# NCERT Solutions for Class 11 Maths Chapter 2 Relations And Functions

NCERT Solutions for Class 11 Maths Chapter 2 Relations And Functions: In the previous chapter you have learnt about sets. This chapter is the continuation of chapter 1 sets. In this article, you will get NCERT solutions for class 11 maths chapter 2 relations and functions. You have many relations in your daily life, like father-son, mother-son, etc. In a similar way, you can form relations in mathematics also. When there is only one output for every input, the relation becomes a function. Important topics like domain, co-domain, and range of functions are covered in this chapter. In solutions of NCERT for class 11 maths chapter 2 relations and functions, questions related to these topics are covered. Also, you will learn different types of specific real-valued functions and their graphs. CBSE NCERT solutions for class 11 maths chapter 2 relations and functions will build your fundamentals of functions which will be helpful in the 12th board exam also. Check all NCERT solutions from class 6 to 12 to learn science and maths. There are three exercises and a miscellaneous exercise in this chapter which are explained below.

Exercise:2.1

Exercise:2.2

Exercise:2.3

Miscellaneous Exercise

A={1,2,3,4,5}, B={2,6,12,20,30}, we can find a relation from A to B by inspection. That is set B is related to set A by the relation $\mathbf{(x,y):y=x^2+x}$ and this is one possible relation from set A to set B

The NCERT solutions for class 11 maths chapter 2 relations and functions start with cartesian products of sets. The main points to remember in these topics are:

1. Cartesian products two sets A and B are set of all ordered pair of the elements from A and B, for example, A={1,2,3}, B= {2,3} then

$A\times B=(1,2),(1,3),(2,2),(2,3),(3,2),(3,3)$

2. If set A contains n elements and set B contains m elements then Cartesian Products A and B contains mn elements.

3. Two ordered pairs are equal if corresponding first and second elements are equal. That is if (a,b) =(c,d) then a=c and b=d

## The main topics under NCERT Solutions for Class 11 Maths Chapter 2 Relations And Functions are listed below

2.1 Introduction

2.2 Cartesian Products of Sets

2.3 Relations

2.4 Functions

The second important topic of this chapter is a relation. A few points to remember are listed below

1. A relation from non-empty set A to non-empty set B is a subset of $A\times B$

2. The set of all first elements of the ordered pairs in a relation R from a set A to a set B is called the domain R and the set of all second elements in a relation R from a set A to a set B is called the range of R., For example, consider a relation R from set A ={1,2,3,4} to Set B={1,2,3.........20} of real numbers defined by R={(1,1),(2,4),(3,9),(4,16)}. Then the set {1,2,3,4} is the domain of R and {1,4,9,16} is the range of R and the whole set B is the codomain.

3. Let the set A has n element and set B has m element then the number of possible relation from A to B is $2^{mn}$

The last topic of this chapter is functions: A relation R from set A to set B is called as a function if every element of set A has only one image in set B.

Algebra of real functions: Let f and g be any real functions then

$\\a)(f + g) (x) = f (x) + g (x)\\ b)(f-g) (x) = f(x) -g(x) \\c)(fg) (x) = f(x) g(x)\\d)\frac{f}{g}(x)=\frac{f(x))}{g(x))}, \ g(x)\neq0$

NCERT solutions for class 11 maths chapter 2 relations and functions-Exercise: 2.1

It is given that
$\left ( \frac{x}{3}+1 , y - \frac{2}{3} \right ) = \left ( \frac{5}{3},\frac{1}{3} \right )$
Since the ordered pairs are equal, the corresponding elements will also be equal
Therefore,
$\frac{x}{3}+1= \frac{5}{3} \ \ \ and \ \ \ y - \frac{2}{3}= \frac{1}{3}$
$\frac{x}{3}= \frac{5}{3}-1 \ \ \ and \ \ \ y = \frac{1}{3}+ \frac{2}{3}$
$\frac{x}{3}= \frac{5-3}{3} \ \ \ and \ \ \ y = \frac{1+2}{3}$
$\frac{x}{3}= \frac{2}{3} \ \ \ and \ \ \ y = \frac{3}{3}$
$x= 2 \ \ \ and \ \ \ y = 1$

Therefore, values of x and y are  2  and 1  respectively

It is given that set A has 3 elements and the elements in set B are 3 , 4 , and 5
Therefore, the number of elements in set B is 3
Now,
Number of elements in $(A \times B)$
= ( Number of elements in set A ) $\times$ ( Number of elements in set B)
= 3 $\times$ 3
= 9
Therefore, number of elements in $(A \times B)$ is 9

It is given that
G = {7, 8} and H = {5, 4, 2}
We know that the cartesian product of two non-empty sets P and Q is defined as
$\times$ Q = {(p,q) , where p $\epsilon$ P , q $\epsilon$ Q }
Therefore,
$\times$ H = {(7,5),(7,4),(7,2),(8,5),(8,4),(8,2)}
And
$\times$ G = {(5,7),(5,8),(4,7),(4,8),(2,7),(2,8)}

FALSE
If  P = {m, n} and Q = { n, m}
Then,
$P \times Q = \left \{ (m,m),(m,n),(n,m),(n,n) \right \}$

It is a  TRUE  statement

$\because$    If A and B are non-empty sets, then A × B is a non-empty set of ordered  pairs (x, y) such that $x \epsilon A$ and $y \epsilon B$

This statement is  TRUE

$\because$    If A = {1, 2}, B = {3, 4}, then

$B\cap\phi=\phi$

There for

$A \times (B \cap \phi ) = \phi$

It is given that
A = {–1, 1}
A is an non-empty set
Therefore,
Lets first find $A \times A$
$A \times A = \left \{ -1,1 \right \} \times \left \{ -1,1 \right \} = \left \{ (-1,-1),(-1,1),(1,-1),(1,1) \right \}$
Now,
$A\times A \times A=A\times (A \times A) = \left \{ -1,1 \right \} \times \left \{ (-1,-1),(-1,1),(1,-1),(1,1) \right \}$$=\left \{ (-1,-1,-1),(-1,-1,1),(-1,1,-1),(-1,1,1),(1,-1,-1),(1,-1,1),(1,1,-1),(1,1,1) \right \}$

It is given that
$A \times B$  = {(a, x),(a , y), (b, x), (b, y)}
We know that the cartesian product of two non-empty set P and Q is defined as

$P \times Q = \left \{ (p,q) : p \ \epsilon \ P , q \ \epsilon \ Q \right \}$
Now, we know that A is the set of all first elements and B is the set of all second elements
Therefore,
$A= \left \{ a,b \right \} \ \ \ and \ \ \ B = \left \{ x , y \right \}$

It is given that
A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}
Now,
$B \cap C = \left \{ 1,2,3,4 \right \} \cap \left \{ 5,6 \right \} = \Phi$
Now,
$A \times ( B \cap C ) = A \times \phi = \phi \ \ \ \ \ \ \ \ \ \ \ \ -(i)$

$A \times B = \left \{ (1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4) \right \}$
And
$A \times C = \left \{ (1,5),(1,6),(2,5),(2,6) \right \}$
Now,
$(A \times B)\cap (A \times C) =\phi \ \ \ \ \ \ \ \ \ \ \ \ -(ii)$
From equation (i) and (ii) it is clear that
$L.H.S. = R.H.S.$
Hence,
$A \times ( B \cap C ) = ( A \times B ) \cap ( A \times C )$

It is given that
A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}
Now,

$A \times C = \left \{ (1,5),(1,6),(2,5),(2,6) \right \}$
And
$B \times D = \left \{ (1,5),(1,6),(1,7),(1,8),(2,5),(2,6),(2,7),(2,8),(3,5),(3,6),(3,7),(3,8),(4,5),(4,6),(4,7),(4,8) \right \}$
We can clearly observe that all the elements of the set $A \times C$ are the elements of the set  $B \times D$
Therefore, $A \times C$ is a subset  of $B \times D$

It is given that
A = {1, 2} and B = {3, 4}
Then,
$A \times B = \left \{ (1,3),(1,4),(2,3),(2,4) \right \}$
$\Rightarrow n\left ( A \times B \right ) = 4$
Now, we know that if C is a set with $n(C) = m$
Then,
$n[P(C)]= 2^m$
Therefore,
The set $A \times B$  has $2^4=16$   subsets.

It is given that
n(A) = 3 and n(B) = 2 and If (x, 1), (y, 2), (z, 1) are in A × B.

By definition of Cartesian product of two non-empty Set P and Q:
$P \times Q = \left \{ (p,q) : p \ \epsilon \ P , q \ \epsilon \ Q \right \}$
Now, we can see that
P = set of all first elements.
And
Q = set of all second elements.
Now,
$\Rightarrow$ (x, y, z) are elements of A and (1,2) are elements of B
As n(A) = 3 and n(B) = 2
Therefore,
A = {x, y, z} and B = {1, 2}

It is given that Cartesian product A × A having 9 elements among which are found (–1, 0) and (0,1).
Now,
Number of elements in (A× B) = (Number of elements in set A) × (Number of elements in B)
$n(A \times A) = n(A) \times n(A)$
It is given that  $n(A \times A) = 9$
Therefore,
$n(A) \times n(A) = 9$
$\Rightarrow n(A) = 3$
Now,
By definition A × A = {(a, a): a ? A}
Therefore,
-1, 0 and 1 are the elements of set A
Now, because, n(A) = 3 therefore, A = {-1, 0, 1}
Therefore,
the remaining elements of set (A × A) are
(-1,-1), (-1,1), (0,0), (0, -1), (1,1), (1, -1) and (1, 0)

## Solutions of NCERT for class 11 maths chapter 2 relations and funcions-Exercise: 2.2

It is given that
$A = \left \{ 1, 2, 3, ..., 14 \right \} \ and \ R = \left \{ (x, y) : 3x - y = 0, \ where \ x, y \ \epsilon \ A \right \}$
Now, the relation R from A to A is given as
$R = \left \{ ( x,y): 3x -y = 0 , where \: \: x , y \epsilon A \right \}$
Therefore,
the relation in roaster form is ,  $,R = \left \{ (1, 3), (2, 6), (3, 9), (4, 12) \right \}$
Now,
We know that  Domain of R = set of all first elements of the order pairs in the relation
Therefore,
Domain of   $R = \left \{ 1, 2, 3, 4 \right \}$
And
Codomain of R = the whole set A
i.e.   Codomain of   $R = \left \{ 1, 2, 3, ..., 14 \right \}$
Now,
Range of R = set of all second elements of the order pairs in the relation.
Therefore,
range of    $R = \left \{ 3, 6, 9, 12 \right \}$

It is given that
$R = \left \{ ( x,y ) : y = x +5 , x$ is a natural number less than $4 ; x , y \epsilon N \left. \right \}$

As x is a natural number which is less than 4.
Therefore,
the relation in roaster form is,  $R = \left \{ (1,6), (2,7), (3,8) \right \}$
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of   $R = \left \{ 1, 2, 3 \right \}$

Now,
Range of R = set of all second elements of the order pairs in the relation.
Therefore,
the range of   $R = \left \{ 6, 7, 8 \right \}$

Therefore,  domain and the range are $\left \{ 1,2,3 \right \} \ \ and \ \ \left \{ 6, 7, 8 \right \}$  respectively

It is given that
A = {1, 2, 3, 5} and B = {4, 6, 9}
And
$R = \left \{ ( x,y ) : the \: \: diffrence \: \: between \: \: x \: \: and \: \: y \: \: is \: \: odd ; x \epsilon A , y \epsilon B \right \}$
Now, it is given that the difference should be odd. Let us take all possible differences.
(1 - 4) = - 3, (1 - 6) = - 5, (1 - 9) = - 8(2 - 4) = - 2, (2 - 6) = - 4, (2 - 9) = - 7(3 - 4) = - 1, (3 - 6) = - 3, (3 - 9) = - 6(5 - 4) = 1, (5 - 6) = - 1, (5 - 9) = - 4
Taking the difference which are odd we get,

Therefore,
the relation in roaster form,  $R = \left \{ (1,4), (1,6), (2,9), (3,4), (3,6), (5,4), (5,6) \right \}$

It is given in the figure that

P = {5,6,7}, Q = {3,4,5}

Therefore,
the relation in set builder form is ,
$R = \left \{ {(x, y): y = x-2; x \ \epsilon \ P} \right \}$
OR
$R = \left \{ {(x, y): y = x-2; \ for \ x = 5, 6, 7} \right \}$

From the given figure. we observe that

P = {5,6,7}, Q = {3,4,5}

And the relation in roaster form is ,  $R = \left \{ {(5,3), (6,4), (7,5)} \right \}$

As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of  $R = \left \{ {5, 6, 7} \right \}$

Now,
Range of R = set of all second elements of the order pairs in the relation.
Therefore,
the range of    $R = \left \{ {3, 4, 5} \right \}$

It is given that
A = {1, 2, 3, 4, 6}
And
$R = \left \{ ( a,b): a ,b \epsilon A , b\: \: is\: \: exactly \: \: divisible\: \: by \: \: a \right \}$

Therefore,
the relation in roaster form is , $R = \left \{ {(1,1), (1,2), (1,3), (1,4), (1,6), (2,2), (2,4), (2,6), (3,3), (3,6), (4,4), (6,6)} \right \}$

It is given that
A = {1, 2, 3, 4, 6}
And
$R = \left \{ ( a,b): a ,b \epsilon A , b\: \: is\: \: exactly \: \: divisible\: \: by \: \: a \right \}$
Now,
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of   $R = \left \{ {1, 2, 3, 4, 6} \right \}$

It is given that
A = {1, 2, 3, 4, 6}
And
$R = \left \{ ( a,b): a ,b \epsilon A , b\: \: is\: \: exactly \: \: divisible\: \: by \: \: a \right \}$
Now,
As the range of R = set of all second elements of the order pairs in the relation.
Therefore,
Range of   $R = \left \{ {1, 2, 3, 4, 6} \right \}$

It is given that
$R = \left \{ ( x , x +5 ): x \epsilon \left \{ 0,1,2,3,,4,5 \right \} \right \}$

Therefore,
the relation in roaster form is , $R = \left \{ {(0,5), (1,6), (2,7), (3,8), (4,9), (5,10)} \right \}$

Now,
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of  $R =\left \{ {0, 1, 2, 3, 4, 5} \right \}$

Now,
As Range of R = set of all second elements of the order pairs in the relation.
Range of    $R =\left \{ {5, 6, 7, 8, 9, 10} \right \}$

Therefore,  the domain and range of the relation R is $\left \{ 0,1,2,3,4,5 \right \} \ \ and \ \ \left \{ {5, 6, 7, 8, 9, 10} \right \}$   respectively

It is given that
$R = \left \{ \right.(x, x^3) : x\: \: is\: \: a\: \: prime\: \: number \: \: less\: \: than\: \: 10\: \: \left. \right \}$
Now,
As we know the prime number less than 10 are 2, 3, 5 and 7.
Therefore,
the relation in roaster form is , $R =\left \{ {(2,8), (3,27), (5,125), (7,343)} \right \}$

It is given that
A = {x, y, z} and B = {1, 2}
Now,
$A \times B = \left \{ {(x,1), (x,2), (y,1), (y,2), (z,1), (z,2)} \right \}$
Therefore,
$n(A \times B) = 6$
Then, the number of subsets of the set $(A \times B) = 2^n = 2^6$

Therefore, the number of relations from A to B is  $2^6$

It is given that
$R = \left \{ ( a,b) : a , b \epsilon Z , a-b\: \: is \: \: an \: \: integer \right \}$
Now, as we know that the difference between any two integers is always an integer.
And
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
The domain of R = Z

Now,
Range of R = set of all second elements of the order pairs in the relation.
Therefore,
range of R = Z

Therefore, the domain and range of R is Z and  respectively

## CBSE NCERT solutions for class 11 maths chapter 2 relations and functions-Exercise: 2.3

Since, 2, 5, 8, 11, 14 and 17 are the elements of domain R having their unique images. Hence, this relation R is a function.
Now,
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of  $R =\left \{ {2, 5, 8, 11, 14, 17} \right \}$

Now,
As Range of R = set of all second elements of the order pairs in the relation.
Therefore,
Range of $R =\left \{ {1} \right \}$

Therefore, domain and range of R are  $\left \{ {2, 5, 8, 11, 14, 17} \right \} \ and \ \left \{ 1 \right \}$  respectively

Since, 2, 4, 6, 8, 10,12 and 14 are the elements of domain R having their unique images. Hence, this relation R is a function.

Now,
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of  $R =\left \{ {2, 4, 6, 8, 10,12, 14} \right \}$

Now,
As Range of R = set of all second elements of the order pairs in the relation.
Therefore,
Range of  $R =\left \{ {1, 2, 3, 4, 5,6, 7} \right \}$

Therefore, domain and range of R are  $\left \{ {2, 4, 6, 8, 10,12, 14} \right \} \ and \ \left \{ {1, 2, 3, 4, 5,6, 7} \right \}$  respectively

Since the same first element 1 corresponds to two different images 3 and 5. Hence, this relation is not a function.

$f (x ) = - |x|$

Given function is
$f (x ) = - |x|$
Now,  we know that

$|x|\left\{\begin{matrix} x &if \ x> 0 \\ -x& if \ x<0 \end{matrix}\right.$
$\Rightarrow f(x)=-|x|\left\{\begin{matrix} -x &if \ x> 0 \\ x& if \ x<0 \end{matrix}\right.$

Now, for a function f(x),
Domain: The values that can be put in the function to obtain real value. For example f(x) = x, now we can put any value in place of x and we will get a real value. Hence, the domain of this function will be Real Numbers.
Range: The values that we obtain of the function after putting the value from domain. For Example: f(x) = x + 1, now if we put x = 0, f(x) = 1. This 1 is a value of Range that we obtained.

Since f(x) is defined for $x \ \epsilon \ R$the domain of f is R.

It can be observed that the range of f(x) = -|x| is all real numbers except positive real numbers. Because will always get a negative number when we put a value from the domain.
Therefore, the range of f is $(-\infty , 0]$

Question:2 (ii)  Find the domain and range of the following real functions:

$f ( x ) = \sqrt { 9- x ^2 }$

Given function is
$f ( x ) = \sqrt { 9- x ^2 }$
Now,
Domain: These are the values of x for which f(x) is defined.
for the given f(x) we can say that, f(x) should be real and for that,9 - x2 ≥ 0 [Since a value less than 0 will give an imaginary value]
$\Rightarrow 3^2-x^2= (3-x)(3+x)\geq 0$
$\Rightarrow -3\leq x\leq 3$
Therefore,
The domain of f(x) is  $[-3,3]$
Now,
If  we put the value of x from  $[-3,3]$ we will observe that the value of function $f ( x ) = \sqrt { 9- x ^2 }$  varies from 0 to 3
Therefore,
Range of f(x) is  $[0,3]$

Given function is
$f(x) = 2x-5$
Now,
$f(0) = 2(0)-5=0-5 = -5$
Therefore,
Value of f(0) is -5

Given function is
$f(x) = 2x-5$
Now,
$f(7) = 2(7)-5=14-5 = 9$
Therefore,
Value of f(7) is 9

Given function is
$f(x) = 2x-5$
Now,
$f(-3) = 2(-3)-5=-6-5 = -11$
Therefore,
Value of f(-3) is -11

Given function is
$t ( C ) = \frac{9 C }{5} + 32$
Now,
$t ( 0 ) = \frac{9 (0) }{5} + 32= 0+ 32 = 32$
Therefore,
Value of t(0) is 32

Given function is
$t ( C ) = \frac{9 C }{5} + 32$
Now,
$t ( 28 ) = \frac{9 (28) }{5} + 32= \frac{252}{5}+ 32 = \frac{252+160}{5}= \frac{412}{5}$
Therefore,
Value of t(28) is  $\frac{412}{5}$

Given function is
$t ( C ) = \frac{9 C }{5} + 32$
Now,
$t ( -10 ) = \frac{9 (-10) }{5} + 32= \frac{-90}{5}+ 32 = -18+32= 14$
Therefore,
Value of t(-10) is  14

Given function is
$t ( C ) = \frac{9 C }{5} + 32$
Now,
$212 = \frac{9 (C) }{5} + 32$
$212 \times 5= {9 (C) } + 160$
${9 (C) } =1060-160$
$C = \frac{900}{9} = 100$
Therefore,
When t(C) = 212 , value of C is  100

Question:5 (i)  Find the range of each of the following functions.

$f (x) = 2 - 3x, x \epsilon R, x > 0.$

Given function is

$f (x) = 2 - 3x, x \epsilon R, x > 0.$
It is given that $x > 0$
Now,
$\Rightarrow 3x > 0$
$\Rightarrow -3x < 0$
Add 2 on both the sides
$\Rightarrow -3x+2 < 0+2$
$\Rightarrow 2-3x < 2$
$\Rightarrow f(x) < 2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because f(x) = 2-3x)$
Therefore,
Range of function $f(x) = 2 -3x$ is  $(-\infty,2)$

Question:5 (ii) Find the range of each of the following functions

$f ( x ) = x ^2 +2$ , x is a real number.

Given function is

$f ( x ) = x ^2 +2$
It is given that  is a real  number
Now,
$\Rightarrow x^2 \geq 0$
Add 2 on both the sides
$\Rightarrow x^2+2 \geq 0+2$
$\Rightarrow f(x) \geq 2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because f(x) = x^2+2)$
Therefore,
Range of function $f ( x ) = x ^2 +2$ is  $[2,\infty)$

Question:5 (iii)  Find the range of each of the following functions.

f (x) = x, x is a real number

Given function is

$f ( x ) = x$
It is given that  is a real  number
Therefore,
Range of function $f ( x ) = x$  is  R

## NCERT solutions for class 11 maths chapter 2 relations and functions-Miscellaneous Exercise

It is given that
$f (x) = \left\{\begin{matrix} x^2 & 0 \leq x\leq 3 \\ 3x &3 \leq x \leq 10 \end{matrix}\right.$
Now,
$f(x) = x^2 \ for \ 0\leq x\leq 3$
And
$f(x) = 3x \ for \ 3\leq x\leq 10$

At x = 3,$f(x) = x^2 = 3^2 = 9$

Also, at x = 3,$f(x) = 3x = 3\times 3 = 9$

We can see that for $0\leq x\leq 10$, f(x) has unique images.

Therefore, By definition of a function, the given relation is function.

Now,
It is given that
$g(x) = \left\{\begin{matrix} x^2 & 0 \leq x\leq 2 \\ 3x &2 \leq x \leq 10 \end{matrix}\right.$
Now,
$g(x) = x^2 \ for \ 0\leq x\leq 2$
And
$g(x) = 3x \ for \ 2\leq x\leq 10$

At x = 2, $g(x) = x^2 = 2^2 = 4$
Also, at x = 2, $g(x) = 3x = 3\times2 = 6$
We can clearly see that element 2 of the domain of relation g(x) corresponds to two different images i.e. 4 and 6. Thus,  f(x) does not have unique images
Therefore, by definition of a function, the given relation is not a function

Hence proved

Given function is
$f(x)= x^2$
Now,
$\frac{f ( 1.1)- f (1)}{(1.1-1)} = \frac{(1.1)^2-1^2}{(1.1-1)} = \frac{1.21-1}{0.1}= \frac{0.21}{0.1}= 2.1$

Therefore, value of $\frac{f ( 1.1)- f (1)}{(1.1-1)}$  is  2.1

Given function is
$f (x) = \frac{x^2 + 2 x +1}{x^2 - 8x + 12 }$
Now, we will simplify it into
$f (x) = \frac{x^2 + 2 x +1}{x^2 - 8x + 12 }$
$= \frac{x^2+2x+1}{x^2-6x-2x+12}$
$= \frac{x^2+2x+1}{x(x-6)-2(x-6)}$
$= \frac{x^2+2x+1}{(x-2)(x-6)}$
Now, we can clearly see that  $x \neq 2 , 6$
Therefore, the Domain of f(x)  is $(R-\left \{ 2,6 \right \})$

Given function is
$f (x) = \sqrt{(x-1)}$
We can clearly see that f(x) is only defined for the values of x ,  $x\geq 1$
Therefore,
The domain of the function $f (x) = \sqrt{(x-1)}$  is  $[1,\infty)$
Now, as
$\Rightarrow x\geq 1$
$\Rightarrow x-1\geq 1-1$
$\Rightarrow x-1\geq 0$
take square root on both sides
$\Rightarrow \sqrt{x-1}\geq 0$
$\Rightarrow f(x)\geq 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because f(x)= \sqrt{x-1})$
Therefore,
Range  of function $f (x) = \sqrt{(x-1)}$  is  $[0,\infty)$

Given function is
$f (x) = |x-1|$
As the given function is defined of all real number
The domain of the function $f (x) = |x-1|$  is  R
Now, as we know that the mod function always gives only  positive values
Therefore,
Range  of function $f (x) = |x-1|$  is  all non-negative real numbers i.e. $[0,\infty)$

Given function is
$f = \left \{ \left ( x , \frac{x^2}{1+ x^2} \right ) : x \epsilon R \right \}$
Range of any function is the set of values obtained after the mapping is done in the domain of the function. So every value of the codomain that is being mapped is Range of the function.
Let's take
$y = \frac{x^2}{1+x^2}$
$\Rightarrow y(1+x^2)=x^2$
$\Rightarrow y+yx^2=x^2$
$\Rightarrow y=x^2(1-y)$
$\Rightarrow x^2= \frac{y}{1-y}$
$\Rightarrow x= \pm \sqrt{\frac{y}{1-y}}$
Now, 1 - y should be greater than zero and y should be greater than and equal to zero for x to exist because other than those values the x will be imaginary
Thus, $1 - y > 0 , y < 1 \ and \ y \geq 0$
Therefore,
Range of given function is  $[0,1)$

It is given that
$f,g : R \rightarrow R$
$f(x)=x+1 \ \ and \ \ g(x) = 2x - 3$
Now,
$(f+g)x = f(x)+g(x)$
$= (x+1)+(2x-3)$
$= 3x-2$
Therefore,
$(f+g)x= 3x-2$

Now,
$(f-g)x = f(x)-g(x)$
$= (x+1)-(2x-3)$
$= x+1-2x+3$
$= -x+4$
Therefore,
$(f-g)x= -x+4$

Now,
$\left ( \frac{f}{g} \right )x = \frac{f(x)}{g(x)} , g(x)\neq 0$
$=\frac{x+1}{2x-3} \ , x \neq \frac{3}{2}$
Therefore, values of $(f+g)x,(f-g)x \ and \ \left ( \frac{f}{g} \right )x$  are $(3x-2),(-x+4) \ and \ \frac{x+1}{2x-3}$   respectively

It is given that
$f =\left \{ {(1,1), (2,3), (0,-1), (-1, -3)} \right \}$
And
$f(x) = ax+b$
Now,
At x = 1 , $f(x) = 1$
$\Rightarrow f(1)= a(1)+b$
$\Rightarrow a+b = 1 \ \ \ \ \ \ \ \ \ \ -(i)$

Similarly,
At $x = 0$ , $f(x) = -1$
$\Rightarrow f(0) = a(0)+b$
$\Rightarrow b = -1$
Now, put this value of  b  in equation (i)
we will get,
$a = 2$
Therefore, values of and  b are and -1  respectively

$( a,a ) \epsilon R ,$ for all $a \epsilon N$

It is given that
$R = \left \{ ( a,b): a,b \epsilon N \: \:and \: \: a = b ^ 2 \right \}$
And
$( a,a ) \epsilon R ,$  for all  $a \epsilon N$
Now, it can be seen that  $2 \ \epsilon \ N$  But, $2 \neq 2^ 2 = 4$
Therefore, this statement is FALSE

$( a,a ) \epsilon R ,$ implies (b,a)  $\epsilon$ R

It is given that
$R = \left \{ ( a,b): a,b \epsilon N \: \:and \: \: a = b ^ 2 \right \}$
And
$( a,b ) \ \epsilon \ R ,$ implies $(b,a) \ \epsilon \ R$
Now , it can be seen that  $( 2,4 ) \ \epsilon \ R ,$  and $4 = 2^2 = 4$ ,  But  $2 \neq 4^2 =16$
Therefore,  $(2,4) \ \notin \ N$
Therefore,  given statement is FALSE

(a,b) $\epsilon$ R, (b,c) $\epsilon$ R implies (a,c) $\epsilon$ R.

It is given that
$R = \left \{ ( a,b): a,b \epsilon N \: \:and \: \: a = b ^ 2 \right \}$
And
$( a,b ) \ \epsilon \ R , (b,c) \ \epsilon \ R$ implies $(a,c) \ \epsilon \ R$
Now, it can be seen that  $(16,4) \ \epsilon \ R , ( 4,2 ) \ \epsilon \ R$  because  $16 = 4^2 = 16$  and $4 = 2^2 = 4$ ,  But  $16 \neq 2^2 =4$
Therefore,  $(16,2) \ \notin \ N$
Therefore, the given statement is FALSE

It is given that
$A =\left \{ {1,2,3,4} \right \}$
$B =\left \{ {1,5,9,11,15,16} \right \}$
and $f =\left \{ {(1,5), (2,9), (3,1), (4,5), (2,11)} \right \}$
Now,
$A \times B =\left \{ (1,1), (1,5), (1,9), (1,11), (1,15), (1,16), (2,1), (2,5), (2,9), (2,11), (2,15), (2,16), (3,1), (3,5), (3,9), (3,11), (3,15), (3,16), (4,1), (4,5), (4,9), (4,11), (4,15), (4,16) \right \}$Now, a relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B
And we can see that f is a subset of $A \times B$
Hence  f is a relation from A to B
Therefore, given statement is TRUE

It is given that
$A =\left \{ {1,2,3,4} \right \}$
$B =\left \{ {1,5,9,11,15,16} \right \}$
and $f =\left \{ {(1,5), (2,9), (3,1), (4,5), (2,11)} \right \}$
Now,
$A \times B =\left \{ (1,1), (1,5), (1,9), (1,11), (1,15), (1,16), (2,1), (2,5), (2,9), (2,11), (2,15), (2,16), (3,1), (3,5), (3,9), (3,11), (3,15), (3,16), (4,1), (4,5), (4,9), (4,11), (4,15), (4,16) \right \}$
As we can observe that same first element i.e. 2 corresponds to two different images that is 9 and 11.
Hence f is not a function from A to B
Therefore, given statement is FALSE

It is given that
$f = \left \{ (ab, a + b) : a, b \epsilon Z } \right \}$
Now, we know that relation f from a set A to a set B is said to be a function only if every element of set A has a unique image in set B
Now,  for value 2, 6, -2, -6 $\epsilon \ Z$
$f = \left \{ (2 �\times 6, 2 + 6), (-2 \times -6, -2 - 6), (2 \times -6, 2 - 6), (-2 \times 6, -2 + 6) \right \}$
$\Rightarrow f = \left \{ (12, 8), (12, -8), (-12, -4), (-12, 4) \right \}$
Now,  we can observe that same first element i.e. 12 corresponds to two different images that are 8 and -8.
Thus,   is not a function

It is given that
A = {9,10,11,12,13}
And
f : A
N be defined by f(n) = the highest prime factor of n.
Now,
Prime factor of 9 = 3

Prime factor of 10 = 2,5

Prime factor of 11 = 11

Prime factor of 12 = 2,3

Prime factor of 13 = 13

f(n) = the highest prime factor of n.

Hence,

f(9) = the highest prime factor of 9 = 3

f(10) = the highest prime factor of 10 = 5

f(11) = the highest prime factor of 11 = 11

f(12) = the highest prime factor of 12 = 3

f(13) = the highest prime factor of 13 = 13

As the range of f is the set of all f(n), where $n \ \epsilon \ A$

Therefore, the range of f is: {3, 5, 11, 13}.

## NCERT solutions for class 11 mathematics

 chapter-1 Solutions of NCERT for class 11 maths chapter 1 Sets chapter-2 NCERT solutions for class 11 maths chapter 2 Relations and Functions chapter-3 CBSE NCERT solutions for class 11 chapter 3 Trigonometric Functions chapter-4 NCERT solutions for class 11 chapter 4 Principle of Mathematical Induction chapter-5 Solutions of NCERT for class 11 chapter 5 Complex Numbers and Quadratic equations chapter-6 CBSE NCERT solutions for class 11 maths chapter 6 Linear Inequalities chapter-7 NCERT solutions for class 11 maths chapter 7 Permutation and Combinations chapter-8 Solutions of NCERT for class 11 maths chapter 8 Binomial Theorem chapter-9 CBSE NCERT solutions for class 11 maths chapter 9 Sequences and Series chapter-10 NCERT solutions for class 11 maths chapter 10 Straight Lines chapter-11 Solutions of NCERT for class 11 maths chapter 11 Conic Section chapter-12 CBSE NCERT solutions for class 11 maths chapter 12 Introduction to Three Dimensional Geometry chapter-13 NCERT solutions for class 11 maths chapter 13 Limits and Derivatives chapter-14 Solutions of NCERT for class 11 maths chapter 14 Mathematical Reasoning chapter-15 CBSE NCERT solutions for class 11 maths chapter 15 Statistics chapter-16 NCERT solutions for class 11 maths chapter 16 Probability

## NCERT solutions for class 11- Subject wise

 Solutions of NCERT for class 11 biology CBSE NCERT solutions for class 11 maths NCERT solutions for class 11 chemistry Solutions of NCERT for class 11 physics

## Benefits of NCERT solutions

• NCERT solutions for class 11 maths chapter 2 relations and functions will build your basics of functions which will be helpful in 12th board exams also.
• All these questions are prepared and explained in a detailed manner so it will be very easy for you to understand the concepts
• CBSE NCERT solutions for class 11 maths chapter 2 relations and functions ll develop yous basic concept which will be helpful in further studies relational algebra, relational calculus, statistics, machine learning, etc.
• Tip- Only reading the solutions won't help, you should try to solve on your own. If you are not able to solve, you can take the help of NCERT solutions for class 11 maths chapter 2 relations and functions.