If $$f(x) = x^{2}$$, $$-1\leq x \leq 4$$ ,$$ g(x) = sec^{-1}x$$, $$x\geq 1$$ then

Range of gof(x) is $$[0, sec^{-1}16]$$

Domain of gof(x) is $$[1, 4] \cup {-1}$$

Domain of gof(x) is $$[1, 4] $$

Range of fog(x) is $$\left ( 0,\frac{\pi^{2}}{4} \right )$$

If $$\displaystyle f\left ( x \right )= \frac{3x+2}{5x-3}$$ , then

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

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

$$\displaystyle f\left ( f{}'\left ( x \right ) \right )=f'\left ( x \right )$$

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

MCQ Questions for Class 12 Commerce Maths Relations And Functions Quiz 9

The graph of y = g(x) in its domain is broken at

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

Let

$g(x) = f(x) - 1$ . If

$f(x) + f(1 - x) = 2 \space \forall \space x \space \epsilon \space R$ , then

$g(x)$ is symmetrical about

0%
Origin
0%
The line $x = \frac{1} {2}$
0%
The point $\left (1, 0 \right )$
0%
The point $\left (\frac {1} {2}, 0 \right )$

Which of the following is not true about

$h_1(x)$ ?

0%
It is periodic function with period $\pi$
0%
Range is [0,1]
0%
Domain if R
0%
None of these

g(f(x)) is not defined if

0%
$a\space \epsilon ( 10, \infty) b\space \epsilon (5, \infty)$
0%
$a\space \epsilon ( 4, 10) b\space \epsilon (5, \infty)$
0%
$a\space \epsilon ( 10, \infty) b\space \epsilon (0, 1)$
0%
$a\space \epsilon ( 4, 10) b\space \epsilon (1, 5)$

Let

$f(x)$ and

$g(x)$ be differentiable for

$0\times < 1$ such that

$f(0)=0, g(0), f(1)=6$ . Let there exist a real number

$c$ in

$(0,1)$ such that

$f'(c)=2g'(c)$ , then the value of

$g(1)$ must be

0%
$1$
0%
$3$
0%
$2.5$
0%
$-1$

For set

$A, B$ and

$C$ , let

$f:A\to B, g:B\to C$ be functions such that

$gof$ is surjective.
Then

$g$ is surjective function.

0%
True
0%
False

Explanation

Suppose that

$g\circ f$ is surjective.

Let

$z∈C$ .

Then since

$g\circ f$ is surjective, there exists

$x\in A$ such that

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

Therefore if we assume

$y=f(x)\in B$ ,

then

$g(y) =z$ .

Thus

$g$ is surjective

Let

$A$ be a finite set. Then, each injective function from

$A$ into itself is not surjective.

0%
True
0%
False

For set

$A, B$ and

$C$ , let

$f:A\to B, g:B\to C$ be functions such that

$gof$ is Injective.
Then

$f$ is injective.

0%
True
0%
False

Explanation

Suppose that

$g\circ f$ is injective;
we show that

$f$ is injective.
To this end, let

$x_1, x_2\in A$
and suppose that

$f(x_1) =f(x_2)$ .
Then

$(g \circ f)(x_1) =g(f(x_1)) =g(f(x_2)) = (g\circ f)(x_2).$
But since

$g\circ f$ is injective,
this implies that

$x_1=x_2.$
Therefore

$f$ is injective.

If g is the inverse of function

$f$ and

$f'(x) = \frac{1}{1 + x}$ , then the value of g'(x) is equal to:

0%
$1 + x^7$
0%
$\frac{1}{1 + [g(x)]^7}$
0%
$1 + [g(x)]^7$
0%
$7x^6$

Explanation

Since g is the inverse of

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

$\therefore f[f^{-1}(x)] = f [g(x)] = x$

$\therefore f'[g(x)] \cdot \frac{d}{dx} [g(x)] = 1$

$\therefore f'[g(x)] \times g'(x) = 1$

$\therefore g'(x) = \frac{1}{f'[g(x)]}$ , where

$f'(x) = \frac{1}{1 + x^7}$

$\therefore g(x) = 1 + [g(x)]^7$

Computers use

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decimal system
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binary system
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base 5 system
0%
base 6 system

From

$] \dfrac{- \pi}{2} , \dfrac{- \pi}{2}[$ which of the following is one - one onto function defined in R

0%
$f(x) = \tan x$
0%
$f(x) = \sin x$
0%
$f(x) = \cos x$
0%
$f(x) = e^{x} + r^{-x}$

Which of the following in one -one function defined from R to R

0%
$f(x) = |x|$
0%
$f(x) = \cos x$
0%
$f(x) =e^{x}$
0%
$f(x) = x^{2}$

Subtraction is an operation on

$Z$ , which is

0%
commutative and associative
0%
associative but not commutative
0%
neither commutative and nor associative
0%
commutative but not associative

Explanation

Let

$a, b \in z$

Commutativity: As we know,

$a-b \neq b-a$

Associativity :

$(a-b)-c\neq a-(b-c)$

So, option (c) is correct.

Let

$f: [ -1,1] \rightarrow [0,2]$ be a linear function which is onto then f(x) is/are

0%
$1 - x$
0%
$1 + x$
0%
$x - 1$
0%
$x - 2$

$n \geq 2$ then the number of surjections that can be defined from

$\{1, 2, 3, ....... n\}$ onto

$\{1, 2\}$ is

0%
$2n$
0%
$^nP_2$
0%
$2^n$
0%
$2^{n}-2$

Explanation

Let

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

$B=\{1,2\}$ .

Therefore, total number of mappings from

$A$ to

$B$ is

$2^n$ of which two functions

$f(x)=1$ for all

$x\in A$ and

$g(x)=2$ for all

$x\in A$ are not surjective.
Thus, the total number of surjections from

$A$ to

$B$ is

$2^n-2$ .

Let

$f : X\rightarrow Y, f (x) = \sin x+ \cos x + 2\sqrt2$ is invertible, then

$X\rightarrow Y$ is

$/$ are

0%
$\left [\displaystyle \frac{\pi }{4},\frac{5\pi}{4} \right ]\rightarrow \left [ \sqrt{2},3\sqrt{2} \right ]$
0%
$\left [ -\displaystyle \frac{\pi }{4},\frac{3\pi}{4} \right ]\rightarrow \left [ \sqrt{2},3\sqrt{2} \right ]$
0%
$\left [\displaystyle \frac{\pi }{4},\frac{3\pi}{4} \right ]\rightarrow \left [ \sqrt{2},3\sqrt{2} \right ]$
0%
$\left [ -\displaystyle \frac{3\pi }{4},\frac{\pi}{4} \right ]\rightarrow \left [ \sqrt{2},3\sqrt{2} \right ]$

Explanation

Let

$f :$

$X\rightarrow Y,f (x) = \sin x \cos x + 2\sqrt2$

$f(x)=\sqrt{2}\sin\left ( x+\dfrac{\pi}{4} \right )+2\sqrt{2}$ or

$f(x)=\sqrt{2}\cos\left ( x-\dfrac{\pi}{4} \right )+2\sqrt{2}\Rightarrow Y=[\sqrt2,3\sqrt2]$
and

$X=[-\dfrac{3\pi}{4},\dfrac{\pi}{4}]$ or

$[\dfrac{\pi}{4},\dfrac{5\pi}{4}].$

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

$-1\leq x \leq 4$ ,

$g(x) = sec^{-1}x$ ,

$x\geq 1$ then

0%
Domain of gof(x) is $[1, 4] \cup {-1}$
0%
Domain of gof(x) is $[1, 4]$
0%
Range of gof(x) is $[0, sec^{-1}16]$
0%
Range of fog(x) is $\left ( 0,\frac{\pi^{2}}{4} \right )$

Explanation

gof(x) exists when

$f(x) \geq 1$ and

$x \epsilon [-1, 4]$

$x^{2}\geq 1\Rightarrow X \epsilon (-\infty,-1]\cup[1,\infty)$

$\Rightarrow X \epsilon [1,4]\cup {-1}$
Range of

$gof(x) = [0, sec^{-1}16]$

$f(x) = x^3- 4x + 1 \forall x > 0$ and

$g(x)$ is image of

$f(x)$ with respect to line

$y = x$ then

0%
$g(1) = 2$
0%
$g'(1)=\displaystyle \frac{1}{8}$
0%
$f''(g(1))=12$
0%
$g''(1)=-\displaystyle \frac{3}{16}$

Explanation

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

$\Rightarrow f(g(x))=x\Rightarrow g'(x)=\frac{1}{f'(g(x))}$

$g'(1)=\frac{1}{f'(g(1))}=\frac{1}{f'(2)}=\frac{1}{8}$

$g"(x)=-\frac{f"g(x).g'(x)}{f'g(x)}^2$

$g"(1)=-\frac{f"(2).g'(1)}{f'g(1)}=\frac{-12\left ( \frac{1}{8} \right )}{8}=\frac{-3}{16}$

Consider the function

$f (x)=\dfrac{x}{x-1}.$ Which of the following statements are correct ?

0%
$f$ has the same domain and range
0%
$f$ has its own inverse
0%
$f$ is not injective
0%
$f$ is neither odd nor even

Explanation

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

$(-\infty,1)\bigcup(1,\infty)$
Now the inverse for the given domain will be

$y(x-1)=x$

$xy-y=x$

$xy-x=y$

$x(y-1)=y$

$x=\dfrac{y}{y-1}$
Hence range is

$(-\infty,1)\bigcup(1,\infty)$
Hence

$f(x)$ is the inverse of itself.
Also

$f(x)+f(-x)\neq0$ and

$f(x)\neq f(-x)$
Thus

$f(x)$ is not an even or odd function.
Hence, options 'A', 'B' and 'D' are correct.

Let

$\displaystyle \mathrm{f}:\mathrm{R}\rightarrow \left[0,\frac{\pi}{2}\right)$ be defined by

$\mathrm{f}(\mathrm{x})=\mathrm{t}\mathrm{a}\mathrm{n}^{-1}(\mathrm{x}^{2}+\mathrm{x}+\mathrm{a})$ . Then the set of values of '

$\mathrm{a}$ ' for which

$\mathrm{f}$ is onto is

0%
$[0,\infty)$
0%
$[2, 1]$
0%
$\displaystyle \left\{ \frac{1}{4}\right\}$
0%
$\displaystyle \left[ \frac{1}{4}, \infty\right)$

Explanation

$f(x)=\tan^{-1}(x^2+x+a)$
For,

$f(x)$ to be onto, codomain should be exactly equal to range.
That is, range of funnction

$\displaystyle f(x) = \left[ 0,\frac { \pi }{ 2 } \right)$
So,

$0\leqslant \tan^{-1}(x^2+x+a)<\dfrac{\pi}{2}$
Now,

$\displaystyle { x }^{ 2 }+x+a={ \left( x+\dfrac { 1 }{ 2 } \right) }^{ 2 }+a-\dfrac { 1 }{ 4 }$
The above expression will take all real values from

$\left[0, \infty \right)$ , only if

$\displaystyle a = \dfrac{1}{4}$
Hence, only for

$a= \dfrac{1}{4}$ , the function is onto.

Let

$\displaystyle {f}({x})=\frac{{a}{x}+{b}}{{c}{x}+{d}}$ , then

$fof(x)={x}$ , provided

0%
$d = -a$
0%
$d = a$
0%
$a = b = c = d = 1$
0%
$a = b = 1$

Explanation

Given,

$f(x)=\dfrac{ax+b}{cx+d}$

$f(f(x))=\dfrac{a\left ( \dfrac{ax+b}{cx+d}\right )+b}{c\left ( \dfrac{ax+b}{cx+d} \right )+d}=x$

$=\dfrac{a(ax+b)+b(cx+d)}{c(ax+b)+d(cx+d)}=x$

$a(ax+b)+b(cx+d)=x[c(ax+b)+d(cx+d)]$

$(a^2+bc)x+(ab+bd)=(ac+cd)x^2+(bc+d^2)x$

matching the coefficients of

$x$ , we get,

$a^2+bc=bc+d^2$

$a^2=d^2$

$d=\pm a$

matching the coefficients of

$x^2$ , we get,

$ac+cd=0$

$\therefore d=-a$

matching constant terms, we get,

$ab+bd=0$

$d=-a$

Which of the following functions is/are injective map(s) ?

0%
$f(x)=x^2+2, x \in (-\infty,\infty)$
0%
$f(x)=|x+2|, x \in [-2,\infty)$
0%
$f(x)=(x-4)(x-5), x \in (-\infty,\infty)$
0%
$f(x)=\dfrac{4x^2 + 3x -5}{4+3x-5x^2}, x\in(-\infty, \infty)$

Explanation

The function

$f(x)=x^2 + 2, x \in (-\infty, \infty)$ is not injective as

$f(1)=f(-1)$ but

$1 \neq -1.$

The function

$f(x)=(x-4)(x-5), x \in (-\infty, \infty)$ is not one-one as

$f(4)=f(5)$ but

$4 \neq 5.$

The functin

$f(x)=\dfrac{4x^2 + 3x -5}{4 + 3x - 5x^2}, x \in (-\infty,\infty)$ is also not injective as

$f(1)=f(-1)$ but

$1 \neq -1$ .

For the function ,

$f(x)=-|x+2|, x\in [-2,\infty)$

Let

$f(x)=f(y),x,y \in [-2,\infty) \Rightarrow |x+2| = |y+2|$

$\Rightarrow x+2 = y+2$

$\Rightarrow x=y$
So ,

$f$ is an injective map.

Which of the following functions is not injective ?

0%
$f:R \rightarrow R, f(x)=2x+7$
0%
$f:[0,\pi]\rightarrow[-1,1],f(x)=\cos x$
0%
$f:\left [ -\dfrac{\pi}{2},\dfrac{\pi}{2} \right ]\rightarrow R, f(x)=2 \sin x +3$
0%
$f:R\rightarrow [-1,1],f(x)=\sin x$

Explanation

$f:R\rightarrow R, f(x)=2x+7$

$\dfrac{dy}{dx}=2>0 \rightarrow$ one-one (Injective)

$f:[0,\pi]\rightarrow [-1,1],f(x)=\cos x$

$\dfrac{dy}{dx}=\sin x=(+ve) \rightarrow$ one-one (Injective)

$f: \left[-\dfrac{\pi}2, \dfrac{\pi}2\right], f(x) = 2\sin x+3$

$\dfrac{dy}{dx}= 2\cos x=+ve \rightarrow$ one-one (Injective)

$f:R\rightarrow [-1,1],f(x)=\sin x$

$\dfrac{dy}{dx}=\cos x=+ve$ &

$-ve \rightarrow$ many one

Hence, option D.

Let

$f(x)=max\left\{1+\sin x,1,1-\cos x \right\}, x\in \left [ 0,2\pi \right ]$ and

$g(x)=max\left\{ 1,\left | x-1 \right |\right\},x\in R$ , then

0%
$g(f(0))=1$
0%
$g(f(1))=1$
0%
$f(g(1))=1$
0%
$f(g(0))=\sin 1$

Explanation

$f(x)=(max [1+sinx,1,1-cos(x)]$ for

$x\epsilon[0,2\pi]$

$f(0)$

$=(max(1+0,1,0))$

$=1$

$f(1)$

$=(max(1+sin(1)),1,1-cos1)$

$=1+sin(1)$

$g(x)=max(1,|x-1|)$
Hence,

$g(f(0))$

$=g(1)$

$=max(1,0)$

$=1$
And,

$g(f(1))$

$=max(1,|1+sin(1)-1|)$

$=max(1,|sin(1)|)$

$=1$

$g(0) = 1$

$f(g(0)) = 1+sin 1$

$f\left( x \right)=\sqrt { 3\left| x \right| -x-2 }$ and

$g(x)=\sin(x)$ , then the domain of the definition of

$f\circ g\left( x \right)$ is

0%
$\displaystyle \left\{ 2n\pi +\frac { \pi }{ 2 } \right\}$
0%
$\displaystyle \left( 2n\pi +\frac { 7\pi }{ 6 } ,2n\pi +\frac { 11\pi }{ 6 } \right)$
0%
$\displaystyle \left\{ 2n\pi +\frac { 7\pi }{ 6 } \right\}$
0%
$\displaystyle \left( 2n\pi +\frac { 7\pi }{ 6 } ,2n\pi +\frac { 11\pi }{ 6 } \right) \bigcup _{ n,m\in I }^{ }{ \left( 2n\pi +\frac { \pi }{ 2 } \right) }$

Explanation

We have

$\displaystyle f\left( x \right) =\sqrt { 3\left| x \right| -x-2 }$ and

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

$\therefore f\circ g\left( x \right) =\sqrt { 3\left| \sin { x } \right| -\sin { x } -2 }$ which is defined,

$3\left| \sin { x } \right| -\sin { x } -2\ge 0$

$\sin { x } >0,$ then

$2\sin { x } -2\ge 0\quad$

$\Rightarrow \sin { x } \ge 1$

$\displaystyle \Rightarrow \sin { x } =1\Rightarrow x=\frac { \pi }{ 2 }$

$\sin { x } <0,$ then

$\displaystyle -4\sin { x } -2\ge 0\Rightarrow -1\le -\frac { 1 }{ 2 }$

$\displaystyle \therefore x\in \left( 2n\pi +\frac { 7\pi }{ 6 } ,2n\pi +\frac { 11\pi }{ 6 } \right) \bigcup _{ n,m\in I }^{ }{ \left( 2n\pi +\frac { \pi }{ 2 } \right) } n,m\in I$

Let

$\displaystyle a > 1$ be a real number and

$\displaystyle f\left ( x \right ) = \log _{a} x^{2}$ for

$\displaystyle x > 0$ . If

$\displaystyle f^{-1}$ is the inverse function of

$f$ and

$b$ and

$c$ are real numbers then

$\displaystyle f^{-1} \left ( b+c \right )$ is equal to

0%
$\displaystyle f^{-1}\left ( b \right ) . f^{-1}\left ( c \right )$
0%
$\displaystyle f^{-1}\left ( b \right ) + f^{-1}\left ( c \right )$
0%
$\displaystyle \frac {1}{f\left ( b+c \right )}$
0%
$\displaystyle \frac {1}{f^{-1}\left ( b \right ) + f^{-1} \left ( c \right )}$

Explanation

$y=\log_{a}(x^2)$

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

$a^{y/2}=x$
Hence

$f^{-1}=a^{x/2}$

$f^{-1}(b+c)$

$=a^{(b+c)/2}$

$=a^{b/2}a^{c/2}$

$=f^{-1}(b)f^{-1}(c)$
Hence, option 'A' is correct.

Suppose

$f(x)=ax+b$ and

$g(x)=bx+a$ , where

$a$ and

$b$ are positive integers. If

$f\left ( g(50) \right )-g\left ( f(50) \right )=28$ then the product

$(ab)$ can have the value equal to

0%
$12$
0%
$48$
0%
$180$
0%
$210$

Explanation

$f(g(x))=a(bx+a)+b$

$=abx+a^2+b$

$g(f(x))=b(ax+b)+a$

$=abx+b^2+a.$
Therefore

$f(g(x))-g(f(x))$

$=abx+a^2+b-(abx+b^2+a)$

$=a^2-b^2+b-a$

$=(a-b)(a+b-1)$
Since it is a constant function.

$(a-b)(a+b-1)=28$

$(a-b)(a+b-1)=4(7)$
Hence

$a-b=4$

$a+b=8$

$a=6$ and

$b=2$
Hence

$ab=12$
Also

$(a-b)(a+b-1)=(1)(28)$

$a-b=1$

$a+b=29$

$a=15$ and

$b=14$
Therefore

$ab=14(15)=210$

Which one of the following functions is not one-one?

0%
$f:(-1,\infty )\rightarrow R$ given by $f(x)={ x }^{ 2 }+2x\quad$
0%
$g:(2,\infty )\rightarrow R$ given by $g(x)={ e }^{ { x }^{ 3 }-3x+2 }\quad$
0%
$h:R\rightarrow R$ given by $h(x)={ 2 }^{ { x }(x-1) }\quad$
0%
$\phi :(-\infty ,0)\rightarrow R$ given by $\phi (x)=\cfrac { { x }^{ 2 } }{ { x }^{ 2 }+1 }$

Explanation

$x(x-1)$ is not a one -one function in whole

$R$ .

This function has a minima at

$x=\dfrac12$ , therefore repeats value after

$x=\dfrac12$ .

We can also see that

$x(x-1) = 0$ for

$x=0,1$ and therefore

$e^{x(x-1)} =1$ for both

$x=0 \&\ 1$ , which rules it out as a one to one function.

$f:R\rightarrow \left [\dfrac {\pi}{6}, \dfrac {\pi}{2}\right ), f(x)=\sin^{-1}\left (\dfrac {x^2-a}{x^2+1}\right )$ is a onto function, then set of values of

$a$ is

0%
$\left \{-\dfrac {1}{2}\right \}$
0%
$\left [-\dfrac {1}{2}, -1\right )$
0%
$(-1, \infty)$
0%
none of these

Explanation

Given,

$f(x)$ is onto.

$\therefore \displaystyle \frac {\pi}{6}\leq \sin^{-1}\left (\frac {-x^2-a}{x^2+1}\right ) < \frac {\pi}{2}$

$\Rightarrow \displaystyle \frac {1}{2}\leq \frac {x^2-a}{x^2+1} < 1$

$\Rightarrow \displaystyle \frac {1}{2}\leq 1-\frac {(a+1)}{x^2+1} < 1, \forall x\epsilon R$

$\Rightarrow a+1 > 0$

$\Rightarrow a\in (-1, \infty)$
Hence, (c) is the correct answer.

$f (x) = x + 2, g (x) = 2 x +3,$ then find gof

0%
$2x -7$
0%
$7x + 2$
0%
$2x + 7$
0%
$7 + 2x$

Which of the function defined below are one-one function(s)?

0%
$f(x)=x+1,(x\geq-1)$
0%
$g(x)=x+\dfrac1x,(x\geq0)$
0%
$h(x)=x^2+4x-5,(x>0)$
0%
$f(x)=e^{-x},(x\geq0)$

Explanation

A function is called one-one if it is monotonic
A.

$f(x)= x+1)\Rightarrow f'(x) =1>0\Rightarrow f$ is monotonically increasing
B.

$g(x) =x+\dfrac{1}{x}\Rightarrow g'(x)=1-\dfrac{1}{x^2},$ which is -ve for

$0\le x<1$ and +ve for

$x>0\Rightarrow f$ is not monotonic
C.

$h(x) =x^2+4x-5\Rightarrow h'(x)=2x+4>0\forall x>0\Rightarrow f$ is monotonically increasing
D.

$f(x) =e^{-x}\Rightarrow f'(x) =-e^{-x}<0\forall x\le 0\Rightarrow f$ is monotonically decreasing

$f(x)=x^3+3x^2+4x+b \sin x+c \cos x, \forall x\in R$ is a one-one function, then the value of

$b^2+c^2$ is

0%
$\geq 1$
0%
$\geq 2$
0%
$\leq 1$
0%
none of these

Explanation

Here,

$f(x)=x^3+3x^2+4x+b \sin x+c \cos x$

$f'(x)=3x^2+6x+4+b \cos x-c \sin x$
Now, for

$f(x)$ to be one-one, the only possibility is

$f'(x)\geq 0, \forall x\in R$
ie,

$3x^2+6x+4+b \cos x-c \sin x\geq 0, \forall x\in R$
ie,

$3x^2+6x+4 \geq c \sin x-b \cos x, \forall x\in R$

As we are approximating lower bound for the function by a number instead of a function, hence we chose the number.
ie,

$3x^2+6x+4 \geq \sqrt {b^2+c^2}, \forall x\in R$
ie,

$\sqrt {b^2+c^2} \leq 3(x^2+2x+1)+1,\forall x\in R$

$\sqrt {b^2+c^2} \leq 3(x+1)^2+1, \forall x\in R$

$\sqrt {b^2+c^2} \leq 1, \forall x \in R$

$\Rightarrow b^2+c^2 \leq 1, \forall x \in R$
Hence, (c) is the correct answer.

$f(x)=2x+|x|, g(x)=\dfrac {1}{3}(2x-|x|)$ and

$h(x)=f(g(x))$ , then domain of

$\sin^{-1}\underset {\text {n times}}{\underbrace {(h(h(h(h.....h(x).....))))}}$ is

0%
$[-1, 1]$
0%
$\left [-1, -\dfrac {1}{2}\right ]\cup \left [\dfrac {1}{2}, 1\right ]$
0%
$\left [-1, -\dfrac {1}{2}\right ]$
0%
$\left [\dfrac {1}{2}, 1\right ]$

Explanation

Since,

$f(x)=\left\{\begin{matrix}2x+x, & x\geq 0\\ 2x-x, & x < 0\end{matrix}\right.=\left\{\begin{matrix}3x, & x\geq 0\\ x, & x < 0\end{matrix}\right.$
and

$g(x)=\dfrac {1}{3}\left\{\begin{matrix}2x-x, & x\geq 0\\ 2x+x, & x < 0\end{matrix}\right.=\left\{\begin{matrix}\dfrac {x}{3}, & x\geq 0\\ x, & x < 0\end{matrix}\right.$

$\therefore f(g(x))=\left\{\begin{matrix}3\left (\dfrac {x}{3} \right ) & x\geq0\\ x & x < 0\end{matrix}\right.$

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

$\therefore h(x)=x$

$\Rightarrow \sin^{-1} (h(h(h....h(x)....)))=\sin^{-1}x$

$\therefore$ Domain of

$\sin^{-1} (h(h(h(h.....h(x)....))))$ is

$[-1, 1]$
Hence, A is the correct answer.

Let

$f:{x, y, z}\rightarrow (a, b, c)$ be a one-one function. It is known that only one of the following statements is true:

(i)

$f(x)\neq b$
(ii)

$f(y)=b$
(iii)

$f(z)\neq a$

0%
$f=\{(x, a), (y, b), (z, c)\}$
0%
$f=\{(x, b), (y, a), (z, c)\}$
0%
$f=\{(x, b), (y, c), (z, c)\}$
0%
$f=\{(x, b), (y, c), (z, a)\}$

Explanation

When (i) is true, then

$f(x) \neq b, f(y) \neq b , f(z) = a$

$\Rightarrow$ Two ordered pair function is possible

$f(x) = a, f(y) = c, f(z) = a$ or

$f(x) = c, f(y) = a, f(z) = a$

But given

$f$ is one-one and only one such function is possible. Hence (i) can't be true.

When (ii) is true, then

$f(y) = b, f(z) =a , f(x) = b$ . This is also not possible.

Clearly if (iii) is true then it is satisfying every condition.
Hence ordered pair of

$f$ is

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

The inverse of the function

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

0%
$\displaystyle \log_{e}\left ( \frac{1+2x}{1-2x} \right )$
0%
$\displaystyle \frac{1}{4}\log_{e}\left ( \frac{1-x}{1+x} \right )$
0%
$\displaystyle \frac{1}{4}\log_{e}\left ( \frac{1+x}{1-x} \right )$
0%
an odd function

Explanation

$\displaystyle y= \dfrac{t-\dfrac{1}{t}}{t+\dfrac{1}{t}}$ where

$\displaystyle t= e^{2x}$

$\displaystyle \therefore y\left ( t^{2}+1 \right )= t^{2}-1$

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

$\displaystyle \Rightarrow \left ( e^{2x} \right )^{2}= \frac{1+y}{1-y}\left ( \therefore t= e^{2x} \right )$

$\displaystyle \Rightarrow e^{4x}= \frac{1+y}{1-y}$

$\displaystyle \Rightarrow 4x=\log _{e}\left ( \frac{1+y}{1-y} \right )$

$\displaystyle x= \frac{1}{4}\log _{e}\left ( \frac{1+y}{1-y} \right )$

$\displaystyle \therefore f^{-1}\left ( x \right )= \frac{1}{4}\log _{e}\left ( \frac{1+x}{1-x} \right )= p\left ( x \right )$

$\displaystyle \therefore p\left ( -x \right )= -p\left ( x \right )$

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

$\Rightarrow$ Thus

$f^{-1}(x)$ is an odd function.

Let

$\displaystyle f(x)=\begin{cases}x^{2} & \mbox{if} \quad0< x< 2\\2x-3 & \mbox{if} \quad2\leq x< 3 \\ x+2 & \mbox{if}\quad x\geq 3\end{cases}$ .
Then

0%
$\displaystyle f\left \{ f\left ( f\left ( \frac{3}{2} \right ) \right ) \right \}=f\left ( \frac{3}{2} \right )$
0%
$\displaystyle 1+f\left \{ f\left ( f\left ( \frac{5}{2} \right ) \right ) \right \}=f\left ( \frac{5}{2} \right )$
0%
$\displaystyle f\left \{ f(0) \right \}=f\left ( 1 \right )=1$
0%
none of these

Explanation

consider option A)

$\displaystyle f\left( {\frac{3}{2}} \right) = \frac{9}{4}$ as

$\displaystyle \frac{3}{2} < 2$
and

$\displaystyle f\left( {\frac{9}{4}} \right) =2 \times \frac{9}{4} - 3 = \frac{3}{2}$ as

$\displaystyle \frac{9}{4} > 2$ .
Therefore

$\displaystyle f\left( {f\left( {f\left( {\frac{3}{2}} \right)} \right)} \right) = f\left( {\frac{3}{2}} \right)$ .
Hence option A is correct answer.

Consider option B)

$\displaystyle f\left( {\frac{5}{2}} \right) = 2 \times \frac{5}{2} - 3 = 2$ as

$\displaystyle \frac{5}{2} > 2$
and

$\displaystyle f\left( 2 \right) = 2 \times 2 - 3 = 1$ .
Also,

$\displaystyle f\left( 1 \right) = 1$
Hence,

$\displaystyle 1+f\left \{ f\left ( f\left ( \frac{5}{2} \right ) \right ) \right \}= 2 =f\left( {\frac{5}{2}} \right)$
Hence option B is also a correct answer .
Since the function is not defined for

$x=0$ option C can be eliminated.

The function

$f$ is one to one and the sum of all the intercepts of the graph is

$5$ . The sum of all the intercept of the graph

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

0%
$5$
0%
$\dfrac15$
0%
$\dfrac25$
0%
$-5$

Explanation

Since the function

$f$ is one-one there exist only one

$x$ -intercept and only one

$y$ -intercept for the function.
Let

$a$ be the

$y$ -intercept of the function

$f$ . Hence

$f\left( 0 \right) = a$ , but then we have

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

$a$ is an

$x$ -intercept of the function

${f^{ - 1}}$ .
Similarly the

$x$ -intercept of the function

$f$ is

$y$ -intercept of the function

${f^{ - 1}}$ , say

$b$ .
Hence the sum of the intercepts of the function

$f$ is same as the sum of the intercepts of the function

${f^{ - 1}}$ which is equal to 5.

Suppose

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

$\displaystyle x\geq -1$ . If

$\displaystyle g(x)$ is the function whose graph is the reflection of the graph of

$\displaystyle f(x)$ with respect to the line

$y=x$ , then

$\displaystyle g(x)$ is equal to

0%
$\displaystyle -\sqrt{x}-1$ for $\displaystyle x\geq 0$
0%
$\displaystyle \frac{1}{\left ( x+1 \right )^{2}}$ for $\displaystyle x > -1$
0%
$\displaystyle \sqrt{x+1}$ for $\displaystyle x > -1$
0%
$\displaystyle \sqrt{x}-1$ for $\displaystyle x\geq 0$

Explanation

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

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

$\sqrt{y}=x+1$

$\sqrt{y}-1=x$

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

$x\geq 0$

Let

$f\left( x \right) =\left\{ \begin{matrix} 1+|x|,\; x<-1 \\ \left[ x \right] ,\; x\ge -1 \end{matrix} \right.$ where

$[\cdot]$ denotes the greatest integer function. Then

$\displaystyle f\left \{f(-2.3) \right\}$ is equal to

0%
$4$
0%
$2$
0%
$-3$
0%
$3$

Explanation

$f\left( x \right) =\left\{ \begin{matrix} 1+|x|,\; x<-1 \\ \left[ x \right] ,\; x\ge -1 \end{matrix} \right.$
.

$f\left( -2.3 \right) =1+\left| -2.3 \right| =3.3\dots ( \because x < -1$ )
.

hence,

$f\left( f\left( -2.3 \right) \right) =f\left( 3.3 \right) =\left[ 3.3 \right] =3$

$f\left( f\left( -2.3 \right) \right)=3$

The inverse function of the function

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

0%
$\displaystyle \frac{1}{2}\log\frac{1+x}{1-x}$
0%
$\displaystyle \frac{1}{2}\log\frac{2+x}{2-x}$
0%
$\displaystyle \frac{1}{2}\log\frac{1-x}{1+x}$
0%
none of these

Explanation

$y=\dfrac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$

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

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

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

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

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

Taking logarithm on both the sides, we get

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

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

Hence

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

Alternatively,

$f\left(x\right)$ is

$\tan h\left(x\right)$

So, it's inverse will be

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

$\displaystyle f \left ( x \right ) = px + q$ and

$\displaystyle f \left ( f\left ( f\left ( x \right ) \right ) \right ) = 8x + 21$ , where

$p$ and

$q$ are real numbers, the

$p + q$ equals

0%
$3$
0%
$5$
0%
$7$
0%
$11$

Explanation

$f(x)=px+q$

$f(f(x))=p(px+q)+q$

$=p^2x+q(p+1)$

$f(f(f(x)))=p(p^2x+q(p+1))+q$

$=p^{3}x+q(p(p+1)+1)$

$=8x+21$
By comparing coefficients, we get

$p^{3}=8$

$p=2$
And

$q(2(3)+1)=21$

$7q=21$

$q=3$
Hence

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

$p+q=2+3=5$

$K(x)$ is a function such that

$K(f(x))=a+b+c+d$ ,
Where,
$$a=\begin{cases}
0 & \text{ if f(x) is even} \\
-1 & \text{ if f(x) is odd} \\
2 & \text{ if f(x) is neither even nor odd}
\end{cases}$$
$$b=\begin{cases}
3 & \text{ if f(x) is periodic} \\
4 & \text{ if f(x) is aperiodic}
\end{cases}$$
$$c=\begin{cases}
5 & \text{ if f(x) is one one} \\
6 & \text{ if f(x) is many one}
\end{cases}$$
$$d=\begin{cases}
7 & \text{ if f(x) is onto} \\
8 & \text{ if f(x) is into}
\end{cases}$$

$h:R\rightarrow R,h(x)=\left ( \displaystyle \frac{e^{2x}+e^{x}+1}{e^{2x}-e^{x}+1} \right )$

On the basis of above information, answer the following questions.

$K(\phi(x))$

0%
$15$
0%
$16$
0%
$17$
0%
$18$

Explanation

$\phi(x)\rightarrow \left(\dfrac{-\pi}{2},\dfrac{\pi}{2}\right)$
Now we know that for the given domain,

$\tan(x)$ is one-one, periodic, odd and an into function.
Hence

$f(\phi(x))$

$=-1+3+5+8$

$=16-1$

$=15$

The total number of injective mappings from a set with

$m$ elements to a set with

$n$ elements,

$m \leq n$ is

0%
$\displaystyle m^{n}$
0%
$\displaystyle n^{m}$
0%
$\displaystyle \frac{n!}{(n-m)!}$
0%
$n!$

Explanation

$a_{1} \in$ A can have

$n$ images in

$B$ , but the element

$a_{2}$ will have only

$(n-1)$ images as the mappings are to be one-one (injective).

Similarly the elements

$a_{3}$ will have

$(n-2)$ images.

Hence the total number of mappings will be,

$n(n-1)(n-2)...(n-\overline{m-1})=n(n-1)(n-2)...(n-m+1)$

Multiply above and below by

$(n-m)(n-m-1)...3.2.1$

$\therefore$ Required numbers is

$\displaystyle \frac{n!}{(n-m)!}$

$f:R\rightarrow R$ and

$g:R\rightarrow R$ are given by

$f(x)=|x|$ and

$g(x)=[x]$ for each

$x\in R,$ then

$\left\{ x\in R:g\left( f\left( x \right) \right) \le f\left( g\left( x \right) \right) \right\} =$

0%
$Z\cup \left( -\infty ,0 \right)$
0%
$\left( -\infty ,0 \right)$
0%
$Z$
0%
$R$

Explanation

$x\in R$

Let

$\displaystyle f:\left[-\frac{\pi}{3},\frac{2\pi}{3}\right]\rightarrow [0,4]$ be a function defined by

$\displaystyle f(x)=\sqrt 3 \sin x-\cos x+2$ then

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

0%
$\displaystyle \frac{2\pi}{3}-\cos ^{-1}\left(\frac{x-2}{2}\right)$
0%
$\displaystyle \sin ^{-1}\left ( \frac{x-2}{2} \right )+\frac{\pi}{3}$
0%
$\displaystyle \sin ^{-1}\left ( \frac{x-2}{2} \right )$
0%
$\displaystyle \sin ^{-1}\left ( \frac{x+2}{2} \right )+\frac{\pi}{6}$

Explanation

$y=2\left(\dfrac{\sqrt{3}}{2}\sin\left(x\right)-\dfrac{1}{2}\cos\left(x\right)\right)+2$

$\Rightarrow \dfrac{y}{2}-1=\dfrac{\sqrt{3}}{2}\sin\left(x\right)-\dfrac{1}{2}\cos\left(x\right)$

$\Rightarrow \dfrac{y-2}{2}=-\cos\left(\dfrac{\pi}{3}+x\right)$

$\Rightarrow \dfrac{y-2}{2}=\cos\left(\pi-\left(\dfrac{\pi}{3}+x\right)\right)$

$\Rightarrow \dfrac{y-2}{2}=\cos\left(\dfrac{2\pi}{3}-x\right)$

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

$x=\dfrac{2\pi}{3}-\cos^{-1}\left(\dfrac{y-2}{2}\right)$
Therefore,

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

Let

$f$ and

$g$ be increasing and decreasing functions respectively from

$\displaystyle \left ( 0,\infty \right )$ to

$\left ( 0,\infty \right )$ and let

$h\left ( x \right )=f\left [ g\left ( x \right ) \right ]$ . If

$h\left ( 0 \right )=0$ then

$h\left ( x \right )-h\left ( 1 \right )$ is

0%
always zero
0%
always negative
0%
always positive
0%
strictly increasing
0%
None of these

Explanation

Let

$\displaystyle F\left ( x \right )=h\left ( x \right )-h\left ( 1 \right )=f\left ( g\left ( x \right ) \right )-h\left ( 1 \right )$

$\displaystyle F'\left ( x \right )=f'\left ( g\left ( x \right ) \right ).g'\left ( x \right )=\left ( + \right )\left ( - \right )=-ive.$ (As f is increasing function

$\displaystyle f'\left ( g\left ( x \right ) \right )$ is +ive and as g is decreasing function

$\displaystyle g'\left ( x \right )$ is-ive.)
Since F'(x) is-ive therefore F(x)i.e.h(x)-h(1) is decreasing function.
Now split the interval

$\displaystyle I=\left [ 0,\infty \right ]$ into two intervals

$\displaystyle I_{1},0\leq x< 1and I_{2}, 1\leq x< \infty ,$
Apply the definition of decreasing function on

$\displaystyle h\left ( x \right )-h\left ( 1 \right ):$ on

$\displaystyle I_{1},0\leq x< 1,\underset{\left ( Big \right )}{h\left ( x \right )}-\underset{\left ( Less \right )}{h\left ( 1 \right )}=+ive$ On

$\displaystyle I_{2},1\leq x< \infty ,\underset{\left ( Less \right )}{h\left ( x \right )}-\underset{\left ( Big \right )}{h\left ( 1 \right )}=-ive$ Hence for

$\displaystyle I,h\left ( x \right )-h\left ( 1 \right )$ is neither always zero nor always +ive nor always-ive,nor strictly increasing throughout.
Hence (v) is the correct answer.

$\displaystyle f\left ( x \right )= \frac{3x+2}{5x-3}$ , then

0%
$\displaystyle f\left ( f{}'\left ( x \right ) \right )=f^{-1}\left ( x \right )$
0%
$\displaystyle f^{-1}\left ( x \right )=f\left ( x \right )$
0%
$\displaystyle f\left ( f{}'\left ( x \right ) \right )=f'\left ( x \right )$
0%
$\displaystyle f\left ( f\left ( x \right ) \right )=-x^{2}$

Explanation

$\displaystyle y=\frac{3x+2}{5x-3}$

$\displaystyle 5xy-3y=3x+2$

$\displaystyle x(5y-3)=3y+2$

$\displaystyle x=\frac{3y+2}{5y-3}$

Hence

$f(x)$ is its own inverse.

$\displaystyle f^{-1}(x)=\frac{3x+2}{5x-3}=f(x)$

$\displaystyle f\left ( x \right )+f\left ( \dfrac1x \right )=0,\; f(e)=1; g\left ( x \right )=f^{-1}\left ( x \right )$ then

$\displaystyle g{}'\left ( x \right )$ equals

0%
$\displaystyle e^{x}$
0%
$\displaystyle x$
0%
$\displaystyle x^{2}$
0%
$\displaystyle e^{-x}$

Explanation

$f(x)+f\left(\dfrac{1}{x}\right)=0$ is satisfied by $f(x)=k \ln x$ , where $k$ is a constant.

Given that $f(e)=1 \Rightarrow k=1$

$f(x)= \ln x$

$\Rightarrow f^{-1}x=e^x$

$\Rightarrow g(x)=e^x$

$\Rightarrow g'(x)=e^x$

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

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

Explanation

Let

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

$\displaystyle \Rightarrow { x }^{ 2 }-xy+1=0\quad \Rightarrow x=\frac { y\pm \sqrt { { y }^{ 2 }-4 } }{ 2 }$

$\because \quad x\in \left( 2,\infty \right)$

$\displaystyle \therefore x=\frac { y+\sqrt { { y }^{ 2 }-4 } }{ 2 }$

Hence,

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

Let

$\displaystyle f\left ( x \right )=\frac{ax^{2}+2x+1}{2x^{2}-2x+1}$ , the value of

$a$ for which

$\displaystyle f:R\rightarrow \left [ -1,2 \right ]$ is onto , is

0%
$\displaystyle \left [ 2,5 \right ]$
0%
$\displaystyle \left [ -5,-2 \right ]$
0%
$\displaystyle \left [ 0,5 \right ]$
0%
None of these.

Explanation

Given

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

Since,

$\displaystyle f:R\rightarrow \left [ -1,2 \right ]$ is onto

$R(f)=Co-domain [-1,2]$

$-1 \le \dfrac{ax^{2}+2x+1}{2x^{2}-2x+1}\le 2$

$\Rightarrow -(2x^{ 2 }-2x+1)\le ax^{ 2 }+2x+1\le 2(2x^{ 2 }-2x+1)$

$\Rightarrow -2x^{ 2 }+2x-1\le ax^{ 2 }+2x+1\le 4x^{ 2 }-4x+2$ ....(1)

$\Rightarrow -2x^{ 2 }+2x-1\le ax^{ 2 }+2x+1$

$\Rightarrow (a+2)x^2+2\ge 0$

So for all

$x\in R$ ,

$a+2\ge 0$

$\Rightarrow a\ge -2$ .....(2)

From the inequality (1), it follows that

$ax^{ 2 }+2x+1\le 4x^{ 2 }-4x+2$

$\Rightarrow (a-4)x^2+6x-1 \le 0$

$\Rightarrow a-4 < 0$ and

$D\le 0$

$\Rightarrow a<4$ and

$36+4a-16 \le 0$

$\Rightarrow a<4$ and

$a\le -5$ ....(3)

From (2) and (3), we get

$a\in (-\infty,-5]\cup [-2,4)$

CBSE Questions for Class 12 Commerce Maths Relations And Functions Quiz 9 - 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 9 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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