NCERT Exemplar Class 12 Maths Solutions Chapter 1 Relations and Functions 2021

Q: Are these solutions helpful for board examinations?

Yes, NCERT exemplar solutions for Class 12 Maths chapter 1 are designed well to prepare you for board examinations.

Q: What are the important topics of this chapter?

The important topics of this NCERT Exemplar Class 12 Maths solutions chapter 1 include Types of Relations, Types of Functions, Binary Operations and Composition of Functions and Invertible Function.

Q: How to remember the formulae well?

The best ways to remember formulae is to keep revising them and solving related problems. Other methods include reading them on a daily basis. You can also prepare charts to stick near your study table or bed to go through in your leisure time.

Q: How many questions are there in this chapter?

There are a total 70 questions in NCERT Exemplar Class 12 Maths solutions chapter 1 based on different concepts.

NCERT Exemplar Class 12 Maths Solutions Chapter 1 Relations and Functions

Edited By Ravindra Pindel | Updated on Sep 15, 2022 05:48 PM IST | #CBSE Class 12th

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NCERT Exemplar Class 12 Maths solutions chapter 1 is crucial to building a base from ground up for the entire academic session. We here will help you to find the solutions of all NCERT questions of the chapter with complete steps and detailed answers. This chapter of NCERT Class 12 Maths Solutions is divided into two parts; relations and functions. We have solutions for questions from the parts explained and solved in detail. NCERT Exemplar Class 12 Maths solutions chapter 1 PDF download is to be available for students for extending learning of the concepts.

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NCERT Exemplar Class 12 Maths Solutions Chapter 1: Exercise-1.3
Subtopics Covered in NCERT Exemplar Solutions for Class 12 Maths Chapter 1
What student will learn in NCERT Exemplar Class 12 Maths Solutions Chapter 1?
NCERT Exemplar Class 12 Maths Solutions
Important Topics To Cover From NCERT Exemplar Class 12 Maths Solutions Chapter 1

NCERT Exemplar Class 12 Maths Solutions Chapter 1: Exercise-1.3

Question:1

Let A = {a, b, c} and the relation R be defined on A as follows:

R = {(a, a), (b, c), (a, b)}.

Then, write the minimum number of ordered pairs to be added in R to make R reflexive and transitive.

Answer:

Here, R = {(a, a), (b, c), (a, b)}

The minimum number of ordered pairs to be added to make R as reflexive is (b, b) and (c, c) to R. Whereas, to make R transitive, the minimum number of ordered pairs to be added is (a, c) to R.

Therefore, we need 3 ordered pairs to add with R to make it reflexive and transitive.

Question:2

Let D be the domain of the real-valued function f defined by $f(x) = \sqrt {25 - x\textsuperscript{2}}$ . Then, write D.

Answer:

$f(x) = \sqrt {25 - x\textsuperscript{2}}$
Therefore, the function can be defined as:
$\\25 - x\textsuperscript{2} \geq 0\\ or, x\textsuperscript{2} \leq 25\\ -5 \leq x \leq 5\\$

Therefore, the given function has a domain of [-5, 5].

Question:3

Let f, g: $R \rightarrow R$ be defined by f(x) = 2x + 1 and $g (x) = x\textsuperscript{2} - 2, \forall x \in R,$ respectively. Then, find gof.

Answer:

Here,
f(x) = 2x + 1
$g (x) = x\textsuperscript{2} -2, \forall x \in R\\$
Therefore, $gof= g(f(x)) = g (2x + 1) = (2x + 1)\textsuperscript{2} -2 = 4x\textsuperscript{2} + 4x + 1 -2 = 4x\textsuperscript{2} + 4x - 1\\$

Question:4

. Let f: $R \rightarrow R$ be the function defined by $f (x) = 2x - 3, \forall x \in R$ . write $f\textsuperscript{-1}$ .

Answer:

Here, $f (x) = 2x - 3, \forall x \in R\\$
Let’s say,
$y = 2x - 3\\ or, x = (y + 3)/ 2$

Therefore,
$f\textsuperscript{-1}(x) = (x + 3)/ 2\\$

Question:5

If $A = \{ a, b, c, d \}$ and the function $f = \{ (a, b), (b, d), (c, a), (d, c) \}$ , write $f\textsuperscript{-1}$ .

Answer:
Here, $A = \{ a, b, c, d \}$ and the function $f = \{ (a, b), (b, d), (c, a), (d, c) \} \\$
Therefore, $f\textsuperscript{-1} = \{ (b, a), (d, b), (a, c), (c, d) \} \\$

Question:6

If $f:R \rightarrow R$ is defined by $f (x) = x\textsuperscript{2} - 3x + 2$ , write $f (f (x))$ .

Answer:

Here, $f (x) = x\textsuperscript{2} - 3x + 2$
Therefore,
$\\ f(f(x)) = f(x\textsuperscript{2} - 3x + 2)\\ = (x\textsuperscript{2} - 3x + 2)\textsuperscript{2} - 3(x\textsuperscript{2} - 3x + 2) + 2,\\ = x\textsuperscript{4} + 9x\textsuperscript{2} + 4 - 6x\textsuperscript{3} + 4x\textsuperscript{2} - 12x - 3x\textsuperscript{2} + 9x - 6 + 2\\ = x\textsuperscript{4} - 6x\textsuperscript{3} + 10x\textsuperscript{2} - 3x\\$
Similarly,
$\\f(f(x)) = x\textsuperscript{4} - 6x\textsuperscript{3} + 10x\textsuperscript{2} - 3x\\$

Question:7

Is $g = \{ (1, 1), (2, 3), (3, 5), (4, 7) \}$ a function? If g is described by $g (x) = \alpha x + \beta$ , then what value should be assigned to $\alpha\: \: and\: \: \beta$ .

Answer:

Here, $g = \{ (1, 1), (2, 3), (3, 5), (4, 7) \}$
Here, each and every element of a domain has a unique image. Therefore, g is a function.
Also, $g (x) = \alpha x + \beta$
Therefore,
$\\g(1) = \alpha (1) + \beta = 1\\ \alpha + \beta = 1 \ldots \ldots .. (1)\\$
Similarly,
$g(2) = \alpha (2) + \beta = 2 \alpha + \beta = 3 \ldots \ldots .. (2)\\$
By solving (1) and (2), we get
$\alpha = 2 and \beta = -1\\$
Therefore, $g (x) = 2x - 1$

Question:8

Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective.
(i) $\{ (x, y): \text{x is a person, y is the mother of x} \} .\\$
(ii) $\{ (a, b): \text{a is a person, b is an ancestor of a} \} .\\$

Answer:
(i)Here, $\{ (x, y): \text{x is a person, y is the mother of x} \} .\\$
Therefore, each person (x) has only one biological mother.
Hence, the given set of ordered pairs make a function.
Therefore, there are more than one person who may have the same mother. Hence, the function is many-one and surjective.
(ii) Here, $\{ (a, b): \text{a is a person, b is an ancestor of a} \} .\\$
It’s seen that any person ‘a’ has more than one ancestor.
Therefore, it is not a function.

Question:9

If the mappings f and g are given by $f = \{ (1, 2), (3, 5), (4, 1) \}$ and $g = \{ (2, 3), (5, 1), (1, 3) \}$ , write $fog.\\$

Answer:
Here,
$f = \{ (1, 2), (3, 5), (4, 1) \}$ and $g = \{ (2, 3), (5, 1), (1, 3) \}$
Therefore,
$\\fog (2) = f(g(2)) = f(3) = 5\\ fog (5) = f(g(5)) = f(1) = 2\\ fog (1) = f(g(1)) = f(3) = 5\\$
So, we can write that:
$\\fog = \{ (2, 5), (5, 2), (1, 5) \} \\$

Question:10

Let C be the set of complex numbers. Prove that the mapping $f: C \rightarrow R$ is given by $f (z) = \vert z \vert , \forall z \in C$ , is neither one-one nor onto.

Answer:

Here, $f: C \rightarrow R$ is given by $f (z) = \vert z \vert , \forall z \in C$
If we assume $z = 4 + 3i$
Then,
$f(4 + 3i) = \vert 4 + 3i \vert = \sqrt (4\textsuperscript{2} + 3\textsuperscript{2})= \sqrt 25 = 5\\$
Similarly, for $z = 4 - 3i$
$f(4 - 3i) = \vert 4 - 3i \vert = \sqrt (4\textsuperscript{2} + 3\textsuperscript{2}) = \sqrt 25 = 5\\$
Therefore, it is clear that f(z) is many-one.
So, $\vert z \vert \geq 0, \forall z \in C,\\$
However, in the question R is the co-domain given.
Hence, f(z) is not onto. So, f(z) is neither one-one nor onto.

Question:11

Let the function $f: R \rightarrow R$ be defined by $f (x) = cos x, \forall x \in R.$ Show that f is neither one-one nor onto.

Answer:
It is given that, $f: R \rightarrow R, f(x) = cos x, \forall x \in R\\$
So, we can write:
$f(x\textsubscript{1}) = f(x\textsubscript{2})\\ or, cos x\textsubscript{1}= cos x\textsubscript{2}\\$
Hence, $x\textsubscript{1 }= 2n \pi \pm x\textsubscript{2}, where n \in Z\\$
It is understandable that for any value of $x\textsubscript{1 }and\: \: x\textsubscript{2}$ , the above equation has an infinite number of solutions.
Therefore, f(x) is a many one function.
We know the range of cos x is [-1, 1] and it is a subset of the given co-domain R.
Hence, the given function is not onto.

Question:12

Let $X = \{ 1, 2, 3 \} and Y = \{ 4, 5 \}$ . Find whether the following subsets of $X \times Y$ are functions from X to Y or not.
(i) $f = \{ (1, 4), (1, 5), (2, 4), (3, 5) \}$
(ii) $g = \{ (1, 4), (2, 4), (3, 4) \} \\$
(iii) $h = \{ (1,4), (2, 5), (3, 5) \}$
(iv) $k = \{ (1,4), (2, 5) \} .\\$

Answer:

Here, $X = \{ 1, 2, 3 \} and Y = \{ 4, 5 \} \\$
Therefore, $X \times Y = \{ (1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5) \} \\$
(i) $f = \{ (1, 4), (1, 5), (2, 4), (3, 5) \} \\$
Here, f(1) = 4 and again f(1) = 5.
Therefore, f is not a function here.
As a result, there is no unique of pre- image ‘1’.
(ii) $g = \{ (1, 4), (2, 4), (3, 4) \} \\$
We can clearly see that g is a function. Here in g, each element of the given domain has a unique image at the given range
(iii) $h = \{ (1,4), (2, 5), (3, 5) \} \\$
It’s clear that h is a function of each pre-image that has a unique image.
Again, h(2) = h(3) = 5
Therefore, the function h is also many-one.
(iv) $k = \{ (1, 4), (2, 5) \} \\$
Here, ‘3’ does not have any image under the mapping. Therefore, k is not a function.

Question:13

If functions $f: A \rightarrow B \: \: and\: \: g: B \rightarrow A \text{ satisfies } g of = IA$ , then show that f is one-one and g is onto.

Answer:

Here, it is given:
$f: A \rightarrow B \: \: and\: \: g: B \rightarrow A \text{ satisfies } g of = IA$
It’s clear here that the function ‘g’ is inverse of ‘f’.
So, ‘f’ has to be both one-one as well as onto.
As a result, ‘g’ is both one-one and onto.

Question:14

Let $f: R \rightarrow R$ be the function defined by $f(x) = 1/(2 - cos x) \forall x \in R$ . Then, find the range off.

Answer:

Here, $f(x) = 1/(2 - cos x) \forall x \in R$
Let’s say,
$\\y = 1/(2 - cos x)\\ \therefore2y - ycos x = 1\\ Or, cos x = (2y - 1)/ y\\ Or, cos x = 2 - 1/y\\$
As the range of cos x is, $-1 \leq cos x \leq 1\\$
Therefore,
$\\-1 \leq 2 - 1/y \leq 1\\ -3 \leq - 1/y \leq -1\\ 1 \leq - 1/y \leq 3\\ 1/3 \leq y \leq 1\\$
Hence, the range of the given function is [1/3, 1].

Question:15

Let n be a fixed positive integer. Define a relation R in Z as follows: $\forall a, b \in Z$ , aRb if and only if a - b is divisible by n. Show that R is an equivalence relation.

Answer:

Here, we have to a relation R in Z as follows: $\forall a, b \in Z$ , aRb if and only if a - b is divisible by n.
Here, aRa $\Rightarrow (a - a)$ is divisible by n and this is true for all integers.
Therefore, R is reflective.
For aRb, $aRb \Rightarrow (a - b)$ is also divisible by n.
or, - (b - a) is divisible by n.
or, (b - a) is divisible by n
Hence, we can write it as bRa.
Therefore, R is symmetric.
For aRb , (a - b) is divisible by n.
For bRc, (b - c) is divisible by n.
Hence, (a - b) + (b-c) is divisible by n.
Or, (a-c) is divisible by n. This can be expressed as aRc.
Therefore, R is transitive.
So, R is an equivalence relation.

Question:16

If A = {1, 2, 3 }, define relations on A which have properties of being:

(a) reflexive, transitive but not symmetric

(b) symmetric but neither reflexive nor transitive

Answer:

Here, $A = \{ 1, 2, 3 \} .\\$
(i) Assume $R\textsubscript{1}= \{ (1, 1), (1, 2), (1, 3), (2, 3), (2, 2), (1, 3), (3, 3) \} \\$
Here, (1, 1), (2, 2) and (3, 3) $\in \ R\textsubscript{1}. R\textsubscript{1}$ is reflexive.
$(1, 2) \in R\textsubscript{1}, (2, 3) \in R\textsubscript{1} \Rightarrow (1, 3) \in R\textsubscript{1}$
. Hence, $R\textsubscript{1}$ is transitive.
Now, $(1, 2) \in R\textsubscript{1 } \Rightarrow (2, 1) \notin R\textsubscript{1}.$
Therefore, $R\textsubscript{1}$ is not symmetric.
(ii) Let say, $R\textsubscript{2}= \{ (1, 2), (2, 1) \}$
So, $(1, 2) \in R\textsubscript{2}, (2, 1) \in R\textsubscript{2}\\$
Therefore, $R\textsubscript{2 }$ is symmetric,
$(1, 1) \notin R\textsubscript{2}$ . Therefore, $R\textsubscript{2 }$ is not reflexive.
$(1, 2) \in R\textsubscript{2}$ , $(2, 1) \in R\textsubscript{2 }$ but $(1, 1) \notin R\textsubscript{2}$ . Hence, $R\textsubscript{2 }$ is not transitive.
(iii) Let $R\textsubscript{3 }$ = $\{ (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3) \} \\$
$R\textsubscript{3 }$ is reflexive as $(1, 1) (2, 2) and (3, 3) \in R\textsubscript{3 }\\$
$R\textsubscript{3 }$ is symmetric as $(1, 2), (1, 3), (2, 3) \in R\textsubscript{3 }$ $\Rightarrow (2, 1), (3, 1), (3, 2) \in R\textsubscript{3 }\\$
Therefore, $R\textsubscript{3 }$ is reflexive, symmetric and transitive.

Question:17

Let R be relation defined on the set of natural number N as follows:
$R = \{ (x, y): x \in N, y \in N, 2x + y = 41 \}$ . Find the domain and range of the relation R. Also verify whether R is reflexive, symmetric, and transitive.

Answer:

Here, $R = \{ (x, y): x \in N, y \in N, 2x + y = 41 \}$
So, the domain $D= \{ 1, 2, 3, \ldots .., 20 \} \\$
And the Range $= \{ 1, 3, 5, \ldots .., 39 \} \\$
Here, $(2, 2) \notin \: \: R\: \: as\: \: 2 \times 2 + 2 \neq 41$ . Therefore, R is not reflexive.
Again, $(1, 39) \in R \: \: but\: \: (39, 1) \: \: \notin R$ . So, R is not symmetric.
Again, $(11, 19) \notin R, (19, 3) \notin R; but (11, 3) \notin R$ . So, Further R is not transitive.
Therefore, R is neither reflexive nor symmetric and nor transitive.

Question:18

Given A = {2, 3, 4}, B = {2, 5, 6, 7}. Construct an example of each of the following:

(a) an injective mapping from A to B

(b) a mapping from A to B which is not injective

Answer:

Here, $A = \{ 2, 3, 4 \} , B = \{ 2, 5, 6, 7 \} \\$
(i) Assume, $f: A \rightarrow B \text{ denote a mapping}$
Then, $f = \{ (x, y): y = x + 3 \} or\\$
$f = \{ (2, 5), (3, 6), (4, 7) \}$ , this is an injective mapping.
(ii) Again assume, $g: A \rightarrow B$ denotes a mapping: $g = \{ (2, 2), (3, 2), (4, 5) \}$ . Hence, it is not an injective mapping.
(iii) Again assume, h: $B \rightarrow A$ denotes a mapping: $h = \{ (2, 2), (5, 3), (6, 4), (7, 4) \}$ . Hence, it is a mapping from B to A.

Question:19

Give an example of a map

(i) which is one-one but not onto

(ii) which is not one-one but onto

(iii) which is neither one-one nor onto.
Answer:
(i) Say, $f: N \rightarrow N,$ be a mapping defined by $f (x) = x\textsuperscript{2}\\$
If, $f(x\textsubscript{1}) = f (x\textsubscript{2})\\$
Then, $x\textsubscript{1}\textsuperscript{2}= x\textsubscript{2}\textsuperscript{2}\\$
So, $x\textsubscript{1 }= x\textsubscript{2 }(as, x\textsubscript{1 }+ x\textsubscript{1 }\text{cannot be } 0)$
$f(x\textsubscript{1 }) = f(x\textsubscript{2 })$ , hence, f(x) is one-one.
However, ‘f’ is not onto, as for $1 \in N$ , therefore, there is no existence of x in N : f(x) = 2x + 1.
(ii) Let $f: R \rightarrow [0, \infty )$ , is a mapping that is defined by $f(x) = \vert x \vert \\$
Then, we can conclude that f(x) is not a one-one as f(2) and f(-2) are the same here.
But $\vert x \vert \geq 0$ , so the range is $[0, \infty ].\\$
Therefore, f(x) is onto.
(iii) Assume, $f: R \rightarrow R$ , be a mapping which is defined by $f(x) = x\textsuperscript{2}\\$
Then we can say that f(x) is not one-one as f(1) and f (-1) are the same.
The range of f(x) is $[0, \infty )$ .
Therefore, f (x) is neither one-one nor onto.

Question:20

Let $A = R - \{ 3 \} , B = R - \{ 1 \}$ . Let $f : A \rightarrow B$ be defined by $f(x) = x - 2/ x - 3 \forall x \in A$ . Then show that f is bijective.

Answer:

Here,
$A = R - \{ 3 \} , B = R - \{ 1 \} \\$
$f : A \rightarrow B$ be defined by $f (x) = x - 2/ x - 3 \forall x \in A\\$
Hence, $f (x) = (x - 3 + 1)/ (x - 3) = 1 + 1/ (x - 3)\\$
Say $f(x1) = f (x2)\\$
$1+\frac{1}{x_{1}-3}=1+\frac{1}{x_{2}-3}$
$\\\frac{1}{x_{1}-3}=\frac{1}{x_{2}-3} \\\\ x_{1}=x_{2}$
So, f (x) is an injective function.
$\\If, y = (x - 2)/ (x -3)\\ Or, x - 2 = xy - 3y\\ Or, x(1 - y) = 2 - 3y\\ Or, x = (3y - 2)/ (y - 1)\\ Or, y \in R - \{ 1 \} = B\\$
So, f (x) is onto or subjective.
Therefore, f(x) is a bijective function.

Question:21

Let A = [-1, 1]. Then, discuss whether the following functions defined on A are one-one, onto, or bijective:
(i) $f(x) = x/2$ (ii) $g(x) = \vert x \vert \\$
(iii) $h(x) = x \vert x \vert$ (iv) $k(x) = x\textsuperscript{2}\\$

Answer:

Here, A = [-1, 1]
(i) $f: [-1, 1] \rightarrow [-1, 1], f (x) = x/2\\$
If, $f(x\textsubscript{1}) = f(x\textsubscript{2})\\$
$x\textsubscript{1}/2 = x\textsubscript{2}/2\\$
So, f(x) is one-one.
Also, $x \in [-1, 1]\\$
$x/2 = f(x) = [-1/2, 1/2]\\$
Hence, the range is a subset of co-domain A
So, f(x) is not onto.
Therefore, f (x) is not bijective.
(ii)
$\\g(x) = \vert x \vert \\ \text{Let} g(x\textsubscript{1}) = g (x\textsubscript{2})\\ \vert x\textsubscript{1} \vert = \vert x\textsubscript{2} \vert \\ x\textsubscript{1}= \pm x\textsubscript{2}\\$
So, g(x) is not one-one.
Also, $g(x) = \vert x \vert \geq 0, \text{for all real x}\\$
Hence, the range is [0, 1], which is subset of co-domain ‘A’
So, f(x) is not onto.
Therefore, f(x) is not bijective.
(iii)
$\\\\h(x) = x \vert x \vert \\ \text{Let } h(x\textsubscript{1}) = h(x\textsubscript{2})\\ x\textsubscript{1} \vert x\textsubscript{1} \vert = x\textsubscript{2} \vert x\textsubscript{2} \vert \\ \text{If }x\textsubscript{1}, x\textsubscript{2}> 0\\ x\textsubscript{1}\textsuperscript{2}= x\textsubscript{2}\textsuperscript{2}\\ x\textsubscript{1}\textsuperscript{2}- x\textsubscript{2}\textsuperscript{2}= 0\\ (x\textsubscript{1} - x\textsubscript{2})(x\textsubscript{1} + x\textsubscript{2}) = 0\\ x\textsubscript{1 }= x\textsubscript{2 }(as x\textsubscript{1}+ x\textsubscript{2} \neq 0)\\ \text{Again }, x\textsubscript{1}, x\textsubscript{2 }< 0, and x\textsubscript{1}= x\textsubscript{2}\\ \text{Therefore }, x\textsubscript{1 }and x\textsubscript{2 }\text{of opposite sign}, x\textsubscript{1 } \neq x\textsubscript{2 }.\\$
Hence, f(x) is one-one.
$\\\text{For x} \in [0, 1], f(x) = x\textsubscript{2} \in [0, 1]\\ \text{For x} < 0, f(x) = -x\textsubscript{2} \in [-1, 0)\\$
Hence, the range is [-1, 1].
So, h(x) is onto.
Therefore, h(x) is bijective.
(iv)
$\\k(x) = x^{2}\\ \text{Let }k (x\textsubscript{1}) = k (x\textsubscript{2})\\ x\textsubscript{1}\textsuperscript{2}= x\textsubscript{2}\textsuperscript{2}\\ x\textsubscript{1 }= \pm x\textsubscript{2}\\$
Therefore, k(x) is not one-one.

Question:22

Each of the following defines a relation on N:
(i) x is greater than y, $x, y \in N\\$
(ii) x + y = 10, $x, y \in N\\$
(iii) x y is square of an integer $x, y \in N\\$
(iv) x + 4y = 10 $x, y \in N\\$
Determine which of the above relations are reflexive, symmetric, and transitive.

Answer:

(i)Here, x is greater than y; $x, y \in N\\$
If $(x, x) \in R$ , then x > x, that does not satisfy for any $x \in N$ .\\
Therefore, R is not reflexive.
Say, $(x, y) \in R$
$\\Or, xRy\\ Or, x > y$
$\Rightarrow y > x \\$ For any $x, y \in N$ , the above condition is not true.
Hence, R is not symmetric.
Again, xRy and yRz
$\\Or, x > y and y > z\\ Or, x > z\\ Or, xRz\\$
Hence, R is transitive.
(ii) x + y = 10; $x, y \in N$
Thus,
$\\R = \{ (x, y); x + y = 10, x, y \in N \} \\ R = \{ (1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1) \} \\$
Therefore, $(1, 1) \notin R\\$
So, R is not reflexive.
Again, $(x, y) \in R \Rightarrow (y, x) \in R\\$
Therefore, R is symmetric.
And, $(1, 9) \in R, (9, 1) \in R, but (1, 1) \notin R\\$
Therefore, R is not transitive.
(iii)Here, xy is square of an integer $x, y \in N$
$R = \{ (x, y) : \text{xy is a square of an integer} x, y \in N \} \\$
So, $(x, x) \in R, \forall x \in N\\$
For any $x \in N, x\textsuperscript{2 }$ is an integer.
Thus, R is reflexive.
If $(x, y) \in R \Rightarrow (y, x) \in R\\$
So, R is symmetric.
Again, if xy and yz both are square of an integer.
Then, $xy = m\textsuperscript{2} \: \: and \: \: yz = n\textsuperscript{2} \text{for some m, n} \in Z\\$
$\\x = m\textsuperscript{2} /y and z = n\textsuperscript{2} /y\\ xz = m\textsuperscript{2} n\textsuperscript{2} / y\textsuperscript{2}$ , this must be the square of an integer.
Therefore, R is transitive.
(iv) x + 4y = 10; $x, y \in N\\$
$\\R = \{ (x, y): x + 4y = 10; x, y \in N \} \\ R = \{ (2, 2), (6, 1) \} \\ Here, (1, 1) \notin R\\$
Hence, R is not symmetric.
$\\(x, y) \in R \Rightarrow x + 4y = 10\\ And, (y, z) \in R \Rightarrow y + 4z = 10\\ Or,\ x - 16z = -30\\ Or,\ (x, z) \notin R\\$
Therefore, R is not transitive.

Question:23

Let A = {1, 2, 3, … 9} and R be the relation in $A \times A$ defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) in $A \times A$ . Prove that R is an equivalence relation and also obtain the equivalent class [(2, 5)].

Answer:

$\\Here, A = \{ 1, 2, 3, \ldots 9 \} \: \: and \: \: (a, b) R (c, d)\: \: if a + d = b + c for (a, b), (c, d) \in A \times A.\\ Say (a, b) R (a, b)\\ Therefore, a + b = b + a, \forall a, b \in A .$
This must be true for any $a, b \in A.\\$
Hence, R is reflexive.
Say, (a, b) R (c, d)
Then,
$\\a + d = b + c\\ c + b = d + a\\ (c, d) R (a, b)\\$
Therefore, R is symmetric.
Let $\\(a, b) R(c, d) ,and, (c, d) R(e, f)$
$\\ a + d = b + c ,and, c + f = d + e\\ a + d = b + c ,and, d + e = c + f\\ Or, (a + d) - (d + e = (b + c) - (c + f)\\ Or, a - e = b - f\\ Or, a + f = b + e\\ Or, (a, b) R(e, f)\\$
So, R is transitive.
Therefore, R is an equivalence relation.

Question:24

Using the definition, prove that the function $f: A \rightarrow B$ is invertible if and only if f is both one-one and onto.

Answer:

Say, $f: A \rightarrow B$ be many-one functions.
If, $f(a) = p \text{ and } f(b) = p\\$
Then $f\textsuperscript{-1}(p) = a \: \: and \: \: f\textsuperscript{-1}(p) = b\\$
In this case, we have two images ‘a and b’ for one pre-image ‘p’. This is because the inverse function is not defined here.
However, to be one-one, f must be invertible.
Say, $f: A \rightarrow B$ is not onto function.
$B = \{ p, q, r \} and \{ p, q \}$ is the range of f.
There is no pre-image for the image r, which will have no image in set A.
And, f must be onto to be invertible.
Thus, to be both one-one and onto f must be invertible
If f is a bijective function, then $f = X \rightarrow Y$ is invertible.

Question:25

Functions $f, g : R \rightarrow R$ are defined, respectively, by $f(x) = x\textsuperscript{2} + 3x + 1, g(x) = 2x - 3$ , find
(i) fog (ii) gof (iii) fof (iv) gog

Answer:

Here,
$f(x) = x\textsuperscript{2} + 3x + 1, g (x) = 2x - 3\\$

$\\(i)fog = f(g(x))\\ = f(2x - 3)\\ = (2x - 3)\textsuperscript{2} + 3(2x - 3) + 1\\ = 4x\textsuperscript{2}\ + 9 - 12x + 6x - 9 + 1\\ = 4x\textsuperscript{2}\ - 6x + 1\\ (ii) gof = g(f(x))\\ = g(x\textsuperscript{2}\ + 3x + 1)\\ = 2(x\textsuperscript{2}\ + 3x + 1) - 3\\ = 2x\textsuperscript{2}\ + 6x - 1\\ (iii) fof = f(f(x))\\ = f(x\textsuperscript{2}\ + 3x + 1)\\ = (x\textsuperscript{2}\ + 3x + 1)2 + 3(x\textsuperscript{2} + 3x + 1) + 1\\ = x4 + 9x\textsuperscript{2}\ + 1 + 6x\textsuperscript{3} + 6x + 2x\textsuperscript{2}\ + 3x\textsuperscript{2}\ + 9x + 3 + 1\\ = x4 + 6x\textsuperscript{3} + 14x\textsuperscript{2}\ + 15x + 5\\ (iv) gog = g(g(x))\\ = g(2x - 3)\\ = 2(2x - 3) - 3\\ = 4x - 6 - 3\\ = 4x - 9\\$

Question:26

Let $\ast$ be the binary operation defined on Q. Find which of the following binary operations are commutative
$\\(i) a \ast b = a - b\: \: \forall a, b \in Q\\ (ii) a \ast b = a^2 + b^2 \: \: \forall a, b \in Q\\ (iii) a \ast b = a + ab \: \: \forall a, b \in Q\\ (iv) a \ast b = (a - b)\textsuperscript{2}\: \: \forall a, b \in Q\\$

Answer:

Here, $\ast$ is a binary operation defined on Q.
(i)
$\\a \ast b = a - b, \forall a, b \in Q \: \: and \: \: b \ast a = b - a\\ So, a \ast b \neq b \ast a\\$
Hence, $\ast$ is not commutative.
(ii)
$\\a \ast b = a\textsuperscript{2}+ b\textsuperscript{2}\\ b \ast a = b\textsuperscript{2}+ a\textsuperscript{2}\\$
Therefore, $\ast$ is commutative.
(iii)
$\\a \ast b = a + ab\\ b \ast a = b + ab\\ Hence, a + ab \neq b + ab\\$
Therefore, $\ast$ is not commutative.
(iv)
$\\a \ast b = (a - b)\textsuperscript{2}, \forall a, b \in Q\\ b \ast a = (b -a)\textsuperscript{2}\\ As, (a - b)\textsuperscript{2} = (b - a)\textsuperscript{2}\\$
Therefore, $\ast$ is commutative.

Question:27

Let $\ast$ be binary operation defined on R by $a \ast b = 1 + ab, \forall a, b \in R$ . Then the operation $\ast$ is
(i) commutative but not associative
(ii) associative but not commutative
(iii) neither commutative nor associative
(iv) both commutative and associative

Answer:

(i) Here,
$\ast$ is a binary operation defined on R by $a \ast b = 1 + ab, \forall a, b \in R$
So, $a \ast b = ab + 1 = b \ast a\\$
Therefore, $\ast$ is a commutative binary operation.
$\\Now, a \ast (b \ast c) = a \ast (1 + bc) = 1 + a (1 + bc) = 1 + a + abc\\ Again, (a \ast b) \ast c = (1 + ab) \ast c = 1 + (1 + ab) c = 1 + c + abc\\ Therefore, a \ast (b \ast c) \neq (a \ast b) \ast c\\$
Hence, $\ast$ is not associative.
Therefore, $\ast$ is commutative but not associative.

Question:28

Let T be the set of all triangles in the Euclidean plane, and let a relation R on T be defined as aRb if a is congruent to $b \forall a, b \in T$ . Then R is
(A) reflexive but not transitive (B) transitive but not symmetric
(C) equivalence (D) none of these

Answer:

(C) equivalence
Here aRb, if a is congruent to b, $\forall a, b \in T.\\$
So, in aRa, a is congruent to a. This must always be true.
Therefore, R is reflexive.
Say, $aRb \Rightarrow a \sim b\\$
$b \sim a \Rightarrow bRa\\$
Therefore, R is symmetric.
Say, aRb and bRc
$\\a \sim b \: \: and\: \: b \sim c\\ a \sim c \Rightarrow aRc\\$
Therefore, R is transitive.
Therefore, R is an equivalence relation.

Question:29

Consider the non-empty set consisting of children in a family and a relation R defined as aRb if a is brother of b. Then R is

(A) symmetric but not transitive (B) transitive but not symmetric

Answer:

(B) transitive but not symmetric

If, aRb means a is brother of b.

Then, it does not mean b is also a brother of a. Because, b can be a sister of a too.

Therefore, R is not symmetric.

If, aRb implies that a is the brother of b.

and bRc implies that b is the brother of c.

Therefore, a must be the brother of c.

Hence, R is transitive.

Question:30

The maximum number of equivalence relations on the set A = {1, 2, 3} are

(A) 1 (B) 2

Answer:

(D) 5
Given, set $A = \{ 1, 2, 3 \} \\$
Now, the number of equivalence relations as follows
$\\R1 = \{ (1, 1), (2, 2), (3, 3) \} \\ R2 = \{ (1, 1), (2, 2), (3, 3), (1, 2), (2, 1) \} \\ R3 = \{ (1, 1), (2, 2), (3, 3), (1, 3), (3, 1) \} \\ R4 = \{ (1, 1), (2, 2), (3, 3), (2, 3), (3, 2) \} \\ R5 = \{ (1, 2, 3) \Leftrightarrow A \times A = A^2 \} \\$
Thus, the maximum number of equivalence relations is ‘5’.

Question:31

If a relation R on the set {1, 2, 3} be defined by R = {(1, 2)}, then R is

(A) reflexive (B) transitive

Answer:

(D) none of these

If, R be defined on the set {1, 2, 3} by R = {(1, 2)}

Then, we can say that R is not reflexive, transitive, and symmetric.

Question:32

Let us define a relation R in R as aRb if $a \geq b.$ Then R is

(a) an equivalence relation (b) reflexive, transitive but not symmetric (c) symmetric, transitive but not reflexive (d) neither transitive nor reflexive but symmetric.

Answer:

(b).
The defined relation R in R as aRb if $a \geq b.\\$
Similarly, aRa implies $a \geq a$ which is true.
Therefore, it is reflexive.
Let $aRb \rightarrow a \geq b, but, b \ngtr a.\\$
Therefore, we can’t write it as Rba
Hence, R is not symmetric.
Now, $a \geq b, b \geq c$ . So, $a \geq c$ , and this is true.
Therefore, R is transitive.

Question:33

Let A = {1, 2, 3} and consider the relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}, then R is

(a) reflexive but not symmetric (b) reflexive but not transitive (c) symmetric and transitive (d) neither symmetric nor transitive.

Answer:

(a)

Given, R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}

Therefore, it can be written as: 1R1, 2R2 and 3R3.

Therefore, R is reflexive.

Here, 1R2 is not the same as 2R1 and 2R3 is not the same as 3R2.

Therefore, R is not symmetric.

Again, 1R1 and $1R2 \rightarrow 1R3\\$

Therefore, R is transitive.

Question:34

The identity element for the binary operation * defined on Q ~ {0} as a * b = 2 ab a, b Q ~ {0} is

(a) 1 (b) 0 (c) 2 (d) None of these

Answer:

(c)
Here: $a \ast b = ab/2 a, b Q \sim \{ 0 \} \\$
Assume ‘e’ as the identity element.
Therefore, $a \ast e = ae/2 = a \rightarrow e = 2 \\$

Question:35

If the set A contains 5 elements and set B contains 6 elements, then the number of one-one and onto mapping from A to B is

(a) 720 (b) 120 (c) 0 (d) None of these

Answer:

Let, the number of elements in A and B set are m and n respectively. Therefore, one-one and onto mapping

from A to B is n! when m = n

and, 0 if m ≠ n

It is given that, m = 5 and n = 6.

As,5 ≠ 6 So, from A to B mapping = 0

Question:36

Let $A = \{ 1, 2, 3, ..., n \}$ and $B = \{ a, b \}$ . Then the number of surjections from A to B is
(a) $\textsuperscript{n}P\textsubscript{2}$
(b) $2\textsuperscript{n} - 2$
(c) $2\textsuperscript{n} - 1$
(d) None of these

Answer:

(d)
It is given that, $A = \{ 1, 2, 3, ..., n \}$ and $B = \{ a, b \} \\$
Say, the number of elements in set A and B are m and n respectively.
Therefore, $\textsuperscript{n}C\textsubscript{m} \times m!$ is the number of surjections from A to B, for $n \ngeq m \\$
Given, m=2.
Therefore, the number of surjections from A to B
$= nC_2 \times 2! = n!/[2! \times (n-2)!] = n(n - 1) = n\textsuperscript{2} - n \\$

Question:37

Let $f : R \rightarrow R$ be defined by $f(x) = 1/x \forall x \in R$ then f is
(a) one-one (b) onto (c) bijective (d) f is not defined

Answer:

(d)
Here, f(x) = 1/x
Say, x = 0. Then f(x) = 1/0 = undefined
Hence, f(x) is not defined.

Question:38

Let $f : R \rightarrow R$ be defined by $f(x) = 3x\textsuperscript{2} - 5$ and $g : R \rightarrow R$ by $g(x) = x/(x\textsuperscript{2}+1)$ , then gof is
$\\(a) (3x\textsuperscript{2}-5)/(9x\textsuperscript{4}-30x\textsuperscript{2}+ 26)\\ (b) (3x\textsuperscript{2}-5)/(9x\textsuperscript{4}-6x\textsuperscript{2}+ 26) \\(c) 3x\textsuperscript{2}/(x\textsuperscript{4}+2x\textsuperscript{2}-4) \\(d) (3x\textsuperscript{2})/(9x\textsuperscript{4}-30\textsuperscript{2}+2)\\$

Answer:

(a)
Given, $f(x) = 3x\textsuperscript{2} - 5 \ and \ g(x) = x/(x\textsuperscript{2}+1)\\$
Therefore,
$\\gof = gof (x) = g(3x\textsuperscript{2}-5) = (3x\textsuperscript{2}-5)/[(3x\textsuperscript{2}-5)\textsuperscript{2}+1]\\ = (3x\textsuperscript{2}-5)/(9x\textsuperscript{4}-30x\textsuperscript{2}+ 26)\\$

Question:39

Which of the following functions from Z to Z are bijections?
$\\(a) f(x) = x\textsuperscript{3 }\\(b) f(x) = x + 2 \\(c) f(x) = 2x + 1 \\(d) f(x) = x\textsuperscript{2} + 1 \\$

Answer:

(b)
Given that $f : Z \rightarrow Z \\$
Say, $x\textsubscript{1}, x\textsubscript{2} \in f(x) \rightarrow f(x\textsubscript{1}) = x\textsubscript{1}+ 2; f(x\textsubscript{2}) = x\textsubscript{2}+ 2;\\$
For, $f(x\textsubscript{1}) = f(x\textsubscript{2})\ \rightarrow x\textsubscript{1}+ 2 = x\textsubscript{2}+\ 2 \\$
or, $x\textsubscript{1 }= x\textsubscript{2}\\$
Therefore, the function f(x) is one-one.
Say, y = x + 2
or, $x = y - 2 \forall y \in Z \\$
Hence, f(x) is onto.
Therefore, the function f(x) is bijective.

Question:40

Let $f : R \rightarrow R$ be the functions defined by $f(x) = x\textsuperscript{3} + 5$ . Then $f\textsuperscript{-1}(x) is$
$\\(a) (x + 5)\textsuperscript{1/3}\\ (b) (x - 5)\textsuperscript{1/3}\\ (c) (5 - x)\textsuperscript{1/3}\\ (d) 5 - x \\$

Answer:

(b)Here,
$\\f(x) = x\textsuperscript{3} + 5\\ Say, y = x\textsuperscript{3} + 5\\ or, x\textsuperscript{3} = y - 5 \\ or, x = (y - 5)\textsuperscript{1/3}\ f\textsuperscript{-1}1(x) = (x - 5)\textsuperscript{1/3} \\$

Question:41

Let $f : A \rightarrow B$ and $g : B \rightarrow C$ be the bijective functions. Then $(gof)\textsuperscript{-1}$ is
$\\(a) f\textsuperscript{-1 }og\textsuperscript{-1} \\(b) fog \\(c) g\textsuperscript{-1}of\textsuperscript{-1}\\ (d) gof \\$
Answer:

(a)
Given that, $f : A \rightarrow B \text{ and } g : B \rightarrow C \\$
Then, $(gof)\textsuperscript{-1} = f\textsuperscript{-1}og\textsuperscript{-1}\\$

Question:42

Let $f: \{ R-3/5 \} \rightarrow R$ be defined by $f(x) = (3x + 2)/(5x - 3)$ then
$\\(a) f(x) = f(x); \\(b) f\textsuperscript{-1}(x) = - f(x) \\(c) (fof)x = - x \\(d) f\textsuperscript{-1}(x) = f(x)/19\\$

Answer:

(a)
Here, $f(x) = (3x + 2)/(5x - 3) \forall x \neq 3/5\\$
$\\So, y = (3x + 2)/(5x - 3)\\ or, y (5x-3) = (3x + 2)\\ or, 5xy-3y = 3x + 2\\ or, x = (3y+2)/(5y-3)\\ Therefore, f\textsuperscript{-1}(x) = (3x+2)/(5x-3)\\ Hence, f\textsuperscript{-1}(x) = f(x)\\$

Question:43

Let $f : [0, 1] \rightarrow [0, 1]$ be defined by

f(x) = { x, if is rational

{ 1-x , if is irrational
Then (fof)x is

(a) constant

(b) 1 + x

(d) None of these

Answer:

Question:44

Let $f : [2, \infty )$ and R be the function defined by $f(x) = x\textsuperscript{2} - 4x + 5$ , then the range of f is
$\\(a) R\\ (b) [1, \infty ) \\(c) [4, \infty ) \\(d) [5, \infty ) \\$
Answer:

(b)
$\\Here, f(x) = x\textsuperscript{2} - 4x + 5 \\ Assume, y = x\textsuperscript{2}\ - 4x + 5 \\ or, x\textsuperscript{2}\ - 4x + 5 - y = 0 \\ or, x = [-4 \pm \sqrt \{ (-4)\textsuperscript{2}- 4(-4)(5-y) \} ]/2x = 2 \pm \sqrt (y-1)\\ \text{When x is real}, y-1 \geq 0\\ Therefore, y \geq 1\\ \text{Hence the range is}[1, \infty )\\$

Question:45

Let $f : N \rightarrow R$ be the function defined by $f(x) = (2x-1)/2$ and $g : Q \rightarrow R$ be another function defined by $g(x) = x + 2$ then, gof(3/2) is
$\\(a) 1 \\(b) - 1 \\(c) 7/2 \\(d) None of these \\$

Answer:

(d)
Given that, $f(x) = (2x-1)/2 \: \: and\: \: g(x) = x + 2 \\$
Therefore, $gof(x) = g[(f(x)] = f(x) + 2 = (2x-1)/2 = (2x+3)/2\\$
Hence, $gof(3/2) = 3 \\$

Question:46

Let f : $R \rightarrow R$ be defined by

$\\f(x) = \{2x, when x> 3 \\ \ \ \ \: \: \: \: \: \: \: \: \ \ \ \ \ \{ x\textsuperscript{2}, when 1< x \leq 3\\ \ \ \ \ \ \ \ \ \ \ \ \{ 3x, when x \leq 1\\ then f(- 1) + f(2) + f(4) is\\$

$\\ (a) 9 \\(b) 14 \\(c) 5 \\(d) None of these \\$

Answer:

(a)
$Here, f(- 1) + f(2) + f(4) = 3(- 1) + (2)\textsuperscript{2} + 2(4) = - 3 + 4 + 8 = 9 \\$

Question:47

If $f : R \rightarrow R$ be given by $f(x) = tan x$ , then f^-1 is
$\\(a) \frac{\pi}{4} \\ (b) (n \pi + \pi /4) where\ n \in Z \\(c) \ does \ not \ exist\\ (d) None\ of \ these \\$
Answer:

(a)
$\\Here, f(x) = tan x \\ Assume, f(x) = y = tan x\\ Then,\ \ x = tan\textsuperscript{-1} (y)\\ or, f\textsuperscript{-1}(x) = tan\textsuperscript{-1}\ (x) \\ or, f\textsuperscript{-1}(1) = tan\textsuperscript{-1} (1)\\ or,\ f\textsuperscript{-1}(1) = tan\textsuperscript{-1} tan ( \pi /4)=\frac{\pi}{4}\\$

Question:48

Let the relation R be defined in N by aRb if 2a + 3b = 30. Then R = ............

Answer:

Here, aRb : 2a + 3b = 30

Or, 3b = 30 – 2a

Or, b = (30 - 2a)/ 3

for a = 3, b = 8

a = 6, b = 6

a = 9, b = 4

a = 12, b = 2

Therefore, R = {(3, 8), (6, 6), (9, 4), (12, 2)}

Question:49

Let the relation R be defined on the set $A = \{ 1, 2, 3, 4, 5 \}$ by $R = \{ (a, b) : \vert a\textsuperscript{2} - b\textsuperscript{2} \vert < 8 \}$ . Then R is given by ............

Answer:

Here
$\\A = \{ 1, 2, 3, 4, 5 \} and R = \{ (a, b) : \vert a\textsuperscript{2} - b\textsuperscript{2} \vert < 8 \} \\ \text{Therefore, we can say}, R = \{ (1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (4, 3) (3, 4), (4, 4), (5, 5) \} \\$

Question:50

Let f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)}. Then gof = ................ and fog = ...................

Answer:

$\\Here, f = \{ (1, 2), (3, 5), (4, 1) \} and g = \{ (2, 3), (5, 1), (1, 3) \} \\ Then, gof(1) = g[f(1)] = g(2) = 3\\ Again, gof(3) = g[f(3)] = g(5) = 1 \\ Again, gof(4) = g[f(4)] = g(1) = 3\\ Therefore, gof = \{ (1, 3), (3, 1), (4, 3) \} \\ Again, fog(2) = f[g(2)] = f(3) = 5\\ Again, fog(5) = f[g(5)] = f(1) = 2\\ Again, fog(1) = f[g(1)] = f(3) = 5\\ Hence, fog = \{ (2, 5), (5, 2), (1, 5) \} \\$

Question:51

Let $f : R \rightarrow R$ be defined by $f(x) = x/ \sqrt (1+x\textsuperscript{2})$ then $(fofof)(x) = ................. \\$

Answer:

$\begin{aligned} (\text { fofof })(x) &=f[f(x))] \\ &=f\left[\begin{array}{l} f\left(\frac{x}{\sqrt{1+x^{2}}}\right) \end{array}\right] \\ &=f\left(\frac{\frac{x}{\sqrt{1+x^{2}}}}{\sqrt{1+\frac{x^{2}}{1+x^{2}}}}\right) \\ &=f\left(\frac{x}{\sqrt{1+2 x^{2}}}\right) \end{aligned}$
$\\\text { fofof }(x) =\frac{\frac{x}{\sqrt{1+2 x^{2}}}}{\sqrt{1+\frac{x^{2}}{1+2 x^{2}}}} \\ =\frac{x}{\sqrt{1+3 x^{2}}}$

Question:52

If, $f(x) = [4 - (x - 7)\textsuperscript{3}]$ , then $f\textsuperscript{-1}$ (x) = .............

Answer:

$\\Here, f(x) = [4 - (x - 7)\textsuperscript{3}]\\ Say, y = [4 - (x - 7)\textsuperscript{3}]\\ Or, (x - 7)\textsuperscript{3} = (4 - y)\\ Or, (x-7) = (4-y)\textsuperscript{1/3}\\ Or, x = 7 + (4-y)\textsuperscript{1/3}\\ Therefore, f\textsuperscript{-1}(x) = 7 + (4 - x)\textsuperscript{1/3}\\$

Question:53

Let R = {(3,1), (1,3), (3,3)} be a relation defined on the set A = {1, 2, 3}. Then R is symmetric, transitive and not reflexive.

Answer:

False Here in the question, R = {(3,1), (1,3), (3,3)} which is defined on the set A = {1, 2, 3}

As (1 ,1) R, R is not a reflexive one.

As (3, 1) R and (1, 3) belongs to R, then R is symmetric.

Again, (1, 3) R, (3, 1) R. However (1, 1) does not belong to R. Then R is not transitive.

Question:54

Let f: R→R be the function defined by $f(x) = sin(3x+2)$ , x R. Then f is invertible.

Answer:

False
Here, $f(x) = sin(3x+2)$ , x R. This function is not a one-one onto function for all $x \in R.\\$
Again, $f(x) = sin(3x+2) = 0\\$
$or, 3x + 2 = n\pi, n \in \ Z$ .
Hence, the function is not invertible.

Question:55

Every relation which is symmetric and transitive is also reflexive.

Answer:

False

Assume, a relation R : R= {(1,2), (2,1), (2,2)} on the set A = {1,2}

Therefore, it is clear that (1,1) ∉ R. Hence, it is not reflexive.

Question:56

An integer m is said to be related to another integer n, if m is an integral multiple of n. This relation Z is reflexive, symmetric, and transitive.

Answer:

False.

Here, the given relation in the question is reflexive and transitive. However, it is not symmetric

Question:57

Let $A = \{ 0,1 \}$ and N be the set of natural numbers. Then the mapping $f: N \rightarrow A$ defined by $f(2n - 1) = 0, f(2n) = 1, n \in N, \\$ is onto.

Answer:

True
Here, $A = \{ 0,1 \} , and\: \: f(2n - 1) = 0, f(2n) = 1, \forall A n \in N\\$
Therefore, the range of f is $\{ 0, 1 \} \\$
Hence, $f: N \rightarrow A$ mapping is onto.

Question:58

The relation R on the set A = {1, 2, 3} defined as R = {(1,1), (1,2), (2,1), (3,3)}

Answer:

False

As, $R = \{ (1,1), (1,2), (2,1), (3,3) \} \\$

Therefore, $(2,2) \notin R\\$

Hence, R is not reflexive.

Question:59

The composition function is cumulative.
Answer:

$\\ False \\ Let, f(x) = x+1, g(x) = x \textsuperscript{2} \\ Therefore, fog(x) = f(g(x))\\ = f(x\textsuperscript{2})\\ \ \ \ \ \ = x\textsuperscript{2 }+ 1\\ Again, gof(x) = g(f(x))\\ =g(x+1)\\ =(x+1)\textsuperscript{2}\\ = (x\textsuperscript{2} + 2x + 1)\\ Hence, fog(x) \neq gof(x) \\$

Question:60

The composition of function is associative.

Answer:

True.

fo(goh)(x) = (fog)oh

Question:61

Every function is invertible.

Answer:

False.

Only bijective functions are invertible.

Question:62

A binary operation on a set always has the identity element.

Answer:

False

‘+’ is a binary operation on the set N but it has no identity element.

Subtopics Covered in NCERT Exemplar Solutions for Class 12 Maths Chapter 1

This chapter is divided into various sub-topics that are related to relations and functions and its operations. They are mentioned below:

Introduction
Types of relations
Types of functions
Composition of functions & invertible functions
Binary operations

What student will learn in NCERT Exemplar Class 12 Maths Solutions Chapter 1?

NCERT Exemplar solutions for Class 12 Maths chapter 1, students will learn about the basic concept of functions and relations. They will also get a better view of domains, co-domains, range etc. They will solve questions that will help in understanding these relations in a much more detailed way with help on lemmas and theorems.
NCERT Exemplar Class 12 Maths solutions chapter 1 will throw light on how to operate these functions by adding, multiplying, subtracting etc. Students will also get a more detailed view of various function types, injectivity and surjectivity. One can understand what all elements and factors comprise a function. Also, it throws light on a topic called invertible functions.
Topics about binary operations of functions using two operands will be covered in Class 12 Maths NCERT Exemplar solutions chapter 1 as well. It will explain in detail how commutativity, symmetry, associativity is used, in binary operations.
We have a team of expert teachers and guides who have solved these questions and have provided the same for the students to understand.
Solving NCERT questions is important, to understand the topic from the very beginning. We have solved the questions in a very simple way so that it is easier for the student to understand and grasp the topic and idea with much ease.

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NCERT Exemplar Class 12 Maths Solutions

Chapter 2 Inverse Trigonometric Functions

Chapter 3 Matrices

Chapter 4 Determinants

Chapter 5 Continuity and Differentiability

Chapter 6 Applications of Derivatives

Chapter 7 Integrals

Chapter 8 Applications of Integrals

Chapter 9 Differential Equations

Chapter 10 Vector Algebra

Chapter 11 Three dimensional Geometry

Chapter 12 Linear Programming

Chapter 13 Probability

Important Topics To Cover From NCERT Exemplar Class 12 Maths Solutions Chapter 1

· This is the very start of high school algebra, which holds the very root of higher education mathematics. In NCERT Exemplar Class 12 Maths chapter 1 solutions, students will learn about relations and functions, which is a type of relation. The relation is nothing but the relationship between two groups or the relationship between input and output.

· The function is a relation, but is used for obtaining the output from a set input. Learning the relations and functions at the very start of 12 Class will help in setting one's roots in calculus mathematics for cracking entrance exams and higher education.

· Several things are covered in Class 12 Maths NCERT Exemplar solutions chapter 1 which goes into the types of functions and relations. One will learn about various relations that will be used, like universal relations, reflective relations, empty relation, symmetric relation, etc.

NCERT Exemplar Class 12 Solutions

NCERT Exemplar Class 12 Chemistry Solutions

NCERT Exemplar Class 12 Physics Solutions

NCERT Exemplar Class 12 Biology Solutions

Also, check NCERT Solutions for questions given in the book:

Chapter 1	Relations and Functions
Chapter 2	Inverse Trigonometric Functions
Chapter 3	Matrices
Chapter 4	Determinants
Chapter 5	Continuity and Differentiability
Chapter 6	Application of Derivatives
Chapter 7	Integrals
Chapter 8	Application of Integrals
Chapter 9	Differential Equations
Chapter 10	Vector Algebra
Chapter 11	Three Dimensional Geometry
Chapter 12	Linear Programming
Chapter 13	Probability

Must Read NCERT Solution subject wise

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Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

1. Are these solutions helpful for board examinations?

Yes, NCERT exemplar solutions for Class 12 Maths chapter 1 are designed well to prepare you for board examinations.

2. What are the important topics of this chapter?

The important topics of this NCERT Exemplar Class 12 Maths solutions chapter 1 include Types of Relations, Types of Functions, Binary Operations and Composition of Functions and Invertible Function.

3. How to remember the formulae well?

The best ways to remember formulae is to keep revising them and solving related problems. Other methods include reading them on a daily basis. You can also prepare charts to stick near your study table or bed to go through in your leisure time.

4. How many questions are there in this chapter?

There are a total 70 questions in NCERT Exemplar Class 12 Maths solutions chapter 1 based on different concepts.

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NCERT Class 12 Maths Notes - Download Chapter wise Free PDFs

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NCERT Books for Class 12 - Download All Subjects PDFs Here

NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12 - Integrals

NCERT Solutions for Exercise 4.1 Class 11 Maths Chapter 4 - Principle of Mathematical Induction

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry

NCERT Exemplar Class 12 Physics Solutions Chapter 8 Electromagnetic Waves

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Can I change my board from CBSE to CHSE Odisha in Class 12th?

Hello,

Yes, you can switch from CBSE to CHSE Odisha in Class 12th, but there are a few conditions:

Eligibility : Check if CHSE Odisha allows lateral entry into Class 12th. Generally, students can transfer from CBSE to state boards with prior approval.
Document Requirements : You may need to provide transcripts, mark sheets, and a transfer certificate.
Subject Alignment : Ensure that the subjects you studied in CBSE align with the CHSE curriculum.
Application Process : Contact the respective CHSE Odisha authority or school for specific transfer procedures and deadlines.

Hope it helps !

Read Complete Answer

Prachi Kumari 30 December,2024

quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer

Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

Read Complete Answer

Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

Read Complete Answer

Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer

Ashvadha 12 July,2024

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Central Board of Secondary Education Class 12th Examination

JEE Test Series

Applications Open

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Exemplar Class 12 Maths Solutions Chapter 1 Relations and Functions

NCERT Exemplar Class 12 Maths Solutions Chapter 1: Exercise-1.3

Subtopics Covered in NCERT Exemplar Solutions for Class 12 Maths Chapter 1

What student will learn in NCERT Exemplar Class 12 Maths Solutions Chapter 1?

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NCERT Exemplar Class 12 Maths Solutions

Important Topics To Cover From NCERT Exemplar Class 12 Maths Solutions Chapter 1

NCERT Exemplar Class 12 Solutions

Also, check NCERT Solutions for questions given in the book:

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Read more NCERT Notes subject wise

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