ((c))

    (lap, map,nap)\rm\left(\frac{l}{\mathrm{ap}}, \frac{\mathrm{~m}}{\mathrm{ap}}, \frac{\mathrm{n}}{\mathrm{ap}}\right)

  4. ((d))

    (pla2,p ma2,pna2)\rm \left(\frac{\mathrm{p}}{l} \mathrm{a}^{2}, \frac{\mathrm{p}}{\mathrm{~m}} \mathrm{a}^{2}, \frac{\mathrm{p}}{\mathrm{n}} \mathrm{a}^{2}\right)

Show Answer
Answer: ((b))

(lpa2,mpa2,npa2)\rm \left(\frac{l}{p} a^{2}, \frac{m}{p} a^{2}, \frac{n}{p} a^{2}\right)

Concept:

For the sphere x2+y2+z2=a2 x^2 + y^2 + z^2 = a^2 , the polar plane of a point (x1,y1,z1) (x_1, y_1, z_1) is

xx1+yy1+zz1=a2 xx_1 + yy_1 + zz_1 = a^2

If this polar plane coincides with the given plane

lx + my + nz = p

then comparing coefficients, we get

x1=a2lp,y1=a2mp,z1=a2np x_1 = \frac{a^2 l}{p}, \quad y_1 = \frac{a^2 m}{p}, \quad z_1 = \frac{a^2 n}{p}

Hence**,** the pole of the plane  lx +my + nz = p with respect to the sphere x2+y2+z2=a2 x^2 + y^2 + z^2 = a^2 is

(lpa2,mpa2,npa2)\rm \left(\frac{l}{p} a^{2}, \frac{m}{p} a^{2}, \frac{n}{p} a^{2}\right)

∴ The correct option is (lpa2,mpa2,npa2)\rm \left(\frac{l}{p} a^{2}, \frac{m}{p} a^{2}, \frac{n}{p} a^{2}\right)

37

The sequence < sn > = (1+1n)n\rm \left(1+\frac{1}{n}\right)^{n}

  1. ((a))

    Is absolutely convergent to a rational number

  2. ((b))

    Converges to an irrational number

  3. ((c))

    Converges to a rational number

  4. ((d))

    Is divergent

Show Answer
Answer: ((b))

Converges to an irrational number

Calculation:

Consider the sequence Sn=(1+1n)n S_n = \left(1+\frac{1}{n}\right)^n

It is a well-known standard result that

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

The number e is known as Euler’s number

It is an established fact that e is an irrational number

Hence, the given sequence converges and its limit is an irrational number

The Correct answer is  Converges to an irrational number

38

What is the differential equation which represents the family of parabolas whose vertices are at the origin and whose axes are along the positive direction of x-axis?

  1. ((a))

    y2+2xydydx=0\rm y^{2}+2 x y \frac{d y}{d x}=0

  2. ((b))

    ydydx=0\rm y \frac{d y}{d x}=0

  3. ((c))

    y22xydydx=0\rm y^{2}-2 x y \frac{d y}{d x}=0

  4. ((d))

    None of the above

Show Answer
Answer: ((c))

y22xydydx=0\rm y^{2}-2 x y \frac{d y}{d x}=0

Calculation:

The equation of the family of parabolas having vertices at the origin and axes along the positive direction of the x-axis is given by:

y2=4ax y^2 = 4ax ... (1)

where a is an arbitrary parameter.

To form the differential equation, differentiate (1)

2ydydx=4a 2y \frac{dy}{dx} = 4a ... (2)

Now, we substitute the value of 4a from equation (2) back into equation (1) to eliminate the arbitrary constant:

y2=(2ydydx)x y^2 = \left( 2y \frac{dy}{dx} \right) x

y2=2xydydx \Rightarrow y^2 = 2xy \frac{dy}{dx}

Rearranging the terms, we get:

y22xydydx=0 y^2 - 2xy \frac{dy}{dx} = 0

∴   The Correct Answer is y22xydydx=0\rm y^{2}-2 x y \frac{d y}{d x}=0

39

The integrating factor of the differential equation (1y2)dxdy+yx=ay,y<1\rm \left(1-y^{2}\right) \frac{d x}{d y}+y x=a y,|y|<1 is:

  1. ((a))

    11y2\rm \frac{1}{1-y^{2}}

  2. ((b))

    1y2\rm \sqrt{1-y^{2}}

  3. ((c))

    (1 - y2)

  4. ((d))

    11y2\rm \frac{1}{\sqrt{1-y^{2}}}

Show Answer
Answer: ((d))

11y2\rm \frac{1}{\sqrt{1-y^{2}}}

Calculation:

A first order linear differential equation in x(y) of the form dxdy+P(y)x=Q(y) \dfrac{dx}{dy} + P(y)x = Q(y) has integrating factor μ(y)=eP(y),dy \mu(y) = e^{\int P(y),dy}

We have  (1y2)dxdy+yx=ay (1-y^2)\dfrac{dx}{dy} + yx = ay , with y<1|y|<1

Divide throughout by 1-y2 we get:

dxdy+y1y2x=ay1y2 \dfrac{dx}{dy} + \dfrac{y}{1-y^2}x = \dfrac{ay}{1-y^2}

So P(y)=y1y2 P(y) = \dfrac{y}{1-y^2}

⇒ μ(y)=ey1y2,dy \mu(y) = e^{\int \frac{y}{1-y^2},dy}

Evaluate the integral using u=1y2du=2y,dyu = 1-y^2 \Rightarrow du = -2y,dy:

y1y2,dy=12duu=12lnu=12ln(1y2) \displaystyle \int \frac{y}{1-y^2},dy = -\frac12 \int \frac{du}{u} = -\frac12 \ln|u| = -\frac12 \ln(1-y^2)

(since y<11y2>0|y|<1 \Rightarrow 1-y^2>0)

∴  μ(y)=e12ln(1y2)=(1y2)1/2=11y2 \mu(y) = e^{-\frac12 \ln(1-y^2)} = (1-y^2)^{-1/2} = \dfrac{1}{\sqrt{1-y^2}}

∴ The correct option is 11y2\rm \frac{1}{\sqrt{1-y^{2}}}

40

The solution of differential equation (1+x) y dx + (1 - y) x dy = 0 is:

  1. ((a))

    xy = Ce-x+y

  2. ((b))

    x - y = Cex+y

  3. ((c))

    xy = Cex-y

  4. ((d))

    (x + y) = Cex-y

Show Answer
Answer: ((a))

xy = Ce-x+y

Calculation:

We are given the differential equation

(1 + x) y dx + (1 − y) x dy = 0

Divide by x y:

((1 + x)/x) dx + ((1 − y)/y) dy = 0

⇒ (1/x + 1) dx + (1/y − 1) dy = 0

Integrate both terms:

∫(1/x + 1) dx + ∫(1/y − 1) dy = 0

⇒ ln|x| + x + ln|y| − y = C

⇒ ln|x| + ln|y| + x − y = C

⇒ ln|xy| + x − y = C

⇒ ln|xy| = C − x + y

Taking exponential on both sides:

xy = C ey − x

∴ The Correct Answer is : xy = Ce− x+y

41

If f(x) is an even function and f'(0) exists, then the value of f'(0) will be:

  1. ((a))

    0

  2. ((b))

    -1

  3. ((c))

    1

  4. ((d))

    2

Show Answer
Answer: ((a))

0

Calculation:

An even function satisfies f(x)=f(x) f(x) = f(-x) for all x

Differentiate both sides with respect to x :

f(x)=f(x) f'(x) = -f'(-x)

Now put x=0 x = 0 :

f(0)=f(0) f'(0) = -f'(0)

⇒ 2f(0)=0 2f'(0) = 0

⇒ f(0)=0 f'(0) = 0

  If f(x) f(x) is even and f(0) f'(0) exists, then f(0)=0 f'(0) = 0

∴ The correct answer is 0

42

Consider the following statements:

I. The general value of log √i is 12\frac{1}{2}(8n + 1)πi.

II. The value of cos (log i-i) is zero.

Which of the above statements is/are correct?

  1. ((a))

    Only I

  2. ((b))

    Only II

  3. ((c))

    Neither I nor II

  4. ((d))

    Both I and II

Show Answer
Answer: ((b))

Only II

Calculation:

Statement I: General value of log √i

First write i in exponential form:

i = e^{ i(π/2 + 2πk) }

Taking square root:

√i = e^{ i(π/4 + πk) }

Now take logarithm:

log √i = ln|√i| + i(π/4 + πk + 2πn)

Since |√i| = 1, ln|√i| = 0

So, log √i = i(π/4 + πm) = πi(4m + 1)/4

This does not match the given form (1/2)(8n + 1)πi.

Hence, Statement I is incorrect.

Statement II: Value of cos(log i-i)

Using principal values, i = eiπ/2

i-i = (eiπ/2)-i = e-i2π/2 = eπ/2

Now, log i-i = log(eπ/2) = π/2

cos(log i-i) = cos(π/2) = 0

Hence, Statement II is correct.

So, Only Statement II is correct.

∴ The correct answer is: Only II

43

In a group (G, ⋆), where G = {0, 1, ± 2, ± 3, ...} and m ⋆ n = m + n + mn ∀ m, n ∈ G, the inverse of n is:

  1. ((a))

    nn+1\rm \frac{n}{n+1}

  2. ((b))

    n+1n\rm -\frac{n+1}{n}

  3. ((c))

    nn+1\rm -\frac{n}{n+1}

  4. ((d))

    n+1n\rm \frac{n+1}{n}

Show Answer
Answer: ((c))

nn+1\rm -\frac{n}{n+1}

Concept Used:

In a group, n−1 satisfies  n ☆ n−1 = e

where e is identity element

Calculation:

Step 1: Identity element

Let identity = e

⇒ n ☆ e = n

⇒ n + e + ne = n

⇒ e(1 + n) = 0

⇒ e = 0

∴ Identity element = 0

Step 2: Inverse of n

Let inverse of n = n−1

⇒ n ☆ n−1 = 0

⇒ n + n−1 + nn−1 = 0

⇒ n−1(1 + n) = −n

⇒ n−1 = nn+1\rm -\frac{n}{n+1}

∴ The correct answer is nn+1\rm -\frac{n}{n+1}

44

If a^,b^,c^\rm \hat{a}, \hat{b}, \hat{c} are unit vectors and a^+b^2=b^+c^2=c^+a^2=4\rm |\hat{a}+\hat{b}|^{2}=|\hat{b}+\hat{c}|^{2}=|\hat{c}+\hat{a}|^{2}=4 then 2a^+3b^+4c^\rm |2 \hat{a}+3 \hat{b}+4 \hat{c}| is equal to :

  1. ((a))

    4√5

  2. ((b))

    8

  3. ((c))

    9

  4. ((d))

    181

Show Answer
Answer: ((c))

9

Calculation:

|a| = |b| = |c| = 1 , |a + b|2 = 4 ,|b + c|2 = 4 ,|c + a|2 = 4

|a + b|2 = 1 + 1 + 2a·b

⇒ 4 = 2 + 2a·b ⇒ a·b = 1 …… (1)

|b + c|2 = 1 + 1 + 2b·c

⇒ 4 = 2 + 2b·c ⇒ b·c = 1 …… (2)

|c + a|2 = 1 + 1 + 2c·a

⇒ 4 = 2 + 2c·a ⇒ c·a = 1 …… (3)

|2a + 3b + 4c|2 = 4|a|2 + 9|b|2 + 16|c|2 +12a·b + 16a·c + 24b·c

= 4 + 9 + 16 + 12(1) + 16(1) + 24(1) = 29 + 52 = 81

⇒ |2a + 3b + 4c| = √81

∴ |2a** + 3b + 4c| = 9**

∴ The Correct Answer is 9

45

An integrating factor of differential equation (1+x2)dydx+2xy=cosx\rm \left(1+x^{2}\right) \frac{d y}{d x}+2 x y=\cos x  is:

  1. ((a))

    x

  2. ((b))

    log (1 + x2)

  3. ((c))

    1 + x2

  4. ((d))

    x2

Show Answer
Answer: ((c))

1 + x2

Given:

(1 + x2) dy/dx + 2x y = cos x

Formula used:

Linear differential equation:

dy/dx + P(x) y = Q(x)

Integrating factor = e∫P(x) dx

Calculation:

⇒ Divide whole equation by (1 + x2)

⇒ dy/dx + (2x / (1 + x2)) y = cos x / (1 + x2)

⇒ P(x) = 2x / (1 + x2)

⇒ IF = e∫ 2x / (1 + x2) dx

⇒ IF = elog(1 + x2)

⇒ IF = 1 + x2

∴ The correct answer is 1 + x2.

46

The equation of the normal at 'θ' on the hyperbola x2a2y2b2\rm \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1 is:

  1. ((a))

    xasecθybtanθ=1\rm \frac{x}{a} \sec \theta - \frac{y}{b} \tan \theta=1

  2. ((b))

    axsinθ by cosθ=a2+b2\rm \frac{a x}{\sin \theta}-\frac{\text { by }}{\cos \theta}=a^{2}+b^{2}

  3. ((c))

    axsecθ+ by tanθ=a2+b2\rm \frac{a x}{\sec \theta}+\frac{\text { by }}{\tan \theta}=a^{2}+b^{2}

  4. ((d))

    axsecθbytanθ=a2b2\rm \frac{a x}{\sec \theta}-\frac{b y}{\tan \theta}=a^{2}-b^{2}

Show Answer
Answer: ((c))

axsecθ+ by tanθ=a2+b2\rm \frac{a x}{\sec \theta}+\frac{\text { by }}{\tan \theta}=a^{2}+b^{2}

Formula Used:

Parametric point: (a sec θ, b tan θ)

Derivative slope: dydx\frac{dy}{dx}

Normal slope = −1 / (slope of tangent)

Calculation:

Given Equation of  Hyperbola :x2a2y2b2\rm \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1

The Parametric coordinates: x = a sec θ,   y = b tan θ

Slope of tangent: dydx=dydθdxdθ\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}

 dydx=bsec2θasecθtanθ=bsecθatanθ\Rightarrow \frac{dy}{dx} = \frac{b\sec^2\theta}{a\sec\theta\tan\theta} = \frac{b\sec\theta}{a\tan\theta}

Slope of normal = − reciprocal of tangent slope

mnormal=atanθbsecθm_{normal} = -\frac{a\tan\theta}{b\sec\theta}

Point on hyperbola = (a sec θ, b tan θ)

Therefore the equation of normal becomes 

ybtanθ=atanθbsecθ(xasecθ) y - b\tan\theta = -\frac{a\tan\theta}{b\sec\theta} (x - a\sec\theta)

Multiply both sides by b sec θ:

⇒ b sec θ y − b2 sec θ tan θ = − a tan θ x + a2 sec θ tan θ

⇒ b y(sec θ ) +  a x(tan θ) x =  a2 sec θ tan θ + b2 sec θ tan θ

Rearranging terms:axsecθ+ by tanθ=a2+b2\rm \frac{a x}{\sec \theta}+\frac{\text { by }}{\tan \theta}=a^{2}+b^{2} 

∴ The correct answer is axsecθ+ by tanθ=a2+b2\rm \frac{a x}{\sec \theta}+\frac{\text { by }}{\tan \theta}=a^{2}+b^{2}

47

If A=i^k^,B=xi^+j^+(1x)k^\rm \vec{A}=\hat{i}-\widehat{k}, \vec{B}=x \hat{i}+\hat{j}+(1-x) \hat{k} and C=yi^+xj^+(1+xy)k^\overrightarrow{\mathrm{C}}=\mathrm{y} \hat{\mathrm{i}}+\mathrm{x} \hat{\mathrm{j}}+(1+\mathrm{x}-\mathrm{y}) \hat{\mathrm{k}}, then [A,B,C]\rm [\vec{A}, \vec{B}, \vec{C}] depends on :

  1. ((a))

    Neither on x nor on y

  2. ((b))

    y only

  3. ((c))

    Both x and y

  4. ((d))

    x only

Show Answer
Answer: ((a))

Neither on x nor on y

Concept Used:

Scalar triple product [A,B,C]\rm [\vec{A}, \vec{B}, \vec{C}]= determinant of matrix 

Calculation:

The given vectors are  

A\vec{A} = î − k̂ = (1, 0, −1) 

B \vec{B} = xî + ĵ + (1 − x)k̂ = (x, 1, 1 − x)

C \vec{C} = yî + xĵ + (1 + x − y)k̂ = (y, x, 1 + x − y)

[A,B,C]\rm [\vec{A}, \vec{B}, \vec{C}]  =101 x11x yx1+xy \begin{vmatrix} 1 & 0 & -1 \ x & 1 & 1 - x \ y & x & 1 + x - y \end{vmatrix}

 = (1 + x − y) − (x − x2) − (x2 − y)

 = 1 + x − y − x+ x2 − x2 + y = 1

No x, y present in final value.

∴ The Correct Answer isoption (1)Neither on x nor on y

48

The transformed equation of the straight line xa+yb\rm \frac{x}{a}+\frac{y}{b} = 2, when origin is shifted to (a, b) is :

  1. ((a))

    xb+ya\rm \frac{x}{b}+\frac{y}{a} = -2

  2. ((b))

    xa+yb\rm \frac{x}{a}+\frac{y}{b} = 0

  3. ((c))

    x + y = 0

  4. ((d))

    xb+ya\rm \frac{x}{b}+\frac{y}{a} = 0

Show Answer
Answer: ((b))

xa+yb\rm \frac{x}{a}+\frac{y}{b} = 0

Concept Used:

Shift of origin transformation

If new origin is (h, k), then:

x = X + h,   y = Y + k

Calculation:

Given Equation of line: xa+yb\rm \frac{x}{a}+\frac{y}{b} = 2

Here, h = a, k = b

⇒ x = X + a

⇒ y = Y + b

Substitute into given equation:

(X+a)a+(Y+b)b=2\frac{(X + a)}{a }+ \frac{(Y + b)}{b }= 2

⇒ Xa+1+Yb+1=2\frac{X }{a } +1+ \frac{Y }{b } +1= 2

Xa+Yb=0\frac{X }{a } + \frac{Y }{b } = 0

Hence, Final Transformed Equation: Xa+Yb=0\frac{X }{a } + \frac{Y }{b } = 0

∴ The Correct answer is xa+yb=0\rm \frac{x}{a}+\frac{y}{b} =0

49

A box contains 4 white and 3 black balls. A man picks up 2 balls at random. What is the probability of both balls being of the same colour ?

  1. ((a))

    57\frac{5}{7}

  2. ((b))

    17\frac{1}{7}

  3. ((c))

    27\frac{2}{7}

  4. ((d))

    37\frac{3}{7}

Show Answer
Answer: ((d))

37\frac{3}{7}

Concept Used:

Probability = (Favourable outcomes) / (Total outcomes)

Both balls same colour = both white OR both black

Formula Used:

nCr = n! / r!(n − r)!

Calculation:

Number of white balls = 4

Number of black balls = 3

Total balls = 7

Balls drawn = 2

⇒ Total ways to choose 2 balls = 7C2 = 21

⇒ Ways to choose 2 white balls = 4C2 = 6

⇒ Ways to choose 2 black balls = 3C2 = 3

⇒ Favourable outcomes = 6 + 3 = 9

⇒ Probability = 9 / 21

⇒ Probability = 3 / 7

Hence, The probability that both balls are of the same colour = 3/7

∴The Correct Answer is 3/7

50

The eccentricity of the ellipse 4x2 + y2 - 8x + 2y + 1 = 0; is:

  1. ((a))

    23\frac{2}{\sqrt{3}}

  2. ((b))

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

  3. ((c))

    23\frac{\sqrt{2}}{3}

  4. ((d))

    34\frac{\sqrt{3}}{4}

Show Answer
Answer: ((b))

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

Formula Used:

eccentricity e: = √(1 − b2/a2)

Ellipse standard form:

(x − h)2/a2 + (y − k)2/b2 = 1

Calculation:

Given Ellipse equation is 4x2 + y2 − 8x + 2y + 1 = 0

⇒ 4x2 − 8x + y2 + 2y + 1 = 0

⇒ 4(x2 − 2x) + (y2 + 2y) + 1 = 0

⇒ 4[(x − 1)2 − 1] + [(y + 1)2 − 1] + 1 = 0

⇒ 4(x − 1)2 + (y + 1)2 − 4 = 0

⇒ 4(x − 1)2 + (y + 1)2 = 4

⇒ (x − 1)2/1 + (y + 1)2/4 = 1

⇒ a2 = 4,   b2 = 1

Now, eccentricity e=(1b2a2)=(114)=34=32e = √(1 − \frac{b2}{a2}) = √(1 − \frac{1}{4}) = √\frac{3}{4} = √\frac{3 }{2}

∴ The Correct Answer is 32\frac{\sqrt{3}}{2}

51

The solution of dydx\rm \frac{d y}{d x} + y = ex, if y(0) = 12\frac{1}{2} is :

  1. ((a))

    y = 12\frac{1}{2}ex

  2. ((b))

    y = ex

  3. ((c))

    y = 2ex + 2

  4. ((d))

    y = 12\frac{1}{2}ex + 4

Show Answer
Answer: ((a))

y = 12\frac{1}{2}ex

Formula Used :

For the differential Equation dy/dx + Py = Q

Integrating Factor (I.F.) = e∫P dx

Solution: y × I.F. = ∫(Q × I.F.) dx + C

Calculation:

Given equation is 

dy/dx + y = ex

Here, P = 1 and Q = ex.

The integrating factor, I(x):

I(x) = e^(∫1 dx) = ex

 Multiply through by the integrating factor:

ex * dy/dx + ex * y = ex * ex

⇒(d/dx)(ex * y) = e2x

Integrate both sides:

⇒∫(d/dx)(ex * y) dx = ∫e2x dx

⇒ex * y = (12\frac{1}{2})e2x + C

⇒y = (12\frac{1}{2})ex + Ce-x

 Apply the initial condition y(0) = 1/2:

12\frac{1}{2} = (12\frac{1}{2})e0 + Ce0

12\frac{1}{2} = 12\frac{1}{2} + C

C = 0

Thus, the solution is:y = 12\frac{1}{2}ex

∴ The correct answer is y = 12\frac{1}{2}ex

52

The value of 01x(1x)n\rm \int_{0}^{1} x(1-x)^{n}  dx is:

  1. ((a))

    n+1n+2\rm \frac{n+1}{n+2}

  2. ((b))

    n+2n+1\rm \frac{n+2}{n+1}

  3. ((c))

    1(n+1)(n+2)\rm \frac{1}{(n+1)(n+2)}

  4. ((d))

    None of the above

Show Answer
Answer: ((c))

1(n+1)(n+2)\rm \frac{1}{(n+1)(n+2)}

Formula Used:

⇒ ∫ xa dx = xa+1 / (a + 1)

Calculation:

⇒ I =01x(1x)ndx \int^1_0 x(1 − x)n dx

⇒ Substitute u = 1 − x

⇒ x = 1 − u, dx = −du

⇒ Limits: x = 0 → u = 1, x = 1 → u = 0

⇒ I =01\int^1_0 (1 − u)un du

⇒ I =  01\int^1_0  (un − un+1) du

un+1n+1un+2n+201| \frac{u^{n+1}}{ n + 1} − \frac{u^{n+2} }{n + 2} \left.\right|_0^1

⇒ I = 1n+11n+2\frac{1}{n+1} - \frac{1}{n+2} 

⇒ I =1(n+1)(n+2)\frac{1}{ (n + 1)(n + 2)}

∴ The Correct Answer is  1(n+1)(n+2)\frac{1}{ (n + 1)(n + 2)}

53

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

List IList II
A.Gradienti.(ix+jy+kz)×f\left(\mathrm{i} \frac{\partial}{\partial \mathrm{x}}+\mathrm{j} \frac{\partial}{\partial \mathrm{y}}+\mathrm{k} \frac{\partial}{\partial \mathrm{z}}\right) \rm \times f
B.Divergenceii.ix+jy+kz\mathrm{i} \frac{\partial}{\partial \mathrm{x}}+\mathrm{j} \frac{\partial}{\partial \mathrm{y}}+\mathrm{k} \frac{\partial}{\partial \mathrm{z}}
C.Curliii.fxi+fyj+fzk\rm \frac{\partial f}{\partial x} \mathrm{i}+\frac{\partial f}{\partial y} \mathrm{j}+\frac{\partial f}{\partial z} \mathrm{k}
D.Deliv.(ix+jy+kz).v\left(\mathrm{i} \frac{\partial}{\partial \mathrm{x}}+\mathrm{j} \frac{\partial}{\partial \mathrm{y}}+\mathrm{k} \frac{\partial}{\partial \mathrm{z}}\right) \rm .v
  1. ((a))

    A-iii, B-i, C-iv, D-ii

  2. ((b))

    A-iii, B-iv, C-i, D-ii

  3. ((c))

    A-i, B-iii, C-iv, D-ii

  4. ((d))

    A-iv, B-iii, C-i, D-ii

Show Answer
Answer: ((b))

A-iii, B-iv, C-i, D-ii

Formula Used:

Del operator: =ix+jy+kz \nabla = i \frac{\partial}{\partial x} + j \frac{\partial}{\partial y} + k \frac{\partial}{\partial z}

Gradient: f=ifx+jfy+kfz \nabla f = i \frac{\partial f}{\partial x} + j \frac{\partial f}{\partial y} + k \frac{\partial f}{\partial z}

Divergence:F=Fxx+Fyy+Fzz \nabla \cdot F = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}

Curl: ∇×F = (ix+jy+kz)×F\left(\mathrm{i} \frac{\partial}{\partial \mathrm{x}}+\mathrm{j} \frac{\partial}{\partial \mathrm{y}}+\mathrm{k} \frac{\partial}{\partial \mathrm{z}}\right) \rm \times F

Calculation:

1) Gradient

∇f =fxi+fyj+fzk\rm \frac{\partial f}{\partial x} \mathrm{i}+\frac{\partial f}{\partial y} \mathrm{j}+\frac{\partial f}{\partial z} \mathrm{k}

⇒ Matches expression (iii)

2) Divergence

∇·v = (ix+jy+kz).v\left(\mathrm{i} \frac{\partial}{\partial \mathrm{x}}+\mathrm{j} \frac{\partial}{\partial \mathrm{y}}+\mathrm{k} \frac{\partial}{\partial \mathrm{z}}\right) \rm .v

⇒ Matches expression (iv)

3) Curl

∇×F = (ix+jy+kz)×f\left(\mathrm{i} \frac{\partial}{\partial \mathrm{x}}+\mathrm{j} \frac{\partial}{\partial \mathrm{y}}+\mathrm{k} \frac{\partial}{\partial \mathrm{z}}\right) \rm \times f

⇒ Matches expression (i)

4) Del

∇ =ix+jy+kz i \frac{\partial}{\partial x} + j \frac{\partial}{\partial y} + k \frac{\partial}{\partial z}

⇒ Matches expression (ii)

Hence, A-iii, B-iv, C-i, D-ii

∴ The Correct  Answer is option: A-iii, B-iv, C-i, D-ii

54

If f(x, y) = x3y5 tan-1(yx)\rm \left(\frac{y}{x}\right), then the value of xfx+yfy\rm x \frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y} is:

  1. ((a))

    9f

  2. ((b))

    3f

  3. ((c))

    8f

  4. ((d))

    5f

Show Answer
Answer: ((c))

8f

Concept Used:

If f(x, y) = xm yn φ(yx)\rm \left(\frac{y}{x}\right), then

x(fx)+y(fy)x (\frac{∂f}{∂x}) + y (\frac{∂f}{∂y}) = (m + n) f(x, y)

Calculation:

given f(x, y) = x3 y5 tan−1(yx)\rm \left(\frac{y}{x}\right)

⇒ m = 3, n = 5

m + n = 3 + 5 = 8

x(fx)+y(fy)x (\frac{∂f}{∂x}) + y (\frac{∂f}{∂y}) = 8  x3 y5tan-1(yx)\rm \left(\frac{y}{x}\right)  = 8f(x, y)

∴ The Correct Answer is  8f

55

The angle between the diagonal and an edge of a cube is:

  1. ((a))

    cos-1(13)\left(\frac{1}{\sqrt{3}}\right)

  2. ((b))

    π3\frac{\pi}{3}

  3. ((c))

    π4\frac{\pi}{4}

  4. ((d))

    sin-1(13)\left(\frac{1}{\sqrt{3}}\right)

Show Answer
Answer: ((a))

cos-1(13)\left(\frac{1}{\sqrt{3}}\right)

Formula  Used:

Angle between two lines using cosine formula:cosθ=(ab)(a×b)cos θ = \frac{(a · b) }{ (|a| × |b|)}

Calculation:

Let the side of the cube  is a

⇒Edge vector = <a, 0, 0>

⇒ Body diagonal vector = <a, a, a>

⇒ Dot product = (a × a) + 0 + 0 = a2

⇒ |Edge| = a

⇒ |Diagonal| = √(a2 + a2 + a2)

⇒ |Diagonal| = a√3

cosθ=a2(a×a3)cos θ =\frac{ a^2 }{ (a × a√3)}

⇒ cos θ = (13)\left(\frac{1}{\sqrt{3}}\right)

⇒ θ = cos−1(13)\left(\frac{1}{\sqrt{3}}\right)

**Hence ,**Angle between diagonal and edge = cos-1(13)\left(\frac{1}{\sqrt{3}}\right) 

∴ The Correct Answer is  cos−1 (13)\left(\frac{1}{\sqrt{3}}\right)

56

The distance of the point (-1, -5, -10) from the point of intersection of the line x23=y+14=z212\rm \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}and the plane x - y + z = 5, is:

  1. ((a))

    13

  2. ((b))

    161\sqrt{161}

  3. ((c))

    12

  4. ((d))

    341\sqrt{41}

Show Answer
Answer: ((a))

13

**Calculation:**​

The Given Equation is x23=y+14=z212\rm \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}

Line parametric form:x = 2 + 3t, y = −1 + 4t, z = 2 + 12t ​

Substitute these values  in the plane equation: x - y + z = 5, is:

⇒ (2 + 3t) − (−1 + 4t) + (2 + 12t) = 5

⇒ 2 + 3t + 1 − 4t + 2 + 12t = 5

⇒ 5 + 11t = 5

⇒ 11t = 0

⇒ t = 0

Intersection point Q:

⇒ x = 2, y = −1, z = 2

Distance PQ:

⇒ PQ = √[(2 + 1)² + (−1 + 5)² + (2 + 10)²]

⇒ PQ = √[(3)² + (4)² + (12)²]

⇒ PQ = √[9 + 16 + 144]

⇒ PQ = √169

⇒ PQ = 13

∴ Distance = 13 units.

57

If f(z) = |z|, then for z ≠ 0, f (z) is :

  1. ((a))

    nowhere differentiable

  2. ((b))

    analytic

  3. ((c))

    differentiable

  4. ((d))

    None of the above

Show Answer
Answer: ((a))

nowhere differentiable

Concept Used:

If z = x + iy

|z| = √(x2 + y2)

Analytic ⇔ C-R equations hold

Calculation:

Let z = x + iy

⇒ f(z) = √(x2 + y2)

⇒ u(x,y) = √(x2 + y2)

⇒ v(x,y) = 0

1) Partial derivatives

⇒ ux =x(x2+y2)\frac{ x }{ √(x^2 + y^2)}

⇒ uyy(x2+y2)\frac{ y }{ √(x^2 + y^2)}

⇒ vx = 0

⇒ vy = 0

2) C-R equations

ux = vy ⇒ x(x2+y2)\frac{ x }{ √(x2 + y2)}≠ 0

uy = −vxy(x2+y2)\frac{ y }{ √(x^2 + y^2)} ≠ 0

⇒ C-R not satisfied for z ≠ 0

Hence, f(z) = |z| is not analytic for z ≠ 0

∴ The correct Answer is f(z) is nowhere differentiable.

58

If a,b,c\rm \vec{a}, \vec{b}, \vec{c} are any three vectors such that (a+b).c=(ab)c=0\rm (\vec{a}+\vec{b}). \vec{c}=(\vec{a}-\vec{b}) \cdot \vec{c}=0, then (a×b)×c\rm (\vec{a} \times \vec{b}) \times \vec{c} is equal to :

  1. ((a))

    b\rm \overrightarrow{{b}}

  2. ((b))

    0\rm \overrightarrow{{0}}

  3. ((c))

    a\rm \overrightarrow{{a}}

  4. ((d))

    None of the above

Show Answer
Answer: ((b))

0\rm \overrightarrow{{0}}

Concept Used:

Vector triple product:(a×b)×c=b(a.c)a(b.c)\rm (\vec{a}\times\vec{b})\times\vec{c}=\vec{b}(\vec{a}.\vec{c})- \vec{a}( \vec{b}.\vec{c})

Calculation:

given conditions:

(a+b)c=0 ac+bc=0...(1) (ab)c=0 acbc=0...(2)(\vec{a} + \vec{b}) \cdot \vec{c} = 0\ \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c} = 0 \quad ...(1)\ (\vec{a} - \vec{b}) \cdot \vec{c} = 0\ \vec{a} \cdot \vec{c} - \vec{b} \cdot \vec{c} = 0 \quad ...(2)

Add (1) and (2)

⇒ 2(a.c)(\vec{a}.\vec{c}) = 0

⇒ (a.c)(\vec{a}.\vec{c}) = 0 ...(3)

Subtract (2) from (1)

⇒ 2(b.c)( \vec{b}.\vec{c})= 0

(b.c)( \vec{b}.\vec{c})= 0 ...(4)

 Apply triple product formula : (a×b)×c=b(a.c)a(b.c)\rm (\vec{a}\times\vec{b})\times\vec{c}=\vec{b}(\vec{a}.\vec{c})- \vec{a}( \vec{b}.\vec{c})

=b\rm \overrightarrow{{b}} (0\rm \overrightarrow{{0}}) - a\rm \overrightarrow{{a}}(0\rm \overrightarrow{{0}}) = 0\rm \overrightarrow{{0}}

 Hence,(a×b)×c\rm (\vec{a} \times \vec{b}) \times \vec{c} = 0\rm \overrightarrow{{0}}

∴The Correct answer is 0\rm \overrightarrow{{0}}

59

If a=2i^+j^+3k^,b=3i^+5j^2k^,\rm \vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}, \vec{b}=3 \hat{i}+5 \hat{j}-2 \hat{k},, then consider the given statements:

I. a×b=17i^+13j^+7k^,a b=5\rm \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=-17 \hat{\mathrm{i}}+13 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}, \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{~b}}=5

II. a×b=507,ab=5\rm |\vec{a} \times \vec{b}|=\sqrt{507}, \vec{a} \cdot \vec{b}=5

III. a×b=71,ab=6\rm |\vec{a} \times \vec{b}|=\sqrt{71}, \vec{a} \cdot \vec{b}=6

Which of the above statements is/are incorrect?

  1. ((a))

    Only III

  2. ((b))

    Only I

  3. ((c))

    All I, II and III

  4. ((d))

    Only II

Show Answer
Answer: ((a))

Only III

Calculation:

1) Dot Product: 

ab \vec{a} \cdot \vec{b} =  a1b1 + a2b2 + a3b3 = (2)(3) + (1)(5) + (3)(−2) 

 =  6 + 5 − 6 = 5

2) Cross Product:

 a×b=i^j^k^ 213 352\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & 3 \ 3 & 5 & -2 \end{vmatrix}

=  î(1×−2 − 3×5) - ĵ(2×−2 − 3×3) + k̂(2×5 − 1×3)

=  î(−2 −15) − ĵ(−4 −9) + k̂(10 −3)

=  −17î +13ĵ +7k̂

3) Magnitude

a×b\rm |\vec{a} \times \vec{b}| = √ (−17)2 + 132 + 72 

= √(289 +169 +49) = √507

Verification:

Statement I:

a×b\rm |\vec{a} \times \vec{b}| = −17î+13ĵ+7k̂ 

ab \vec{a} \cdot \vec{b}= 5

⇒Statement 1 is Correct.

Statement II:

a×b\rm |\vec{a} \times \vec{b}|= √507 

ab \vec{a} \cdot \vec{b} = 5

⇒Statement 2 is Correct.

Statement III:

a×b\rm |\vec{a} \times \vec{b}|= √71 

 ab \vec{a} \cdot \vec{b}= 6 

⇒Statement 3 is incorrect.

Hence, Only III is incorrect.

∴The Correct Answer is Only III

60

If ω is the complex cube root of unity, then ω100 + ω17 + 1 is equal to :

  1. ((a))

    1 + ω

  2. ((b))

    ω + ω

  3. ((c))

    -1

  4. ((d))

    0

Show Answer
Answer: ((d))

0

Calculation:

Given Expression: ω100 + ω17 + 1

⇒((ω)3 )33 ω  + ((ω))5  ω2 + 1 = ω + ω2 + 1

Using identity: 1 + ω + ω2 = 0

⇒ ω100 + ω17 + 1 =0 

∴ Hence, The Correct Answer is 0

61

Consider the following statements with reference to conics:

I. 5x2 - 6xy + 5y2 + 10x - 6y+ 10 = 0 represents an ellipse.

II. The centre of the conic 5x2 - 6xy + 5y2 - 4x - 4y + 5 = 0 is (1, 2).

Which of the above statements is/are correct?

  1. ((a))

    Only II

  2. ((b))

    Neither I nor II

  3. ((c))

    Only I

  4. ((d))

    Both I and II

Show Answer
Answer: ((c))

Only I

Concept Used:

General second-degree equation of conic:

Ax2 + 2Hxy + By2 + 2Gx + 2Fy + C = 0

Discriminant condition:

H2 − AB < 0 ⇒ Ellipse

Centre found by solving:

∂/∂x = 0 and ∂/∂y = 0

Calculation:

Statement I:

General second-degree equation of conic:

Ax2 + 2Hxy + By2 + 2Gx + 2Fy + C = 0

⇒ A = 5,   2H = −6 ⇒ H = −3,   B = 5

⇒ H2 − AB = (−3)2 − (5 × 5) = 9 − 25 = −16 < 0

⇒ Conic represents an ellipse

Hence, Statement I is correct.

Statement II:

Given equation of conic: 5x2 − 6xy + 5y2 − 4x − 4y + 5 = 0

Differentiate partially  w.r.t x   we get,

∂/∂x = 10x − 6y − 4 = 0  

⇒ ∂/∂x = 5x − 3y = 2   ----(1)

Differentiate partially w.r.t y  we get,

∂/∂y = −6x + 10y − 4 = 0  

⇒ ∂/∂y = −3x + 5y = 2   -----(2)

On Solving equations(1) &(2)we get x = 1 & y = 1

⇒ Centre = (1, 1)

Given centre (1, 2) is incorrect.

Hence, Statement II is false.

∴The Correct Answer is  Only  I.

62

Match list I with list II and choose the correct answer using the codes given below the lists: 

List I (Cyclic Group)List II (Number of Generators)
A.(ℤ, +)i.1
B.(ℤ8, +)ii.2
C.(e2nπi7,.)\rm (e^\frac{2n\pi i}{7} ,.)iii.4
D.(ℤ2, +)iv.6
  1. ((a))

    A-ii, B-iii, C-iv, D-i

  2. ((b))

    A-iii, B-ii, C-iv, D-i

  3. ((c))

    A-i, B-ii, C-iii, D-iv

  4. ((d))

    A-iv, B-iii, C-ii, D-i

Show Answer
Answer: ((a))

A-ii, B-iii, C-iv, D-i

Concept Used:

A cyclic group of order n has φ(n) generators,

where φ(n) is Euler’s Totient Function.

Calculation:

Group A: (ℤ, +) is an Infinite cyclic group

⇒ Generators = {1, −1}

⇒ Number of generators = 2 (counted unique)

⇒ A → ii

Group B: (ℤ8, +)

⇒ Order = 8

⇒ φ(8) = 4

⇒ Number of generators = 4

⇒ B → iii

Group C: (e2πi/7, ·)

⇒ Order = 7

⇒ φ(7) = 6

⇒ Number of generators = 6

⇒ C → iv

Group D: (ℤ2, +)

⇒ Order = 2

⇒ φ(2) = 1

⇒ Number of generators = 1

⇒ D → i

Final Matching:

A → i,   B → iii,   C → iv,   D → i

∴ The Correct answer is: A-i, B-iii, C-iv, D-i

63

If y = ax+bcx+d\rm \frac{a x+b}{c x+d}, then d3ydx3\rm \frac{d^{3} y}{d x^{3}} is :

  1. ((a))

    (1)33!(bcad)c2(cx+d)4\rm (-1)^{3} 3!(b c-a d) c^{2}(c x+d)^{-4}

  2. ((b))

    (1)33!(bcad)(cx+d)4\rm (-1)^{3} 3!(b c-a d)(c x+d)^{-4}

  3. ((c))

    (1)33!(bcad)(cx+d)3\rm (-1)^{3} 3!(b c-a d)(c x+d)^{-3}

  4. ((d))

    (1)33!c2(cx+d)3\rm (-1)^{3} 3!c^{2}(c x+d)^{-3}

Show Answer
Answer: ((a))

(1)33!(bcad)c2(cx+d)4\rm (-1)^{3} 3!(b c-a d) c^{2}(c x+d)^{-4}

Calculation:

Step 1: First Derivative

y  =   ax+bcx+d\rm \frac{a x+b}{c x+d}

let u = ax + b,   v = cx + d

⇒ u′ = a,   v′ = c

⇒ dy/dx = [(cx + d)a − (ax + b)c] / (cx + d)2

⇒ dy/dx = (acx + ad − acx − bc) / (cx + d)2

⇒ dy/dx = (ad − bc)  (cx + d) - 2

Step 2: Second Derivative

dy/dx = (ad − bc)(cx + d)−2

⇒ d2y/dx2 = (ad − bc)(−2)(cx + d)−3 × c

⇒ d2y/dx2 = −2c(ad − bc) (cx + d)−3 

Step 3: Third Derivative

d2y/dx2 = −2c(ad − bc)(cx + d)−3

⇒ d3y/dx3 = −2c(ad − bc)(−3)(cx + d)−4 × c

⇒ d3y/dx3 = 6c2(ad − bc) (cx + d) 

⇒ d3y/dx3 = (-1)3 3! c2(bc-ad) (cx + d) - 4 

∴The Correct Answer is   (1)33!(bcad)c2(cx+d)4\rm (-1)^{3} 3!(b c-a d) c^{2}(c x+d)^{-4}

64

If the equation x2 - y2 + z2 - 4x + 2y + 6z + λ = 0 represents a cone, then the value of λ is:

  1. ((a))

    12

  2. ((b))

    10

  3. ((c))

    8

  4. ((d))

    14

Show Answer
Answer: ((a))

12

Concept  Used:

If S(x, y, z) = 0 is quadratic, then for cone → constant term = 0 after removing linear terms.

Calculation:

given Equation of Cone is x2 − y2 + z2 − 4x + 2y + 6z + λ = 0  ...(1)

⇒ x2 − 4x = (x − 2)2 − 4

⇒ −y2 + 2y = −[(y − 1)2 − 1]

⇒ z2 + 6z = (z + 3)2 − 9

Substitute in (1):

⇒ (x − 2)2 − 4− (y − 1)2 + 1+ (z + 3)2 − 9 + λ = 0

⇒ (x − 2)2− (y − 1)2+ (z + 3)2+ (λ − 12) = 0

For cone, constant term = 0

⇒ λ − 12 = 0

⇒ λ = 12

∴The Correct Answer is  λ = 12.

65

Given below are two statements, one is labelled as Assertion (A) and the other as Reason (R). Select the correct answer using the options given below:

Assertion (A) : The order of a finite group is divisible by the order of its subgroup.

Reason (R) : Every finite group contains an element of every order that divides the order of the group.

  1. ((a))

    (A) is true, but (R) is false.

  2. ((b))

    Both (A) and (R) are true and (R) is the correct explanation of (A).

  3. ((c))

    (A) is false, but (R) is true.

  4. ((d))

    Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Show Answer
Answer: ((a))

(A) is true, but (R) is false.

Calculation:

By Lagrange’s Theorem,

The order of subgroup  divides the order of finite group.

Hence, Assertion (A) is TRUE.

Reason (R) claims existence of element of every divisor of group order.

This is not always true.

For Example: G = A4

⇒ |G| = 12

Divisors: 1, 2, 3, 4, 6, 12

⇒ No element of order 6

Hence, Reason (R) is FALSE

∴ The Correct Answer is  (A) is true, but (R) is false.

66

The order of the element -i of the multiplicative group G ={1,-1, i, -i} is:

  1. ((a))

    2

  2. ((b))

    1

  3. ((c))

    4

  4. ((d))

    3

Show Answer
Answer: ((c))

4

Calculation:

⇒ (−i)1 = −i ≠ 1

⇒ (−i)2 = (−i) × (−i) = −1 ≠ 1

⇒ (−i)3 = (−1) × (−i) = i ≠ 1

⇒ (−i)4 = i × (−i) = 1

The smallest n such that (−i)n = 1 is n = 4

∴ The Correct answer is 4.

67

For which function f(x, y) does lim(x,y)(0,0)\rm \lim _{(x, y) \rightarrow(0,0)} f(x, y) exist?

  1. ((a))

    f(x, y) = xyx2+y2\rm \frac{x y}{x^{2}+y^{2}}

  2. ((b))

    f(x, y) = x2y2x2y2+(xy)4\rm \frac{x^{2} y^{2}}{x^{2} y^{2}+(x-y)^{4}}

  3. ((c))

    f(x, y) = x2y2x2+y2\rm \frac{x^{2}-y^{2}}{x^{2}+y^{2}}

  4. ((d))

    f(x, y) = xyx2+y2\frac{x y}{\sqrt{x^{2}+y^{2}}}

Show Answer
Answer: ((b))

f(x, y) = x2y2x2y2+(xy)4\rm \frac{x^{2} y^{2}}{x^{2} y^{2}+(x-y)^{4}}

Calculation:

Option 1: f(x,y) =xyx2+y2\rm \frac{x y}{x^{2}+y^{2}} 

Along y = x ⇒ f =x22x2\rm \frac{x ^2}{2x^{2}} =12\frac{1}{2}

Along y = 0 ⇒ f = 0

⇒ Different values ⇒ Limit does NOT exist

Option 2: f(x,y) =x2y2x2y2+(xy)4\rm \frac{x^{2} y^{2}}{x^{2} y^{2}+(x-y)^{4}}

⇒ Numerator → 0 as (x,y)→(0,0)

⇒ Denominator → positive value

⇒ Limit = 0

⇒ Limit exists

Option 3: f(x, y) =x2y2x2+y2\rm \frac{x^{2}-y^{2}}{x^{2}+y^{2}}

Along y = 0 ⇒ f = 1

Along x = 0 ⇒ f = −1

⇒ Different values ⇒ Limit does NOT exist

Option 4: f(x, y) =xyx2+y2\frac{x y}{\sqrt{x^{2}+y^{2}}}

Along y = mx ⇒ value depends on m

⇒ Different values ⇒ Limit does NOT exist

∴  Limit exists only for Option 2

Hence, the correct answer is f(x, y) = x2y2x2y2+(xy)4\rm \frac{x^{2} y^{2}}{x^{2} y^{2}+(x-y)^{4}}

68

Which of the following pairs is not correctly matched?

  1. ((a))

    The projection of the vector i^+j^+k^\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}} in the j^\hat{\mathrm{j}} direction : 1

  2. ((b))

    If  ,a+b=ab\rm |\vec{a}+\vec{b}|=|\vec{a}-\vec{b}| ,then angle between a\rm \vec{a} and b\rm \vec{b} is : π4\frac{\pi}{4}

  3. ((c))

    If |a\rm \vec{a}| = 3, |b\rm \vec{b}| = 4 and |a\rm \vec{a} + b\rm \vec{b}| = 1, then |a\rm \vec{a} - b\rm \vec{b}| is : 7

  4. ((d))

    The non-zero vectors a\rm \vec{a}b\rm \vec{b} and c\rm \vec{c} are related bya=8b\rm \vec{a}= 8\rm \vec{b} ,c=7b\rm \vec{c}=-7\rm \vec{b} , then angle between a\rm \vec{a} and c\rm \vec{c} is : π

Show Answer
Answer: ((b))

If  ,a+b=ab\rm |\vec{a}+\vec{b}|=|\vec{a}-\vec{b}| ,then angle between a\rm \vec{a} and b\rm \vec{b} is : π4\frac{\pi}{4}

Calculation:

Statement 1:

Projection of (i + j + k) on j = (i + j + k)·j = 1

The statement is Correct.

Statement 2:

a+b=ab\rm |\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|

 By Squaring

a+b2=ab2\rm |\vec{a}+\vec{b}|^2=|\vec{a}-\vec{b}|^2

⇒ 2a\rm \vec{a}·b\rm \vec{b} = −2a\rm \vec{a}·b\rm \vec{b}

a\rm \vec{a}·b\rm \vec{b} = 0

⇒ θ = π/2

 But Given θ = π/4

∴ This  statement is  Incorrect.

Statement 3:

|a\rm \vec{a}| = 3, |b\rm \vec{b}| = 4, |a\rm \vec{a} + b\rm \vec{b}| = 1

|a\rm \vec{a} + b\rm \vec{b}|= |a\rm \vec{a}| +  |b\rm \vec{b}| + 2a\rm \vec{a}·b\rm \vec{b}

⇒ 12 = 9 + 16 + 2a\rm \vec{a}·b\rm \vec{b}

a\rm \vec{a}·b\rm \vec{b} = −12

|a\rm \vec{a} − b\rm \vec{b}|2 = 9 + 16 − 2(−12) = 49

⇒ |a\rm \vec{a} − b\rm \vec{b}| = 7

The statement is Correct.

Statement 4:

 a=8b\rm \vec{a}= 8\rm \vec{b}  c=7b\rm \vec{c}=-7\rm \vec{b}

a\rm \vec{a} and c\rm \vec{c}are  in opposite direction

⇒ θ = π

The statement is Correct.

∴ Statement 2 is not correctly matched.

Hence, the correct answer is option 2.

69

Which term of the progression 5, √5, 1, ... is 1625\frac{1}{625} ?

  1. ((a))

    10th

  2. ((b))

    12th

  3. ((c))

    11th

  4. ((d))

    9th

Show Answer
Answer: ((c))

11th

Formula Used:

an = a × rn−1

Calculation:

Given Progression: 5, √5, 1, ...

⇒ First term, a = 5

⇒ Common ratio, r = 55\frac{\sqrt5}{5}15\frac{1}{\sqrt5}

⇒ an = 5 × (15)(\frac{1}{\sqrt5})n-1

⇒ 1625\frac{1}{625} = 5 × (5−1/2)n−1

1625\frac{1}{625}  = 5×5(n1)2 5 \times 5^{- \frac{(n−1)}{2}}

1625\frac{1}{625}  = 51(n1)2 5^{1 − \frac{(n−1)}{2}}

⇒ 54=51(n1)25^{-4} = 5^{1 − \frac{(n−1)}{2}}

1(n1)21 − \frac{(n−1)}{2}= −4

(n1)2 \frac{(n−1)}{2} = 5

⇒ n − 1 = 10

⇒ n = 11

∴ The Correct Answer  is 11th term.

70

A problem of mathematics in given to three students A, B and C, whose probability of solving it is 12,34\frac{1}{2}, \frac{3}{4} and 14 \frac{1}{4}respectively. The probability that the problem will be solved is:

  1. ((a))

    332\frac{3}{32}

  2. ((b))

    38\frac{3}{8}

  3. ((c))

    12\frac{1}{2}

  4. ((d))

    2932\frac{29}{32}

Show Answer
Answer: ((d))

2932\frac{29}{32}

Calculation:

Problem given to three students A, B, C

P(A) = 12\frac{1}{2} , P(B) = 34\frac{3}4 , P(C) = 14 \frac{1}{4}

⇒ P(A not solve) = 1 − 12\frac{1}{2}= ​12\frac{1}{2}

⇒ P(B not solve) = 1 − 34\frac{3}4 = 14 \frac{1}{4}

⇒ P(C not solve) = 1 − 14 \frac{1}{4} = 34\frac{3}4

⇒ P(none solves) = (12\frac{1}{2}) × (14 \frac{1}{4}) × (34\frac{3}4) = 332\frac{3}{32}

⇒ P(at least one solves)= 1 − 332\frac{3}{32}2932\frac{29}{32}

**∴ The Correct Answer is **2932\frac{29}{32}

71

Consider the following statements in the context of eigenvalues of a matrix:

I. The eigenvalues of a Hermitian matrix are all real.

II. If λ is an eigenvalue of an orthogonal matrix A, then λ-1is also an eigenvalue of A.

Which of the above statement(s) is/are correct?

 

 

Continue

  1. ((a))

    Only I

  2. ((b))

    Neither I nor II

  3. ((c))

    Only II

  4. ((d))

    Both I and II

Show Answer
Answer: ((d))

Both I and II

Calculation:

Statement I:

A Hermitian matrix satisfies A = A*

Eigenvalues of Hermitian matrices are always real

⇒ Statement I is True

Statement II:

For an orthogonal matrix, ATA = I⇒ A-1 = AT

If Aν = λν, then A-1ν = (1/λ)ν

Therefore, λ-1 is also an eigenvalue

⇒ Statement II is True

∴ Both Statement I and Statement II are correct.

72

If x2 + 3xy + y2 = 1, then dydx\rm \frac{d y}{d x} is equal to :

  1. ((a))

    2x+3y3x+2y\rm -\frac{2 x+3 y}{3 x+2 y}

  2. ((b))

    2x+3y3x+4y\rm \frac{-2 x+3 y}{3 x+4 y}

  3. ((c))

    x+3yy+2x\rm \frac{x+3 y}{y+2 x}

  4. ((d))

    None of the above

Show Answer
Answer: ((a))

2x+3y3x+2y\rm -\frac{2 x+3 y}{3 x+2 y}

Calculation:

 Given x2 + 3xy + y2 = 1

Differentiate both sides w.r.t. x

⇒2x + 3(y + x·dydx\frac{dy}{dx}) + 2y·dydx\frac{dy}{dx} = 0

⇒ 2x + 3y + 3x· dydx\frac{dy}{dx}+ 2y· dydx\frac{dy}{dx}= 0

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

⇒  dydx\frac{dy}{dx}(3x + 2y) = −(2x + 3y) 

⇒  dydx\frac{dy}{dx} =  2x+3y3x+2y\rm -\frac{2 x+3 y}{3 x+2 y} \() 

∴  The Correct Answer is :   2x+3y3x+2y\rm -\frac{2 x+3 y}{3 x+2 y}

73

If limx1f(x)2f(x)+2\rm \lim _{x \rightarrow 1} \frac{f(x)-2}{f(x)+2} = 0, then limx1f(x)\rm \lim _{x \rightarrow 1} f(x) is equal to :

  1. ((a))

    2

  2. ((b))

    1

  3. ((c))

    -2

  4. ((d))

    -1

Show Answer
Answer: ((a))

2

Calculation:

given : limx1f(x)2f(x)+2\rm \lim _{x \rightarrow 1} \frac{f(x)-2}{f(x)+2}  = 0

⇒ f(x) − 2 → 0

⇒ f(x) → 2

Hence, limx→1 f(x) = 2

∴ The Correct Answer is  limx→1 f(x) = 2

74

Given below are two statements, one is labelled as Assertion (A) and the other as Reason (R). Select the correct answer using the options given below:

Assertion (A) : The sum of infinite geometric series n=1(1)n\sum_{n=1}^{\infty}(-1)^{n} cannot be found.

Reason (R): The sum of infinite geometric series n=1rn\sum_{n=1}^{\infty} r^{n}  is 11r \frac{1}{1-r} provided that -1 < r < 1.

  1. ((a))

    (A) is true, but (R) is false.

  2. ((b))

    Both (A) and (R) are true, but (R) is not the correct explanation of (A).

  3. ((c))

    (A) is false, but (R) is true.

  4. ((d))

    Both (A) and (R) are true and (R) is the correct explanation of (A).

Show Answer
Answer: ((b))

Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Calculation:

Given series: (−1) + 1 − 1 + 1 − …

⇒ First term, a = −1

⇒ Common ratio, r = −1

Condition for convergence: −1 < r < 1

Here, r = −1 (condition NOT satisfied)

Hence, the series does not converge

Assertion (A) is TRUE

Formula in Reason (R) is correct for

|r| < 1

⇒ Reason (R) is TRUE

But Reason (R) does not directly explain

why r = −1 fails to converge

Hence, Both (A) and (R) are true, but (R) is not the correct explanation of (A).

∴ The Correct Answer is option 2

75

The value of m so that the function 2x - x2 + my2 is harmonic is:

  1. ((a))

    1

  2. ((b))

    2

  3. ((c))

    0

  4. ((d))

    -1

Show Answer
Answer: ((a))

1

Formula used:

For harmonic function:

2fx2\frac{∂^2f}{∂x^2} +2fy2\frac{∂^2f}{∂y^2}  = 0x=b±b24ac2ax = {-b \pm \sqrt{b^2-4ac} \over 2a}

Calculation:

f = 2x − x2 + m y2

2fx2\frac{∂^2f}{∂x^2} = −2

2fy2\frac{∂^2f}{∂y^2}= 2m

⇒ −2 + 2m = 0

⇒ 2m = 2

⇒ m = 1

∴ The Correct Answer is m = 1.

76

The equation of a line of intersection of two planes \(\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{n}}{1}=\mathrm{q}{1}\) and \(\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{n}}{2}=\mathrm{q}{2}\) is:

  1. ((a))

    \(\rm \vec{r} \times\left(\vec{n}{1} \times \vec{n}{2}\right)=q_{2} \vec{n}{1}-q{1} \vec{n}_{2}\)

  2. ((b))

    \(\rm \vec{r} \times\left(\vec{n}{1} \times \vec{n}{2}\right)=\vec{n}{1}+\vec{n}{2}\)

  3. ((c))

    \(\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{n}}{1}=\mathrm{q}{1} \overrightarrow{\mathrm{n}}_{2}\)

  4. ((d))

    \(\rm \vec{r} \times \vec{n}{2}=q{2} \ \vec{n}_{1}\)

Show Answer
Answer: ((a))

\(\rm \vec{r} \times\left(\vec{n}{1} \times \vec{n}{2}\right)=q_{2} \vec{n}{1}-q{1} \vec{n}_{2}\)

Concept Used:

Direction of line of intersection = n1×n2\vec{ n_1} × \vec{n_2}

Vector identity: r×(n1×n2)=n1(rn2)n2(rn1)\vec{r }× {(\vec{n_1 }×\vec{ n_2})}= \vec{n_1 }(\vec{r }· \vec{n_2}) − \vec{n_2 }(\vec{r} · \vec{n_1})

Calculation:

The Given Conditions are

 \(\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{n}}{1}=\mathrm{q}{1}\)

\(\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{n}}{2}=\mathrm{q}{2}\)

The equation of a line of intersection of two planes is :

 r×(n1×n2)=n1(rn2)n2(rn1)\vec{r }× {(\vec{n_1 }×\vec{ n_2})}= \vec{n_1 }(\vec{r }· \vec{n_2}) − \vec{n_2 }(\vec{r} · \vec{n_1})

= \(q_{2} \vec{n}{1}-q{1} \vec{n}_{2}\) 

∴ The Correct answer is \(\rm \vec{r} \times\left(\vec{n}{1} \times \vec{n}{2}\right)=q_{2} \vec{n}{1}-q{1} \vec{n}_{2}\)

77

Given below are two statements, one is labelled as Assertion (A) and the other as Reason (R). Select the correct answer using the options given below:

Assertion (A) : The sum of all coefficients in the binomial expansion of (x2 + x - 3)19  is - 1.

Reason (R): The sum of coefficients in the expansion of (x + y)n  is 2n.

  1. ((a))

    Both (A) and (R) are true and (R) is the correct explanation of (A).

  2. ((b))

    (A) is true, but (R) is false.

  3. ((c))

    (A) is false, but (R) is true.

  4. ((d))

    Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Show Answer
Answer: ((d))

Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Calculation:

Assertion (A) : The sum of all coefficients in the binomial expansion of (x2 + x - 3)19  is - 1.

f(x) = (x2 + x − 3)19

⇒ f(1) = (1 + 1 − 3)19  = (−1)19 = −1

⇒ (A) is true

Reason (R): The sum of coefficients in the expansion of (x + y)n  is 2n.

(x + y)n

Put x = 1, y = 1

⇒ (1 + 1)n = 2n

⇒ (R) is true

Explanation check

⇒ (A) uses f(1) method but Not directly using 2n

⇒ (R) does not explain (A)

Hence**,** Both (A) and (R) are true but (R) is not correct explanation

∴ The Correct option is Both (A) and (R) are true, but (R) is not correct explanation.

78

Choose the correct answer from the given code :

The rank of the Matrix A, where

A = [12223242 22324252 32425262 42526272]\left[\begin{array}{cccc} 1^{2} & 2^{2} & 3^{2} & 4^{2} \ 2^{2} & 3^{2} & 4^{2} & 5^{2} \ 3^{2} & 4^{2} & 5^{2} & 6^{2} \ 4^{2} & 5^{2} & 6^{2} & 7^{2} \end{array}\right] is :

  1. ((a))

    2

  2. ((b))

    3

  3. ((c))

    1

  4. ((d))

    4

Show Answer
Answer: ((b))

3

Concept Used:

If rows are linearly dependent then rank < order

Calculation:

The given Matrix A = [12223242 22324252 32425262 42526272]\left[\begin{array}{cccc} 1^{2} & 2^{2} & 3^{2} & 4^{2} \ 2^{2} & 3^{2} & 4^{2} & 5^{2} \ 3^{2} & 4^{2} & 5^{2} & 6^{2} \ 4^{2} & 5^{2} & 6^{2} & 7^{2} \end{array}\right]

⇒ aij = (i+j−1)2 = i2 + 2i(j−1) + (j−1)2

⇒ Each entry is form: f(i,j) = A(i) + B(i)(j) Depends on i, j up to power 2

⇒ Columns are combinations of 1, j, j2

So maximum independent columns = 3

Therefore rank ≤ 3

Check 3×3 minor ≠ 0

⇒ Rank ≥ 3

Hence,Rank (A) = 3

∴ The Correct Answer is 3

79

If u = exyz, then 3uxyz\rm \frac{\partial^{3} u}{\partial x \partial y \partial z} is equal to :

  1. ((a))

    u[1 + 2xyz + x2y2z2]

  2. ((b))

    u[1 + 3xyz + 2x2y2z2]

  3. ((c))

    u[1 + 3xyz + x2y2z2]

  4. ((d))

    None of the above

Show Answer
Answer: ((c))

u[1 + 3xyz + x2y2z2]

Calculation:

given u = exyz

Differentiate w.r.t x

ux\rm \frac{\partial^{} u}{\partial x} =  exyz × yz =  yz exyz ...(1)

Differentiate (1) w.r.t y

2uyx\rm \frac{\partial^{2} u}{\partial y \partial x}  =  z exyz + yz( xz exyz )

= z exyz + xyz2 exyz

= exyz( z + xyz2 ) ...(2)

Differentiate (2) w.r.t z

3uzyx\rm \frac{\partial^{3} u}{\partial z\partial y \partial x} =  exyz(1 + 2xyz) + xyz exyz( z + xyz2 )

⇒ exyz[1 + 2xyz + x2y2z2]

Hence,3uxyz\rm \frac{\partial^{3} u}{\partial x \partial y \partial z}= exyz[1 + 2xyz + x2y2z2]

∴ The Correct Answer is : u[1 + 2xyz + x2y2z2]

80

Which of the following test refers that ∑ un is convergent if:

\(\rm \lim {n \rightarrow \infty}\left[n\left(\frac{u{n}}{u_{n+1}}-1\right)\right]>1:\)

  1. ((a))

    4th Root test

  2. ((b))

    Ratio test

  3. ((c))

    De-Morgan's test

  4. ((d))

    Raabe's test

Show Answer
Answer: ((d))

Raabe's test

Concept Used:

Raabe's Test (Higher Ratio Test)

Let L = limn→∞ n[(un/un+1) − 1]

If L > 1 ⇒ Series convergent

If L < 1 ⇒ Series divergent

If L = 1 ⇒ Test fails

Calculation:

Given condition:

limn→∞ n[(un/un+1) − 1] > 1

This matches Raabe's criterion

Therefore series is convergent by Raabe's Test

∴The  Correct Answer is Raabe's test

81

If f(x, y) = \(\rm \left{\begin{array}{ll} \rm \frac{x}{|x|} \sqrt{x^{2}+y^{2}} & ; \rm x \neq 0 \ \rm 0 & ; \rm x=0 \rm \end{array}\right.\) then fx(0, 0) + fy(0, 0) is equal to

  1. ((a))

    0

  2. ((b))

    -1

  3. ((c))

    1

  4. ((d))

    2

Show Answer
Answer: ((c))

1

Concept Used:

Partial derivative by definition

fx(0,0) = limh0[f(h,0)f(0,0)]hlimh→0 \frac{[f(h,0) − f(0,0)]}{h}l

fy(0,0) = limk0[f(0,k)f(0,0)]klimk→0\frac{[f(0,k) − f(0,0)]}{k}

Calculation:

Given f(x, y) = \(\rm \left{\begin{array}{ll} \rm \frac{x}{|x|} \sqrt{x^{2}+y^{2}} & ; \rm x \neq 0 \ \rm 0 & ; \rm x=0 \rm \end{array}\right.\)

⇒ f(0,0) = 0

For h ≠ 0, f(h,0) =hh×(h2)\frac{ h}{|h|} × √(h^2)

⇒ f(h,0) = h

⇒ fx(0,0) =limh0(h0)h limh→0 \frac{(h − 0)}{h}

= limh→0 1 = 1

To Find fy(0,0)

For x = 0, f(0,k) = 0

⇒ fy(0,0) = limk0(00)klimk→0 \frac{(0 − 0)}{k}

⇒ fy(0,0) = 0

⇒ fx(0,0) + fy(0,0) = 1 + 0 = 1 

∴The Correct Answer is  1

82

For this question, two statements are given one labelled as Assertion (A) and the other as Reason (R). Choose the correct answer from the options given below.

Assertion (A) : The series of positive terms ∑ Un; where Un = 5n4n+5n\rm \frac{5^{n}}{4^{n}+5^{n}} is divergent.

Reason (R) : \(\rm \lim {n \rightarrow \infty} U{n} \neq 0\).

  1. ((a))

    (A) is true, but (R) is false.

  2. ((b))

    Both (A) and (R) are true, but (R) is not the correct explanation of (A).

  3. ((c))

    (A) is false, but (R) is true.

  4. ((d))

    Both (A) and (R) are true and (R) is the correct explanation of (A).

Show Answer
Answer: ((d))

Both (A) and (R) are true and (R) is the correct explanation of (A).

Concept Used:

If lim Un ≠ 0 ⇒ series divergent(Test for divergence)

Calculation:

Assertion (A) : The series of positive terms ∑ Un; where Un = 5n4n+5n\rm \frac{5^{n}}{4^{n}+5^{n}} is divergent.

⇒ Un5n4n+5n\rm \frac{5^{n}}{4^{n}+5^{n}}

Divide Numerator and Denominator by 5n

⇒ Un = 1[(45)n+1]\frac{1 }{[(\frac{4}{5})^n + 1]}

Since (4/5)n → 0

⇒ lim Un = 1(0+1)\frac{1}{(0+1)}

⇒ lim Un = 1 ≠ 0

⇒ Necessary condition fails

⇒∑Un is divergent. This is also True and 

Reason (R) :\(\rm \lim {n \rightarrow \infty} U{n} \neq 0\) 

This is also True and give Explanation of  Assertion.

 ∴ The Correct Answer is Both (A) and (R) are true and (R) is correct explanation.

83

Which of the following statements is / are correct ? 

I. Every square matrix can be uniquely expressed as P + i Q where P and Q are Hermitian.

II. If A and B are square matrices of same order, then adj (AB) = adj (B). adj (A).

Select the correct answer using the code given below :

  1. ((a))

    Only II

  2. ((b))

    Both I and II

  3. ((c))

    Only I

  4. ((d))

    Neither I nor II 

Show Answer
Answer: ((b))

Both I and II

Given:

I. A = P + iQ, P,Q Hermitian

II. adj(AB) = adj(B) × adj(A)

Formula Used:

P = (A + A*)/2

Q = (A − A*)/(2i)

adj(AB)=adj(B)×adj(A)

Calculation:

Statement I: A = P + iQ, P,Q Hermitian

Let A be any square matrix

⇒P = (A + A*)/2 ......(1)

and Q = (A − A*)/(2i) ...(2)

⇒ P* = P  and Q* = Q

⇒ P,Q are Hermitian

⇒ A = P + iQ

⇒ Representation is unique

Hence,Statement I is true

Statement II: adj(AB) = adj(B). adj(A)

According to the  Property of adjoint: adj(AB)=adj(B).adj(A)

This is True for all square matrices

Hence,Statement II is true

∴ The Correct answer is  Both I and II

84

Which of the following pairs is not correctly matched?

  1. ((a))

    div r^\rm \hat{{r}} =  2r

  2. ((b))

    If r\rm \overrightarrow{{r}} = xi^+yj^+zk^\rm x \hat{i}+y \hat{j}+z \hat{k} तथा r = r\rm \overrightarrow{{r}}, then curl r\rm \overrightarrow{{r}} is  0\rm \overrightarrow{{0}}

  3. ((c))

    div r\rm \overrightarrow{{r}} =  3

  4. ((d))

    div (curl r\rm \overrightarrow{{r}}) = 0

Show Answer
Answer: ((a))

div r^\rm \hat{{r}} =  2r

Formula Used:

  1. div V = Vxx+Vyy+Vzz\frac{\partial V_x}{\partial x} + \frac{\partial V_y}{\partial y} + \frac{\partial V_z}{\partial z}
  2. curl V =i^j^k^ xyz VxVyVz\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ V_x & V_y & V_z \end{vmatrix} 
  3. div(curl V) = 0

Calculation:

We analyze each option step-by-step:

Step 1: Verify div r (Option 3)

⇒ div r\rm \overrightarrow{{r}} = ∂(x)/∂x + ∂(y)/∂y + ∂(z)/∂z

⇒ div r\rm \overrightarrow{{r}} = 1 + 1 + 1

⇒ div r\rm \overrightarrow{{r}} = 3

Hence, the pair div r\rm \overrightarrow{{r}}=  3 is correctly matched.

Step 2: Verify curl r (Option 2)

⇒ curl r =i^j^k^ xyz VxVyVz\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ V_x & V_y & V_z \end{vmatrix} 

= î(0−0) − ĵ(0−0) + k̂(0−0)

= 0

Hence, the pair curl r : 0 is correctly matched.

Step 3: Verify div(curl r) (Option 4)

Since curl r = 0

⇒ div(0) = 0

Alternatively, ∇·(∇×V) = 0 is a standard identity.

Hence, the pair div(curl r) : 0 is correctly matched.

Step 4: Verify div r : 2r (Option 1)

From Step 1, div r = 3.

The option states div r = 2r.

Since 3 ≠ 2r, this is false.

Hence, this pair is not correctly matched.

∴ The incorrectly matched pair is first option: div r = 2r.

85

The value of 1e1x2e(1+lnx)\rm \int_{1}^{e} \frac{1}{x^{2}} e^{(1+\ln x)} dx is :

  1. ((a))

    e

  2. ((b))

    1

  3. ((c))

    0

  4. ((d))

    1e\rm \frac{1}{e}

Show Answer
Answer: ((a))

e

Calculation:

given integration  I = 1e1x2e(1+lnx)\rm \int_{1}^{e} \frac{1}{x^{2}} e^{(1+\ln x)}dx

As e1+ln x = e × eln x 

⇒ e1+ln x = e × x

⇒ I = 1e1x2ex\rm \int_{1}^{e} \frac{1}{x^{2}} e x dx

⇒ I = e1e1xe\rm \int_{1}^{e} \frac{1}{x^{}}  dx

⇒ I = e [ln x]1e

⇒ I = e (ln e − ln 1)

⇒ I = e (1 − 0)

⇒ I = e

∴ I = e

∴ The Correct Answer is e.

86

Equation of the sphere passing through the four points (4, -1, 2), (0, -2, 3), (1, -5, -1) and (2, 0, 1) is:

  1. ((a))

    x2 + 2y2 + z2 - 4x + 6y - 2z + 5 = 0

  2. ((b))

    x2 + y2 + z2 - 4x + 6y - 2z + 5 = 0

  3. ((c))

    x2 + y2 + 2z2 - 4x + 6y - 2z + 5 = 0

  4. ((d))

    2x2 + y2 + z2 - 4x + 6y - 2z + 5 = 0

Show Answer
Answer: ((b))

x2 + y2 + z2 - 4x + 6y - 2z + 5 = 0

Formula Used:

Equation of the sphere=  x2+y2+z2 +2ux+2vy+2wz+d=0

Calculation:

Put (4,−1,2)

⇒ 21+8u−2v+4w+d=0 ...(1)

Put (0,−2,3)

⇒ 13−4v+6w+d=0 ...(2)

Put (1,−5,−1)

⇒ 27+2u−10v−2w+d=0 ...(3)

Put (2,0,1)

⇒ 5+4u+2w+d=0 ...(4)

By Solving these equations ,we get 

u=−2, v=3 , w=−1,d=5

∴ The Correct Answer  is x2+y2+z2 −4x+6y−2z+5=0

87

The value of 23x21\rm \int_{-2}^{3}\left|x^{2}-1\right| dx is:

  1. ((a))

    283\frac{28}{3}

  2. ((b))

    183\frac{18}{3}

  3. ((c))

    293\frac{29}{3}

  4. ((d))

    143\frac{14}{3}

Show Answer
Answer: ((a))

283\frac{28}{3}

Concept Used:

|f(x)| = f(x), f(x) ≥ 0

|f(x)| = −f(x), f(x) < 0

Calculation:

given I = 23x21dx\rm \int_{-2}^{3}\left|x^{2}-1\right| dx

⇒ x2 − 1 = 0

⇒ x = ±1

∴ Split interval at −1,1

I=21(x21),dx+11(1x2),dx+13(x21),dxI = \int_{-2}^{-1} (x^2 - 1),dx + \int_{-1}^{1} (1 - x^2),dx + \int_{1}^{3} (x^2 - 1),dx

=  \(\left[\frac{x^3}{3} - x\right]{-2}^{-1} + \left[x - \frac{x^3}{3}\right]{-1}^{1} + \left[\frac{x^3}{3} - x\right]_{1}^{3}\)  

=(13+1)(83+2)+(113)(1+13)+(93)(131) =23(23)+23(23)+6(23) =43+43+203 =283=(-\tfrac{1}{3} + 1) - (-\tfrac{8}{3} + 2) + (1 - \tfrac{1}{3}) - (-1 + \tfrac{1}{3}) + (9 - 3) - (\tfrac{1}{3} - 1)\ = \tfrac{2}{3} - (-\tfrac{2}{3}) + \tfrac{2}{3} - (-\tfrac{2}{3}) + 6 - (-\tfrac{2}{3})\ = \tfrac{4}{3} + \tfrac{4}{3} + \tfrac{20}{3}\ = \frac{28}{3}   

∴ The Correct answer is 283\frac{28}{3}

88

Match List I with List II and choose the correct answer using the codes given below the lists:

List IList II
A.3x2 + 4y2 - 18x - 24y + 47 = 0i.Ellipse
B.(4x + 3y)2 - 256x - 142y + 849 = 0ii.Circle
C.9x2 - 16y2 - 72x + 96y - 144 = 0iii.Hyperbola
D.2x2 + 2y2 + 6x + 4y - 5 = 0iv.Parabola

 

<br>
  1. ((a))

    A-i, B-ii, C-iv, D-iii

  2. ((b))

    A-i, B-iv, C-iii, D-ii

  3. ((c))

    A-i, B-iv, C-ii, D-iii 

  4. ((d))

    None of the above

Show Answer
Answer: ((b))

A-i, B-iv, C-iii, D-ii

Formula Used:

General equation: ax2 + 2hxy + by2 + 2gx + 2fy + c = 0

  1. Circle: a = b and h = 0
  2. Parabola: h2 - ab = 0
  3. Ellipse: h2 - ab < 0 and a ≠ b
  4. Hyperbola: h2 - ab > 0

Calculation:

For A: 3x2 + 4y2 - 18x - 24y + 47 = 0

⇒ a = 3, b = 4, h = 0

⇒ h2 - ab = 0 - (3 × 4) = -12 < 0

⇒ Since h2 - ab < 0, it is an Ellipse (i).

For B: (4x + 3y)2 - 256x - 142y + 849 = 0

⇒ 16x2 + 24xy + 9y2 - 256x - 142y + 849 = 0

⇒ a = 16, b = 9, 2h = 24 ⇒ h = 12

⇒ h2 - ab = 122 - (16 × 9) = 144 - 144 = 0

⇒ Since h2 - ab = 0, it is a Parabola (iv).

For C: 9x2 - 16y2 - 72x + 96y - 144 = 0

⇒ a = 9, b = -16, h = 0

⇒ h2 - ab = 0 - (9 × -16) = 144 > 0

⇒ Since h2 - ab > 0, it is a Hyperbola (iii).

For D: 2x2 + 2y2 + 6x + 4y - 5 = 0

⇒ a = 2, b = 2, h = 0

⇒ Since a = b and h = 0, it is a Circle (ii).

Matching the results: A-i, B-iv, C-iii, D-ii

∴ The correct option is A-i, B-iv, C-iii, D-ii.

89

A line x+24=y+93=z85\rm \frac{x+2}{4}=\frac{y+9}{3}=\frac{z-8}{-5} cuts a sphere x2 + y2 + z2 = 49 at P and Q. Then the distance PQ is:

  1. ((a))

    5

  2. ((b))

    5√5

  3. ((c))

    5√3

  4. ((d))

    5√2

Show Answer
Answer: ((d))

5√2

Calculation:

Given Equation of line is x+24=y+93=z85\rm \frac{x+2}{4}=\frac{y+9}{3}=\frac{z-8}{-5}

Parametric coordinates of any point on the line are 

x = 4r - 2, y = 3r - 9, z = -5r + 8

Given Equation of sphere is 

x2 + y2 + z2 = 49

Substitute x, y, z in sphere equation:

⇒ (4r - 2)2 + (3r - 9)2 + (8 - 5r)2 = 49

⇒ (16r2 - 16r + 4) + (9r2 - 54r + 81) + (25r2 - 80r + 64) = 49

⇒ 50r2 - 150r + 149 = 49

⇒ 50r2 - 150r + 100 = 0

⇒ r2 - 3r + 2 = 0

⇒ (r - 1)(r - 2) = 0 ⇒ r = 1, 2

Finding coordinates of P and Q:

⇒ At r = 1: P = (2, -6, 3)

⇒ At r = 2: Q = (6, -3, -2)

Calculating distance PQ:

⇒ PQ = √[(6 - 2)2 + (-3 + 6)2 + (-2 - 3)2]

⇒ PQ = √[42 + 32 + (-5)2]

⇒ PQ = √[16 + 9 + 25]

⇒ PQ = √50

⇒ PQ = 5√2

∴ The Correct Answer is 5√2.

90

Match List I and List II and choose the correct answer using the codes given below the lists: 

List IList II
A.The mean of cubes of first 10 natural numbers isi.10-67
B.The mean of first 100 natural numbers isii.18
C.The median of 25, 36, 18, 17, 17, 29, 16 isiii.50-5
D.If the mode of distribution is 20, mean is 6, then median isiv.302-5
  1. ((a))

    A-iv, B-iii, C-i, D-ii

  2. ((b))

    A-iii, B-ii, C-i, D-iv

  3. ((c))

    A-iii, B-i, C-ii, D-iv

  4. ((d))

    A-iv, B-iii, C-ii, D-i

Show Answer
Answer: ((d))

A-iv, B-iii, C-ii, D-i

Formula Used:

  1. Sum of cubes of n natural numbers: S = [n(n+1)2]2[\frac{n(n + 1) }{2}]^2
  2. Mean of n natural numbers:(n+1)2 \frac{(n + 1) }{2}
  3. Empirical relation:Mode = 3 × Median - 2 × Mean

Calculation:

For A: Mean of cubes of first 10 numbers

⇒ n = 10

⇒ Sum =[10×112]2 [10 × \frac{11 }{2}]^2 = (55)2 = 3025

⇒ Mean = 302510\frac{3025 }{ 10} = 302.5

⇒ A matches with (iv)

For B: Mean of first 100 natural numbers

⇒ n = 100

⇒ Mean = (100+1)2\frac{(100 + 1) }{2}

⇒ Mean = 50.5

⇒ B matches with (iii)

For C: Median of 25, 36, 18, 17, 17, 29, 16

Ascending order is  16, 17, 17, 18, 25, 29, 36

⇒ n = 7 (odd)

⇒ Median = (n+1)2 \frac{(n + 1) }{2} th term

⇒ Median = 4th term = 18

⇒ C matches with (ii)

For D: Mode = 20, Mean = 6

⇒ 20 = 3 × Median - 2 × 6

⇒ 20 = 3 × Median - 12

⇒ 3 × Median = 32

⇒ Median = 10.67

⇒ D matches with (i)

∴ The CorrectAnswer is  A-iv, B-iii, C-ii, D-i.

91

If a set A has 3 elements, then the total number of reflexive relations is:

  1. ((a))

    4

  2. ((b))

    128

  3. ((c))

    16

  4. ((d))

    64

Show Answer
Answer: ((d))

64

Formula Used:

Total relations = 2n2

Reflexive relations = 2n2−n

Calculation:

Set A has 3 elements :n(A) = 3

⇒ n2 = 9

⇒ n2 − n = 9 − 3  = 6

⇒ Reflexive relations = 26 = 64

∴ The Correct Answer is 64

92

The points of discontinuity of the greatest integer function defined by f(x) = [x], where [x] denotes the greatest integer less than or equal to x are:

  1. ((a))

    Only positive integral points

  2. ((b))

    All real numbers

  3. ((c))

    Only negative integral points

  4. ((d))

    Every integral number

Show Answer
Answer: ((d))

Every integral number

Calculation:

Given function f(x) = [x], where [x] denotes the greatest integer less than or equal to x 

Let's test the continuity at an integer point 'k'.

Calculate RHL at x = k:

⇒ limh→0 [k + h] = k (since k+h > k)

Calculate LHL at x = k:

⇒ limh→0 [k - h] = k - 1 (since k-h < k)

Here, LHL ≠ RHL (k - 1 ≠ k).

⇒ Limit does not exist at integer points.

⇒ So, f(x) is discontinuous at all integers.

∴ The Correct Answer is every integral number.

93

Which of the following series is convergent?

  1. ((a))

    n=1(121n)n\rm \sum_{n=1}^{\infty}\left(\frac{1}{2}-\frac{1}{n}\right)^{n}

  2. ((b))

    n=1(321n)n\rm \sum_{n=1}^{\infty}\left(\frac{3}{2}-\frac{1}{n}\right)^{n}

  3. ((c))

    n=1(521n)n\rm \sum_{n=1}^{\infty}\left(\frac{5}{2}-\frac{1}{n}\right)^{n}

  4. ((d))

    None of the above

Show Answer
Answer: ((a))

n=1(121n)n\rm \sum_{n=1}^{\infty}\left(\frac{1}{2}-\frac{1}{n}\right)^{n}

Given:

Infinite series options:

  1. n=1(1/2 - 1/n)n
  2. n=1(3/2 - 1/n)n
  3. n=1(5/2 - 1/n)n

Formula Used:

Cauchy's Root Test: Let ∑ an be a series with non-negative terms and let L = limn→∞ (an)1/n. If L < 1, the series converges.

Calculation:

Checking convergence condition using Root Test:

For (option 1): n=1(121n)n\rm \sum_{n=1}^{\infty}\left(\frac{1}{2}-\frac{1}{n}\right)^{n}

⇒ an = (121n)n\left(\frac{1}{2}-\frac{1}{n}\right)^{n}

⇒ limn→∞ (an)1/n = limn→∞ (121n)\left(\frac{1}{2}-\frac{1}{n}\right) = 12\frac{1}{2} - 0 = 12\frac{1}{2}

⇒ Since 12\frac{1}{2} < 1, Series (1) converges.

For (option 2): n=1(321n)n\rm \sum_{n=1}^{\infty}\left(\frac{3}{2}-\frac{1}{n}\right)^{n}

⇒ limn→∞ (an)1/n = 32\frac{3}{2} > 1. Series (2) diverges.

For (option 3): n=1(521n)n\rm \sum_{n=1}^{\infty}\left(\frac{5}{2}-\frac{1}{n}\right)^{n}

⇒ limn→∞ (an)1/n = 52\frac{5}{2} > 1. Series (3) diverges.

The Correct Answer is Option 1.

94

Equation of Cylinder with axis as z-axis and passing through curve, x2 + y2 = 2z, x + y + z = 1 is:

  1. ((a))

    x2 + y2 - 2x - 2y = 2

  2. ((b))

    x2 - y2 - 2x + 2y = 2

  3. ((c))

    x2 + y2 + 2x + 2y = 2

  4. ((d))

    None of the above

Show Answer
Answer: ((c))

x2 + y2 + 2x + 2y = 2

Concept Used:

To find the equation of a cylinder with axis parallel to the z-axis, eliminate variable z from the guiding curve equations.

Calculation:

Given equation of plane: x + y + z = 1

⇒ z = 1 - x - y ....(1)

Equation of surface: x2 + y2 = 2z

Substitute (1) in the surface equation:

⇒ x2 + y2 = 2(1 - x - y)

⇒ x2 + y2 = 2 - 2x - 2y

⇒ x2 + y2 + 2x + 2y = 2

Hence, the equation of the cylinder is:x2 + y2 + 2x + 2y = 2

∴ The correct Answer is x2 + y2 + 2x + 2y = 2

95

Consider the following two statements for the cone x2 + y2 - z2 = 0

<br>

I. The cone has three mutually perpendicular tangent planes.

II. The cone as self-reciprocal.

Which of the above statements is/are correct?

  1. ((a))

    Only I

  2. ((b))

    Both I and II

  3. ((c))

    Neither I nor II

  4. ((d))

    Only II

Show Answer
Answer: ((d))

Only II

Given:

Equation of cone: x2 + y2 - z2 = 0

Statement I: Presence of 3 perpendicular tangent planes.

Statement II: The cone is self-reciprocal.

Concept Used:

  1. Condition for 3 perpendicular tangent planes:

For a cone Ax2 + By2 + Cz2 = 0, the sum of

cofactors (A+B+C) of the reciprocal cone must be 0.

  1. Reciprocal Cone: The cone formed by lines

perpendicular to the tangent planes of the original.

Calculation:

For Statement I:

General form of Cone : ax2 + by2 + cz2 = 0

Gven equation of cone: x2 + y2 - z2 = 0

⇒ Here a = 1, b = 1, c = -1

Condition for ⊥ tangent planes: 1a+1b+1c=0\frac{1}{a}+ \frac{1}{b} + \frac{1}{c}= 0

⇒ 11+11+1(1)\frac{1}{1}+ \frac{1}{1} + \frac{1}{(-1)} = 1 + 1 - 1 = 1

Since 1 ≠ 0, Statement I is Incorrect.

For Statement II

Reciprocal cone of ax2+by2+cz2=0 is

x2a+y2b+z2c=0 \frac{x^2}{a }+ \frac{y^2}{b} + \frac{z^2}{c} = 0\

x21+y21+z2(1)=0 \frac{x^2}{1 }+ \frac{y^2}{1} + \frac{z^2}{(-1)}=0 

⇒ x2 + y2 - z2 = 0

This is identical to the original cone.

Thus, the cone is self-reciprocal.

Hence,Statement II is Correct.

∴ Only Statement II is correct.

96

The shortest distance between the lines r=a+t b\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\mathrm{t} \overrightarrow{\mathrm{~b}}  and r=c+Δd\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{c}}+\Delta \overrightarrow{\mathrm{d}} is:

  1. ((a))

    [c, b, d][\overrightarrow{\mathrm{c}}, \overrightarrow{\mathrm{~b}}, \overrightarrow{\mathrm{~d}}]

  2. ((b))

    a,b,d[b×d]\rm \frac{\vec{a}, \vec{b}, \vec{d}}{[\vec{b} \times \vec{d}]}

  3. ((c))

    [ca, b, d[ b×d]]\left[\frac{\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{~b}}, \overrightarrow{\mathrm{~d}}}{[\overrightarrow{\mathrm{~b}} \times \overrightarrow{\mathrm{d}}]}\right]

  4. ((d))

    [ca, b, d][\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{~b}}, \overrightarrow{\mathrm{~d}}]

Show Answer
Answer: ((c))

[ca, b, d[ b×d]]\left[\frac{\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{~b}}, \overrightarrow{\mathrm{~d}}}{[\overrightarrow{\mathrm{~b}} \times \overrightarrow{\mathrm{d}}]}\right]

Given:

Line 1 : r=a+tb\vec r = \vec a + t\vec b

Line 2 : r=c+λd\vec r = \vec c + \lambda \vec d

Direction vectors = b,d\vec b , \vec d

Formula used:

Shortest Distance between skew lines

S.D=(ca)(b×d)b×d\text{S.D}=\dfrac{|(\vec c-\vec a)\cdot(\vec b \times \vec d)|}{|\vec b \times \vec d|}

Scalar Triple Product:

(ca)(b×d)=[ca,b,d](\vec c-\vec a)\cdot(\vec b \times \vec d)= [\vec c-\vec a,\vec b,\vec d]

Calculation:

Vector joining points = (ca)(\vec c-\vec a)

Common perpendicular direction = b×d\vec b \times \vec d

⇒ S.D = Projection of (ca)(\vec c-\vec a) on (b×d)(\vec b \times \vec d)

S.D=(ca)(b×d)b×d\text{S.D}=\dfrac{|(\vec c-\vec a)\cdot(\vec b \times \vec d)|}{|\vec b \times \vec d|}

S.D=[ca,b,d]b×d\text{S.D}=\dfrac{|[\vec c-\vec a,\vec b,\vec d]|}{|\vec b \times \vec d|}

∴ The correct answer is option (3).

97

Given below are two statements, one is labelled as Assertion (A) and the other as Reason (R). Select the correct answer using the options given below:

Assertion (A) : The equation x39+5x+6\rm \frac{x^{3}}{9}+\frac{5}{x+6} = x can be solved by using solution of quadratic equations.

Reason (R) : The equation can be reduced to quadratic equation by putting x2 + 3x = y. Options:

  1. ((a))

    Both (A) and (R) are true and (R) is the correct explanation of (A).

  2. ((b))

    (A) is false, but (R) is true.

  3. ((c))

    Both (A) and (R) are true, but (R) is not the correct explanation of (A).

  4. ((d))

    (A) is true, but (R) is false

Show Answer
Answer: ((a))

Both (A) and (R) are true and (R) is the correct explanation of (A).

Calculation:

The given equation:x39+5x+6\rm \frac{x^{3}}{9}+\frac{5}{x+6} =x

Multiply both sides by 9(x + 6):

⇒ x3(x + 6) + 45 = 9x(x + 6)

⇒ x4 + 6x3 + 45 = 9x2 + 54x

Rearrange terms to form a polynomial:

⇒ x4 + 6x3 - 9x2 - 54x + 45 = 0

To use the substitution from Reason (R), check :

(x2 + 3x)2 = x4 + 6x3 + 9x2

Rewrite the main equation to fit this form:

⇒ (x4 + 6x3 + 9x2) - 18x2 - 54x + 45 = 0

Factor out -18 from the middle terms:

⇒ (x2 + 3x)2 - 18(x2 + 3x) + 45 = 0

Apply the substitution x2 + 3x = y:

⇒ y2 - 18y + 45 = 0

This is a quadratic equation in terms of y.

Thus, the original equation can be solved using quadratic methods.

Hence, Assertion (A) is true because the equation is solvable via quadratic methods.

Reason (R) is true and correctly provides the substitution to reduce it.

∴The Correct Answer is  Both (A) and (R) are true and (R) is the correct explanation of (A).

98

There are 6 red, 4 white and 5 black balls in a bag 3 balls are drawn successively without replacement. The probability of balls being red, white and black in that order is:

  1. ((a))

    491\frac{4}{91}

  2. ((b))

    591\frac{5}{91}

  3. ((c))

    588\frac{5}{88}

  4. ((d))

    None of the above

Show Answer
Answer: ((a))

491\frac{4}{91}

Given:

Red balls = 6

White balls = 4

Black balls = 5

Total balls = 6 + 4 + 5 = 15

3 balls drawn without replacement

Order: Red, White, Black

Calculation:

Given Red balls = 6

White balls = 4

Black balls = 5

Total balls = 6 + 4 + 5 = 15

3 balls are drawn without replacement in Order: Red, White, Black

⇒P(R1) = 615\frac{6}{15}

After 1 red, remaining = 14

⇒P(W2) = 414\frac{4}{14}

After 1 white, remaining = 13

⇒P(B3) = 513\frac{5}{13}

⇒ Required P = (615\frac{6}{15}) × (414\frac{4}{14}) × (513\frac{5}{13})

= 1202750\frac{120}{2750}

491\frac{4}{91}

∴ The Correct Answer is 491\frac{4}{91}.

99

The series n=1(1)n11np\rm \sum_{n=1}^{\infty}(-1)^{n-1} \frac{1}{n p} is convergent only if :

  1. ((a))

    p = 0

  2. ((b))

    p > 1

  3. ((c))

    p > 0

  4. ((d))

    p < 0

Show Answer
Answer: ((c))

p > 0

⇒ Series: ∑ (-1)n-1 / np

⇒ General term an = (-1)n-1 / np

Concept Used: 

Leibnitz (Alternating Series) Test :An alternating series ∑ (-1)n bn converges if:

  1. bn ≥ 0

  2. bn is decreasing

  3. lim n→∞ bn = 0

Calculation:

Given Series is n=1(1)n11np\rm \sum_{n=1}^{\infty}(-1)^{n-1} \frac{1}{n p}

⇒ bn = 1 / np

Case 1: p ≤ 0

⇒ np ≤ 1

⇒ 1 / np does not → 0 ❌

⇒ Divergent

Case 2: p > 0

⇒ np → ∞

⇒ 1 / np → 0 ✅

⇒ Also decreasing for p > 0

⇒ All Leibnitz conditions satisfied.

⇒ The series converges only if p > 0

100

The sum of the infinite series 21!+43!+65!+87!+\frac{2}{1!}+\frac{4}{3!}+\frac{6}{5!}+\frac{8}{7!}+\ldots is:

  1. ((a))

    e + 1

  2. ((b))

    e

  3. ((c))

    1e\frac{1}{\mathrm{e}}

  4. ((d))

    e - 1

Show Answer
Answer: ((b))

e

Formula Used:

General term: Tn = 2n / (2n − 1)!

e = 1 + 1/1! + 1/2! + 1/3! + ...

Calculation:

 Given Series = 21!+43!+65!+87!+\frac{2}{1!}+\frac{4}{3!}+\frac{6}{5!}+\frac{8}{7!}+\ldots

General term: Tn = 2n / (2n − 1)!

Let S = Σ 2n / (2n − 1)! , n = 1 → ∞

⇒ S = Σ [(2n − 1) + 1] / (2n − 1)!

⇒ S = Σ (2n − 1)/(2n − 1)! + Σ 1/(2n − 1)!

⇒ (2n − 1)/(2n − 1)! = 1/(2n − 2)!

⇒ S = Σ 1/(2n − 2)! + Σ 1/(2n − 1)!

Put n = 1,2,3,...

⇒S = 1/0! + 1/2! + 1/4! + ...      + 1/1! + 1/3! + 1/5! + ..

⇒ S = 1/0! + 1/1! + 1/2! + 1/3! + ...

⇒ S = e

∴ Sum of the series = e

∴ The Correct Answer is e.

101

The differential coefficient of sin2 x with respect ecos x is:

  1. ((a))

    2cosxecosx-\rm \frac{2 \cos x}{e^{\cos x}}

  2. ((b))

    2cosxsinxecosx\rm \frac{2 \cos x \sin x}{e^{\cos x}}

  3. ((c))

    cosxecosx\rm \frac{\cos x}{e^{\cos x}}

  4. ((d))

    2sinxecosx\rm \frac{2 \sin x}{e^{\cos x}}

Show Answer
Answer: ((a))

2cosxecosx-\rm \frac{2 \cos x}{e^{\cos x}}

Concept Used:

Derivative of one function w.r.t. another ,If y = f(x), z = g(x)

Then dydz=(dydx)(dzdx)\frac{dy}{dz }= (\frac{dy}{dx})(\frac{dz}{dx})

Formula Used:

(1) ddx(sin2x)\frac{d}{dx }(sin^2x) = 2sinx cosx

(2) ddx(eu)\frac{d}{dx }(e^u) = eu · dudx\frac{du}{dx }

Calculation:

Let y = sin2x

⇒ dydx\frac{dy}{dx } = 2sinx cosx

Let z = ecos x 

⇒  dzdx\frac{dz}{dx }= ecos x × ddx(cosx)\frac{d}{dx }(cos x)

dzdx\frac{dz}{dx } = ecos x × (−sinx)

dzdx\frac{dz}{dx } = −ecos x sinx

Now,

⇒ dydz=(dydx)(dzdx)\frac{dy}{dz }= (\frac{dy}{dx})(\frac{dz}{dx})

⇒ = (2sinx)(cosx)ecosx(sinx)\frac{(2sinx)( cosx)}{−e^{cos x}(sinx)}

⇒ = 2cosxecosx\frac{-2cosx}{e^{cos x}}

∴The Correct Answer is :  2cosxecosx\frac{-2cosx}{e^{cos x}}

102

Match List I and List II and choose the correct answer using the codes given below the lists:

List IList II
A.Polar form of -1 + i√3 isi.2(cos2π3+isin2π3)\rm 2\left(\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right)
B.The values of (1)1/4 areii.2πi
C.The period of ez isiii.i sin θ
D.sinh (i θ) is equal toiv.+1, ±i
  1. ((a))

    A-i, B-iii, C-iv, D-ii

  2. ((b))

    A-i, B-iv, C-ii, D-iii

  3. ((c))

    A-i, B-iv, C-iii, D-ii

  4. ((d))

    A-iv, B-iii, C-ii, D-i

Show Answer
Answer: ((b))

A-i, B-iv, C-ii, D-iii

Formula Used:

1) z = r(cosθ + i sinθ)

2) xn = 1 roots

3) ez period is 2πi

4). sinh(iθ) = i sinθ

Calculation:

For A: Polar form of -1 + i√3 is

⇒ r = √[(-1)2 + (√3)2] = 2

⇒ θ = π - tan-1(√3) = 2π3\frac{2π}{3}

⇒ The Polar Form is  2(cos2π3+isin2π3)\rm 2\left(\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right) → (i)

For B: The values of (1)1/4 are

⇒ Roots: ±1, ±i → (iv)

For C: Period of ez

⇒ ez + 2πi = ez → (ii)

For D: sinh(iθ) is equal to 

⇒sinh (i θ) = (eiθeiθ)2\frac{(e^{iθ }- e^{-iθ})}{2} = i sinθ → (iii)

Matching: A-i, B-iv, C-ii, D-iii

∴ The correct option is A-i, B-iv, C-ii, D-iii.

103

Consider the following equations:

I. sinh-1x = log (x + 1+x2\rm \sqrt{1+x^{2}})

II. tanh-1 x = 12log1x1+x\rm \frac{1}{2} \log \frac{1-x}{1+x}

Which of the above equations is/are true?

  1. ((a))

    Only I

  2. ((b))

    Both I and II

  3. ((c))

    Neither I nor II

  4. ((d))

    Only II

Show Answer
Answer: ((a))

Only I

Formula Used:

(1) sinh-1x = log(x + √(1 + x2))

(2) tanh-1x = (1/2) log((1 + x)/(1 − x))

Calculation:

Statement I:

Given: sinh-1x  = log(x + √(1 + x2))

⇒ Standard formula matches exactly

Hence, Statement I is true

Statement II:

Given: tanh-1 x = 12log1x1+x\rm \frac{1}{2} \log \frac{1-x}{1+x}

⇒ Standard: 12log1+x1x\rm \frac{1}{2} \log \frac{1+x}{1-x}

⇒ Given expression is reciprocal

As log(a/b) = −log(b/a)

So given 12log1x1+x\rm \frac{1}{2} \log \frac{1-x}{1+x}= −tanh-1x

⇒ Hence Statement II is false

∴ The Correct Answer is Only I

104

For this question, two statements are given, one labelled as Assertion (A) and the other as Reason (R). Choose the correct answer from the options given below:

Assertion (A): The straight lines 2x - 3y + 1 = 0; x + y - 2 = 0 and x + y - 3 = 0 are concurrent.

Reason (R): If Δ = a1b1c1 a2b2c2 a3b3c3\rm \left|\begin{array}{lll} a_{1} & b_{1} & c_{1} \ a_{2} & b_{2} & c_{2} \ a_{3} & b_{3} & c_{3} \end{array}\right| = 0, then the straight lines a1x + b1y + c1 = 0, a2x + b2y + c2 = 0 and a3x + b3y + c3 = 0 must be concurrent.

  1. ((a))

    (A) is true, but (R) is false.

  2. ((b))

    Both (A) and (R) are true and (R) is the correct explanation of (A).

  3. ((c))

    (A) is false, but (R) is true.

  4. ((d))

    Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Show Answer
Answer: ((c))

(A) is false, but (R) is true.

Calculation:

Given Lines: L1: 2x − 3y + 1 = 0

L2: x + y − 2 = 0

L3: x + y − 3 = 0

Δ = \(\rm \left|\begin{array}{lll} 2 & -3 & 1\ 1&1 & -2 \1 & 1 &-3 \end{array}\right|\) = 0

⇒ Δ = 2[(1×−3) − (1×−2)] − (−3)[(1×−3) − (1×−2)] + 1[(1×1) − (1×1)]

= 2[−3 + 2]  − (−3)[−3 + 2] + 1[0]

= 2(−1) − (−3)(−1)

 = −2 − 3  = −5 ≠ 0

⇒ Lines are not concurrent

⇒ Assertion (A) is false

Reason (R) states correct condition

Hence (R) is true

∴ The Correct Answer is (A) is false, but (R) is true.

105

The equation of a sphere for which the circle x2 + y2 + z2 + 7y - 2z + 2 = 0, 2x + 3y + 4z - 8 is a great circle, will be:

  1. ((a))

    x2 + y2 + z + 4x - 2y - 6z - 10 = 0

  2. ((b))

    x2 + y2 + z2 - 2x + 4y - 6z + 10 = 0

  3. ((c))

    x2 + y2 + z2 + 2x - 4y + 6z + 8 = 0

  4. ((d))

    x2 + y2 + z2 + 2x + 4y + 6z -10 = 0

Show Answer
Answer: ((b))

x2 + y2 + z2 - 2x + 4y - 6z + 10 = 0

Concept Used:

Great circle plane passes through centre

Sphere: x2+y2+z2+2ux+2vy+2wz+d=0

Centre = (−u, −v, −w)

Calculation:

Let S1: x2 + y2 + z2 + 7y − 2z + 2 = 0

Compare S1 with standard form

⇒ Centre C = (0, −7/2, 1)

 

Now ,Check plane passes through C

⇒ 2(0)+3(−7/2)+4(1)−8

= 0 −21/2 +4 −8

= −21/2 −4

= −29/2 ≠ 0

⇒ So plane not passes through centre

⇒ Hence given sphere not required

⇒ Required sphere = S1 + λ(plane)=0

⇒ x2+y2+z2+7y −2z+2 +λ(2x+3y+4z−8)=0

Compare with standard form

⇒ Centre = (−λ, −(7+3λ)/2, −(−2+4λ)/2)

 

Since great circle, plane passes centre

Substitute in plane:

⇒ 2(−λ)+3[−(7+3λ)/2] +4[(2−4λ)/2]−8=0

⇒ −2λ −3(7+3λ)/2 +2(2−4λ)−8=0

By Solving  λ = −1

Substitute λ = −1

⇒ Required sphere: x2+y2+z2+7y−2z+2 −(2x+3y+4z−8)=0

⇒ x2+y2+z2−2x+4y−6z+10=0

∴ The Correct Answer is x2+y2+z2−2x+4y−6z+10=0

106

The whole area surrounded by the curve with the equations x = a cos3 t, y = b sin3 t is:

  1. ((a))

    34\frac{3}{4}π ab

  2. ((b))

    38\frac{3}{8}​​π ab

  3. ((c))

    38\frac{3}{8} π a2b

  4. ((d))

    38\frac{3}{8} ab

Show Answer
Answer: ((b))

38\frac{3}{8}​​π ab

Concept Used:

Area = 4 × area in 1st quadrant

Area = ∫ y (dx/dt) dt

Formula Used:

(1) A = ∫ y (dx/dt) dt

(2) ∫0π/2 sinmx cosnx dx = ( (m−1)!! (n−1)!! ) / ( (m+n)!! ) × π/2

Calculation:

Given x = a cos3t

⇒ dx/dt = a × 3cos2t(−sin t)

⇒ dx/dt = −3a cos2t sin t

Given y = b sin3t

Area (A)  = 4 ∫0π/2 y(−dx/dt) dt

⇒ A = 4 ∫0π/2 b sin3t × 3a cos2t sin t dt

⇒ A = 12ab ∫0π/2 sin4t cos2t dt

⇒ Let I = ∫0π/2 sin4t cos2t dt

⇒ Using formula:

0π/2 sinmx cosnx dx

= ( (m−1)!! (n−1)!! ) / ( (m+n)!! ) × π/2

⇒ Here m=4, n=2

⇒ I = (3×1 × 1) / (6×4×2) × π/2

⇒ I = 3 / 48 × π/2

⇒ I = 3π / 96

⇒ I = π / 32

⇒ A = 12ab × π/32

⇒ A = 3πab / 8

∴ The Correct Answer is 38\frac{3}{8}πab

107

For the statistical data, Mode is related to Median and Mean by the relation

  1. ((a))

    3 Median - 2 Mean = Mode

  2. ((b))

    2 Median - 3 Mean = Mode

  3. ((c))

    3 Median + 2 Mean = Mode

  4. ((d))

    None of the above

Show Answer
Answer: ((a))

3 Median - 2 Mean = Mode

Concept Used:

The relation between Mean, Median and Mode is based on the concept of Skewness of a distribution.

In statistics:

Mean is influenced by all observations.

Median divides the data into two equal parts.

Mode is the value with highest frequency.

In a perfectly symmetrical distribution:

Mean = Median = Mode

But in a skewed distribution, they are not equal.

Explanation of Skewness:

1. Symmetrical Distribution:

Mean = Median = Mode

No skewness.

2. Positively Skewed Distribution:

Mean > Median > Mode

Right tail is longer.

Mean shifts toward larger values.

3. Negatively Skewed Distribution:

Mode > Median > Mean

Left tail is longer.

Mean shifts toward smaller values.

Empirical Relation (Karl Pearson):

For a moderately skewed distribution:

Mean − Mode = 3(Mean − Median)

Rearranging the formula:

Mode = 3Median − 2Mean

Conclusion:

The relation between Mean, Median and Mode is based on the concept of skewness in a moderately skewed distribution.

∴  The Correct Answer is Mode = 3Median − 2Mean

108

The intercept form of plane is:

  1. ((a))

    xaybzc=1\rm \frac{x}{a}-\frac{y}{b}-\frac{z}{c}=1

  2. ((b))

    ax + by + cz = 1

  3. ((c))

    xa+yb+zc=1\rm \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1

  4. ((d))

    None of the above

Show Answer
Answer: ((c))

xa+yb+zc=1\rm \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1

Concept Used:

Equation of plane cutting axes at a, b, c on x, y, z axes.

If plane cuts intercepts:(a,0,0), (0,b,0), (0,0,c)

Calculation:

Plane cuts x-axis at a ,So x-intercept = a

Plane cuts y-axis at b, So y-intercept = b

Plane cuts z-axis at c, So z-intercept = c

As Standard intercept form:xa+yb+zc=1\rm \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1

⇒The intercept form of plane  is: xa+yb+zc=1\rm \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1

∴ The Correct Answer is xa+yb+zc=1\rm \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1

109

Consider the following statements :

I. A series un=xn(n+1)n\rm \sum u_{n}=\sum \frac{x^{n}}{(n+1)^{n}} is convergent for all values of x.

II. The series 2+32+43+54+2+\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\ldots is convergent.

Which of the above statements is/are correct?

  1. ((a))

    Both I and II

  2. ((b))

    Only II

  3. ((c))

    Neither I nor II

  4. ((d))

    Only I

Show Answer
Answer: ((d))

Only I

Given:

I. un=xn(n+1)n\sum u_n = \sum \dfrac{x^n}{(n+1)^n}

Check convergence for all x.

II. 2+32+43+54+2 + \dfrac{3}{2} + \dfrac{4}{3} + \dfrac{5}{4} + \dots

Check convergence.

Concept Used:

  1. Ratio Test for infinite series.

  2. nth term test for divergence.

  3. If limun0\lim u_n \ne 0, series diverges.

Calculation:

Statement I: un=xn(n+1)n\sum u_n = \sum \dfrac{x^n}{(n+1)^n}

un=xn(n+1)nu_n = \dfrac{x^n}{(n+1)^n}

⇒ Apply Ratio Test:

un+1un=xn+1(n+2)n+1×(n+1)nxn\left|\dfrac{u_{n+1}}{u_n}\right| = \left|\dfrac{x^{n+1}}{(n+2)^{n+1}} \times \dfrac{(n+1)^n}{x^n}\right|

=x×(n+1)n(n+2)n+1= |x| \times \dfrac{(n+1)^n}{(n+2)^{n+1}}

=x×(n+1n+2)n×1n+2= |x| \times \left(\dfrac{n+1}{n+2}\right)^n \times \dfrac{1}{n+2}

Take limit as nn \to \infty

(n+1n+2)ne1\left(\dfrac{n+1}{n+2}\right)^n \to e^{-1}

1n+20\dfrac{1}{n+2} \to 0

⇒ Limit = x×e1×0=0|x| \times e^{-1} \times 0 = 0

0<10 < 1 for all x

⇒ Series converges x\forall x

Statement I is correct.

Statement II:

⇒ General term: an=n+1na_n = \dfrac{n+1}{n}

an=1+1na_n = 1 + \dfrac{1}{n}

limnan=1\lim_{n \to \infty} a_n = 1

⇒ Since limit 0\ne 0

⇒ Series diverges

Statement II is false.

∴ The Correct Answer is Only I is correct.

110

The locus of a point z satisfying Re (1z)\rm \left(\frac{1}{z}\right) = λ (λ is non-zero real number) is :

  1. ((a))

    Hyperbola

  2. ((b))

    Ellipse

  3. ((c))

    Straight line

  4. ((d))

    Circle

Show Answer
Answer: ((d))

Circle

Concept Used:

Let z = x + iy

1/z = conjugate(z) / |z|2

Formula Used:

(1) z = x + iy

(2) |z|2 = x2 + y2

(3) 1/z = (x − iy)/(x2 + y2)

Calculation:

Let z = x + iy

⇒ 1/z = (x − iy)/(x2 + y2)

⇒ Re(1/z) = x/(x2 + y2)

Given Re(1/z) = λ

⇒ x/(x2 + y2) = λ

⇒ x = λ(x2 + y2)

⇒ x = λx2 + λy2

⇒ λx2 + λy2 − x = 0

⇒ x2 + y2 − x/λ = 0

⇒ x2 − x/λ + y2 = 0

Complete square in x

⇒ x2 − x/λ = (x − 1/2λ)2 − 1/4λ2

⇒ (x − 1/2λ)2 − 1/4λ2 + y2 = 0

⇒ (x − 1/2λ)2 + y2 = 1/4λ2

Equation represents a circle with Centre = (1/2λ , 0)

and Radius = 1/2|λ|

∴ The Correct Answer is  circle.

111

A sample space has two events A and B such that P(A∩B) = 12\frac{1}{2}, P(A̅) = 13\frac{1}{3}, P(B̅) = 13\frac{1}{3}, Then P(A∪B):

  1. ((a))

    812\frac{8}{12}

  2. ((b))

    1112\frac{11}{12}

  3. ((c))

    1012\frac{10}{12}

  4. ((d))

    912\frac{9}{12}

Show Answer
Answer: ((c))

1012\frac{10}{12}

Given: P(A ∩ B) = 12 ,P(A) = 13 ,P(B) = 13

Formula Used:

P(A) = 1 − P(A) ,P(B) = 1 − P(B)

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

Calculation:

P(A) = 1 − 13 = 23

P(B) = 1 − 13 = 23

 P(A ∪ B) = 23 + 2312

 = 4312 =  56

∴ P(A ∪ B) = 56 = 1012 

The correct option is 10⁄12

112

An example of a function of a complex variable which is not continuous everywhere is given by:

  1. ((a))

    exp z

  2. ((b))

    log z

  3. ((c))

    sin z

  4. ((d))

    None of the above

Show Answer
Answer: ((b))

log z

Concept Used:

A complex function is continuous at a point if it is differentiable in a neighborhood of that point. Entire functions (analytic everywhere in the complex plane) are continuous everywhere.

Explanation:

The function exp z is entire (analytic for all complex z).

Therefore, exp z is continuous everywhere in ℂ.

The function sin z is also entire.

Therefore, sin z is continuous everywhere in ℂ.

The function log z is not defined at z = 0 and is multi-valued in the complex plane.

It has a branch cut (commonly taken along the negative real axis).

Therefore, log z is not continuous everywhere.

Hence,The function log z is not continuous everywhere in the complex plane.

∴ The Correct Answer is  log z.

113

Match List I with List II. Condition is that the line y = mx + c will be tangent to the curve:

 

List I (Curve)List II (Condition)
A.x2 + y2 = a2i.c2 = a2m2 + b2
B.y2 = 4axii.c2 = a2m2 + a2
C.x2a2+y2b2=1\rm \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 iii.c=am\rm c=\frac{a}{m}
D.x2 - y2 = a2iv.c2 = a2m2 - a2
  1. ((a))

    A-i, B-iii, C-ii, D-iv

  2. ((b))

    A-iv, B-iii, C-ii, D-i

  3. ((c))

    A-iv, B-iii, C-i, D-ii

  4. ((d))

    A-ii, B-iii, C-i, D-iv

Show Answer
Answer: ((d))

A-ii, B-iii, C-i, D-iv

Concept Used:

  1. Line is tangent if Discriminant = 0

Calculation:

A. x2 + y2 = a2

Since y = mx + c

⇒ x2 + (mx + c)2 = a2

⇒ (1 + m2)x2 + 2mcx + (c2 - a2) = 0

Tangent ⇒ D = 0

⇒ (2mc)2- 4(1 + m2)(c2 - a2) = 0

⇒ c2 = a2(1 + m2)

⇒ c2 = a2m2 + a2

∴A ⇒ ii

B. y2 = 4ax

For parabola, tangent form:c=am\rm c=\frac{a}{m}

∴B ⇒ iii

C. x2a2+y2b2=1\rm \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1

Substitute y = mx + c

x2a2+(mx+c)2b2=1\rm \frac{x^{2}}{a^{2}}+\frac{(mx+ c)^2}{b^{2}}=1

Tangent ⇒ D = 0

⇒ c2 = a2m2 + b2

∴C ⇒ i

D. x2 - y2 = a2

Substitute y = mx + c

⇒ x2- (mx + c)2 = a2

Tangent ⇒ D = 0

⇒ c2 = a2m2 - a2

∴D ⇒ iv

Hence, A-ii, B-iii, C-i, D-iv

∴ The Correct Answer  is A-ii, B-iii, C-i, D-iv.

114

The function defined by \(\rm \oint(x)=\left{\begin{array}{l} \rm 1+x \text { for } 0<x \leq 2 \ \rm 5-x \text { for } 3>x>2 \rm \end{array}\right.\) is:

  1. ((a))

    Differentiable at every point in (0, 3)

  2. ((b))

    Not continuous at x = 2

  3. ((c))

    Not differentiable at x = 2

  4. ((d))

    None of the above

Show Answer
Answer: ((c))

Not differentiable at x = 2

Given:

f(x) = 1 + x,  0 < x ≤ 2

f(x) = 5 - x,  2 < x < 3

Concept Used:

Continuity at x = a ⇒  LHL = RHL = f(a)

Differentiable at x = a ⇒ Left derivative = Right derivative

Calculation:

1. Continuity at x = 2

⇒ f(2) = 1 + 2 = 3

LHL: lim x→2- f(x) = 1 + 2 = 3

RHL: lim x→2+ f(x) = 5 - 2 = 3

⇒ LHL = RHL = f(2)

⇒ Continuous at x = 2

2. Differentiability at x = 2

For 0 < x < 2:

⇒ f'(x) = 1

For 2 < x < 3:

⇒ f'(x) = -1

Left derivative at 2 is 1

Right derivative at 2 is -1

⇒ 1 ≠ -1

⇒ Not differentiable at x = 2

Function is continuous at x = 2

Not differentiable at x = 2

∴ The Correct Answer is Not differentiable at x = 2

115

The differential equation yx' = y - 1, y(0) = 1  has:

  1. ((a))

    Infinitely many solutions

  2. ((b))

    Finitely many solutions

  3. ((c))

    No solution

  4. ((d))

    Unique solution

Show Answer
Answer: ((a))

Infinitely many solutions

Given: yx' = y - 1,y(0) = 1

Concept Used:

Existence & Uniqueness theorem : y' = f(x,y) ,If f, ∂f/∂y continuous then solution is unique.

Calculation:

yx' = y - 1

⇒ x dy/dx = y - 1

⇒ dy/dx = (y - 1)/x  (1)

Here,f(x,y) = (y - 1)/x

At (0,1):  f(0,1) = (1 - 1)/0

⇒ 0/0  undefined

Thus f is not continuous at x = 0

Rewrite (1):  x dy/dx = y - 1

If y = 1:

⇒ LHS = x × 0 = 0

⇒ RHS = 1 - 1 = 0

⇒ y = 1 satisfies equation

Now solve generally:

⇒ dy/(y - 1) = dx/x

⇒ ln|y - 1| = ln|x| + C

⇒ y - 1 = Cx

⇒ y = 1 + Cx

Apply y(0) = 1:

⇒ 1 = 1 + C(0)

⇒ 1 = 1

⇒ True for all C

∴ Infinite values of C

Hence,The Correct Answer is infinitely many solutions.

116

If the line y - 1 = m(x - 1) cuts the circle x2 + y2 = 4 at two real points, then the number of possible values of m will be:

  1. ((a))

    Infinite

  2. ((b))

    2

  3. ((c))

    1

  4. ((d))

    None of the above

Show Answer
Answer: ((a))

Infinite

Given: Line: y - 1 = m(x - 1)

Circle: x2 + y2 = 4

Concept Used:

Line cuts circle at two points ⇒ Distance from centre < radius

Calculation:

From the equation of line y = mx - m + 1

Standard form: mx - y - m + 1 = 0

Centre = (0,0)

Radius = 2

Distance from centre:

m+1(m2+1)\frac{ | -m + 1 |}{√(m^2 + 1)}

For two real points:

m+1(m2+1)\frac{ | -m + 1 |}{√(m^2 + 1)}< 2

Square both sides:

⇒ (1 - m)2< 4(m2 + 1)

⇒ 1 - 2m + m2 < 4m2 + 4

⇒ 0 < 3m2+ 2m + 3

Discriminant: (2)2 - 4×3×3

⇒ 4 - 36 = -32 <0  and the leading coefficient (3) is positive, the expression 3m2+ 2m + 3 is always positive  for all real values of m .

∴The Correct Answer  is  Infinite.

117

Which of the following pairs is nor correctly matched?

  1. ((a))

    limxπ2(sinx)tanx\rm \lim _{x \rightarrow \frac{\pi}{2}}(\sin x)^{\tan x} : 0

  2. ((b))

    limx0axbxx\rm \lim _{x \rightarrow 0} \frac{a^{x}-b^{x}}{x} : logab\rm \log \frac{a}{b}

  3. ((c))

    limx0xlogx\rm \lim _{x \rightarrow 0} x \log x : 0

  4. ((d))

    limxπ2(secxtanx)\rm \lim _{x \rightarrow \frac{\pi}{2}}(\sec x-\tan x) : 0

Show Answer
Answer: ((a))

limxπ2(sinx)tanx\rm \lim _{x \rightarrow \frac{\pi}{2}}(\sin x)^{\tan x} : 0

Calculation:

(Option 1)limxπ2(sinx)tanx\rm \lim _{x \rightarrow \frac{\pi}{2}}(\sin x)^{\tan x}

Let x = π/2 - h :h → 0

⇒ sin x = cos h ≈ 1 - h2/2

⇒ tan x = cot h ≈ 1/h

⇒ (sin x)tan x

= (1 - h2/2)1/h

⇒ ln L = (1/h)

ln(1 - h2/2) ≈ (1/h)(-h2/2)

⇒ -h/2 → 0

⇒ L = e0 = 1

Given 0 ⇒ Wrong

(Option 2)limx0axbxx\rm \lim _{x \rightarrow 0} \frac{a^{x}-b^{x}}{x}

Using expansion:

ax = 1 + x ln a

bx = 1 + x ln b

⇒ ax - bx = x(ln a - ln b)

Divide by x:

⇒ ln a - ln b = ln(a/b) Correct

(Option 3)limx0xlogx\rm \lim _{x \rightarrow 0} x \log x

Put x = 1/t

⇒ (1/t) log(1/t)

= -(log t)/t ⇒ 0 Correct

(Option 4)limxπ2(secxtanx)\rm \lim _{x \rightarrow \frac{\pi}{2}}(\sec x-\tan x)

Let x = π/2 - h

⇒ sec x = cosec h ≈ 1/h

⇒ tan x = cot h ≈ 1/h

⇒ sec x - tan x≈ 0 Correct

∴ The Correct Answer is limxπ2(sinx)tanx\rm \lim _{x \rightarrow \frac{\pi}{2}}(\sin x)^{\tan x}

118

If X. [1 5 -3] = [3159 2106 153]\left[\begin{array}{rrr} 3 & 15 & -9 \ 2 & 10 & -6 \ -1 & -5 & 3 \end{array}\right], then X is:

  1. ((a))

    [3 2 1]\left[\begin{array}{r} 3 \ 2 \ -1 \end{array}\right]

  2. ((b))

    [0 2 1]\left[\begin{array}{r} 0 \ 2 \ -1 \end{array}\right]

  3. ((c))

    [2 3 1]\left[\begin{array}{r} 2 \ 3 \ -1 \end{array}\right]

  4. ((d))

    [3 2 2]\left[\begin{array}{r} 3 \ 2 \ -2 \end{array}\right]

Show Answer
Answer: ((a))

[3 2 1]\left[\begin{array}{r} 3 \ 2 \ -1 \end{array}\right]

Given:

X [ 1  5  -3 ] = [3159 2106 153]\left[\begin{array}{rrr} 3 & 15 & -9 \ 2 & 10 & -6 \ -1 & -5 & 3 \end{array}\right]

Calculation:

X=[x1 x2 x3]andR=[153]  X = \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix} \quad \text{and} \quad R = \begin{bmatrix} 1 & 5 & -3 \end{bmatrix}\

 ⇒ x1 × R = [3 15 -9]

⇒ x1 × 1 = 3

⇒ x1 = 3

Check: 3 × 5 = 15 

3 × (-3) = -9 

Similarly 

 x2 × R= [2 10 -6]

⇒ x2 × 1 = 2

⇒ x2 = 2

Check: 2 × 5 = 10 

2 × (-3) = -6 

x3 × R = [-1 -5 3]

⇒ x3 × 1 = -1

⇒ x3 = -1

Check:  -1 × 5 = -5 

-1 × (-3) = 3

X=[x1 x2 x3] X = \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix}  =[3 2 1]\left[\begin{array}{r} 3 \ 2 \ -1 \end{array}\right]

∴ The Correct Answer is[3 2 1]\left[\begin{array}{r} 3 \ 2 \ -1 \end{array}\right]

119

The series (n+1)nxnnn+1\rm \sum \frac{(n+1)^{n} x^{n}}{n^{n+1}} at x = 1 is :

  1. ((a))

    Divergent

  2. ((b))

    Oscillatory

  3. ((c))

    Convergent

  4. ((d))

    None of the above

Show Answer
Answer: ((a))

Divergent

Given:(n+1)nxnnn+1\rm \sum \frac{(n+1)^{n} x^{n}}{n^{n+1}}

Calculation:(n+1)nxnnn+1\rm \sum \frac{(n+1)^{n} x^{n}}{n^{n+1}}

At x = 1⇒ (n+1)nnn+1\rm \sum \frac{(n+1)^{n} }{n^{n+1}}

an =(n+1)nnn+1 \frac{(n+1)^{n} }{n^{n+1}}

⇒ an = 1/n× (1 + 1/n)n

Sinc (1 + 1/n)n → e

⇒ an ≈ e/n

Compare with Σ 1/n

Σ 1/n diverges

∴ Given series diverges.

Hence, The Correct Answer is Divergent.

120

Consider the following two statements for two subgroups H1 and H2 of a group G:

I. H1 ∩ H2 is not necessarily a subgroup of G.

II. H1 ∪ H2 is not necessarily a subgroup of G.

Select the correct statement using the code given below:

  1. ((a))

    Both Statements I and II are incorrect.

  2. ((b))

    Both Statements I and II are correct.

  3. ((c))

    Statement II is correct, but Statement I is incorrect.

  4. ((d))

    Statement I is correct, but Statement II is incorrect.

Show Answer
Answer: ((c))

Statement II is correct, but Statement I is incorrect.

Given:

H1, H2 are subgroups of G

Statement I:H1 ∩ H2 not necessarily subgroup

Statement II:H1 ∪ H2 not necessarily subgroup

Concept Used:

Subgroup test: Non-empty + closed + inverse

Intersection property: Intersection of subgroups is always a subgroup

Union property: Union of two subgroups is a subgroup if and only if one is contained in the other. Thus, it is not necessarily a subgroup.

Calculation:

Since H1, H2 subgroups

⇒ e ∈ H1, e ∈ H2

⇒ e ∈ H1 ∩ H2

Non-empty

Let a,b ∈ H1 ∩ H2

⇒ a,b ∈ H ⇒ ab-1 ∈ H1

⇒ a,b ∈ H2 ⇒ ab-1 ∈ H2

⇒ ab-1 ∈ H1 ∩ H2

Closed under operation

⇒ H1 ∩ H2 is subgroup

Hence Statement I is false, Statement II is true

∴ The Correct Answer is Statement II is correct, but Statement I is incorrect.

121

For what value of 'a', the angle between the normals to the planes ax - y + z = 5 and x + y + 2z = 7 is π3\frac{\pi}{3}?

  1. ((a))

    1

  2. ((b))

    2

  3. ((c))

    0

  4. ((d))

    -1

Show Answer
Answer: ((b))

2

Concept Used:

Angle between two normals equals angle between planes

cosθ = (n₁·n₂) ÷ (|n₁||n₂|)

Calculation:

Normal to Plane 1  (n₁)= (a, −1, 1)

Normal to Plane 2 (n₂)= (1, 1, 2)

⇒ n₁·n₂ = a − 1 + 2 = a + 1

⇒ |n₁| = √(a2 + (−1)2 + 12)

⇒ |n₁| = √(a2 + 2)

⇒ |n₂| = √(12 + 12 + 22)

⇒ |n₂| = √6

⇒ cos(π/3) = (a + 1) ÷ (√(a2 + 2) × √6)

⇒ 1/2 = (a + 1) ÷ √(6(a2 + 2))

⇒ 2(a + 1) = √(6(a2 + 2))

Square both sides:

⇒ 4(a + 1)2 = 6(a2 + 2)

⇒ 4(a2 + 2a + 1) = 6a2 + 12

⇒ 4a2 + 8a + 4 = 6a2 + 12

⇒ 0 = 2a2 − 8a + 8

⇒ a2 − 4a + 4 = 0

⇒ (a − 2)2 = 0

⇒ a = 2

∴The Correct Answer is  a = 2.

122

The infinite series 2x+3x28+4x327++(n+1)n3xn+\rm 2 x+\frac{3 x^{2}}{8}+\frac{4 x^{3}}{27}+\ldots+\frac{(n+1)}{n^{3}} x^{n}+\ldots is convergent if:

  1. ((a))

    x > 1

  2. ((b))

    x ≤ 1

  3. ((c))

    x < 1

  4. ((d))

    None of the above

Show Answer
Answer: ((b))

x ≤ 1

Calculation:

The given series is 2x + (3x2)/8 + (4x3)/27 + …

⇒ Tn =  (n+1)xnn3\frac{(n + 1)x^n} {n^3} 

⇒ Tn+1 = (n+2)xn+1(n+1)3\frac{(n + 2)x^{n+1} }{ (n + 1)^3}

Tn+1Tn\frac{Tn+1 }{ Tn} = (n+2)x(n+1)×n3(n+1)3\frac{(n + 2)x }{ (n + 1)} × \frac{ n^3 }{ (n + 1)^3}

⇒ limn→∞ Tn+1Tn\frac{Tn+1 }{ Tn} = |x|   (1)

⇒ For convergence: |x| < 1

Endpoint Check:

 At x = 1, series = Σ (n+1)n3\frac{(n + 1)} {n^3} → convergent

At x = −1, series is alternating → convergent

Hence, convergence condition  is |x| ≤ 1

∴ The correct Answer is  x ≤ 1.

123

The value of 0πxtanxsecx+tanx\rm \int_{0}^{\pi} \frac{x \tan x}{\sec x+\tan x} dx is:

  1. ((a))

    π2\frac{\pi}{2}(π-2)

  2. ((b))

    π(π-2)

  3. ((c))

    π(π-1)

  4. ((d))

    π2\frac{π}{2}(π-1)

Show Answer
Answer: ((a))

π2\frac{\pi}{2}(π-2)

Formula Used:

sec²x − tan²x = 1

∫ u dv = uv − ∫ v du

Calculation:

I = 0πxtanxsecx+tanx\rm \int_{0}^{\pi} \frac{x \tan x}{\sec x+\tan x} dx

Multiply numerator & denominator by (sec x − tan x)

0π(xtanx)(secxtanx)(secx+tanx)(secxtanx)\rm \int_{0}^{\pi} \frac{(x \tan x)(\sec x-\tan x)}{(\sec x+\tan x)(\sec x-\tan x)}dx

(secx+tanx)(secx−tanx)=1

⇒ I = 0π\int^π_0 x(secx tanx − tan²x) dx

⇒ I = ∫ x secx tanx dx − ∫ x tan²x dx

Let u = x,  du =dx

v=secx   ,dv = secx tanx dx

⇒ ∫ x secx tanx dx = x secx − ∫ secx dx

⇒ = x secx − ln|secx + tanx|

As  tan²x = sec²x − 1

⇒ ∫ x tan²x dx = ∫ x(sec²x − 1) dx

⇒ x tan x + ln|cos x| − x22\frac{x^2}{2}

⇒ I = x sec x − x tan x − ln|sec x + tan x| − ln|cos x| + x22\frac{x^2}{2} 

Apply Limits:

At x = π → sec π = −1, tan π = 0

⇒ Value = −π + π22\frac{π²}{2}

At x = 0 → sec 0 = 1, tan 0 = 0

⇒ Value = 0

⇒ I = (−π + π22\frac{π²}{2} − 0

⇒I = π2\frac{\pi}{2}(π-2) 

The Correct answer is π2\frac{\pi}{2}(π-2)

124

If f(x, y, z) = x2y + y2x + z2, then the value of ∇f at the point (1, 1, 1) is:

  1. ((a))

    3i^+3j^+2k^\rm 3 \hat{i}+3 \hat{j}+2 \hat{k}

  2. ((b))

    0

  3. ((c))

    22\sqrt{22}

  4. ((d))

    3i^+3j^+3k^3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}

Show Answer
Answer: ((a))

3i^+3j^+2k^\rm 3 \hat{i}+3 \hat{j}+2 \hat{k}

Formula Used:

∇f = fx\frac{∂f}{∂x} î + fy\frac{∂f}{∂y} ĵ + fz\frac{∂f}{∂z}

Calculation:

f(x, y, z) = x2y + y2x + z2

fx\frac{∂f}{∂x}= 2xy + y2

⇒  fy\frac{∂f}{∂y}= x2 + 2xy

⇒  fz\frac{∂f}{∂z}= 2z

Substitute (x, y, z) = (1, 1, 1)

⇒  fx\frac{∂f}{∂x}= 2(1)(1) + 12 = 3

fy\frac{∂f}{∂y}= 12 + 2(1)(1) = 3

⇒ fz\frac{∂f}{∂z} = 2(1) = 2

⇒∇f(1, 1, 1) = 3 î + 3 ĵ + 2 k̂

∴  The correct answer is  3 î + 3 ĵ + 2 k̂

125

If z = x + iy lies in the third quadrant, then zˉz\rm \frac{\bar{z}}{z} also lies in the third quadrant if:

  1. ((a))

    x > y > 0

  2. ((b))

    x < y < 0

  3. ((c))

    y < x < 0

  4. ((d))

    y > x > 0

Show Answer
Answer: ((c))

y < x < 0

Given:

z = x + iy lies in III quadrant

∴ x < 0 , y < 0

Find condition so that zˉz\dfrac{\bar z}{z} also lies in III quadrant

Formula used:

zˉ=xiy\bar z = x - iy

a+ibc+id=(a+ib)(cid)c2+d2\dfrac{a+ib}{c+id} = \dfrac{(a+ib)(c-id)}{c^2+d^2}

III quadrant ⇒ Real < 0 and Imaginary < 0

Calculation:

zˉz=xiyx+iy\dfrac{\bar z}{z} = \dfrac{x-iy}{x+iy}

(xiy)(xiy)x2+y2\dfrac{(x-iy)(x-iy)}{x^2+y^2}

x2y22ixyx2+y2\dfrac{x^2 - y^2 - 2ixy}{x^2+y^2}

Real part = x2y2x2+y2\dfrac{x^2-y^2}{x^2+y^2}

Imaginary part = 2xyx2+y2\dfrac{-2xy}{x^2+y^2}

Condition 1:

 2xy<0-2xy < 0

xy>0xy > 0

Since x < 0 , y < 0 ⇒ xy > 0 

Condition 2:

 x2y2<0x^2 - y^2 < 0

x2<y2x^2 < y^2

x<y|x| < |y|

Since x < 0 , y < 0

⇒ y < x < 0

∴ Correct option is (3) y < x < 0

126

For this question, two statements are given one labelled as Assertion (A) and the other as Reason (R). Choose the correct answer from the options given below :

Assertion (A) : If f(x) is an odd function, then f'(x) is an even function.

Reason (R) : If f'(x) is an even function, then f(x) is an odd function.

  1. ((a))

    Both (A) and (R) are true, but (R) is not the correct explanation of (A).

  2. ((b))

    Both (A) and (R) are true and (R) is the correct explanation of (A).

  3. ((c))

    (A) is true, but (R) is false.

  4. ((d))

    (A) is false, but (R) is true.

Show Answer
Answer: ((c))

(A) is true, but (R) is false.

Concept :-

If f(x) is an even function, then f(-x) = f(x)

If f(x) is an odd function, then f(-x) =  - f(x)

Assertion (A) :

.If f(x) is an even function, then f(-x) = f(x)

Differentiating with respect to x, we get

⇒ -f′(-x) = f′(x)

⇒ f′(-x) = -f′(x)

Hence f′(x) is odd function.

Reason (R) : If f'(x) is an even function, then f(x) is an odd function. 

 if  f′(x) is an even function.

Then f '( - x)  = f'(x)

By integrating f'(x) yields an odd function with some  constant. 

 

 for example 

∫x2dx=x33∫x2dx=x33 integral of x squared d x equals the fraction with numerator x cubed and denominator 3 end-fraction

𝑥2𝑑𝑥=𝑥33

is not true, but rather

∫(even)dx=odd+C∫(even)dx=odd+C integral of open paren even close paren d x equals odd plus cap C

(even)𝑑𝑥=odd+𝐶

⇒ The derivative of an even function is an odd function, and the antiderivative of an even function is an odd function with  constant .

Hence Reason R is False .

∴ The Correct Answer is (A) is true, but (R) is false.

127

A coin is tossed twice. The probability of getting head both the times is:

  1. ((a))

    12\frac{1}{2}

  2. ((b))

    14\frac{1}{4}

  3. ((c))

    1

  4. ((d))

    34\frac{3}{4}

Show Answer
Answer: ((b))

14\frac{1}{4}

Calculation:

When the coin is tossed twice ,

Possible outcomes = {HH, HT, TH, TT}

⇒ Total outcomes = 4

⇒ Favorable outcome ( HH) =  1

⇒ Probability = 14\frac{1}{4}

∴ The Correct Answer is  14\frac{1}{4}

1

128

If a function f : [2, +∞) → A defined by f(x) = x2 - 4x + 5 is one-one and onto, then the set A is:

  1. ((a))

    (-∞, 1)

  2. ((b))

    (0,∞)

  3. ((c))

    (0, 1)

  4. ((d))

    [1, +∞)

Show Answer
Answer: ((d))

[1, +∞)

Given:

f : [2, +∞) → A

f(x) = x2 − 4x + 5

Function is one–one and onto

Concept Used:

A function is onto if Codomain = Range

Range of quadratic depends on vertex

Calculation:

⇒ f(x) = x2 − 4x + 5

⇒ Complete square:

⇒ f(x) = (x − 2)2 + 1   (1)

⇒ Minimum value occurs at x = 2

⇒ f(2) = (2 − 2)2 + 1 = 1

⇒ Since domain x ≥ 2

⇒ (x − 2)2 ≥ 0

⇒ f(x) ≥ 1

⇒ Range = [1, +∞)

For onto function, A = Range

∴ Hence,The Set A = [1, +∞).

∴ The Correct Answer is  [1, +∞).

129

What is the value of  limn(1+1n)n\rm \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n} for n ∈ N?

  1. ((a))

    e

  2. ((b))

    1

  3. ((c))

    1e\frac{1}{\mathrm{e}}

  4. ((d))

    None of the above

Show Answer
Answer: ((a))

e

Formula Used:

limn→∞ (1 + 1/n)n = e   (1)

Calculation:

Let L = limn→∞ (1 + 1/n)n

This is the defining limit of e

∴  The correct answer is  e.

130

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

List IList II
A.Right-circular conei.lx + my + nz = p
B.Planeii.2x + 3y = 0 = z - 3
C.Straight lineiii.x2 + y2 = z2 tan2 α
D.Circleiv.x2 + z2 = 9, y = 4

 

<br>
  1. ((a))

    A-iv, B-i, C-ii, D-iii

  2. ((b))

    A-iii, B-i, C-ii, D-iv

  3. ((c))

    A-i, B-iii, C-ii, D-iv

  4. ((d))

    A-i, B-ii, C-iv, D-iii

Show Answer
Answer: ((b))

A-iii, B-i, C-ii, D-iv

Calculation:-

Right-circular cone equation  is 

x2 + y2 = z2 tan2α

⇒ A → iii

Plane equation is 

lx + my + nz = p

⇒ B → i

Straight line is intersection of planes

2x + 3y = 0 and z = 3

⇒ C → ii

Circle in xz-plane at y = 4 is

x2 + z2 = 9

⇒ D → iv

∴ The Correct matching is A-iii, B-i, C-ii, D-iv.

131

Consider the following statements:

I. If f(x) = cosx10 10cosx 0cosx1\rm \left|\begin{array}{ccc} \rm\cos x & 1 & 0 \ \rm 1 & 0 & \rm \cos x \ \rm 0 &\rm \cos x & 1 \rm\end{array}\right|, then f(π6)\rm f^{\prime}\left(\frac{\pi}{6}\right) is equal to 98\frac{9}{8}.

II. If f(x) = cosx10 12cosx1 012cosx\rm \left|\begin{array}{ccc} \rm \cos x & 1 & 0 \ \rm 1 & 2 \rm \cos x & 1 \ \rm 0 & 1 & 2 \rm \cos x \rm \end{array}\right|, then 0π2\rm \int_{0}^{\frac{\pi}{2}} f(x) dx is equal to -13\frac{1}{3}.

Which of the above statements is/are correct?

  1. ((a))

    Only I

  2. ((b))

    Only II

  3. ((c))

    Neither I nor II

  4. ((d))

    Both I and II

Show Answer
Answer: ((d))

Both I and II

Calculation:

Statement I:

f(x) = cosx(0 − cosx × cosx)− 1(1 − 0×cosx)

⇒ f(x) = −cos3x − 1

⇒ f′(x) = 3cos2x sinx

⇒ f′(π/6) = 3×(√3/2)2×(1/2)

 = 3×3/4×1/2 = 9/8

Hence, Statement I is correct.

Statement II:

f(x) = cosx(4cos2x − 1) − 1(2cosx)

⇒ f(x) = 4cos3x − 3cosx

⇒ f(x) = cos3x

By integrating

 ∫0π/2 cos3x dx = [ (1/3)sin3x ]0π/2

 = (1/3)(−1 − 0)

 = −1/3

Hence, Statement II is correct.

∴ The correct answer is: Both I and II are correct.

132

If f(x) = xn, then the value of f(1)+f(1)1!+f(1)2!+f(1)3!++fn(1)n!\rm f(1)+\frac{f^{\prime}(1)}{1!}+\frac{f^{\prime \prime}(1)}{2!}+\frac{f^{\prime \prime \prime}(1)}{3!}+\ldots+\frac{f^{n}(1)}{n!} is:

  1. ((a))

    2n+1

  2. ((b))

    2n

  3. ((c))

    2n-1

  4. ((d))

    2-n

Show Answer
Answer: ((b))

2n

Concept Used:

Taylor Series Expansion at x = 1

Formula Used:

f(1 + h) = f(1) +f(1)h1!+f(1)h22!\frac{ f′(1)h}{1!} +\frac{ f″(1)h^2}{2!}+ …

Calculation:

given : f(x) = xn

⇒ f(1 + 1) = (1 + 1)n

⇒ f(2) = 2n

Using Taylor expansion at x = 1 with h = 1:

⇒ f(2) = f(1) +f(1)1!+f(1)2!++f(n)(1)n!\frac{ f′(1)}{1! }+\frac{ f″(1)}{2! }+ … + \frac{f(n)(1)}{n!}

⇒ Required Sum = 2n

∴ The Correct Answer is 2n

133

The volume generated by the revolution of the loop of the curve x = t2, y = t - 13\frac{1}{3} t3 about the x-axis is:

  1. ((a))

    π2\frac{\pi}{2}

  2. ((b))

    π4\frac{\pi}{4}

  3. ((c))

    3π2\frac{3\pi}{2}

  4. ((d))

    π

Show Answer
Answer: ((c))

3π2\frac{3\pi}{2}

​Formula Used:

V = π ∫ y2 (dx/dt) dt

Calculation:

Given  x = t⇒ dx/dt = 2t

given y = t − (1/3)t3

⇒ y2 = (t − (1/3)t3)2

⇒ y2 = t2 − (2/3)t4 + (1/9)t6

Loop exists for y = 0

⇒ t − (1/3)t3 = 0

⇒ t(1 − t2/3) = 0

⇒ t = 0, ±√3

when t =-√3 to √3  and the function is odd. Then volume become

\(\begin{align*} V &= 2\pi \int_{0}^{\sqrt{3}} y^2 \cdot 2t , dt \[0.5em] &\Rightarrow V = 4\pi \int_{0}^{\sqrt{3}} \left(t^3 - \tfrac{2}{3}t^5 + \tfrac{1}{9}t^7\right) dt \[0.5em] &\Rightarrow V = 4\pi \Biggl[\tfrac{1}{4}t^4 - \tfrac{1}{9}t^6 + \tfrac{1}{72}t^8 \Biggr]_{0}^{\sqrt{3}} \[0.5em] &\Rightarrow V = 4\pi \Biggl[\tfrac{1}{4}\cdot 9 - \tfrac{1}{9}\cdot 27 + \tfrac{1}{72}\cdot 81 \Biggr] \[0.5em] &\Rightarrow V = 4\pi(9)\left[\tfrac{1}{4} - \tfrac{1}{3} + \tfrac{1}{8}\right] \[0.5em] &\Rightarrow V = \tfrac{36\pi}{24} \end{align*} \)

∴ Volume = 3π/2

∴ The correct answer is  3π/2

134

If f : R → R is defined by f(x) = x24x2+1\rm \frac{x^{2}-4}{x^{2}+1}, then f is:

  1. ((a))

    Onto but not one-one

  2. ((b))

    One-one but not onto

  3. ((c))

    Neither one-one nor onto function

  4. ((d))

    One-one and onto both

Show Answer
Answer: ((c))

Neither one-one nor onto function

Calculation:

Step 1: One-one test

Let f(x1) = f(x2)

⇒ (x12 − 4)/(x12 + 1) = (x22 − 4)/(x22 + 1)

⇒ (x12 − 4)(x22 + 1)=(x22 − 4)(x12 + 1)

⇒ x12 = x22

⇒ x1 = ±x2

⇒ x1 ≠ x2 always

⇒ Function is not one-one

Step 2: Onto test

Let y = (x2 − 4)/(x2 + 1)

⇒ y(x2 + 1) = x2 − 4

⇒ yx2 + y = x2 − 4

⇒ x2(1 − y) = y + 4

⇒ x2 = (y + 4)/(1 − y)

 x2 ≥ 0

⇒ (y + 4)/(1 − y) ≥ 0

⇒ −4 ≤ y < 1

⇒ Range ≠ R

⇒ Function is not onto

∴ The given function is neither one-one nor onto.

135

Unit vectors â and b̂ are inclined at an angle 2θ and |â - b̂| < 1. If 0 ≤ θ ≤ π, then θ belongs to:

  1. ((a))

    [π6,5π2]\left[\frac{\pi}{6}, \frac{5 \pi}{2}\right]

  2. ((b))

    [0,π6)\left[0, \frac{\pi}{6}\right)

  3. ((c))

    [0,3π4]\left[0, \frac{3 \pi}{4}\right]

  4. ((d))

    (π6,5π6)\left(\frac{\pi}{6}, \frac{5 \pi}{6}\right)

Show Answer
Answer: ((b))

[0,π6)\left[0, \frac{\pi}{6}\right)

Formula Used:

|a − b| = √(a² + b² − 2a·b)

Calculation:

â &𝒃̂ are unit vectors

∴ |â| = |𝒃̂| = 1

⇒ a·b = cos(2θ)

Also |â − 𝒃̂|² = 1² + 1² − 2cos(2θ)

⇒ |â − 𝒃̂|² = 2 − 2cos(2θ)

⇒ |â − 𝒃̂| = √[2(1 − cos2θ)]

⇒ |â − 𝒃̂| < 1

⇒ √[2(1 − cos2θ)] < 1

⇒ 2(1 − cos2θ) < 1

⇒ 1 − cos2θ < 1/2

⇒ cos2θ > 1/2

⇒ −π/3 < 2θ < π/3

⇒ −π/6 < θ < π/6

Given 0 ≤ θ ≤ π

⇒ 0 ≤ θ < π/6

∴ The Correct answer is [0, π/6).

136

From a pack of well-shuffled cards, three cards are drawn one-by-one without replacement. The probability of first card being ace, second, king and third, queen is:

  1. ((a))

    12197\frac{1}{2197}

  2. ((b))

    716575\frac{7}{16575}

  3. ((c))

    816575\frac{8}{16575}

  4. ((d))

    None of the above

Show Answer
Answer: ((c))

816575\frac{8}{16575}

Calculation:

A standard deck has 52 cards.

Number of Aces = 4

Number of Kings = 4

Number of Queens = 4

⇒ P(1st card is Ace) = 4/52

⇒ P(2nd card is King) = 4/51

⇒ P(3rd card is Queen) = 4/50

⇒ Required probability

⇒ = (4/52) × (4/51) × (4/50)

⇒ = 64 / 132600

⇒ = 8 / 16575

∴  The correct answer is 8/16575.

137

Matrix [224 134 123]\left[\begin{array}{rrr} 2 & -2 & -4 \ -1 & 3 & 4 \ 1 & -2 & -3 \end{array}\right] is:

  1. ((a))

    Nilpotent Matrix

  2. ((b))

    Idempotent Matrix

  3. ((c))

    Periodic Matrix

  4. ((d))

    Orthogonal Matrix

Show Answer
Answer: ((b))

Idempotent Matrix

Concept Used:

A matrix A is idempotent if A2 = A

Calculation:

A2[224 134 123]\left[\begin{array}{rrr} 2 & -2 & -4 \ -1 & 3 & 4 \ 1 & -2 & -3 \end{array}\right]

 

⇒ A2 = A

∴ The Correct answer is Idempotent Matrix.

138

The mean of the following data is : 

Numbers8101520
Frequency5884
  1. ((a))

    26.8

  2. ((b))

    12.8

  3. ((c))

    12.5

  4. ((d))

    26.5

Show Answer
Answer: ((b))

12.8

Formula Used:

Mean = Σ(fx) / Σf

Calculation:

x : 8, 10, 15, 20

f : 5, 8, 8, 4, 

fx = 8×5, 10×8, 15×8, 20×4

⇒ fx = 40, 80, 120, 80

Σfx =  40 + 80 + 120 + 80

⇒ Σfx = 320

Σf = 5 + 8 + 8 + 4

⇒ Σf = 25

Mean = 320 / 25

Mean = 12.8

∴ The  correct  Answer is 12.8

139

The angle between the straight lines represented by the equation y2 - xy - 6x2 = 0 is:

  1. ((a))

    45°

  2. ((b))

    60°

  3. ((c))

    30°

  4. ((d))

    None of the above

Show Answer
Answer: ((a))

45°

Calculation:

given equation is :

y2 − xy − 6x2 = 0

On factorization

⇒ (y − 3x)(y + 2x) = 0

⇒ Slopes are m1 = 3, m2 = −2

Angle between lines:

⇒ tan θ = |(m1 − m2) / (1 + m1m2)|

⇒ tan θ = |(3 − (−2)) / (1 − 6)|

⇒ tan θ = 5/5 = 1

⇒ θ = 45°

∴ The Correct Answer is 45°.

140

If a, b, c are in a geometrical progression (G.P.) and a1/x = b1/y = c1/z, then x, y, z are in:

  1. ((a))

    Harmonic progression (H.P.)

  2. ((b))

    Geometric progression (G.P.)

  3. ((c))

    Arithmetic progression (A.P.) 

  4. ((d))

    None of the above

Show Answer
Answer: ((c))

Arithmetic progression (A.P.) 

Calculation:

Let a1/x = b1/y = c1/z = k

⇒ a = kx

⇒ b = ky

⇒ c = kz

Since a, b, c are in G.P.

⇒ b2 = a × c

⇒ (ky)2 = kx × kz

⇒ k2y = kx+z

⇒ 2y = x + z    ...(1)

⇒ x, y, z are in Arithmetic Progression

∴ The Correct Answer is  Arithmetic Progression (A.P.)

141

If a = b = c, then the rank of the matrix A =[111 b+cc+aa+b bccaab]\rm \left[\begin{array}{ccc} \rm 1 & 1 & 1 \ \rm b+c & \rm c+a & \rm a+b \ \rm b c & \rm c a & \rm a b \rm \end{array}\right] is:

  1. ((a))

    3

  2. ((b))

    2

  3. ((c))

    1

  4. ((d))

    4

Show Answer
Answer: ((c))

1

Concept Used:

Rank of a matrix depends on linear dependence of rows.

If one row is a scalar multiple of another, rank reduces.

Calculation:

Given a = b = c

⇒ b + c = a + a = 2a

⇒ c + a = a + a = 2a

⇒ a + b = a + a = 2a

⇒ bc = a × a = a2

⇒ ca = a × a = a2

⇒ ab = a × a = a2

Matrix A becomes

\(A = \begin{bmatrix} 1 &1 & 1 \[0.3em] 2a & 2a &2a \[0.3em] a^2 &a^2 & a^2 \end{bmatrix}\)

⇒ Row2 = 2a × Row1

⇒ Row3 = a2 × Row1

All rows are linearly dependent

⇒Rank of matrix A = 1

∴ The Correct answer is 1

142

The sum up to infinity of the series 17+272+173+274++\frac{1}{7}+\frac{2}{7^{2}}+\frac{1}{7^{3}}+\frac{2}{7^{4}}+\ldots+\infty is:

  1. ((a))

    15\frac{1}{5}

  2. ((b))

    116\frac{1}{16}

  3. ((c))

    124\frac{1}{24}

  4. ((d))

    316\frac{3}{16}

Show Answer
Answer: ((d))

316\frac{3}{16}

Formula Used:

Sum to infinity of G.P. = a / (1 − r), |r| < 1

Calculation:

given Series:

17+272+173+274++\frac{1}{7}+\frac{2}{7^{2}}+\frac{1}{7^{3}}+\frac{2}{7^{4}}+\ldots+\infty

Separate the series

⇒ S117+173++\frac{1}{7}+\frac{1}{7^{3}}+\ldots+\infty

and S2  = 272+274++\frac{2}{7^{2}}+\frac{2}{7^{4}}+\ldots+\infty

In S1 ,

a = 1/7 

r = (1/73) / (1/7) = 1/49

S1 = a / (1 − r)

⇒ S1 = 1/7 / (1 − 1/49)

⇒ S1 = 7/48

proceeding in same way 

S2 = 1 / 24

Total sum: S = S1 + S2

⇒ S = 7/48 + 1/24

⇒ S = 9/48

⇒ Simplified value = 3/16

∴ The correct answer is 3/16.

143

The lines x = ay + b, z = cy + d and x = αy + β, z = γy + δ are perpendicular, if:

  1. ((a))

    aα + bβ + cγ + 1 = 0

  2. ((b))

    bβ + cγ + 1 = 0

  3. ((c))

    aα + bβ + 1 = 0

  4. ((d))

    aα + cγ + 1 = 0

Show Answer
Answer: ((d))

aα + cγ + 1 = 0

Formula Used: 

If two lines are perpendicular, then 

l1l2 + m1m2 + n1n2 = 0

Calculation:

From x = ay + b ⇒ x − ay − b = 0

From z = cy + d ⇒ z − cy − d = 0

Direction Ratios by considering y as a parameter = (a, 1, c)

From x = αy + β

⇒ x − αy − β = 0

From z = γy + δ

⇒ z − γy − δ = 0

Direction Ratios by considering y as a parameter = (α, 1,  δ )

For perpendicular lines

aα + 1×1 + c δ  = 0

⇒ aα + cγ + 1 = 0

∴ The Correct Answer is aα  + cγ + 1 = 0

144

If the arithmetic mean of the following frequency distribution is 39, find the missing term:

Daily wages (in ₹)2530506075
No. of Labourers10-854
  1. ((a))

    22

  2. ((b))

    13

  3. ((c))

    14

  4. ((d))

    15

Show Answer
Answer: ((a))

22

Given:

Daily wages (₹): 25, 30, 50, 60, 75

No. of labourers: 10, x, 8, 5, 4

Arithmetic mean = 39

Find the missing frequency x

 

Formula Used:

Mean = Σ(fx) / Σf

Calculation:

Daily wages (₹) (x): 25, 30, 50, 60, 75

No. of labourers (f): 10, x, 8, 5, 4

Σf = 10 + x + 8 + 5 + 4

⇒ Σf = 27 + x

Σ(fx) = 25×10 + 30×x

    + 50×8 + 60×5 + 75×4

⇒ Σ(fx) = 250 + 30x + 400

    + 300 + 300

⇒ Σ(fx) = 1250 + 30x

Mean = (1250 + 30x)/(27 + x)

⇒ 39 = (1250 + 30x)/(27 + x) …(1)

⇒ 39(27 + x) = 1250 + 30x

⇒ 1053 + 39x = 1250 + 30x

⇒ 39x − 30x = 1250 − 1053

⇒ 9x = 197

⇒ x = 197/9 ≈ 21.9

∴ The required missing term is 22.

145

If R is the set of real numbers and f : R → R be the function defined by f(x) = (5 - x4)14\frac{1}{4}, then (fof) (x) is equal to :

  1. ((a))

    x

  2. ((b))

    x116\frac{1}{16}

  3. ((c))

    x14\frac{1}{4}

  4. ((d))

    x132\frac{1}{32}

Show Answer
Answer: ((a))

x

Calculation:

f(x) = (5 − x4)1/4 …(1)

f(f(x)) = f[(5 − x4)1/4]

⇒f(f(x)) = (5 − {(5 − x4)1/4}4)1/4

=  (5 − (5 − x4))1/4

=  (x4)1/4

=  |x|

For x ≥ 0, |x| = x

∴ (f∘f)(x) = x

Hence, the correct answer is x

146

The value of sinh (x + y) cosh (x - y) is equal to :

  1. ((a))

    12\frac{1}{2}(sinh 2x - cosh 2y)

  2. ((b))

    12\frac{1}{2}(sinh 2x + sinh 2y)

  3. ((c))

    12\frac{1}{2}(sinh 2x + cosh 2y)

  4. ((d))

    12\frac{1}{2}(sinh 2x – sinh 2y)

Show Answer
Answer: ((b))

12\frac{1}{2}(sinh 2x + sinh 2y)

Formula Used:

sinh A cosh B = 1/2 [sinh(A + B) + sinh(A − B)]

Calculation:

Let A = x + y

Let B = x − y

⇒ A + B = (x + y) + (x − y) = 2x

⇒ A − B = (x + y) − (x − y) = 2y

By using formula 

sinh(x + y) cosh(x − y)

= 12\frac{1}{2} [sinh(2x) + sinh(2y)]

∴ The correct answer is  12\frac{1}{2}(sinh 2x + sinh 2y)

147

If A = x2yz i - 2xz3j + xz2k, B = 2zi + yj - x2k, then the value of 2xy\rm \frac{\partial^{2}}{\partial x \partial y} (A × B) at (1, 0, -2) is:

  1. ((a))

    -8i - 4j

  2. ((b))

    -8i - 8j

  3. ((c))

    -4i - 8j

  4. ((d))

    -4i - 4j

Show Answer
Answer: ((a))

-8i - 4j

Formula Used:

A × B =ijk AxAyAz BxByBz \begin{vmatrix} i& j& k\ A_x & A_y & A_z \ B_x& B_y & B_z \end{vmatrix}

Calculation:

A = x2yz i - 2xz3j + xz2k,

B = 2zi + yj - x2k,

⇒ A × B = i(AyBz − AzBy) − j(AxBz − AzBx) + k(AxBy − AyBx)

 = i[(−2xz3)(−x2) − (xz2)y] − j[(x2yz)(−x2) − (xz2)(2z)] + k[(x2yz)(y) − (−2xz3)(2z)]

⇒ A × B = i(2x3z3 − xyz2) − j(−x4yz − 2xz3) + k(x2y2z + 4xz4)

 By Taking ∂/∂y:

⇒ = i(−xz2) − j(−x4z) + k(2x2yz)

 Take ∂/∂x:

⇒ = i(−z2) − j(−4x3z) + k(2yz)

Substitute (1, 0, −2):

⇒ = i(−4) − j(8) + k(0)

2xy\rm \frac{\partial^{2}}{\partial x \partial y} (A × B) = −4i − 8j

∴ The correct answer  is  −4i − 8j.

148

Which of the following is a group?

  1. ((a))

    (Z, ⋆), where a ⋆ b = ab ∀ a, b ∈ Z

  2. ((b))

    (R-, ⋆), where a ⋆ b = a - b ∀ a, b ∈ R-

  3. ((c))

    (R+, ⋆), where a ⋆ b = ab ∀ a, b ∈ R+

  4. ((d))

    (Z, ⋆), where a ⋆ b = a - b ∀ a, b ∈ Z

Show Answer
Answer: ((c))

(R+, ⋆), where a ⋆ b = ab ∀ a, b ∈ R+

Calculation:

Check (i): (Z, ☆), a ☆ b = ab

⇒ Closure: true

⇒ Identity = 1 ∉ Z (for inverse)

⇒ Inverse of 2 is 1/2 ∉ Z

⇒ Not a group

Check (ii): (R*, ☆), a ☆ b = a − b

⇒ a ☆ (b ☆ c) ≠ (a ☆ b) ☆ c

⇒ Not associative

⇒ Not a group

Check (iii): (R+, ☆), a ☆ b = ab

⇒ Closure: ab ∈ R+

⇒ Associative: a(bc) = (ab)c

⇒ Identity = 1 ∈ R+

⇒ Inverse of a = 1/a ∈ R+

⇒ Satisfies all axioms

Check (iv): (Z, ☆), a ☆ b = a − b

⇒ Not associative

⇒ No identity element

⇒ Not a group

∴  The correct option is option (3).

149

Which of the following statements is incorrect ?

  1. ((a))

    If f(x) = \(\rm \left{\begin{array}{l} \rm 5 x-4,0<x<1 \ \rm 4 x^{2}+3 b\rm x, 1<x<2 \rm \end{array}\right.\) is continuous at x = 1, then the value of b is -1.

  2. ((b))

    The number of points where f(x) = |x| + |x - 1| is not differentiable is 2.

  3. ((c))

    The number of points at which f(x) = |x| + |x - x2|, x ∈ [-1, 1] is discontinuous, is 0.

  4. ((d))

    The number of points, where f(x) = 1logx\rm \frac{1}{\log |x|} is discontinuous is 1.

Show Answer
Answer: ((d))

The number of points, where f(x) = 1logx\rm \frac{1}{\log |x|} is discontinuous is 1.

Calculation:

Statement (i):

 f(x) =\(\rm \left{\begin{array}{l} \rm 5 x-4,0<x<1 \ \rm 4 x^{2}+3 b\rm x, 1<x<2 \rm \end{array}\right.\)  

Continuity at x = 1 ⇒ LHL = RHL

LHL = 5(1) − 4 = 1

RHL = 4(1)2 + 3b(1)

⇒ 1 = 4 + 3b

⇒ b = −1

⇒ Statement (i) is correct

Statement (ii):

f(x) = |x| + |x − 1|

⇒ Non-differentiable at x = 0, 1

⇒ Total points = 2

⇒ Statement (ii) is correct

Statement (iii):

f(x) = |x| + |x - x2|

As |x| and |x − x2| are continuous

⇒ Sum of continuous functions is continuous

⇒ No discontinuity in [−1, 1]

⇒ Statement (iii) is correct

Statement (iv):

f(x) = 1logx\rm \frac{1}{\log |x|}

log|x| = 0 at x = 1, −1

⇒ Discontinuous at 2 points

But Statement says 1 point

⇒ Statement (iv) is false

∴ The correct answer is option (iv).

150

For the curve y2 = (x - a) (x - b) (x - c), where a > 0, b > 0, c > 0, the incorrect statement is :

  1. ((a))

    The curve is not passing through the origin. 

  2. ((b))

    The curve cuts the y-axis.

  3. ((c))

    The curve cuts the axis of x at points (a, 0), (b, 0), (c, 0).

  4. ((d))

    The curve is symmetric about x-axis.

Show Answer
Answer: ((b))

The curve cuts the y-axis.

**Given:**The curve is y2 = (x − a)(x − b)(x − c) where a > 0, b > 0, c > 0

Concept Used:

Properties of curves, intercepts and symmetry

Calculation:

Statement (i):

The curve is not passing through origin

 At origin: x = 0, y = 0

RHS = (−a)(−b)(−c) = −abc ≠ 0

LHS = 0

⇒ Equation not satisfied

⇒ Statement (i) is correct

Statement (ii):

The curve cuts the y-axis

Put x = 0

⇒ y2 = −abc < 0

⇒ No real y exists

⇒ Curve does not cut y-axis

⇒ Statement (ii) is incorrect

Statement (iii):

The curve cuts x-axis at (a,0), (b,0), (c,0)

Put y = 0

⇒ (x − a)(x − b)(x − c) = 0

⇒ x = a, b, c

⇒ Statement (iii) is correct

Statement (iv):

The curve is symmetric about x-axis

⇒ y appears as y2

⇒ Replacing y by −y gives same equation

⇒ Statement (iv) is correct

∴ The Correct Answer is Option 1

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