MCQ Questions for Class 11 Commerce Applied Mathematics Functions Quiz 8

$f(x) = \sqrt {4 - {x^2}} + \dfrac{1}{{\sqrt {\left| {\sin x} \right| - \sin x} }}$ , then the domain of f(x) is

0%
[-2,0]
0%
(0,2]
0%
[-2,2}
0%
[ $\dfrac{\pi}{2}$ , 2]

$f(x)= sin^{2}x$ and the composite functions g{f(x)}=|sin x|, then the function g(x)=

0%
$\sqrt{x-1}$
0%
$\sqrt{x}$
0%
$\sqrt{x+1}$
0%
$-\sqrt{x}$

the equation

$|x+1$^{log}(x+1)^{3+2x-x^{2}}= (x-3)|x|$ has

0%
Uniques solution
0%
Two solution
0%
No solutions
0%
more thatn two solution

if f(x)=

$\sqrt {4 - {x^2}} + \dfrac{1}{{\sqrt {|\sin x| - \sin x} }}$ , then the domain of f(x) is

0%
$[-2,0)$
0%
$[0,2]$
0%
$[-2,2]$
0%
$None\ of\ these$

Explanation

$f ( x ) = \sqrt { 4 - x ^ { 2 } } + \dfrac { 1 } { \sqrt { | \sin x | } - \sin x }$

$\Rightarrow 4 - x ^ { 2 } \geqslant 0\ and\ \sqrt{(|\sin x|)} - \sin x > 0$

$\Rightarrow x ^ { 2 } \leq 4 \quad 4 \quad \sin x < | \sin x |$

$\Rightarrow \quad x \in [ - 2,2 ] \quad \& \quad \sin x < 0$

$\Rightarrow \quad x \in [ 2,2 ] + x \in [ - 2,0 )$

$\Rightarrow \quad x \in [ - 2,0 )$

The domain of the function

$f ( x ) = \log _ { 1 / 2 } \left( - \log _ { 2 } \left( 1 + \dfrac { 1 } { \sqrt [ 4 ] { x } } \right) - 1 \right)$ is

0%
$0 < x < 1$
0%
$0 < x \leq 1$
0%
$x \geq 1$
0%
null set

Explanation

$f\left( x \right)=\log _{2}^{{}}\left( -\log _{1/2}^{{}}\left( 1+\frac{1}{{{x}^{1/4}}} \right)-1 \right)$

$\,it\,is\,defined\,for$

$\,\left( -\log _{1/2}^{{}}\left( 1+\frac{1}{{{x}^{1/4}}} \right)-1 \right)>0$

$\Rightarrow \left( \log _{1/2}^{{}}\left( 1+\frac{1}{{{x}^{1/4}}} \right)+1 \right)<0$

$\Rightarrow \left( \log _{1/2}^{{}}\left( 1+\frac{1}{{{x}^{1/4}}} \right) \right)<-1$

$\Rightarrow \left( 1+\frac{1}{{{x}^{1/4}}} \right)>{{\left( \frac{1}{2} \right)}^{-1}}$

$\Rightarrow 1+\frac{1}{{{x}^{1/4}}}>2$

$\Rightarrow \frac{1}{{{x}^{1/4}}}>1$

$\Rightarrow 0<x<1$

Find the domain of function

$\sin^{-1}\left[\dfrac {1+x^{2}}{2x}\right].$

0%
$[-1,1]$
0%
$(-1,1)$
0%
$\left\{-1,1\right\}$
0%
$\left\{0\right\}$

Explanation

${{\sin }^{-1}}\left( \frac{1+{{x}^{2}}}{2x} \right)$

$-1\le \frac{1+{{x}^{2}}}{2x}\le 1$

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

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

$\Rightarrow x\left( {{x}^{2}}+1+2x \right)\ge 0$

$\Rightarrow x{{\left( x+1 \right)}^{2}}\ge 0$

$which\,is\,true\,for\,x\ge 0$

$x=-1$

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

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

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

$which\,is\,true\,for\,x\le 0,x=1$

$Hence\,domain:\left\{ \left. -1,1 \right\} \right.$

The domain of

$f(x)= e^{sin (x-|x|)}+|x|cos (\dfrac{x}{|x+1|})$ , where [.] represents greatest integer function , is

0%
R
0%
R-[-1, 0]
0%
R-[0,1]
0%
R-[-1)

The equation

$|x-1|+|a|=4$ can have real solution for

$x$ if a belongs to the interval

0%
$(-\infty,4]$
0%
$(4,\infty)$
0%
$(-4,4)$
0%
$(\infty,-4)\cup (4,\infty)$

$f ( x ) = \sin ^ { - 1 } ( \sin x ) + \cos ^ { - 1 } ( \sin x ) \text { and } \phi ( x ) = f ( f ( f ( x ) ) )$ 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%
$44$

Number of positive integers in the domain of the function

$f(x)=\sqrt {\log_{0.5}\log_{6}\left(\dfrac {x^{2}+x}{x+4}\right)}$ is

0%
$5$
0%
$6$
0%
$7$
0%
$8$

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.

The domain of the function

$f\left(x\right)=\sin^{-1}{\left(1+{e}^{x}\right)^{-1}}$ is

0%
$R$
0%
$[-1,0]$
0%
$[0,1]$
0%
$[-1,1]$

The interval on which the function

$f(x)=2x^3+9x^2+12x-1$ is decreasing is?

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

Explanation

$f(x) = 2x^{3} + 9x^{2} + 12x - 1$

$f'(x) = 6x^{2} + 18x +12$

$= 6(x^{2} + 3x + 2)$

$= 6 (x+2)(x+1)$

$\therefore f(x)$ decreases in interval

$xt [-2,-1]$

1188002_1295156_ans_e1c2ef81ccbf4390acb6471f3021bb7a.png

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

The domain of the function

$f(x) = \frac {1} {\sqrt {^{10}C_{x - 1} - 3 \times ^{10} C_x}}$ contains the points

0%
9, 10, 11
0%
9, 10, 12
0%
all natural numbers
0%
None of these

Explanation

Given function is defined if

$^{10} C_{x + 1} > 3 ^{10}C_x$

$\Rightarrow \dfrac{1} {11 - x} > \dfrac {3} {4} \Rightarrow 4x > 33$

$\Rightarrow x \geqslant 9 \ but \ x \leq 10 \Rightarrow x = 9, 10$

$f:( - 1,1) \to B$ , 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.

The sum of all real values of

$x$ satisfying the equation

${\left( {{{\rm{x}}^{{\rm{2 - }}}}{\rm{5x + 5}}} \right)^{{{\rm{x}}{{\rm{^2 + 4x - 60}}}}}}{\rm{ = 1}}$ is

0%
$-4$
0%
$6$
0%
$5$
0%
$3$

Explanation

$(x^2 - 5x + 5)^{x^2 + 4x - 60} = 1$

$\Rightarrow (x^2 - 5x + 5)^0$

$\Rightarrow x^2 + 4x - 60 = 0$

Solving for x.

$x = -10, 6$

And,

$x^2 - 5x + 5$

$1^a = 1$ or

$b^o = 1$ or

$(-1)^x = 1$

$\therefore x^2 - 5x + 5 = 1$

$x^2 - 5x + 4 = 0$

$(x - 4) (x - 1) = 0$

$x = 4$ or

$1$

$x^2 - 5x + 5 = -1$

$x^2 - 5x + 6 = 0$

$(x - 3) (x - 2) = 0$

$x = 3, 2$

2 will satisfy not 3 because power will be odd

Hence

$x = \{-10, 1, 2, 4, 6\}$

$\therefore$ sum of solutions

$=3$

$f \left( \dfrac { x + y } { 2 } \right) = \dfrac { f ( x ) + f ( y ) } { 2 }$ for all

$x , y \in R$ and

$f ^ { \prime } ( o ) = - 1 , f ( o ) = 1$ then

$f(2)=$

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

Explanation

let

$f(x)=ax+b$

$f(0)=1\implies b=1$

$f'(0)=-1 \implies a=-1$

$\implies f(x)=1-x$

$\implies f(2)=-1$

The domain of

$f(x) = \sqrt{2 - log_3 (x - 1)}$ is

0%
$(2, 12]$
0%
$(\infty, 10]$
0%
$(3, 12]$
0%
$(1, 10]$

Explanation

$\sqrt{2-log_3(x-1)}$

$x-1 > 0$

$\Rightarrow x > 1$ …………

$(1)$

Also,

$2-log_3(x-1)\geq 0$

$log_3(x-1)\leq 2$

$(x-1)\leq 3^2$

$\Rightarrow x-1\leq 9$

$x\leq 10$ …………..

$(2)$

Intersection of

$(1)$ &

$(2)$

$x\in (1, 10]$ .

1126250_1291070_ans_5a1438baad964871a122428b9be8a5db.JPG

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

$f(x)$ is

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

Let

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

$F=\left\{1,2\right\}$ . Then the number of onto functions from

$E$ to

$F$ is

0%
$14$
0%
$16$
0%
$12$
0%
$8$

Explanation

The total number of ways of distributing

$(1,2)$ considering them as 2 objects among

$(1,2,3,4)$ considering them as 4 people is equal to total no. of Onto functions from E to F.

There can be two ways of distributing

$(1,2),$

Either we give both to one people

Or give

$1$ to one people and

$2$ to another.

So, Total no. of ways

$=4×1+^4C_2×2=16$

The domain of the function,

$f(x)=\dfrac{|x|-2}{|x|-3}$ is

0%
$R$
0%
$R-\{2,3\}$
0%
$R-\{2,-2\}$
0%
$R-\{-3,3\}$

Explanation

$f(x)=\dfrac{|x|-2}{|x|-3}$

Here,

$x$ cannot be

$\pm 3$ .

(f is not defined when

$x=\pm 3$ )

Therefore,

Domain

$= R-\{-3,3\}$

$f\left( x \right) =\sqrt { { x }^{ 2 }-4 }$ and

$g\left( x \right) =\dfrac { x-1 }{ x-3 }$ then number of integer elements, which are not in the domain of the function

$(f.g)(x)$ equals

0%
3
0%
4
0%
5
0%
None of these

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

$f(x)=\dfrac {4^{x}}{4^{x}+2}$ , then the value of

$f(x)+f(1-x)$ is

0%
$0$
0%
$-1$
0%
$1$
0%
$can't\ be\ determined$

Explanation

Given

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

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

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

$\Rightarrow f{\left( 1 - x \right)} = \cfrac{4}{4 + {4}^{x} \cdot 2} = \cfrac{2}{2 + {4}^{x}}$

Therefore,

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

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

Domain of the function

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

0%
$-\sqrt { 3 } \le x\le \sqrt { 3 }$
0%
$-1-\sqrt { 3 } \le x\le -1+\sqrt { 3 }$
0%
$-2\le x\le 2$
0%
$-2+\sqrt { 3 } \le x\le -2-\sqrt { 3 }$

Explanation

Given,

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

$\sqrt{f\left(x\right)}\quad \Rightarrow \quad \:f\left(x\right)\ge 0$

$2-2x-x^2\ge \:0$

$-\left(x+1\right)^2+3\ge \:0$

$-\left(x+1\right)^2\ge \:-3$

$\left(x+1\right)^2\le \:3$

$-\sqrt{3}\le \:x+1\le \sqrt{3}$

$-\sqrt{3}\le \:x+1\quad \mathrm{and}\quad \:x+1\le \sqrt{3}$

$-\sqrt{3}\le \:x+1\quad :\quad x\ge \:-\sqrt{3}-1$

$x+1\le \sqrt{3}\quad :\quad x\le \sqrt{3}-1$

$\Rightarrow -\sqrt{3}-1\le \:x\le \sqrt{3}-1$

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

The domain of the function

$\sin^{-1} (log_2(\frac{x}{3}))$ is-

0%
[ $\frac{1}{2},3$ ]
0%
[ $\frac{1}{2},3$ ]
0%
[ $\frac{3}{2},6$ ]
0%
[ $\frac{1}{2},2$ ]

Domain of function

$f(x)=\dfrac{|x|-x}{2x}$ is

0%
$\mathbb{R}$
0%
$\mathbb{R}-\{0\}$
0%
$\mathbb{Z}$
0%
$\mathbb{N}$

The domain of

$f(x)= e^{\sqrt{x}}+cos x$ is

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

The domain of function

$f ( x ) = \dfrac { x ^ { 2 } - 10 x + 26 } { x ^ { 4 } \left( x ^ { 2 } - 9 \right) \left( 1 + 27 x ^ { 2 } \right) }$

0%
$\mathbf { x } \in \mathbf { R } - \{ 0,\pm3 \}$
0%
$\mathbf { x } \in \mathrm { R } - \{ 0,3 \}$
0%
$x \in R$
0%
none

Explanation

$f(x)=\dfrac{x^2-10x+26}{x^4(x^2-9)(1+27x^2)}$

The function is defined only when

$x^4(x^2-9)(1+27x^2)\neq 0\\\implies x\neq 0,x^2-9\neq 0\\x^2\neq 9\\x\neq \pm 3$

The domain is

$R-\{0,\pm3\}$

The domain of the function

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

0%
$(2,\infty )$
0%
$(1,\infty )$
0%
$[-1,1] \cup (2,\infty)$
0%
none of these

Explanation

Given,

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

$\dfrac{x^2-1}{x-2}\ge \:0:\quad -1\le \:x\le \:1\quad \mathrm{or}\quad \:x>2$

undefined singularity point:

$x=2$

combined domain of function,

$-1\le \:x\le \:1\quad \mathrm{or}\quad \:x>2$

$\left[-1,\:1\right]\cup \left(2,\:\infty \:\right)$

The domain of the function

$f ( x ) = \log _ { 2 } x^2$ is

0%
$\mathbf { x } \in \mathbf { R }$
0%
$x \in [ 0 , \infty )$
0%
$x \in (- \infty ,0)\cup( 0 , \infty )$
0%
$x \in R - \{ x | x \in 1 \}$

Domain of the function

$f(x) = \dfrac{x^2-3x+2}{x^2+x-6}$ is

0%
{ $x:x \epsilon R, x \neq -3$ }
0%
{ $x:x \epsilon R, x \neq 2$ }
0%
{ $x:x \epsilon R$ }
0%
{ $x:x \epsilon R, x \neq 2, x \neq -3$ }

Explanation

Given,

$f(x)=\dfrac{x^2-3x+2}{x^2+x-6}$

$x^2+x-6=0$

roots of above equation are:

$x=-3,2$

So,

$x$ can't be equal to

$-3 \, or \, 2$

The above points are undefined

The function domain is

$x<-3,-3<x<2,x>2$

$\left \{ x:x\in R,x\neq 2,x\neq -3 \right \}$

The function

$y=\dfrac { x }{ 1+{ x }^{ 2 } }$ has its domain as

0%
$x \in R$
0%
$x\in R-(-1,1)$
0%
$x \in \left( 0,\infty \right)$
0%
$x \in \left( -\infty ,-1 \right)$

Explanation

The function

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

is defined only when

$1+x^2\neq 0$

$x^2\neq -1$

$\implies x\in R$

The domain of the function,

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

$-\dfrac{1}{\sqrt{9-x^{2}}}$ is

0%
(-3,1)
0%
[-3,1]
0%
(-3,2)
0%
(-3,2]

Explanation

Given,

$f\left(x\right)=\sqrt{2-x}-\dfrac{1}{\sqrt{9-x^2}}$

$\sqrt{f\left(x\right)}\quad \Rightarrow \quad \:f\left(x\right)\ge 0\:$

$2-x\ge \:0:\quad x\le \:2$

$9-x^2\ge \:0:\quad -3\le \:x\le \:3$

Combined interval:

$-3\le \:x\le \:2$

Undefined singularity points:

$x=3,\:x=-3$

Combining real regions and undefined singularity points:

$-3<x\le \:2$

$(-3,\:2]$

$f : R \rightarrow R , f ( x ) = e ^ { | x | } - e ^ { - x }$ 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

$B$

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

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

Domain of the function

$f(x)=$

$\dfrac { x-3 }{ (x-1)\sqrt { { x }^{ 2 }-4 } }$

0%
$(1,2)$
0%
$(-\infty,-2)\cup(2,\infty)$
0%
$(-\infty,-2)\cup(1,\infty)$
0%
$(-x,x)-{(t\pm2)}$

Explanation

Given,

$f(x)=\dfrac{x-3}{\left(x-1\right)\sqrt{x^2-4}}$

$\sqrt{f\left(x\right)}\quad \Rightarrow \quad \:f\left(x\right)\ge 0$

$x^2-4\ge \:0:\quad x\le \:-2\quad \mathrm{or}\quad \:x\ge \:2$

Undefined singularity points

$x<-2\quad \mathrm{or}\quad \:x>2$

Domain

$x<-2\quad \mathrm{or}\quad \:x>2$

So, the domain is

$(-\infty,-2)\cup(2,\infty)$

In a set

$A=\left\{1,2,3,4\right\}$ , the relation R is defined as

$x\quad R\quad y\quad \Longleftrightarrow \quad x\le y$ , then the domain of the inverse relation is

0%
$\left\{1,2,3\right\}$
0%
$\left\{3,4,5,6\right\}$
0%
$\left\{1,2,3,4\right\}$
0%
$\left\{4,5,6\right\}$

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

$f:R\rightarrow R,f\left( x \right) =\dfrac { { ax }^{ 2 }+6x-8 }{ a+6x-{ 8x }^{ 2 } }$ is onto, then

$\alpha \in$

0%
$\left( 1,\infty \right)$
0%
$\left( 0,\infty \right)$
0%
$\left( 2,12 \right)$
0%
$\left[ 2,14 \right]$

If f :

$R\rightarrow S$ , defined by f(x) =sin x -

$\sqrt{3}$ cos x +1, is onto, then the interval of S is

0%
[0, 3]
0%
[-1, 1]
0%
[0, 1]
0%
[-1, 3]

$f : R \rightarrow R$ 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

The domain of the function

$f(x) =$

$\dfrac{\sin^{-1}(x-3)}{\sqrt{9-x^2}}$ is

0%
$[1, 2]$
0%
$[2, 3)$
0%
$[2, 3]$
0%
$[1, 2)$

Explanation

Given the function is

$f(x) =$

$\dfrac{\sin^{-1}(x-3)}{\sqrt{9-x^2}}$ .

Now,

$f(x)$ will be defined if

$|x-3|\le 1$

or,

$-1\le x-3\le 1$

or,

$2\le x\le 4$ ......(1),

and

${{9-x^2}}\gt 0$

or,

$x^2<9$

or,

$-3<x<3$ .....(2).

$f(x)$ will be defined in the common portion of (1) and (2).

And the common protion is

$2\le x\lt 3$ or,

$x\in [2,3)$ .

CBSE Questions for Class 11 Commerce Applied Mathematics Functions Quiz 8 - 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 8 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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