If $$\displaystyle f(x)= \frac{3x+2}{5x-3}$$ then

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

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

$$\displaystyle fo(f(x))= -x $$

$$\displaystyle f^{-1}(x)= -\frac{1}{19}f(x)$$

If $$f(x) = \sqrt{| x-1|}$$ and $$g(x) = \sin x$$, then $$(fog) (x)$$ equals

$$\left|\sin x + \cos x\right|$$

MCQ Questions for Class 12 Commerce Maths Relations And Functions Quiz 4

$f:\left( 3,6 \right) \rightarrow \left( 1,3 \right)$ is a function defined by

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

$($ where

$\left[ . \right]$ denotes the greatest integer function

$),$ then

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

0%
$x-1$
0%
$x+1$
0%
$x$
0%
none of these

Explanation

Given that

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

where

$\displaystyle f:\left( 3,6 \right) \rightarrow \left( 1,3 \right)$

Now, inside the given domain, we will always have

$\displaystyle \left[ \frac { x }{ 3 } \right] =1$

$\displaystyle \therefore f\left( x \right) =x-1$ throughout its domain

$\displaystyle \Rightarrow y=x-1$

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

$\displaystyle \therefore { f }^{ -1 }\left( x \right) =x+1$

Let

$\displaystyle g(x)=1+x-[x]$ and

$\displaystyle f(x)=\left\{\begin{matrix}{-1}\quad {x< 0} \\ {0} \quad {x=0}\\{1} \quad {x> 0} \end{matrix}\right.$ Then for all

$\displaystyle x, f\left \{ g\left ( x \right ) \right \}$ is equal to

0%
$x$
0%
$1$
0%
$\displaystyle f(x)$
0%
$\displaystyle g(x)$

Explanation

Let

$g(x) = 1+x-[x] = 1+\{x\}> 0$

since

$\{x\}\in [0,1) \forall x\in R$

Hence

$f\{g(x)\} = 1$

The inverse of the function

$\displaystyle f(x) = \log_{2}(x+\sqrt{x^{2}+1})$ is

0%
$\displaystyle 2^{x}+2^{-x}$
0%
$\displaystyle \frac{2^{x}+2^{-x}}{2}$
0%
$\displaystyle \frac{2^{-x}-2^{x}}{2}$
0%
$\displaystyle \frac{2^{x}-2^{-x}}{2}$

Explanation

We have to find the inverse of

$f(x)$ which means

$f$ is one one and onto function
Let

$\displaystyle f\left ( x \right )= y= \log _{2}\left ( x+\sqrt{x^{2}+1} \right )$

$\displaystyle \therefore 2^{y}= x+\sqrt{x^{2}+1}$ ...(i)

$\displaystyle 2^{-y}= \frac{1}{x+\sqrt{x^{2}+1}}= \frac{\sqrt{x^{2}+1}-x}{1}$ ...(ii)

By subtracting (i) and (ii) we have

$\therefore 2^{y}-2^{-y}= 2x$
or

$\displaystyle x= \frac{1}{2}\left ( 2^{y}-2^{-y} \right )$

$\displaystyle \therefore f^{-1}\left ( y \right )= \frac{1}{2}\left ( 2^{y}-2^{-y} \right )$
Replace

$\displaystyle y\rightarrow x$

$\displaystyle \therefore f^{-1}\left ( x \right )= \frac{1}{2}\left ( 2^{x}-2^{-x} \right )$

$f : \{1,2,3,...\} \rightarrow \{0, \pm 1, \pm 2,...\}$ is defined by

$\displaystyle y=f(x)=\begin{cases} \displaystyle \frac { x }{ 2 } \quad \quad \text{ if x is even } \\ -\displaystyle \frac { (x-1) }{ 2 } \quad ,\text{ if x is odd } \end{cases}$ , then

$f^{-1}(100)$ is

0%
Function is not invertible as it is not onto
0%
$199$
0%
$201$
0%
$200$

Explanation

Since

$100$ is even

$f^{-1}(x)=2x$

Hence

$f^{-1}(100)=200$

$\displaystyle f(y)=\frac{y}{\sqrt{1-y^2}}$ ;

$\displaystyle g(y)=\frac{y}{\sqrt{1+y^2}}$ then

$(fog)y$ is equal to

0%
$\displaystyle \frac{y}{\sqrt{1-y^2}}$
0%
$\displaystyle \frac{y}{\sqrt{1+y^2}}$
0%
$y$
0%
$2f(x)$

Explanation

Given

$\displaystyle f(y)=\frac{y}{\sqrt{1-y^2}}$ and

$\displaystyle g(y)=\frac{y}{\sqrt{1+y^2}}$

$\therefore \displaystyle (fog)y = f(g(y))=\dfrac{\dfrac{y}{\sqrt{1+y^2}}}{\sqrt{1-\dfrac{y^2}{1+y^2}}}=y$

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

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

$\displaystyle {g}'(3)$

0%
equals $1$
0%
equals $0$
0%
equals $3$
0%
equals $4$

Explanation

Simplifying,

$f(x)$ we get

$f(x)=2x$

Hence

$f(f(x))=2(f(x))$

$=2(2x)$

$=4x$

Hence

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

$=4x$

Thus

$g'(x)=4$

Hence

$g'(3)=4$ .

$\displaystyle f\left ( x \right )=x+\tan x$ and

$\displaystyle g^{-1}=f$ then

$\displaystyle g{}'\left ( x \right )$ equals

0%
$\displaystyle \frac{1}{2+\left [ g\left ( x \right )+x \right ]^{2}}$
0%
$\displaystyle \frac{1}{1+\left [ g\left ( x \right )-x \right ]^{2}}$
0%
$\displaystyle \frac{1}{2+\left [ g\left ( x \right )-x \right ]^{2}}$
0%
$\displaystyle \frac{1}{2-\left [ g\left ( x \right )-x \right ]^{2}}$

Explanation

$f\left( x \right) =x+\tan { x } \dashrightarrow \left( i \right) \\ { g }^{ -1 }(x)=f(x)\dashrightarrow \left( ii \right) \\ f\left( g(x) \right) =g(x)+\tan { g(x) } \quad \left( from\quad equation\quad \left( i \right) \quad \right) \\ { g }^{ -1 }\left( g(x) \right) =g(x)+\tan { g(x) } \left( from\quad equation\quad \left( ii \right) \quad \right) \\ x=g(x)+\tan { g(x) } \quad \dashrightarrow \left( iii \right) \quad \left( { \because \quad g }^{ -1 }\left( g(x) \right) =x \right) \\$
Differentiating the above equation w.r.t x,

$\\ 1=g^{ ' }\left( x \right) +(\sec ^{ 2 }{ g(x) } )g^{ ' }\left( x \right) \\ g^{ ' }\left( x \right) =\dfrac { 1 }{ 1+\sec ^{ 2 }{ g(x) } } =\dfrac { 1 }{ 2+\tan ^{ 2 }{ g\left( x \right) } } \quad \left( \because \sec ^{ 2 }{ g(x) } =1+\tan ^{ 2 }{ g\left( x \right) } \quad \right) \\ from\quad eqn\quad \left( iii \right) ,\quad \tan { g(x) } =x-g\left( x \right) \\ \therefore g^{ ' }\left( x \right) =\dfrac { 1 }{ 2+{ (x-g\left( x \right) ) }^{ 2 } }$

Let

$f:[4,\infty )\rightarrow [4,\infty )$ be a function defined by

$f\left( x \right)={ 5 }^{ x\left( x-4 \right) }$ , then

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

0%
$2-\sqrt { 4+\log _{ 5 }{ x } }$
0%
$2+\sqrt { 4+\log _{ 5 }{ x } }$
0%
$\displaystyle { \left( \frac { 1 }{ 5 } \right) }^{ x\left( x-4 \right) }$
0%
None of these

Explanation

Let

$y={ 5 }^{ x\left( x-4 \right) }\Rightarrow x\left( x-4 \right) =\log _{ 5 }{ y } \Rightarrow { x }^{ 2 }-4x-\log _{ 5 }{ y } =0$

$\displaystyle \Rightarrow x=\frac { 4\pm \sqrt { 16+4\log _{ 5 }{ y } } }{ 2 } =\left( 2\pm \sqrt { 4+\log _{ 5 }{ y } } \right)$

But

$x\ge 4$ , so

$x=\left( 2+\sqrt { 4+\log _{ 5 }{ y } } \right)$

$\therefore f^{ -1 }\left( y \right) =2+\sqrt { 4+\log _{ 5 }{ y } }$

$\therefore f^{ -1 }\left( x \right) =2+\sqrt { 4+\log _{ 5 }{ x } }$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct

Explanation

For Onto function

codomain

$\displaystyle =\left [ 0, \frac{\pi}{2} \right )=$ Range

Which is possiable where

$\displaystyle x^{2}+x+a\geq 0$

Now using fact

$\displaystyle f(x)\geq 0 \Rightarrow D\leq 0$

$\displaystyle \left ( A=1=b \right )\ and \ C=a$

$\displaystyle \Rightarrow 1-4a\leq 0$

$\displaystyle \Rightarrow a\geq \frac{1}{4}$

$\displaystyle \therefore$ Assertion is false, Reason is true

Let

$\displaystyle f:N\rightarrow Y$ be a function defined as

$f(x)=4x+3$ where

$\displaystyle Y=\left \{ y \in N:y=4x+3 \right \}$ for some

$\displaystyle x\in N$ such that

$f$ is invertible then its inverse is

0%
$\displaystyle g\left ( y \right )=4+\frac{y+3}{4}$
0%
$\displaystyle g\left ( y \right )=\frac{2y-3}{4}$
0%
$\displaystyle g\left ( y \right )=\frac{3y+4}{3}$
0%
$\displaystyle g\left ( y \right )=\frac{y-3}{4}$

Explanation

Let

$\displaystyle y=f(x)\Rightarrow x= f^{-1}(y)$ ........(A)

$\displaystyle \therefore y=4x+3$

$\displaystyle \Rightarrow 4x=y-3 \Rightarrow x=\frac{y-3}{4}$

$\displaystyle \Rightarrow f^{-1}(y)=\frac{y-3}{4}$ .......(using A)

$\displaystyle \Rightarrow g(y)=\frac{y-3}{4}$

$\displaystyle f\left ( x \right )=\left\{\begin{matrix} x^{2} x \geq 0\\ x x < 0 \end{matrix}\right.$
then

$\displaystyle (f o f)(x)$ is given by

0%
$x^{2}$ for $x\geq 0$ and $x$ for $x< 0$
0%
$\displaystyle x^{4}$ for $\displaystyle x\geq 0$ and $x^{2}$ for $x< 0$
0%
$\displaystyle x^{4}$ for $\displaystyle x\geq 0$ and $-x^{2}$ for $x < 0$
0%
$\displaystyle x^{4}$ for $x\geq 0$ and $x$ for $x< 0$

Explanation

For

$x\ge0$ ,we have

$\displaystyle f \circ f\left( x \right)= {\left( {{x^2}} \right)^2}=\left( {{x^4}} \right)$
For

$x<0$ ,we have

$\displaystyle f \circ f\left( x \right)=x$

$\displaystyle f(x)= \frac{3x+2}{5x-3}$ then

0%
$\displaystyle f^{-1}(x)= -f(x)$
0%
$\displaystyle f^{-1}(x)= f(x)$
0%
$\displaystyle fo(f(x))= -x$
0%
$\displaystyle f^{-1}(x)= -\frac{1}{19}f(x)$

Explanation

Let

$y=\displaystyle \frac{3x+2}{5x-3}$

$\Rightarrow \displaystyle x=\frac{3y+2}{5y-3}=f^{-1}(y)$

$\displaystyle \therefore f^{-1}(x)=\frac{3x+2}{5x-3}=f(x)$

Let

$\displaystyle f:R \rightarrow R$ be defined as

$\displaystyle f(x)= x^{2}+5x+9$ then

$\displaystyle f^{-1}(8)$ equals to

0%
$\displaystyle \left \{ \frac{-5+\sqrt{20}}{2},\frac{-5-\sqrt{21}}{2} \right \}$
0%
$\displaystyle \left \{ \frac{-5+\sqrt{21}}{2} ,\frac{-5-\sqrt{21}}{2}\right \}$
0%
$\displaystyle \left \{ \frac{5-\sqrt{21}}{2} ,\frac{21-\sqrt{5}}{2}\right \}$
0%
Does not exist.

Explanation

Let

$\displaystyle f^{-1}(8)= x,$ then

$\displaystyle f(x)= 8$

$\implies x^{2}+5x+9= 8$

$\implies x^{2}+5x+1= 0$

$\displaystyle \Rightarrow x= \frac{-5\pm \sqrt{25-4}}{2}$

$\displaystyle x= \frac{-5+\sqrt{21}}{2},\frac{-5-\sqrt{21}}{2}\in R$

Hence

$\displaystyle f^{-1}(8)= \left \{ \frac{-5+\sqrt{21}}{2} ,\frac{-5-\sqrt{21}}{2}\right \}$

Which of the following functions have inverse defined on the ranges

0%
$\displaystyle f\left ( x \right )=x^{2}, x$ $\displaystyle \in$ R
0%
$\displaystyle f\left ( x \right )=x^{3},x$ $\displaystyle \in$ R
0%
$\displaystyle f\left ( x \right )=e^{x},x$ $\displaystyle \in$ R
0%
$\displaystyle f(x)=\sin x,$ $\displaystyle 0< x< 2\pi$

Explanation

We know that only one-one onto functions are invertible.

In (A) it is not one-one and hence not invertible.

In (B) and (C)both are one-one onto and both are invertible.

In (D) Since

$\displaystyle f\left ( \frac{\pi}{4} \right )=f\left ( \frac{3\pi }{4} \right )=\frac{1}{\sqrt{2}}$

$\displaystyle \therefore$ (D) is not invertible.

Let f(x)=tan x, x

$\displaystyle \epsilon \left [ -\frac{\pi }{2},\frac{\pi }{2} \right ]$ and

$\displaystyle g\left (x \right )=\sqrt{1-x^{2}}$ Determine

$g o f(1)$ .

0%
1
0%
0
0%
-1
0%
not defined

Find

$\displaystyle \phi \left [ \Psi \left ( x \right ) \right ]$ and

$\displaystyle \Psi \left [ \phi \left ( x \right ) \right ]$ if

$\displaystyle \phi \left ( x \right )=x^{2}+1$ and

$\displaystyle \Psi \left ( x \right )=3^{x}.$

0%
$\displaystyle \Psi \left [ \phi \left ( x \right ) \right ]=3^{x^{2}+1}.$
0%
$\displaystyle \phi \left [ \Psi \left [ x \right ] \right ]=3^{2x}+1$
0%
$\displaystyle \Psi \left [ \phi \left ( x \right ) \right ]=3^{x^{3}+1}.$
0%
$\displaystyle \phi \left [ \Psi \left [ x \right ] \right ]=3^{x}+1$

Explanation

$\displaystyle \phi \left [ \Psi \left [ x \right ] \right ]=\phi \left ( 3^{x} \right )=\left ( 3^{x} \right )^{2}+1=3^{2x}+1$

and

$\displaystyle \Psi \left [ \phi \left ( x \right ) \right ]=\Psi \left ( x^{2} +1\right )=3^{x^{2}+1}.$

The inverse of the function

$\displaystyle f(x)=(1-(x-5)^{3})^{1/5}$ is

0%
$5-(1-x^{5})^{1/3}$
0%
$5+(1-x^{5})^{1/3}$
0%
$5+(1+x^{5})^{1/3}$
0%
$5-(1+x^{5})^{1/3}$

Explanation

Let

$y=(1-(x-5)^{3})^{\frac{1}{5}}$

$y^{5}=1-(x-5)^{3}$

$(x-5)^{3}=1-y^{5}$

$x-5=(1-y^{5})^{\frac{1}{3}}$

$x=5+(1-y^{5})^{\frac{1}{3}}$

Replacing

$x$ with

$y$ gives us

$f^{-1}(x)=5+(1-x^{5})^{\frac{1}{3}}$

$\displaystyle f\left ( x \right )=\frac{ax+b}{cx+d}$ and

$\displaystyle \left ( fof \right )x=x,$ then d=?

0%
$a$
0%
$-a$
0%
$b$
0%
$-b$

Explanation

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

$\displaystyle f\left [ \frac{ax+b}{cx+d} \right ]=x$ or

$\displaystyle\frac{a\left [ \frac{ax+b}{cx+d}\right ]+b}{c\left [\frac{ax+b}{cx+d} \right ]+d}=x$

$\displaystyle \therefore \frac{x\left ( a^{2}+bc \right )+b\left ( a+d \right )}{cx\left ( a+d \right )+\left ( bc+d^{2} \right )}=x$ Clearly if

$\displaystyle a+d=0$ or

$\displaystyle d=-a$ then in that case

$\displaystyle L.H.S.=\frac{x\left ( a^{2}+bc \right )+0}{0+\left ( bc+a^{2} \right )}=x$

$\displaystyle \because d=-a$

The inverse of the function

$\log_{e}x$ is

0%
$10^{x}$ .
0%
$e ^{x}$
0%
$10^{e}$ .
0%
$x^{e}$ .

Explanation

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

$\Rightarrow x=f\left( y \right) =\log _{ e }{ y }$

$\Rightarrow y={ e }^{ x }$

The total number of injective mappings from a set with

$m$ elements to a set with

$n$ elements,

$\displaystyle m\leq n,$ is

0%
$\displaystyle m^{n}$
0%
$\displaystyle n^{m}$
0%
$\displaystyle \frac{n!}{\left ( n-m \right )!}$
0%
$\displaystyle n!$

Explanation

Let

$A=\{a_1,a_2, a_3.....a_m\}$
and

$B=\{b_1,b_2, b_3.....b_n\}$ where

$m \le n$
Given

$f:A\rightarrow B$ be an injective mapping.
So, for

$a_1 \in A$ , there are

$n$ possible choices for

$f(a_1)\in B$ .
For

$a_2 \in A$ , there are

$(n-1)$ possible choices for

$f(a_2)\in B$ .
Similarly for

$a_m \in A$ , there are

$(n-m-1)$ choices for

$f(a_m)\in B$
So, there are

$n(n-1)(n-2).....(n-m-1)=\dfrac{n!}{(n-m)!}$ injective mapping from

$A$ to

$B.$

$\displaystyle A= \left \{ a,b,c,d \right \}, B= \left \{ 1,2,3 \right \}$ find whether or not the following sets of ordered pairs are relations from

$A$ to

$B$ or not.

$\displaystyle R_{1}= \left \{ \left ( a,1 \right ), \left ( a,3 \right ) \right \}$

$\displaystyle R_{2}= \left \{ \left ( a,1 \right ), \left ( c,2 \right ), \left ( d,1 \right ) \right \}$

$\displaystyle R_{3}= \left \{ \left ( a,1 \right ), \left ( b,2 \right ), \left ( 3,c \right ) \right \}.$

0%
$R_{1}$ $R_{2}$ are relations but $R_{3}$ is not a relation.
0%
$R_{1}$ $R_{3}$ are relations but $R_{2}$ is not a relation.
0%
All are relations
0%
none of these

Explanation

Given,

$A= \left \{ a,b,c,d \right \}, B= \left \{ 1,2,3 \right \}$ and

$R_{1}= \left \{ \left ( a,1 \right ), \left ( a,3 \right ) \right \}$

$R_{2}= \left \{ \left ( a,1 \right ), \left ( c,2 \right ), \left ( d,1 \right ) \right \}$

$R_{3}= \left \{ \left ( a,1 \right ), \left ( b,2 \right ), \left ( 3,c \right ) \right \}.$

We know that every relation from

$A$ to

$B$ will be a subset of

$A \times B$ .
Thus, both

$R_{1}$ and

$R_{2}$ are subsets of

$A\times B$ and hence are relations but

$R_{3}$ is not a relation because

$R_{3}$ is not a subset of

$A\times B$ because the element

$(3,c) \notin (A\times B).$ It belongs to

$B\times A.$

Are the following sets of ordered pairs functions? If so, examine whether the mapping is surjective or injective :

{(x, y): x is a person, y is the mother of x}

0%
injective (one- one ) and surjective (into)
0%
injective (one- one ) and not surjective (into)
0%
not injective (one- one ) and surjective (into)
0%
not injective (one- one ) and not surjective (into)

Given

$\displaystyle f\left ( x \right )=\log \left ( \frac{1+x}{1-x} \right )$ and

$\displaystyle g\left ( x \right )=\frac{3x+x^{3}}{1+3x^{2}}, fog (x)$ equals

0%
$-f(x)$
0%
$3f(x)$
0%
$\displaystyle \left [ f\left ( x \right ) \right ]^{3}$
0%
none of these

Explanation

Given

$f(x)\, =\, \log\, \left( \displaystyle \dfrac{1\, +\, x}{1\, -\, x} \right)$ and

$g(x)\, =\, \displaystyle \dfrac{3x\, +\, x^{3}}{1\, +\, 3x^{2}}$

$\therefore fog (x)\, =\, \log \left( \displaystyle \dfrac{1\, +\, \dfrac{3x\, +\, x^{3}}{1\, +\, 3x^{2}}}{1\, -\,\displaystyle \dfrac{3x\, +\, x^{3}}{1\, +\, 3x^{2}}} \right)$

$=\, \log\, \left(\displaystyle \dfrac{(1\, +\, x)^{3}}{(1\, -\, x)^{3}} \right)\, =\, 3\, \log\, \left(\displaystyle \dfrac{1\, +\, x}{1\, -\, x} \right)\, =\, 3\, f(x)$

Let

$R$ be a relation from a set

$A$ to a set

$B$ ,then

0%
$\displaystyle R=A\cup B$
0%
$\displaystyle R=A\cap B$
0%
$\displaystyle R\subseteq A\times B$
0%
$\displaystyle R\subseteq B\times A$

Explanation

$R$ is a relation from

$A$ to

$B$ then

$R$ is the subset of

$A \times B$ because it contains all the possible mappings from

$A$ to

$B$ .

$A=\{a,b,c,d\}, B=\{p,q,r,s\}$ , then which of the following are relations from

$A$ to

$B$ ?

0%
$\displaystyle R_{1}= \left \{ \left ( a,p \right ), \left ( b,r \right ), \left ( c,s \right ) \right \}$
0%
$\displaystyle R_{2}= \left \{ \left ( q,b \right ), \left ( c,s \right ), \left ( d,r \right ) \right \}$
0%
$\displaystyle R_{3}= \left \{ \left ( a,p \right ), \left ( a,q \right ), \left ( d,p \right ), \left ( c,r \right ), \left ( b,r \right ) \right \}$
0%
$\displaystyle R_{4}= \left \{ \left ( a,p \right ), \left ( q,a \right ), \left ( b,s \right ), \left ( s,b \right ) \right \}$

Explanation

From then definiation of Relation

Let

$R$ be

$a$ relation from

$A$ to

$B$

i.e.,

$R\subseteq A\times B,$ then

Domain of {

$a:a\in A,\left( a,b \right) \in R$ for some

$b\in B$ }

Range of

$R=$ {

$b:b\in B,\left( a,b \right) \in R$ for some

$a\in A$ }.

Therefore,

(A)

$\displaystyle R,=\left\{ \left( a,p \right) ,\left( b,s \right) ,\left( c,s \right) \right\}$ is a relation as

$\left\{ a,b,c \right\} \subseteq A$ and

$\left\{ p,r,s \right\} \subseteq B.$

(B)

${ R }_{ 2 }=\left\{ \left( q,b \right) ,\left( c,s \right) ,\left( d,r \right) \right\}$ is not a relation as

$q$ is not from set

$A.$

(C)

${ R }_{ 3 }=\left\{ \left( a,p \right) ,\left( a,q \right) ,\left( d,p \right) ,\left( c,r \right) ,\left( b,r \right) \right\}$ is a relation as

$\displaystyle \left\{ a,d,c,b \right\} \subseteq A$ and

$\displaystyle \left\{ p,q,r \right\} \subseteq B.$

(D)

${ R }_{ 4 }=\left\{ \left( a,p \right) ,\left( q,a \right) ,\left( b,s \right) ,\left( s,b \right) \right\}$ is not a relation as

$q$ and

$s$ does not belong to set

$A.$

$f:R\rightarrow R$ and

$g:R\rightarrow R$ are functions defined by

$f(x)=3x-1; g(x)=\sqrt{x+6}$ , then the value of

$(g\circ f^{-1})(2009)$ is

0%
$26$
0%
$29$
0%
$16$
0%
$15$

Explanation

Given

$f(x)=3x-1, g(x)=\sqrt{x+6}$

Let

$f(x)=y$

$\Rightarrow y=3x-1\Rightarrow y+1-3x\Rightarrow x=\cfrac{y+1}{3}$

Now,

$f^{-1}(y)=x=\cfrac{y+1}{3}$

$\Rightarrow f^{-1}(x)=\cfrac{x+1}{3}$ and

$g(x)=\sqrt {x+6}$

Consider,

$(g\circ f^{-1})(2009)=g[{f^{-1}(2009)}]$

$=g\left(\cfrac{2009+1}{3}\right)=g\left(\cfrac{2010}{3}\right)$

$=\sqrt{\cfrac{2010}{3}+6}=\sqrt{\cfrac{2028}{3}}=\sqrt{676}=26$

Hence, option

$A$ is correct.

$\displaystyle X= \left \{ 1,2,3,4,5 \right \}, Y= \left \{ 1,3,5,7,9 \right \}$ determine which of the following sets are mappings, relations or neither from A to B:

(i)

$\displaystyle F= \left \{ \left ( x,y \right ) \because y= x+2, x \in X, y \in Y \right \}$

0%
It is clearly a one-one onto mapping i.e. a bijection. It is also a relation.
0%
It is clearly a many-one onto mapping. It is also a relation.
0%
It is clearly a one-one but not onto mapping. It is also a relation.
0%
It is not a mapping but a relation

Explanation

Given def of F as

$\displaystyle F= \left \{ \left ( x,y \right ) \because y= x+2, x \epsilon X, y \epsilon Y \right \}$
where

$\displaystyle X= \left \{ 1,2,3,4,5 \right \}, Y= \left \{ 1,3,5,7,9 \right \}$

So, we get

$\left \{ \left ( 1,3 \right ), \left ( 2,4 \right ), \left ( 3,5 \right ), \left ( 4,6 \right ), \left ( 5,7 \right ) \right \}$
Since,

$\displaystyle y \in Y$ and in the above 4,6 do not belong to Y.

$\displaystyle \therefore$ we exclude the ordered pair (2,4) and (4,6)

$\displaystyle \therefore \displaystyle F=\left \{ \left ( 1,3 \right ), \left ( 3,5 \right ), \left ( 5,7 \right ) \right \}$
F is not a mapping because the elements 2,4

$\displaystyle \epsilon X$ do not have any image.

$\displaystyle F\subset X\times Y$ and hence F is a relation from X to Y.

Let

$f:[2,\infty)\rightarrow [1,\infty)$ defined by

$f(x)=2^{x^{4}-4x^{2}}$ and

$\displaystyle g:\left[ \frac{\pi}{2},\pi \right] \rightarrow A$ defined by

$\displaystyle g(x)=\frac {\sin x+4}{\sin x-2}$ be two invertible functions, then

$f^{-1}(x)$ is equal to

0%
$\sqrt{2+\sqrt{4-\log_{2}x}}$
0%
$\sqrt{2+\sqrt{4+\log_{2}x}}$
0%
$\sqrt{2-\sqrt{4+\log_{2}x}}$
0%
None of these

Explanation

$\because ff^{-1}(x)=x$

$2^{(f^{-1}(x))^{4}}-4^{(f^{-1}(x))^{2}}=x$

$\Rightarrow (f^{-1}(x))^{4}-4(f^{-1}(x))^{2}=\log _{2}x=0$

$\therefore (f^{-1}(x))^{2}=2+\sqrt{4+\log _{2}x}$

$\therefore$ Range of

$f^{-1}(x)$ is

$(2, \infty )$ .

$\therefore f^{-1}(x)=\sqrt{2+\sqrt{4+\log _{2}x}}$

$f_{0}(x)\, =\, \dfrac{x}{(x\, +\, 1)}$ and

$f_{n\, +\, 1}\, =\, f_{0}\circ f_{n}(x)$ for

$n = 0, 1, 2,\cdots$ then

$f_{n}(x)$ is

0%
$\displaystyle \frac{x}{(n\, +\, 1) x\, +\, 1}$
0%
$f_{0}(x)$
0%
$\displaystyle \frac{nx}{nx\, +\, 1}$
0%
$\displaystyle \frac{x}{nx\, +\, 1}$

Explanation

Given

$f_{0}\left(x\right)=\dfrac{x}{ \left(x\, +\, 1\right)}\cdots\cdots\left(1\right)$

Also given

$f_{n\, +\, 1}\, =\, f_{0}\, o\, f_{n}$ for

$(n = 0, 1, 2,\cdots)\ldots\left(2\right)$

Put

$n=0$ in eqn (2)

$f_1=f_{0} o f_0$

$f_1\left(x\right)=f_{0}[f_{0}\left(x\right)]$

$\Rightarrow f_{1}\left(x\right)=f_{0}\left(\dfrac{x}{x+1}\right)$

$=\dfrac{\dfrac{x}{x+1}}{\dfrac{x}{x+1}+1}$

$=\dfrac{x}{2x+1}$

$\Rightarrow f_1\left(x\right)=\dfrac{x}{2x+1}$

Put

$n=1$ in eqn

$(2)$

$f_2=f_{0} o f_1$

$f_2\left(x\right)=f_{0}[f_{1}\left(x\right)]$

$\Rightarrow f_{2}\left(x\right)=f_{0}\left(\dfrac{x}{2x+1}\right)$

$=\dfrac{\dfrac{x}{2x+1}}{\dfrac{x}{2x+1}+1}$

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

$\Rightarrow f_2\left(x\right)=\dfrac{x}{3x+1}$

Hence,

$f_n\left(x\right)=\dfrac{x}{\left(n+1\right)x+1}$

Let

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

$g(x)=f(f(x)-1)+f(5-f(x)),$ then

0%
$g(x)<0,\forall x\in R$
0%
$g(x)<0$ for some $x\in R$
0%
$g(x)\leq 0$ for some $x\in R$
0%
$g(x)\geq 0,\forall x\in R$

Explanation

Given,

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

Now,

$f(f(x) - 1) \\ = f(x^2 - 2x - 1 ) \\ = (x^2 - 2x - 1 )^2 - 2 (x^2 - 2x - 1 )$

And,

$f(5 - f(x) ) \\ = f(5 - x^2 + 2x ) \\ = (5 - x^2 + 2x )^2 - 2 (5 - x^2 + 2x )$

Hence,

$g(x) = f(f(x) - 1) + f (5 - f(x)) \\ \therefore g(x) = (x^2 - 2x - 1)^2 - 2 (x^2 - 2x -1 ) + (5 - x^2 + 2x )^2 - 2(5 - x^2 + 2x ) \\ \Rightarrow g(x) = (x^2 - 2x - 1)^2 + (5- x^2 + 2x)^2 - 2(x^2 - 2x - 1 + 5 - x^2 + 2x) \\ \therefore g(x) = (x^2 + 2x - 1 )^2 + ( 5 - 2x ^2 + 2 x^2)^2 - 8$

Since the first two terms are in square,

$\therefore$ it can not be negative and if x = 9 then we also get positive value.

$\therefore g(x) \geq 0 \, \, \forall x \in R$

$\therefore$ option D is correct

$f(x)=\begin{cases} 2x+3\quad \quad x\le 1 \\ a^{ 2 }x+1\quad x>1 \end{cases}$ , then the values of

$a$ for which

$f(x)$ is injective.

0%
$-3$
0%
$1$
0%
$0$
0%
none of these

Explanation

A function is called injective (one-one) if it is monontonic.
Clearly for

$x<1$ f is increasing and it's maximum values is

$2(1)+3=5$
Hence for f to be monotonic

$(x>1)$ ,

$f(1) \geq 5$

$\Rightarrow a^2(1)+1\geq 5 \Rightarrow a^2\geq 2 \Rightarrow a\in R-(-2,2)$

Which of the functions defined below are NOT one-one function(s)

0%
$f(x)\, =\, 5(x^{2}\, +\, 4),\, (x\, \in\, R)$
0%
$g(x)\, =\, 2x\, +\, \dfrac1x$
0%
$h(x)\, =\, ln(x^{2}\, +\, x\, +\, 1)\,, (x\, \in\, R)$
0%
$f(x)\, =\, e^{-x}$

$g(x)=1+\sqrt { x }$ and

$f(g(x))=3+2\sqrt { x } +x$ , then

$f(x)=$

0%
$1+2{ x }^{ 2 }$
0%
$2+{ x }^{ 2 }$
0%
$1+x$
0%
$2+x$

Explanation

We have

$g\left( x \right)=1+\sqrt { x }$ and

$f(g(x))=3+2\sqrt { x } +x$ ...(1)

Also,

$f\left( g\left( x \right) \right) =f\left( 1+\sqrt { x } \right)$ ...(2)

From (1) and (2), we get

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

Let

$1+\sqrt { x } =y\Rightarrow x={ \left( y-1 \right) }^{ 2 }$

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

Let

$X=\left\{ 1,2,3,4 \right\}$ and

$Y=\left\{ 1,2,3,4 \right\}$ . Which of the following is a relation from

$X$ to

$Y$ .

0%
${R}_{1}=\left\{ (x,y)| y=2+x, x\in X, y\in Y \right\}$
0%
${R}_{2}=\left\{ (1,1),(2,1),(3,3),(4,3),(5,5) \right\}$
0%
${R}_{3}=\left\{ (1,1),(1,3),(3,5),(3,7),(5,7) \right\}$
0%
${R}_{4}=\left\{ (1,3),(2,5),(2,4),(7,9) \right\}$

Explanation

Given

$X=\left\{ 1,2,3,4 \right\}$ and

$Y=\left\{ 1,2,3,4 \right\}$

$X\times Y=\{ (1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)$

$(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(4,4)\}$

Now,

$R_1=\{(1,3),(2,2)\}$

Since,

$R_1\subset X\times Y$

Hence,

$R_1$ is a relation from

$X$ to

$Y$

${R}_{2}=\left\{ (1,1),(2,1),(3,3),(4,3),(5,5) \right\}$

Since,

$(5,5) \in R_2$ but

$\notin X\times Y$

So,

$R_2$ is not a relation from

$X$ to

$Y$

${R}_{3}=\left\{ (1,1),(1,3),(3,5),(3,7),(5,7) \right\}$

Since,

$(3,5),(3,7),(5,7) \in R_3$ but

$\notin X\times Y$

So,

$R_3$ is not a relation from

$X$ to

$Y$

${R}_{4}=\left\{ (1,3),(2,5),(2,4),(7,9) \right\}$

Since,

$(2,5),(7,9) \in R_4$ but

$\notin X\times Y$

So,

$R_4$ is not a relation from

$X$ to

$Y$

Find inverse

$f(x)=\log_{e}(x+\sqrt{x^{2}+1})$

0%
$\sinh(x)$
0%
$\cosh(x)$
0%
$\tanh(x)$
0%
$\coth(x)$

Explanation

$y=\log(x+\sqrt{x^{2}+1})$

$\Rightarrow e^{y}=x+\sqrt{x^{2}+1}$

$\Rightarrow e^{y}-x=\sqrt{x^{2}+1}$

$\Rightarrow x^{2}-2xe^{y}+e^{2y}=x^{2}+1$

$\Rightarrow e^{2y}=1+2xe^{y}$

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

$\Rightarrow x=\dfrac{e^{y}-e^{-y}}{2}$
Replacing x with y

$f^{-1}(x)=\dfrac{e^{x}-e^{-x}}{2}$

$=\sinh(x)$

Let f : {x,y,z}

$\rightarrow$ {a,b,c} be a one-one function. It is known that only one of the following statment is true, and only one such function exists :

find the function f (as ordered pair).(i) f(x)

$\neq$ b
(i) f(y) = b

(ii) f(z)

$\neq$ a

0%
{(x,b), (y,a), (z,c)}
0%
{(x,a), (y,b), (z,c)}
0%
{(x,b), (y,c), (z,a)}
0%
{(x,c), (y,a), (z,b)}

Explanation

When (i) is true, then

$f(x) \neq b, f(y) \neq b , f(z) = a,b,c$

$\Rightarrow$ Two ordered pair function is possible

$f(x) = a, f(y) = c, f(z) = b$ or

$f(x) = c, f(y) = a, f(z) = b$

But given

$f$ is one-one and only one such function is possible. Hence (i) can't be true.

When (ii) is true, then

$f(y) = b, f(z) =c , f(x) = a\Rightarrow f(x) = a, f(y) = b, f(z) = c$

Clearly if (ii) is true then it is satisfying every condition.

Hence ordered pair of

$f$ is

$\{(x,a), (y,b), (z,c)\}$

Suppose f and g both are linear function with

$\displaystyle f(x)=-2x+1$ and

$\displaystyle f \left ( g\left ( x \right ) \right )=6x-7$ then slope of line

$y=g(x)$ is

0%
$3$
0%
$-3$
0%
$6$
0%
$-2$

Explanation

Given,

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

$\therefore f(g(x))=-2g(x)+1=6x-7$

$\Rightarrow g(x)=-3x+4$

$\Rightarrow g'(x)=-3$

Hence slope of

$g(x)$ is

$-3$

from the given statement

$N$ denotes the natural number and

$W$ denotes the whole number, so which statement in the following is correct

0%
N=W
0%
N $\subset$ W
0%
W $\subset$ N
0%
N $\cong$ W

Explanation

Natural numbers are naturally occurring counts of an object

$1, 2, 3, 4,…$ forming an infinite list in which the first number is

$1$ and every next number is equal to the preceding number

$+1.$

Whole number include the set of all Natural numbers and also

$0.$

Let

$N$ and

$W$ elements of natural and whole numbers.

$N=\{1,2,3,4,....\infty\}$

$W=\{0,1,2,3,4,...\infty\}$

So, here we can say

$N$ is subset of

$W.$

$\therefore$

$N\subset W$ is correct statement.

$f(x)=\begin{cases} x+1,\quad \quad if\quad x\, \leq \, 1 \\ 5-x^{ 2 }\quad \quad if\quad x>1 \end{cases},g(x)=\begin{cases} x\quad \quad if\quad x\leq 1 \\ 2-x\quad if\quad x>1 \end{cases}$

and

$x\, \in\, (1, 2)$ , then

$g(f(x))$ is equal to

0%
$x^{2}\, +\, 3$
0%
$x^{2}\, -\, 3$
0%
$5\, -\, x^{2}$
0%
$1 - x$

Explanation

$x\, \epsilon\, (1, 2)$

Hence for x >1,

$f(x) = 5-x^2$ and

$f(x) \epsilon (1,4)$

$f(x) >1$

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

$g(x) = 2x + 1$ and

$h(x) = 4x^{2} + 4x + 7$ , find a function

$f$ such that

$f o g = h$

0%
$f(x) = x^{3} - 6$
0%
$f(x) = x^{2} + 6$
0%
$f(x) = x^{2} - 6$
0%
$f(x) = (2x+1)^2 + 6$

Explanation

$f(g(x)) = h = 4x^2+4x+7=(2x+1)^2+6=(g(x))^2+6$

$\Rightarrow f(x) =x^2+6$

Let

$X = \left\{1,2,3,4\right\}$ and

$Y = \left\{1,3,5,7,9\right\}$ . Which of the following is relations from

$X$ to

$Y$

0%
$R_1 = \left\{(x,y) | y = 2x+1, x \in X, y \in Y\right\}$
0%
$R_2 = \left\{(1,1),(2,1),(3,3),(4,3),(5,5)\right\}$
0%
$R_3 = \left\{(1,1),(1,3),(3,5),(3,7),(5,7)\right\}$
0%
$R_4 = \left\{(1,3),(2,5), (2,4), (7,9)\right\}$

Explanation

We have

$X = \left\{1,2,3,4\right\}$ and

$Y = \left\{1,3,5,7,9\right\}$ .

$X \times Y=\{(1,1),(1,3),(1,5),(1,7),(1,9), (2,1),(2,3),(2,5),$

$(2,7),(2,9), (3,1),(3,3),(3,5),(3,7),(3,9),(4,1),(4,3),(4,5),(4,7),(4,9)\}$

Let's take option A,

$R_1 = \left\{(x,y) | y = 2x+1, x \in X, y \in Y\right\}$

$R_{1}= \{(1,3),(2,5),(3,7), (4,9)\}$
Since,

$R_1 \subseteq X\times Y$
So,

$R_1$ is a relation from

$X$ to

$Y$

Clearly,

$R_2$ is not a relation from

$X$ to

$Y$ as

$(5,5) \notin X\times Y$
Similarly,

$R_3$ is not a relation from

$X$ to

$Y$ as

$(5,7) \notin X\times Y$
In the same manner,

$R_4$ is also not a relation from

$X$ to

$Y$ as

$(7,9) \notin X\times Y$

Which of the following are two distinct linear functions which map the interval

$[-1, 1]$ onto

$[0, 2]$

0%
$f(x) = 1 + x$ or $1 - x$
0%
$f(x) = 1 + 2x$ or $1 - x$
0%
$f(x) = 1 + x$ or $1 - 2x$
0%
$f(x) = 1 + x$ or $2 - x$

Explanation

Out of two linear functions one will be increasing and other will be decreasing.

Let

$f(x) = ax+b$

For increasing:

$f(-1)=0=> b-a=0\Rightarrow b=a$ and

$f(1) = 2\Rightarrow a+b=2\Rightarrow a=b=1\therefore f(x) = 1+x$

For decreasing:

$f(-1)=2=> b-a=2$ and

$f(1) = 0\Rightarrow a+b=2\Rightarrow b=1, a=-1\therefore f(x) = 1-x$

$f(x) =\ln {\displaystyle \frac { 1+x }{ 1-x } }$ and

$g(x)=\displaystyle \frac {3x+x^3}{1+3x^2}$ , then

$f[g(x)]$ equals.

0%
$f(x)$
0%
$[f(x)]^3$
0%
$3f(x)$
0%
${f(x)}^2$

Explanation

Given

$f(x)=\ln {\displaystyle \frac { 1+x }{ 1-x } }$ and

$g(x)=\displaystyle \frac {3x+x^3}{1+3x^2}$

then

$f[g(x)]=\ln {\displaystyle \frac { 1+\displaystyle\frac { 3x+x^{ 3 } }{ 1+3x^{ 2 } } }{ 1-\displaystyle\frac { 3x+x^{ 3 } }{ 1+3x^{ 2 } } } }$

$=\ln {\displaystyle \frac { 1+3x^{ 2 }+3x+x^{ 3 } }{ 1+3x^{ 2 }-3x-x^{ 3 } } }$

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

$=3\ln {\displaystyle \frac { 1+x }{ 1-x } }$

$\Rightarrow f[g(x)]=3f(x)$

Hence, option C.

Let

$f(x) = e^{3x}, g(x) = \log_ex, x > 0$ , then

$fog (x)$ is

0%
$3x$
0%
$x^3$
0%
$\log_{10}3x$
0%
$\log3x$

Explanation

Given

$f(x) = e^{3x}, g(x) = \log_ex, x > 0$

$fog(x)=f(\log_ex)=e^{3\log_ex}=x^3$

Hence, option B.

$f(x)\, >\, x;\, \forall\, x\, \epsilon\, R.$ The equation

$f (f(x)) -x = 0$ has

0%
Atleast one real root
0%
More than one real root
0%
No real root if f(x) is a polynomial & one real root if f(x) is not a polynomial
0%
No real root at all

Explanation

Given

$f(x) > x$

$\forall x \epsilon R$

Hence,

$f(x)-x$ has no real roots, because

$f(x)-x >0$

Since,

$x \epsilon R \implies f(x) \epsilon R$

$f(f(x))>x$

$\implies f(f(x))-x>0$

Hence it will have no real root.

$f : R \rightarrow R, f(x) = (x + 1)^2$ and

$g : R \rightarrow R, g(x) = x^2 + 1$ then

$(fog)(3)$ is equal to

0%
$121$
0%
$144$
0%
$112$
0%
$11$

Explanation

Given,

$f : R \rightarrow R, f(x) = (x + 1)^2$ and

$g : R \rightarrow R,g(x) = x^2 + 1$

$g(3)=3^2+1=10$

$f(10)=(10+1)^2=121$

$\therefore fog(3)=f\left(g(3)\right)=f(10)=121$

Hence, option A.

$f(x) = \sqrt{| x-1|}$ and

$g(x) = \sin x$ , then

$(fog) (x)$ equals

0%
$\sin \sqrt{| x-1|}$
0%
$\left|\sin\dfrac{x}{2} - \cos\dfrac{x}{2}\right|$
0%
$\left|\sin x + \cos x\right|$
0%
$\left|\sin\dfrac{x}{2} + \cos\dfrac{x}{2}\right|$

Explanation

$\displaystyle fog\left( x \right) =f\left( g\left( x \right) \right) =\sqrt { \left| \sin { x } -1 \right| }$

$=\sqrt { \left| 1-\cos { \left( \dfrac { \pi }{ 2 } -x \right) } \right| } =\sqrt { 2 } \left| \sin { \left( \dfrac { \pi }{ 4 } -\dfrac { x }{ 2 } \right) } \right| =\left| \sin { \dfrac { x }{ 2 } } -\cos { \dfrac { x }{ 2 } } \right|$

$f(x) = \log x$ ,

$g(x) = x^3$ , then

$f[g(a)] + f[g(b)]$ equals

0%
$f[g(a) + g(b)]$
0%
$3f(ab)$
0%
$g[f(ab)]$
0%
$g[f(a) + f(b)]$

Explanation

We have

$f(x) = \log x$

$g(x) = x^3$

$g(a) = a^3$

$f(g(a)) = \log (a^3)$

$f(g(a)) =3 \log a$

Similarly,

$f(g(b)) = 3 \log b$

$\Rightarrow f(g(a)) + f(g(b)) = 3 \log a + 3 \log b$

$\Rightarrow f(g(a)) + f(g(b)) = 3 \log ab$

$\Rightarrow f(g(a)) + f(g(b)) = 3 f(ab)$

$f(x) = x^3$ and

$g(x) = sin2x$ , then

0%
$g[f(1)] = 1$
0%
$f(g(\pi/12) = 1/8$
0%
$g{f(2)} = \sin 2$
0%
none of these

Explanation

Given,

$f(x)=x^3$ and

$g(x)=\sin 2x$

Clearly,

$f(1)=1 \Rightarrow g[f(1)]=\sin 2$

$g(\cfrac{\pi}{12}) =\dfrac{1} {2} \Rightarrow f[g(\dfrac{\pi}{12})]=\dfrac{1}{8}$

and,

$f(2)=8 \Rightarrow g[f(2)]= \sin 16$

So, the correct option is (B).

$f(x) = (a x^n)^{1/n},$ where

$\ n \in N$ , then

$f\{f(x)\}$ equals

0%
$0$
0%
$x$
0%
$x^n$
0%
none of these

Explanation

We have,

$f(x) = (a x^n)^{1/n}=a^{1/n}x=kx$ where

$k = a^{1/n}$

$\therefore f\{f(x)\} = kf(x) = k^2x =a^{2/n} x$

CBSE Questions for Class 12 Commerce Maths Relations And Functions Quiz 4 - 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 4 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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