 # Extra questions on relations and functions class 12

In this page we have extra questions on relations and functions class 12. Hope you like them and do not forget to like , social share and comment at the end of the page.

## Multiple Choice Questions

Question 1
Let R be a relation on the set N of natural numbers defined by nRm if n divides m. Then R is
(a) Equivalence
(b) Reflexive and symmetric
(c) Transitive and symmetric
(d) Reflexive, transitive but not symmetric

Since n divides n, for all n in N, R is reflexive. R is not symmetric since for 3, 6 in N, 3 R 6 $\neq$ 6 R 3. R is transitive since for n, m, r whenever n/m and m/r ⇒ n/r, i.e., n divides m and m divides r, then n will devide r

Question 2
Let P ={1,2,4,5}. The total number of distincts relations that can be defined on P is
(a) 16
(b) $2^{16}$
(c) 32
(d) 256

(b)
This is by formula

Question 3
The function $f:[0,\infty)$ -> R given by $f(x)= \frac {x}{x+1}$ is
(a) one -one and onto
(b) one-one but not onto
(c) onto but not one-one
(d) Neither one-one nor onto

$f(x)= \frac {x}{x+1}$
$\frac {df(x)}{dx}= \frac {1}{(x+1)^2}$
This is positive, So f(x) is a increasing function and it is one-one function
for onto
$y=\frac {x}{x+1}$
$x = \frac {y}{1-y}$
So y cannot be 1,So Range is subset of codomain
Hence not onto function

Question 4
Let f : R -> R be defined by $f(x) = 2x - 1$. Then $f^{–1}(x)$ is given by
(a) $\frac {x-2}{2}$
(b) $\frac {x+1}{2}$
(c) $\frac {x-1}{2}$
(d) $\frac {2x-2}{3}$

Let y=2x- 1
$y+1=2x$
$x= \frac {y+1}{2}$
Therefore
$f^{–1}(x)= \frac {x+1}{2}$

Question 5
The function f:R-> R given by $f(x)= \frac {x^2 -8}{x^2+2}$ is
(a) one -one and onto
(b) one-one but not onto
(c) onto but not one-one
(d) Neither one-one nor onto

for one-one
$f(x_1)=f(x_2)$
$\frac {x_1^2 -8}{x_1^2+2}=\frac {x_2^2 -8}{x_2^2+2}$
$10x_1^2= 10x_2^2$
$x_1= \pm x_2$
Hence not a one-one function
For onto
$y=\frac {x^2 -8}{x^2+2}$
Solving this for x
$x= \sqrt {\frac {2y+8}{1-y}}$
So, y cannot assume 1
So not onto

Question 6
let A={1,2,3} Which of the following relation defined on A is a reflexive relation?
(a) {(1, 1), (1, 2), (2, 1), (2, 2)}
(b) {(1, 2), (2, 1)}
(c) {(1, 1), (2, 2), (3, 3)}
(d) {(1, 2), (2, 3), (3, 1)}

{(1, 1), (2, 2), (3, 3)}

Question 7
Which of the following is an onto function f:R -> R?
(a) $f(x) = x^2$
(b) $f(x) = 2x + 1$
(c) $f(x) = |x|$
(d) $f(x) = x^2 + 1$

f(x) = 2x + 1

Question 8
For real numbers x and y, define xRy if and only if $x – y + \sqrt 2$ is an irrational number. Then the relation R is
(a) symmetric
(b) Reflexive
(c) Transitive
(d) None of these

Question 9
$f(x) =x^2 +1$ and $g(x) = sin x$, then the value of $g \circ f$ is
(a) $sin^2x + 1$
(b) $sin (x^2 +1)$
(c) $x sin x + 1$
(d) None of these

$g \circ f= g(f(x)) = g(x^2 +1) =sin (x^2 +1)$

Question 10
let f(x)=[x] and g(x) =|x|, the the value of $(g \circ f) (\frac {-5}{3}) - (f \circ g)(\frac {-5}{3})$ is
(a) 1/2
(b) 1
(c) -1
(d) -2

$(g \circ f) (\frac {-5}{3}) - (f \circ g)(\frac {-5}{3})= g[f(\frac {-5}{3}] - f[g(\frac {-5}{3})]=g(-2) -f(\frac {5}{3}) =2 -1=1$

## Fill in the blanks

Question 11
(i) Let the relation R be defined in N by aRb if 2a + 3b = 30. Then R = _____
(ii) if $f(x)=x^2 +2$ and $g(x)= 1 - \frac{1}{1-x}$, the $f \circ g$ is _____
(iii) The inverse of the function $f(x)= \frac {x}{x+2}$ is _______

(i) {(3,8) , (6,6) , ((9,4), (12,2)}
(ii) $\frac {3x^2 -4x+2}{(1-x)^2}$
(iii) $\frac {2x}{1-x}$

Question 12
Let f : W -> W be defined as
$f(n)=\begin{cases} & \text{ n-1 if } n \ is \ odd \\ & \text{ n+1 if } n \ is \ even \end{cases}$
Then show that f is invertible.Also, find the inverse of f.

Here,
f : W -> W
is such that, if n is odd,
$f \circ f (n)= f (f (n)) =f (n -1)= n-1 +1= n$
and if n is even,
$f \circ f (n)= f (f (n)) =f (n +1)= n+ 1 =1= n$
Hence

$f \circ f =I$ This implies that f is invertible and $f^{-1} = f$

Question 13
Show that f : [– 1, 1] -> R , given by $f(x) =\frac {x}{x+2}$ is one-one.
Find the inverse of the function f : [– 1, 1] -> Range of f

For one-one
$f(x_1) = f(x_2)$
$\frac {x_1}{x_1+2}=\frac {x_2}{x_2+2}$
$x_1=x_2$
Hence it is one-one function
For onto
let $y= \frac {x}{x+2}$
or
$x=\frac {2y}{1-y}$
From this we see that,y cannot take the value 1
So it is not onto on R
But we can make it onto function, by defining the function as
f : [– 1, 1] -> Range of f , given by $f(x) =\frac {x}{x+2}$
Also
$f(x) = f(\frac {2y}{1-y})= y$
Now Since this is both one-one and onto, we can have the inverse and inverse will be
$f^{-1} (x) = \frac {2x}{1-x}$

Question 14
Show that the relation R in the set N X N defined by (a,b)R(c,d) if $a^2+d^2 =b^2+ c^2$ for all a,b,c,d in N is an equivalence relation

Let (a,b) in N x N
Now as
$a^2 + b^2 = b^2 + a^2$
Thre (a,b)R(a,b)
Hence, R is reflexive.

Let (a, b), (c, d) in N x N be such that (a, b)R(c, d)
$a^2+d^2 =b^2+ c^2$
$c^2 + b^2=a^2+d^2$
So ,(c, d)R(a, b)
Hence, R is symmetric.

Let (a, b), (c, d), (e, f ) N N be such that (a, b)R(c, d), (c, d)R(e, f ).
$a^2+d^2 =b^2+ c^2$
$c^2 + f^2=d^2 +e^2$
$a^2 + f^2= e^2 + b^2$
(a,b) R (e,f)
Here , R is transitive also

Question 15
Functions f , g : R -> R are defined, respectively, by $f(x) = x^2 + 3x + 1$,g(x) = 2x – 3, find
(i) $f \circ g$
(ii) $g \circ f$
(iii)$f \circ f$
(iv) $g \circ g$

(i) $f \circ g = f(g(x)) = f(2x-3) = (2x-3)^2 + 3(2x-3) +1=4x^2 -6x +1$
(ii)$g \circ f = g(f(x))= g(x^2 + 3x + 1) = 2(x^2 + 3x +1) -3 = 2x^2 + 6x-1$
(iii) $f \circ f = f(f(x)) = f(x^2 + 3x +1) = (x^2 + 3x+1)^2 + 3(x^2 + 3x+1) + 1=x^4 + 6x^3 + 14x^2 + 15x + 5$
(iv) $g \circ g= g(g(x) = g(2x-3) = 2(2x-3) -3 = 4x-9$

Question 16
Show that the relation R in the set A X A defined by (a,b)R(c,d) if $a+d =b+c$ for all a,b,c,d in A is an equivalence relation. Here A={1,2,3...10}

Let (a,b) in A x A
Now as
$a + b = b + a$
Thre (a,b)R(a,b)
Hence, R is reflexive.

Let (a, b), (c, d) in N x N be such that (a, b)R(c, d)
$a+d =b+ c$
$c + b=a+d$
So ,(c, d)R(a, b)
Hence, R is symmetric.

Let (a, b), (c, d), (e, f ) N N be such that (a, b)R(c, d), (c, d)R(e, f ).
$a+d =b + c$
$c+ f=d +e$
$a + f= e + b$
(a,b) R (e,f)
Here , R is transitive also

Question 17
If $f(x) =\frac {4x+3}{6x-4}$ and $x \ne \frac {-2}{3}$ , show that fof(x) = x for all x $x \ne \frac {-2}{3}$ . Also, find the inverse of f

$f(x) =\frac {4x+3}{6x-4}$
Now
$f \circ f (x) = f(f(x)) = f(\frac {4x+3}{6x-4}) = \frac { 4\frac {4x+3}{6x-4} + 3}{6\frac {4x+3}{6x-4} - 4} = \frac { 16x + 12 +18x -12}{24x + 18 -24x+ 16} = x$
For inverse
$y=\frac {4x+3}{6x-4}$
Solving
$x= \frac {4y+3}{6y-4}$
So inverse is $\frac {4x+3}{6x-4}$

Question 18
Let A= R -{-1} and * be the binary operation on A defined by
a*b = a+b+ab
(i) Show that * is commutative and Associative
(ii) Find the identity element for * on A
(iii) Prove that every element of A is invertible

(i)Commutative:
a∗b=a+b+ab
b*a =b+a +ab=a +b +ab
So , ∗ is commutative.
Associative:
a∗(b *c)= a *(b + c +bc)
=a + b + c +bc + a(b + c +bc) = a+b+c +ab + ab+ac + bc +abc
(a∗b) *c)=(a +b + ab) * c
=a +b + ab + c + (a +b + ab) c= a+b+c +ab + ab+ac + bc +abc
Hence ∗ is associative.
(ii) Identity element:
Let (e) be the identity element
then by definition
a* e = a = e*a
a+ e + ae=a=e+a + ae So e+ ae=0 e(1+a)=0 Hence e=0
So (0,0) is the identity element
(iii)Inverse:
Let i is inverse of element a ∈ A
Now by definition.
a* i= 0 =i* a
a+i + ai=0 = i+ a+ia
So, $i = \frac {a}{1+a}$
So we have inverse for all elements

Question 19
Prove that the function f:[0, $\infty$) -> R given by $f(x) = 9x^2+ 6x – 5$ is not invertible. Modify the codomain of the function f to make it invertible, and hence find $f^{–1}$

$f(x) = 9x^2+ 6x – 5$
for one-one
$f(x_1) = f(x_2)$
$9x_1^2 + 6x_1 -5= 9x_2^2 + 6x_2 -5$
$9(x_1^2 - x_2^2) + 6(x_1 - x_2) =0$
$9(x_1 -x_2)(x_1 + x_2) + 6(x_1 - x_2) =0$
$((x_1 - x_2)[9(x_1 + x_2) +6]=0$
either $x_1=x_2$
or $9(x_1 + x_2) + 6=0$
Since $x \geq 0$, this cannot be true.
so $x_1=x_2$
This is one-one function
For onto
$f(x) =y= 9x^2+ 6x – 5$
$9x^2+ 6x – 5 -y=0$
$x=\frac {-1 \pm \sqrt {6+y}}{3}$
Since $x \geq 0$
$x=\frac {-1 + \sqrt {6+y}}{3}$
So y cannot have value lessor than -6. So it is not onto function
Now $6 + y \geq$ or $y \geq -6$,
Also $\sqrt {6+y} \geq 1$ as $x \geq 0$
so, $6+y \geq 1$ or $y \geq -5$
So, Range of the function will be (-5,$\infty$) .
So we can define the function as
function f:[0, $\infty$) -> (-5,$\infty$) given by $f(x) = 9x^2+ 6x – 5$
Now for any value of y in (-5,$\infty$) , there exists an value $\frac {-1 + \sqrt {6+y}}{3}$ in [0, $\infty$) such that
$f(x) = 9(\frac {-1 + \sqrt {6+y}}{3})^2 + 6(\frac {-1 + \sqrt {6+y}}{3}) -5= y$
So this is both one-one and onto .hence invertible
Inverse is given by
$f^{–1} (x) =\frac {-1 + \sqrt {6+x}}{3}$

Question 20
Check whether the relation R in the set R of real numbers, defined by R = {(a, b) : 1 + ab > 0}, is reflexive, symmetric or transitive
Question 21
Show that the relation R on the set Z of integers, given by R = {(a, b) : 2 divides (a – b)} is an equivalence relation
Question 22
Prove that the relation R in the set A = {1, 2, 3, 4, 5, 6, 7} given by R = {(a, b) : |a – b| is even} is an equivalence relation
Question 23
Show by examples that the relation R in , defined by
R = {(a, b) : $a \le b^3$} is neither reflexive nor transitive.
Question 24
Let R be a relation on the set A of ordered pairs of positive integers defined by (x, y) R (u, v) if and only if xv = yu. Show that R is an equivalence relation

Clearly, (x, y) R (x, y), ∀ (x, y) A, since xy = yx. This shows that R is reflexive.
Further, (x, y) R (u, v) ⇒ xv = yu ⇒ uy = vx and hence (u, v) R (x, y). This shows that R is symmetric.
Similarly, (x, y) R (u, v) , xv = yu or v/u=y/x,
(u, v) R (a, b), ub = va or va=ub or v/u=b/a
So y/x=b/a or ay=bx or xb=ya
so (x,y) R (a, b)
This shows that R is transitive
Therefore R is an equivalence relation

Go back to Class 12 Main Page using below links