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

If $$\displaystyle f(x)=3x-5$$ then $$f^{-1}(x)=$$

$$\displaystyle \frac{x+5}{3}$$

$$\displaystyle \frac{1}{3x-5}$$

does not exist because $$f$$ is not one-one

does not exist because $$f$$ is not onto

MCQ Questions for Class 12 Commerce Maths Relations And Functions Quiz 3

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

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

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

Set

$A$ has

$n$ elements. The number of functions that can be defined from

$A$ into

$A$ is:

0%
$n^2$
0%
$n!$
0%
$n^n$
0%
$n$

Explanation

The number of functions that can be defined from a set into another set can be found by

${ \text{(Number of elements in co-domain)} }^{ \text{Number of elements in domain} }$

Since, set

$A$ has

$n$ elements, we have

Number of elements in co-domain

$=$ Number of elements in domain

$= n (A) = n$ .

Therefore, number of functions that can be defined from

$A$ into

$A$

$=n(A)^{n(A)}= n^{n}$ .

$n\geq 1$ is any integer,

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

$n$ , then for any prime number

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

0%
$1$
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

$3$ .

Let

$\displaystyle f\left( x \right)={ x }^{ 2 }-x+1,x\ge \left( \frac { 1 }{ 2 } \right)$ then the solution of the equation

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

0%
$x=1$
0%
$x=2$
0%
$\displaystyle x=\frac{1}{2}$
0%
None of these

Explanation

$\displaystyle y={ x }^{ 2 }-x+1\Rightarrow { x }^{ 2 }-x+\left( 1-y \right) =0$

Here,

$a=1,b=-1$ and

$c=1-y$

$\displaystyle x=\frac { 1\pm \sqrt { 1-4\left( 1-y \right) } }{ 2 } \quad \quad \left[ \because x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \right]$

$\displaystyle \because x>\frac { 1 }{ 2 } ,\quad \therefore x=\frac { 1 }{ 2 } +\sqrt { y-\frac { 3 }{ 4 } }$

$\displaystyle \Rightarrow f^{ -1 }\left( x \right) =\frac { 1 }{ 2 } +\sqrt { x-\frac { 3 }{ 4 } }$

Now,

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

Since the graphs of the original and inverse functions can intersect only on the straight line

$y=x,$ therefore

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

$\Rightarrow { x }^{ 2 }-2x+1=0$

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

$\Rightarrow x=1$

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

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

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

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

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

$=32-3$

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

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

$f$ :

$R\rightarrow R$ 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) $0$

0%
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

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

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

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

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

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

$=4$

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

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

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

0%
${f}({x})=\sin^{2} {x},{g}({x})=\sqrt{{x}}$
0%
${f}({x})=\sin x,g({x})=|{x}|$
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

$\Rightarrow$

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

$\Rightarrow$

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

$f(x)=x-x^{2}+x^{3}-x^{4}+\ldots..\infty$ when

$|x|<1$ , then the ascending order of the following is
a)

$f(1/2)$
b)

$f^{-1}(1/2)$
c)

$f(-1/2)$
d)

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

0%
a, b, c, d
0%
c, d, a, b
0%
b, a, d, c
0%
d, c, a, b

Explanation

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

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

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

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

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

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

$y+xy=x$

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

$f^{-1}x=\dfrac{-x}{x-1}$

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

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

Let

$f$ be an injective function with domain

$\{x, y, z\}$ and range

$\{1,2,3\}$ such that exactly one of the follwowing statements is correct and the remaining are false :

${f}({x})=1,{f}({y})\neq 1$ ,

${f}({z})\neq 2$ ,

then the value of

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

0%
$x$
0%
$y$
0%
$z$
0%
none

Explanation

It gives three cases

Case (1) When

$f(x)=1$ is true.

In this case remaining two are false

$f(y)=1$ and

$f(z)=2$

This means

$x$ and

$y$ have the same image, so

$f(x)$ is not an injective, which is a contradiction.

Case (2) when

$f\left( y \right)\neq 1$ is true

$f\left( y \right)\neq 1$ is true then the remaining statements are false.

$\therefore f\left( x\right)\neq 1$ and

$f(z)=2$

i.e., both

$x$ and

$y$ are not mapped to

$1.$

So, either both associate to

$2$ or

$3,$

Thus, it is not injective.

Case (3) When

$f(z) \neq 2$ is true

$f(z) \neq2$ is true then remaining statements are fales

$\therefore$ If

$f(x) \neq1$ and

$f(y)=1$

But

$f$ is injective

Thus, we have

$f(x)=2, f(y)=1$ and

$f(z)=3$

Hence,

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

$f : R \rightarrow R$ 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%
$\displaystyle \frac{x}{\sqrt{1-x^2}}$
0%
$\displaystyle \frac{x}{\sqrt{1+x^2}}$
0%
$\displaystyle \frac{1-x^2}{\sqrt{1-x^2}}$
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

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

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

$=f[g(ab)]$

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

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

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

Explanation

$R_1={(1,3),(2,4),(3,5),(4,6),(5,7)}$

Since

$4$ and

$6$ do not belong to

$Y$

$(2,4),(4,6)∉R_1$

$R_1={(1,3),(3,5),(5,7)}⊂A×B$

Hence

$R_1$ is a relation but not a mapping as the elements

$2$ and

$4$ do not have any image.

$R_2$ : It is certainly a mapping and since every mapping is a relation, it is a relation as well.

$R_3$ : It is a relation being a subset of

$A×B$ but the elements

$1$ and

$3$ do not have a unique image and hence it is not mapping.

$R_4$ : It is both a mapping and a relation. Each element in A has a unique image. It is also one-one and onto mapping and hence a bijection

$f:R \rightarrow R$ 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%
$1, 2$
0%
$-1, 2$
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

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

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

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

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

$ABC\sim PQR$ , then AB:PQ =

0%
$AC:PB$
0%
$AC:PR$
0%
$AB:PR$
0%
$AC:RQ$

$f : R \rightarrow R$ 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
0%
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?

0%
$f:R\rightarrow R$ given by $f(x)={ 2x }^{ 2 }+1$ for all $\quad x\in R$
0%
$g:Z\rightarrow Z$ given by $g(x)={ x }^{ 4 }$ for all $\quad x\in R$
0%
$h:R\rightarrow R$ given by $h(x)={ x }^{ 3 }+4$ for all $\quad x\in R$
0%
$\phi :C\rightarrow C$ 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

0%
$f({ x }_{ 1 })\neq f({ x }_{ 2 })\$ for all ${ x }_{ 1 },{ x }_{ 2 }\in X$
0%
$f({ x }_{ 1 })=f({ x }_{ 2 })\Rightarrow { x }_{ 1 }={ x }_{ 2 }$ for all ${ x }_{ 1 },{ x }_{ 2 }\in X$
0%
${ x }_{ 1 }={ x }_{ 2 }\Rightarrow f({ x }_{ 1 })=f({ x }_{ 2 })$ for all ${ x }_{ 1 },{ x }_{ 2 }\in X$
0%
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

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

$b^2-4ac=0$

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

$f:R\rightarrow R$ 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,

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

Let R be the relation in the set N given by

$=\left \{ (a, b):a=b-2, b >6 \right \}$ . Choose the correct answer.

0%
$(2, 4)\in R$
0%
$(3, 8)\in R$
0%
$(6, 8)\in R$
0%
$(8, 7)\in R$

Explanation

$(a,b)∈R$ only if

$a=b−2$ and

$b>6$

$(6,8)∈R$ as

$6=8−2$ and

$8 >6$

Hence

$(c)$ is the correct alternative

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

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

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

Which of the following function is one-one?

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

$f:R^+\rightarrow A$ , Where

$A=\{x:-5<x<\infty\}$ be defined by

$f(x)=x^2-5$ . Then

$f^{-1}(7)=$

0%
only $-2{\sqrt3}$
0%
only $2{\sqrt3}$
0%
both options A and B
0%
none

Explanation

The value

$f^{-1}(7)$ will be the roots of the equation

$x^2-5= 7\Rightarrow x^2 = 12\Rightarrow x = \pm 2\sqrt{3}$

But only positive.

$\therefore x = 2\sqrt{3}$

If the function

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

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

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

0%
$\displaystyle \left ( \frac{1}{2} \right )^{x(x-1)}$
0%
$\displaystyle \frac{1}{2}\left ( 1+\sqrt{1+4\log_{3}x}\right)$
0%
$\displaystyle \frac{1}{2}\left ( 1-\sqrt{1+4\log_{3}x}\right)$
0%
not defined

Explanation

For the given domain, the function is one one.

Thus,

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

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

$lny=(x^{2}-x)ln2$

$log_{2}y=x^{2}-x$

$log_{2}y=(x-\dfrac{1}{2})^{2}-\dfrac{1}{4}$

$log_{2}y+\dfrac{1}{4}=(x-\dfrac{1}{2})^{2}$

$\dfrac{\pm 1}{2}(\sqrt{1+4log_{2}y})=x-\dfrac{1}{2}$

Since

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

Thus domain is

$[1,\infty)$

Hence domain is positive so negative sign will be ignored.

$So$

$x-\dfrac{1}{2}=\dfrac{1}{2}(\sqrt{1+4log_{2}y})$

$x=\dfrac{1}{2}[1+\sqrt{1+4log_{2}y}]$

Replacing x and y, gives us

$f^{-1}(x)=\dfrac{1}{2}[1+\sqrt{1+4log_{2}x}]$

If the function

$f:R\rightarrow R$ be such that

$\displaystyle f(x)=x-[x],$ where

$[y]$ denotes the greatest integer less than or equal to

$y$ , then

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

0%
$\displaystyle \frac{1}{x-[x]}$
0%
$\displaystyle [x]-x$
0%
not defined
0%
none of these

Explanation

$f\left( x \right) = x - \left[ x \right] = \left\{ x \right\}$ is not one-one function from R to R. Hence

${f^{ - 1}}\left( x \right)$ can not be defined.

$\displaystyle f:\left ( 3,4 \right )\rightarrow \left ( 0,1 \right )$ is defined by

$\displaystyle f\left ( x \right )=x-\left [ x \right ]$ where

$\displaystyle [x]$ denotes the greatest integer function then

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

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

Explanation

Note that

$\displaystyle f\left ( x \right )=x-\left [ x \right ] =\left\{ x \right\}$ . Hence every number

$x \in \left( {3,4} \right)$ is mapped to its fractional part.
Hence, the inverse of the function can be written by adding the integer part of the number i.e.,

$3$ to each of the fraction part.
Hence the function

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

Let

$\displaystyle f:(-\infty,1]\rightarrow (-\infty,1]$ such that

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

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

0%
$1+\sqrt{1-x}$
0%
$1-\sqrt{1-x}$
0%
$\sqrt{1-x}$
0%
none of these

Explanation

We have,

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

$y=x(2-x)$

$-y=x(x-2)$

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

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

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

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

Now

$f:(-\infty,1] \rightarrow (-\infty,1]$

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

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

Hence

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

$f(x)=ax+b$ and

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

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

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

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.

$\displaystyle f(x)=3x-5$ then

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

0%
$\displaystyle \frac{1}{3x-5}$
0%
$\displaystyle \frac{x+5}{3}$
0%
does not exist because $f$ is not one-one
0%
does not exist because $f$ is not onto

Explanation

$f(x)=3x-5$

Since

$f(x)$ is a purely linear function (straight line), it is therefore a one-one function.

Hence

$f(x)$ is revertible for all real

$x.$

$y=3x-5$

$y+5=3x$

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

Replacing

$x$ by

$y$ , gives us

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

$f$ 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

0%
$\displaystyle f\left ( x \right )=\sqrt{x}, g\left ( x \right )=\cos x$
0%
$\displaystyle f\left ( x \right )=x^{3}, g\left ( x \right )=x+1$
0%
$\displaystyle f\left ( x \right )=x-1, g\left ( x \right )=x^{2}+1$
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 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.

Let

$S$ be a set containing

$n$ elements. Then the total number of binary operations on

$S$ is

0%
$n^n$
0%
${2^n}^{2}$
0%
$n^{n^2}$
0%
$n^2$

Explanation

No of binary operations imply

$S\times S\rightarrow S$

$=n(S)^{[n(S\times S)]}$

$=n^{n\times n}$

$=n^{n^2}$

Let

$f: R\rightarrow R$ be defined by

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

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

0%
$\dfrac{1}{3}(x+4)$
0%
$\dfrac{1}{3}(x-4)$
0%
$3x+4$
0%
not defined

Explanation

$f(x)$ is invertible for all real values of

$x$
Hence

$y=3x-4$

$y+4=3x$

$\dfrac{y+4}{3}=x$

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

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

$a=-1$

A function

$y=f\left ( x \right )$ is invertible only when

0%
$y=f\left ( x \right )$ is monotonic increasing
0%
$y=f\left ( x \right )$ is bijective
0%
$y=f\left ( x \right )$ is monotonic decreasing
0%
invertible

Explanation

Invertible function is defined as, the function

$f$ applied to an input

$x$ gives a result of

$y$ , then applying its inverse function

$g$ to

$y$ gives the result

$x$ , and vice versa. i.e.,

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

$\Rightarrow g\left ( y \right )$ =

$x$ .
And bijective function has the same definition as that of an Invertible function

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.

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

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

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

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

Explanation

Let

$y = x+\cfrac{1}{x}$

$\Rightarrow x^2-yx+1=0\Rightarrow x = \cfrac{y+\sqrt{y^2-4}}{2}=f^{-1}(y)$
Hence

$f^{-1}(x) = \cfrac{x+\sqrt{x^2-4}}{2}$
Note: -ve term in step two is discarded using given domain and co-domain.

$f:R\rightarrow R$ is a function defined by

$f(x)=10x-7$ . If

$g=f^{-1}$ , then

$g(x)$ is equals

0%
$\dfrac{1}{10x-7}$
0%
$\dfrac{1}{10x+7}$
0%
$\dfrac{x+7}{10}$
0%
$\dfrac{x-7}{10}$

Explanation

$y=10x-7$

It is a one one function over the given domain.

$y+7=10(x)$

$x=\dfrac{y+7}{10}$

Hence

$f^{-1}$ will be

$g(x)$

$g(x)=\dfrac{x+7}{10}$

Hence, option 'C' is correct.

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%
$64$

$f:R\rightarrow R$ such that

$f(x)=log_3 x$ then

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

0%
$log x^3$
0%
$3^x$
0%
$3^{-x}$
0%
$3^{1/x}$

Explanation

$f(x)$ is invertible for all positive

$x$ .

$y=log_{3}(x)$

Taking antilogarithm on both sides, we get

$3^{y}=x$ .

Hence,

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

Hence, option 'B' is correct.

The inverse of

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

0%
$\displaystyle \frac{1}{6}\log_{10}\left ( \frac{1+x}{1-x} \right )$
0%
$\displaystyle \frac{1}{6}\log_{10}\left ( \frac{x}{1-x} \right )$
0%
$\displaystyle \frac{1}{6}\log_{e}\left ( \frac{1+x}{1-x} \right )$
0%
$\displaystyle \frac{1}{6}\log_{e}\left ( \frac{1-x}{1+x} \right )$

Explanation

$\displaystyle y=\frac{e^{3x}-e^{-3x}}{e^{3x}+e^{-3x}}$

$\Rightarrow \displaystyle y=\frac{e^{6x}-1}{e^{6x}+1}$

$\Rightarrow \displaystyle \frac{y+1}{y-1}=\frac{e^{6x}-1+e^{6x}+1}{e^{6x}-1-e^{6x}-1}$

$\Rightarrow \displaystyle \frac{y+1}{y-1}=\frac{2e^{6x}}{-2}$

$\Rightarrow \displaystyle \frac{1+y}{1-y}=e^{6x}$

Taking

$\log_e$ we get

$\displaystyle \log_{e}\left(\frac{1+y}{1-y}\right)=6x$

$\displaystyle x=\frac{1}{6}\left(\log_{e}\left(\frac{1+y}{1-y}\right)\right)$

Hence

$f^{-1}$

$\displaystyle =\frac{1}{6}\left(\log_{e}\left(\frac{1+x}{1-x}\right)\right)$

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

$\displaystyle Y= \left \{ 1,3,5,7,9 \right \}$ then which of the following sets are relation from

$X$ to

$Y$

0%
$\displaystyle R_{1}= \left \{ (x,a):a= x+2,x\in X,a\in Y \right \}$
0%
$\displaystyle R_{2}= \left \{ (1,1),(2,1),(3,3),(4,3),(5,5) \right \}$
0%
$\displaystyle R_{3}= \left \{ (1,1),(1,3),(3,5),(3,7),(5,7) \right \}$
0%
$\displaystyle R_{4}= \left \{ (1,3),(2,5),(4,7),(5,9),(3,1) \right \}$

Explanation

${ R }_{ 1 }=\left\{ \left( 1,3 \right) ,\left( 2,4 \right) ,\left( 3,5 \right) ,\left( 4,6 \right) ,\left( 5,7 \right) \right\}$
since

$4$ and

$6$ do not belong to

$y$ .

$\therefore \left( 2,4 \right) ,\left( 4,6 \right) \notin { R }_{ 1 }$

$\therefore { R }_{ 1 }=\left\{ \left( 1,3 \right) ,\left( 3,5 \right) ,\left( 5,7 \right) \right\} \subset X\times Y$
Hence,

${ R }_{ 1 }$ is a relation but not a mapping as the elements

$2$ and

$4$ do not have any image.

${ R }_{ 2 }:$ It is certainly a mapping and since every mapping is a relation, it is a relation as well.

${ R }_{ 3 }:$ It is a relation being a subset of

$X\times Y$ .

${ R }_{ 4 }:$ It is both mapping and a relation. Each element in

$X$ has a unique image. It is also one-one and on-to mapping and hence a bijection.

CBSE Questions for Class 12 Commerce Maths Relations And Functions Quiz 3 - 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 3 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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