NVS TGT Mathematics 2019 Shift 1 Official Paper (Previous Year Paper)

Family chart:

Possible Family Chart:

So, Kannan is the Son of Swathi.

Hence, the correct answer is Son

Pattern followed here;

So, the wrong term in the given series is 140

Hence, the correct option is 2.

Pattern followed here;

Hence, the correct option is 187.

In 343 ∶ 6

6 + 1 = 7

7³ = 343

Similarly;

7 + 1 = 8

8³ = 512

Hence, the correct answer is 512 ∶ 7

Pattern followed here,

In 3/8 : 73

⇒3² + 8² 

⇒9 + 64 

⇒73

Similarly;

In 7/9

⇒ 7² + 9²

⇒ 49 + 81

⇒ 130

Hence. the correct option is 130

261 ⇒ have multiple factors (1, 3, 9, 29, 87, 261.)

277 ⇒ Prime Number

271 ⇒ Prime Number

233 ⇒ Prime Number

So, option 1 is odd one.

Hence, the correct option is 261.

Figure 1 2 and 3 follow the same pattern.

But from above it is clear that figure 4 don't follow the same pattern.

Hence, the correct option is 4.

Additional Information

Figure 4 is the water image of the figure 3.

Statement:

All beggars are poor.

In the above statement it is clearly mention that "All beggars are poor".

It means If someone is a beggar, they are poor.

So, Rest don't follow the logic.

Hence, Only conclusion I follows.

13 ⇒ Prime Number

7  ⇒ Prime Number

9  ⇒ It can be divided by 3 as well as 1 and 9. 

11  ⇒ Prime Number

Hence, the correct option is 3.

Statement: 

1. All hill stations have a sunset point. 2. A is a hill station.

Thus statement talks about all the Hill stations which have sunset points. It is said that A is a Hill station thus we can say it will definitely have the sunset point.

Conclusion:

(1) A has a sunset point → True → According to the statement explained.

(2) Places other than hill stations do NOT have sun-set points → False → It is nothing mentioned in the statement. It does not directly follow the given statement.

Hence, Option (2 ) is correct.

Additional Information 

1. If there are two or more sentences that are used to frame a statement, then, the sentences must be interrelated, and mutual contradiction should be there.
2. Do not look for truthful notions. The information provided in the statement is the only requirement for a student to answer the question. No assumptions must be made.
3. Read the statement carefully and look for keywords that are common between the statement and the conclusions.
4. If there is more than one conclusion that is applicable to the statement, students must ensure that the conclusions they opt for have some relation with each other.

The current option is buffer stock.

Key Points

• Buffer stock is a system in which a commodity is stored in bulk amounts when its supply is in surplus to maintain its price from falling below a particular level.
• This storage is utilised when the supply of the commodity is scarce, to maintain sudden price hikes and to prevent instability in the economy.
• In India, it is mainly used in the case of food grains, particularly wheat and rice.
• In India buffer stock is stored by the Food Corporation of India.
• The main objective of this system is **to maintain price stability in the economy.**​

Additional Information

• The concept of a buffer stock was first introduced during the 4^th Five Year Plan (1969-74)
• Reserve stock is the amount of stock or materials which are ready to use or allocate in the business in a particular span of time.

The correct answer is SIx.

Key Points

• There are six schools of Indian philosophy known as Shad Darshanas.
• Each philosophical system has its distinctive way of observing things in the world.
• The six principal Hindu darshans are Samkhya, Yoga, Nyaya, Vaisheshika, Mimamsa, and Vedanta.
• In Indian philosophy, Darshan means the vision of a noble person, the way in which he observes and identifies the world.

Additional Information

• Shankaracharya is known as the father of Indian Philosophy.
• He was the prominent exponent of the Advaita Vedanta school of philosophy.
• Darshan teaches us how to get the realization of ourselves.

The correct answer is Rome.

Key Points

• One of the oldest cities in the world Rome is named after Romulus.
• It is said that the city of Rome was founded by twin brothers or demigods Romulus and Remus.
• It was found on 21 April 753 BCE.
• During the fight to rule the city, Romulus killed Remus and put the name of the city after himself **Rome.**​

Additional Information

• ​Some stories claim that the city was named after a woman, Roma, who travelled with Aeneas a Trojan prince and son of the Greek goddess Aphrodite.
• Rome was situated on the bank of the Tiber river.
• It is also known as the City of seven hills or Eternal City.
• Currently, Rome has 280 fountains and more than 900 churches.
• Rome became the capital city of unified Italy in 1870.

The correct answer is 1990.

Key Points

• First Interstate Council was formed in the year 1990.
• Article 263 of the Indian Constitution empowers the President to constitute the Interstate council.
• In 1990, the Janta Dal government established an interstate council following the recommendations of the Sarkaria Commission.
• The interstate council comprises of :
• The chairman: Prime Minister of India.
• The chief ministers of all the states and Union territories of Delhi and Puducherry.
• The administrators of UTs that do not have Legislative Assemblies.
• Governors of the state under the President's rule.
• Central Home Minister and five other cabinet ministers nominated by the Prime Minister.
• The interstate council can take up the issues (defined by the President) related to interstate disputes and policy formulation and coordination.
• While inquiring into the interstate disputes the council performs an advisory role only and cannot take binding decisions like the Supreme Court.
• Interstate council secretariat set up in 1991, assists the interstate council and zonal council as well.

The correct answer is 86^th Constitutional Amendment Act, 2002.

Key Points 

• 11^th fundamental duty Provide opportunities for education to his child or ward between the age of six and fourteen years.
• This duty was added by the 86^th Constitutional Amendment Act, 2002.
• The 11^th fundamental duty which was added to this list is:
• To provide opportunities for education to children between 6-14 years of age, and the duty as parents to ensure that such opportunities are being awarded to their child.
• The 86^th amendment to the constitution of India in 2002, provided the Right to Education as a fundamental right in part III of the Constitution.

Important Points

• The same amendment inserted Article 21A which made the Right to Education a fundamental right for children between 6-14 years.
• The Fundamental Duty added by the Constitution (86^th Amendment) Act, 2002 is that it is the duty of a parent or guardian to provide opportunities for education to his child or, as the case may be ward between the age of six and fourteen years
• Today, there are 11 Fundamental Duties described under Article 51-A, of which 10 were introduced by the 42^nd Amendment and the 11^th was added by the 86^th Amendment in 2002 The Fundamental Rights, Directive Principles of State Policy and Fundamental Duties.
• Fundamental duties are sections of the Constitution of India that prescribe the fundamental obligations of the states to its citizens and their duties.

Additional Information 

• The rights of the citizens to the StateThe Fundamental Duties were inserted in Constitution by 42^nd Amendment in 1976:
• To remind its citizens that though they enjoy basic fundamental rights, they are supposed to observe certain basic norms of democratic conduct and behaviour.
• The Constitution of India was adopted by the Constituent Assembly on 26^th November 1949 and came into action on 26^th January 1950.

The correct answer is Telangana

Key Points

• Indian archaeologists recently unearthed a rare life-sized stucco sculpture from a Buddhist site in Telangana.
• The Buddhist site name is Phanigiri in the Surya pet district of Telangana.
• The stucco was found in April 2019.
• Department of Heritage estimates that the figurine is associated with Bodhisattva in Jathaka Chakra
• The stucco was about 1.73 metres in height and 35 cm.

Additional Information

• The Phanigiri site is one of the oldest sites of Buddhists, established about 1 century BCE.
• It is said that footprints of Gautam Buddha were found on this site
• Jatakas are the stories of Buddhist literature in which Bodhisattvas are mentioned to the person who seeks enlightenment or Moksha.

The correct answer is 104.

Key Points 

• ISRO created history by launching 104 satellites in a single flight in Feb 2017.
• Indian Space Research Organisation's workhorse rocket PSLV launched 104 satellites in a single mission from the space centre at Sriharikota, Andhra Pradesh.
• Polar Satellite Launch Vehicle or PSLV, in its 39^th flight (PSLV-C37), launched the 714 kg Cartosat-2 series satellite for earth observation along with 103 co-passenger satellites, together weighing about 664 kg at liftoff.
• ISRO the co-passenger satellites comprise 101 nano-satellites, one each from Israel, Kazakhstan, the Netherlands, Switzerland, United Arab Emirates (UAE), and 96 from the United States of America (USA), as well as two from India.
• The two Indian nano-satellites are ISRO's INS-1A and INS-1B.

Additional Information 

• Initially, the planned mission was to launch 83 satellites in one go in the last week of January, which would have been a record.
• ISRO scientists then added 20 more satellites to the launch, rescheduling the launch date to the first week of February.
• ISRO was formed in 1969 with a vision to develop and harness space technology in national development while pursuing planetary exploration and space science research.
• ISRO is the space agency under the Department of Space of the Government of India, headquartered in the city of Bengaluru, Karnataka.
• Its vision is to harness space technology for national development while pursuing space science research and planetary exploration.
• Antrix Corporation Limited (ACL) is a Marketing arm of ISRO for the promotion and commercial exploitation of space products, technical consultancy services, and transfer of technologies developed by ISRO.
• ISRO replaced its predecessor, INCOSPAR (Indian National Committee for Space Research), established in 1962 by India's first Prime Minister.
• PSLV full Form - Polar Satellite Launch Vehicle

​The correct answer is Sanjay Gandhi National Park.

Key Points 

• In May 2019, the last captive White tiger in Mumbai died.
• The name of the tiger is Bajirao.
• The last captive white tiger inside Sanjay Gandhi National Park (SGNP) died of age-related complications. The animal was 18 years old.
• Bajirao was the eldest male in their enclosure.
• Bajirao’s carcass was burnt after the autopsy as per National Tiger Conservation Authority guidelines. There are six remaining Royal Bengal tigers at the park– four female and two male, all above 10 years of age.

Additional Information 

• Bajirao was born in SGNP in 2001 to tigress Renuka and tiger Siddharth, both white tigers brought from Aurangabad Zoo in 1999.
• An expert advisory committee, comprising of senior veterinarians, both current and retired from SGNP and the Bombay Veterinary College (BVC), had been monitoring Bajirao’s health over the past four years.
• The tiger’s autopsy carried out by the pathology department of BVC indicated multiple organ failures along with old age as the probable cause of death.
• “The post-mortem findings were suggestive of lesions (rotting or putrefied tissues) in visceral organs such as the kidney, heart, and liver,”

Important Points

• The following National parks are the most famous in India:

The correct answer is Assamese

Key Points

• Bohag Bihu is the celebration of the Assamese new year in the state of Assam and other parts of Northeast India.
• It is also known as Rongali Bihu.
• It is celebrated in the middle of the month of April every year.
• The new year starts with the month of Bohag.
• It is a harvest-related festival.
• It is a three-day festival and on all three days, **musical groups visit houses and perform the Bihu dance in open.**​

Additional Information

• Manipuri new year is celebrated in the form of the Sajibu Nongma Pan festival
• Phoela Baisakh is the Bengali new year festival.
• Pana Sankranti also known as Vishnu Sankranti is the Odia new year festival.

The correct answer is Saurabh Chaudhary.

Key Points 

• Saurabh Chaudhary is an Indian sports ace pistol shooter who won a gold medal in the 10 m Air Pistol at the 2018 Asian Games.
• He was the Asian Games' youngest gold medalist from India.
• In the ISSF Junior World Cup in Suhl, Germany, he had already won gold medals and set a new junior world record.
• He is the first Indian shooter to have won gold in the ISSF World Championship, ISSF World Cup, Youth Olympic Games, Asian Games, and Asian Air Gun Championship.
• In January 2020, he earned 63^rd gold in men's 10-meter air pistol at the 63^rd National Shooting Championship.

​

Important Points 

• The Chinese city of Hangzhou has been confirmed as the host of the 2022 Asian Games by the Olympic Council of Asia.
• China has hosted the Asian Games twice before, at Beijing in 1990 and Guangzhou in 2010, and has already been awarded the Winter Olympics in 2022.
• The first Asian Games were held from 4 March to 11 March 1951 in New Delhi.​
• India has hosted the Asian Games twice, that is in 1951 and in 1982.
• The first edition of the games which took place in 1951 was hosted by India.

Additional Information

• The Asian Games:
• It is a continental multi-sport event held every four years among athletes from all over Asia.
• The Games are organized by the Olympic Council of Asia (OCA).
• The Games are recognized by the International Olympic Committee (IOC) and are described as the second-largest multi-sport event after the Olympic Games.
• The most recent edition (18^th edition) was held in Jakarta and Palembang, Indonesia from 18 August to 2 September 2018.

The learning of social sciences is not a new one. The learning of social sciences has become a part of the education system from the first day of the creation of civilization in this world in one way or another.

• At the upper primary stage, the subject-area of Social Science drawing its content from history, geography, political science, and economics will be introduced.
• Simultaneously, the child may be introduced to contemporary issues and problems.
• Contemporary issues may be looked at from multiple perspectives introducing the child to the social and economic problems of society.
• Emphasis needs to be given to issues like poverty, illiteracy, child and bonded labour, class, caste, gender, and environment.
• Geography and Economics may together help in developing a proper perspective related to the issue concerning the environment, resources and development at different levels from local to global.
• Similarly, Indian history will be taught emphasizing the concepts of plurality and change.
• The child will be introduced to the formation and functioning of governments at the local, state, and central level and the democratic processes of participation.

Thus, from above-mentioned points, it is clear that History teaches the concepts of 'plurality' and 'change'.

Creativity is the ability to see things in a new and unusual light, to see problems that no one else may even realize its existence, and then to come up with new, unusual, and effective solutions.

Intelligence can be defined as the ability to grasp information, understanding concept, and solving problems. Different theories have been given by different philosophers on intelligence. J.P. Guilford, an American psychologist, has proposed the concept of 'convergent thinking'  and 'divergent thinking' which tells the difference between intelligence and creativity.

Important Points

• Convergent thinking: It refers to an ability which relates to followup logical steps to arrive at one correct solution. It doesn't require creativity, hence, is equivalent to intelligence.
• Divergent thinking: It refers to a way of solving problems by more than one approach. It is a creative process which includes out of the box thinking. Thus, it is equivalent to creativity.

Hence, we conclude that intelligence and creativity are two independent functions of a human personality.

The arts education comprises of visual and performing arts including heritage crafts. For activity-based subjects like arts and craft, schools should allocate block periods, which is two periods of 40-45 minutes each.

• On an average, schools have 40 periods per week (for primary) and 48 periods in upper primary. Out of this,1/4 time should be allocated for arts education.
• Schools should assure that every student at this level is given an equal opportunity to participate in art activities, take active part in the art-making or doing processes and should be encouraged to express themselves.

Key Points​ The objectives of learning in arts education are to:

• Accustom students with basics of visual and performing arts.
• Encourage students for free expression and creativity.
• Acquaint students with basic elements /principles of design.
• Develop an insight towards sensibility and aesthetic appreciation.
• Make children understand cultural diversity by recognising different traditional art forms prevalent in the country.

Hence, According to the guidelines by NCERT, One-fourth of the total time should be allocated for art education in schools.

India made a Constitutional commitment to provide free and compulsory education to all children up to the age of 14 nearly sixty years ago. The goal, which was expected to be achieved by 1960, has remained elusive even now.

• Yet, one has to admit that developments in recent years have made significant impact on the situation, raising the hope that universal basic education could be a reality within a reasonable period of time.
• The National Sample Survey Organisation (NSSO) has been set up by the Government of India in 1950 to collect socio-economic data employing scientific sampling methods. The NSSO conducts regular surveys as part of its “rounds”, each round being normally of a year's duration and covering more than one subject of study.

Key Points

• The findings of the National Sample Survey Organization (NSSO), conducted in 1986-87 (42^nd Round) showed that 35 percent (17 percent in the case of urban males and 47 percent among rural females) of 5-29 years old had never attended any institution in their life.
• It suggests that more boys than girls, both in rural and urban areas, reportedly could not attend school as they had to work and contribute to household income; this is more pronounced among the older age group, whereas more girls than boys typically reported that they could not attend school because of their engagement in domestic chores.
• “Education considered not necessary” has been the more important reason for non-enrolment of girls in rural areas since 17 percent girls of 5-14 years age group were found not attending any educational institution due to this reason; as against this only 10 percent of their counterpart in an urban area could not attend school due to the same reason.
• Further, only 8 percent boys of 5-14 years age group were found to have never attended school due to this reason in the urban areas, showing a clear bias against female participation in schooling.

Hence, it is clear that according to the National Sample Survey conducted in 1986-87, rural females only could never enroll as students since their priority is attending to household chores.

Algebra is generalized arithmetic where letters are used as symbols to represent numbers. Every number is a constant and every symbol can be assigned different values in different situations. Algebraic expressions are formed using symbols and constants. Four fundamental operations on the symbols and constants to form expressions are used. Terms are parts of an expression that are separated by a '+' or '_' sign. It may be a constant, a variable, or a combination of both. The algorithm (method) involved in solving real-life problems is to:

• Understand the situation expressed in the word problem
• Choose a symbol and substitute it for the unknown to be determined
• Write an equation from the given relation in the problem
• Solve the equation and find the value of the unknown
• Verify the correctness of the solution

Hence, we conclude that the above statement is of algebra.

'"It is more useful to know how to Mathematise than to know a lot of Mathematics”: David Wheeler.

• Perhaps this sums up the entire problem of Mathematics teaching-learning succinctly. Knowing a lot of Mathematics implies being able to do a variety of computations according to a set procedure that is learnt.
• Mathematization, on the other hand, involves the ability to apply Mathematics according to the need of the situation, to be able to think mathematically. Descriptive problems in Mathematics could offer this kind of opportunity to mathematize.
• Very often, when a word problem is given to children, they get confused about which operation to use. Hence, if we try and understand why children might be getting confused, we would be in a position to acknowledge the challenges children might be facing in solving descriptive problems.

Hence, it can be concluded that the above-mentioned statement is said by David Wheeler.

Distance Learning: It is a method or a set of techniques for teaching and learning. It is a mode of education that refers to a transactional strategy of different methods. The teaching and learning is through multimedia instructional packages, e-learning, blended learning, etc.

• It is far from a new phenomenon, but it continues to reach new heights as the developments in technology advance.
• The earliest model of distance education is correspondence i.e., a form of distance education given that the teacher and students are physically separated. It is defined as “a method of providing education for nonresident students, primarily adults, who receive lessons and exercises through the mail, or some other device, and, upon completion, return them for analysis, criticism, and grading".
• It progresses to the use of parcel post, to radio, then to television, and finally to online education.

Hence, it can be concluded that the earliest mode of distance education is education through correspondence.

Activity-based learning: As the name suggests, actively involves learners in the construction and re-construction of knowledge based on his or her individual experiences. ​Activity-based approach helps to achieve hands-on experience effectively. 

In the above situation, the best way for the teacher to ensure that the students have achieved the intended learning from the activity is to allow individual student reflections through discussion with peers or with the teacher. As, reflection is a powerful tool that builds onto knowledge they have acquired throughout the lesson. 

Benefits of Reflection

• Significance: It allows students to see the importance of their own learning process.
• Process Recognition: Students can identify what they did well, what they failed at, what they need to change.
• Solutions/Strategies: Provides students opportunities to come up with solutions and strategies to improve on their learning.
• Motivation: Reflection provides students with the motivation to learn and enjoy the process of learning. This motivation comes from them reflecting on their thoughts, feelings and emotions.
• Analysis: The most important benefit of reflection in the classroom is for students to be able to know why they needed to learn these concepts, theories, and ideas.

Hence, the best technique in the above situation is to allow individual student reflections through discussion with peers or with the teacher.

The draft of New Education Policy 2016 focuses on addressing gender discrimination, the creation of educational tribunals, and a common curriculum for Science, Mathematics and English. Revision of no-detention policy, promotion of Sanskrit, and bringing back Class 10 board examinations are some of the key changes that the policy talks about.

Protection of Rights of the Child & Adolescent Education that is included in this policy:

• Self-learning online programmes on child rights will be developed for the benefit of students, teachers and parents.
• Framework and guidelines for ensuring school safety and security of children will be developed.
• Every Principal and teacher will be made aware of the provisions of the relevant Acts, Rules, Regulations, etc.
• The Adolescent Education Programme and National Population Education Programme will be integrated into the curriculum of schools in a phased manner.
• Adolescent Education will be included in pre and in-service training programmes of secondary school teachers.
• Schools will engage trained counsellors to confidentially advise parents and teachers on adolescence problems faced by growing boys and girls.

Hence, Child rights is the focus of self-learning online courses for teachers in the draft of the New Education Policy, 2016.

Gender is a social construct. It refers to the socially and culturally constructed system that attributes meaning to what it means to be a male or a female in a particular society. A society comprises males and females. 

• Gender inequality, gender biases, and gender stereotypes are common ill practices in a society.
• Gender bias refers to the tendency to prioritize one gender over another. It should be discouraged in the classroom.

Key Points

Role of the teacher in creating a gender-fair classroom:

• The teacher should avoid sexist bias in the use of their language they must give equal weightage to all genders by using gender-neutral language.
• They should consciously monitor their own conduct and the conduct of children to identify gender bias and discrimination in order to avoid such occurrences.
• They should avoid assigning gendered jobs and conduct them in a gender-fair manner both inside and outside the classroom.
• They should place equal expectations on boys and girls and discuss them with the class and make it clear that gender will have no bearing on them.

Hence, we conclude that by giving equal weightage to all genders, we can avoid gender bias while using language for teaching social studies in a classroom. 

Hint

• Only showing pictures without using any language will not help children understand the concepts properly.
• A teacher should have to be gender sensitive while managing a classroom and must avoid giving different adjectives to different students.

National Adult Education Programme (NAEP) - Launched on October 2, 1978. It was a macro program to eradicate illiteracy through a project approach. It was a massive program aimed at educating 100 million non-literate adults in the age group of 15-35 years within a time frame of five years.

Key Points

Objectives of NAEP:

• ​To promote the literacy rate of the country among those who live in economically and socially disadvantaged areas of society.
• To create an awareness among illiterate adults about their helplessness due to education.
• Raising the functional capabilities of the adults in their careers and vocations.
• To develop self-dependency in adults. To relate education with the work and life situations of the adults.

Hence, we conclude that the creation of employment is not an objective of the national adult education program.

Here, a teacher is asking her students to read the news about recent events in the state. Then the students are asked to classify the events into different 'ruling systems.' The purpose of this learning activity is to understand the topic.

When we refer to bloom's taxonomy that has been shown below. Understand includes interpreting, exemplifying, classifying, summarizing, inferring, comparing, explaining

NOTE:

• Recall and revise are the objectives under 'Remember'.
• Research is included in 'create'.

​Hence, the purpose of the above activity is to understand.

Migration is defined as a geographical movement of people involving a change from their usual place of residence. But it is distinguished from temporary and very short-distance moves. Migration can be internal (within the national boundaries) or international (across the international borders).

Migration presents both opportunities and challenges for societies, communities and individuals. Migration alters the structure of families. While it is true that economic factors are major drivers, migration involves highly diverse groups of people, including girls, boys, women, men, and better-off as well as poorer people.

• A child growing up in a migrating family faces different opportunities and challenges, i.e., marginalization and discrimination in the country of settlement, barriers to accessing social services, challenges to the rights to citizenship and identity, parents' economic insecurity.
• But, compared to a child growing up in one place, a child growing up in a migrating family is typically able to ​connect with people.
• None of these are necessary outcomes, however, most migrant children flourish and contribute positively to their new communities; and policies and programs can be devised to support and protect those children who may become vulnerable.
• According to the Convention on the Rights of the Child, it is the duty of any country to ensure that all children enjoy their rights, irrespective of their migration status or that of their parents.

Hence, we can conclude that as compared to a child growing up in one place, a child growing up in a migrating family is typically able to ​connect with people.

The National curriculum framework (NCF) 2005 is India's 4^th national curriculum framework which was released in 2005, by NCERT. Its predecessors were released in 1975, 1988, and 2000 respectively.

• It serves as a guideline for syllabus, textbooks, and teaching practices for the schools in India.

Key Points

Gender equality and inclusion

• One of the many messages received by NCERT in the course of deliberations over the national curriculum framework review came from a teenage girl, who suggested that specific measures are needed to inculcate greater self-awareness among boys regarding their behavior towards girls.
• This idea could be extended to cover all aspects of a culturally inclusive classroom and social policy.

Hence, we conclude that option 4 is the correct answer.

Communitarianism is a philosophy that places a strong emphasis on the relationship between the individual and the community.

Key Points

• The general consensus of Communitarianism is that the group is more significant than the individual.
• In exchange for loyalty, the group offers support and protection. Individuals are never put ahead of the group.
• Principle of Communitarianism:
• Praise and reward group performance.
• Don't praise individuals publicly.
• Allow people to involve others in decision-making.
• Avoid showing favoritism.

Hence, we conclude that communitarianism is a school of philosophy that praises and rewards group performance.

Additional Information

• Universalism: Laws, regulations, moral principles and duties are highly valued by people. They attempt to treat individuals equitably based on these standards but rules come first before relationships.
• Individualism: People have a belief in individual success and independence. They hold the view that an individual is responsible for his own welfare and make his own decisions.
• Particularism: People think that their behavior is dictated by their circumstances and relationships depending on the circumstances.

Concept:

If first terms and common difference of AP are 'a' and 'd' respectively, then

n^th term of Ap:

T_n = a + (n - 1)d

Sum of n terms of AP:

\(S_n \ =\ \frac{n}{2}[2a\ +\ (n\ -\ 1)d]\)

Note: If d > 0, AP will be increasing and d < 0, AP will be decreasing.

Calculation:

Given series is 

\(20 + 19 \frac{1}{3} + 18\frac{2}{3} + 18 + ...\)

This can be written as 

\(\frac{60}{3} + \frac{58}{3} + \frac{56}{3} + \frac{54}{3} + ...\)

Clearly we can see that, this is an AP series.

First term a = \(\frac{20}{3}\) and common difference

d = \(\frac{58}{3}\ -\ \frac{60}{3}\ = \ -\frac{2}{3}\)

Therefore, this is the decreasing series and sum will be maximum if its last term is positive or zero.

Let there are n positive terms, where T_n is n^th term then

T_n = a + (n - 1)d ≥  0

⇒ 20 + (n - 1)(\(-\frac{2}{3}\)) ≥  0

⇒ 2(n - 1) ≤ 60

⇒ n ≤ 31 

Hence, we will take 31 terms for maximum sum.

Hence, sum of  31 terms of AP  is given by

\(S_{31}\ =\ \frac{31}{2}[2\times 20\ +\ (31\ -\ 1)\frac{-2}{3}]\)

\(\Rightarrow \ S_{31}\ =\ \frac{31}{2}[40\ -\ 20]\)

⇒ S_n = 310

Hence, option 2 will be correct.

Concept:

Let us consider sequence a1, a2, a3 …. an is an A.P.

1. Common difference “d”= a2 – a1 = a3 – a2 = …. = an – an – 1
2. nth term of the A.P. is given by an = a + (n – 1) d
3. Sum of the first n terms = Sn =\(\rm \frac n 2\) [2a + (n − 1) × d] = \(\rm \frac n 2\)(a + l)

Where, a = First term, d = Common difference,

n = number of terms, an = nth term and

 l = Last term

Calculation:

We have

p time the pth term of an AP = q times the qth term

We know that, n^th term of AP is given by

a_n = a + (n - 1)d     -----(1)

⇒ p[a + (p - 1)d] = q[a + (q - 1)d]

⇒ a(p - q) = [q(q - 1) - p(p - 1)]d

⇒ a(p - q) = [q² - p² - (q - p)]d

⇒ a(p - q) = [(q - p)(q + p) - (q - p)]d

⇒ a(p - q) = (q - p)(q + p - 1)d

⇒ a = - d(q + p - 1)

⇒ a + (p + q - 1)d  = 0  -----(2)

On comparing equation (1) & (2),

⇒ a_{p + q} = 0

Hence, option 3 is correct.

Here,

Total number of distribution in frequency 1 and 2 = 5, 5

Sum of all distribution in frequency 1 = 20 + 15 + 10 + x + y = 45 + x + y

Sum of all distributions in frequency 2 = 4 + 8 + 4 + 2x + y = 16 + 2x + y

Mean of frequency 1 = (1025 + 45x + 55y)/(45 + x + y) = 25.4

⇒1025 + 45x + 55y = 1143 + 25.4x + 25.4y

⇒ 19.6x + 29.6y = 118__(i)

Mean of frequency 2 = (400 + 90x + 55y)/(16 + 2x + y) = 32.5

⇒ 400 + 90x + 55y = 520 + 65x + 32.5y

⇒ 25x + 22.5y = 120______(ii)

From eq.(i) and (ii), we get 

x = 3, y = 2

∴ Option 2 is correct.

Concept:

• A Linear Diophantine equation( in two variables) is an equation of the general form ax + by = c where a, b, c are given integers and x, y are unknown integers.
• If a = b then all solutions will have the same sum x + y.

Calculations:

Let S = 2x + 3y   ......(1)

⇒ S = x + x + y + y + y

The product of each of these terms gives us x2y3 

S will have least value only when x = y

So, using this condition in  x2y³ = 32 we get,

x2x3 = 32             put x = y in x2y3 = 32

⇒ x⁵ = 32

⇒ x = 2                taking fifth root on both sides

⇒ y = 2    ∵ x = y

Substitute these values in (1) we get

S = 2(2) + 3(2)

⇒ S = 10 is the least values of  2x + 3y

Hence, the correct answer is option 4)

Concept:

The inequality of arithmetic and geometric means states that the arithmetic mean of a list of non-negative real numbers is greater than or equal to the geometric mean of the same list i.e., AM ≥ GM

where for any two positive real numbers x and y AM is \(\frac{x+y}{2}\) and GM is (xy)^1/2 

Calculation:

Given real numbers a-5, a-4, 3a-3, 1, a8 and a10 with a > 0 which means a-5, a-4, 3a-3, 1, a8, a10 > 0

The inequality of arithmetic and geometric means states that the arithmetic mean of a list of non-negative real numbers is greater than or equal to the geometric mean of

the same list i.e AM ≥ GM

AM of the given numbers = \(\frac{\frac{1}{a^5}+\frac{1}{a^4}+\frac{1}{a^3}+\frac{1}{a^3}+\frac{1}{a^3}+1+a^8+a^{10}}{8}\)

GM of given numbers= \((\frac{1}{a^5}\times\frac{1}{a^4}\times\frac{1}{a^3}\times\frac{1}{a^3}\times\frac{1}{a^3}\times1\times a^8\times a^{10})^\frac{1}{8}\)

Now, putting the value in the relation: AM ≥ GM

\(\frac{\frac{1}{a^5}+\frac{1}{a^4}+\frac{1}{a^3}+\frac{1}{a^3}+\frac{1}{a^3}+1+a^8+a^{10}}{8}\)  ≥   \((\frac{1}{a^5}\times\frac{1}{a^4}\times\frac{1}{a^3}\times\frac{1}{a^3}\times\frac{1}{a^3}\times1\times a^8\times a^{10})^\frac{1}{8}\)

⇒ \(\frac{\frac{1}{a^5}+\frac{1}{a^4}+\frac{1}{a^3}+\frac{1}{a^3}+\frac{1}{a^3}+1+a^8+a^{10}}{8}≥ (1)^\frac{1}{8}\) 

⇒ \(\frac{1}{a^5}+\frac{1}{a^4}+\frac{1}{a^3}+\frac{1}{a^3}+\frac{1}{a^3}+1+a^8+a^{10}≥8\) 

⇒ a^-5 + a^-4 + 3a^-3 + 1 + a⁸ + a¹⁰ ≥ 8

Thus, the minimum value of the sum of real numbers a-5, a-4, 3a-3, 1, a8 and a10 with a > 0 is 8

Hence, the correct answer is option (4).

Formula used: 

For a quadratic equation ax² + bx + c = 0

Sum of roots = -b/a

Product of roots = c/a

Calculation:

Given that, α, β are zeros of x2 - 6x + k

Then, sum of roots

⇒ α + β = 6  

⇒ 2α + 2β = 12       -----(1)

And product of root

αβ = k                   -----(2)

Also, according to question,

3α + 2β = 20       -----(3)

Substracting equation (1) from equation (3), we will get

α = 8

From equation (1)

2 × 8 + 2β = 12

⇒ β = -2

Therefore, 

k = αβ = 8 × (-2) = -16 

Hence, option 4 is correct.

Given:

Ratio of sides = 1 : 6 : 8

Formula used:

The volume of cube = (Side)3

Surface area of cube = 6(side)²

Calculation:

Volume of new cube = Sum of the volume of three small cube

Since, the volume of cube = (Side)3

⇒ Volume of new cube = 63 + 83 + 13 

⇒ Volume of new cube = 729 cm3

Since, the side of cube = (volume)^1/3

⇒ Side of new cube = (729)1/3 = 9 cm

We know that, the surface area of cube = 6(side)2

⇒ Surface area of new cube = 6 × 92 = 486 cm2 

∴ Half of the surface area = 243 cm2

Concept:

\(\displaystyle \lim_{x\rightarrow a^-} f(x) = \displaystyle \lim_{h\rightarrow 0} f(a-h)\)

Calculation:

Given:

\(\rm f(x) = \left{ \begin{matrix}\rm \frac{|x -4| }{(x - 4)}; &\rm x \ne 4 \\ 0; & \rm x = 4 \end{matrix} \right.\)

L.H.L = \(\displaystyle \lim_{x\rightarrow 4^-} f(x) = \displaystyle \lim_{h\rightarrow 0} f(4-h)\)

=  \(\displaystyle \lim_{h\rightarrow 0} \frac {|(4-h)-4|}{((4-h)-4)}\) = \(\frac {|-h|}{-h}\)

= \(\frac h{-h}\)

= -1

Concept:

For two independent events X & Y, 

P(X∩Y) = P(X) × P(Y)

Also, \(P(\frac{X}{Y})=\frac {P(X​​ ​​∩Y)}{P(Y)}\)

Calculation:

Given:

P(A) > 0 and P(B) ≠ 1

\(P \left( \frac{\bar A }{B} \right)\) = \(\frac {P(\bar {A}​​ ​​∩B)}{P(B)}\) = \(\frac {P(\bar {A}​​ ​​) \times P(B)}{P(B)}\)

\(P \left( \frac{\bar A }{B} \right)\) = P(\(\bar A\))

\(P \left( \frac{\bar A }{B} \right)\) = 1 - P(A)

Now, simplifying options,

option 1

\(1 - P \left( \frac{\bar A }{B} \right)\) = \(1 - \left( \frac{P(\bar A)\times P(B) }{P(B)} \right)\) = P(A)

option 4

\(1-P \left( \frac{ A }{\overline B} \right)\) = \(1 - \left( \frac{P( A)\times P(\bar B) }{P(\bar B)} \right)\) = 1 - P(A)

Hence, option 4 is the correct answer.

Given polynomial

√3x2 - 8x + 4√3 = 0

√3x2 - 6x - 2x + 4√3 = 0

√3x(x - 2√3) - 2(x - 2√3) = 0

(x - 2√3)(√3x - 2) = 0

\(x=2\sqrt 3\)

\(x=\frac{2}{\sqrt3}\)

Hence, option 3 is correct answer.

Concept:

Range of a function: The range of a function may refer to either of two closely related concepts: The codomain of the function, The image of the function Given two sets X and Y, a binary relation f between X and Y is a function if for every x in X there is exactly one y in Y such that f relates x to y.

(or) The range of a function is the set of all output values (Y-values).

Calculation:

Given f(x) = \(\rm \log _e \sqrt{4 - x^2} \) 

Since the logarithmic function is a strictly increasing function as shown in the below figure.

here we have two cases

Case 1): \(\sqrt{4 - x^2 }≥ 0 \) 

squaring on both sides we get,

⇒ 4 - x² ≥ 0

⇒ x²- 4 ≤ 0

⇒ x² ≤ 4

⇒ x ≤ ± 2

⇒ x ∈ [- 2, 2]

Case 2): \(\sqrt{4-x^2}>0\)

Squaring on both sides

⇒ 4 - x² > 0

⇒ x2 < 4

⇒ x < ± 2

⇒ x ∈ (- 2, 2)

So the intersection of these two intervals i.e., x ∈ [- 2, 2] ∩ x ∈ (- 2, 2) = x ∈ (- 2, 2) is the domain of f(x).

Now, f(x) = \(\rm \log _e \sqrt{4 - x^2} \) has maximum value if \(\sqrt{4-x^2}\) is of maximum value

and \(\sqrt{4-x^2}\) will be maximum at x = 0

∴ f(x) = \(\rm \log _e \sqrt{4 - 0^2} \) = \(\rm \log _e 2\)  or ln 2.( maximum value of f)

Similarly, f(x) = \(\rm \log _e \sqrt{4 - x^2} \) has minimum value if \(\sqrt{4-x^2}\) is of minimum value

and \(\sqrt{4-x^2}\) will be minimum if \(\sqrt{4-x^2}\) = 0

∴ f(x) = \(\rm \log _e 0 = -∞\) ( minimum value of f)

Thus the range of a function f(x) is (-∞ , ln 2].

Hence, the correct answer is option 3)

Given:

3^{(x + y)} = 81 and 81^{(x - y)}  = 3

Calculation:

3(x + y) = 81 and 81(x - y)  = 3

3(x + y) = 81 can be written as:

3(x + y) = (3)⁴

∴ x + y = 4  ----(1)

Now, 81(x - y)  = 3 can be written as:

3⁴(x - y)  = (3)¹

∴ 4(x - y) = 1

4x - 4y = 1 -----(2)

On solving equations (1) and (2) we'll get:

\(x=\frac{17}{8}\)

Concept:

• A common tangent is called a direct common tangent*(DCT)*** if both the circles lie on the same side of it.
• The point of intersection of DCT´s (here S₁) will divide the line joining centers in the ratio of radii (externally).

Calculations:

From the figure, we have two circles C1 and C₂ with the radii 3cm(say r₁) and 2cm(say r₂) respectively, and O₁ and O₂ are the centers of the circles C₁ and C₂ respectively and the distance between their centers is 6 cm.

So from the figure we have

\(\bigtriangleup S_1O_2Q \ \cong \ \bigtriangleup S_1O_1P\)

⇒ \(\frac{S_1Q}{S_1P}=\frac{S_1O_2}{S_1O_1}=\frac{O_2Q}{O_1Q}=\frac{r_2}{r_1}\) 

As we need to find the length O2S₁ = x(say)

we shall consider, 

\(\frac{S_1O_2}{S_1O_1}=\frac{r_2}{r_1}\)

⇒ \(\frac{x}{x + 6}=\frac{2}{3}\)      | S₁O₁ = x + 6

⇒ 3x = 2(x + 6)

⇒ 3x - 2x = 12

⇒ x = 12cm

⇒ O2S₁ = 12cm

Hence, the correct answer is option 4)

Concept:

Section formula: The section formula helps in determining the coordinates of a point which facilitates the division of the line joining two points in a ratio. This takes place either internally or externally.

So, Co-ordinate of a point P(x, y) that divides the line segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m:n is 

\(P(x, y) = \bigg(\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n}\bigg) \) . and 

Co-ordinate of a point P(x, y) that divides the line segment joining A(x1, y1) and B(x2, y2) externally in the ratio m:n is 

\(P(x, y) = \bigg(\frac{mx_2-nx_1}{m-n} , \frac{my_2-ny_1}{m-n}\bigg) \) . 

Calculation:

Let A = (1, - 2) and B = (- 3, 4) be the given points.

Let the points of trisection be P and Q as shown in the below diagram

Then AP = PQ = QB = λ(say)

∴ PB = PQ + QP = 2λ 

and AQ = AP + PQ = 2λ 

⇒ AP : PB = λ : 2λ = 1 : 2

and AQ : QB = 2λ : λ = 2 : 1

So P divides AB internally in the ratio 1 : 2 

So, by using the section formula \(P(x, y) = \bigg(\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n}\bigg) \) we have,

\(P(x, y) = \bigg(\frac{1\times(-3)+2\times1}{1+2} , \frac{1\times4+2\times(-2)}{1+2}\bigg) \) or \((\frac{-1}{ \ \ 3}, 0)\)

while Q divides AB internally in the ratio 2 : 1 

So, by using the section formula \(P(x, y) = \bigg(\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n}\bigg) \) we have,

\(Q(x, y) = \bigg(\frac{2\times(-3)+1\times1}{2+1} , \frac{2\times4+1\times(-2)}{2+1}\bigg) \) or \((\frac{-5}{ \ \ 3}, 2)\)

Hence the points of trisection are \((\frac{-5}{ \ \ 3}, 2)\) and \((\frac{-1}{ \ \ 3}, 0)\) .

Hence, the correct answer is option 4)

Concept:

• Range of a function: The range of a function may refer to either of two closely related concepts: The codomain of the function, The image of the function Given two sets X and Y, a binary relation f between X and Y is a function if for every x in X there is exactly one y in Y such that f relates x to y.

(or) The range of a function is the set of all output values (Y-values).

• AM ≥ GM ≥ HM where, AM - Arithmetic mean, GM - Geometric mean, HM - harmonic mean

Calculations:

Given, |a| ≤ 1 and a + b = 1, (a, b ∈ R)

We know that AM ≥ GM

⇒ \(\frac{a+b}{2}\geq(ab)^\frac{1}{2}\)

Since a + b = 1 and squaring on both sides we get

⇒ \(\frac{1}{4}\geq ab\) ( is the maximum value)

And also we have  |a| ≤ 1

⇒ -1 ≤ a ≤ 1

If a = 1, then b = 0 | ∵ a + b = 1

If a = - 1, then b = 1 - a 

b = 2                      | ∵ a + b = 1

and in this case ab = -1 × 2 = - 2 ( is the minimum value)

Hence the range of ab is \(\left[ -2, \frac{1}{4} \right]\)

Hence, the correct answer is option 3)

Concept:

Binomial expansion: The binomial expansion formulas are used to find the expansion when a binomial is raised to a number

Formula used:

​\((x+y)^n=\displaystyle\sum_{k=0}^{n} \bigg(\begin{array}{c} n \ k \end{array}\bigg)x^{n-k}y^k =\displaystyle\sum_{k=0}^{n} \bigg(\begin{array}{c} n \ k \end{array}\bigg)y^{n-k}x^k\) 

Where  \( \bigg(\begin{array}{c} n \ k \end{array}\bigg)\) is called the binomial coefficient of each term in the expansion

\(\left( \begin{array}{c} n \ k \end{array} \right) = \frac{n!}{k!(n-k)!}\)

 

Calculation:

Given  the 4^th term in the expansion of \(\rm (ax + \frac{1}{x})^n \) is \(\frac{5}{2}\) for all x ∈ R 

Let us write the expansion of  \(\rm (ax + \frac{1}{x})^n \) , we have

\((ax+\frac{1}{x})^n= \bigg(\begin{array}{c} n \ 0 \end{array}\bigg)(ax)^{n-0}({\frac{1}{x}})^0 +\bigg(\begin{array}{c} n \ 1 \end{array}\bigg)(ax)^{n-1}({\frac{1}{x}})^1 +\bigg(\begin{array}{c} n \ 2 \end{array}\bigg)(ax)^{n-2}({\frac{1}{x}})^2 +\bigg(\begin{array}{c} n \ 3 \end{array}\bigg)(ax)^{n-3}({\frac{1}{x}})^3 +...+\bigg(\begin{array}{c} n \ n \end{array}\bigg)(ax)^{n-n}({\frac{1}{x}})^n \)

So, the fourth term in the expansion is, 

\( \bigg(\begin{array}{c} n \ 3 \end{array}\bigg)(ax)^{n-3}({\frac{1}{x}})^3\)  

But given that the fourth term is \(\frac{5}{2}\) which is independent of x 

So, the degree of x in the fourth term is n - 3 - 3 = 0

⇒ n - 6 = 0

⇒ n = 6

Now to find the value of an in the fourth term consider the coefficients of x in \( \bigg(\begin{array}{c} n \ 3 \end{array}\bigg)(ax)^{n-3}({\frac{1}{x}})^3\) we get,

\( \bigg(\begin{array}{c} 6 \ 3 \end{array}\bigg)(a)^{6-3}=\frac{5}{2} \)    | given

⇒ \( \frac{6!}{3!(6-3)!}.a^3 = \frac{5}{2}\)       ∵ \(\left( \begin{array}{c} n \ k \end{array} \right) = \frac{n!}{k!(n-k)!}\)

⇒ \( \frac{6!}{3! \ . \ 3!}.a^3 = \frac{5}{2}\)

⇒ \(20.a^3 = \frac{5}{2}\)               ∵  \( \frac{6!}{3! \ .\ 3!} = 20 \)

⇒ \(a^3 = \frac{5}{40}\) 

⇒ \(a^3 = \frac{1}{8}\) 

⇒ \(a = \frac{1}{2}\) 

Thus n = 6 and a = 1/2 

Hence, the correct answer is option 1)

Concept:

When a tangent touches a circle at any point of its circumference then the angle between its radius and the tangent line formed is 90°.

Calculation:

Given O is the center of the circle, PQ is the tangent to the circle at point C, and ∠OAC = 30° = ∠OCA 

OA = OC = radius of the circle

Clearly, AC is a chord of a circle

Since ∠OAC = 30° and Tangent PQ is making an angle of 90° with the radius of a circle we have ∠OCQ = 90°

But the acute angle between AC and the tangent PQ at C is

∠ACQ = ∠OCQ - ∠OCA

= 90° - 30°

= 60°

∠ACQ = 60° is the acute angle between AC and the tangent PQ at C.

Hence, the correct answer is option 1).

Concept:

Let ax2 + bx + c = 0 be the quadratic equation and if α and β are the roots of the quadratic equation then we have  \(\alpha +β = \frac{- b}{a}\) and  \(α.β = \frac{c}{a}\) .

Formula used:

\(tan(A+B)=\frac{tanA+tanB}{1-tanA.tanB} \)

Calculation:

Given quadratic equation x2 + px + q = 0 and 

the roots of the quadratic equation x2 + px + q = 0 are tan 30° and tan 15°

then we have  tan 30° + tan 15° = - p

and  tan 30° . tan 15° = q

Now using the formula \(tan(A+B)=\frac{tanA+tanB}{1-tanA.tanB} \) 

We have \(tan(30°+15°)=\frac{tan30°+tan15°}{1-tan30°.tan15°} \)

⇒ \(tan(45°)=\frac{-p}{1-q} \) 

∵ tan45° = 1

⇒ 1 - q = - p   

⇒ q - p = 1

Add 2 on both sides 

⇒ 2 + q - p = 1 + 2

⇒ 2 + q - p = 3

Hence, the correct answer is option 2)

Concept:

Section formula:

If point P(x, y) divides the line joining the point A(x₁, y₁) and B(x₂, y₂) in the ratio of k : 1

Then coordinates of P(x, y) is

\((\frac{kx_2\ +\ x_1}{k\ +\ 1}, \ \frac{ky_2\ +\ y_1}{k\ +\ 1})\)

Calculation: 

Let required ratio is k : 1

Using the section formula

 \((\frac{kx_2\ +\ x_1}{k\ +\ 1}, \ \frac{ky_2\ +\ y_1}{k\ +\ 1})\)

P(m, 6) = \((\frac{2k\ +\ (-4)}{k\ +\ 1}, \ \frac{8k\ +\ 3}{k\ +\ 1})\)

Comparing the x and y coordinates of point P, 

\(\frac{2k\ +\ (-4)}{k\ +\ 1}\ =\ m\)    -----(1)

\(\frac{8k\ +\ 3}{k\ +\ 1}\) = 6     

⇒ 6k + 6 = 8k + 3

⇒ k = 3/2

From equation (1)

\(\frac{2\ \times \ \frac{3}{2}\ +\ (-4)}{\frac{3}{2}\ +\ 1}\ =\ m\)

⇒ m = -2/5

Hence, k : 1 =  3 : 2 and m = \(\frac{-2}{5}\) respectively.

Formula used:

\(tan(A\ +\ B)\ =\ \frac{tan\ A\ +\ tan\ B}{1\ -\ tan\ A.tan\ B}\)

Calculation:

Given that,

A + B = π/4

⇒ tan(A + B) = tan(π/4)

But we know that,

 \(tan(A\ +\ B)\ =\ \frac{tan\ A\ +\ tan\ B}{1\ -\ tan\ A.tan\ B}\)

\(⇒ \ \frac{tan\ A\ +\ tan\ B}{1\ -\ tan\ A.tan\ B}\ = \ 1\)     [∵ tan(π/4) = 1]

⇒ tan A + tan B = 1 - tan A tan B

⇒ 1 + tan A + tan B + tan A tan B = 2

⇒ 1 + tan A + tan B(1 + tan A) = 2

⇒ (tan A + 1)(tan B + 1) = 2

Hence, option 1 is correct.

Calculation:

\(\frac{11^6}{7}\) = \(\frac{(7\ +\ 4)^6}{7}\)

Hence, remainder of \(\frac{11^6}{7}\) will be same as remainder of  \(\frac{(4)^6}{7}\)

⇒ Remainder of \(\frac{11^6}{7}\) = remainder of \(\frac{64\ \times\ 64}{7}\)

⇒ Remainder of \(\frac{11^6}{7}\) = remainder of  \(\frac{1\ \times\ 1}{7}\) = \(\frac{1}{7}\)

Hence, required remainder will be 1.

Concept:

Standard deviation: A standard deviation (or σ) is a measure of how dispersed the data is in relation to the mean.

It is calculated by using the formula \(σ =\sqrt{\displaystyle\sum_{i=1}^{N} \frac{(x-x_i)^2}{N}} \) 

where, σ  - the standard deviation

x - each entry in the given data

x_i - mean of the given data entries

N - number of entries in the given data.

and \(\displaystyle\sum_{i=1}^{N} \frac{(x-x_i)^2}{N}\) - is the Variance of the given data

Calculation:

So, Given observation data is 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25 

N = 11

So its mean(x_i) = \(\frac{15 + 16 + 17 + 18 + 19 + 20 + 21 + 22 + 23 + 24 + 25}{11} \) = \(\frac{220}{11}\) = 20

∴ x_i = 20

Next,

consider \(\displaystyle\sum_{i=1}^{N} \frac{(x-x_i)^2}{N}\) 

= \(\displaystyle\sum_{i=1}^{11} \frac{(15-20)^2+(16-20)^2+(17-20)^2+(18-20)^2+(19-20)^2+(20-20)^2+(21-20)^2+(22-20)^2+(23-20)^2+(24-20)^2+(25-20)^2}{11}\) 

= \(\displaystyle\sum_{i=1}^{11} \frac{(-5)^2+(-4)^2+(-3)^2+(-2)^2+(-1)^2+(0)^2+(1)^2+(2)^2+(3)^2+(4)^2+(5)^2}{11}\) 

= \(\displaystyle\sum_{i=1}^{11} \frac{25 +16+9+4+1+0+1+4+9+16+25}{11}\)

= \(\frac{110}{11} \) 

= 10

\(\displaystyle\sum_{i=1}^{N} \frac{(x-x_i)^2}{N}=10\) (Variance)

Now take the square root of the variance, we get the required standard deviation i.e.,

\(σ =\sqrt{\displaystyle\sum_{i=1}^{N} \frac{(x-x_i)^2}{N}} =\sqrt{10} \)

\(\sigma =\sqrt{10} \) is the standard deviation of the given data.

Hence, the correct answer is option 1).

Concept:

If P(x1, y1) and Q(x2, y2) are two points in a 2D plane then the distance between the points P(x1, y1) and Q(x2, y2) is given by:

\(PQ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \)

Calculation:

Let, P = (4, p) and Q = (1, 0)

PQ = 5 (Given)

As we know that, the distance between the points P(x1, y1) and Q(x2, y2) is given by:

\(PQ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \)

\(5= \sqrt {{{\left( {1 - 4} \right)}^2} + {{\left( {0 - { p} } \right)}^2}} \)

25 = 9 + p²

p² = 16

p = ±4

Concept:

Dividend = Divisor × Quotient + Remainder

Calculation:

Given:

Divident = 6x3 + 13x2 + x - 2; Divisor = 2x + 1;

Dividend = Divisor × Quotient + Remainder

6x3 + 13x2 + x - 2 = (2x + 1) × (3x² + 5x - 2) + 0

Quotient = 3x2 + 5x - 2, Remainder = 0

Given:

The given equations are

kx + 2y = 5

3x + y = 1

Concept used:

If (a1/a2) ≠ (b1/b2), then the system of equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 has a unique solution.

Calculation:

kx + 2y - 5 = 0      ----(1)

a1 = k, b1 = 2, c1 = -5

3x + y - 1 = 0      ----(2)

a2 = 3, b2 = 1, c2 = -1

According to the question,

(a1/a2) ≠ (b1/b2)

⇒ (k/3) ≠ (2/1)

⇒ 6 ≠ k

⇒ k ≠ 6

∴ The system has a unique solution if k ≠ 6.

Concept:

• A parallelogram is a geometrical figure that has four sides formed by two pairs of parallel lines. In a parallelogram, the opposite sides are equal in length, and opposite angles are equal in measure.
• To find the area of the parallelogram, multiply the base of the perpendicular by its height. i.e., A = b × h

 

Calculation:

Given ABCD is a parallelogram AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm

Area of given parallelogram ABCD = AB × AE = 16 × 8 = 128cm²

Again area of parallelogram ABCD = AD × CF = AD × 10 

or, AD × 10  = 128

⇒ AD = 12.8 cm

Hence, the correct answer is option 1)

Formula used:

Mean: The mean is the arithmetic average of a set of given numbers. It is the ratio of sum of terms and total number of terms.

Sum of n terms of AP: If a is first term and d is common difference of AP, then sum of n terms of AP

\(S_n \ =\ \frac{n}{2}[2a\ +\ (n\ -\ 1)d]\)

Calculation:

we know that first n odd numbers are 

1, 3, 5, ...........n

This form an AP with first term 1 and common difference 2

Sum of n odd natural numbers are 

S_n = 1 + 3 + 5 + ......+ n

\(⇒ \ S_n\ =\ \frac{n}{2}[2\ \times \ 1\ +\ (n\ -\ 1)2]\)

⇒ S_n = n²     -----(1)

Hence, mean of n odd natural number

\(\bar X\ =\ \frac{1\ +\ 2\ +\ 3\ +\ \dots n}{n}\)

\(⇒ \bar X\ =\ \frac{n^2}{n}\)

⇒ X̅ = n

But according to question, mean is \(\frac{n^2}{81}\)

⇒ \(\frac{n^2}{81}\ =\ n\)

⇒ n = 81

Hence option 4 is correct.

Concept:

(a + b) (a - b) = a² - b² 

Rational number: A number which can be written in the form of \(\frac p q\), where q ≠ 0.

Calculation:

Given:

A = \(\sqrt[8]{6} - \sqrt[8]{5} \), B = \(\sqrt[8]{6} + \sqrt[8]{5} \), C = \(\sqrt[6]{6} - \sqrt[6]{5} \), D = \(\sqrt[4]{6} + \sqrt[4]{5} \), E = \(\sqrt{6} + \sqrt{5} \)

We have to check options one by one

option 1

ABDE = (\(\sqrt[8]{6} - \sqrt[8]{5} \)) × (\(\sqrt[8]{6} + \sqrt[8]{5} \)) × (\(\sqrt[4]{6} + \sqrt[4]{5} \)) × (\(\sqrt{6} + \sqrt{5} \))

= \((\sqrt[4]{6} - \sqrt[4]{5} )\) × (\(\sqrt[4]{6} + \sqrt[4]{5} \)) × (\(\sqrt{6} + \sqrt{5} \))

= \((\sqrt[]{6} - \sqrt[]{5} )\) × (\(\sqrt{6} + \sqrt{5} \))

= 6 - 5

= 1 ⇒ rational

Concept:

Median: The median average is the middle number in a set of data, when the data has been written in ascending size order.

Case 1: if n is odd, the median is the \((\frac{n\ +\ 1}{2})^{th}\) observation.

Case 2: if n is even, then the median will be the average of the  \((\frac{n}{2})^{th}\ \ and\ \ (\frac{n\ +\ 1}{2})^{th}\) observations. 

Calculation:

Given terms are \(\rm x + 4,\ x - \frac{7}{2},\ x - \frac{5}{2},\ x - 3,\ x -2,\ x + \frac{1}{2},\ x - \frac{1}{2}, \ x + 5,\ (x > 0)\)

By arranging them in ascending order

\(\rm\ x - \frac{7}{2},\ x - 3, x - \frac{5}{2},\ \ x -2,\ \ x - \frac{1}{2}, x + \frac{1}{2}, x + 4, \ x + 5,\ (x > 0)\)

We can see, there are 8 terms (even)

Hence, will be average of \((\frac{n}{2})^{th}\ \ and\ \ (\frac{n\ +\ 1}{2})^{th}\) term

⇒ Median = \(\frac{(x\ -\ 2)\ +\ x\ -\ \frac{1}{2}}{2}\)

⇒ Median = \(\frac{4x\ -\ 5}{4}\)

⇒ Median = \(\rm x - \frac{5}{4}\)

Hence, option 3 is correct.

Concept:

Let a pair of linear equation in two variable a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

The condition of parallel lines or inconsistent equations.

\(\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \frac{c_1}{c_2}\)

Calculation:

Given equation of lines

3x + 2ky = 2 and 2x + 5y + 1 = 0

a1 = 3; a2 = 2

b1 = 2k; b2 = 5

c1 = -2; c2 = 1

Here, given lines are parallel

So that,

\(\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \frac{c_1}{c_2}\)

\(\therefore \frac{3}{2} = \frac{2k}{{5}}\left( {\because{c_1} \ne {c_2}} \right)\)

∴ \(k=\frac{15}{4}\)

Additional Information

(I) If \(\frac{{{a_1}}}{{{a_1}}} \ne \frac{{{b_1}}}{{{b_2}}}\)

Then the graph will be a pair of lines interesting at a unique point. Which is the solution of the pair of equations.

(II) If 

\(\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{c_1}{{{c_2}}}\) then

Then graph will be a pair of coincident lines

Given:

The ratio of the mode and the median of a distribution is 6 : 5

Concept used:

Empirical formula

Mode = 3 × Median – 2 × Mean

Calculation:

Let the mode be 6x

So, The median = 5x

Mode = 3 × median – 2 × mean

⇒ 6x = 3 × 5x – 2 × mean

⇒ 2 × mean = 15x – 6x

⇒ mean = 4.5x

Now, Mean:Median = 4.5x:5x = 9:10 = 9:10

∴ The Mean:Median is 9:10

Formula used: 

For an AP, with first term 'a' and common difference 'd', n^th term is given by

T_n = a + (n - 1)d

Calculation: 

Given that, first term 'a' and common difference 'd' then

T_m = a + (m - 1)d     

But according to question, T_m = 1/n

⇒ a + (m - 1)d = 1/n        -----(1)

T_n = a + (n - 1)d

But according to question, Tn = 1/m

⇒ a + (n - 1)d = 1/m       -----(2)

Substracting equation (1) and (2), we will get

a + (n - 1)d - [a + (m - 1)d] = 1/m - 1/n

⇒ d(n - m) = \(\frac{n\ -\ m}{mn}\)

⇒ d = 1/mn      

From equation (1) 

a + (m - 1)(1/mn) = 1/n

⇒ a = \(\frac{1}{n}\ -\ \frac{(m \ -\ 1)}{mn}\) 

⇒ a = \(\frac{m\ -\ m\ +\ 1}{mn}\ =\ \frac{1}{mn}\) 

Therefore 

a - d = \(\frac{1}{mn}\ -\ \frac{1}{mn}\ =\ 0\)

Hence, option 1 is correct.

Concept:

Step to find the minimum or maximum value of function:

Step 1: Calculate f'(x)

Step 2: Put the f'(x)  = 0 and find the critical values, let x = a

Step 3: Calculate f''(x)

Step 4: Find f''(a), if it is positive, i.e. function has minima at x = a and if f''(a) is negative, i.e. function has maxima at x = a

Formula used:

For a quadratic equation ax² + bx + c = 0, 

Sum of roots = -b/a

Product of roots = c/a

a² + b² = (a + b)² - 2ab

Calculation:

Given that, x2 - (a - 2)x - a - 1 = 0

Let the root of quadratic equations are α and β 

⇒ α  + β = a - 2 and 

αβ = - (a + 1)

We know that,  a2 + b2 = (a + b)2 - 2ab

⇒ α² + β² = (a - 2)² - 2[-(a + 1)]  

⇒ α2 + β2 = (a - 2)² + 2(a + 1) = f(a)

Differentiating the function f(a) with respect to 'a'

⇒ f'(a) = 2(a - 2) + 2

For minima, f'(a) = 0

⇒ 2(a - 2) + 2 = 0

⇒ a = 1

Hence, option 4 is correct.

Concept:

Slant height of a frustum is given by

l = \(\sqrt {h^2 + (R - r)^2}\)

Where, h = Height of frustum; R = Radius of bigger end; r = Radius of smaller end;

Calculation:

Given:

D = 10 cm; d = 6 cm; h = 5 cm;

R = 5 cm; r = 3 cm;

Using formula

l = \(\sqrt {h^2 + (R - r)^2}\)

l = \(\sqrt {5^2 + (5 - 3)^2} = \sqrt {25 + 4}\)

l = \(\sqrt {29}\)

Formula used:

For quadratic equation ax² + bx + c = 0 

1. Sum of root = -b/a
2. Product of root = c/a
3. (a - b)² = (a + b)² - 4ab
4. (a - b)(a + b) = a² - b²

Calculation:

Let the roots of equation x2 + ax + b = 0 are P and Q

⇒ P + Q = - a         

PQ = b                   

⇒ (P - Q)² = (P + Q)² - 4PQ

⇒ P - Q = \(\sqrt{a^2\ -\ 4b}\)          -----(1)

Similarly let roots of equation x2 + bx + a = 0 are R and S

⇒ R + S = - b          

RS = a                    

⇒ R - S = \(\sqrt{b^2\ -\ 4a}\)       -----(2)

According to question, P - Q and R - S are same. Therefore

\(\sqrt{a^2\ -\ 4b}\) = \(\sqrt{b^2\ -\ 4a}\)

⇒ a² - b² = -4(a - b)

⇒ (a + b)(a - b) = -4(a - b)

⇒ a + b + 4 = 0

Hence, option 1 is correct.

Formula used: 

Volume of cube having side 'a' is given by V = a³

Sum of its side or edges = 12a

Total surface area A = 6a²

Calculation:

Let, each side of cube = a

Hence, sum of its all side = 12a

and volume V = a³

According to question, volume of cube = sum of its edges. Therefore

a³ = 12a

⇒ a(a² - 12) = 0,  a ≠ 0

⇒ a² = 12       -----(1)

Hence, the total surface area

A = 6a²

⇒ A = 6 × 12 = 72        [From (1)]

Hence, total surface area is 72 square units.

Given:

u = a1x + b1y + c1 = 0

v = a2x + b2y + c2 = 0 and

 \(\rm \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\),

Calculation:

u = a1x + b1y + c1 = 0     ----(1)

v = a2x + b2y + c2 = 0     ----(2)

Let λ = \(\rm \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\)  

⇒ a₁ = λa₂, b₁ = λb₂, c₁ = λc₂

⇒ u = a1x + b1y + c1 = λ( a2x + b2y + c2) 

⇒ u = λv       -----(3) 

According to question,

u + kv = 0 

⇒ \(u\ +\ k(\frac{u}{\lambda})\ =\ 0\)         [From equation (3)]

⇒ \(u[1\ +\ k(\frac{1}{\lambda})]\ =\ 0\)

⇒ u = 0

Hence, option 2 is correct.

Concept:

1. If λ₁ and λ₂ are the characteristic roots of square matrix such that |A| ≠ 0, then the characteristic roots of A^-1 are 1/λ₁ and 1/λ₂  
2. If A is non-singular matrix, then, AA^-1 = I 

Calculation:

Let A = \(\begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\)and B = \(\begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}\)

|A| = 3 × 5 - 7 × 2

⇒ |A| = 1 ≠ 0  

⇒ AB = \(\begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\)(\begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}\)

⇒ AB = \(\begin{bmatrix} 15\ -\ 14 & -21\ +\ 21 \\ 10\ -\ 10 & -14\ +\ 15 \end{bmatrix}\)

⇒ AB = \(\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}\) = I

Therefore, B = A^-1

According to question, characteristic roots of A are λ1 and λ2 

From the above concept, characteristic roots of A^-1 are \(\frac{1}{\lambda_1}, \frac{1}{\lambda_2}\)

Hence, option 1 is correct.

Concept:

Distance between two points:

The distance between any two points \(\rm (x_1, y_1)\) and \(\rm(x_2, y_2)\) is given by the distance formula as follows: \(d = \sqrt {(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

The inverse of the Pythagorean theorem:

If a triangle with sides a, b and c satisfies the equation \(\rm c^2=a^2+b^2\) where is the longest side, then the triangle is always right-angled triangle.

(-4, 0), (4, 0), (0, 3)

Calculation:

First we will calculate the distance between all pairs of points from the given pairs.

First consider the points (-4, 0) and (4, 0).

The distance is given as follows:

\(\rm d_1 = \sqrt{(4-(-4))^2+(0-0)^2}\ = \sqrt{64}\ = 8 \)

Now consider the pair (4, 0) and (0, 3).

The distance is given as follows:

\(\rm d_2 = \sqrt{(0-4)^2+(3-0)^2}\= \sqrt{16+9}\= \sqrt {25}= 5\)

Now consider the pair (-4, 0) and (0, 3).

The distance is given as follows:

\(\begin{align*} d_3 &= \sqrt{(0-(-4))^2+(3-0)^2}\ &= \sqrt{16+9}\ &=\sqrt{25} = 5 \end{align*}\)

Therefore, two distances are equal thus, the triangle is isosceles.

Hence, option 3 is correct

Concept: 

Identification of termination decimal:

1  The decimal expansion of the rational number terminates if it can be expressed in form 2^m5ⁿ , where m,n ∈ N.

2. The rational number has a non-terminating recurring decimal expansion if it cannot be expressed in form 2m5n , where m,n ∈ N.

Calculation:

Option 1: \(\frac{17}{2^4 \times 5^2}\)

We can see that, denominator is 2⁴5² which is 2m5n form, hence it has terminating decimal.

Option 2: \(\frac{125}{3^3 \times 7^2}\)

Denominator 3³7², hence it is has non-terminating decimal.

Option 3: \(\frac{68}{2^2 \times 5^2\times 7^2}\)

Since denominator contain 7² along with 2²5², hence it has non-terminating decimal.

Option 4: \(\frac{25}{3^2 \times 2^3}\)

Denominator is not the form of 2^m5ⁿ, hence it has non-terminating decimal.

Concept:

Base 5 number: Any number which contain digit b/w 0 to 5, i.e. 0, 1, 2, 3, 4 called base 5 number.

Base 10 number: In base 10, each digit of a number can have an integer value ranging from 0 to 9 (10 possibilities) depending on its position. 

Step of conversion of base 5 to base 10 number:

Step-1  Let the number is 'abc.de'  in the number left to the point, multiply with the increasing power of 5 and starting from 5⁰.

On the right side of the point, multiply with decreasing power of 5. For example

a × 5² b × 5¹ c × 5⁰  d × 5^-1 e × 5^-2

Step-2 Add all numbers and the result will be the base 10 numbers.

a × 52 + b × 51 + c × 50 + d × 5-1 + e × 5-2

calculation:

As per above discussion,

(4312)5 =  (4 × 5³ + 3 × 5² + 1× 5¹ + 2 × 5⁰)₁₀

⇒ (4312)5 =  (500 + 75+ 5 + 2)10

⇒ (4312)5 =  (582)10

Hence (4312)₅ can be written as (582)₁₀

Concept:

• A straight line is a figure formed when two points A(x₁, y₁) and B(x₂, y₂) are connected with minimum distance between them, and both the ends extended to infinity.
• The equation of a straight line is given by (y - y₁) = m(x - x₁) where m is the slope of the line.
• The general form of a straight line is given by ax+by=cax+by=c.

 

​Calculation:

Given lines are 

x - y + 1 = 0     and

3x + y - 5 = 0   and 

The equation of the straight line passing through the point of intersection of x - y + 1 = 0  and 3x + y - 5 = 0 is (x - y + 1) + λ (3x + y - 5) = 0 for some λ 

i.e., x(3λ + 1) + y(λ - 1) -(5λ  - 1) = 0   .........(1)

from (1) we get the slope of the line ,m = \(\frac{3λ +1}{λ -1}\)

and it is perpendicular to 3x + y - 5 = 0

∴ \(- 3 × \Big(-\frac{3λ+1}{λ -1}\Big)\)  = - 1 

⇒ \(λ =\frac{-1}{5}\)

On substituting λ value in equation (1) we get,

x - 3y + 5 = 0 as the equation of required line.

Hence, the correct answer is option 2).

Concept:

Poisson distribution is given by:

\(P(x) = \frac{(e^{-λ} λ^x)}{x!}\)

Where x = 0, 1, 2, 3 ---- and λ = mean

Calculation:

Given:

λ = 3; 

∵ for only x = 3 value of |X - 3| < 1

∴ P(|X-3| < 1) = P(x = 3)

Using

P(x) = \(\frac {e^{-λ } λ ^x}{x!}\)

P(x = 3) = \(\frac {e^{-3 } \ 3 ^3}{3!}\)

P(x = 3) = \(\frac {e^{-3 }\ \times \ 27}{6}\)

P(x = 3) = \(\frac {9\ e^{-3 }}{2}\)

Concept:

Greatest Integer Function: (Floor function)

The function f (x) = [x] is called the greatest integer function and means greatest integer less than or equal to x i.e [x] ≤ x.

Domain of [x] is R and range is I.

Fractional part function: It is defined as difference between a number and its integral value.

{f} = f - [f]

The binomial expansion of (x + y)n:

(x+ y)n = nC0( xn) + nC1 (xn - 1) y + nC(xn- 2). y2 +....... + nCnyn 

Calculation: 

Given that

R = (5√5 + 11)2n+1      -----(1)

According to question, f is a fractional part of R

⇒ f = R - [R]        ------(2)

Let f₁ =  (5√5 - 11)2n+1    where 0 < f₁ < 1    

R - f₁ =  (5√5 + 11)2n+1 -  (5√5 - 11)2n+1     

[R] + f - f1 =  (5√5 + 11)2n+1 -  (5√5 - 11)2n+1        ------(3)

Expand this Binomially and substract both terms, we will get

(5√5 + 11)2n+1 - (5√5 - 11)2n+1  =

2[^{2n + 1}C₁(5√5)²ⁿ (11)+ 2n + 1C₃(5√5)2n-2(11)³ + ....... +2n +1C_2n+1(11)²ⁿ⁺¹] 

We can see that, √5 has power of multiple 2 hence, 

[R] + f - f1 = even positive integer

We know that, [R] is integer, therefore to make the  [R] + f - f1 integer, 

f - f1 = 0 ⇒ f = f₁

Therefore, 

Rf = Rf₁ =  (5√5 + 11)2n+1.(5√5 - 11)2n+1

⇒ Rf = 4^{2n+ 1}

Hence, option 1 is correct.

Given:

The polynomial = x3 + 4x2 - px + 8

The polynomial is divisible by (x - 2)

Concept used:

If polynomial P(x) is divided by x - a, then remainder is P(a).

If the remainder P(a) is 0, then P(x) is divisible by x - a.

Calculation:

Here, P(x) = x3 + 4x2 - px + 8

If x - 2 = 0 then x = 2

So, remainder = P(2)

But according to the question, P(2) = 0

⇒ 23 + 4 × 22 - p × 2 + 8 = 0

⇒ 8 + 16 - 2p + 8 = 0

⇒ k = 16

∴ The value of K is 16.

Concept: 

Pythagoras Theorem:

In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides.

Calculation:

In a Δ AOP 

\(tan60^∘\ =\ \frac{OP}{OA}\)

\(⇒ \ √3\ =\ \frac{OP}{OA}\)              (∵ tan60∘ = √3) 

\(⇒ \ OA\ =\ \frac{OP}{√ 3}\)       ------(1)

Similarly, in ΔBOP

\(tan30^∘\ =\ \frac{OP}{OB}\)

\(⇒\ \frac{1}{√ 3}=\ \frac{OP}{OB}\)                 (∵ tan30^∘ = 1/√3) 

⇒ OB = OP√3       -----(2)         

∵  AB = 3m, Therefore

in an Δ AOB, using Pythagoras' theorem

OB2 = AO2 + AB2

\(⇒ \ 3^2\ +\ (\frac{OP}{√ 3})^2\ =\ (OP√ 3)^2 \)

\(⇒ \ 3^2\ =\ -(\frac{OP}{√ 3})^2\ +\ (OP√ 3)^2 \)

\(⇒ \ 9\ =\ \frac{8}{ 3}​​​​(OP)^2\)

⇒ (OP) = \(\sqrt{\frac{27}{8}} \) = ​\(\rm \frac{3\sqrt 3}{2\sqrt2} m\) = \(\rm \frac{3\sqrt 6}{4} m\)

​

Concept:

If a vector belongs to the linear span of S then the determinant of vectors in S is zero.

Calculation:

\(\left| {\begin{array}{*{20}{c}} -1&0&1\2&1&4\ 3k+2&3&10 \end{array}} \right|\) = 0

• 1(10 - 12) - 0 + 1(6-3k-2) = 0 ⇒ 2 + 4 - 3k = 0 ⇒ 6 = 3k

⇒ k = 2

Hence, the correct answer is option 3)

Formula used:

1. Area of sector

A = \((\frac{θ}{360^\circ})π R^2\)

Where,

θ = angle of the sector, R = radius of the circle

2. Curved surface area of a cone

A= πrl

Where,

r and l are radius and slant height of cone respectively.

Calculation:

Given that,

Radius of circle R = 15 cm

The angle of sector or cone = 120°

Let height and radius of the cone be h, and r respectively. 

The curved surface area of the cone = Area of the sector of a circle.

πrl  = \(π R^2(\frac{120^\circ}{360^\circ})\)

But slant height l = R

⇒ r = 5 cm

In an ΔOAB, using Pythagoras theorem,

15² = h² + 5²

⇒ h = 10√2 

Hence, height of cone will be 10√2.

Given:

The given number is 105000

Concept:

Let K be the composite number, then

K = x^a × y^b × z^c

Number of total factors = (a + 1) × (b + 1) × (c + 1)

Calculation:

10500 = 2² × 3¹ × 5³ × 7¹

⇒ Number of total factors (divisors) = (2 + 1) × (1 + 1) × (3 + 1) × (1 + 1) = 3 × 2 × 4 × 2 = 48

∴ Number of factors except 1 and itself = 48 – 2 = 46

Concept:

Sum of angle of a triangle  = π  

cos X = \(2\ \cos^2 \frac {X}{2} -1 \)

sin X = \(2 \ \sin \frac {X}{2} \ \cos \frac {X}{2}\)

cos (X + Y) = cos X cos Y - sin X sin Y

Calculation:

Given:

cos 3A + cos 3B + cos 3C = 1         ----(1)

A + B + C = π 

multiply both sides with 3

3A + 3B + 3C = 3π 

3C = 3π - (3A + 3B)

Taking cosine

cos 3C = cos (3π - (3A + 3B))

cos 3C = - cos (3A + 3B)

Using equation (1), We can write it as;

1 - (cos 3A + cos 3B) = - cos (3A + 3B)

cos 3A + cos 3B - cos (3A + 3B) = 1

cos 3A + cos 3B - cos 3A cos 3B + sin 3A sin 3B = 1

cos 3A [1 - cos 3B] + cos 3B + sin 3A sin 3B = 1

cos 3A [1 - cos 3B] + cos 3B - 1 + sin 3A sin 3B = 0

cos 3A (1 - cos 3B) -1 (1 - cos 3B) + sin 3A sin 3B = 0

• (1 - cos 3B)(1 - cos 3A) + sin 3A sin 3B = 0

sin 3A sin 3B = (1 - cos 3B)(1 - cos 3A)

\(\frac {\sin 3A}{1-\cos 3B} = \frac {1-\cos3B}{\sin 3B}\)

\(\frac {2 \ \sin \frac {3A}{2} \cos \frac {3A}{2}}{1+2\ \cos^2 \frac {3A}{2} -1} = \frac {1+2\ \cos^2 \frac {3B}{2} -1}{2 \ \sin \frac {3B}{2} \cos \frac {3B}{2}}\)

\(\frac {2 \ \sin \frac {3A}{2} }{2\ \cos \frac {3A}{2} } = \frac {2\ \cos \frac {3B}{2} }{2 \ \sin \frac {3B}{2}}\)

\(\tan \frac {3A}{2} = \cot \frac {3B}{2}\)

\(\tan \frac {3A}{2} = \tan \left [ \frac {π }{2} - \frac {3B}{2} \right ]\)

\(\frac {3A}{2} = \frac {π - 3B}{2}\)

3A + 3B = π 

A + B = π / 3

⇒ C = π - π / 3

C = 2π / 3 = 120°

Given:

a, b, c are positive real numbers such that a + b + c = p.

Concept used:

Arithmetic mean is always greater than or equal to geometric mean

Calculation:

Here,

We take, 

(p - a) (p - b) (p - c) 

A.M≥ G.M

⇒ {(p - a) + (p - b) + (p - c)}/3≥ [(p - a) (p - b) (p - c) ]^1/3

⇒ [3p - (a + b + c)]/3≥ [(p - a) (p - b) (p - c) ]1/3

⇒ (3p - p)/3 ≥ [(p - a) (p - b) (p - c) ]1/3

⇒ (2p/3)³≥ (p - a) (p - b) (p - c) 

⇒ [(p - a) (p - b) (p - c) ] ≤ 8p³/27

∴ Option 2 is correct.

Concept:

A recurrence relation is an equation that recursively defines a sequence or multidimensional array of values, once one or more initial terms of the same function are given; each further term of the sequence or array is defined as a function of the preceding terms of the same function.

Calculation:

We have

 \(T(2^k)=3T(2^{k-1})+1\) 

\(=3^2T(2^{k-2})+1+3\)

\(=3^3T(2^{k-3})+1+3+9\)

................ k steps of recursion

\(=3^kT(2^{k-k})+(1+3+9+27+...+3^{k-1})\)

\(= 3^k+(\frac{3^k-1}{2})\)

\(= (\frac{3^{k+1}-1}{2})\)

∴ The solution to the recurrence equation \(T(2^k)=3T(2^{k-1})+1\) is \((\frac{3^{k+1}-1}{2})\)

Hence, the correct answer is option 3)

Given-

A wire is in the shape of a circle of radius 21 cm

Formula Used- 

The perimeter of a Square = 4 × Side

The perimeter of a Circle = 2πr

Calculation- 

The radius of the circle be 21 cm

According to the question-

Perimeter of Square = Perimeter of Circle

⇒ 4 × Side = 2 × (22/7) × 21

⇒ Side = 33 cm

∴ The radius of the circle is 33 cm.