If $$f(x)=\dfrac{e^{x}+e^{-x}}{2}$$, then the inverse of $$f(x)$$ is

$$\log_{e}\left(\dfrac {x+\sqrt{x^{2}-1}}{2}\right)$$

If $$f(x)=\dfrac{x+1}{x-1}(x\neq 1)$$ then $$fofofof(x)=$$

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

If $$f(x)=\dfrac{x}{\sqrt{1+x^{2}}}$$ then $$fofof(x)=$$

$$\dfrac{2x}{\sqrt{1+2x^{2}}}$$

If $$f(x)=\displaystyle \frac{2^{x}+2^{-x}}{2^{x}-2^{-x}}$$, then $$f^{-1}(x)=$$

$$\displaystyle\frac{1}{2}\log_{2}\left ( \frac{x+1}{x-1} \right )$$

$$\displaystyle\frac{1}{2}\log_{2}\left ( \frac{x-1}{x+1} \right )$$

$$\displaystyle\frac{1}{2}\log_{2}\left ( \frac{x+1}{x-2} \right )$$

$$\displaystyle\frac{1}{2}\log_{2}\left ( \frac{x-2}{x-1} \right )$$

MCQ Questions for Class 12 Commerce Maths Relations And Functions Quiz 2

$n (A) = 4$ and

$n(B) = 6$ , then the number of surjections from

$A$ to

$B$ is

0%
$4^{6}$
0%
$6^{4}$
0%
$0$
0%
$24$

Explanation

$n(A)<n(B)$

$\therefore$ Co-domain can never be equal to range.

$\therefore$ No of surjections is zero

The number of injections that are possible from

$A$ to itself is

$720,$ then

$n (A) =$

0%
$5$
0%
$6$
0%
$7$
0%
$8$

Explanation

first element of

$A$ in domain can be mapped to
any of the

$n$ elements of

$A$ in range.

$2^{nd}$ element of

$A$ in domain can be mapped to any of

$(n-1)$ elements of

$A$ in range.

$\therefore$ Total no of injections possible is

$n\times (n-1)\times (n-2)...\times 3\times 2\times 1=n!$

$\therefore n!=720$

$\Rightarrow n=6$

Let

$A=\{1,2,3\}, B =\{a, b, c\}$ and If

$f=\{(1,a),(2,b),(3,c)\}, g=\{(1,b),(2,a),(3,b)\}, h=\{(1,b)(2,c),(3,a)\}$ then

0%
$g$ and $h$ are injections
0%
$f$ and $h$ are injections
0%
$f$ and $g$ injections
0%
$f,g$ and $h$ are injections

Explanation

$g(1)=g(3)=b \Rightarrow$ g is not one-one.
domain of

$f =$ domain of

$h = A.$
range of

$f =$ range of

$h = B$

$n(A)=n(B)=3$

$\therefore f$ and

$h$ are injections.

The number of one-one functions that can be defined from

$A = \left \{ 1,2,3 \right \}$ to

$B = \left \{ a,e,i,o,u \right \}$ is

0%
$3^{5}$
0%
$5^{3}$
0%
${_{}}^{5}P_{3}$
0%
$5!$

Explanation

You can correlate this to permutation and combination problems.

We have to arrange 3 people on 5 places.

So total no of permutations

$={_{}}^{5}P_{3}$

The number of non-surjective mappings that can be defined from

$A = \left \{ 1,4,9,16 \right \}$ to

$B=\left \{ 2,8,16,32,64 \right \}$ is

0%
$1024$
0%
$20$
0%
$505$
0%
$625$

Explanation

Here

$n(B)>n(A)$ , no function can be a surjective.

No of functions

$=$ No of non-surjectives

Any elements of

$A$ can be mapped to any of

$5$ elements in

$B$
So non surjective mapping from

$A$ to

$B$

$=5\times5\times5\times5$

$=625$

$f:A\rightarrow B$ is a constant function which is onto then

$B$ is

0%
a singleton set
0%
a null set
0%
an infinite set
0%
a finite set

Explanation

$f(x)$ is a constant fucntion

$\Rightarrow$ Range of

$f(x)$ is a singleton set.
For

$f$ to be an onto function, Co domain

$B$ should be equal to Range.

$\therefore B$ is a singleton set.

$f:A\rightarrow B$ is a bijection then

$f^{-1} of =$

0%
$fof^{-1}$
0%
$f$
0%
$f^{-1}$
0%
an identity

Explanation

Since

$f$ is bijective, its inverse will also be bijective.

$f(x)=y$

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

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

$y=f(f)^{-1}(y)=f(x)=y$
Hence, it is an identity.

The number of injections possible from

$A=\{1,3,5,6\}$ to

$B =\{2,8,11\}$ is

0%
$8$
0%
$64$
0%
$2^{12}$
0%
$0$

Explanation

co-domain

$n(B) <$ domain

$n(A)$
There can't be any one-one function from

$A$ to

$B.$

The number of possible surjection from

$A=\{1,2,3,...n\}$ to

$B = \{1,2\}$ (where

$n \geq 2)$ is

$62$ , then

$n=$

0%
$5$
0%
$6$
0%
$7$
0%
$8$

Explanation

Each element in

$A$ can be mapped onto any of two elements of

$B$

$\therefore$ Total possible functions are

$2^{n}$
For the

$f^{{n}'s}$ to be surjections , they shouldn't be mapped alone to any of the two elements.

$\therefore$ Total no of surjections

$= 2^{n}-2$

$2^{n}-2=62$

$\Rightarrow n=6$

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

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

$(gof)(x)=$

0%
$\cos 2x$
0%
$x^{2}\cos x$
0%
$\cos x^{2}$
0%
$\cos^{2} x^{2}$

Explanation

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

$=\cos x^{2}$

$f:R\rightarrow R$ is defined by

$\displaystyle f(x)={\dfrac{2x+1}{3}}$ then

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

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

Explanation

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

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

$\Rightarrow 2x+1=3y$

$2x=3y-1$

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

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

$\Rightarrow f^{-1}\left ( y \right )=\dfrac{3y-1}{2}$

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

Let

$f(x)=\dfrac{Kx}{x+1}(x\neq -1)$ then the value of

$K$ for which

$(fof)(x)=x$ is

0%
$1$
0%
$-1$
0%
$2$
0%
$\sqrt{2}$

Explanation

$f\left ( f\left ( x \right ) \right )=f\left ( \dfrac{kx}{x+1} \right )$

$=\dfrac{k\left ( \dfrac{kx}{x+1} \right )}{\dfrac{kx}{x+1}=1}=\dfrac{k^{2}x}{\left ( k+1 \right )x+1}$

$\therefore$ for

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

$\Rightarrow k+1=0$

$k^{2}=1$

$\therefore k=-1$

$f:\left ( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right )\rightarrow \left ( -\infty ,\infty \right )$ defined by

$f(x)=1+3x$ is

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

Explanation

$f(x_{1})=f(x_{2}) \Rightarrow 1+3x_{1}=1+3x$

$x_{1}=x_{2}$

$\therefore f(x_{1})=f(x_{2})\Rightarrow x_{1}=x_{2}$

$\therefore$ f is one-one.

$f$ lies b/w

$(1-\dfrac{3\pi }{2},\dfrac{1+3\pi }{2})$

$\therefore$ co-domain

$(-\infty ,\infty )$ is not equal to range

$(\dfrac{1-3\pi }{2},\dfrac{1+3\pi }{2})$

$\Rightarrow$ f is not onto.

$\therefore$ f is one-one but not onto.

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

$f(x)=4x-1,g(x)=x^{3}+2,$ then

$(gof)\left(\dfrac{a+1}{4}\right)=$

0%
$43$
0%
$4a^3-1$
0%
$a^{3}+2$
0%
$64a^3 - 8a^{2}-1$

Explanation

$g\left ( f\left ( \dfrac{a+1}{4} \right ) \right )=g\left ( 4\left ( \dfrac{a+1}{4} \right )-1 \right )$

$=g\left ( a \right )$

$=a^{3}+2$

The function

$f:(0,\infty )\rightarrow (-\infty ,\infty )$ is defined by

$f(x)=\log_{3} x$ then

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

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

Explanation

$f\left ( x \right )=\log _{3}x$

$f^{-1}:\left ( -\infty ,\infty \right )\rightarrow \left ( 0, \infty \right )$
Let

$\log _{3}x=y=f\left ( x \right )$

$\Rightarrow x=3^{y}$

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

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

$f:R\rightarrow R,f(x)=3x-2$ then

$(fof)(x)+2=$

0%
$f(x)$
0%
$2f(x)$
0%
$3f(x)$
0%
$-f(x)$

Explanation

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

$=f(3x-2)$

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

$=9x-8$

$fof(x)+2=9x-6=3(3x-2)$

$=3f(x)$

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

$g(x)=x^{2}+1$ then

$(go(fof))(2)=$

0%
$112$
0%
$122$
0%
$12$
0%
$124$

Explanation

$(go(fof))\left( x \right) =g\left( f\left( f\left( x \right) \right) \right)$

$=g\left ( f\left ( 2x+1 \right ) \right )$

$=g\left ( 2\left ( 2x+1 \right )+1 \right )$

$=g\left ( 4x+3 \right )$

$=\left ( 4x+3 \right )^{2}+1$

$(go(fof))\left ( 2 \right )=\left ( 11 \right )^{2}+1=122$

$f(x)=\dfrac{1}{x}, g(x)=\sqrt{x}$ and

$(go\sqrt{f})(16)=$

0%
$2$
0%
$1$
0%
$\dfrac{1}{2}$
0%
$4$

Explanation

$\sqrt{f\left ( x \right )}=\dfrac{1}{\sqrt{x}}$

$g\left ( \sqrt{f\left ( x \right )} \right )=g\left ( \dfrac{1}{\sqrt{x}} \right )=\dfrac{1}{\sqrt{\sqrt{x}}}=\dfrac{1}{x^{\frac{1}{4}}}$

$\left ( go\sqrt{f} \right )(16)=\dfrac{1}{\left ( 16 \right )^{\dfrac{1}{4}}}=\dfrac{1}{2}$

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

$h(x)=x+1$ then

$(hogof)(x)$ is equal to

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

Explanation

Given

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

$h(x)=x+1$

$hogof(x)$

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

$=h\left ( g\left ( x \right ) \right )\dots\dots \left[ \because f\left ( x \right )=x \right ]$

$=h\left ( 2x^{2}+1 \right )\dots\dots\left [ \because g\left ( x \right )=2x^{2}+1 \right ]$

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

$=2x^2+2$

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

$\therefore hogof(x)=2(x^2+1)$

$f(x)=\dfrac{e^{x}+e^{-x}}{2}$ , then the inverse of

$f(x)$ is

0%
$\log_{e}(x+\sqrt{x^{2}+1})$
0%
$\log_{e}\sqrt{x^{2}-1}$
0%
$\log_{e}\left(\dfrac {x+\sqrt{x^{2}-1}}{2}\right)$
0%
$\log_{e}(x+\sqrt{x^{2}-1})$

Explanation

Let

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

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

$e^{x}+e^{-x}=2y$

$e^{2x}-2y e^{x}+1=0$

$\Rightarrow e^{x}=\dfrac{2y \pm \sqrt{4y^{2}-4}}{2}$

$e^{x}=\dfrac{y\pm \sqrt{y^{2}-1}}{1}$
Range of

$f$ is

$\left ( -\infty ,\infty \right )\Rightarrow e^{x}=\dfrac{y+\sqrt{y^{2}-1}}{1}$
As if we take

$e^{x}=\dfrac{y-\sqrt{y^{2}-1}}{1}$ , which is always small

$\therefore x=\log _{e}\left ( \dfrac{y+\sqrt{y^{2}-1}}{1} \right )$

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

$f:(-\infty ,\infty )\rightarrow (-\infty ,\infty )$ is defined by

$f(x)=5x-6$ , then

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

0%
$\dfrac{x+5}{6}$
0%
$\dfrac{x-5}{6}$
0%
$\dfrac{x-6}{5}$
0%
$\dfrac{x+6}{5}$

Explanation

$f(x)=5x-6=y$

$\Rightarrow 5x=y+6$

$x=\dfrac{y+6}{5}$

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

$\therefore f^{-1}\left ( y \right )=\dfrac{y+6}{5}$

$f(x)=\dfrac{5x+6}{7x+9}$ then

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

0%
$\dfrac{y+6}{7y+9}$
0%
$\dfrac{7y+9}{5y+6}$
0%
$\dfrac{9y-6}{-7y+9}$
0%
$\dfrac{9y-6}{-7y+5}$

Explanation

Let

$f\left ( x \right )=y=\dfrac{5x+6}{7x+9}$

$7xy+9y=5x+6$

$\Rightarrow x\left ( 7y-5 \right )=6-9y$

$x=\dfrac{9y-6}{-7y+5}$

$x=f^{-1}\left ( y \right )=\dfrac{9y-6}{-7y+5}$

$f$ from

$R$ into

$R$ is defined by

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

$f^{-1}\left \{ -2,0,7 \right \}=$

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

Explanation

$f$ is invertible throughout

$R$

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

$\Rightarrow x=\sqrt[3]{y+1}=f^{-1}\left ( y \right )$

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

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

$f^{-1}\left ( 7 \right )=2$

$\therefore f^{-1}\left \{ -2,0,7 \right \}=\left \{ -1,1,2 \right \}$

$f(x)=3x-1$ and

$g(x)=5x+6$ then

$(g^{-1}of^{-1})(2)=$

0%
$10$
0%
$-1$
0%
$11$
0%
$12$

Explanation

$f\left ( x \right )=3x-1=y$

$\Rightarrow x=\displaystyle\frac{y+1}{3}$

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

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

$g\left ( x \right )=5x+6=z$

$\Rightarrow x=\displaystyle \frac{z-6}{5}$

$g^{-1}\left ( z \right )=\displaystyle \frac{z-6}{5}$

$\Rightarrow g^{-1}\left ( x \right )=\displaystyle \frac{x-6}{5}$

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

$=\displaystyle \frac{\dfrac{x+1}{3}-6}{5}$

$\displaystyle =\frac{x-17}{15}$

$\displaystyle \therefore g^{-1}\left ( f^{-1}\left ( 2 \right ) \right )=\frac{2-17}{15}=-1$

$f(x)=e^{5x+13}$ then

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

0%
$\dfrac{13-\log y}{5}$
0%
$\dfrac{-13+\log y}{5}$
0%
$\dfrac{5+\log y}{13}$
0%
$\dfrac{5-\log y}{13}$

Explanation

Let

$f\left ( x \right )=e^{5x+13}=y$

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

$5x+13=\ln y$

$5 x=-13+\log y$

$x=\dfrac{-13+\log y}{5}$

$\therefore f^{-1}\left ( y \right )=\dfrac{-13+\log y}{5}$

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

$f(x)=x+\dfrac{1}{x}$ , then

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

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

Explanation

Let

$f\left ( x \right )=x+\dfrac{1}{x}=y$

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

$x^{2}-yx+1=0$
Solving

$x^{2}-yx+1$ , we get

$x^{2}-yx+1=0$

$x=\dfrac{y\pm \sqrt{y^{2}-4}}{2}$

$\therefore f^{-1}=\dfrac{x+\sqrt{x^{2}-4}}{2}$

$\because$

$f$ is defined from

$\left ( 1,\infty \right )\rightarrow \left ( 2,\infty \right )$ negative part is discarded.

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

$f(n)=\begin{cases} \dfrac{n}{2} & \text{ if } n \space is \space even \\-\left (\dfrac{n-1}{2} \right ) & \text{ if } n \space is \space odd \end{cases}$ then

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

0%
Function is not invertible.
0%
$199$
0%
$201$
0%
$200$

Explanation

$f(n)$ is positive if

$n$ is even & negative if

$n$ is odd.

$\therefore f^{-1}\left ( -100 \right )=-2x+1$

$=-2\left ( -100 \right )+1$

$=200+1$

$=201$

$f:R\rightarrow R$ is defined by

$f(x)=x^{2}+4$ then

$f^{-1}(13)=$

0%
$\left \{ -3,3 \right \}$
0%
$\left \{ -2,2 \right \}$
0%
$\left \{ -1,1 \right \}$
0%
Not invertible

Explanation

$f\left ( x \right )=x^{2}+4=y$

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

$x=\pm \sqrt{y-4}$

$\therefore f^{-1}\left ( 13 \right )=\pm \sqrt{13-4}=\pm 3$

$f(3)=13\,\&\,f(-3)=13$
Thus image

$13$ has two pre-images i.e,

$3$ and

$-3$

$\therefore$

$f$ is not invertible

$f(x)=2+x^{3}$ , then

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

0%
$\sqrt[3]{x}+2$
0%
$\sqrt[3]{x}-2$
0%
$\sqrt[3]{x-2}$
0%
$\sqrt[3]{x+2}$

Explanation

Let

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

$\Rightarrow 2+x^{3}=y$

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

$\Rightarrow x=\left ( y-2 \right )^{1/3}$

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

$=\left ( y-2 \right )^{{1}/{3}}$

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

The solution of

$8x\equiv 6(mod \ 14)$ is

0%
$\{8, 6\}$
0%
$\{6,14\}$
0%
$\{6,13\}$
0%
$\{8,14,6\}$

Explanation

Since,

$8\mathrm{x}\equiv 6(mod \ 14)$
i.e.,

$8\mathrm{x}-6=14\mathrm{k}$ for

$\mathrm{k}\in \mathrm{I}$ .

$\Rightarrow 8x = 14k+6$

$\Rightarrow 4x = 7k+3$
The values

$6$ and

$13$ satisfy this equation (when

$k =3$ and

$k=7$ ),

while

$8, 14$ and

$16$ do not.

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

$g(x)= \ln(x)$ then the domain of

$(gof)(x)$ is

0%
$(-\infty ,2)$
0%
$(-1,1)$
0%
$(-\infty ,1]$
0%
$(-\infty ,1)$

Explanation

Given

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

$g(x)=ln(x)$

$gof(x)$

$=g(f(x))$

$=\ln\left ( 1-x \right )^{1/2}$

$=\dfrac{1}{2}\ln\left ( 1-x \right )$

$\therefore$ For the composite function to be defined

$1-x>0$

$x<1$

$\therefore$ Domain is

$\left ( -\infty ,1 \right )$

$f:R^{+}\rightarrow R$ such that

$f(x)=\log_{5} x$ then

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

0%
$\log_{x}10$
0%
$5^{x}$
0%
$3^{-x}$
0%
$3^{1/x}$

Explanation

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

$\Rightarrow \log _{5}x=y$

$\Rightarrow \dfrac{\log x}{\log 5}=y$

$\Rightarrow \log x=y\log 5$

$\Rightarrow e^{\log x}=e^{y\log 5}$

$\Rightarrow e^{\log x} =e^{\log 5^{y}}$

$[\because a \log x = \log x^{a}$ ]

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

$\therefore f^{-1}\left ( x \right )=5^{x}$

$f(x)=\dfrac{x+1}{x-1}(x\neq 1)$ then

$fofofof(x)=$

0%
$f(x)$
0%
$2\left ( \dfrac{x+1}{x-1} \right )$
0%
$\dfrac{x-1}{x+1}$
0%
$x$

Explanation

$fofofof\left(x\right)=fofof\left(\dfrac{x+1}{x-1}\right)$

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

$=fof\left(x\right)$

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

$F(n)=(-1)^{k-1}(n-1), G(n)=n-F(n)$ then

$(GoG)(n)=$ (where

$k$ is odd)

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

Explanation

$G(n)=n-(-1)^{k-1}(n-1)$

$GoG(n)=G(n-(-1)^{k-1}(n-1))$

$=n-(-1)^{k-1}(n-1)-(-1)^{k-1}((n-1)-(-1)^{k-1}(n-1))$

$=n-(n-1)$

$=1$

$f:[1,\infty )\rightarrow B$ defined by the function

$f(x)=x^{2}-2x+6$ is a surjection, then

$B$ is equals to

0%
$[1,\infty )$
0%
$[5,\infty )$
0%
$[6,\infty )$
0%
$[2,\infty )$

Explanation

$f(x)=x^2-2x+6$ is a surjection.
So the range of

$f(x)$ will be equal to its codomain.

$f(x)=x^2-2x+6$

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

$=2(x-1)$

$f(x)$ will be increasing when

$x\geqslant 1$ .

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

$=5$

$\therefore B=[5, \infty)$

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

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

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

0%
$\displaystyle \left(\dfrac{1}{3}\right)^{-x}$
0%
$3^{x}$
0%
$\displaystyle \log_{1/3}$ $x$
0%
$\displaystyle \log_{x}\left(\dfrac{1}{3}\right)$

Explanation

Let

$\displaystyle f(x)=\left(\frac{1}{3}\right)^{x}=y$
Taking logarithm on both sides,

$\displaystyle x\log \frac{1}{3}=\log y$

$\displaystyle \Rightarrow x=\frac{\log y}{\log \frac{1}{3}}$

$\displaystyle \Rightarrow x=\log _ { 1/3 } y\quad \quad \quad \left[ \because \frac { \log b }{ \log a } =\log _{ a }{ b } \right]$

$\therefore f^{-1}\left ( x \right )=\log _{1/3}x$

$X =\{1, 2,3,4,5\}$ and

$Y =\{1,3,5,7,9\}$ , determine which of the following sets represent a relation and also a mapping?

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$R_{1}= \{(x,y)$ : $y=x+2, x \in Y,y \in Y\}$
0%
$R_{2}=\{(1,1), (1,3), (3,5), (4,7), (5,9)\}$
0%
$R_{3}=\{(1,1), (2,3), (3,5), (3,7), (5,7)\}$
0%
$R_{4}=\{(1,3), (2,5), (4,7), (5,9), (3,1)\}$

Explanation

Here we will check all options one by one:

Option A:

$R_1 = \{(x,y): y= x+2, x\in X, y\in Y\}$

$\implies \mathrm{R}_{1}=\{(1,3),(2, 4),(3,5),(4,6),(5,7)\}$

Since

$4$ amd

$6$ are the images of

$2$ and

$4$ respectively but

$4$ and

$6$ do not belong to

$Y$

$\therefore (2, 4),(4,6)\not\in X \times Y$

Hence

$R_{1}$ is not a relation as well as not a mapping.

Option B:

$R_{2}$ : It is a relation but not a mapping because the element

$1$ has two different images.

Option C:

$R_{2}$ : It is a relation but not a mapping because the element

$3$ has two different images.

Option D:

$R_{4}$ : It is both a relation and a mapping because every element in

$X$ is mapped to the elements in

$Y$ . Also, every element of

$X$ has a one and only one image in

$Y$ and every element in

$Y$ has its pre-image in

$X$ . Hence, it is also one-one and onto mapping and hence it is a bijection.

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

$fofof(x)=$

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

Explanation

$fofof\left(x\right)=fof\left(\dfrac{x}{\sqrt{1+x^{2}}}\right)$

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

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

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

$=\dfrac{x}{\sqrt{1+3x^{2}}}$

If A

$=$ {

$x : x^{2}-3x+2= 0$ }, and

$R$ is a universal relation on

$A$ , then

$R$ is

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$\{(1,1),(2, 2)\}$
0%
$\{(1,1)\}$
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$\phi$
0%
$\{(1,1),(1, 2)(2,1),(2,2)\}$

Explanation

Consider,

$x^2-3x+2=0$

$\implies x^2-2x-x+2=0$

$\implies (x-2)(x-1)=0$

$\implies x=1,2$

$\therefore A=\{1,2\}$
Also R is universal relation on set A, then every element of set A is related every other element of A

$R= \{(1,1), (1,2), (2,1), (2,2)\}$ .

Assertion(A): If

$X=\left \{ x:-1\leq x\leq 1 \right \}$ and

$f:X\rightarrow X$ defined by

$f(x)=\sin \pi x; \forall x\in A$ is not invertible function

Reason (R): For a function

$f$ to have inverse, it should be a bijection

0%
Both A and R are true and R is the correct explanation of A
0%
Both A and R are true but R is not correct explanation of A
0%
A is true but R is false
0%
A is false but R is true

Explanation

$f(-\pi )=f(0)=f(\pi )=0$

$\therefore f$ is not a bijection

$\therefore \sin\pi x$ is not invertible.

$f(x)=\displaystyle \dfrac{x}{\sqrt{1-x^{2}}},g(x)=\displaystyle \dfrac{x}{\sqrt{1+x^{2}}}$ , then

$(fog)(x)=$

0%
$x$
0%
$\dfrac{x}{\sqrt{1+x^{2}}}$
0%
$\sqrt{1+x^{2}}$
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$2x$

Explanation

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

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

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

$f(x)=1+x+x^{2}+x^{3}+\ldots\ldots$ for

$\left | x \right |<1$ then

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

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

Explanation

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

$=\dfrac{1}{1-x}$
Let

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

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

$x=1-\dfrac{1}{y}=\dfrac{y-1}{y}$
But

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

If the function is

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

$f(x)=2x+3, g(x)=x^{2}+7$ and

$f[g(x)]=25$ then

$x=$

0%
$f(x)$
0%
$\pm 2$
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$\pm 3$
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$\pm 4$

Explanation

$f(g(x))=f(x^{2}+7) =25$

$=2(x^{2}+7)+3$

$=2x^{2}+17$

$\Rightarrow 2x^{2}=8$

$x^{2}=4 \Rightarrow x=\pm 2$

$f(x)=\displaystyle \frac{2^{x}+2^{-x}}{2^{x}-2^{-x}}$ , then

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

0%
$\displaystyle\frac{1}{2}\log_{2}\left ( \frac{x-1}{x+1} \right )$
0%
$\displaystyle\frac{1}{2}\log_{2}\left ( \frac{x+1}{x-1} \right )$
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$\displaystyle\frac{1}{2}\log_{2}\left ( \frac{x+1}{x-2} \right )$
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$\displaystyle\frac{1}{2}\log_{2}\left ( \frac{x-2}{x-1} \right )$

Explanation

Let

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

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

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

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

$\Rightarrow \displaystyle x=f^{-1}\left(y\right)=\frac{1}{2}\log_{2}\left(\frac{y+1}{y-1}\right)$
So,

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

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

$(fof)(x)=$

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

Explanation

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

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

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

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

$f:R\rightarrow R$ is defined by

$f(x)=x^{2}-10x+21$ then

$f^{-1}(-3)$ is

0%
$\left \{ -4,6 \right \}$
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$\left \{ 4,6 \right \}$
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$\left \{ -4, 4, 6 \right \}$
0%
Not Invertible

Explanation

Let

$f^{-1}(-3)=t$

$\Rightarrow f(t)=-3$

$t^{2}-10t+21=-3$

$t^{2}-10t+24=0$

$t^{2}-6t-4t+24=0$

$t(t-6)-4(t-6)=0$

$\Rightarrow t=6,4 = f^{-1}(-3)$

I: If

$f:A\rightarrow B$ is a bijection only then does

$f$ have an inverse function
II: The inverse function

$f:R^{+}\rightarrow R^{+}$ defined by

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

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

0%
only I is true
0%
only II is true
0%
both I and II are true
0%
neither I nor II true

Explanation

$f$ is a bijection

$\Rightarrow$

$f$ is one one and onto.

$\therefore$ there exists on inverse value for every value in co-domain.

$\therefore$

$f$ is invertible if and only if

$f$ is a bijection

$f:R^{+}\rightarrow R^{+} ; f(x)=x^{2}=y$

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

$f(x)=\sin^{-1}\left \{ 3-(x-6)^{4} \right \}^{1/3}$ then

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

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$6+\sqrt[4]{3+\sin^{3}x}$
0%
$6+\sqrt[4]{3-\sin^{3}x}$
0%
$6+\sqrt[4]{3+\sin x}$
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$6+\sqrt[4]{3-\sin x}$

Explanation

Let

$f(x)=y=\sin^{-1}(3-(x-6)^{4})^{\frac{1}{3}}$

$\sin y=(3-(x-6)^{4})^{\frac{1}{3}}$

$\sin^{3}y=3-(x-6)^{4}$

$(x-6)^{4}=3-\sin^{3}y$

$x-6=(3-\sin^{3}y)^{\frac{-1}{4}}$

$x=(3-\sin^{3}y)^{\frac{1}{4}}+6$

$x=f^{-1}(y)=(3-\sin^{3}y)^{\frac{1}{4}}+6.$

Which of the following functions defined from

$(-\infty ,\infty )$ to

$(-\infty ,\infty )$ is invertible ?

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$f(x) = \sin (2x+3)$
0%
$f(x) = x^{2} + 4$
0%
$f (x) =x^{3}$
0%
$f (x) = \cos x$

Explanation

$f(x)=\sin(2x+3)$ lies only from

$[-1,1]$ and hence is not onto.
So,

$f(x)$ is not bijective

$\therefore$ it is not invertible

$f(x)=x^{2}+4$ is always positive and so not onto.
So,

$f(x)$ is not bijective

$\therefore$ It is also not invertible.

$f(x)=x^{3}$ varies from

$(-\infty ,\infty )$ as

$x$ varies from

$(-\infty ,\infty )$
So,

$f(x)$ is bijective

$\therefore$ It is invertible.

$f(x)=\cos x$ always lies between

$[-1,1]$ and so is into function.
So,

$f(x)$ is not bijective
Hence, not invertible.

${f}\left({x}\right)=\sin^{2}{x}+\sin^{2}\left({x}+\displaystyle \dfrac{\pi}{3}\right)+ \cos x \cos \left({x}+\displaystyle \dfrac{\pi}{3}\right)$ and

${g}\left(\displaystyle\dfrac{5}{4}\right)=1$ ,

$g\left(1\right) = 0$ then

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

0%
$1$
0%
$0$
0%
$\sin x$
0%
Data is insufficient

Explanation

$\displaystyle {f}\left({x}\right)=\sin^{2}{x}+\left(\sin{x}\cos\frac{\pi}{3}+ \cos{x} \displaystyle \sin\frac{\pi}{3}\right)^{2}+ {c}{o}{s} {x}\left(\displaystyle \cos {x}\cos\frac{\pi}{3}-\sin {x}\sin\frac{\pi}{3}\right)$

$\displaystyle =\sin ^{ 2 }{ x } +{ \left[ \frac { \sin{ x } }{ 2 } +\frac { \sqrt { 3 } \cos{ x } }{ 2 } \right] }^{ 2 }+\frac { \cos^{ 2 }{ x } }{ 2 } -\frac { \sqrt { 3 } }{ 2 } \cos x \sin x$

$\displaystyle =\sin ^{ 2 }{ x } +\frac { \sin ^{ 2 }{ x } }{ 4 } +\frac { 3 }{ 4 } \cos ^{ 2 }{ x } +\frac { \sqrt { 3 } }{ 2 } \sin x\cos{ x }+\frac { \cos ^{ 2 }{ x } }{ 2 } -\frac { \sqrt { 3 } }{ 2 } \sin { x\cos { x } }$

$=\displaystyle \frac{5}{4}\left(\sin^{2}{x}+\cos^{2}{x}\right)=\frac{5}{4}$

$\displaystyle \therefore$

$[{g}{o}{f}]\left({x}\right)={g}[{f}\left({x}\right)]={g}\left(\displaystyle \frac{5}{4}\right)=1$

CBSE Questions for Class 12 Commerce Maths Relations And Functions Quiz 2 - 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 2 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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