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 $$\left ( 0,\frac{\pi^{2}}{4} \right )$$

MCQ Questions for Class 11 Commerce Applied Mathematics Functions Quiz 10

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.

Which of the following functions from

$Z$ to

$Z$ are bijection?

0%
$f(x)=x^3$
0%
$f(x)=x+2$
0%
$f(x)=2x^2+1$
0%
$f(x)=x^2+1$

Explanation

Here,

$f(x)=x+2\Rightarrow f(x_1)=f(x_2)$

$x_1 +2=x_2+2\Rightarrow x_1=x_2$
Let

$y=x+2$

$x=y-2\in Z, \forall \in x$
Here,

$f(x)$ is one-one is onto.

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%
$-1$

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

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

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

Which one of the following is onto function define R to R .

0%
$f(x) = |x|$
0%
$f(x) = e^{x}$
0%
$f(x) = x^{3}$
0%
$f(x) = \sin x$

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

Which of the following is onto function-

0%
$f : Z \rightarrow Z , f (x) = |x|$
0%
$f : N \rightarrow N , f (x) = |x|$
0%
$f : R_{0} \rightarrow R^{+} , f (x) = |x|$
0%
$f : C \rightarrow R , f (x) = |x|$

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%
$f(x) = e^{x} + r^{-x}$

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%
$x - 2$

$f(x) = x^{2} + x$ and

$g(x) = \sqrt {x}$ , then the value of

$f(g(3))$ is

0%
$1.73$
0%
$3.46$
0%
$4.73$
0%
$7.34$
0%
$12.00$

Explanation

Given,

$f(x)=x^2+x, g(x)=\sqrt x$

Then we need to find the value of

$f(g(3))$

$\Rightarrow g(3) = \sqrt3$ ..... given

$g(x) = \sqrt x$

$\Rightarrow f(g(3)) = f(\sqrt3) = 3 + \sqrt3$ ..... given

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

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

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

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

Domain of the function,

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

0%
$[-1, 1]$
0%
$[-2, 2]$
0%
$[-\pi, -\sqrt{2}] \cup [\sqrt{2}, \pi]$
0%
$[-\sqrt{2}, \sqrt{2}]$

Explanation

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

$-1 \leq \sin x \leq 1$

$\forall x \in R$

$\therefore \cos(\sin x) > 0$

$\forall x \in R$

$\therefore$ domain of

$\sqrt{\cos(\sin x)}$ is

$R$ .....(i)

$-1 \leq x^2-1 \leq 1$ i.e.

$0 \leq x^2 \leq 2$

i.e.

$x \in [-\sqrt{2}, \sqrt{2}]$

$\therefore$ domain of

$\sin^{-1}(x^2-1)$ is

$[-\sqrt{2}, \sqrt{2}]$ ....(ii)

From (i) and (ii)

Domain of

$f(x)$ is

$[-\sqrt{2}, \sqrt{2}]$

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

$-1\leq x \leq 4$ ,

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

$x\geq 1$ 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

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

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

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

The domain of the function

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

0%
$\displaystyle (-\infty, \infty)$
0%
$\displaystyle (-\infty, -\sqrt 2]\cup [\sqrt 2, \infty)$
0%
$\displaystyle (-\infty , -\sqrt 2]\cup [\sqrt 2, \infty)\cup \left \{0\right \}$
0%
none of these

Explanation

As,

$\sin^2x+\sin x+1 > 0, \forall x\in R$

$\displaystyle \therefore \frac {1}{\sqrt {\sin^2x+\sin x+1}}$ is always exists.

$\displaystyle \therefore$ For

$\sin^{-1}\left (\dfrac {1}{|x^2-1|}\right )$ to exists,

$\displaystyle 0 < \frac {1}{|x^2-1|} \leq 1\Rightarrow |x^2-1|\geq 1$

$\displaystyle \Rightarrow x^2-1 \leq -1$ or

$x^2-1\geq 1$

$\displaystyle \Rightarrow x^2\leq 0$ or

$x^2\geq 2$

$\displaystyle \Rightarrow x=0$ or

$(x\leq -\sqrt 2$ or

$x\geq \sqrt 2)$

$\displaystyle \therefore x\in (-\infty, -\sqrt 2]\cup [\sqrt 2, \infty)\cup \left \{0\right \}$
Hence, (c) is the correct answer.

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.

The domain of the function

$f(x)=\log_3 \log_{1/3}(x^2+10x+25)+\dfrac {1}{[x]+5}$ where [.] denotes the greatest integer function) is

0%
$(-4, -3)$
0%
$(-6, -5)$
0%
$(-6, 4)$
0%
None of these

Explanation

For given function to be defined,
(i)

$x^2+10x+25>0\Rightarrow (x+5)^2>0\Rightarrow x\neq -5=D_1$
(ii)

$\log_{1/3}(x^2+10x+25)>0\Rightarrow x^2+10x+25<1$

$\Rightarrow x^2+10x+24<0\Rightarrow (x+4)(x+6)<0\Rightarrow x\in(-6,-4)=D_2$
and
(iii)

$[x]+5\neq 0\Rightarrow x\notin [-5,4)=D_3$
Hence domain of f is

$D_!\cap D_2\cap D_3 =(-6,-5)$

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.

$l= \lim_{x\rightarrow \alpha}\displaystyle \frac{f(x)}{x(x-\alpha)(x-2)}$ is

0%
positive
0%
negative
0%
$0$
0%
sign of $l$ depends upon $\alpha\pi$

Explanation

For the quadratic function

$f(x) = ax^2 + bx +c$ , it's given that

$a, b, c$ are non-zero.

Hence,

$f(0) = c \neq 0$

It's given that

$f:[0,2] \rightarrow [0,2]$ is bijective.

Since

$f(0) \neq 0$ ,

$f(0) = 2$ for it to be bijective.

$\Rightarrow f(2) = 0$

So, three cases possible as shown in the figure, which will satisfy the second condition as well.

For cases 1 and 2,

$\displaystyle \lim_{x\rightarrow \alpha}\displaystyle \frac{f(x)}{x(x-\alpha)(x-2)}$

$\displaystyle =\lim _{ x\rightarrow \alpha } \frac { 1 }{ x\left( x-2 \right) } \lim _{ x\rightarrow \alpha } \frac { f(x) }{ x-\alpha } \\ \displaystyle =\frac { 1 }{ \alpha \left( \alpha -2 \right) } \lim _{ x\rightarrow \alpha } \frac { f'(x) }{ 1 } =\frac { f'(\alpha ) }{ \alpha \left( \alpha -2 \right) }$

$f'(\alpha)$ is positive for both cases.

$\alpha( \alpha -2)$ is also positive for both cases. Hence, the limit value is also positive.

For the case in figure-3,

$\displaystyle \lim _{ x\rightarrow \alpha } \frac { f(x) }{ x(x-\alpha )(x-2) } =\lim _{ x\rightarrow \alpha } \frac { 1 }{ x } \lim _{ x\rightarrow \alpha } \frac { f(x) }{ { \left( x-\alpha \right) }^{ 2 } } \\ \displaystyle =\frac { 1 }{ \alpha } \lim _{ x\rightarrow \alpha } \frac { f'(x) }{ 2\left( x-\alpha \right) } =\frac { 1 }{ \alpha } \lim _{ x\rightarrow \alpha } \frac { f''(x) }{ 2 } =\frac { 1 }{ \alpha } \frac { f''\left( \alpha \right) }{ 2 }$

$f''(\alpha) > 0$ since the graph is facing upwards. Also

$\alpha$ is positive.

Hence, in this case also , the value of limit is positive. Hence, option A is correct.

Domain of the function

$f(x)=\dfrac {1}{\sqrt {4x-|x^2-10x+9|}}$ , is

0%
$(7-\sqrt {40}, 7+\sqrt {40})$
0%
$(0, 7+\sqrt {40})$
0%
$(7-\sqrt {40}, \infty)$
0%
none of these

Explanation

Here,

$f(x)=\dfrac {1}{\sqrt {4x-|x^2-10x+9|}}$ would exist, if

$\displaystyle 4x-|x^2-10x+9| > 0$

ie,

$\displaystyle |x^2-10x+9| < 4x$ ,

where

$\displaystyle |x^2-10x+9|=\begin{cases}x^2-10x+9 & x\leq 1 \text{ or } x \geq 9\\ -(x^2-10x+9) &1 < x < 9 \end{cases}$

Case I: When

$x\leq 1$ or

$x\geq 9$

$\therefore x^2-10x+9 < 4x$

$\Rightarrow x^2-14x+9 < 0 \Rightarrow (x-7)^2 < 40$

$\Rightarrow x\in (7-\sqrt {40}, 7+\sqrt {40})$ (But

$x\leq 1$ or

$x\geq 9$ )

$\Rightarrow x\in (7-\sqrt {40}, 1]\cup [9, 7+\sqrt {40})$ .....(i)
Case II: When

$1 < x < 9$

$-x^2+10x-9 < 4x\Rightarrow x^2-6x+9 > 0$

$\Rightarrow (x-3)^2 > 0$ which is always true except

$x=\left \{3\right \}$

$\therefore x\in (1, 9)=\left \{3\right \}$ .....(ii)
From Eqs. (i) and (ii), domain of

$f(x)\in (7-\sqrt {40}, 7+\sqrt {40})-\left \{3\right \}$
Hence, (d) is the correct answer.

$f:R\rightarrow \left [\dfrac {\pi}{6}, \dfrac {\pi}{2}\right ), f(x)=\sin^{-1}\left (\dfrac {x^2-a}{x^2+1}\right )$ 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.

In the following functions defined from

$[-1, 1]$ to

$[-1, 1]$ , then functions which are not bijective are

0%
$\displaystyle \sin (\sin^{-1} \: x)$
0%
$\displaystyle \frac {2}{\pi} \: \sin^{-1} (\sin \: x)$
0%
$\displaystyle (sgn \: x) \ lne^x$
0%
$\displaystyle x^3 \: sgn \: x$

Explanation

$sin\sin ^{ -1 }{ x } = x$ hence it is bijective.

$\dfrac { 2 }{ \pi } \sin ^{ -1 }{ sinx }$ =

$\dfrac { 2x }{ \pi }$ . Hence it is one one but not onto, hence not bijective

$sgn(x).x=\dfrac { \left| x \right| }{ x }.x= |x|, x\neq$ 0
Hence it is not one one or onto, hence not bijective

${ x }^{ 3 }sgnx={ x }^{ 3 }\dfrac { |x| }{ x } ={ x }^{ 2 }|x|={ |x| }^{ 3 },x\neq 0$ . Hence function is not one one or onto, hence not bijective.

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.

Find the domain of the function

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

0%
All real numbers except for $0$
0%
All real numbers greater than or equal to $1$
0%
All real numbers less than or equal to $1$
0%
All real numbers greater than or equal to $-1$ but less than or equal to $1$
0%
All real numbers less than or equal to $-1$

Explanation

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

$f(x)$ to exist, the expression in the square root must be non - negative and the expression in denominator should not be equal to

$0$ .
So,

$x-1 \ge 0$ and

$x \ne 0$

$x \ge 1$ and

$x \ne 0$ , which implies

$x \ge 1$ .

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

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

$g(f(1))$

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

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

$=1$

$g(0) = 1$

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

The domain of function

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

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

Explanation

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

$x-\sqrt{1-x^{2}} \geq 0$ and

$1 -x^{2}\geq0$

$\Rightarrow x\geq\sqrt{1-x^{2}}$ and

$-1\leq x \leq1$ ... (i)

Now,

$x^2\geq(\sqrt{1-x^{2}})^2$ , for positive

$x$

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

$x\geq\dfrac{1}{\sqrt{2}}$ ... (ii)

From (i) and (ii), we get

$x\in\displaystyle \left [ -1,-\frac{1}{\sqrt{2}} \right ]\cup \left [ \frac{1}{\sqrt{2}},1 \right ]$

Hence, option 'D' is correct.

The domain of the function

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

0%
$\displaystyle \left \{ x|x< 1 \right )$
0%
$\displaystyle \left \{ x|x> -1 \right \}$
0%
$[0,1]$
0%
$[-1,1]$

Explanation

For real range of

$f(x)$

$1-x^{2}\geq 0$

$x^2\leq 1$

$|x|\leq 1$

$-1 \leq x \leq 1$

$x\in [-1,1]$

Hence the domain for the above function is

$[-1,1]$

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.

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 largest set of real values of

$x$ for which

$\displaystyle f(x)=\sqrt{(x+2)(5-x)}-\frac{1}{\sqrt{x^{2}-4}}$ is a real function is

0%
$\displaystyle [1,2)\cup (2,5]$
0%
$\displaystyle (2,5]$
0%
$\displaystyle [3,4]$
0%
$none\:of\:these$

Explanation

$\sqrt { \left( x+2 \right) \left( 5-x \right) }$ is defined for

$\left( x+2 \right) \left( 5-x \right) \ge 0\\ \Rightarrow \left( x+2 \right) \left( x-5 \right) \le 0\\ \Rightarrow x\in \left[ -2,5 \right] .........(1)$
.
Similarly,

$\dfrac { 1 }{ \sqrt { { x }^{ 2 }-4 } }$ is defined for

${ x }^{ 2 }-4>0\\ x\in \left( -\infty ,-2 \right) \bigcup \left( 2,\infty \right) ......(2)$
.
From

$(1)$ and

$(2)$ , the domain of

$f(x)$ is

$x \in \left( 2, 5 \right]$

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

$p+q=2+3=5$

The value of

$(a + b)$ is equal to

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

Explanation

$x^{2}-2x-1$ is an upwards facing parabola. Therefore, it has a point of minimum

$=-\dfrac{b}{2a}=\dfrac12\times$ Sum of roots
After the point of minimum the curve slopes up and the function no longer remain one-one. Hence the largest interval for it to be injecvtive is upto the point of minimum.
The roots of the quadratic are,

$\rightarrow x=1+\sqrt{2}$ and

$x=1-\sqrt{2}$
So

$a=\dfrac{1+\sqrt{2}+1-\sqrt{2}}{2}$

$=1$

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

$=-2$

$=b$
Hence

$a+b$

$=1-2$

$=-1$

$K(x)$ 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

$f(\phi(x))$

$=-1+3+5+8$

$=16-1$

$=15$

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

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

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.

$f:R\rightarrow R$ 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

$f$ 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.

$\displaystyle f(x)=27x^{3}+\frac{1}{x^{3}}$ and

$\alpha,\beta$ are the roots of

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

0%
$f(\alpha)=f(\beta)$
0%
$f(\alpha)=10$
0%
$f(\beta)=-10$
0%
none of these

Explanation

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

Cubing the above equation, we get

$\left(3x+\dfrac{1}{x}\right)^3=2^3$

$\therefore 27x^3+\dfrac{1}{x^3}+3(3x)\left(\dfrac{1}{x}\right)\left(3x+\dfrac{1}{x}\right)=8$

$\therefore 27x^3+\dfrac{1}{x^3}+9\left(3x+\dfrac{1}{x}\right)=8$

$\therefore 27x^3+\dfrac{1}{x^3}+9(2)=8$

$\therefore 27x^3+\dfrac{1}{x^3}=-10$

$\alpha$ and

$\beta$ are roots of above equation.

$\therefore 27\alpha^3+\dfrac{1}{\alpha^3}=-10$ ...(1)

and

$27\beta^3+\dfrac{1}{\beta^3}=-10$ ...(2)

$f(x)=27x^3+\dfrac{1}{x^3}$

$f(\alpha)=27\alpha^3+\dfrac{1}{\alpha^3}$

$\implies f(\alpha) =-10$ ...(from 1)

Similarly,

$f(\beta)=27\beta^3+\dfrac{1}{\beta^3}$

$\implies f(\beta)=-10$ ...(from 2)

$\therefore f(\alpha)=f(\beta)=-10$

So, the correct options are option (A) and (C)

The value of

$x$ satisfying the equation

$\displaystyle \left | x-1 \right |^{\log_{3}x^{2}-2\log_{9}x}= (x-1)^7$ is

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

Explanation

$\displaystyle (i) \log_{a}b$ hold good if

$\displaystyle a> 0,a\neq 1,a> 1$

i.e.

$a-1> 0$

$\therefore \left | a-1 \right |= a-1$

$(ii) a^{b}> 0\,\,\forall\, b\in R$

Now from given

$\displaystyle \left | x-1 \right |^{\log_{3}x^{2}-2\log_{9}x}= (x-1)^{7}$

$\displaystyle \Rightarrow \left ( x-1 \right )^{\log_{3}x^{2}-2\log_{9}x}= (x-1)^{7}$

$\displaystyle\Rightarrow 2\log_{3}x-2\log^{x}{9}= 7$ taking log at base

$(x -1)$ bothsides.

$log_{x}{3}=7,x=3^7$

The domain of

$\displaystyle f(x)=\frac{1}{\sqrt{|\cos\:x|+\cos\:x}}$ is

0%
$[-2n\pi,2n\pi]$
0%
$(2n\pi,\overline{2n+1}\pi)$
0%
$\displaystyle \left ( \frac{(4n+1)\pi}{2} ,\frac{(4n+3)\pi}{2}\right )$
0%
$\displaystyle \left ( \frac{(4n-1)\pi}{2} ,\frac{(4n+1)\pi}{2}\right )$

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

$g(x)=3x$ , calculate the value of

$g(f(-2))-f(g(2))$ .

0%
$-480$
0%
$-384$
0%
$0$
0%
$384$
0%
$480$

Explanation

$f(-2) = 2 \times { (-2) }^{ 3 } = -16$
$g(2) = 3 \times 2 =6$
$g(f(-2)) = g(-16) = 3 \times -16 = -48$
$f(g(2)) = f(6) = 2 \times { (6) }^{ 3 } = 432$
$g(f(-2)) - f(g(2)) = -48-432 = -480$

Find the maximum value of

$g(f(x))$ if:

$f(x) = x + 4$ and

$g(x) = 6 - x^{2}$

0%
$-6$
0%
$-4$
0%
$2$
0%
$4$
0%
$6$

Explanation

Given,

$f(x)=x+4, g(x)=6-x^2$

$\therefore g(f(x)) = g(x+4) = 6-{(x+4)}^{2}$
The minimum value of

${(x+4)}^{2}$ is

$0$ which occurs at

$x=-4$ .
So, the maximum value of

$g(f(x))$ is

$6-0 = 6$ .

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

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

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

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

CBSE Questions for Class 11 Commerce Applied Mathematics Functions Quiz 10 - 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 10 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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