If $$f(x) = 3x - 5, \,\,then\,\, f^{-1}(x)$$.

is given by $$\dfrac {x+5}{3}$$

is given by $$\dfrac {1}{3x-5}$$

does not exist because f is not one-one

does not exist because f is not onto

MCQ Questions for Class 12 Commerce Maths Relations And Functions Quiz 5

$f(x) =\dfrac {1}{1-x}, x \neq 0, 1$ then the graph of the function

$y = f[f\{f(x)\}]$ for

$x > 1$ is

0%
a straight line
0%
a circle
0%
an ellipse
0%
a pair of straight lines

Explanation

Given

$f(x) = \cfrac{1}{1-x}$

$\displaystyle \Rightarrow f\{f(x)\} = \frac{1}{1-f(x)} = \frac{1}{1-\frac{1}{1-x}}=\frac{1-x}{1-x-1}=1-\frac{1}{x}$

$\displaystyle \therefore f[f\{f(x)\}] = \frac{1}{1- f\{f(x)\}}=\frac{1}{1-1+\frac{1}{x}}=x$ , which is a straight line.

Let

$f : R \rightarrow R, g : R \rightarrow R$ be two function such that

$f(x) = 2x- 3, g(x) = x^3 + 5$
The function

$(fog)^{1}(x)$ is equal to.

0%
$\left ( \dfrac {x+7}{2} \right )^{1/3}$
0%
$\left (x- \dfrac {7}{2} \right )^{1/3}$
0%
$\left ( \dfrac {x-2}{7} \right )^{1/3}$
0%
$\left ( \dfrac {x-7}{2} \right )^{1/3}$

Explanation

Given

$f(x) = 2x-3$ and

$g(x) = x^3+ 5$

$(fog)(x) = fg(x) = f(x^3+5) = 2(x^3+5) -3 = 2x^3 +7$

Let

$(fog)(x) = y = 2x^3+7$

$y-7 = 2x^3$

$x = (\dfrac{y-7}{2})^{\dfrac{1}{3}}$

$\therefore (fog)^{-1}(x) = (\dfrac{x-7}{2})^{\dfrac{1}{3}}$

$f(x) = 3x - 5, \,\,then\,\, f^{-1}(x)$ .

0%
is given by $\dfrac {1}{3x-5}$
0%
is given by $\dfrac {x+5}{3}$
0%
does not exist because f is not one-one
0%
does not exist because f is not onto

Explanation

Given,

$f(x)= 3x-5$

Let

$y=f^{ -1 }\left( x \right)$

$\Rightarrow x=f\left( y \right) =3y-5$

$\Rightarrow y=\dfrac { x+5 }{ 3 }$

$f : [0, \Pi ] \rightarrow [-1, 1]$ , f(x) = cosx, then f is.

0%
one-one
0%
onto
0%
one-one onto
0%
none of these

Explanation

f takes all values from

$[-1,1]$ while travelling from

$[0,\pi]$ . None of the values are repeated either.

Hence it is one-one and onto.

$f(x) = \left\{\begin{matrix} 1&x \in Q \\ 0 &x \notin Q\end{matrix}\right.$ then

$fof(\sqrt 3 )$ is equal to

0%
$0$
0%
$1$
0%
$\sqrt 3$
0%
none of these

Explanation

We know

$\sqrt{3}$ is an irrational number

Thus

$f(\sqrt{3}) =0$ and

$0$ is a rational number

Hence

$fof(\sqrt 3 )= f(0) = 1$

If functions

$f\left ( x \right )$ and

$g\left ( x \right )$ are defined on

$R\rightarrow R$ such that

$f(x)=x+3, x$

$\in$ rational

$=4x, x$

$\in$ irrational

$g(x)=x+\sqrt{5}$ , x

$\in$ irrational

$=-x, x$

$\in$ rational
then

$\left ( f-g \right )\left ( x \right )$ is

0%
one-one & onto
0%
neither one-one nor onto
0%
one-one but not onto
0%
onto but not one-one

Explanation

$f\left( x \right) =x+3, x\epsilon$ rational

$=4x, x\epsilon$ irrational

$g\left( x \right) =x+\sqrt { 5 } , x \epsilon$ irrational

$= -x,x\epsilon$ rational

$(f-g)(x)=2x+3, x\ \epsilon$ rational

$+3x-\sqrt { 5\quad } x\quad \epsilon$ irrational

For one-one;

We know that one one function is a

${ f }^{ \underline { n } }$ for which every element of the range of the function corresponds to exactly one element of the domain.

But in this case this is not true as for all rational nos. the

${ f }^{ \underline { n } }$ is one one but for irrational

${ f}^{ \underline { n } }$ , every elements of the range does not corresponds to exactly one element of the domain.

$(f-g)(x)\neq (f-g)({ x }^{ \prime })$ when

$x \epsilon$ rational and

$x \epsilon$ irrational

for onto:

The

${ f }^{ \underline { n } }(f-g)(x)$ does not cover the whole range of the function.

$\displaystyle f\left ( x \right )=\left ( 1-x^{3} \right )^{\frac{1}{3}}$ , then find

$fof(x)$

0%
$\dfrac1x$
0%
$x$
0%
$x^2$
0%
$x^3$

Explanation

Given,

$\displaystyle f\left ( x \right )=\left ( 1-x^{3} \right )^{\frac{1}{3}}$
Therefore,

$\displaystyle fof\left ( x \right )=f\left [ f\left ( x \right ) \right ]$

$\displaystyle =f\left ( \left ( 1-x^{3} \right )^{\frac{1}{3}} \right )=\left [ 1-\left \{ \left ( 1-x^{3} \right )^{\frac{1}{3}} \right \}^{3} \right ]^{\frac{1}{3}}$

$\displaystyle =\left [ 1-\left ( 1-x \right )^{3} \right ]^{\frac{1}{3}}=\left ( x^{3} \right )^{\frac{1}{3}}=x$

Let

$f(x) =\frac {ax+b} {cx+d}$ . Then fof(x) = x provided that.

0%
d =- a
0%
d = a
0%
a = b = c = d = 1
0%
a = b = 1

Explanation

$f(f(x))=\dfrac{af(x) + b}{cf(x)+d}$

$f(f(x))=\dfrac{a\dfrac{ax+b}{cx+d} + b}{c\dfrac{ax+b}{cx+d}+d}$

$f(f(x))=\dfrac{a^2x + ab+bcx+bd}{acx+bc+d^2+dcx}=x$

$\Rightarrow a^2x+ab+bcx+bd=acx^2+bcx+d^2x+dcx^2$

Given

$a=-d$ , the above equation can also be verified

$\displaystyle f(x)=2x^{3}+7x-5\, and\, g(x)=f^{-1}(x)$ then g' (4) is equal to

0%
$\displaystyle \frac{1}{13}$
0%
$\displaystyle \frac{1}{103}$
0%
$\displaystyle \frac{1}{4}$
0%
non existent

Explanation

Given,

$f(x)=2x^3+7x-5$

$\therefore f'(x)= 6x ^2+7$

$g(x) =f^{-1}(x)\Rightarrow g\{f(x)\}=x$
Differentiating both sides, w.r.t

$x$

$\displaystyle {g}'\left ( f(x) \right )=\frac{1}{f'\left ( x \right )}=\frac{1}{6x^{2}+7}$
when

$x = 1,f(x) = 4$

$\therefore \displaystyle g'\left ( 4 \right )=\frac{1}{13}$

$\displaystyle R=R^{-1}$ then the relation R is ________

0%
reflexive
0%
symmetric
0%
anti-symmetric
0%
transitive

Let

$\displaystyle f\left ( x \right )=\frac{3}{2}+\sqrt{x-\frac{3}{4}}$ be a function and

$g\left ( x \right )$ be another function such that

$g\left ( f\left ( x \right ) \right )=x,$ then the value of

$g\left ( 20 \right )$ will be

0%
$333$
0%
$335$
0%
$338$
0%
$343$

Explanation

$g\left ( x \right )$ is inverse of

$f\left ( x \right )$

$\Rightarrow$

$\displaystyle x=\frac{3}{2}+\sqrt{y-\frac{3}{4}}$

$\Rightarrow$

$\displaystyle y=\left ( x-\frac{3}{2} \right )^2+\frac{3}{4}=f^{-1}(x)=g(x)$

$\therefore$

$\displaystyle g\left ( 20 \right )=\left ( 20-\frac{3}{2} \right )^{2}+\frac{3}{4}=343$

If f(x) + f(1-x) = 10 then the value of

$\displaystyle f\left ( \frac{1}{10} \right )+f\left ( \frac{2}{10} \right )+.........+f\left ( \frac{9}{10} \right )$

0%
is 45
0%
is 50
0%
is 90
0%
Cannot be determined

Explanation

$f(\dfrac {1}{10}) + f(\dfrac {2}{10}) + f(\dfrac {3}{10}) + f(\dfrac {4}{10}) + f(\dfrac {5}{10}) + f(\dfrac {6}{10}) + f(\dfrac {7}{10}) + f(\dfrac {8}{10}) + f(\dfrac {9}{10})$

$= [f(\dfrac {1}{10}) + f(\dfrac {9}{10}) ] + [ f(\dfrac {2}{10}) + + f(\dfrac {8}{10}) ] + [ f(\dfrac {3}{10}) + + f(\dfrac {7}{10}) ] + [ f(\dfrac {4}{10}) + f(\dfrac {6}{10}) ] + f(\dfrac {5}{10}) ]$

$= 10 + 10 + 10 + 10 + 5 = 45$

$f (x) = 2x - 1$ and

$g (x) = 3x + 2$ , then find

$(fog) (x)$ :

0%
$2 (3x + 1)$
0%
$2 ( 3x + 2)$
0%
$3 (2x + 1 )$
0%
$3 ( 3x + 1 )$

Explanation

$fog(x)=2(3x+2)-1=6x+4-1=6x+3=3(2x+1)$

$f(x) = -x^2+1, g(x) = -\sqrt[3]{x}$ then (gofogofogogog) (x) is.

0%
an odd function
0%
an even function
0%
a polynomial function
0%
an identity function

Explanation

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

$f(x) = -x^2+1, g(x) = -\sqrt[3]{x}$ ........

$\Rightarrow$

$f$ is even and

$g$ is odd

$\therefore$

$gogog$ is even

$\Rightarrow$

$fogogog$ is even

$\Rightarrow$

$gofogofogogog$ is even.

If the function

$f(x) = x^3 + e^{x/2}$ and

$g(x) = f^{-1}(x)$ ; then the value of g'(1) is

0%
$2$
0%
$-2$
0%
$1$
0%
$0$

Explanation

$f(x) =x^3+e^{x/2}$

$\displaystyle f'(x) = 3x^2+\frac{1}{2}e^{x/2}$
Given g is inverse of f

$\Rightarrow g(f(x))=x$
Differentiating both sides w.r.t

$x$

$\displaystyle g'(f(x)).f'(x) =1\Rightarrow g'(f(x))=\frac{1}{f'(x)}$
Clearly

$\displaystyle f'(0)=1\therefore g'(1)=g'(f(0))=\frac{1}{f'(0)}=2$

If f(x)=2x-1 and g(x)=3x+2 then find (fog) (x)

0%
2(3x+1)
0%
2(3x+2)
0%
3(2x+1)
0%
3(3x+1)

Explanation

$(fog)(x) = f(g(x)) = 2(3x+2) -1 = 6x+4-1 = 6x+3 = 3(2x+1)$

If f(x) = 2x+1 and g(x) = 3x-5 then find

$\left ( fog \right )^{-1}\left ( 0 \right )$

0%
5/3
0%
3/2
0%
2/3
0%
3/5

Explanation

$(fog)(x) = f(g(x)) = 2(3x-5) + 1 = 6x-10 + 1 = 6x-9$

Let

$(fog)(x)= y$
Then, to find

$(fog)^{-1} (x)$ , we find

$x$ in terms of

$y$

So,

$6x-9 = y$

$=> x = \dfrac {y+9}{6}$

$=> (fog)^{-1} (x)= \dfrac {y+9}{6}$

$=> (fog)^{-1} (0)= \dfrac {0+9}{6} = \dfrac {3}{2}$

If X = {2,3,5,7,11} and Y = {4,6,8,9,10} then find the number of one-one functions from X to Y

0%
720
0%
120
0%
24
0%
12

Explanation

Both sets X and Y have the same number of elements.

When two sets have same number of

$n$ elements, then the number of one to one functions from one set to the other is

$n!$

So, number of one to one functions from X to Y

$= 5 ! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

Find

$\left( f\circ g \right) \left( 3 \right)$ when

$f\left( x \right) =7x-6$ and

$g\left( x \right) =5{ x }^{ 2 }-7x-6$ .

0%
$-36$
0%
$1014$
0%
$-90$
0%
$120$

Explanation

We have,

$f(x)=7x-6$ and

$g(x)=5x^2-7x-6$

Then,

$(f\circ g)(x)$

$=f(g(x))$

$=f(5x^2-7x-6)$

$=7(5x^2-7x-6)-6$

$=35x^2-49x-42-6$

$=35x^2-49x-48$

$\therefore (f\circ g)(3)$

$=35(3)^2-49(3)-48$

$=315-147-48$

$=120$

So,

$\text{D}$ is the correct option.

$R$ is a relation from a set

$A$ to the set

$B$ and

$S$ is a relation from

$B$ to

$C,$ then the relation

$SoR$

0%
is from $C$ to $A$
0%
does not exist
0%
is from $A$ to $C$
0%
None of these

Explanation

Since

$R\subseteq A\times B$ and

$S\subseteq B\times C$ , we have
So

$R\subseteq A\times C$ .

$\therefore$ So R is a relation from

$A$ to

$C.$

If f = {(1,3) (2,1) (3,4) (4,2)} and g = {(1,2) (2,3) (3,4) (4,1)} then find n(fog)

0%
12
0%
16
0%
4
0%
5

Explanation

we can find

$n(fog)$ by finding the range of the domain of

$f$
In

$(1,3)$ as the range

$3$ is mapped to

$4$ in

$g$ , the ordered pair here for

$fog$ is

$(1,4)$
In

$(2,1)$ as the range

$1$ is mapped to

$2$ in

$g$ , the ordered pair here for

$fog$ is

$(2,2)$
In

$(3,4)$ as the range

$4$ is mapped to

$1$ in

$g$ , the ordered pair here for

$fog$ is

$(3,1)$
In

$(4,2)$ as the range

$2$ is mapped to

$3$ in

$g$ , the ordered pair here for

$fog$ is

$(4,3)$

As we have

$4$ ordered pairs for

$fog$ , we have

$n(fog) = 4$

What is the relation for the following diagram?

0%
$R =\{(2, 5), (2, 6), (3, 7)\}$
0%
$R = \{(1, 5), (2, 6), (3, 7)\}$
0%
$R =\{(1, 5), (2, 6), (1, 7)\}$
0%
$R = \{(1, 5), (2, 6), (2, 7)\}$

Explanation

$R:A\to B$

$A\times B = \{(1,5),(2,6),(3,7)\}$

Which of the following do(es) not belong to

$A \times B$ for the sets

$A = \{1, 2\}$ and

$B =\{0, 2\}$ ?

0%
$R = \{(1, 0), (2, 2)\}$
0%
$R = \{(1, 1), (2, 1)\}$
0%
$R = \{(1, 0), (1, 2)\}$
0%
$R = \{(1, 2), (2, 2)\}$

Explanation

A relation for the sets

$A$ and

$B$ from

$A$ to

$B$ will be a subset of the

$A\times B.$
Now the ordered pair belonging to

$A\times B$ are

$\{1,2\}\times\{0,2\}$

$=\{(1,0),(1,2),(2,0),(2,2)\}$
Hence,

$\{(1,1)\}$ and

$\{(2,1)\}$ do not belong to

$A\times B$ .

What is the relation for the set

$A =\{-1, 0, 3\}$ and

$B =\{1, 2, 3\}$ ?

0%
$R =\{(-1, 1), (1, 2), (3, 3)\}$
0%
$R = \{(-1, 1), (0, 2), (3, 3)\}$
0%
$R = \{(-1, 1), (2, 2), (3, 3)\}$
0%
$R = \{(-1, 1), (0, 2), (-1, 3)\}$

Explanation

$A\times B=\{(-1,1),(-1,2),(-1,3),(0,1),(0,2),(0,3),(3,1),(3,2),(3,3)\}$

R is subset of

$A\times B$

So option B is ans.

$f: R\rightarrow R$ be given by

$f(x) = (3 - x^{3})^{ {1}/{3}}$ , then find

$f(f (x))$ is

0%
$x^{{1}/{3}}$
0%
$x^{3}$
0%
$x$
0%
$3 - x^{3}$

Explanation

$f(x)= (3-x^3)^{\dfrac{1}{3}}\\ f(f(x))=f\left((3-x^3)^{\frac{1}{3}}\right)$

$= (3-((3-x^3)^{\frac{1}{3}})^3)^{\frac{1}{3}}=x$

$f:R\rightarrow R$ and

$g:R\rightarrow R$ are defined by

$f\left( x \right) =\left| x \right|$ and

$g\left( x \right) =\left[ x-3 \right]$ for

$x\in R$ , then

$g\left( f\left( x \right) \right) :\left\{ -\dfrac { 8 }{ 5 } < x < \dfrac { 8 }{ 5 } \right\}$ is equal to

[.] is Greatest integer function

0%
$\left\{ 0,1 \right\}$
0%
$\left\{ 1,2 \right\}$
0%
$\left\{ -3,-2 \right\}$
0%
$\left\{ 2,3 \right\}$

Explanation

Given that,

$f\left( x \right) =\left| x \right|$ and

$g\left( x \right) =\left[ x-3 \right]$
For

$-\dfrac { 8 }{ 5 } < x < \dfrac { 8 }{ 5 } , 0\le f\left( x \right) < \dfrac { 8 }{ 5 }$
Now, for

$0 < f \left( x \right) < 1$ ,

$g\left( f\left( x \right) \right) = \left[ f\left( x \right) -3 \right]$

$=-3\quad \because -3\le f\left( x \right) -3 < -2$
Again, for

$1 < f\left( x \right) < 1.6$

$g\left( f\left( x \right) \right) =-2$

$\because -2\le f\left( x \right) -3 < -1.4$
Hence, required set is

$\left\{ -3,-2 \right\}$

$R$ be the set of all real numbers and

$f:R\rightarrow R$ is given by

$f\left( x \right)=3{ x }^{ 2 }+1$ . Then, the set

$f^{ -1 }\left( \left[ 1,6 \right] \right)$ is

0%
$\left\{ -\sqrt { \dfrac { 5 }{ 3 } } ,0,\sqrt { \dfrac { 5 }{ 3 } } \right\}$
0%
$\left[ -\sqrt { \dfrac { 5 }{ 3 } } ,\sqrt { \dfrac { 5 }{ 3 } } \right]$
0%
$\left[ -\sqrt { \dfrac { 1 }{ 3 } } ,\sqrt { \dfrac { 1 }{ 3 } } \right]$
0%
$\left( -\sqrt { \dfrac { 5 }{ 3 } } ,\sqrt { \dfrac { 5 }{ 3 } } \right)$

Explanation

Given,

$f\left( x \right) =3{ x }^{ 2 }+1$
Let

$y=3{ x }^{ 2 }+1$

$\Rightarrow 3{ x }^{ 2 }=y-1\Rightarrow { x }^{ 2 }=\dfrac { y-1 }{ 3 }$

$\Rightarrow x=\pm \sqrt { \dfrac { y-1 }{ 3 } }$

$\therefore f^{ -1 }\left( x \right) =\pm \sqrt { \dfrac { x-1 }{ 3 } }$
When

$x\in \left[ 1,6 \right]$ , then,

$f^{ -1 }\left( x \right) \in \left[ -\sqrt { \dfrac { 5 }{ 3 } } ,\sqrt { \dfrac { 5 }{ 3 } } \right]$

Let

$R$ be the set of real numbers and the functions

$f: R \rightarrow R$ and

$g: R\rightarrow R$ be defined by

$f(x) = x^{2} + 2x - 3$ and

$g(x) = x + 1$ . Then the value of

$x$ for which

$f(g(x)) = g(f(x))$ is

0%
$-1$
0%
$0$
0%
$1$
0%
$2$

Explanation

According to the question,

$f(g(x)) = g(f(x))$

$\Rightarrow f(x + 1) = g(x^{2} + 2x - 3)$

$\Rightarrow (x + 1)^{2} + 2 (x + 1) - 3 = x^{2} + 2x - 3 + 1$

$\Rightarrow x^{2} + 1 + 2x + 2x + 2 - 3 = x^{2} + 2x - 2$

$\Rightarrow x^{2} + 4x = x^{2} + 2x - 2$

$\Rightarrow x^{2} + 4x - x^{2} - 2x + 2 = 0$

$\Rightarrow 2x + 2 = 0$

$\Rightarrow 2x = -2$

$\Rightarrow x = -1$

Let

$f:R\rightarrow R$ be such that

$f$ is injective and

$f(x)f(y)=f(x+y)$ for all

$x,y\in R$ , if

$f(x), f(y)$ and

$f(z)$ are in GP, then

$x,y$ and

$z$ are in

0%
AP always
0%
GP always
0%
AP depending on the values of $x,y$ and $z$
0%
GP depending on the values of $x,y$ and $z$

Explanation

Let the funtion

$f(x)={a}^{kx}$
which define in

$f:R\rightarrow R$ and injective also.

Now, we have

$f(x)f(y)=f(x+y)$

$\Rightarrow$

${a}^{kx}.{a}^{ky}={a}^{k(x+y)}$

$\Rightarrow$

${a}^{k(x+y)}={a}^{k(x+y)}$

$\because$

$f(x), f(y)$ and

$f(z)$ are in GP

$\therefore$

$f({y}^{2})=f(x).f(z)$

$\Rightarrow$

${a}^{2ky}={a}^{kx}.{a}^{kz}$

$\Rightarrow$

${e}^{2ky}={e}^{k(x+z)}$

On comparing, we get

$2ky=k(x+z)$

$\Rightarrow$

$2y=x+z$

$\Rightarrow$

$x,y$ and

$z$ are in AP

Find the number of binary operations on the set

$\left \{a, b\right \}$

0%
$10$
0%
$16$
0%
$20$
0%
$8$

Explanation

Considering 'n' elements in a set, the no. of binary operations on that set will be

$=n^{n^{2}}$

Applying this concept to the above set gives us

$=2^{2^{2}}$ ...................

$\because n=2$

$=2^{4}$

$=16$ .

Let Q be the set of all rational numbers in [0, 1] and

$f : [0, 1]\rightarrow [0, 1]$ be defined by

$f(x)=\begin{cases}x&for&x\in Q\\ 1-x&for&x\notin Q\end{cases}$
Then the set

$S=\{x\in [0, 1]: (f\, o \, f)(x)=x\}$ is equal to

0%
[0, 1]
0%
Q
0%
[0, 1] - Q
0%
(0, 1)

Explanation

Let

$x\in Q$ , then,

$f(x)=x$ where

$x\in Q$

So,

$fof(x)=f(f(x))=f(x)=x$ as

$x\in Q$

$\therefore fof(x)=x$ when

$x \in Q$

Now,

Let

$x\notin Q$ then

$f(x) =1-x$

$\therefore fof(x)=1-(1-x) = x$

$1-x\notin Q$ as

$x\notin Q$

where

$x\notin Q$

$fof(x)=\begin{cases}x \,where\, x\in Q \,\&\, x\in [0, 1] \\ x\, where x\notin Q\, \&\, x\in [0, 1] \end{cases}$

$\therefore$ the set

$S = [0, 1]$

$f(x)={2}^{100}x+1, g(x)={3}^{100}x+1$ , then the set of real numbers

$x$ such that

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

0%
empty
0%
a singleton
0%
a finite set with more than one element
0%
infinite

Explanation

Given

$f(x)={2}^{100}x+1, g(x)={3}^{100}x+1$

Now

$fo{g(x)}=x$

$\Rightarrow$

$f({3}^{100}.x+1)=x\\$

$\Rightarrow$

${2}^{100}({3}^{100}.x+1)+1=x\\$

$\Rightarrow$

${6}^{100}.x+{2}^{100}+1=x\\$

$\Rightarrow$

$x(1-{6}^{100})=(1+{2}^{100})$

$\Rightarrow$

$x=\cfrac{1+{2}^{100}}{1-{6}^{100}}$

Hence

$fog(x)=x$ represent a singleton set.

In three element group

$\{e, a, b\}$ where

$e$ is the identity,

$a^5b^4$ is equal to

0%
$a$
0%
$e$
0%
$ab$
0%
$b$

Explanation

Given group

$G\equiv\{e,a,b\}$

It is given that

$e$ is an identity.

Therefore only possibilities we have are

Either

$a\cdot a = e$ and

$b\cdot b = e$ .... (i)

$a\cdot b = b\cdot a = e$ ...(ii)

Case I:

$a\cdot a = e$ and

$b\cdot b = e$

$a^5b^4 = a\cdot a\cdot a\cdot a\cdot a\cdot b\cdot b\cdot b\cdot b$

$\Rightarrow a^5b^4 = e\cdot a\cdot a\cdot a\cdot b\cdot b\cdot e$

$\Rightarrow a^5b^4 = e\cdot a\cdot e$

$\Rightarrow a^5b^4 = a$

Case II:

$a\cdot b = b\cdot a = e$

$a^5b^4 = a\cdot a\cdot a\cdot a\cdot a\cdot b\cdot b\cdot b\cdot b$

$\Rightarrow a^5b^4 = a\cdot a\cdot a\cdot a\cdot e\cdot b\cdot b\cdot b$

$\Rightarrow a^5b^4 = a\cdot a\cdot a\cdot e\cdot b\cdot b$

$\Rightarrow a^5b^4 = a\cdot a\cdot e\cdot b$

$\Rightarrow a^5b^4 = a\cdot e$

$\Rightarrow a^5b^4 = a$

Hence, both cases resulted in

$a$ .

Find the value of

${f}^{-1}(1.5)$ if

$f(x)=\sqrt [ 3 ]{ { x }^{ 3 }+1 }$ .

0%
$3.4$
0%
$2.4$
0%
$1.3$
0%
$1.5$

Explanation

As per the condition,

$1.5 = \sqrt[3]{x^3+1}$
Taking cube roots on both the sides,

$1.5^3=x^3+1$

$3.375 - 1 = x^3$

$2.375 = x^3$

$x \approx 1.3$

$f(x)\, =\, (p\, -\, x^n)^{1/n},\, p\, >\, 0$ and

$n$ is a positive integer, then

$f(f(x)) =$

0%
$x$
0%
$x^n$
0%
$p^{1/n}$
0%
$p\, -\, x^n$

Explanation

Given,

$f(x)=(p-x^{n})^{\dfrac{1}{n}}$

$\therefore f(f(x))=(p-((p-x^{n})^{\dfrac{1}{n}})^{n})^{\dfrac{1}{n}}$

$=(p-(p-x^{n}))^{\dfrac{1}{n}}$

$=(x^{n})^{\dfrac{1}{n}}$

$=x$

Which of the following is a subgroup of the group

$G = \left \{1, 2, 3, 4, 5, 6\right \}$ under

$\otimes_{7}$ .

(

$\otimes_7$ : under multiplication modulo 7)

0%
$\left \{2, 6, 1\right \}$
0%
$\left \{1, 2, 4\right \}$
0%
$\left \{5, 4, 2\right \}$
0%
$\left \{2, 3, 1\right \}$

Explanation

$'1'$ is not present
(d) can not be a subgroup of

$2\otimes_{7} 3 = 6\not {\epsilon} \left \{1, 2, 3\right \}$
(a) can not be a subgroup as

$2\otimes_{7} 6 = 5\not {\epsilon} \left \{2, 6, 1\right \}$

$\therefore$ (b) is a subgroup

$f: R\rightarrow R^{+}$ and

$g: R^{+} \rightarrow R$ are such that

$g(f(x)) = |\sin x|$ and

$f(g(x)) = (\sin \sqrt {x})^{2}$ , then a possible choice for f and g is

0%
$f(x) = x^{2} , g(x) = \sin \sqrt {x}$
0%
$f(x) = \sin x, g(x) = |x|$
0%
$f(x) = \sin^{2}x, g(x) = \sqrt {x}$
0%
$f(x) = x^{2}, g(x) = \sqrt {x}$

Explanation

As f returns only positive values and takes any real number it can be mod or square hence option 2 is eliminated

For g only positive values has to be given and it will return any real number

so on trying the option only option

$A$ and

$C$ results in the given equation ie

$g(f(x)) = |sin x|$

Now function g should take positive values and return real values which is satisfied by only option

$A$ as in option

$C$ square root of positive number will always result in positive number while

$sin\sqrt{x}$ will give -ve real numbers as well.

$\displaystyle { Q }_{ 1 }$ is the set of all relations other than

$1$ with the binary operation

$\displaystyle \ast$ defined by

$\displaystyle a\ast b=a+b-ab$ for all

$a, b \in \displaystyle Q_1$ , then the identity in

$\displaystyle { Q }_{ 1 }$ with respect to

$\displaystyle { Q }_{ 1 }$ is

0%
$1$
0%
$0$
0%
$-1$
0%
$2$

Explanation

Let

$b$ be the identity

$a*b=a$

$a+b-ab=a$

$b-ab=0$

$b(1-a)=0$

$b=0$ and

$a=1$
However,

$1$ is excluded.
Hence

$b=0$
Hence

$0$ is the identity element

$Z$ , the set of all integers, the inverse of

$-7$ w.r.t. defined by

$a\times b=a+b+7$ for all

$a, b, \in Z$ is :

0%
$-14$
0%
$7$
0%
$14$
0%
$-7$

Explanation

Consider the identity element of 'b' for any element

$a\in Z$

$a\times b=a$

$a+b+7=a$

$b+7=0$

$b=-7$ .
Hence identity element is -7.
Hence

$e=-7$ .
Now for the inverse element

$a\times a^{-1}=e$

$a+a^{-1}+7=-7$ ...

$(e=-7)$

$a^{-1}=-14-a$
Hence

$7^{-1}=-14-(-7)$

$=-7$ .

$f(x) = x^{2} + d$ and

$g(x) = 2x^{2}$ , where d is a constant. If

$\dfrac {f(g(2))}{f(2)} = 4$ , find the value of

$d$ .

0%
$16$
0%
5
0%
22
0%
18

Explanation

Given,

$f(x)=x^2+d$ and

$g(x)=2x^2$

$g(2)=2(2^{2})=8$

Hence,

$f(g(2))=f(8)=64+d$
Therefore

$\dfrac{f(g(2))}{f(2)}=\dfrac{64+d}{4+d}=4$

Hence,

$64+d=16+4d$

$\Rightarrow 48=3d$

$\Rightarrow d=16$

$f(g(a)) = 0$ where

$g(x) = \dfrac {x}{4} + 2$ and

$f(x) = |x^{2} - 3|$ , find the possible value of

$a.$

0%
$-8+4\sqrt{3}$
0%
$-(8+4\sqrt{3})$
0%
$6$
0%
$18$

Explanation

Given

$g(a)=\dfrac{a}{4}+2,f(x)=|x^2|-3$ and

$f(g(a))=0$

Now,

$f(g(a))=|g(a)^{2}-3|=|(\dfrac{a}{4}+2)^{2}-3|$

$=|\dfrac{a^{2}}{16}+a+4-3|=|\dfrac{a^{2}}{16}+a+1|=\dfrac{|a^{2}+16a+16|}{16}$

but

$f(g(a))=0$

$\Rightarrow\dfrac{|a^2+16a+16|}{16}=0$

$\Rightarrow a^{2}+16a+16=0$

$[\because |x|=0\Rightarrow x=0]$

$\Rightarrow a= \dfrac{-16\pm\sqrt{16^2-4(1)(64)}}{2}$

$\Rightarrow a=\dfrac{-16\pm\sqrt{256-64}}{2}=\dfrac{-16\pm8\sqrt{3}}{2}$

Hence,

$a=-(8+4\sqrt{3})$ and

$a=-8+4\sqrt{3}$ .

$p(x) = \dfrac{x}{x-2}$ and

$q(x) = \sqrt{9-x}$ , find the value of

$(p\circ q)(5)$

0%
$0$
0%
$\dfrac{8}{7}$
0%
$2$
0%
Undefined

Explanation

Given,

$p = \dfrac{x}{x-2}$

$q=\sqrt{9-x}$
For

$(p\circ q)$ , substitute the value

$p$ in

$q$
ie.,

$\dfrac{\sqrt{9-x}}{\sqrt{9-x}-2}(5)$ is a undefined.

$f(x) =$

$\sqrt{x}$ and

$g(x) =$

$\sqrt{x^2+4}$ , calculate the value of

$f(g(2))$ .

0%
$0$
0%
$1.41$
0%
$1.68$
0%
$2.45$
0%
$2.83$

Explanation

Given functions are

$f(x)=\sqrt {x}$ and

$g(x)=\sqrt {x^2+4}$ .

Now

$g(2)=\sqrt{2^2+4}=2\sqrt2$

Then

$f(g(2))=\sqrt { g(2) } = \sqrt { 2\sqrt { 2 } } ={ 2 }^{ 3/4 }=1.681$

$f(x) = 2x$ and

$f(f(x)) = x + 1$ , then the value of

$x$ is

0%
$\dfrac {1}{3}$
0%
$1$
0%
$2$
0%
$3$
0%
$5$

Explanation

Given

$f(x)=2x$ and

$f(f(x))=x+1$

Then

$f(f(x))=x+1$

$\Rightarrow 2[f(x)]=x+1$

$\Rightarrow 2(2x)=x+1$

$\Rightarrow 4x=x+1$

$\Rightarrow x=\dfrac {1}{3}$

Find the value of

$g(f(2))$ , if

$f(x) = e^{x}$ and

$g(x) = \dfrac {x}{2}$

0%
$2.7$
0%
$3.7$
0%
$4.2$
0%
$5.4$
0%
$6.1$

Explanation

Given,

$f(x)=e^x, g(x)=\dfrac {x}{2}$

$\Rightarrow f(2) = { e }^{ 2 }=7.389$

$\Rightarrow g(f(2)) = g(7.389) =\dfrac { 7.389}{2} = 3.7$

$h(x)={x}^{3}+x$ and

$g(x)=2x+3$ , then calculate

$g(h(2))$ .

0%
$7$
0%
$10$
0%
$17$
0%
$19$
0%
$23$

Explanation

Given,

$h(x)=x^3+x$

$\therefore h(2)=2^3+2$

$\Rightarrow h(2)=8+2=10$

Also given,

$g(x)=2x+3$

$\Rightarrow g(h(2))=g(10)$

$\because h(2)=10$

Then the value of

$g(10)=2\times 10+3$

$\Rightarrow g(10)=20+3=23$

$f(x) = x^{2} - 10$ and

$g(x) = 4x + 3$ , calculate the value of

$f(g(2))$ .

0%
$-24$
0%
$-21$
0%
$12$
0%
$27$
0%
$111$

Explanation

Given,

$f(x)=x^2-10$ and

$g(x)=4x+3$

$\therefore g(2) = 4 \times 2 + 3 = 8+3 = 11$
Then,

$f(g(2)) = f(11) = { 11 }^{ 2 }-10=121-10=111$

$f(x) =$

$x^2$ and

$g(x) = 2x$ , calculate the value of

$f(g(-3))-g(f(-3))$ .

0%
54
0%
18
0%
0
0%
-18
0%
-54

Explanation

Given

$f(x)=x^2$ and

$g(x)=2x$ .

Then

$f(-3)={(-3)}^2=9$ and

$g(-3)=2(-3)=-6$

So,

$f(g(-3))={[g(-3)]^2}={[-6]}^2=36$

and

$g(f(-3))=2*f(-3)=2*9=18$

The value of

$f(g(-3))-g(f(-3))$ is

$36-18=18$

The above figure shows the graph of the function

$f(x)$ , the value of

$f(f(3))$ is:

0%
$-4$
0%
$-2$
0%
$0$
0%
$1$
0%
$3$

Explanation

From the graph shown, we need to find

$f(f(3))$

The value of the function at

$x = 3$ is given by

$f(3) = -2$

Next,

$f(f(3)) = f(-2) = 0$

Find the correct expression for

$\displaystyle f\left( g\left( x \right) \right)$ if

$\displaystyle f(x)=4x+1$ and

$\displaystyle g\left( x \right) ={ x }^{ 2 }-2$

0%
$\displaystyle -{ x }^{ 2 }+4x+1$
0%
$\displaystyle { x }^{ 2 }+4x-1$
0%
$\displaystyle 4{ x }^{ 2 }-7$
0%
$\displaystyle 4{ x }^{ 2 }-1$
0%
$\displaystyle 16{ x }^{ 2 }+8x-1$

Explanation

$f(x)=4x+1$ and

$g(x)=x^2-2$

$\Rightarrow f(g(x))=f(x^2-2)$

$\Rightarrow (4(x^2-2)+1$

$\Rightarrow 4x^2-8+1$

$\Rightarrow 4x^2-7$

CBSE Questions for Class 12 Commerce Maths Relations And Functions Quiz 5 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Relations And Functions Quiz 5 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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