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 $$fofof(x)=$$

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

lf $$g(f(x)) =|\sin \mathrm{x}|,f(g(x)) =(\sin\sqrt{\mathrm{x}})^{2}$$, then

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

$${f}({x})=\sin x,g({x})=|{x}|$$

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

$${f}, {g}$$ cannot be determined

MCQ Questions for Class 11 Commerce Applied Mathematics Functions Quiz 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

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

$\Rightarrow k+1=0$

$k^{2}=1$

$\therefore k=-1$

$n (A) = 4$ 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

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

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.

$f:R\rightarrow R, g:R\rightarrow R$ 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

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

$=g\left ( a \right )$

$=a^{3}+2$

$f(x)=2x+1$ 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

$h(x)=x+1$

$hogof(x)$

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

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

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

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

$=2x^2+2$

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

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

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

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

$=\cos x^{2}$

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

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

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%
$\pm 2$
0%
$\pm 3$
0%
$\pm 4$

Explanation

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

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

$=2x^{2}+17$

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

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

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

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

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

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

$=n-(n-1)$

$=1$

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

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

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

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

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

$f:[1,\infty )\rightarrow B$ 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

$x\geqslant 1$ .

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

$=5$

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

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

$(fof)(x)+2=$

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

Explanation

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

$=f(3x-2)$

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

$=9x-8$

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

$=3f(x)$

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

$fofof(x)=$

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

Explanation

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

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

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

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

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

$f(n+1)=f(n)$ for all

$n\in N, f(7)=5$ then

$f(35)=$

0%
$25$
0%
$49$
0%
$35$
0%
$5$

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

$(fog)(x)=$

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

Explanation

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

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

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

The domain of

$\sqrt{\log_{e/3}\mathrm{x}+1}$ is

0%
$\left(0,\displaystyle \frac{3}{\mathrm{e}}\right]$
0%
$\left(-\displaystyle \infty\frac{3}{\mathrm{e}}\right]$
0%
$\left[\displaystyle \frac{3}{\mathrm{e}}\infty\right)$
0%
$R$

Explanation

$\log_{e/3}{x}+1\geq 0$ and

${x}>0$

$\log_{e/3}x\geq-1$

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

$\Rightarrow x\in \left(0, \dfrac{3}{e}\right]$
Hence, option 'A' is correct.

The domain of

${ f }({ x })=\tan ^{ -1 } (5x)$ is

0%
$(-\infty,\infty)$
0%
$(0,\infty)$
0%
$(-\infty,0)$
0%
$\left(\displaystyle -\frac{1}{5},\frac{1}{5}\right)$

Explanation

$\tan^{-1}x$ is defined for all

$x$ as range of

$\tan x$ is from

$(-\infty ,\infty )$ .

$\therefore$ domain of

$\tan^{-1}5x$ is

$(-\infty ,\infty )$ .
Hence, option 'A' is correct.

The domain of

$\displaystyle f(x)= { \sin }^{ -1 }\left(\dfrac { 2x-3 }{ 5 } \right)$ is

0%
$[1,3]$
0%
$[1,4]$
0%
$[2, 14]$
0%
$[-1,4]$

Explanation

$\sin^{-1}x$ is defined if

$x\in [-1,1]$

$\Rightarrow -1\leq \dfrac{2x-3}{5}\leq 1$

$-5\leq 2x-3\leq 5$

$-2\leq 2x\leq 8$

$-1\leq x\leq 4$

$f:R\rightarrow R$ is defined by

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

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

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

Explanation

Let

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

$\Rightarrow f(t)=-3$

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

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

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

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

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

$f:[-2,2]\rightarrow R$ is defined as

$f(x)=\left\{\begin{array}{l}-1,-2\leq x\leq 0\\x-1,0\leq x\leq 2\end{array}\right.$ then

$\{x\in[-2,2]:x\leq 0,\; f(|x|)=x\}=$

0%
$\{-1\}$
0%
$\{0\}$
0%
$\displaystyle \left\{\frac{-1}{2}\right\}$
0%
$\phi$

The domain of

${f}({x})=\log|x^{2}-9|$ is

0%
$\mathrm{R}- \{-3,3\}$
0%
$(-\infty,-3)$
0%
$(3,\infty)$
0%
$(-\infty,\infty)$

Explanation

$f$ is defined for all

$[x^{2}-9]>0 \Rightarrow x\neq \pm 3$

$\therefore$ domain of

$f$ is

$R-\left \{ -3,3 \right \}$
Hence, option 'A' is correct.

The domain of

$\displaystyle \mathrm{f}(\mathrm{x})=\cos^{-1}\left(\frac{2}{2+\sin \mathrm{x}}\right)$ contained in

$[0,2\pi]$ is

0%
$\left[0,\displaystyle \frac{\pi}{2}\right]$
0%
$\left[\displaystyle \frac{\pi}{2},\pi \right]$
0%
$[0, \pi]$
0%
$\left[\displaystyle \frac{-\pi}{2},\frac{\pi}{2}\right]$

Explanation

Domain of the function is value of

$x$ in

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

Since value of

$\cos x$ is between

$-1$ to

$1$ thus

$-1\displaystyle \leq\frac{2}{2+\sin{x}}\leq 1$

$\Rightarrow \sin x\geq0$

$\Rightarrow x\in[0, \pi]$

Hence, option 'C' is correct.

The domain of

$f(x)=\sqrt{\log\left ( \dfrac{7x-x^{2}}{12} \right )}$ is

0%
$(-\infty ,\infty )$
0%
$(-\infty ,4]$
0%
$[3 ,\infty )$
0%
$[3,4]$

Explanation

Domain of a function is value of

$x$ for which function is defined

$f(x)=\sqrt{\log\left ( \dfrac{7x-x^{2}}{12} \right )}$
For log to be always positive, value into log should be greater than

$1$ so

$\displaystyle \frac{7{x}-{x}^{2}}{12}\geq 1$

$x^2-7x+12=0$
So solving the inequality, we get

$x\in[3,4]$
Hence, option 'D' is correct.

The domain of the function

$\displaystyle \mathrm{f}({x})=\frac{1}{\sqrt{x-2}}+\frac{1}{\sqrt{5-x}}$ is

0%
$[2,5]$
0%
$(2,5)$
0%
$[2,5 )$
0%
$(2,5 ]$

Explanation

Given,

$\displaystyle \mathrm{f}({x})=\frac{1}{\sqrt{x-2}}+\frac{1}{\sqrt{5-x}}$

For the

$f(x)$ to be defined.

$x-2>0 \Rightarrow x>2$

$5-x>0 \Rightarrow x<5$

$\Rightarrow x\in (2,5)$

The domain of

$\mathrm{f}({x})=\mathrm{c}\mathrm{o}\mathrm{t}^{-1} \left(\displaystyle \frac{x}{3}\right)$ is

0%
$(-\infty,\infty)$
0%
$(0,\infty)$
0%
$(1,\infty)$
0%
$\left(\displaystyle -\frac{1}{3},\frac{1}{3}\right)$

Explanation

$\cot^{-1}$ is defined for all

$x$

$\therefore$ Domain of

$f$ is

$(-\infty ,\infty )$

The domain of

$\displaystyle \mathrm{f}({x})=\frac{1}{\sqrt{9-x^{2}}}+\sqrt{x^{2}-4}$ is

0%
$(-4,-2)\cup (2,4)$
0%
$(-3,-2]\cup [2,3)$
0%
$(-\infty,-3)\cup (2,\infty)$
0%
$(-\infty,\infty)$

Explanation

$f$ is defined if both terms under square root exist.

$\Rightarrow 9-x^{2}> 0$ and

$x^{2}-4\geq 0$

$\Rightarrow (3-x)(3+x)>0$ and

$(x-2)(x+2)\geq 0$

$\Rightarrow (x-3)(3+x)<0$ and

$(x-2)(x+2)\geq 0$

$\Rightarrow -3<x<3$ and

$x<-2$ and

$x>2$

$x\in (-3,3)$ and

$x\in [-\infty ,-2]\cup [2,\infty ]$

Intersection of both gives the following solution

$\Rightarrow x\in (-3,-2]\cup [2,3)$

The domain of

$\mathrm{f}({x})=\mathrm{e}^{\sqrt{{x}}}+\cos x$ is

0%
$(-\infty,\infty)$
0%
$[0, \infty)$
0%
$(0, 1)$
0%
$(1, \infty)$

Explanation

$\cos x$ is defined for all

$x$

$\therefore$ For

$f(x)$ to be defined.

$\sqrt{x}\geq 0$

$\Rightarrow x\geq 0 \Rightarrow x\in [0,\infty )$

Let

$S$ be set of all rational numbers. The functions

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

$f(x)=\begin{cases} 0, & x \in S \\ 1, & x \notin S \end{cases}$

$g(x)=\begin{cases} -1 & x\in S \\ 0 & x\notin S \end{cases}$
then,

$(fog) (\pi)+(gof)(e)=$

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

Explanation

$f(x)=\begin{cases} 0, & x \text{ is rational} \\ 1, & x \text{ is irrational} \end{cases}$

$g(x)=\begin{cases} -1 & x\text { is rational} \\ 0& x\text { is irrational } \end{cases}$

$fog(\pi )=f(g(\pi ))=f(0)=0$
so

$gof(e)=g(f(e))=g(1)=-1$

$\therefore fog(\pi )+gof(e)=-1$

The domain of

$f(x)= \text{cosec} x-\cot x$ is

0%
$(-\infty,\infty)$
0%
$R-\{n\pi$ : $n\in Z\}$
0%
$R-\displaystyle \{(2n+1)\frac{\pi}{2}$ : $n\in Z\}$
0%
$R-\displaystyle \left\{\frac{n\pi}{2}:n\in Z\right\}$

Explanation

$\text{cosec} x$ &

$\cot x$ are not defined at

$x=n\pi \ \ \forall n\in Z$

$\therefore$ Domain of

$f$ is

$R- \{ n\pi\} , n\in Z$

The domain of

$\mathrm{f}({x})=\mathrm{c}\mathrm{o}\mathrm{s}^{-1} (\sqrt{3x})$ is

0%
$[-1, 1]$
0%
$\left[0,\displaystyle \frac{1}{3}\right]$
0%
$[0,1]$
0%
$[0,3 ]$

Explanation

For

$f$ to be defined

$\sqrt{3x}$ has to exist.

$\Rightarrow 3x\geq 0$

$x\geq 0$

$\cos^{-1}$ is defined only if

$-1\leq \sqrt{3x}\leq 1$
As

$x\geq 0$

$\Rightarrow \sqrt{3x}\leq 1$

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

$\therefore x\in \left[0,\dfrac{1}{3}\right]$

$f:[-6,6]\rightarrow \mathbb{R}$ is defined by

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

$x\in \mathbb{R}$ then

$(fofof)(-1)+(fofof)(0)+(fofof)(1)=$

0%
$f(4\sqrt{2})$
0%
$f(3\sqrt{2})$
0%
$f(2\sqrt{2})$
0%
$f(\sqrt{2})$

Explanation

$fofof(-1)=fof(-2)=f(1)=-2$

$fofof(1)=fof(-2)=f(1)=-2$

$fofof(0)=fof(-3)=f(0)=33$

$\therefore fofof(-1)+fofof(1)+fofof(0)=29$

$=32-3$

$=(4\sqrt{2})^{2}-3$

$\Rightarrow f(4\sqrt{2})=(4\sqrt{2})^{2}-3$

The domain of

$\mathrm{f}({x})=\sqrt{x^{2}-9x+18}$ is

0%
$[3,6]$
0%
$(-\infty,3]\cup[6,\infty)$
0%
$[3,\infty)$
0%
$[6, \infty)$

Explanation

For given function to be defined,

$x^2-9x+18\ge 0$

$\Rightarrow (x-6)(x-3)\ge 0\Rightarrow x\in (-\infty,3]\cup [6,\infty) =$ required domain
Hence option 'B' is correct

The domain of

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

0%
$\left(-\displaystyle \infty,\frac{1}{2}\right)$
0%
$\left(\displaystyle \frac{1}{2},\infty\right)$
0%
$\left[0,\displaystyle \frac{1}{2}\right]$
0%
$(0,\infty)$

Explanation

$f$ is defined if

$\sqrt{1-2x}$ exist

$\sum cos^{-1}(1-2x)$ is define.

$\Rightarrow 1-2x\geq 0 -1\leq 1-2x\leq 1$

$\Rightarrow x\leq \dfrac{1}{2} -2\leq -2x\leq 0$

$0\leq 2x\leq 2$

$\Rightarrow 0\leq x\leq 1$

$\therefore$ For

$x\in [0,\dfrac{1}{2}],$

$f$ is defined.

The domain of the function

$\displaystyle \mathrm{f}({x})=\dfrac{\sqrt{2+x}+\sqrt{2-x}}{x}$ is

0%
$(-2,2)$
0%
$[-2,0)\cup (0,2]$
0%
$[-2,2]$
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$(-\infty,2)$

Explanation

Given,

$\displaystyle \mathrm{f}({x})=\dfrac{\sqrt{2+x}+\sqrt{2-x}}{x}$

For

$f(x)$ to be defined

$x\neq 0$

$2+x\geq 0 \Rightarrow x\geq -2$

$2-x\geq 0 \Rightarrow x\leq 2$

$\therefore x\in [-2,0)\cup (0,2]$

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

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

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

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

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

Explanation

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

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

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

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

$\displaystyle \therefore$

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

$n\geq 1$ is any integer,

$\mathrm{d}(n)$ denotes the number of positive factors of

$n$ , then for any prime number

$\mathrm{p},\ \mathrm{d}(\mathrm{d}(\mathrm{d}(\mathrm{p}^{7})))=$

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

Explanation

For a number being

$p^{n}$ the total number of positive factors excluding

$1$ and the number itself will be

$n-1$ .
Therefore total number of positive factors will be

$n-1+2$

$=n+1$
Hence for

$p^{7}$ we will have

$8$ factors.
Now for

$8=2^{3}$ .
Hence

$8$ will have

$4$ factors.
Now

$4=2^{2}$ , hence the number of factors will be

$3$ .

Set of values of

$x$ for which the function

$\displaystyle \mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}}+2^{\sin^{-1}\mathrm{x}}+\frac{1}{\sqrt{\mathrm{x}-2}}$ exists is

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$R$
0%
$\mathrm{R}-\{\mathrm{0}\}$
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$\phi$
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$\mathrm{R}-\{1\}$

Explanation

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

$x\neq 0$ for

$\dfrac{1}{x}$

$-1\leqslant x\leqslant1$ for

$\sin^{-1}x$

$\sqrt{x-2}\neq 0$ and

$x-2>0$ for

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

$x>2$
n combining, we get domain of

$f(x)$ is

$\phi$ .

The domain of the function

$\displaystyle \mathrm{f}(\mathrm{x})=\frac{1}{\sqrt{|\mathrm{x}|-\mathrm{x}}}$ is

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$(-\infty, \infty)$
0%
$(0, \infty)$
0%
$(-\infty, 0)$
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$(-\infty, \infty)-\{0\}$

Explanation

Given function is

$\displaystyle \mathrm{f}(\mathrm{x})=\frac{1}{\sqrt{|\mathrm{x}|-\mathrm{x}}}$

$\Rightarrow$

$\mathrm{f}(\mathrm{x})$ is defined if

$|x| > x$

$= |\mathrm{x}|>\mathrm{x}$

$= \mathrm{x}<0$

So domain of

$\mathrm{f}(\mathrm{x})$ is

$(-\infty, 0)$ .

$f$ :

$R\rightarrow R$ is defined by

$f(x)=\left\{\begin{array}{l}x+4 & x<-4\\3x+2 & -4\leq x<4\\x-4 & x\geq 4\end{array}\right.$
then the correct matching of list I to List II is.

List - I	List - II
$\mathrm{A}) f(-5)+f(-4)=$	$\mathrm{i}) 14$
$\mathrm{B}) f(\|f(-8)\|)=$	ii $) 4$
$\mathrm{C}) f(f(-7)+f(3))=$	$\mathrm{i}\mathrm{i}\mathrm{i})-11$
$\mathrm{D}) f(f(f(f(0)))+1=$	$\mathrm{i}\mathrm{v})-1$
	v) $1$
	vi) $0$

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A-iii , B-vi , C-ii , D- v
0%
A-iii , B-iv , C-ii , D- vi
0%
A-iv , B-iii , C-ii , D- i
0%
A-ii , B-vi , C-v , D- ii

Explanation

$f(-5)=-5+4=-1 ; f(-4)=3(-4)+2=-10$

$\therefore f(-5)+f(4)=-11$

$f(|f(-8)|)=f((-8+4))=f(4)=4-4=0$

$f(f(-7)) tf(3) = f(-7+4)+3(3)+2$

$=3(-3)+2+3(3)+2$

$=4$

$f(f(f(f(0))))+1=f(f(f(2)))+1=f(f(8))+1$

$=f(4)+1=0+1=1$

Set A has 3 elements and set B has 4 elements. The number of injections that can be defined from A into B is :

0%
144
0%
12
0%
24
0%
64

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

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${f}({x})=\sin^{2} {x},{g}({x})=\sqrt{{x}}$
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${f}({x})=\sin x,g({x})=|{x}|$
0%
${f}({x})={x}^{2},{g}({x})=\sin\sqrt{{x}}$
0%
${f}, {g}$ cannot be determined

Explanation

$g(f(x))=|\sin x|=\sqrt{\sin^{2}x}$ .......(i)

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

Comparing

$i$ and

$ii$

$\Rightarrow$

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

$\Rightarrow$

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

And

$\Rightarrow$

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

$\Rightarrow$

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

Therefore

$\Rightarrow$

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

$\Rightarrow$

$f(x)=\sin^{2}x$

The domain of

$\sqrt{(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)}$ is

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$(1,2)$
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$[1,2]$
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$(1,2)\cup (3, \infty)$
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$[1,2] \cup [3, \infty)$

Domain of definition of the function

$\displaystyle{ f }({ x })=\sqrt { \sin ^{ -1 } (2{ x })+\frac { \pi }{ 6 } }$ for real valued

$x$ , is

0%
$[-\displaystyle \frac{1}{4}, \displaystyle \frac{1}{2}]$
0%
$[-\displaystyle \frac{1}{2}, \displaystyle \frac{1}{2}]$
0%
$(-\displaystyle \frac{1}{2}, \displaystyle \frac{1}{9})$
0%
$[-\displaystyle \frac{1}{4}, \displaystyle \frac{1}{4}]$

Explanation

Here

$\displaystyle f\left( x \right) =\sqrt { \sin ^{ -1 }{ \left( 2x \right) } +\frac { \pi }{ 6 } }$ to find domain we must have

$\displaystyle \sin ^{ -1 }{ \left( 2x \right) } +\frac { \pi }{ 6 } \ge 0$ ( but

$\displaystyle -\frac { \pi }{ 2 } \le \sin ^{ -1 }{ \theta } \le \frac { \pi }{ 2 }$ )
As

$\displaystyle -\frac { \pi }{ 6 } \le \sin ^{ -1 }{ \left( 2x \right) } \le \frac { \pi }{ 2 } \\\Rightarrow \sin { \left( -\cfrac { \pi }{ 6 } \right) } \le 2x\le \sin { \left( \cfrac { \pi }{ 2 } \right) }$

$\displaystyle \Rightarrow -\cfrac { 1 }{ 2 } \le 2x\le 1\\\Rightarrow -\cfrac { 1 }{ 4 } \le x\le \cfrac { 1 }{ 2 } \\\Rightarrow x\in \left[ -\cfrac { 1 }{ 4 } ,\cfrac { 1 }{ 2 } \right]$

Find the domain of

$e^x$ .

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N
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R
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Z
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All of the above

CBSE Questions for Class 11 Commerce Applied Mathematics Functions Quiz 2 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Functions Quiz 2 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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