If $$f\left( x \right) = \frac{{x - 1}}{{x + 1}}$$, then $$f^{-1}\left( x \right)$$ is

$$\frac{{3f\left( x \right) + 1}}{{f\left( x \right) + 3}}$$

$$\frac{{f\left( x \right) + 1}}{{f\left( x \right) + 3}}$$

$$\frac{{f\left( x \right) + 3}}{{f\left( x \right) + 1}}$$

$$\frac{{f\left( x \right) + 3}}{{3f\left( x \right) + 1}}$$

MCQ Questions for Class 12 Commerce Maths Relations And Functions Quiz 8

Let

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

$g\left( x \right) = {2^x}$ . Then the solution of the equation

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

0%
$R$
0%
$\left\{ {0} \right\}$
0%
$\left\{ {0,\,2} \right\}$
0%
none

Explanation

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

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

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

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

Given:

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

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

Since bases are same we can equate the powers

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

$\Rightarrow\,{x}^{2}-2x=0$

$\Rightarrow\,x\left(x-2\right)=0$

$\Rightarrow\,x=0,2$

When

$x=0,\,f{g\left(x\right)}=g{f\left(x\right)}\Rightarrow\,{2}^{0}={2}^{0}=1$

When

$x=2,\,f{g\left(x\right)}=g{f\left(x\right)}\Rightarrow\,{2}^{2\times 2}={2}^{{2}^{2}}\Rightarrow\,16=16$

Hence

$x=\left\{0,2\right\}$

Let

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

$f(x)=\begin{cases} -1\quad if\quad x<0 \\ 0\quad \quad if\quad x=0 \\ 1\quad \quad if\quad x>0 \end{cases}$ , then

$\forall \:x,fog(x)$ equals

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

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

$a > 0$ and

$n$ is a positive integer then

$( f o f ) ( x )$ is

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

Explanation

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

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

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

$fof(n)=x$

1099941_1179199_ans_0ac1100e807c473ea490bfc05967bd4e.jpg

The inverse of the function

$f(x)=\dfrac {e^{x}-e^{-x}}{e^{x}+e^{-x}}+2$ is given by

0%
$\log { e{ \left( \dfrac { x-1 }{ x+1 } \right) }^{ -2 } }$
0%
$\log { e{ \left( \dfrac { x-2 }{ x+1 } \right) }^{ 1/2 } }$
0%
$\log { e{ \left( \dfrac { x }{ 2-x } \right) }^{ 1/2 } }$
0%
$\log { e{ \left( \dfrac { x-1 }{ 3-x } \right) }^{ 1/2 } }$

$f:R \rightarrow R$ such that

$f(x)=\ell n(x+\sqrt {x^{2}+1})$ . Another function

$g(x)$ is defined such that

$gof(x)=x\ \forall\ x \in\ R$ . Then

$g(2)$ is -

0%
$\dfrac {e^{2}+e^{-2}}{2}$
0%
$e^{2}$
0%
$\dfrac {e^{2}-e^{-2}}{2}$
0%
$e^{-2}$

Explanation

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

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

$g(2)=?$

$(x+ \sqrt{x^{2}+1})=y$

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

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

$e^{2y}-2e^{y}x-1=o \Rightarrow$

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

$=\dfrac{(e^{y}- e^{-y})}{2}$

$\therefore g(x) = \dfrac{e{^y}- e^{-y}}{2} \Rightarrow g (2)= \dfrac{e^{2}-e^{-2}}{2}$

Let

$f:R\rightarrow R$ is a function satisfying

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

$f(20-x)=f(x)\forall x\in R$
If

$f(0)=5$ then the minimum possible no. of values of

$x$ satisfying

$f(x)=5$ for

$x=[0.,70]$ , is

0%
$21$
0%
$12$
0%
$11$
0%
$22$

Explanation

Given,

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

$\therefore f(x)$ is symmetric about

$x = 2$

$f(20 - x) = f(x) ..(ii)$

$x \rightarrow x + 10$

$f(10 - x) = f(10 + x)$

$f(x)$ is symmetric about

$x = 10$

$x \rightarrow x + 2$

$f(18 - x) = f(x + 2) .. (ii)$

$f(2 - x) = f(18 - x)$

$x \rightarrow -x$

$f(2 + x) = f(18 + x)$

$\therefore x + 2 \rightarrow x$

$\therefore f(x) = f(x + 16)$

$f(x)$ has period =

$16$

$f(x) = 5 , \, x \in [0, 170]$

put

$x = 2$ in (i)

$f(0) = f(4)$

put

$x = 4$ in (ii)

$f(4) = f(16)$

when

$x \in [0, 16] \rightarrow f(x)$ has two solution

when

$x \in [0, 160]$ has

$20$ solution.

When

$x in [0, 170]$ has

$'21'$ solution

1238432_1195475_ans_7bd229d0e548486197d5b8b994113248.JPG

All values of a for which f : R

$\to R$ defined by f(x)=

${x^3} + a{x^2} + 3x + 100$ is a one one functions, are

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

Explanation

$f ^ { \prime } ( x ) = 3 x ^ { 2 } + 2 a x + 3$

$f ^ { \prime } ( x ) = 0$ and f ( x ) > 0

$f ^ { \prime } ( x ) < 0$

$3 x ^ { 2 } + 2 a x + 3 = 0$

$x = \dfrac { - 2 a \pm \sqrt { 4 a ^ { 2 } - 36 } } { 6 }$

$4 a ^ { 2 } < 36$

$a \in ( - 3,3 )$

Let

$A = \{ 1,2,3 \}$ . Which of the following functions on A is invertible?

0%
$f = \{ ( 1,1 ) , ( 2,1 ) , ( 3,1 ) \}$
0%
$f = \{ ( 1,2 ) , ( 2,3 ) , ( 3,1 ) \}$
0%
$f = \{ ( 1,2 ) , ( 2,3 ) , ( 3,2 ) \}$
0%
$f = \{ ( 1,1 ) , ( 2,2 ) , ( 3,1 ) \}$

Explanation

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

$\therefore$

$A$ is invertible when obviously

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

$f ( x ) = \sin ^ { - 1 } ( \sin x ) + \cos ^ { - 1 } ( \sin x ) \text { and } \phi ( x ) = f ( f ( f ( x ) ) )$ then

$\phi ^ { \prime } ( x )$

0%
1
0%
$\sin x$
0%
0
0%
none of these

Explanation

$f(x)=\sin ^{ -1 }{ \left( \sin { x } \right) } +\cos ^{ -1 }{ \left( \sin { x } \right) } =\cfrac { \pi }{ 2 } \left[ \because \sin ^{ -1 }{ \theta } +\cos ^{ -1 }{ \theta } =\cfrac { \pi }{ 2 } \right]$

$f(f(x))=f\left(\cfrac { \pi }{ 2 }\right )=\cfrac { \pi }{ 2 }$

$\phi (x)=f(f(f(x)))=f\left(\cfrac { \pi }{ 2 }\right )=\cfrac { \pi }{ 2 }$

$\phi (x)=\cfrac { \pi }{ 2 }$

$\phi '(x)=0$

$f\left( x \right) = 3x + 2$ ,

$g\left( x \right) = {x^2} + 1$ ,then the values of

$\left( {f_og} \right)\left( {{x^2} - 1} \right)$

0%
$3{x^4} - 6{x^2} + 8$
0%
$3{x^4} + 3x + 4$
0%
$6{x^4} + 3{x^2} - 2$
0%
$6{x^4} + 3{x^2} + 2$

Explanation

$f\left(x\right)=3x+2$

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

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

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

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

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

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

$=3{x}^{4}-6{x}^{2}+3+5$

$=3{x}^{4}-6{x}^{2}+8$

Let A = {1,2,3,4,5} and B={1,2,3,4,5}. If

$f:A\rightarrow B$ is an one-one function and

$f(x)=x$ holds only for one value of

$x\epsilon \{ 1,2,3,4,5\} ,$ then the number of such possible function is

0%
$120$
0%
$36$
0%
$45$
0%
$44$

Difference between the greatest and the least values of the function

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

$[1, e^{2}]$ is

0%
$2$
0%
$e$
0%
$e^{2}$
0%
$1$

The function

$f :\left[-\dfrac{1}{2}, \dfrac{1}{2}\right]\rightarrow \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ defined by

$f(x)=\sin^{-1}(3x-4x^{3})$ is

0%
both one-one and onto
0%
onto but not one-one
0%
one-one but not onto
0%
neither one-one nor onto

Explanation

$f:[-\cfrac{1}{2},\cfrac{1}{2}]\rightarrow[-\cfrac{\pi}{2},\cfrac{\pi}{2}]\\f(x)=\sin^{-1}(3x-4x^3)$

Range of

$\sin^{-1}(3x-4x^3)$ is $$[-\cfrac{\pi}{2},\cfrac{\pi}{2}]$$

And

$-\cfrac{\pi}{2}\le\sin^{-1}(3x-4x^3)\le\cfrac{\pi}{2}$

$\Rightarrow \sin(\cfrac{-\pi}{2})\le3x-4x^3\le\sin\cfrac{\pi}{2}$

$\Rightarrow-1\le3x-4x^3\le1$

$3x-4x^3\le-1$

$\Rightarrow 4x^3-3x-1\le0$

$\Rightarrow x\le\cfrac{1}{2}$

and

$3x-4x^3\le1\Rightarrow 4x^3-3x+1\le0\\ \Rightarrow x\le-\cfrac{1}{2}\\ \therefore x\epsilon [-\cfrac{1}{2},\cfrac{1}{2}]\\f{x}=\sin^{-1}(3x-4x^3)\\ \Rightarrow f^1(x)=\cfrac{1}{\sqrt{1-(3x-4x^3)^2}}\times(3-12x^2)\\ \Rightarrow f^1(x)=3-12x^2\\ \quad=-(12x^2-3)\\x^2\le0\Rightarrow 12x^2\le0\Rightarrow 12x^2-3\le-3\\-3(12x^2-3)\le3\\ \therefore f^1(x)\le0$

$\therefore f^1(x)$ is a decreasing function.

Hence

$f(x)$ is one-one and range of function

$=$ its co-domain.

Hence it is both one-one and onto.

$f\left( x \right) = \frac{{x - 1}}{{x + 1}}$ , then

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

0%
$\frac{{f\left( x \right) + 1}}{{f\left( x \right) + 3}}$
0%
$\frac{{3f\left( x \right) + 1}}{{f\left( x \right) + 3}}$
0%
$\frac{{f\left( x \right) + 3}}{{f\left( x \right) + 1}}$
0%
$\frac{{f\left( x \right) + 3}}{{3f\left( x \right) + 1}}$

Let g be the inverse function of differentiable function f and

$G\left( x \right) =\frac { 1 }{ g\left( x \right) } if\quad f\left( 4=2 \right)$ and

$f'\left( 4 \right) =\frac { 1 }{ 16 }$ , then the value of

${ \left( G'\left( 2 \right) \right) }^{ 2 }$ equals to:

0%
1
0%
4
0%
16
0%
64

$f:( - 1,1) \to B$ , is a function defined by

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

$B$ when

$f(x)$ is both one-one and onto function.

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

Explanation

For

$x \epsilon (-1,1)$ , we have

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

Substituting

$x=\tan\theta$ in above equation.

Therefore,

$f(\tan \theta)=\tan^{-1}\left[\dfrac {2\tan \theta}{1-\tan^2\theta}\right]$

$=\tan^{-1}\tan (2\theta)=2\theta$

$=2\tan ^{-1}x$

Thus

$-\dfrac {\pi}{2}<\tan ^{-1}\left[\dfrac {2x}{1-x^2}\right]<\dfrac {\pi}{2}$

Thus option B is correct.

$f(x)=x^{3}+x^{2}f'(1)+xf''(2)+f'''(3)\ \forall x\ \epsilon \ R$ , then

$f(x)$ is

0%
one-one and onto
0%
one-one and into
0%
many-one and onto
0%
non-invertible

The multiplicative inverse of the product of the additive inverse of x+1 is ________________.

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

Let S be a non-empty set and P(S) be the power set of set S. Find the identity element for the union

$(\cup)$ as a binary operation on

$P(S)$ .

0%
$\phi$
0%
$1$
0%
$0$
0%
None of these

Explanation

We observe that

$A \cup \phi = A = \phi \cup A$ for every subset A of set S.

$A \cup \phi = A =\phi \cup A$ for all

$A \in P(S)$

$\phi$ is the identity element for union

$(\cup)$ on

$P(S)$ .

$\begin{bmatrix} \sin { \left( \dfrac { \pi }{ 2 } \right) } & \cos { \left( \dfrac { \pi }{ 3 } \right) } \\ 2\tan { \left( \dfrac { \pi }{ 4 } \right) } & 2k \end{bmatrix}$ is not invertible, then

$k=$

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

Explanation

We have,

$\begin{bmatrix} \sin { \left( \dfrac { \pi }{ 2 } \right) } & \cos { \left( \dfrac { \pi }{ 3 } \right) } \\ 2\tan { \left( \dfrac { \pi }{ 4 } \right) } & 2k \end{bmatrix}$

Since,

$|A| = 0$ if it is not invertible,

$1(2k) - ((\cos 60) ( 2 \tan 45 )) = 0$

$2k - 1 = 0$

$k = \dfrac{1}{2}$

Hence, this is the answer.

Let

$N$ be the set of natural numbers and two functions

$f$ and

$g$ be defined as

$f,g : N\to N$ such that :

$f (n)= \begin{cases}\dfrac{n+1}{2}& \text{if n is odd}\\ \dfrac{n}{2} & \text{in n is even} \end{cases}$
and

$g(n) = n - (-1)^n$ . The fog is:

0%
Both one-one and onto
0%
One-one but not onto
0%
Neither one-one nor onto
0%
onto but not one-one

Explanation

$fx = \begin{cases} \dfrac{n+1}{2}& \text{n is odd} \\\dfrac{n}{2}& \text{n is even}\end{cases}$

$g(x) = n -(-1)^n \begin{cases}n+1; \text{n is odd} \\n-1; \text{n is even} \end{cases}$

$f(g(n)) = \begin{cases}\dfrac{n}{2}; & \text{n is even}\\ \dfrac{n+1}{2}; &\text{n is odd} \end{cases}$

$\therefore$ onto but not one-one

The numbers system which uses alphabets as well as numbers is-

0%
Binary numbers system
0%
Octal numbers system
0%
Decimal numbers system
0%
Hexadecimal numbers system

$f : R \rightarrow R , f ( x ) = e ^ { | x | } - e ^ { - x }$ is many-one into function.

0%
True
0%
False

Explanation

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

For every

$x$ , there will be different

$f(x)$

$\therefore$ It is a one - one function

$\therefore False$

Number of one-one functions from A to B where

$n(A)=4, n(B)=5$ .

0%
$4$
0%
$5$
0%
$120$
0%
$90$

Explanation

$n(A)=4$ and

$n(B)=5$

For one-one mapping
4 elements can be selected out of 5 elements of set B in

${}^5C_4$ ways
and then those 4 selected elements can be mapped with 4 elements of set A in

$4!$ ways.

Number of one-one mapping from

$A$ to

$B$

$={}^5C_4\times4!={}^5P_4=\dfrac{5!}{(5-4)!}=5!=120$

Let

$f(x)=x^ {135}+x^ {125}-x^ {115}+x^ {5}+1$ . If

$f(x)$ divided by

$x^ {3}-x$ , then the remainder is some function of

$x$ say

$g(x)$ . Then

$g(x)$ is an:-

0%
one-one function
0%
many one function
0%
into function
0%
onto function

$f : R \rightarrow R , f ( x ) = 2 x + | \sin x |$ is one-one onto.

0%
True
0%
False

Explanation

$f(x) = 2x + \left | sin x \right |$

for query x,there will be a defferent f(x)

$\therefore one-one$ & domain

$\rightarrow$ R

$\therefore onto$

$\therefore True.$

1233697_1503521_ans_de71ccb7a78a4dfb8f79a13c81a06fd5.JPG

$f : R \rightarrow R$ be given by

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

$fof(x)$ is

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

Explanation

Given,

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

Now,

$fof(x)$

$=f[f(x)]$

$=\left(3-[f(x)]^{3}\right)^{\dfrac{1}{3}},$

$=\left(3-[(3-x^3)^{\dfrac{1}{3}}]^{3}\right)^{\dfrac{1}{3}},$

$=\left(3-[(3-x^3)]\right)^{\dfrac{1}{3}},$

$=[x^3]^{\dfrac{1}{3}}$

$=x$ .

Let :

$R\rightarrow R$ defined as

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

0%
odd and one-one
0%
even and one-one
0%
many to one and even
0%
many to one and neither even nor odd

If a binary operation is defined

$a\star b=a^b$ then 2

$\star 2$ is equal to:

0%
$4$
0%
$2$
0%
$9$
0%
$8$

Explanation

$a \star b = {a}^{b} \quad \left( \text{Given} \right)$

Therefore,

$2 \star 2 = {2}^{2} = 4$

Hence the correct answer is

$4$ .

If is a binary operation such that

$a * b = a^2 + b^2$ then

$3 * 5$ is

0%
34
0%
9
0%
8
0%
25

Explanation

$a \star b = {a}^{2} + {b}^{2} \quad \left( \text{Given} \right)$

Therefore,

$3 \star 5 = {3}^{2} + {5}^{2} = 9 + 25 = 34$

Hence the correct answer is

$34$ .

Consider

$f(x) = \dfrac{x^2}{1 + x^3}$ ;

$g(t) = \displaystyle \int f(t) dt$ . If

$g(1) = 0$ then

$g(x)$ equals

0%
$\dfrac{1}{3} ln(1 + x^3)$
0%
$\dfrac{1}{3} ln\left ( \dfrac{1 + x^3}{2} \right )$
0%
$\dfrac{1}{2} ln\left ( \dfrac{1 + x^3}{3} \right )$
0%
$\dfrac{1}{3}l n\left ( \dfrac{1 + x^3}{3} \right )$

Let f :

$R\rightarrow R$ be a function defined by f(x) =

${ x }^{ 3 }+{ x }^{ 2 }+3x+sin\times .$ Then f is.

0%
one-one & onto
0%
one-one & into
0%
many one & onto
0%
many one & into

Explanation

$f(x)=x^{3}+x^{2}+3 x+\sin x$

$Suppose$

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

$x_{1}^{3}+x_{1}^{2}+3 x_{1}+\sin x_{1}=x_{2}+x_{2}+\sin x-\sin x_{2}$

$Equating\;above\;equations\;to \;0$

$x_{1}^{3}+x_{1}^{2}+3 x_{1}+\sin x_{1}=x_{2}+x_{2}+\sin x-\sin x_{2}=0$

$Solving\;them$

$It\; will\; not\; prove\; that$

$x_{1}=x_{2}$

$So,\;f(x)\;is\;one-one$

$Range\;f(x)=Real$

$Co-domain\;f(x)=Real$

$Range=Co-domain$

$Hence,\; f(x) \;is \;onto$

$So,\;option\; C \;is\;correct.$

A function

$f$ from the set of natural numbers to integers defined by

$f(n)=\begin{cases} \cfrac { n-1 }{ 2 } ,\quad \text{when n is odd} \\ -\cfrac { n }{ 2 } ,\quad \text{when n is even} \end{cases}$ is

0%
neither one-one nor onto
0%
one-one but not onto
0%
onto but not one-one
0%
one-one and onto both

Explanation

$one-one$ test of

$f:$

Let

$x_1$ and

$x_2$ be any two elements in the domain

$(N).$

$Case\,I:$ When both

$x_1$ and

$x_2$ are even.

Let

$f(x_1)=f(x_2)$

$\Rightarrow$

$\dfrac{-x_1}{2}=\dfrac{x_2}{2}$

$\Rightarrow$

$-x_1=-x_2$

$\Rightarrow$

$x_1=x_2$

$Case\,II:$ When both

$x_1$ and

$x_2$ are odd.

Let

$f(x_1)=f(x_2)$

$\Rightarrow$

$\dfrac{x_1-1}{2}=\dfrac{x_2-1}{2}$

$\Rightarrow$

$x_1-1=x_2-1$

$\Rightarrow$

$x_1=x_2$

$Case\,III:$ When

$x_1$ be even and

$x_2$ be odd.

Then,

$f(x_1)=\dfrac{-x_1}{2}$ and

$f(x_2)=\dfrac{x_2-1}{2}$

Then clearly,

$\Rightarrow$

$x_1\ne x_2$

$\Rightarrow$

$f(x_1)\ne f(x_2)$

From, all the cases, we can say that,

$f$ is one-one.

$onto$ test of

$f:$

Co-domain of

$f=Z=\{....,-3,-2,-1,0,1,2,3,...\}$

Range of

$f=\left\{...,\dfrac{-2-1}{2},\dfrac{-(-2)}{2},\dfrac{-1-1}{2},\dfrac{0}{2},\dfrac{1-1}{2},\dfrac{-2}{2},\dfrac{3-1}{2},...\right\}$

Range of

$f=\{...,-2,1,-1,0,0,-1,1,..\}$

$\Rightarrow$ Co-domain of

$f=$ Range of

$f$

$\therefore$

$f$ is onto.

Let

$f:[2,\infty )\rightarrow X$ be defined by

$f(x)=4x-{x}^{2}$ . Then,

$f$ is invertible, if

$X=$

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

Explanation

Since,

$f$ is invertible, range of

$f=$ Co-domain of

$f=X$

So, we need to find the range of

$f$ to find

$X.$

Foe finding the range, let

$f(x)=y$

$\Rightarrow$

$4x-x^2=y$

$\Rightarrow$

$x^2-4x=-y$

$\Rightarrow$

$x^2-4x+4=4-y$ [ Adding

$4$ on both sides ]

$\Rightarrow$

$(x-2)^2=4-y$

$\Rightarrow$

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

$\Rightarrow$

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

This is defined only when,

$4-y\ge 0$

$\Rightarrow$

$y\le 4$

$X=$ Range of

$f=(-\infty , 4]$

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

$\cfrac { 1 }{ 2 } (g\circ f(x))=2{ x }^{ 2 }-5x+2$ , then

$f(x)$ is equal to

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

Explanation

We will solve this problem by the trial and error method.

Let us check option

$A$ first.

$f(x)=2x-3$

$g(x)=x^2+x-2$ [ Given ]

$\Rightarrow$

$\dfrac{1}{2}(g\circ f)(x)=g[f(x)]$

$=\dfrac{1}{2}g(2x-3)$

$=\dfrac{1}{2}[(2x-3)^2+(2x-3)-2]$

$=\dfrac{1}{2}[4x^2+9-12x+2x-3-2]$

$=\dfrac{1}{2}[4x^2-10x+4]$

$=2x^2-5x+2$

The given condition is satisfied by

$A.$

$g(x)=x^2+x-1$ and

$(gof)(x)=4x^2-10x+5$ , then

$f\left(\dfrac{5}{4}\right)$ is equal to:

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

Explanation

$g(x)=x^2+x-1$

$g\left(f\left(\dfrac{5}{4}\right)\right)=4\left(\dfrac{5}{4}\right)^2-10\dfrac{5}{4}+5=-\dfrac{5}{4}$

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

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

$f^2\left(\dfrac{5}{4}\right)+f\left(\dfrac{5}{4}\right)+\dfrac{1}{4}=0$

$\left(f\left(\dfrac{5}{4}\right)+\dfrac{1}{2}\right)^2=0$

$\boxed{f\left(\dfrac{5}{4}\right)=\dfrac{-1}{2}}$

The inverse function of

$f(x) = \dfrac{8^{2x}-{8^{-2x}}} {8^{2x}+8^{-2x}}\in (-1, 1)$ , is ________.

0%
$\dfrac{1}{4} \log_e\left(\dfrac{1-x}{1+x}\right)$
0%
$\dfrac{1}{4} \log_e\left(\dfrac{1+x}{1-x}\right)$
0%
$\dfrac{1}{4} (\log_e)\log_e\left(\dfrac{1-x}{1+x}\right)$
0%
$\dfrac{1}{4} (\log_e)\log_e\left(\dfrac{1+x}{1-x}\right)$

Explanation

$f(x) = \dfrac{8^{2x} - 8^{-2x}}{8^{2x}+8^{-2x}}$

$y =\dfrac{8^{2x} - 8^{-2x}}{8^{2x}+8^{-2x}}$

$\dfrac{1+y}{1-y} = \dfrac{8^{2x}}{8^{-2x}}$

$8^{4x} = \dfrac{1+y}{1-y}$

$4x = \log_8\left(\dfrac{1+y}{1-y}\right)$

$x = \dfrac{1}{4} \log_8\left(\dfrac{1+y}{1-y}\right)$

$f^{-1}(y) = \dfrac{1}{4} \log_8\left(\dfrac{1+y}{1-y}\right)$

$\therefore \boxed {f^{-1}(x) = \dfrac{1}{4} \log_8 \left(\dfrac{1+x}{1-x}\right)}....Answer$

Hence option

$'B'$ is the answer.

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

$f(f(x))$ is equal to

0%
$x$
0%
$0$
0%
$-x$
0%
$1$

Explanation

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

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

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

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

$=\dfrac{2x}{2}$

$=x$

Let

$f : x \rightarrow y$ be such that

$f(1) = 2$ and

$f(x + y) = f(x) f(y)$ for all natural numbers x and y. If

$\displaystyle \sum_{k= 1}^n f(a + k) = 16 (2^n - 1)$ , then a is equal to

0%
$3$
0%
$4$
0%
$5$
0%
$6$
0%
$7$

Explanation

We have,

$f(1)=2$ and

$f(x+y)=f(x).f(y)$

Now,

$f(2)=f(1+1)=f(1).f(1)=2.2=2^2$

$f(3)=f(2+1)+f(2).f(1)=2^2.2=2^3$

and so on

$\therefore f(x)=2^n$ ......(i)

Now, we have

$\displaystyle \sum^n_{k=1}f(a+k)=16(2^n-1)$

$\Rightarrow f(a+1)+f(a+2)+----f(a+n)=16(2^n-1)$

$\Rightarrow f(a).f(1)+f(a).f(2)+-----f(a).f(n)=16(2^n-1)$

$\Rightarrow f(a)=[f(1)+f(2)+---f(n)]=16(2^n-1)$

$\Rightarrow f(a)[2+2^2+----+2^n]=16(2^n-1)$

$\Rightarrow f(a).[2\dfrac{(2^n-1)}{2-1}]=16(2^n-1)$

$\Rightarrow 2f(a).(2^n-1)=16.(2^n-1)\Rightarrow f(a)=8$

$\Rightarrow 2^a=8[\because f(x)=2^n\Rightarrow f(a)=2^a]$

$\Rightarrow 2^a=2^3=a=3$

$f(x)=\dfrac{(4x+3)}{(6x-4)}, x\neq \dfrac{2}{3}$ then

$(f o f)(x)=?$

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

$f(x)=\sqrt[3]{3-x^3}$ then

$(f o f)(x)=?$

0%
$x^{1/3}$
0%
$x$
0%
$(1-x^{1/3})$
0%
None of these

Let

$f : R \rightarrow R : f(x) =x +1$ and

$g : R \rightarrow R : g(x) = 2x -3$ .
Find

$(f +g) (x)$ .

0%
$3x -2$
0%
$4x -5$
0%
$3x -4$
0%
$2x -3$

Explanation

Given ,

$f(x)=x+1,g(x)=2x-3$

$\implies (f+g)x=f(x)+g(x)=x+1+2x-3=3x-2$

$\displaystyle f(x) = | x - 2 |$ and

$g(x) = fof\,(x)$ , then for

$x > 20 , {g}\,'(x) =$

0%
$2$
0%
$1$
0%
$3$
0%
None of these

$\displaystyle {f}'(x) = g\,(x)$ and

$\displaystyle {g}'(x) = - f\,(x)$ for all

$x$ and

$f\,(2) = 4 = {f}'(2)$ then

$\displaystyle f^{2}\,(19) + g^{2} \,(19)$ is

0%
$16$
0%
$32$
0%
$64$
0%
None of these

The value of f(0), so that the function
f(x) =

$\dfrac{2x-sin^{-1}x}{2x+tan^{-1}x}$ is continuous at each point in its domain, is equal to

0%
2
0%
1/3
0%
2/3
0%
-1/3

Explanation

The function f is clearly continuous at each point in its domain except possibly at x=0 Given that f(x) is continuous at x=0
Therfore,f (0) =

$\underset{x\rightarrow 0}{lim}f(x)$

$=\underset{x\rightarrow 0}{lim}\dfrac{2x-sin^{-1}x}{2x+tan^{-1}x}$

$= \underset{x\rightarrow 0}{lim}\dfrac{2-\frac{(sin^{-1}x)}x}{2+\frac{(tan^{-1}x)}x}$

$\dfrac{2-1}{2+1}=\dfrac13$

let

$f(x) = sin^2 x/2 + cos ^2 x/2$ and

$g(x) = sec^2 x - tan ^2 x.$ The two functions are equal over the set

0%
$\phi$
0%
$R$
0%
$R-{ x:x (2n+1) \frac{\pi}{2}, n\in1}$
0%
None of these

Let

$f(n)$ denote the number of different ways in which the positive integer

$n$ can be expressed as the sum of

$1s$ and

$2s$ . For example,

$f(4) = 5$ , since

$4 = 2 + 2 = 2 + 1 + 1 = 1 + 2 + 1 = 1 + 1 + 2 = 1 + 1 + 1 + 1$ . Note that order of

$1s$ and

$2s$ is important.

$f : N\rightarrow N$ is

0%
One-one and onto
0%
One-one and into
0%
Many-one and onto
0%
Many-one and into

Explanation

$6 = 0(2) + 6(1) = 1(2) + 4(1) = 2(2) + 2(1) = 3(2) + 0(1)$

Number of $2s$	Number of $1s$	Number of permutations
$0$	$6$	$1$
$1$	$4$	$\dfrac {5!}{4!} = 5$
$2$	$2$	$\dfrac {4!}{2!2!} = 6$
$3$	$0$	$\dfrac {3!}{3!} = 1$
		$Total = 13$

$\therefore f(6) = 13$

Now,

$f(f(6)) = f(13)$

Number of $1s$	Number of $2s$	Number of permutations
$13$	$0$	$1$
$11$	$1$	$\dfrac {12!}{11!} = 12$
$9$	$2$	$\dfrac {11!}{9!2!} = 55$
$7$	$3$	$\dfrac {10!}{7!3!} = 120$
$5$	$4$	$\dfrac {9!}{5!4!} = 126$
$3$	$5$	$\dfrac {8!}{3!5!} = 56$
$1$	$6$	$\dfrac {7!}{6!} = 7$
		$Total = 377$

$\therefore f(f(6)) = f(13) = 377$

$f(1) = 1(1)$

$f(2) = 2 (1, 1$ or

$2)$

$f(3) = 3(1, 1, 1\ or\ 2, 1\ or\ 1, 2)$

$f(4) = 5$ (explained in the paragraph)

By taking higher value of

$n$ in

$f(n)$ , we always get more value of

$f(n)$ . Hence,

$f(x)$ is one-one. Clearly,

$f(x)$ is into.

The function

$f(x)= \dfrac{(3^{x}-1^{})^2}{\sin x. \ln(1+x)}, x\neq 0$ , is continuous at

$x=0$ . Then the value of

$f(0)$ is

0%
2log 3
0%
$(\log_{e}3)^{2}$
0%
$\log_{e} 6$
0%
None of these

Explanation

Given f(x) is continuous at

$x=0$

$\Rightarrow \underset{x\rightarrow 0}{\lim}f(x)=f(0)$

$\Rightarrow \underset{x\rightarrow 0}{\lim}\dfrac{(3^{x}-1)^{2}}{\sin x\ln(1+x)}=f(0)$

$\Rightarrow f(0)=\underset{x\rightarrow 0}{\lim} \dfrac{\bigg({\dfrac{3^x-1}{x}}\bigg)^2}{\dfrac{\sin x}{x}\dfrac{\ln(1+x)}{x}}=$

$(\log_e3)^{2}$

Let f(x)= max { 1+sinx, 1, 1 -cosx},

$x \epsilon [0, 2 \pi]$ and g(x)= max {1, |x-1|}

$x \epsilon R$ , then

0%
g(f(0))=1
0%
g(f(1))=1
0%
f(f(1))=1
0%
f(g(0))=1+sin1

If f:

$R\rightarrow R$ be given by

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

$a_n = A + Bx$ , then which of the following is not true?

0%
A + B + 1 = $2^{2n + 1}$
0%
| A - B| = 1`
0%
$lim_{n \to \infty} \dfrac{A}{B} = -1$
0%
None of these

CBSE Questions for Class 12 Commerce Maths Relations And Functions Quiz 8 - 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 8 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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