If $$g(f(x) ) = |\sin x |$$ and $$f(g(x))=(\sin\sqrt x)^2$$ , then

$$f(x) = \sin^2x. g(x) =\sqrt x$$

$$f(x) = x^2, g(x) = \sin \sqrt x$$

MCQ Questions for Class 12 Commerce Maths Relations And Functions Quiz 7

$f(x)=ax+b$ and

$g(x)=cx+d$ , then

$f\left( g(x) \right) =g\left( f(x) \right) \Leftrightarrow$

0%
$f(a)=g(c)$
0%
$f(b)=g(b)$
0%
$f(d)=g(b)$
0%
$f(x)=g(a)$

Explanation

We have

$f(x)=ax+b,g(x)=cx+d$
Therefore,

$f\left\{ g(x) \right\} =g\left\{ f(x) \right\} \Leftrightarrow f(cx+d)=g(ax+b)$

$\Leftrightarrow a(cx+d)+b=c(ax+b)+d$

$\Leftrightarrow ad+b=cb+d\Leftrightarrow f(d)=g(b)$

The inverse of the function

$f(x) = \log (x^{2} + 3x + 1), x\epsilon [1, 3]$ , assuming it to be an onto function, is

0%
$\dfrac {-3 + \sqrt {5 + 4e^{x}}}{2}$
0%
$\dfrac {-3 \pm \sqrt {5 + 4e^{x}}}{2}$
0%
$\dfrac {-3 - \sqrt {5 + 4e^{x}}}{2}$
0%
None of the above

Explanation

Given,

$f(x) = \log (x^{2} + 3x + 1)$
Therefore,

$f'(x) = \dfrac {2x + 3}{(x^{2} + 3x + 1)} > 0\forall x\epsilon [1, 3]$
which is a strictly increasing function. Thus,

$f(x)$ is injective, given that

$f(x)$ is onto. Hence, the given function

$f(x)$ is invertible.
Now,

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

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

$\Rightarrow \left \{f^{-1}(x)\right \}^{2} + 3\left \{f^{-1}(x)\right \} + 1 - e^{x} = 0$
Thus

$f^{-1}(x) = \dfrac {-3\pm \sqrt {9 - 4.1 (1 - e^{x})}}{2}$

$= \dfrac {-3\pm \sqrt {5 + 4e^{x}}}{2}$

$= f^{-1}(x) = \dfrac {-3\pm \sqrt {5 + 4e^{x}}}{2} [\because f^{-1}(x) \epsilon [1, 3]]$
Hence,

$f^{-1}(x) = \dfrac {-3 + \sqrt {5 + 4e^{x}}}{2}$

Let

$f(x)={ x }^{ 3 }-3x+1$ . The number of different real solutions of

$f(f(x))=0$

0%
$2$
0%
$4$
0%
$5$
0%
$7$

$f\left( x \right)$ and

$g\left( x \right)$ are two functions with

$g\left( x \right) =x-\dfrac { 1 }{ x }$ and

$f\circ g\left( x \right) ={ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } }$ , then

$f^{ ' }\left( x \right)$ is equal to

0%
$3{ x }^{ 2 }+3$
0%
${ x }^{ 2 }-\dfrac { 1 }{ { x }^{ 2 } }$
0%
$1+\dfrac { 1 }{ { x }^{ 2 } }$
0%
$3{ x }^{ 2 }+\dfrac { 3 }{ { x }^{ 4 } }$

Explanation

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

${ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } }$ using

${ \left( a-b \right) }^{ 3 }={ a }^{ 3 }-{ b }^{ 3 }-3ab\left( a-b \right)$ , we have

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

$\Rightarrow { x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } ={ \left( x-\dfrac { 1 }{ x } \right) }^{ 3 }+3\left( x-\dfrac { 1 }{ x } \right)$
We have,

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

$g\left( x \right) =x-\dfrac { 1 }{ x }$ , this yields

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

$x-\dfrac { 1 }{ x } =t$ , we get

$f\left( t \right) ={ t }^{ 3 }+3t$
Thus,

$f\left( x \right) ={ x }^{ 3 }+3x$
and

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

If $f:R\rightarrow R$ , $g:R\rightarrow R$ are defined by $f(x)=5x-3$ , $g(x)=x^{2}+3$ , then $(gof^{-1})(3)$ =

0%
$\dfrac{25}{3}$
0%
$\dfrac{111}{25}$
0%
$\dfrac{9}{25}$
0%
$\dfrac{25}{111}$

Explanation

$f:R\rightarrow R$

$g:R\longrightarrow R$

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

$g\left( x \right) ={ x }^{ 2 }+3$

$\left(gof^{-1}\right)\left(3\right)$

$f\left(x\right)={5}{x}-{3}$

Let

$f\left(x\right)=y$

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

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

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

$\Rightarrow f^{-1}\left(y\right)=\dfrac{y+3}{5}$

$f^{-1}\left(x\right)=\dfrac{x+3}{5}$ and

$g\left(x\right)=x^{2}+{3}$

$\left(gof^{-1}\right)\left(3\right)=g\left[f^{-1}\left(3\right)\right]=g\left(\dfrac{3+3}{5}\right)=g\left(\dfrac{6}{5}\right)$

$=\left[\left(\dfrac{6}{5}\right)^{2}+3\right]=\dfrac{36}{25}+{3}=\dfrac{111}{25}$

Hence,

$\dfrac{111}{25}$ is the correct answer.

The inverse of the function

$y = 5^{\ln\ x}$ is

0%
$x = y^{\frac {1}{\ln 5}}, y > 0$
0%
$x = y^{\ln 5}, y > 0$
0%
$x = y^{\frac {1}{\ln 5}}, y < 0$
0%
$x = 5\ln y, y > 0$

Explanation

Given that

$y = 5^{\ln x}$

$\ln y = \ln (5^{\ln x})$

$= (\ln x) (\ln 5)$

$\therefore \ln x = \dfrac{\ln y}{\ln 5}$

$x = e^{\frac{\ln y}{\ln 5} }$

$= (e^{\ln y})^{\frac{1}{\ln 5}}$

$= y^{\frac{1}{\ln 5}}$

Since the domain of the natural log function ln() is the set of positive real numbers, it is required here that

$y > 0$ .

The inverse function:

$x = y^{\frac{1}{\ln 5}}, \; \; y > 0$

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

$f^{-1}(4)$ is

0%
Equal to $1$
0%
Equal to $2$
0%
Equal to $1/3$
0%
Non existent

Explanation

$f\left( x \right) =2{ x }^{ 3 }+7x-5$

$x=2\left( { f^{ -1 }\left( x \right) }^{ 3 } \right) +7f^{ -1 }\left( x \right) -5$

$f^{ -1 }\left( 4 \right) =?$

$4=2\left( { f^{ -1 }\left( 4 \right) }^{ 3 } \right) +7f^{ -1 }\left( 4 \right) -5$

$2\left( { f^{ -1 }\left( 4 \right) }^{ 3 } \right) +7f^{ -1 }\left( 4 \right) -9=0$

Let

$f^{ -1 }\left( 4 \right) =t$

$2{ t }^{ 3 }+7t-9=0$

$2{ t }^{ 3 }-2+7t-7=0$

$2\left( { t }^{ 3 }-1 \right) +7\left( t-1 \right) =0$

$2\left( t-1 \right) \left( { t }^{ 2 }+t+1 \right) +7t-1=0$

$\left( t-1 \right) \left( 2{ t }^{ 2 }+2t+2+7 \right) =0$

$t=1\quad 2{ t }^{ 2 }+2t+9=0$

So,

$f^{ -1 }\left( 4 \right) =1$

Answer A

$f,g:R\rightarrow R$ are functions such that

$f(x)=3x-\sin { \left( \cfrac { \pi x }{ 2 } \right) } ,g(x)={ x }^{ 3 }+2x-\sin { \left( \cfrac { \pi x }{ 2 } \right) }$
The value of

$\cfrac { d }{ dx } { f }^{ -1 }{ \left( { g }^{ -1 }(x) \right) }_{ x=12 }$ is equal to

0%
$\cfrac { 2 }{ 30+x }$
0%
$\cfrac { 2 }{ 30-x }$
0%
$\cfrac { 2 }{ 3\left( 28-\pi \right) }$
0%
$\cfrac { 2 }{ 3\left( 28+\pi \right) }$

Explanation

Given that,

$g:R \to R$

$f\left( x \right) = 3x - \sin \left( {\dfrac{{\pi x}}{2}} \right)$

$g\left( x \right) = {x^3} + 2x - \sin \left( {\dfrac{{\pi x}}{2}} \right)$

$\therefore \dfrac{{d{f^{ - 1}}}}{{dx}}{\left( {g'\left( x \right)} \right)_{x = 12}} = \dfrac{2}{{30 + x}}$

Hence, the option

$(A)$ is correct.

$f: R\rightarrow R$ and

$g: R\rightarrow R$ are defined

$f(x) = x - [x]$ and

$g(x) = [x]\forall x\epsilon R, f(g(x))$ .

0%
$x$
0%
$0$
0%
$f(x)$
0%
$g(x)$

Explanation

We have,

$f(x)=x-[x]=\{x\}\dots (1)$

where

$\{x\}$ is the fractional part of

$x$

$g(x)=[x]$ , where

$[x]$ is the greatest integer part of

$x$ , then

The value of

$g(x)$ will always be an integer.

And

$\therefore f(g(x))=f([x])=0$

$\because$ the fractional part of

$g(x)$ i.e

$[x]$ is always

$0$ .

Let

$f : A \rightarrow B$ be a function defined as

$f(x) = \dfrac {x - 1}{x - 2}$ , where

$A = R - \left \{2\right \}$ and

$B = R - \left \{1\right \}$ . Then

$f$ is :

0%
invertible and $f^{-1}(y) = \dfrac {2y + 1}{y - 1}$
0%
invertible and $f^{-1}(y) = \dfrac {3y - 1}{y - 1}$
0%
not invertible
0%
invertible and $f^{-1}(y) = \dfrac {2y - 1}{y - 1}$

Explanation

Let

$y = f(x)$

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

$\Rightarrow yx - 2y = x - 1$

$\Rightarrow (y-1)x = 2y-1$

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

So on the given domain the function is invertible and its inverse can be computed as shown above.

So, option D is the correct answer.

$f(x)$ is a real valued function, then which of the following is one-one function?

0%
$f(x)=e^{|x|}$
0%
$f(x)=|e^x|$
0%
$f(x)=\sin x$
0%
$f(x)=|\sin x|$

Explanation

$(1)$

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

Since

$f(x)=f(-x)$

$f(x)$ is many one.

$(2)$

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

so,

$f(x)=e^x$

and as

$e^x$ is one-one

$f(x)$ is one-one.

$(3)$

$f(x)=\sin x$

Since

$\sin x$ is many one

$f(x)$ is also many-one.

$(4)$

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

Since

$\sin x$ is many one

$f(x)$ is also many-one.

$A =\{1, 2, 3\}$ and

$B = \{4, 5\}$ then the number of function

$f : A \rightarrow B$ which is not onto is ______

0%
$2$
0%
$6$
0%
$8$
0%
$4$

Explanation

Total number of function

$f:A\rightarrow B$ is the number of relations from

$A$ to

$B$ such that all the elements of

$A$ are in the first elements of function

$f$ .

Hence, total number of functions are

$=2\times 2\times 2 =2^3=8$

There are two functions for which the function is

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

$\{(1,5),(2,5),(3,5)\}$ .

Hence there are two non-onto functions.

$f\,: R \rightarrow R, g: R \rightarrow R\,$ are defined by

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

$(gof^{-1})(3) =$

0%
$\displaystyle \frac{25}{3}$
0%
$\displaystyle \frac{111}{25}$
0%
$\displaystyle \frac{9}{25}$
0%
$\displaystyle \frac{25}{111}$

Explanation

$f(x)=5x-3,$

$g(x)=x^2+3$

Let

$f(x)=y$

$\Rightarrow$

$5x-3=y$

$\Rightarrow$

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

Now,

$f(x)=y$

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

$f^{-1}(y)=\dfrac{y+3}{5}$ [ From ( 1 ) ]

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

Next,

$(gof^{-1})(3)$

$\Rightarrow$

$g[f^{-1}(3)]$

$\Rightarrow$

$g\left[\dfrac{3+3}{5}\right]$

$\Rightarrow$

$g\left[\dfrac{6}{5}\right]$

$\Rightarrow$

$\left(\dfrac{6}{5}\right)^2+3$ [ Since,

$g(x)=x^2+3$ ]

$\Rightarrow$

$\dfrac{36}{25}+3$

$\Rightarrow$

$\dfrac{36+75}{25}$

$\Rightarrow$

$\dfrac{111}{25}$

Let

$f:A \to b$ be a function defined by f(x) =

$\sqrt {1 - {x^2}}$

0%
f(x) is one-one if A =[0,1]
0%
f(x) is onto if B = [0,1]
0%
f(x) is one-one if A =[-1 , 0]
0%
f(x) is onto if B = [-1,1]

Explanation

$f:A\rightarrow B$

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

When

$x=0$

$f(0)=\sqrt { 1-0 }$

$=1$

$f(1)=\sqrt { 1-{ 1 }^{ 2 } }$

$=0$

$f(x)$ is one - one function when

$A=[0,1]$

$f:R\rightarrow R,f(x)=\begin{cases} 1\quad \quad x>0 \\ 0\quad \quad x=0 \\-1\quad x<0 \end{cases}$ and

$g:R\rightarrow R,g(x)=\left[ x \right]$ , then

$\left( f\circ g \right) \left( \pi \right)$ is:

0%
$\pi$
0%
$0$
0%
$1$
0%
$-1$

Explanation

Given that

$g(x)=[x]$

$\Rightarrow$

$g(\pi )=g(3.141....)$

i.e

$g(\pi )=3$

$(f\circ g)(x)=f(g(x))$

$\Rightarrow$

$(f\circ g)(\pi ))=f(g(\pi ))$

But

$g(\pi)=3$

$\Rightarrow f(3)=1$ ......

$[\because f(x)=1 \text{ for } x>0]$

The inverse of the function

$y=\large{\frac{e^x-e^{-x}}{e^x+e^{-x}}}$ is

0%
$\large{\frac{1}{2}}$ $\log \large{\frac{1+x}{1-x}}$
0%
$\large{\frac{1}{2}}$ $\log \large{\frac{2+x}{2-x}}$
0%
$\large{\frac{1}{2}}$ $\log \large{\frac{1-x}{1+x}}$
0%
$2 \log (1+x)$

Explanation

Given

$y=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}$

or,

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

or,

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

or,

$x=\dfrac{1}{2}\log \dfrac{1+y}{1-y}$

So the inverse is

$\dfrac{1}{2}\log \dfrac{1+x}{1-x}$ .

Let

$f\left( x \right) = {x^2}$ and

$g\left( x \right) = \sqrt x$ (where

$x > 0$ ),then

0%
$f\left( {g\left( x \right)} \right) = x$
0%
$g\left( {f\left( x \right)} \right) = x$
0%
The least value of $f\left( {g\left( x \right)} \right) + {1 \over {g\left( {f\left( x \right)} \right)}}$ is $2$
0%
The least value of $g\left( {f\left( x \right)} \right) + {1 \over {f\left( {g\left( x \right)} \right)}}$ is $2$

Explanation

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

$f(g(x))=(\sqrt{x})^{2}=x$

$g(f(x))=\sqrt{x^{2}}=x$

$f(g(x))+\dfrac{1}{g(f(x))}=g(f(x))+\dfrac{1}{f(g(x))}=x+\dfrac{1}{x}\ge 2$

Hence, all options are correct.

The solution of

$( 3 * 4) * 3$ , when

$*$ is a binary operation on Z such that:

$a * b = a + b$ , is.

0%
$10$
0%
$-10$
0%
$-\dfrac16$
0%
$-6$

Explanation

Given

$( 3 * 4) * 3$

$(3*4)*3=(3+4)+3$ (as

$(a \ast b)=(a+b)$ )

$=7+3$

$=10$

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

$\frac{1}{2}gof\left( x \right) = 2{x^2} + 5x + 2$ , then

$f\left( x \right)$ is

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

Explanation

$g(x)={ x }^{ 2 }+x-2\\ \cfrac { 1 }{ 2 } g(f(x))=2{ x }^{ 2 }+5x+2\\ \Rightarrow f^{ 2 }\left( x \right) +f(x)-2=4{ x }^{ 2 }-10x+4\\ \Rightarrow f^{ 2 }\left( x \right) +f(x)-(4{ x }^{ 2 }-10x+6)=0\\ f(x)=\cfrac { -1\pm \sqrt { 1+4(4{ x }^{ 2 }-10x+6) } }{ 2 } \\ f(x)=\cfrac { -1\pm \sqrt { 1+16{ x }^{ 2 }-40x+24 } }{ 2 } \\ f(x)=\cfrac { -1\pm (4x-5) }{ 2 } =(2x-3)$

Let f :

$R \to R$ and g :

$R \to R$ be two one-one and onto functions such that they are the mirror images of each other about the line y =If h(x) = f(x) + g(x), then h(0) equal to

0%
2
0%
4
0%
0
0%
1

Explanation

Given

$g(x)$ and

$f(x)$ are mirror images about

$y=2$

$\implies g(x)-2=2-f(x)\implies f(x)+g(x)=4\implies h(x)=4$

$h(0)=4$

$f: A\rightarrow A$ defined by

$f(x) =\dfrac{4x +3}{6x-4}$ where

$A= R-\dfrac{2}{3}$ . Find

$f^{-1}$

0%
$2x$
0%
$\dfrac {4x+3}{6x-4}$
0%
$\dfrac x2$
0%
None of these

Explanation

$f\left( x \right) =\cfrac { 4x+3 }{ 6x-4 } \\ let\quad y=\cfrac { 4x+3 }{ 6x-4 } \\ \Rightarrow 6xy-4y=4x+3\\ \Rightarrow 6xy-4x=3+4y\\ \Rightarrow x(6y-4)=3+4y\\ \Rightarrow x=\cfrac { 3+4y }{ 6y-4 } \\ Replace\quad x\quad by\quad y\quad \& \quad y\quad by\quad x.\\ y=\cfrac { 3+4x }{ 6x-4 } \\ f^{ ' }\left( x \right)=\cfrac { 4x+3 }{ 6x-4 }$

If the binary operation

$*$ is defined on a set of integers as

$a * b = a + 3 b ^{2}$ , then the value of

$2 * 3$ is

0%
$27$
0%
$29$
0%
$2$
0%
None of these

Explanation

$a\ast b =a+3b^{2}$

$\Rightarrow 2\ast 3=2+3(3)^{2}$

$=2+27$

$=29$

Let

$f$ ,

$g:R\rightarrow {R}$ be two functions defined as

$f\left( x \right) =\left| x \right| +x$ ,

$g\left( x \right) =\left| x \right| -x, \forall x\in R$ . Then, find

$fog(x)$

0%
$||x|-x|-|x|-x$
0%
$||x|-x|+|x|-x$
0%
$||x|-x|-|x|+x$
0%
None of thesse

Explanation

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

$g\left( x \right) = \left| x \right| - x$

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

$= f\left( {\left| x \right| - x} \right)$

$= \left| {\left| x \right| - x} \right| + \left| x \right| - x$

Consider set

$A={1,2,3,4}$ and set

$B={0,2,4,6,8}$ , then the number of one-one function from set

$A$ to set

$B$ is ?

0%
$5$
0%
$24$
0%
$120$
0%
None of these

Explanation

Number of one-one function from

$\underset{(m)}{A}$ to

$\underset{(n)}{B} = \left\{\begin{matrix} \, ^nP_m, & if \, n \ge m \\ 0, & if \ n < m \end{matrix}\right.$

$m = 4, \ n = 5$

One-one function

$=\, ^5P_4 = \dfrac{5!}{1!} = 120$

The function

$*$ on

$N$ as

$a * b = ( a - b)^{2}$ is a binary operator

0%
True
0%
False

Explanation

$\ast$ is relation of

$N: a\ast b=(a-b)^{2}$

Checking at

$a=b, a,b\in N,\Rightarrow a\ast b=(a-b)^{2}=0 \notin N$

$\Rightarrow \ast$ is not a binary opeation.

If the binary operation

$*$ is on set of integers

$Z$ is defined as

$a * b = a + 2b ^{2}$ , then the value of

$(8 * 3) * 2$

0%
$26$
0%
$22$
0%
$32$
0%
$34$

Explanation

$a\ast b=a+2b^{2}$

$\Rightarrow (8\ast 3)\ast 2=(8+2(3)^{2})\ast 2$

$=(26\ast 2)$

$=26+2(2)^{2}$

$=26+8$

$=34$

$f(x)=2x+5$ and

$g(x)=x^2+1$ be two real function , then value of

$fog$ at x=1

0%
$9$
0%
$6$
0%
$5$
0%
$4$

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

$f(g(x))=(\sin\sqrt x)^2$ , then

0%
$f(x) = \sin^2x. g(x) =\sqrt x$
0%
$f(x) = \sin x , g(x) =|x|$
0%
$f(x) = x^2, g(x) = \sin \sqrt x$
0%
f and g can not be determined

Explanation

$g(f(x))=|sinx|=\sqrt{(sinx)^2}\\f(g(x))=(sin\sqrt x )^2=sin^2\sqrt x\\\therefore f(x)=sin^2x \>and \>g(x)=\sqrt x$

Let

$f : R \rightarrow R$ be defined by

$f(x) = x^2 - 3x + 4$ for all

$x \epsilon R$ , then

$f^{-1}(2)$ is

0%
2
0%
1
0%
Not defined
0%
$\dfrac{1}{2}$

Explanation

$F:R\to R$

$f(x)=x^2-3x+4$

$f(1)=1-3+4$

$f(1)=2$

$\boxed {f^{-1}(2)=1}$

Let

$f(x+\dfrac{1}{x})=x^2+\dfrac{1}{x^2}(x\neq 0)$ , then

$f(x)=$

0%
$x^2$
0%
$x^2-1$
0%
$x^2-2$
0%
N.O.T

Explanation

We are given

$f(x+ \dfrac{1}{2})= x^{2}+ \dfrac{1}{x^{2}}$ (where

$x \neq 0$ )

$= x^{2}+ \dfrac{1}{x^{2}}+2-2$ .

$=x^{2}+ \dfrac{1}{x^{2}}+ 2 (x^{2}) \left( \dfrac{1}{x^{2}} \right)-2$ .

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

So, simply put

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

we get

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

The set onto which the derivative of the function

$f(x)=x(\log x-1)$ maps the range

$[1,\infty )$ is

0%
$\left[1,\infty \right)$
0%
$\left( e,\infty \right)$
0%
$\left[e,\infty \right)$
0%
$\left( 0,0 \right)$

Explanation

Given

$f (x) = x [\log x-1] x \in [+, \infty]$

$f (x) = x \left[ \dfrac{d}{dx} (\log x-1) \right]+ (\log x-1) \dfrac{d}{dx} (x)$

$=x. \dfrac{1}{x}+ (\log x-1).1$

$=1 -1 \log x-1$

$f(x)= \log x$

$\therefore$ The set of

$f(x)= \log x$ that maps

$[1, \infty )$ is

$[e, \infty )$

Let

$E=\{1, 2, 3, 4\}$ and

$F=\{1, 2\}$ then the number of onto functions from E to F is

0%
$14$
0%
$16$
0%
$12$
0%
$8$

Explanation

Number of onto function from

$E$ to

$F.$

$E=\{1,2,3,4\}$

$F=\{1,2\}$

Number of function from

$E$ to

$F=2\times 2\times 2\times 2=16$

We have to exclude functions where

$f(x)=1$ &

$f(x)=2$ .

$\therefore$ Total number of onto function

$=16-2=14$ .

Let

$f\left( x \right) ={ x }^{ 2 },g\left( x \right) ={ 2 }^{ x }$ , then solution set of

$fog\left( x \right) =gof\left( x \right)$ is

0%
R
0%
$\left\{ 0 \right\}$
0%
$\left\{ 0,2 \right\}$
0%
None of these

$f\left( x \right) =\begin{cases} 2+x,\quad x\ge 0 \\ 2-x,\quad x<0 \end{cases}$ then

$f\left( f\left( x \right) \right)$ is given by

0%
$f\left( f\left( x \right) \right) =\begin{cases} 2+x,\quad x\ge 0 \\ 4-x,\quad x<0 \end{cases}$
0%
$f\left( f\left( x \right) \right) =\begin{cases} 2+x,\quad x\ge 0 \\ 2-x,\quad x<0 \end{cases}$
0%
$f\left( f\left( x \right) \right) =\begin{cases} 4+x,\quad x< 0 \\ x,\quad x\ge 0 \end{cases}$
0%
$f\left( f\left( x \right) \right) =\begin{cases} 4+x,\quad x\ge 0 \\ x,\quad x<0 \end{cases}$

Let

$f\left[-1, \dfrac{-1}{2}\right] \to [-1, 1]$ is defined by

$f(x) = 4x^3 - 3x$ , then

$f^{-1} (x) =$ ____ .

0%
$\cos \left(\dfrac{1}{3}\cos^{-1} x\right)$
0%
$\cos \left(3\cos^{-1} x\right)$
0%
$\sin \left(\dfrac{1}{3}\sin^{-1} x\right)$
0%
$\cos \left(\dfrac{2\pi}{3}+\dfrac{1}{3} \cos^{-1} x\right)$

Explanation

$f(x)=4x^3-3x$

$f\left[-1,\cfrac{-1}{2}\right]\longrightarrow \left[-1,1\right]$

let,

$x=\cos\theta$

$\because y=\cos 3\theta$

$f(\cos\theta)=4cos^3\theta-3\cos\theta$

$=\cos3\theta$

$\because 3\theta=\cos^{-1}y$

$\theta=\cfrac{1}{3}\cos^{-1}y$

$f^{-1}(\cos3\theta)=\cos\theta$

$\because y=f(x)$

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

$\because f^{-1}(\cos3\theta)=\cos\theta$

$f^{-1}(y)=\cos\left(\cfrac{1}{3}\cos^{-1}y\right)$

$f^{-1}(x)=\cos\left(\cfrac{1}{3}\cos^{-1}x\right)$

If :

$f(x) = 5 {x}^{2}$ ,

$g(x) = 3x^{4}$ , then :

$(fog) (-1) =$

0%
$45$
0%
$-54$
0%
$-32$
0%
$-64$

Let

$f:X \to \left[ {1,\,27} \right]$ be a function by

$f\left( x \right) = 5\sin x + 12\cos x + 14$ . The set

$X$ so that

$f$ is one-one and onto is

0%
$\left[ { - \pi /2,\pi /2\,} \right]$
0%
$\left[ {0,\,\pi } \right]$
0%
$\left[ {0,\,\pi /2} \right]$
0%
non of these

For

$a,\ b\ \in \ R-\left\{ 0 \right\}$ , let

$f(x)=ax^{2}+bx+a$ satisfies

$f\left(x+\dfrac{7}{4}\right)=f\left(\dfrac{7}{4}-x\right) \forall \ x\ \in\ R$ .
Also the equation

$f(x)=7x+a$ has only one real distinct solution. The minimum value of

$f(x)$ in

$\left[0,\dfrac{3}{2}\right]$ is equal to

0%
$\dfrac{-33}{8}$
0%
$0$
0%
$4$
0%
$-2$

$f\left( x \right) = (1 - x)$ ,

$x \in \left[ { - 3,3} \right]$ , then the domain of

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

0%
$\left[ { - 2,3} \right]$
0%
$\left( { - 2,3} \right)$
0%
$\left[ { - 2,3]} \right.$
0%
$( - 2,3]$

Explanation

$f\left(x\right)=1-x$

$f\left(f\left(x\right)\right)=f\left(1-x\right)=1-1+x=x$

And

$x\in\left[-3,3\right]$

Domain of

$f\left(x\right)$ is

$\left[-3,3\right]$

Domain of

$f\left(f\left(x\right)\right)=$ Domain of

$f\left(1-x\right)$

Domain of

$f\left(f\left(x\right)\right)=\left[1-3,3\right]=\left[-2,3\right]$

If f(g(x))=5x+2 and g(x)=8x then f(x)=

0%
$\frac{5}{8}x+2$ .
0%
$\frac{8}{5}x+2$ .
0%
$\frac{5}{8}x-2$ .
0%
8x-2
0%
5x-2

Let

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

$f\left( x \right) =\begin{cases} -1,x<0 \\ 0,x=0 \\ 1,x>0 \end{cases}$ Then for all

$x,f\left( g\left( x \right) \right)$ is equal to (where

$\left[ . \right]$ represents the greatest integer function)

0%
$x$
0%
$1$
0%
$f\left( x \right)$
0%
$g\left( x \right)$

Let

$f:(2,3) \rightarrow (0,1)$ be defined by

$f(x)=x-[x]$ then

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

0%
$x-2$
0%
$x+1$
0%
$x-1$
0%
$x+2$

$f(x)=\frac{x}{\sqrt{1-x^{2}}}$ and g(x) =

$f(x)=\frac{x}{\sqrt{1+x^{2}}}$ , then (fog)(x) =

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

Explanation

Given:-

$f{\left( x \right)} = \cfrac{x}{\sqrt{1 - {x}^{2}}}$

$g{\left( x \right)} = \cfrac{x}{\sqrt{1 + {x}^{2}}}$

To find:-

$fog{\left( x \right)} = ?$

$fog{\left( x \right)}$

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

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

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

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

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

$= \cfrac{\left( \cfrac{x}{\sqrt{1 + {x}^{2}}} \right)}{\left( \cfrac{1}{1 + {x}^{2}} \right)}$

$= \cfrac{x}{\sqrt{1 + {x}^{2}}}$

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

Hence the correct answer is

$\cfrac{x}{\sqrt{1 + {x}^{2}}}$ .

Let

$f:X\rightarrow Y$ be an invertible function. Then f has unique inverse.

0%
True
0%
False

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

$g; R \rightarrow R, g(x)=x^{2}+2$ , then

$(gof)(x)$ equals-

0%
$2x^{2}-1$
0%
$(2x-1)^{2}$
0%
$2x^{2}+3$
0%
$4x^{2}-4x+3$

Explanation

$f(x)=2x-1$

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

$g(f(x))=(2x-1)^2+2=4x^2-4x+3$ .

1191545_1155623_ans_574198b726b647c79ac862bd2b5b5edb.jpg

Let

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

$x\ge 1$ , and

$g(x)$ is its reflection in the line mirror

$y=x$ , then function

$h(x)=\begin{cases} f\left( x \right) & x\ge 1 \\ g\left( x \right) & 0<x<1 \end{cases}$ , is

0%
derivable at $x=1$
0%
continuous at $x=1$
0%
not derivable at $x=1$
0%
not continuous at $x=1$

Explanation

Let

$\left(t, \dfrac{1}{t^2} \right)$ be a point on f(x)

$\therefore$ reflection in line mirror y = x

$\therefore t = \gamma$

$\therefore \dfrac{\gamma - \dfrac{1}{t^2}}{1} = \dfrac{x - t}{-1} = -2 \dfrac{(-t + \dfrac{1}{t^2})}{2}$

$\therefore \gamma = \dfrac{1}{t^2} + t - \dfrac{1}{t^2}$

$\therefore \gamma = t$

$\therefore x = \dfrac{1}{t^2}$

$\therefore x = \dfrac{1}{y^2}$

$\therefore xy^2 = 1$

$=9(x)$

$\therefore h'(x)$ at

$x = 1$ for

$x \ge 1 = f'(x)$

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

$\therefore h'(x)$ at x = 1 for

$x \le 1 = 9'(x)$

$= \dfrac{-2}{13} = -2$

derivable at

$x = 1$ .

$f(x)=\begin{cases} x+1 & x\epsilon \left[ -1,0 \right] \\ { x }^{ 2 }+1 & x\epsilon \left( 0,1 \right) \end{cases}$ , then the value of

$\dfrac{f^{-1}(0)+f^{-1}(1)+f^{-1}(2)}{f(-1)+f(0)+f(1)}$ is-

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

Explanation

$f(x) = \begin{cases} x +1 & x \in (-1, 0) \\ x^2 + 1 & x \in (0,1) \end{cases}$

$\therefore x = \begin{cases} f^{-1} (x + 1) & x \in [-1, 0] \\ f^{-1} (x^2 + 1) & x \in (0, 1)\end{cases}$

$f^{-1} (0)$ :- put

$x = -1$

$\therefore f^{-1} (0) = x = -1$

$\therefore f^{-1} (1)$ put

$x = 0$

$f^{-1} = 0$

$f^{-1} (2)$ :- put

$x = 1$

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

$\therefore f(0) = 1$

$\therefore f(1) = 1 + 1 = 2$

$\therefore f(-1) = -1 + 1 = 0$

$\therefore \dfrac{f^{-1} (0) + f^{-1} (1) + f^{-1} (2)}{f(-1) + f(0) + f(1)} = \dfrac{-1 + 0 + 1}{3}$

$= \dfrac{0}{3}$

$= 0$ .

The last three digits, if

$(12345956)_{10}$ is expressed in binary system.

0%
$110$
0%
$210$
0%
$100$
0%
$010$

$f ( x ) = \left( a - x ^ { n } \right) ^ { 1 / n }$ where

$a > 0$ and }

$n$ is a positive integer then

$( f o f ) ( x )$ is

0%
$f ( x )$
0%
x
0%
0
0%
1

Explanation

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

$\therefore f(f(x))=(a-(f(x))^n))^{1/n}$

$\therefore f(f(x))=(a-((a-x^n)^{1/n)^n)^{1/n}}$

$\therefore f(f(x))=(a-((a-x^n)))^{1/n}$

$\therefore f(f(x))=(a-a+x^n)^{1/n}$

$\therefore f(f(x))=(x^n)^{1/n}$

$\therefore(f(x))=x$

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

$g\left( x \right) = \dfrac{{3x + {x^3}}}{{1 + 3{x^2}}}$ then

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

0%
$- f\left( x \right)$
0%
$3f\left( x \right)$
0%
${\left( {f\left( x \right)} \right)^3}$
0%
$f\left( {3x} \right)$

Explanation

$f(x)=\log\bigg(\dfrac{1+x}{1-x}\bigg)\ and\ g(x)=\dfrac{3x+x^{3}}{1+3x^{2}}$

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

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

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

$f(g(x))=\log\bigg(\dfrac{(1+x)^3}{(1-x)^{3}}\bigg)$

$f(g(x))=\log\bigg(\dfrac{1+x}{1-x}\bigg)^{3}$

$f(g(x))=3\log\bigg(\dfrac{1+x}{1-x}\bigg)$

$f(g(x))=3f(x)$

CBSE Questions for Class 12 Commerce Maths Relations And Functions Quiz 7 - MCQExams.com

0:0:2

Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Relations And Functions Quiz 7 - MCQExams.com

0:0:2

Practice Class 12 Commerce Maths Quiz Questions and Answers

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