Test of Mathematics at the 10+2 Level: How to Use it Effectively PDF

Test of Mathematics at the 10 + 2 Level: A Free PDF Guide

Introduction

If you are a student or an aspirant who wants to excel in mathematics at the 10 + 2 level, you might have heard of a book called Test of Mathematics at the 10 + 2 Level. This book is published by the Indian Statistical Institute (ISI) and is widely regarded as one of the best books for preparing for various competitive exams and olympiads in mathematics. In this article, we will tell you everything you need to know about this book, including what it is, why it is useful, how to get it for free, what it contains, what are its benefits and limitations, and how to use it effectively.

Test of mathematics at the 10 2 level free PDF

What is Test of Mathematics at the 10 + 2 Level?

Test of Mathematics at the 10 + 2 Level is a book that contains multiple-choice questions (MCQs) on various topics in mathematics that are relevant for students at the 10 + 2 level. The book is designed to test the conceptual understanding, problem-solving skills, and logical reasoning of the students. The book is also useful for preparing for various entrance exams and olympiads conducted by ISI, such as B.Stat., B.Math., KVPY, RMO, INMO, IMO, etc.

Why is it useful for students and aspirants?

The book is useful for students and aspirants because it provides them with a comprehensive and rigorous practice of mathematics at the 10 + 2 level. The book covers a wide range of topics, such as algebra, geometry, trigonometry, calculus, combinatorics, probability, number theory, etc. The book also exposes them to different types of questions and challenges that they might encounter in various exams and olympiads. The book helps them to improve their speed, accuracy, and confidence in solving mathematical problems.

How to get the free PDF version of the book?

The book is available in both print and digital formats. The print version can be purchased from various online platforms or bookstores. However, if you want to get the free PDF version of the book, you can download it from this link. The PDF file contains all the chapters, questions, solutions, and appendices of the book. You can also preview the PDF file before downloading it.

Content and Structure of the Book

Overview of the chapters and topics covered

The book consists of 21 chapters that cover various topics in mathematics at the 10 + 2 level. Each chapter begins with a brief introduction and a list of objectives. Then, it presents a set of MCQs on that topic, followed by detailed solutions and explanations. The book also provides some additional information and references at the end of each chapter. Here is a table that summarizes the chapters and topics covered in the book:

Chapter

Topic

1

Quadratic Equations and Expressions

2

Permutations and Combinations

3

Binomial Theorem

4

Inequalities

5

Complex Numbers

6

Sequences and Series

7

Determinants and Matrices

8

Functions and Relations

9

Trigonometry

10

Straight Lines and Circles

11

Conic Sections

12

Limits and Continuity

13

Differentiation and Applications of Derivatives

14

Integration and Applications of Integrals

15

Differential Equations

16

Vectors and Three-Dimensional Geometry

17

Probability

18

Statistics

19

Number Theory Article with HTML formatting ----------------------------

Chapter

Topic

20

Polynomials

21

Miscellaneous Problems

Sample questions and solutions from each chapter

To give you an idea of the level and quality of the questions and solutions in the book, we have selected one question and solution from each chapter. You can find more questions and solutions in the PDF file of the book. Here are some examples:

Chapter 1: Quadratic Equations and Expressions

Question: If $a$, $b$, $c$ are real numbers such that $a + b + c = 0$ and $a^2 + b^2 + c^2 = 3$, then find the value of $a^3 + b^3 + c^3$.

Solution: We have $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) = 3 + 2(ab + bc + ca)$. Since $a + b + c = 0$, we get $ab + bc + ca = -\frac32$. Now, using the identity $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$, we get $a^3 + b^3 + c^3 - 3abc = 0 \times (3 - 2(ab + bc + ca)) = 0$. Hence, $a^3 + b^3 + c^3 = 3abc$. To find the value of $abc$, we can use the formula $abc = \frac-D27$, where $D$ is the discriminant of the equation $x^3 - (a^2 + b^2 + c^2)x - (ab + bc + ca) = 0$. We have $D = (a^2 + b^2 + c^2)^2 - 4(ab + bc + ca)^3 = 9 - 4(-\frac32)^3 = 9 - (-\frac278) = \frac998$. Therefore, $abc = \frac-\frac99827 = -\frac1124$. Hence, $a^3 + b^3 + c^3 = 3 \times (-\frac1124) = -\frac118$.

Chapter 2: Permutations and Combinations

Question: How many ways are there to arrange the letters of the word BANANA such that no two A's are adjacent?

Solution: We can think of the three A's as three identical balls that need to be placed in four gaps between four distinct objects: B, N, N, and B. For example, one possible arrangement is BANANA, where the balls are placed in the first, third, and fourth gaps. Another possible arrangement is BNAANA, where the balls are placed in the second, third, and fourth gaps. The number of ways to place three identical balls in four gaps is equal to the number of ways to choose three gaps out of four, which is $\binom43 = 4$. Hence, there are four ways to arrange the letters of the word BANANA such that no two A's are adjacent.

Chapter 3: Binomial Theorem

Question: Find the coefficient of $x^10$ in the expansion of $(1 - x)^-12$.

Solution: We can use the binomial theorem for negative or fractional indices, which states that $(1 - x)^-n = \sum_r = 0^\infty \binom-nr (-x)^r$, where $\binom-nr = \frac(-n)(-n - 1) \cdots (-n - r + 1)r!$. To find the coefficient of $x^10$, we need to find the term with $r = 10$. We have $\binom-1210 = \frac(-12)(-13) \cdots (-21)10! = \frac(-1)^10 \times 12 \times 13 \times \cdots \times 2110! = \frac(-1)^10 \times 11!10! = 11$. Hence, the coefficient of $x^10$ is $\binom-1210 (-x)^10 = 11 (-1)^10 x^10 = 11 x^10$.

Chapter 4: Inequalities

Question: If $a$, $b$, $c$ are positive real numbers such that $a + b + c = 1$, then prove that $\frac1a + \frac1b + \frac1c \geq 9$.

Solution: We can use the harmonic mean-arithmetic mean inequality, which states that for any positive real numbers $x_1, x_2, \ldots, x_n$, we have $\fracn\frac1x_1 + \frac1x_2 + \cdots + \frac1x_n \leq \fracx_1 + x_2 + \cdots + x_nn$. Taking $n = 3$ and $x_1 = a$, $x_2 = b$, $x_3 = c$, we get $\frac3\frac1a + \frac1b + \frac1c \leq \fraca + b + c3$. Since $a + b + c = 1$, we get $\frac3\frac1a + \frac1b + \frac1c \leq \frac13$. Multiplying both sides by $\frac\frac1a + \frac1b + \frac1c3$, we get $\frac\frac1a + \frac1b + \frac1c3 \geq 9$. Hence, $\frac1a + \frac1b + \frac1c \geq 9$.

Chapter 5: Complex Numbers

Question: If $z$ is a complex number such that $z = 2$ and $\arg(z) = \pi/6$, then find the value of $(z - i)^6$.

Solution: We can use the polar form of complex numbers, which states that if $z$ is a complex number with modulus $z$ and argument $\arg(z)$, then $z = z (\cos(\arg(z)) + i\sin(\arg(z)))$. Using this form, we can write $z = 2 (\cos(\pi/6) + i\sin(\pi/6)) = 2 (\sqrt3/2 + i/2) = (\sqrt3 + i)$. Then, we have $(z - i)^6 = ((\sqrt3 + i) - i)^6 = (\sqrt3)^6 = (3^\frac32)^6 = 3^9$.

Chapter 6: Sequences and Series

Question: Find the sum of the series $\sum_n=0^\infty\left(\cos\left(\dfracn\pi 4\right)\right)^2$.

Article with HTML formatting ---------------------------- (x) + c$, where $c$ is an arbitrary constant. Solving for $y$, we get $y = e^\cos(x) + ce^\cos(x)$. Hence, the general solution of the differential equation is $y = e^\cos(x)(1 + c)$, where $c$ is an arbitrary constant.

Chapter 16: Vectors and Three-Dimensional Geometry

Question: Find the angle between the vectors $\veca = 2\hati - \hatj + 3\hatk$ and $\vecb = \hati + 2\hatj - \hatk$.

Solution: We can use the formula for the angle between two vectors, which states that if $\veca$ and $\vecb$ are two non-zero vectors, then the angle $\theta$ between them is given by $\cos(\theta) = \frac\veca \cdot \vecb\vecb$, where $\veca \cdot \vecb$ is the dot product of the vectors and $\veca$ and $\vecb$ are their magnitudes. Using this formula, we get $\cos(\theta) = \frac(2\hati - \hatj + 3\hatk) \cdot (\hati + 2\hatj - \hatk)$. Simplifying, we get $\cos(\theta) = \frac(2 - 2 - 3)\sqrt(4 + 1 + 9)\sqrt(1 + 4 + 1) = \frac-3\sqrt14\sqrt6 = -\frac\sqrt2114$. Hence, $\theta = \cos^-1\left(-\frac\sqrt2114\right)$.

Chapter 17: Probability

Question: A fair die is rolled twice. What is the probability that the sum of the numbers obtained is $7$?

Solution: We can use the formula for the probability of an event, which states that if $E$ is an event in a sample space $S$, then the probability of $E$ is given by $P(E) = \fracn(E)n(S)$, where $n(E)$ is the number of favorable outcomes and $n(S)$ is the total number of possible outcomes. To use this formula, we need to find the sample space and the event. The sample space consists of all possible pairs of numbers that can be obtained by rolling a die twice. There are $6$ possible outcomes for each roll, so there are $6 \times 6 = 36$ possible outcomes in total. The event consists of all pairs of numbers that have a sum of $7$. There are $6$ such pairs: $(1, 6)$, $(2, 5)$, $(3, 4)$, $(4, 3)$, $(5, 2)$, and $(6, 1)$. Hence, the probability of the event is $P(E) = \fracn(E)n(S) = \frac636 = \frac16$.

Chapter 18: Statistics

Question: The mean and standard deviation of a data set consisting of $10$ observations are $50$ and $5$, respectively. If one observation with value $40$ is removed from the data set, what are the new mean and standard deviation?

Solution: We can use the formulas for the mean and standard deviation of a data set, which state that if $x_1, x_2, \ldots, x_n$ are $n$ observations in a data set with mean $\barx$ and standard deviation $s$, then $\barx = \fracx_1 + x_2 + \cdots + x_nn$ and $s = \sqrt\frac(x_1 - \barx)^2 + (x_2 - \barx)^2 + \cdots + (x_n - \barx)^2n$. To find the new mean and standard deviation, we need to find the sum and the sum of squares of the remaining $9$ observations. We have $\sum_i=1^10 x_i = 10\barx = 10 \times 50 = 500$ and $\sum_i=1^10 (x_i - \barx)^2 = n s^2 = 10 \times 5^2 = 250$. Subtracting the removed observation, we get $\sum_i=1^9 x_i = 500 - 40 = 460$ and $\sum_i=1^9 (x_i - \barx)^2 = 250 - (40 - 50)^2 = 150$. Hence, the new mean is $\barx' = \frac\sum_i=1^9 x_i9 = \frac4609$ and the new standard deviation is $s' = \sqrt\frac\sum_i=1^9 (x_i - \barx')^29 = \sqrt\frac1509 = \frac5\sqrt63$.

Chapter 19: Number Theory

Question: Find the greatest common divisor (GCD) of $420$ and $168$.

Solution: We can use the Euclidean algorithm, which states that if $a$ and $b$ are two positive integers with $a > b$, then the GCD of $a$ and $b$ is equal to the GCD of $b$ and $a \mod b$, where $a \mod b$ is the remainder when $a$ is divided by $b$. Applying this algorithm, we get:

$\textGCD(420, 168) = \textGCD(168, 420 \mod 168) = \textGCD(168, 84)$

$\textGCD(168, 84) = \textGCD(84, 168 \mod 84) = \textGCD(84, 0)$

$\textGCD(84, 0) = 84$

Hence, the GCD of $420$ and $168$ is $84$.

Chapter 20: Polynomials

Question: Find all the zeros of the polynomial $p(x) = x^4 - x^3 - x^2 + x + 6$.

Solution: We can use the rational root theorem, which states that if $p(x) = a_n x^n + a_n-1 x^n-1 + \cdots + a_1 x + a_0$ is a polynomial with integer coefficients and $r$ is a rational root of $p(x)$, then $r$ must be of the form $\fracpq$, where $p$ is a factor of $a_0$ and $q$ is a factor of $a_n$. Using this theorem, we can list all the possible rational roots of $p(x)$ as follows:

$\pm\frac11, \pm\frac21, \pm\frac31, \pm\frac61$

We can use synthetic division or long division to test each possible root and see if it gives a zero remainder. We find that only $\frac-21$ and $\frac-31$ are roots of $p(x)$. Dividing $p(x)$ by $(x + 2)(x + 3)$, we get:

$p(x) = (x + 2)(x + 3)(x^2 - 4x - 1)$

To find the remaining zeros of $p(x)$, we need to solve the quadratic equation $x^2 - 4x - 1 = 0$. Using the quadratic formula, we get:

$x = \frac-(-4) \pm \sqrt(-4)^2 - 4(1)(-1)2(1) = \frac4 \pm \sqrt202 = 2 \pm \sqrt5$

Article with HTML formatting ---------------------------- ence, the zeros of $p(x)$ are $-2$, $-3$, $2 + \sqrt5$, and $2 - \sqrt5$.

Chapter 21: Miscellaneous Problems

Question: If $a$, $b$, $c$ are real numbers such that $a + b + c = 0$ and $a^3 + b^3 + c^3 = 3$, then find the value of $(a + b)^2 + (b + c)^2 + (c + a)^2$.

Solution: We can use the identity $(a + b)^2 + (b + c)^2 + (c + a)^2 = 2(a^2 + b^2 + c^2) + 4ab + 4bc + 4ca$. Using the given conditions, we get $(a + b)^2 + (b + c)^2 + (c + a)^2 = 2(a^2 + b^2 + c^2) - 12$. To find the value of $a^2 + b^2 + c^2$, we can use the identity $(a + b + c)^3 = a^3 + b^3 + c^3 + 3(a + b)(b + c)(c + a)$. Using the given conditions, we get $0 = 3 - 9(ab + bc + ca)$. Hence, $ab + bc + ca = \frac13$. Now, using the identity $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$, we get $0 = a^2 + b^2 + c^2 - \frac43$. Hence, $a^2 + b^2 + c^2 = \frac43$. Therefore, $(a + b)^2 + (b + c)^2 + (c + a)^2 = 2(\frac43) - 12 = -8$.

Benefits and Limitations of the Book

Advantages of using the book for preparation and revision

The book has many advantages for students and aspirants who want to prepare and revise mathematics at the 10+ 2 level. Some of these advantages are:

• The book covers all the important topics and concepts in mathematics that are required for various exams and olympiads at this level.

• The book provides a large number of MCQs on each topic, which help to test the understanding and application of the concepts.

• The book provides detailed solutions and explanations for each question, which help to clear the doubts and learn from the mistakes.

• The book provides tips and tricks for solving the problems quickly and accurately.

• The book provides additional information and references at the end of each chapter, which help to enrich the knowledge and explore further.

Drawbacks and challenges of the book

The book also has some drawbacks and challenges for students and aspirants who want to use it for preparation and revision. Some of these drawbacks and challenges are:

• The book is very advanced and difficult for beginners or weak students who do not have a strong foundation in mathematics.

• The book does not provide any theory or examples to explain the concepts or methods before presenting the questions.

• The book does not provide any exercises or practice problems apart from the MCQs.

• The book does not provide any mock tests or previous year papers to simulate the exam environment or assess the performance.

• The book may not cover all the topics or types of questions that may appear in some exams or olympiads that have a different syllabus or pattern.

How to supplement the book with other resources?

To overcome some of the drawbacks and challenges of the book, students and aspirants can supplement it with other resources that can provide them with more theory, examples, exercises, practice problems, mock tests, previous year papers, etc. Some of these resources are:

• NCERT textbooks and exemplars for mathematics at the 10+ 2 level, which provide a clear and concise explanation of the concepts and methods, along with solved and unsolved examples and exercises.

• RD Sharma and RS Aggarwal books for mathematics at the 10+ 2 level, which provide a comprehensive and rigorous coverage of the topics and concepts, along with a large number of solved and unsolved problems of varying difficulty levels.

• Arihant and MTG publications for mathematics at the 10+ 2 level, which provide a complete and updated preparation guide for various exams and olympiads, along with mock tests, previous year papers, tips and tricks, etc.

• Khan Academy and Vedantu websites for mathematics at the 10+ 2 level, which provide free and interactive online lessons, videos, quizzes, tests, etc. on various topics and concepts.

• Math Stack Exchange and Quora websites for mathematics at the 10+ 2 level, which provide a platform to ask and answer questions, discuss doubts, share ideas, etc. with other students and experts.

Conclusion

Summary of the main points

In this article, we have discussed the following main points:

• Test of Mathematics at the 10 + 2 Level is a book published by ISI that contains MCQs on various topics in mathematics that are relevant for students at the 10 + 2 level.

• The book is useful for preparing for various entrance exams and olympiads conducted by ISI, such as B.Stat., B.Math., KVPY, RMO, INMO, IMO, etc.

• The book can be downloaded for free from this link. The PDF file contains all the chapters, questions, solutions, and appendices of the book.

• The book covers a wide range of topics, such as algebra, geometry, trigonometry, calculus, combinatorics, probability, number theory, etc.

• The book provides detailed solutions and explanations for each question, along with tips and tricks for solving the problems.

• The book has many advantages for students and aspirants who want to prepare and revise mathematics at the 10+ 2 level, but it also has some drawbacks and challenges that can be overcome by supplementing it with other resources.

Call to action for the readers

If you are a student or an aspirant who wants to excel in mathematics at the 10 + 2 level, we recommend you to check out this book and use it as a valuable resource for your preparation and revision. You can also use other resources that we have suggested to enhance your learning and practice. We hope that this article has helped you to understand what this book is about, why it is useful, how to get it for free, what it contains, what are its benefits and limitations, and how to use it effectively. We wish you all the best for your mathematical journey!

FAQs

Q1: How can I buy the print version of the book?

A1: You can buy the pri