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 11 Commerce Applied Mathematics Functions Quiz 3

$f : R \rightarrow R$ is defined by

$f(x)=2x-2,$ then

$(f\circ f) (x) + 2 =$

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

Explanation

Let

$f : R \rightarrow R$ is defined by

$f(x)=2x-2$ . Then

$(f\circ f)(x)+2= f[f(x)]+2$

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

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

$=2f(x)$

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

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

0%
$\displaystyle \frac{x}{\sqrt{1-x^2}}$
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$\displaystyle \frac{x}{\sqrt{1+x^2}}$
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$\displaystyle \frac{1-x^2}{\sqrt{1-x^2}}$
0%
$x$

Explanation

Let

$x= \tan \theta$

$(f\circ g)(x)=f[g(x)] = f[g(\tan \theta)]$ .
Since

$\displaystyle g(x) = \frac{x}{\sqrt{1+x^2}}$ , we have

$f[g(\tan \theta)]=\displaystyle f \left [ \frac{\tan \theta}{\sqrt{1+\tan^2 \theta}} \right ] = f \left [ \frac{\tan \theta}{\sec \theta} \right ]=f (\sin \theta)$
Also since

$f(x)=\cfrac{x}{\sqrt{1-x^2}}$ , we have

$\displaystyle f (\sin \theta) = \frac{\sin \theta}{\sqrt{1-\sin^2 \theta}}$

$=\tan \theta =x$

$f(x)=\log x, g(x) = x^3$ then

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

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

Explanation

Given that

$g(x)=x^3$ . Therefore,

$f[g(a)]+f[g(b)]=f(a^3)+f(b^3)$ .

Since

$f(x)=\log x$ , we have

$f(a^3)+fb^3=\log a^3+\log b^3$

$=\log (a^3b^3)=\log ((ab)^{3})$

$=f[g(ab)]$

Find the domain of

$x^2+2$

0%
R
0%
N
0%
Z
0%
All of the above

The domain of

$f(x) = \displaystyle \frac{3}{4-x^2}+\log_{10} (x^3-x)$ is:

0%
$(1, 2)$
0%
$[-1, 1) \cup (1, 2)$
0%
$(1, 2) \cup (2, \infty)$
0%
$(-1, 0) \cup (1,2) \cup (2, \infty)$

Explanation

$f(x)$ is defined if either

(1)

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

$\Rightarrow x\neq \pm 2$

or (2)

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

$\Rightarrow x\left( { x }^{ 2 }-1 \right) >0$

$\Rightarrow \quad x(x+1)(x-1)>0$

$\Rightarrow x>0, x>1, x>-1$

Therefore, we have

$x>-1, x>0, x>1$ and

$x\neq 2, x\neq -2$ .

$\therefore x$ lies between

$(-1,0), (1,2), (2,\infty )$ .

$\therefore$ Domain is

$(-1,0)\cup (1,2)\cup (2,\infty )$ .

$f:R \rightarrow R$ and

$g : R \rightarrow R$ are defined by

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

$g(x)=x^2+7$ , then the values of

$x$ such that

$g(f(x)) =8$ are:

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

Explanation

Since

$f(x)=2x+3$ ,

$g[f(x)]=8 \Rightarrow g(2x+3)=8$ .
Also since

$g(x)=x^2+7$ ,

$g(2+3x)=8\Rightarrow (2x+3)^2+7=8$

$\Rightarrow 2x+3 = \pm 1$

$\Rightarrow x=-1$ or

$-2$ .

$f(x) = \dfrac{2x+5}{x^{2} + x + 5}$ , then

$f\left [ f(- 1 ) \right ]$ is equal to

0%
$\dfrac{149}{155}$
0%
$\dfrac{155}{147}$
0%
$\dfrac{155}{149}$
0%
$\dfrac{147}{155}$

Explanation

Given expression is

$f(x)=\dfrac { 2x+5 }{ { x }^{ 2 }+x+5 }$

$\therefore f(-1)=\dfrac { 2\times (-1)+5 }{ (-1)^{ 2 }+(-1)+5 } =\dfrac { 3 }{ 5 }$

$\therefore f\left( f\left( -1 \right) \right) =\dfrac { 2\times \dfrac { 3 }{ 5 } +5 }{ \left( \dfrac { 3 }{ 5 } \right) ^{ 2 }+\dfrac { 3 }{ 5 } +5 } =\dfrac { 155 }{ 149 }$

Which one of the following relation is a function

0%
0%
0%
0%
All of these

$f : R \rightarrow R$ and

$g :R \rightarrow R$ are defined by

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

$g(x) = [x]$ for

$x \in R$ , where

$[x]$ is the greatest integer not exceeding

$x$ , then for every

$x \in R, f(g(x)) =$

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

Explanation

Since

$f(x)=x-[x]$ , we have

$f[g(x)] =g(x)-[g(x)]$
Also since

$g(x)=[x]$ ,we have

$g(x)-[g(x)]=[x] - [[x]]$ .
Here,

$[x]$ is the greatest integer not exceeding

$x$ . Therefore,

$[[x]]=[x]$ and hence,

$[x]-[[x]]=[x]-[x]=0$

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

$f(y)=$

0%
$x$
0%
$y$
0%
$2y-1$
0%
$y-2$

Explanation

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

$=\displaystyle \frac{4x-2-x+2}{2x-1-2x+4} = \frac{3x}{3} = x$

$f(g(x))$ is one-one function, then

0%
g(x) must be one-one
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f(x) must be one-one
0%
f(x) may not be one-one
0%
g(x) may not be one-one

Explanation

It is fundamental concept that, If

$f(g(x))$ is one-one function then,

$f(x)$ must be one-one function and

$g(x)$ may be one-one or many one

Which of the following functions are one-one?

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$f:R\rightarrow R$ given by $f(x)={ 2x }^{ 2 }+1$ for all $\quad x\in R$
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$g:Z\rightarrow Z$ given by $g(x)={ x }^{ 4 }$ for all $\quad x\in R$
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$h:R\rightarrow R$ given by $h(x)={ x }^{ 3 }+4$ for all $\quad x\in R$
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$\phi :C\rightarrow C$ given $\phi (z)={ 2z }^{ 6 }+4$ for all $\quad x\in R$

Explanation

For all even powered function,

$f(x_{1})=f(x_{2})$ where

$x_{1}=\pm(x_{2})$
Hence, the only possible one-one function is

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

$f(x)$ is a strictly increasing function. So, it is one-one.

A mapping function

$f:X\rightarrow Y$ is one-one, if

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$f({ x }_{ 1 })\neq f({ x }_{ 2 })\$ for all ${ x }_{ 1 },{ x }_{ 2 }\in X$
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$f({ x }_{ 1 })=f({ x }_{ 2 })\Rightarrow { x }_{ 1 }={ x }_{ 2 }$ for all ${ x }_{ 1 },{ x }_{ 2 }\in X$
0%
${ x }_{ 1 }={ x }_{ 2 }\Rightarrow f({ x }_{ 1 })=f({ x }_{ 2 })$ for all ${ x }_{ 1 },{ x }_{ 2 }\in X$
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none of these

Explanation

For a function to be one one

$f(x_{1})=f(x_{2})$
Implies that

$x_{1}=x_{2}$
Hence the answer is option B

Find the domain of

$x$ if

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

0%
$x\in R-(-2, 2)$
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$x\in R$
0%
$x\in R-(0, 2)$
0%
None of these

Explanation

Case I

$x<0$
Hence

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

$=\sqrt{x^2+x-2}$

$=\sqrt{(x+2)(x-1)}$
Therefore the domain will be

$(-\infty,-2]$ ....

$(D_{1})$
Case II

$x>0$
Hence

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

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

$=\sqrt{(x-2)(x+1)}$
Therefore the domain will be

$[2,\infty)$ ....

$(D_{2})$
Hence the domain of the function

$f(x)$ from

$D_{1}$ and

$D_{2}$ will be

$=(-\infty,-2]\cup[2,\infty)$

$f:R\rightarrow R$ given by

$f(x)={ x }^{ 3 }+({ a+2)x }^{ 2 }+3ax+5$ is one-one, then

$a$ belongs to the interval

0%
$(-\infty ,1)$
0%
$(1 ,\infty)$
0%
$(1 ,4)$
0%
$(4 ,\infty)$

Explanation

$f(x)$ is one one then

$f'\left( x \right)$ should be either positive or negative for all

$x$ .

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

$\Rightarrow D = 4{ (a+2) }^{ 2 }-4\times3\times3a<0$

$\Rightarrow$

${ a }^{ 2 }-5a+4<0$

$\Rightarrow$

$(a-1)(a-4)<0$

Hence,

$a\in \left( 1,4 \right)$

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

Domain of

$g(x)$ :

$g(x)$ is defined if

$3-x\ge 0$ and

$(x-1)(x-2)(x-2)\neq 0$

$\Rightarrow x\le3$ and

$x\neq 1,2,3$

$\therefore$ Domain of

$g(x)=(\infty,3)-{1,2,3}$
Domain of

$h(x):$

$\displaystyle h\left( x \right) =\sin ^{ -1 }{ \left[ \frac { 3x-2 }{ 2 } \right] } \Rightarrow -1\le \left[ \frac { 3x-2 }{ 2 } \right] \le 1$
Case I:
If

$\displaystyle \left[ \frac { 3x-2 }{ 2 } \right] =-1\Rightarrow -1\le \frac { 3x-2 }{ 2 } <0\Rightarrow -2\le 3x-2<0\Rightarrow 0\le x<\frac { 2 }{ 3 }$ ...(1)
Case II:

$\displaystyle \left[ \frac { 3x-2 }{ 2 } \right] =0\Rightarrow 0\le \frac { 3x-2 }{ 2 } <1\Rightarrow 0\le 3x-2<2$

$\displaystyle \Rightarrow 2\le 3x<4\Rightarrow \frac { 2 }{ 3 } \le x<\frac { 4 }{ 3 }$ ...(2)
Case III:
If

$\displaystyle \left[ \frac { 3x-2 }{ 2 } \right] =1\Rightarrow 1\le \frac { 3x-2 }{ 2 } <2\Rightarrow 2\le 3x-2<4\Rightarrow \frac { 4 }{ 3 } \le x<2$ ...(3)
Thus, from (1),(2) and (3), we have
Domain of

$h(x)=[0,2)$

$\therefore$ Domain of

$f=[2,0)-{1}$

Which of the following function is one-one?

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$f:R\rightarrow R$ given by $f(x)=|x-1|$ for all $x\in R$
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$g:\left[ -\dfrac{\pi }{ 2 },\dfrac{ \pi }{ 2 } \right] \rightarrow R$ given by $g(x)=|sinx|$ for all $x\in \left[ \dfrac{ -\pi }{ 2 },\dfrac{ \pi }{ 2 } \right]$
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$h:\left[ \dfrac{ -\pi }{ 2 },\dfrac{ \pi }{ 2 } \right] \in R$ given by $h(x)=sinx$ for all $x\in \left[ \dfrac{ -\pi }{ 2 },\dfrac{ \pi }{ 2 } \right]$
0%
$\phi :R\rightarrow R$ given by $f(x)={ x }^{ 2 }-4$ for all $x\in R$

Explanation

Option A - Consider

$x = 2$ and

$x =0$ , the value of

$f(x)$ is same. Hence it is not one-one
Option B - If we replace

$x$ by

$-x$ , then the value of

$g(x)$ remains the same. Hence it is not one-one
Option C -

$h(x)$ is an increasing function for the given values of

$x$ . Hence it is one-one function
Option D -

$f(x)$ is an even function. So it is not one-one for the given values of

$x$

$f$ and

$g$ are one-one functions from

$R\to R$ , then

0%
$f+g$ is one-one
0%
$fg$ is one-one
0%
$fog$ is one-one
0%
none of these

Explanation

Let

${x_1},{x_2} \in R$ be two distinct elements, then

$g\left( {{x_1}} \right) \ne g\left( {{x_2}} \right)$ , as

$g$ is one-one function. Similarly

$f\left( {g\left( {{x_1}} \right)} \right) \ne f\left( {g\left( {{x_2}} \right)} \right)$ as

$f$ is also one-one function.
Hence,

$f\circ g$ is one-one function.
Note that

$f+g$ and

$f\cdot g$ may not one-one functions even if

$f$ and

$g$ are one one. For example consider

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

$g\left( x \right) =- x$ , then

$f+g$ and

$f\cdot g$ are not one-one.

Let

$\displaystyle f:R\rightarrow A=\left \{ y: 0\leq y< \dfrac{\pi}{2} \right \}$ be a function such that

$\displaystyle f(x)=\tan^{-1}(x^{2}+x+k),$ where

$k$ is a constant. The value of

$k$ for which

$f$ is an onto function is

0%
$1$
0%
$0$
0%
$\displaystyle \frac{1}{4}$
0%
none of these

Explanation

For

$f$ to be an onto function, Range and co-domain of

$f$ should be equal

$\Rightarrow f(x)\ge 0 \forall x\in R$

$\Rightarrow \tan^{-1}(x^2+x+k)\ge 0$

For the above equation to be valid for all

$x$

We must have, the discriminant of

$x^2+x+k=0$ , is zero

$b^2-4ac=0$

$\Rightarrow 1-4k=0\Rightarrow k =\dfrac{1}{4}$

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

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

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

The domain of the function

$\displaystyle f(x)=\sqrt{x^{2}-[x]^{2}},$ where

$[x]=$ the greatest integer less than or equal to

$x$ , is

0%
$R$
0%
$[0,+\infty)$
0%
$(-\infty,0]$
0%
none of these

Explanation

$x \ge 0$ , then

$\left[x\right]$ is an integer part of

$x$ .

Hence

${x^2} \ge {\left[ x \right]^2}$ and

$\displaystyle f\left( x \right) = \sqrt {{x^2} - {{\left[ x \right]}^2}}$ is well-defined.

$x < 0$ say

$x=-a$ where

$a$ is a positive number,

then

$\displaystyle \left[ x \right] =- \left( {\left[ a \right] + 1} \right)$ .

Hence,

${x^2} < {\left[ x \right]^2}$ and the function

$\displaystyle f\left( x \right) = \sqrt {{x^2} - {{\left[ x \right]}^2}}$ is not defined.

Therefore the domain of the function is

$\left[ {0,\infty } \right)$ .

$f(x)=ax+b$ and

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

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

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

Explanation

$f(g(x))=a(cx+d)+b$

$=acx+ad+b$ ...(i)

$g(f(x))=c(ax+b)+d$

$=ac(x)+bc+d$ ...(ii)

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

From (i) and (ii)

$ac(x)+bc+d=acx+ad+b$

$cb+d=ad+b$

$g(b)=f(d)$

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

$\displaystyle y=f\left \{ f(f(x)) \right \},x> 1,$ is

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

Explanation

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

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

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

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

$=x$

$y=x$ is an equation of straight line.

Hence, 'C' is correct.

Let

$\displaystyle f:\left \{ x,y,z \right \}\rightarrow \left \{ a,b,c \right \}$ be a one-one function and only one of the conditions

$(i)f(x)\neq b, (ii)f(y)=b,(iii)f(z)\neq a$ is true then the function

$f$ is given by the set

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$\displaystyle \left \{ (x,a),(y,b),(z,c)\right \}$
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$\displaystyle \left \{ (x,a),(y,c),(z,b)\right \}$
0%
$\displaystyle \left \{ (x,b),(y,a),(z,c)\right \}$
0%
$\displaystyle \left \{ (x,c),(y,b),(z,a)\right \}$

Explanation

$f:\left\{ x,y,z \right\} \rightarrow \left\{ a,b,c \right\}$ is a one-one function

$\Rightarrow$ each element in

$\left\{ x,y,z \right\}$ will have exactly one image in

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

and no two elements of

$\left\{ x,y,z \right\}$ will have same image in

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

Coming to the given 3 conditions, only one is true.

1) if

$f\left( x \right) \neq b$ is true then

$f\left( y \right) =b$ is false which makes

$f\left( z \right) \neq a$ true

$\Longrightarrow f\left( x \right) \neq b$ is false.

2) if

$f\left( y \right) =b$ is true then

$f\left( x \right) \neq b$ will also be true

$\Longrightarrow f\left( y \right) =b$ is false

$\therefore f\left( z \right) \neq a$ is the true condition and remainig two are false conditions.

$\therefore f\left( x \right) =b, f\left( y \right) =a, f\left( z \right) =c$

hence

$f=\left\{ \left( x,b \right) ,\left( y,a \right) ,\left( z,c \right) \right\}$

Let

$f:R \rightarrow R$ and

$g:R \rightarrow R$ be defined by

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

$(gof) (x)=$

0%
$3x^2+6x-13$
0%
$3x^2-6x-13$
0%
$3x^2+6x+13$
0%
$-3x^2+6x-13$

Explanation

Given that,

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

$g(x)=3x-4$

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

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

$(gof) (x)=3x^2+6x-13$

Hence, option A.

The domain of the function

$\displaystyle f(x)=\log_{10}\log_{10}(1+x^{3})$ is

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

Explanation

Given,

$\displaystyle f(x)=\log_{10}\log_{10}(1+x^{3})$

The value inside a log function must not be negative.
Hence

$\log_{10}(1+x^3)>0$

$1+x^{3}>10^{0}$ ....(by taking anti-log).

$x^{3}>0$

$x>0$
Hence the domain for

$f(x)$ is

$x>0$ .

$f$ and

$g$ are two functions such that

$\displaystyle \left ( fg \right )\left ( x \right )=\left ( gf \right )\left ( x \right )$ for all

$x$ . Then

$f$ and

$g$ may be defined as

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$\displaystyle f\left ( x \right )=\sqrt{x}, g\left ( x \right )=\cos x$
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$\displaystyle f\left ( x \right )=x^{3}, g\left ( x \right )=x+1$
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$\displaystyle f\left ( x \right )=x-1, g\left ( x \right )=x^{2}+1$
0%
$\displaystyle f\left ( x \right )=x^{m}, g\left ( x \right )=x^{n}$ where $m, n$ are unequal integers

Explanation

$)f(x)=$

$\sqrt{x},g(x)=cosx then (fg)(x) = \sqrt{\cos x} and (gf)(x) = \cos \sqrt{x}$ .They are not equal.

$f(x)=x^{3},g(x)=x+1 then (fg)(x) = (x+1)^{3} and (gf)(x)= x^{3}+1$ .They are not equal.

$f(x)=x1,g(x)=x^{2}+1 then (fg)(x)=x^{2} and (gf)(x)=x^{2}-2x$ . They are not equal.

$f(x)=x^{m}, g(x)=x^{n} then (fg)(x)=x^{n+m} and (gf)(x)=x^{n+m}$ .Hence

$(fg)(x)= (gf)(x)$

$\displaystyle f(x)=x^{n},n\in N$ and

$(gof)(x)=ng(x)$ then

$g(x)$ can be

0%
$n\:|x|$
0%
$3.\sqrt[3]{x}$
0%
$e^{x}$
0%
$\log\:|x|$

Explanation

$f(x)=x^{n}$

$g(f(x))=ng(x)$ ...

$(i)$

$\log(f(x))=n\log(x)$ ...

$(ii)$

Taking

$\log(|x|)$ as

$g(x)$ the above expression is reduced to eq

$i$ .

Hence

$g(x)=\log(|x|)$ .

The domain of

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

0%
$(\sqrt{2},+\infty)$
0%
$(0,+\infty)$
0%
$(1,+\infty)$
0%
none of these

Explanation

$\log_{x^{2}-1}(x)\geq 0$

Hence,

$x\geq 1 \quad \dots (1)$

Now,

$x^{2}-1>1$

$x^{2}>2$

$x<-\sqrt{2}$ and

$x>\sqrt{2} \quad \dots (2)$

Hence from

$(1)$ and

$(2)$ ,

$x\in(\sqrt{2},\infty)$

The composite mapping

$fog$ of the map

$f: R\rightarrow R,f(x)=\sin x$ and

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

0%
$x^2 \sin x$
0%
$(\sin x)^2$
0%
$\sin x^2$
0%
$\dfrac{ \sin x}{x^2}$

Explanation

Composite mapping will be

$f(g(x))$

$=f(x^2)$

$=\sin(x^2)$

Hence option

$'C'$ is the answer.

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

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

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

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

Let

$\displaystyle f(x)=\frac{ax}{x+1}$ , where

$\displaystyle x\neq -1$ . Then for what value of

$\displaystyle a$ is

$\displaystyle f( f(x))=x$ always true

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

Explanation

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

Since,

$\displaystyle f\left( {f\left( x \right)} \right) =x$ , we have,

$\displaystyle\dfrac{{{a^2}x}}{{ax + x + 1}}=x$ .

Simplifying the equation we get,

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

$\therefore \left( {a + 1} \right){x^2} + \left( {1 - {a^2}} \right)x = 0$

$\left( {a + 1} \right)x\left( {x + 1 - a} \right) = 0$

Hence the only possible value is

$a=-1$

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

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

$g'(3)=4$ .

Set

$A$ has

$3$ elements and set

$B$ has

$4$ elements. The number of injections that can be defined from

$A$ to

$B$ is

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

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

$B.$

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

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

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

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

$\displaystyle \because d=-a$

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

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

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

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

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

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)

$f:R\rightarrow R$ 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.

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

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

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

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

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

$-3$

If the function

$f : R \rightarrow$ A given by

$f(x)\, =\, \displaystyle \frac{x^{2}}{x^{2}\, +\, 1}$ is a surjection, then A is

0%
$R$
0%
$[0, 1]$
0%
$(0, 1]$
0%
$[0, 1)$

Explanation

As 'f' is surjective,

Range of f = co-domain of 'f'

$\implies$ A = range of 'f'

Since,

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

$\implies$

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

$\implies$

$y(x^2+1)=x^2$

$\implies$

$(y-1)x^2+y=0$

$\implies$

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

$\implies$

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

$\implies$

$\dfrac{y}{1-y} \geq 0$

$\implies$

$y \epsilon [0,1)$

$\implies$

$A=[0,1)$

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

$(x>1)$ ,

$f(1) \geq 5$

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

CBSE Questions for Class 11 Commerce Applied Mathematics Functions Quiz 3 - 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 3 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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