Extra questions on relations and functions class 12
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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 Answers
Answer is (d)
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 Answers
(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 Answers
Answer is (b)
$f(x)= \frac {x}{x+1}$
$\frac {df(x)}{dx}= \frac {1}{(x+1)^2}$
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) $\frac {2x-2}{3}$ Answers
Answer is (b)
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 Answers
Answer is (d)
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)} Answers
Answer: (c)
{(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) $f(x) = x^2 + 1$ Answers
Answer: (b)
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 Answers
Answer is (b)
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 Answers
$g \circ f= g(f(x)) = g(x^2 +1) =sin (x^2 +1)$
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 Answers
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 _______ Answers
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. Answers
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 Answers
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 Answers
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 ).
$a^2+d^2 =b^2+ c^2$
$c^2 + f^2=d^2 +e^2$
Adding both
$a^2 + f^2= e^2 + b^2$
(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) $g \circ g$ Answers
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} Answers
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 ).
$a+d =b + c$
$c+ f=d +e$
Adding both
$a + f= e + b$
(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 Answers
$f(x) =\frac {4x+3}{6x-4}$
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 Answers
(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 $f^{–1}$ Answers
$f(x) = 9x^2+ 6x – 5$
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$
Solving the quadratic equation
$x=\frac {-1 \pm \sqrt {6+y}}{3}$
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 Answers
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
