# Relations and Functions

## Class 11 NCERT

### NCERT

1   If$(\dfrac{x}{3}+1,y-\dfrac{2}{3})=(\dfrac{5}{3},\dfrac{1}{3}),$ find the values of x and y.

##### Solution :

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

2   If the set $A$ has $3$ elements and the set $B =\{3, 4,5 \}$, then find the number of elements in $( A * B )$ ?

##### Solution :

It is given that set $A$ has $3$ elements and the elements of set $B$ are $3, 4,$ and $5.$$\\ \Rightarrow Number of elements in set B = 3$$\\$ Number of elements in $(A * B )$$\\ = (Number of elements in A) *(Number of elements in B)\\ = 3 *3 = 9$$\\$ Thus, the number of elements in $( A * B )$ in $9.$

3   If $G = \{7, 8\}$ and $H = \{5, 4, 2\},$ find $G * H$ and $H * G$ .

##### Solution :

$G = \{7, 8\}$ and $H = \{5, 4, 2\},$$\\ We know that the Cartesian product P * Q of two non-empty sets P and Q is defined as\\ P * Q -\{ (p , q) : p \in P , q \in Q \}\\ \therefore G*H=\{(7,5),(7,4),(8,5),(8,4),(8,2)\}\\ H*G=\{(5,7),(5,8),(4,7),(4,8),(2,7),(2,8)\}$$\\$

4   State whether each of the following statement are true or false. If the statement is false, rewrite the given statement correctly.$\\$ (i) If P = {m, n} and Q = {n, m}, then P *Q =$\{$( m , n )( n , m )$\}$ .$\\$ (ii) If A and B are non-empty sets, then A *B is a non-empty set of ordered pairs (x, y) such that x $\in$ A and y $\in$ B .$\\$ (iii) If A =$\{ 1, 2 \}$ , B $\{3, 4 \}$ , then A *$\{$ B $\cap \oslash \}=\oslash$

If is known that for any non-empty set $A,A*A*A$ is defined as$\\$ $A*A*A=\{(a,b,c):a,b,c\in A\}$ It is given that $A=\{-1,1\}$$\\ \therefore A*A*A=\{(-1,-1,-1),(-1,-1,1),(-1,1,-1),(-1,1,1),\\ (1,-1,-1),(1,-1,1),(1,1,-1),(1,1,1)\} 5 If A=\{-1,1\}, find A*A*A. ##### Solution : If is known that for any non-empty set A,A*A*A is defined as\\ A*A*A=\{(a,b,c):a,b,c\in A\} It is given that A=\{-1,1\}$$\\$ $\therefore A*A*A=\{(-1,-1,-1),(-1,-1,1),(-1,1,-1),(-1,1,1),\\ (1,-1,-1),(1,-1,1),(1,1,-1),(1,1,1)\}$

6   .If $A*B=\{(a,x),(a,y),(b,x),(b,y)\}$ . Find A and B.

If is given that $A*B=\{(a,x),(a,y),(b,x),(b,y)\}$ $\\$ We know that the Cartesian product of two non-empty sets P and Q is defined as$\\$ $P*Q=\{(p,q):p\in P,q\in Q\}$$\\ \therefore A is the set of all first elements and B is the set of all second elements.\\ Thus,A=\{a,b\} and B=\{x,y\} 7 Let A=\{1,2\},B=\{1,2,3,4\},C=\{5,6\} \text{and} D=\{5,6,7,8\}. Verify that \\ (i) A*(B\cap C)=(A*B)\cap(A*C)$$\\$ $(ii) A*C$ is a subset of $B*D$$\\ ##### Solution : (i)To verify:A*(B\cap C)=(A*B)\cap(A*C)$$\\$ We have $B\cap C=\{1,2,3,4\}\cap \{5,6\}=\oslash$$\\ \therefore L>H>S>=A*(B\cap C)=A*\oslash =\oslash$$\\$ $A*B=\{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)\}$$\\ A*C=\{(1,5),(1,6),(2,5),(2,6)\}$$\\$ $\therefore R.H.S.=(A*B)\cap (A*C)=\oslash$$\\ \therefore L.H.S.=R.H.S.$$\\$ Hence,$A*(B\cap C)=(A*B)\cap (A*C)$$\\ (ii) To verify: A * C is a subset of B * D$$\\$ $A*C=\{(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)\}$$\\ We can observe that all the elements of set A * C are the elements of set B * D . Therefore, A * C is a subset of B * D . 8 Let A =\{ 1, 2 \} and B=\{ 3, 4 \} . Write A * B . How many subsets will A * B have? List them. ##### Solution : A=\{1,2\} \text{and} B=\{3,4\}\\ \therefore A*B = \{(1,3),(1,4),(2,3),(2,4)\}\\ \Rightarrow n(A*B) =4 We know that if C is a set with n (C)= m , then n [ P (C )] =2^ m .\\ Therefore, the set A * B has 2^4 = 16 subsets. These are\\ \oslash , \{(1,3)\},\{(1,4)\},\{(2,3)\},\{(2,4)\},\\ \{(1,3)(1,4)\},\{(1,3),(2,3)\},\{(1,3),(2,4)\},\{(1,4),(2,3)\},\{(1,4),(2,4)\},\{(2,3),(2,4)\}\\ \{(1,3),(1,4),(2,3)\},\{(1,3),(1,4),(2,4)\},\{(1,3),(2,3),(2,4)\}\\ \{(1,4),(2,3),(2,4)\},\{(1,3),(1,4),(2,3),(2,4)\} 9 Let A and B be two sets such that n ( A )=3 and n( B)= 2 . If (x ,1 ), ( y , 2 ) , ( z ,1 ) are in A * B , find A and B, where x , y and z are distinct elements. ##### Solution : It is given that n(A)=3 and n (B )= 2; and ( x ,1 ) , ( y , 2 ) , ( z ,1 ) are in A * B .$$\\$ We know that$\\$ $A =$ Set of first elements of the ordered pair elements of $A * B$$\\ B = Set of second elements of the ordered pair elements of A * B .$$\\$ $\therefore x , y ,$ and $z$are the elements of $A$; and$1$and$2$are the elements of$B.$$\\ Since n (A )= 3 and n ( B )=2,$$\\$ It is clear that $A=\{x , y , z \}$ and $B=\{ 1, 2 \}$ .

10   The Cartesian product $A * A$ has $9$ elements among which are found $(-1,0 )$ and$(0, 1).$ Find the set $A$ and the remaining elements of $A * A$.

##### Solution :

We know that if $n (A )= p$ and $n (B )= q$, then $n = (A * B )= pq .$$\\ \therefore n ( A * A)= n ( A )* n ( A )$$\\$ It is given that $n (A *A )= 9$$\\ \therefore n ( A )* n ( A )= 9$$\\$ $\Rightarrow n ( A )= 3$$\\ The ordered pairs (- 1,0 ) and (0, 1) are two of the nine elements of A * A .$$\\$ We know that $A * A =\{ (a , a ) : a \in A \}$ . Therefore, $-1,0,$ and $1$ are elements of $A.$$\\ Since n ( A )= 3, it is clear that A =\{- 1,0,1 \}$$\\$ . The remaining elements of set $A * A$ are $\\$ $(- 1,-1 ) , (- 1,1) , (0, - 1 ) , ( 0,0 ) , (1, - 1 ) , (1,0 ) ,$ and $(1,1 ) .$

11   Let A =$\{1, 2,3...14 \}$ . Define a relation R from A to A by R =$\{( x , y) : 3 x - y = 0 \}$ , where x , y $\in$ A . Write down its domain, codomain and range.

##### Solution :

The relation R from A to A is given as R =$\{( x , y ) : 3 x - y = 0,$ where x , y $\in$ A $\}$ $\\$ i.e., R =$\{( x , y ) : 3 x= y$ , where x , y $\in$ A $\}$ $\\$ $\therefore$ R =$\{(1,3 ) , ( 2,6 ) , (3,9 ) , (4,12 )\}$ $\\$ The domain of R is the set of all first elements of the ordered pairs in the relation. $\\$ $\therefore$ Domain of R =$\{ 1, 2,3, 4 \}$ $\\$ The whole set A is he codomain of the relation R. $\\$ $\therefore$ Codomain of R = A = 1, 2,3....14 $\\$ The range of R is the set of all second elements of the ordered pairs in the relation. $\\$ $\therefore$ Range of R =$\{ 3,6,9,12 \}$ $\\$

12   Define a relation R on the set N of natural numbers by R = {( x , y ) : y= x + 5, x isa naturalnumber less than 4; x , y $\in$N } . Depict this relationship using roster form. Write down the domain and the range.

##### Solution :

R ={ x , y ) : y = x + 5, x isa natural number less than 4, x , y $\in$N }$\\$ The natural numbers less than 4 are 1, 2, and 3.$\\$ $\therefore$ R ={( 1,6 ),( 2,7 ),( 3,8 )}$\\$ The domain of R is the set of all first elements of the ordered pairs in the relation.$\\$ $\therefore$ Domain of R ={ 1, 2,3 }$\\$ The range of R is the set of all second elements of the ordered pairs in the relation.$\\$ $\therefore$Range of R ={ 6,7,8 }$\\$

13   A ={1, 2,3,5 } and B ={ 4,6,9 } . Define a relation R from A to Bby R = {( x , y ) : thedifference between x and y isodd; x $\in$ A , y $\in$ B } . Write R in roster form.

##### Solution :

A ={ 1, 2,3,5 } and B ={ 4,6,9 }$\\$ R ={( x , y ) :thedifferencebetween x and y isodd; x $\in$ A , y $\in$ B }$\\$ $\therefore$ R ={( 1, 4 ) , ( 1,6 ) , ( 2,9 ) , ( 3, 4 ) , (3,6 ) , ( 5, 4 ) , ( 5,6 ) }

14   The given figure shows a relationship between the sets P and Q. Write this relation$\\$ (i) in set-builder form$\\$ (ii) in roster form.$\\$ What is its domain and range?$\\$

##### Solution :

According to the given figure, P ={ 5,6,7 } , Q ={ 3, 4,5 } $\\$ (i) R ={( x , y ) : y = x - 2; x $\in$ P } or R ={( x , y ) : y = x - 2for x = 5,6,7 } $\\$ (ii) R ={( 5,3 ) , ( 6, 4 ) , ( 7,5 )} $\\$ Domain of R ={ 5,6,7 } $\\$ Range of R ={ 3, 4,5 } $\\$

15   Let A ={ 1, 2,3, 4,6 }. Let R be the relation on A defined by {( a , b ) : a , b $\in$ A , b isexactlydivisible bya} . $\\$ (i) Write R in roster form $\\$ (ii) Find the domain of R $\\$ (iii) Find the range of R. $\\$

##### Solution :

A ={ 1, 2,3, 4,6 } , R ={ a , b } : a , b $\in$ A , b is exactly divisible by a } $\\$ (i) R ={( 1,1 ) , (1, 2 ) , ( 1,3 ) , ( 1, 4 ) , ( 1,6 ) , ( 2, 2 ) , ( 2, 4 ) , ( 2,6 ) , ( 3,3 ) , ( 3,6 ) , ( 4, 4 ) , ( 6,6 )} $\\$ (ii) Domain of R ={ 1, 2,3, 4,6 } $\\$ (iii) Range of R ={ 1, 2,3, 4,6 } $\\$

16   Determine the domain and range of the relation R defined by R ={( x , x ) 5 ) : x $\in${ 0,1, 2,3, 4,5 }} .

##### Solution :

R ={( x , x )+ 5 ) : x $\in${ 0,1, 2,3, 4,5 }}$\\$ $\therefore$ R ={( 0,5 ) , ( 1,6 ) , ( 2,7 ) , ( 3,8 ) , (4,9 ) , (5,10 )}$\\$ $\therefore$ Domain of R ={ 0,1, 2,3, 4,5 }$\\$ Range of R ={ 5,6,7,8,9,10}$\\$

17   Write the relation R ={( x , x $^3$) : x isa prime number less than10 } in roster form.

##### Solution :

R={( x , x$^3$) : x isa prime number less than10 } . The prime numbers less than 10 are 2, 3, 5 and 3 7.$\\$ $\therefore$ R ={(2,8 ) , ( 3, 27 ) , (5,125 ) , ( 7,343 )}

18   Let A ={ x , y , z } and B ={ 1, 2 } . Find the number of relations from A to B.

##### Solution :

It is given that A={ x , y , z } and B ={ 1, 2} .$\\$ A * B={( x ,1 ) , ( x , 2) , ( y ,1 ) , ( y , 2) , (z ,1 ) , ( z , 2 )}$\\$ Since n ( A * B )= 6 , the number of subsets of A * B is 2$^6$ .$\\$ Therefore, the number of relations from A to B is 2$^ 6$ .

19   Let R be the relation on Z defined by R ={ (a , b ) : a , b $\in$ Z , a - b isan integer } . Find the domain and range of R.

##### Solution :

R ={( a , b ) : a , b $\in$ Z , a - b isan integer }$\\$ It is known that the difference between any two integers is always an integer.$\\$ $\therefore$ Domain of R = Z$\\$ Range of R = Z$\\$

20   Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range. $\\$ (i) {( 2,1 ) , (5,1 ) , ( 8,1 ) , ( 11,1 ) , ( 14,1 ) , ( 17,1 )} $\\$ (ii){( 2,1 ) , ( 4, 2 ) ,( 6,3 ) , ( 8, 4 ) , ( 10,5 ) , ( 12,6 ) , ( 14,7 )} $\\$ (iii){( 1,3 ), ( 1,5 ) , ( 2,5)} $\\$

##### Solution :

{( 2,1 ) , (5,1 ) , ( 8,1 ) , ( 11,1 ) , ( 14,1 ) , ( 17,1 )} $\\$ Since 2, 5, 8, 11, 14 and 17 are the elements of the domain of the given relation having their unique images, this relation is a function. $\\$Here, domain ={2,5,8,11,14,17 } and range ={ 1 } $\\$ (ii) {( 2,1 ) , ( 4, 2 ) , ( 6,3 ) , ( 8, 4 ) , ( 10,5) , (12,6 ) , ( 14,7 )} $\\$ Since 2, 4, 6, 8, 10, 12 and 14 are the elements of the domain of the given relation having their unique images, this relation is a function. $\\$ Here, domain ={2, 4,6,8,10,12,14 } and range ={( 1, 2,3, 4,5,6,7 )} $\\$ (iii) {( 1,3 ) , ( 1,5 ) , ( 2,5 )} $\\$ Since the same first element i.e., 1 corresponds to two different images i.e., 3 and 5, this relation is not a function.

21   Find the domain and range of the following real function:$\\$ (i) f ( x )-| x|$\\$ (ii) f ( x )=$\sqrt{ 9 - x ^2}$

(i) f(x)=-|x|,x$\in$R$\\$ we know that |x|= \begin{cases} x, & \quad \text{if } x \geq 0\\ -x & \quad \text{if } x < 0 \end{cases}$\\$ $\therefore$ f(x)=-|x|=\begin{cases} -x, & \quad \text{if } x \geq 0\\ x & \quad \text{if } x < 0 \end{cases}$\\$ Since f(x) is defined for x$\in$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.$\\$ $\therefore$ The range of f is (-$\infty$,0 ] .$\\$ (ii)f(x)=$\sqrt{9-x^2}$$\\ Since \sqrt{9 -x^ 2} is defined for all real numbers that are greater than or equal to - 3 and less than or equal to 3, the domain of f { x } is {x : - 3 \leq x \leq 3 } or [- 3,3 ] .\\ For any value of x such that - 3 \leq x \leq 3 , the value of f ( x )will lie between 0 and 3..\\ \therefore The range of f ( x ) is {x : 0 \leq x \leq 3 } or [ 0,3 ] . 22 A function f is defined by f ( x )= 2 x - 5 .\\ (i) f (0 ) ,\\ (ii) f ( 7 )\\ (iii) f (- 3 )\\ ##### Solution : The given function is f ( x )= 2 x - 5\\ Therefore,\\ (i) f ( 0 )= 2 * 0 - 5 = 0 - 5 =- 5\\ (ii) f ( 7 )= 2 * 7 - 5 = 14 - 5 = 9\\ (iii) f (- 3 )= 2 *(- 3 )- 5 =-6 - 5 =- 11\\ 23 The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by t ( C )=\dfrac{9C}{5}+32.Find\\ (i)t(0)\\ (ii)t(28)\\ (iii)t(-10)\\ (iv)The value of C, when t(C)=212 ##### Solution : The given function is t(C)=\dfrac{9C}{5}+5.$$\\$ Therefore,$\\$ (i)$t(0)=\dfrac{9*0}{5}+32=32$$\\ (ii) t(28)=\dfrac{9*28}{5}+32=\dfrac{252+160}{5}=\dfrac{412}{5}=82.4 \\ (iii)t(-10)=\dfrac{9*(-10)}{5}+32=9*(-2)+32=-18+32=14 \\ (iv) It is given that t ( C )= 212\\ \therefore 212 = \dfrac{9 C}{5} + 32$$\\$ $\implies\dfrac{9 C}{5}= 212- 32$$\\ \implies\dfrac{9 C}{5}= 180$$\\$ $\implies 9 C = 180 * 5$$\\ C=\dfrac{180*5}{9}=100$$\\$ Thus, the value of t, when t ( C )= 212 , is 100.

24   Find the range of each of the following functions. $\\$ (i) f (x )= 2- 3 x , x $\in$ R, x > 0 . $\\$ (ii) f ( x )=x $^2$ + 2, x , is a real number. $\\$ (iii) f ( x )= x , x is a real number. $\\$

##### Solution :

(i) f (x )= 2- 3 x , x $\in$ R, x > 0 . $\\$ The values of f ( x )for various values of real numbers x > 0 can be written in the tabular form as $\\$ $\begin{array}{|c|c|} \hline x&0.001&0.1&0.9&1&2&2.5&4&5&... \\ \hline f(x)&1.97&1.7&-0.7&-1&-4&-5.5&-10&-13&...\\ \hline \end{array}$ $\\$ Thus, it can be clearly observed that the range of f is the set of all real numbers less than 2. i.e., range of $f =(-\infty,, 2 )$ $\\$ Alter: Let x > 0 $\\$ 3 x > 0 $\\$ 2 - 3 x < 2 $\\$ f (x ) < 2 $\\$ Range of f =(-$\infty$ , 2 ) $\\$ (ii) f ( x )= x $^2$+ 2, x , is a real number $\\$ The values of f (x ) for various of real numbers x can be written in the tabular form as $\\$ $\begin{array}{|c|c|} \hline x&0&\pm0.3&\pm0.8&\pm1&\pm2&\pm 3&... \\ \hline f(x)&2&2.09&2.64&3&6&11&...\\ \hline \end{array}$ $\\$ Thus, it can be clearly observed that the range of f is the set of all real numbers greater than 2. i.e., range of f=[2, $\infty$] $\\$ Alter: $\\$ Let x be any real number. Accordingly, $\\$ x $^2$$\leq$ 0 $\\$ $\\$ x $^2$2 + 0$\leq$ 2 $\\$ x $^2$ 2 $\leq$ 2 $\\$ f ( x ) $\leq$ 2 $\\$ Range of f =[ 2, $\infty$] $\\$ (iii) f ( x )= x , x is a real number $\\$ It is clear that the range of f is the set of all real numbers. $\\$ Range of f = R .