NCERT Exemplar Class 11 Maths Solutions Chapter 2 Relations and Functions

While travelling in a taxi, have you ever noticed that the app shows the estimated time to reach the destination? It is possible due to the study of the relation between total distance and the average speed of the taxi, this is known as Relation. Also you have seen the fare for different distances is different. This shows that for every different distance, the fare changes as a variable function depending on distance. Let’s see the definition of Relation and Function. A relation is a relationship between sets of values. In math, the relation is between the $x$-values and $y$-values of ordered pairs. The set of all primary elements of the ordered pairs is called a domain of $R$, and the set of all second elements of the ordered pairs is called a range of $R$. A relation ' $f$ ' is said to be a function if every element of a non-empty set $X$ has only one image or range to a non-empty set Y.

This article on NCERT Math Class 11 Chapter 2 is briefly about relations and functions. This article contains NCERT Class 11 Maths Chapter 2 exemplar solutions with step-by-step explanations. NCERT Exemplar solutions for other subjects and classes can be downloaded by clicking on NCERT Exemplar solutions.

NCERT Exemplar Class 11 Maths Solutions Chapter 2 Relations and Functions

Question:1

Let $A=\{-1,2,3\}$ and $B=\{1,3\}$. Determine
i) $A\times B$ ii)$B\times A$ iii) $B\times B$ iv) $A\times A$

Answer:

Given data: $A=\{-1,2,3\}$and$B=\{1,3\}$.

Now, let’s solve the problems one by one.

i) $A\times B=\{-1,2,3\}\times \{1,3\}=\{(-1,1),(-1,3),(2,1),(2,3),(3,1),(3,3)\}$

ii) $B\times A=\{1,3\}\times \{-1,2,3\}=\{(1,-1),(3,-1),(1,2),(3,2),(1,3),(3,3)\}$

iii) $B\times B=\{1,3\}\times \{1,3\}=\{(1,1),(1,3),(3,1),(3,3)\}$

iv)$A\times A=\{-1,2,3\}\times \{-1,2,3\}=\{(-1,-1),(-1,2),(-1,3),(2,-1),(2,2),(2,3),(3,-1),(3,2),(3,3)\}$

Question:2

If $P=\{x:x<3,x\in N\}$, $Q=\{x:x\le 2,x\in W\}$. Find $(P\cup Q)\times (P\cap Q)$, where W is the set of whole numbers.
Answer:
Given data:
$P=\{x:x\le 3,x\in N\}\Rightarrow P=\{1,2\}$
$Q=\{x:x\le 2,x\in W\}\Rightarrow Q=\{0,1,2\}$

Now, $(P\cup Q)=0,1,2\text{ and}(P\cap Q)=1,2$

Thus, $(P\cup Q)\times (P\cap Q)=\{0,1,2\}\times \{1,2\}=\{(0,1),(0,2),(1,1),(1,2),(2,1),(2,2)\}$

Question:3

If $A=\{x:x\in W,x<2\}$,$B=\{x:x\in N,1<x<5\}$, $C=\{3,5\}$ find

I) $A\times (B\cap C)$ (ii) $A\times (B\cup C)$

Answer:

Given data:

$C=\{3,5\}$

Now, we can find,

$(B\cap C)=\{3\}$ &

$(B\cup C)=\{2,3,4,5\}$
i)$A\times (B\cap C)=\{0,1\}\times \{3\}=\{(0,3),(1,3)\}$
ii)$A\times (B\cup C)=\{0,1\}\times \{2,3,4,5\}=\{(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)\}$

Question:4

In each of the following cases, find a and b.
(i) $(2a+b,a-b)=(8,3)$
(ii) $(a/4,a-2b)=(0,6+b)$

Answer:

i) Given data:
$(2a+b,a-b)=(8,3)$
If & only if the corresponding coordinates are equal, the two ordered pairs will be equal.

Thus, $2a+b=8$ …… (i)

& $a-b=3$ ……. (ii)

From (i) & (ii), we get,

$a=\frac{11}{3}$ & $b=\frac{2}{3}$

ii) Given:

$(a/4,a-2b)=(0,6+b)$

If & only if the corresponding coordinates are equal, the two ordered pairs will be equal.

Thus, $a/4=0\to a=0$

& $a-2b=6+b\to a+3b=6$

$0-3b=6$

$b=-2$

Thus, a = 0 & b = -2

Question:5

Given $A=\{1,2,3,4,5\}$, $S=\{(x,y):x\in A,y\in A\}$. Find the ordered pairs which satisfy the conditions given below:
(i) $x+y=5$
(ii) $x+y<5$
(iii) $x+y>8$

Answer:

Given data:

$A=\{1,2,3,4,5\}$ &

$S=\{(x,y):x\in A,y\in A\}$

Now,

• The ordered pairs that satisfy the given condition, $x+y=5$ are, $(1,4),(4,1),(2,3)\mathrm{\&}(3,2)$
• The ordered pairs that satisfy the given condition, $x+y<5$ are, $(1,1),(1,2),(2,1),(1,3),(2,2),(3,1)$.
• The ordered pairs that satisfy the given condition, $x+y>8$ are, $(4,5),(5,4),(5,5)$.

Question:6

Given $R=\{(x,y):x,y\in W,x^2+y^2=25\}$. Find the domain and Range of R.

Answer:

Given:

$R=\{(x,y):x,y\in W,x^2+y^2=25\}$

$(0,5),(3,4),(5,0)\mathrm{\&}(4,3)$, are the ordered pairs satisfying the condition $x^2+y^2=25$.

Therefore, here,

$Domain=\{0,3,4,5\}$ &

$Range=\{0,3,4,5\}$

Question:7

If $R1=\{(x,y)|y=2x+7,wherex\in Rand-5\le x\le 5\}$ is a relation. Then find the domain and Range of $R1$.

Answer:

Given data: $R1=\{(x,y)/y=2x+7\text{ where}x\in R\text{ and}-5\le x\le 5\}$

Thus, domain of R₁ will be $\{x:-5\le x\le 5=[-5,5]\}$

Domain = $\{-5,-4,-3,-2,-1,0,1,2,3,4,5\}$

& $y=2x+7$

Thus, the values of y will be

$\{-3,-1,1,3,5,7,9,11,13,15,17\}$

Thus, domain of $R_1=[-5,5]$

& range of $R_1=[-3,17]$

Question:8

If $R_2=\{(x,y)|xandyareintegersand{x}^2+y^2=64\}$ is a relation. Then find $R_2$.

Answer:

Given data:$x^2+y^2=64$, where x & y $\in$ Z

Thus, it is clear that 64 is the sum of the squares of 2 integers

Thus, for, x=0

Y will be = $\pm 8$ & vice-versa.

Therefore, $R_2=\{(0,8),(0,-8),(8,0),(-8,0)\}$

Question:9

If $R3=\{(x,|x|)|xisarealnumber\}$is a relation. Then, find the domain and range of R3.

Answer:

Given data: $R_3=\{(x,|x|\}$ where x is a real no.

Thus, the domain of R₃ will be equal to that of R & its range will be

$R_3=(0,\mathrm{\infty })...(since,\Vert x\Vert =R_{+}]$

Question:10

Is the given relation a function? Give reasons for your answer.

(i) $h=\{(4,6),(3,9),(-11,6),(3,11)\}$

(ii) $f=\{(x,x)|xisarealnumber\}$

(iii) $g=n,(1/n)|nisapositiveinteger$

(iv)$s=\{(n,n2)|nisapositiveinteger\}$

(v) $t=\{(x,3)|xisarealnumber\}$.

Answer:

(i) Given data: $h=\{(4,6),(3,9),(-11,6),(3,11)\}$

‘h’ is not a function since there are two images- 9 & 11 for the relation 3.

(ii) $f=\{(x,x)/xisarealno.\}$

Here, f is a function because every element of the domain has a unique image.

(iii) $g=\{(n,1/n)/nisapositiveinteger\}$

‘g’ is a function because there is a unique image, ‘1/n’, for every element in the domain.

(iv)
$S=\{(n,n2)/nisapositiveinteger.\}$

‘S’ is a function since the square of any integer is a unique number. & thus, for every element in the domain, there is a unique image.

(V) $T=\{(x,3)/xisarealnumber\}$

Here, ‘t’ is a constant function since we can observe that there is a constant no. 3 for every real element in the domain.

Question:11

If f and g are real functions defined by $f(x)=x^2+7$ and $g(x)=3x+5$, find each of the following
(a) $f(3)+g(-5)$

(b) $f\left(\frac{1}{2}\right)\times g(14)$

(c) $f(-2)+g(-1)$

(d) $f(t)-f(-2)$

(e) $\frac{(f(t)-f(5))}{(t-5)},ift\ne 5$

Answer:

Given data: $f(x)=x^2+7$ & $g(x)=3x+5$

1. $f(3)+g(-5)=[(3{)}^2+7]+[3(-5)+5]$

$=16-10$

$=6$

2. $f(\frac{1}{2})\times g(14)=[{\left(\frac{1}{2}\right)}^2+7]\times [3\times 14+5]$

$=\frac{29}{4}\times 47$

$=\frac{1363}{4}$

3. $f(-2)+g(-1)=[(-2{)}^2+7]+[(-1)+5]$

$=11+2=13$

4. $f(t)-f(-2)=(t^2+7)-[(-2{)}^2+7]$

$=t^2-4$

5. $f(t)-f(5)/t-5$, (t is not equal to 5) = $(t^2+7)-(5^2+7)/t-5$

$=\frac{(t^2+7-32)}{t–5}$

$=\frac{(t^2-25)}{(t-5)}$

$=(t-5)\frac{(t+5)}{(t-5)}$

$=t+5$

Question:12

Let f and g be real functions defined by $(x)=2x+1$ and $g(x)=4x-7$.
(a) For what real numbers x, $f(x)$ = $g(x)$?

(b) For what real numbers x, $f(x)<g(x)$?

Answer:

Given data: $f(x)=2x+1$& $g(x)=4x-7$

i) Now, for, f(x) = g(x),

$2x+1=4x-7$

$-2x=-8$

Thus, $x=4$, viz, the required real no.

ii) For, f(x) < g(x),

$2x+1<4x-7$

$-2x<-8$

Thus, $2x>8$

Thus, $x>4$, viz., the required real no.

Question:13

If f and g are two real-valued functions defined as $f(x)=2x+1$, $g(x)=x^2+1$, then find.

(i) f + g (ii) f - g (iii) fg (iv)$\frac{f}{g}$

Answer:

Given data: $f(x)=2x+1$ & $g(x)=x^2+1$
i) $f+g=f(x)+g(x)$

$=2x+1+x^2+1$

$=x^2+2x+2$
ii) $f-g=f(x)-g(x)$

$=(2x+1)-(x^2+1)$

$=2x-x^2$

iii) $f.g=f(x).g(x)$

$=(2x+1)(x^2+1)$

$=2x^3+x^2+2x+1$

iv) $\frac{f}{8}=\frac{f(x)}{g(x)}$

$=2x+\frac{1}{x^2}+1$

Question:14

Express the following functions as a set of ordered pairs and determine their range.

$f:X\to R,f(x)=x^3+1,whereX=\{-1,0,3,9,7\}$

Answer:

Given data: $F:x\to R,f(x)=x^3+1,where,x=\{-1,0,3,9,7\}$

We know that, here, $x=\{-1,0,3,9,7\}$

For x = -1,

$f(-1{)}^3+1=0$

For x = 0,

$f(0)=(0{)}^3+1=1$

For x = 3,

$f(3)=(3{)}^3+1=28$

For x = 9,

$F(9)=(9{)}^3+1=730$

For x = 7,

$F(7)=(7{)}^3+1=344$

Thus, $(-1,0),(0,1),(3,28),(7,344)\mathrm{\&}(9,730)$ are the ordered pairs &

$Range=\{0,1,28,344,730\}.$

Question:15

Find the values of x for which the functions

$f(x)=3x^2-1andg(x)=3+x$ are equal.

Answer:

Given data: $f(x)=3x^2-1andg(x)=3+x$

Now, it is given that - $f(x)=g(x),$

Thus, $3x^2-1=3+x$

$3x^2-x-4=0$

$3x^2-4x+3x-4=0$

$x(3x-4)+1(3x-4)=0$

$(3x-4)(x+1)=0$

$3x-4=0orx+1=0$

$3x=4orx=-1$

$x=4/3$

Thus, -1 & 4/3 are the values of x.

Question:16

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

Answer:

Given data:

$g=\{(1,1),(2,3),(3,5),(4,7)\}$

Here, ‘g’ is a function since every element of the domain has a unique image.

$g(x)=\alpha x+\beta ...(given)$

Now, for (1,1)

$g(1)=\alpha (1)+\beta =1$

$\alpha +\beta =1.......(i)$

& for (2,3),

$g(2)=\alpha (2)+\beta =3$

$2\alpha +\beta =3........(ii)$

On solving (i) & (ii), we get,

$\alpha =2and\beta =-1.$

$g(x)=2x-1$

It is satisfying for other values of x; hence, it is a function.

Question:17

Find the domain of each of the following functions given by

i) $f(x)=\frac{1}{\sqrt{1-\cos x}}$

ii) $f(x)=\frac{1}{\sqrt{x+|x|}}$

iii)$f(x)=x|x|$

iv) $f(x)=\frac{x^3-x+3}{x^2-1}$

v) $f(x)=\frac{3x}{2x-8}$

Answer:

i) Given data: $f(x)=\frac{1}{\sqrt{1-\cos x}}$
Now, we know that,
$-1\le cosx\le 1$
$1\ge -cosx\ge -1$
$1+1\ge 1-cosx\ge -1+1$
$2\ge 1-cosx\ge 0$
$0\le 1-cosx\le 2$
Now, for the real value of the domain,
$1-cosx\ne 0,cosx\ne 1$
But, $x\ne 02n\pi \mathrm{\forall }n\in Z$
Thus, domain of $f=R-\{2n\pi ,n\in Z\}$

ii)Given data: $f(x)=\frac{1}{\sqrt{x+|x|}}$
Now, $x+|x|=x+x=2x......ifx\ge 0$
&$x+|x|=x-x=0......ifx<0$
Now, $x<0$ is not defined so far; hence,
The domain = $R^{+}$
iii)Given data:$f(x)=x|x|$
For all x $\in$ R, f(x) is defined
Thus, the domain of f = R.
iv)Given data:$f(x)=\frac{x^3-x+3}{x^2-1}$
Here, only if$x^2-1\ne 0$, f(x) is defined,
$(x-1)(x+1)\ne 0,$
$Thus,x\ne 1$
$andx\ne -1$
Therefore, domain of $f=R-\{-1,1\}$
v)Given data: $f(x)=\frac{3x}{2x-8}$

F(x) is only defined at $2x-8\ne 0,x\ne 4$,

Thus, the domain = $R-\{4\}$.

Question:18

Find the range of the following functions given by

i) $f(x)=\frac{3}{2-x^2}$

ii)$f(x)=1-|x-2|$

iii) $f(x)=|x-3|$

iv) $f(x)=1+3\cos 2x$

Answer:

i)Given data: $f(x)=\frac{3}{2-x^2}$
Let us consider that, y = f(x)
Thus, $y=3/(2-x^2)$
Thus,
$y(2-x^2)=3$
$2y-y{x}^2=3$
$y{x}^2=2y-3$
$x^2=2-3/y$
x is real, if $2y-3\ge 0andy\ge 0$
Thus, $y\ge 3/2$
Therefore,
Range of $f=(3/2,\mathrm{\infty })$
ii)Given data: $f(x)=1-|x-2|$
Now, we know that,
$|x-2|=-(x-2),ifx<2,and|x-2|=(x-2),if\ge 2$
Thus, $|-x-2|\ge 0$
Thus, $1-|x-2|\le 1$
Therefore,
Range of$f=(-\mathrm{\infty },1)$
iii)Given data: $f(x)=|x-3|$

Now, we know that,

$|x-3|\ge 0$

Thus, $f(x)=0$
Therefore, range of $f=(0,\mathrm{\infty })$
iv)Given data: $f(x)=1+3\cos 2x$

Now, we know that,

$-1\le cos2x\le 1$

Thus, $-3\le 3cos2x\le 3$

$-3+1\le 1+3cos2x\le 3+1$

Thus, $-2\le 1+3cos3x\le 4$

$-2\le f(x)\le 4$

Therefore, range of $f=[-2,4].$

Question:19

Redefine the function $f(x)=|x-2|+|2+x|,-3\le x\le 3$
Answer:

$\text{Given data: }f(x)=|x-2|+|2+x|,-3\le x\le 3$

$|x-2|=\{\begin{array}{ll}x-2,& x\ge 2\\ -(x-2),& x<2\end{array}$

$|2+x|=\{\begin{array}{ll}-(2+x),& x<-2\\ (2+x),& x\ge -2\end{array}$

$f(x)=\{\begin{array}{ll}-(x-2)-(2+x),& -3\le x<-2\\ -(x-2)+(2+x),& -2\le x<2\\ (x-2)+(2+x),& 2\le x\le 3\end{array}$

$f(x)=\{\begin{array}{ll}-2x,& -3\le x<-2\\ 4,& -2\le x<2\\ 2x,& 2\le x\le 3\end{array}$

Question:20

If $f(x)=\frac{x-1}{x+1}$, then show that:

i)$f(\frac{1}{x})=-f(x)$ ii) $f(-\frac{1}{x})=\frac{-1}{f(x)}$

Answer:

Given data: $f(x)=\frac{x-1}{x+1}$
i) $f\left(\frac{1}{x}\right)=\frac{\frac{1}{x}-1}{\frac{1}{x}+1}$
$=\frac{1-x}{1+x}$
= $-\frac{(x-1)}{x+1}=-f(x)$
Thus, $f\left(\frac{1}{x}\right)=-f(x)$
ii) $f(-1/x)=\frac{\frac{-1}{x}-1}{\frac{-1}{x}+1}$
$=\frac{1+x}{1–x}$
$=\frac{1}{1+x/1-x}$
$=\frac{-1}{f(x)}$
Thus, $f\left(\frac{-1}{x}\right)=\frac{-1}{f(x)}$.

Question:21

Let$f(x)=\sqrt{x}andg(x)=x$ be two functions defined in the domain $R^{+}\cup \{0\}$. Find

(i)$(f+g)(x)$

(ii) $(f-g)(x)$

(iii) $(fg)(x)$

(iv) $\left(\frac{f}{g}\right)(x)$

Answer:

Given data: $f(x)=\sqrt{x}andg(x)=x$ are the two functions which are defined in the domain $R^{+}\cup \{0\}$

Now, (i) $(f+g)(x)=f(x)+g(x)$

$=\sqrt{x}+x$

ii) $(f-g)(x)=f(x)-g(x)$

$=\sqrt{x}-x$

iii)$(fg)(x)=f(x)\cdot g(x)$

$=\sqrt{x}.x$

$=x^{\frac{3}{2}}$

iv) $\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}$

$=\frac{\sqrt{x}}{x}$

$=\frac{1}{\sqrt{x}}$

Question:22

Find the domain and Range of the function $f(x)=1/\sqrt{(x-5)}$.
Answer:

Given: $f(x)=\frac{1}{\sqrt{(x-5)}}$

When, $x-5>0,i.e.,x>5$, f(x) is real

Thus, $domain=(5,\mathrm{\infty })$

Now, to find the range, we will put

$y=f(x)=\frac{1}{\sqrt{x-5}}$

Thus, $\sqrt{x-5}=1/y$

$x-5=\frac{1}{y^2}$

$x=\frac{1}{y^2}+5$

Therefore, for $x\in (5,\mathrm{\infty }),y\in R$.

Therefore, the range of $=R^{+}$

Question:23

If $f(x)=y=\frac{ax-b}{cx-a}$ then prove that $f(y)=x$

Answer:

Given: $f(x)=y=\frac{ax-b}{cx-a}$

Now, let us put $x=y$ in $f(x)$,

$f(y)=\frac{ay-b}{cy-a}$

$=\frac{a\frac{|ax-b|}{|cx-a|}-b}{c\frac{|ax-b|}{|cx-a|}-a}$

Thus, $f(y)=\frac{a^{2}x-ab-bcx+ab}{cax-bc-cax+a^2}$

$=\frac{(a^{2}x)-bcx}{a^2-bc}$

$=\frac{x(a^2-bc)}{a^2-bc}$

$=x$

Therefore, $f(y)=x$.

Question:24

Let n(A) = m, and n(B) = n. Then the total number of non-empty relations that can be defined from A to B is
(a) $m^n$
(b) $n^m-1$
(c) $mn–1$
(d) $2^{mn}-1$