If $$g(f(x) ) = |\sin x |$$ and $$f(g(x))=(\sin\sqrt x)^2$$ , then

$$f(x) = \sin^2x. g(x) =\sqrt x$$

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

If $$f(x)=\frac{1}{x},g(x)=\frac{1}{x^{2}} $$ and h $$(x)= x^{2},$$ then

$$fog(x)=x^{2},x\neq 0,h(g(x))= (g(x))^{2},x\neq 0$$

$$fog (x)= x^{2}, x\neq \bar{0}, h(g(x))= \frac{1}{x^{2}}$$

$$h(g(x))= \frac{1}{x^{2}}x\neq 0, fog(x)= x^{2}$$

MCQ Questions for Class 11 Commerce Applied Mathematics Functions Quiz 7

Let f :

$R \to R$ and g :

$R \to R$ be two one-one and onto functions such that they are the mirror images of each other about the line y =If h(x) = f(x) + g(x), then h(0) equal to

0%
2
0%
4
0%
0
0%
1

Explanation

Given

$g(x)$ and

$f(x)$ are mirror images about

$y=2$

$\implies g(x)-2=2-f(x)\implies f(x)+g(x)=4\implies h(x)=4$

$h(0)=4$

$c \to c\,\,is\,defined\,as\,f\left( x \right) = \frac{{ax + b}}{{cx + d}}\,\,bd \ne 0$ .then f is a constant function when

0%
a=c
0%
b=d
0%
ad=bc
0%
ab=cd

Let

$A$ be a set of

$4$ elements and

$B$ has

$3$ elements . From the set of all functions from

$A$ to

$B$ , the probability that it is an onto function is

0%
$\dfrac{4}{9}$
0%
$0$
0%
$\dfrac{29}{32}$
0%
$1$

Explanation

No .of functions from

$A$ to

$B$ is

$3^4=81$ elements

No.of onto functions

$4\times 3^2=36$

Probability

$\dfrac{36}{81}=\dfrac 49$

$f : A \rightarrow B$ defined as

$f(x) = x^2+2x+\frac{1}{1+(x+1)^2}$ is onto function, then set B is equal to

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

Explanation

$f:A\rightarrow B,\quad f(x)={ x }^{ 2 }+2x+\cfrac { 1 }{ 1+{ (x+1) }^{ 2 } } \\ \because \cfrac { 1 }{ 1+{ (x+1) }^{ 2 } } >0\\ \therefore f(x)={ x }^{ 2 }+2x+\cfrac { 1 }{ 1+{ (x+1) }^{ 2 } } \\ a=1,\quad b=2\\ D=4-4\times 1\times (\cfrac { 1 }{ +ve } )\quad \ge 0\\ \because x\epsilon R$

Find the domain on which the function

$f(x) = 2x^2 -5$ and

$g(x) = 5x - 2$ are equal.

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

Explanation

$f(x)=2x^{2}-5$

$g(x)=5x-2$

Find the values of

$x$ for which the above two functions are equal.

That is,

$f(x)=g(x)$

$2x^{2}-5=5x-2$

$2x^{2}-5x-3=0$

$(2x+1)(x-3)=0$

$x=3$ or

$x=-\frac{1}{2}$

The domain of the function

$f(x) = \dfrac{1}{\log_{10}(1-x)} + \sqrt{x+2}$ is equal to

0%
$[-3, -2.5) \cup (-2.5, 2]$
0%
$[-2, 0) \cup (0, 1]$
0%
$[0,1]$
0%
None of these

$g(f(x) ) = |\sin x |$ and

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

0%
$f(x) = \sin^2x. g(x) =\sqrt x$
0%
$f(x) = \sin x , g(x) =|x|$
0%
$f(x) = x^2, g(x) = \sin \sqrt x$
0%
f and g can not be determined

Explanation

$g(f(x))=|sinx|=\sqrt{(sinx)^2}\\f(g(x))=(sin\sqrt x )^2=sin^2\sqrt x\\\therefore f(x)=sin^2x \>and \>g(x)=\sqrt x$

Let

$f$ ,

$g:R\rightarrow {R}$ be two functions defined as

$f\left( x \right) =\left| x \right| +x$ ,

$g\left( x \right) =\left| x \right| -x, \forall x\in R$ . Then, find

$fog(x)$

0%
$||x|-x|-|x|-x$
0%
$||x|-x|+|x|-x$
0%
$||x|-x|-|x|+x$
0%
None of thesse

Explanation

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

$g\left( x \right) = \left| x \right| - x$

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

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

$= \left| {\left| x \right| - x} \right| + \left| x \right| - x$

The set of all values of

$x$ for which

$\dfrac { \sqrt { -{ x }^{ 2 }+5x-6 } }{ \sqrt { 1-2\left\{ x \right\} } } \ge 0$ is (where{.} denotes the fractional part function)

0%
$\left[2,\dfrac{5}{2}\right]\cup{3}$
0%
$(2, 3)$
0%
$\left[\dfrac{5}{2},\ 3\right]$
0%
$\left[2,\dfrac{5}{2}\right]\cup\left[\dfrac{5}{2}, 3\right]$

Consider set

$A={1,2,3,4}$ and set

$B={0,2,4,6,8}$ , then the number of one-one function from set

$A$ to set

$B$ is ?

0%
$5$
0%
$24$
0%
$120$
0%
None of these

Explanation

Number of one-one function from

$\underset{(m)}{A}$ to

$\underset{(n)}{B} = \left\{\begin{matrix} \, ^nP_m, & if \, n \ge m \\ 0, & if \ n < m \end{matrix}\right.$

$m = 4, \ n = 5$

One-one function

$=\, ^5P_4 = \dfrac{5!}{1!} = 120$

The set onto which the derivative of the function

$f(x)=x(\log x-1)$ maps the range

$[1,\infty )$ is

0%
$\left[1,\infty \right)$
0%
$\left( e,\infty \right)$
0%
$\left[e,\infty \right)$
0%
$\left( 0,0 \right)$

Explanation

Given

$f (x) = x [\log x-1] x \in [+, \infty]$

$f (x) = x \left[ \dfrac{d}{dx} (\log x-1) \right]+ (\log x-1) \dfrac{d}{dx} (x)$

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

$=1 -1 \log x-1$

$f(x)= \log x$

$\therefore$ The set of

$f(x)= \log x$ that maps

$[1, \infty )$ is

$[e, \infty )$

$f(x)=2x+5$ and

$g(x)=x^2+1$ be two real function , then value of

$fog$ at x=1

0%
$9$
0%
$6$
0%
$5$
0%
$4$

$f ( x ) = \left( a - x ^ { n } \right) ^ { 1 / n }$ 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 domain of the function,

$f(x) = \dfrac{\sqrt{\sin x}}{\sqrt{(x - 2)(8 - x)}}$ is

0%
$[0, \pi] \cup [2\pi, 8)$
0%
$(2, \pi] \cup [2\pi, 8)$
0%
$(2, 8)$
0%
$(0, 8)$

Explanation

Given the function

$f(x) = \dfrac{\sqrt{\sin x}}{\sqrt{(x - 2)(8 - x)}}$ .

Now

$f(x)$ is defined for

$\sin x\ge 0$ .......(1) and

$(x-2)(8-x)\gt 0$ or,

$(x-2)(x-8)\lt 0$ .......(2).

From (2) we get,

$x\in (2,8)$ ......(3)

Form (1) we get,

$x\in [0,\pi], [2\pi,3\pi],....$ .......(4).

Combining (3) and (4) we get,

$x\in (2,\pi]\cup [2\pi, 8)$ .

Let

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

$F=\{1, 2\}$ then the number of onto functions from E to F is

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

Explanation

Number of onto function from

$E$ to

$F.$

$E=\{1,2,3,4\}$

$F=\{1,2\}$

Number of function from

$E$ to

$F=2\times 2\times 2\times 2=16$

We have to exclude functions where

$f(x)=1$ &

$f(x)=2$ .

$\therefore$ Total number of onto function

$=16-2=14$ .

Let

$f\left( x \right) ={ x }^{ 2 },g\left( x \right) ={ 2 }^{ x }$ , then solution set of

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

0%
R
0%
$\left\{ 0 \right\}$
0%
$\left\{ 0,2 \right\}$
0%
None of these

Let

$f(x+\dfrac{1}{x})=x^2+\dfrac{1}{x^2}(x\neq 0)$ , then

$f(x)=$

0%
$x^2$
0%
$x^2-1$
0%
$x^2-2$
0%
N.O.T

Explanation

We are given

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

$x \neq 0$ )

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

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

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

So, simply put

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

we get

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

If a relation

$'R'$ is defined by

$R = \left\{ {\left( {x,y} \right)/2{x^2} + 3{y^2} \le 6} \right\},$ then the domain of

$R(x,y)$ is

0%
$x\in \left[ { - 3,3} \right]$
0%
$x\in \left[ { - \sqrt 3 ,\sqrt 3 } \right]$
0%
$y\in \left[ { - \sqrt 2 ,\sqrt 2 } \right]$
0%
$y\in \left[ { - 2,2} \right]$

Explanation

$2x^2+3y^2 \leq 6$

$\implies \cfrac{x^2}{3} + \cfrac{y^2}{2} \leq 1$

This is an ellipse with ends being (

$-\sqrt3,0$ ) , (

$\sqrt3,0$ ) , (

$0,-\sqrt2$ ) , (

$0,\sqrt2$ )

Hence the domain for

$x =(-\sqrt3 , \sqrt3)$

Hence the domain for

$y =(-\sqrt2, \sqrt2)$

$f\left( x \right) =\begin{cases} 2+x,\quad x\ge 0 \\ 2-x,\quad x<0 \end{cases}$ then

$f\left( f\left( x \right) \right)$ is given by

0%
$f\left( f\left( x \right) \right) =\begin{cases} 2+x,\quad x\ge 0 \\ 4-x,\quad x<0 \end{cases}$
0%
$f\left( f\left( x \right) \right) =\begin{cases} 2+x,\quad x\ge 0 \\ 2-x,\quad x<0 \end{cases}$
0%
$f\left( f\left( x \right) \right) =\begin{cases} 4+x,\quad x< 0 \\ x,\quad x\ge 0 \end{cases}$
0%
$f\left( f\left( x \right) \right) =\begin{cases} 4+x,\quad x\ge 0 \\ x,\quad x<0 \end{cases}$

$(x)$ is defined on

$(0,1)$ then the domain of defination of

$f[({e^x})]$

$+f\left( {ln\,\left| x \right|} \right)$ is subset of

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

Explanation

$x$ is defined prom

$(0,1)$

then

$f(e^{x})+ f(ln |x|)$ domain is

$0 < e^{x} < 1$ &

$0 < in |x| < 1$

$$- \infty < x < 0 \n \ x \in (-e,-i) v (i,e)$$

$\downarrow$

$x \in (-e, -1)$

$A$ is correct

If f(x)=sin log

$\left( {\sqrt {4 - {x^2}} /\left( {1 - x} \right)} \right)$ then the domain and range f are (respectively)

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

Explanation

consider the given function ,

$f\left( x \right)=\sin {{\log }_{e}}\left( \dfrac{\sqrt{4-{{x}^{2}}}}{1-x} \right)$

Now ,

for $\log_e\dfrac{\sqrt{4x^2}}{1-x}$ to be defined $\dfrac{\sqrt{4-{{x}^{2}}}}{1-x}>0$

$\dfrac{\sqrt{4-{{x}^{2}}}}{1-x}>0$

$\sqrt{4-{{x}^{2}}}>0$ , $x\not=1$ and $1-x>0$

$4-{{x}^{2}}>0$ $1>x$

$4>{{x}^{2}}$ $x<1$

$x^2-4<0$

$(x+2)(x-2)<0$

$-2<x<2$

combining the two we get

$-2<x<1$

$x\epsilon(-2,1)$

Hence , the domain of the function is $(-2,1)$

Hence, this is the range of the function $[-1, 1]$

If f(g(x))=5x+2 and g(x)=8x then f(x)=

0%
$\frac{5}{8}x+2$ .
0%
$\frac{8}{5}x+2$ .
0%
$\frac{5}{8}x-2$ .
0%
8x-2
0%
5x-2

Let

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

$f\left( x \right) =\begin{cases} -1,x<0 \\ 0,x=0 \\ 1,x>0 \end{cases}$ Then for all

$x,f\left( g\left( x \right) \right)$ is equal to (where

$\left[ . \right]$ represents the greatest integer function)

0%
$x$
0%
$1$
0%
$f\left( x \right)$
0%
$g\left( x \right)$

If :

$f(x) = 5 {x}^{2}$ ,

$g(x) = 3x^{4}$ , then :

$(fog) (-1) =$

0%
$45$
0%
$-54$
0%
$-32$
0%
$-64$

Let

$f:X \to \left[ {1,\,27} \right]$ be a function by

$f\left( x \right) = 5\sin x + 12\cos x + 14$ . The set

$X$ so that

$f$ is one-one and onto is

0%
$\left[ { - \pi /2,\pi /2\,} \right]$
0%
$\left[ {0,\,\pi } \right]$
0%
$\left[ {0,\,\pi /2} \right]$
0%
non of these

The distinct linear functions which maps from

$[-1,1]$ onto

$[0,2]$ are

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

Explanation

$[-1,1]\rightarrow [0,2]$ onto are

$x1!=f(x)$

$f(-1)=0$

$f(1)=2$

$-1+2=f(x)$

$f(-1)=2$

$f(1)=0$

For

$a,\ b\ \in \ R-\left\{ 0 \right\}$ , let

$f(x)=ax^{2}+bx+a$ satisfies

$f\left(x+\dfrac{7}{4}\right)=f\left(\dfrac{7}{4}-x\right) \forall \ x\ \in\ R$ .
Also the equation

$f(x)=7x+a$ has only one real distinct solution. The minimum value of

$f(x)$ in

$\left[0,\dfrac{3}{2}\right]$ is equal to

0%
$\dfrac{-33}{8}$
0%
$0$
0%
$4$
0%
$-2$

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

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

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

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

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

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

$\therefore f(f(x))=(x^n)^{1/n}$

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

The identity function on real numbers given by

$f(x)=x$ is continuous at every real numbers.

0%
True
0%
False

The set of all

$x$ for which the functions are not defined

$f\left( x \right) =\log _{ [(x-2)/( x+3)]}2$ and

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

0%
$\left[ -3,2 \right]$
0%
$\left[ -3,2 \right)$
0%
$(-3,2]$
0%
$\left( -3,-2 \right)$

Explanation

$g(x) = \dfrac{1}{\sqrt{x^2 - 9}}$

$x^2 - 9 > 0$

$x>3,x<-3$

but we require

$-3 \leq x \leq 3$

$\Rightarrow x \in [-3, 3]$ as

$g(x)$ will not be defined in this interval

$f(x) = \log \left(\dfrac{x - 2}{x + 3} \right)$

$\dfrac{x - 2}{x + 3} > 0, 1$

$x > 2$

but we require

$x \leq 2$

$\Rightarrow x \in [-\infty,2]$ as

$f(x)$ will not be defined in this interval

Taking common of both interval we get

$-3 \leq x \leq 2$

$\therefore x \in [-3 , 2]$

From given options option A is appropriate

$f:R \rightarrow R, f(x)=2x-1$ and

$g; R \rightarrow R, g(x)=x^{2}+2$ , then

$(gof)(x)$ equals-

0%
$2x^{2}-1$
0%
$(2x-1)^{2}$
0%
$2x^{2}+3$
0%
$4x^{2}-4x+3$

Explanation

$f(x)=2x-1$

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

$g(f(x))=(2x-1)^2+2=4x^2-4x+3$ .

1191545_1155623_ans_574198b726b647c79ac862bd2b5b5edb.jpg

The domain of definition of

$f\left( x \right) =\sqrt { \log _{ 0.4 }{ \frac { x-1 }{ x+5 } } } \times \dfrac { 1 }{ { x }^{ 2 }-36 }$ , is

0%
$\left( -\infty ,0 \right) -\left\{ -6 \right\}$
0%
$\left( 0,\infty \right) -\left\{ 1,6 \right\}$
0%
$\left( 1,\infty \right) -\left\{ 6 \right\}$
0%
$\left( 1,\infty \right) +\left\{ 6 \right\}$

$f\left( x \right) = (1 - x)$ ,

$x \in \left[ { - 3,3} \right]$ , then the domain of

$f\left( {f\left( x \right)} \right)$ is

0%
$\left[ { - 2,3} \right]$
0%
$\left( { - 2,3} \right)$
0%
$\left[ { - 2,3]} \right.$
0%
$( - 2,3]$

Explanation

$f\left(x\right)=1-x$

$f\left(f\left(x\right)\right)=f\left(1-x\right)=1-1+x=x$

And

$x\in\left[-3,3\right]$

Domain of

$f\left(x\right)$ is

$\left[-3,3\right]$

Domain of

$f\left(f\left(x\right)\right)=$ Domain of

$f\left(1-x\right)$

Domain of

$f\left(f\left(x\right)\right)=\left[1-3,3\right]=\left[-2,3\right]$

$\log_2\left ( 3^{2x+2}+7 \right )=2+\log_2\left ( 3^{x-1}+1 \right )$ , then number of real values of

$x$ is/are

0%
$0$
0%
$1$
0%
$2$
0%
$4$

Explanation

Given,

$log_{2}(3^{2x+2}+7)= 2+log_{2}(3^{x-1}+1)$

$\Rightarrow log_{2}(3^{2x+2}+7)= log_{2}4+log_{2}(3^{x-1}+1)$

$= log_{2}(4)(3^{x-1}+1)$

$[\because log\,ab= loga+ logb]$

As both bases are same,

$\Rightarrow 3^{2x+2}+7=4(3^{x-1}+1)$

$9.3^{2x}-\dfrac{4}{3}.3^{x}+6=0$

Let

$3^{x}=u, \Rightarrow 9u^{2}-\dfrac{4}{3}u+6=0$

$27u^{2}-4u+18=0$

$\Rightarrow u= \dfrac{4\pm \sqrt{16-4(18)(27)}}{2(27)}$

$D= 16-4(18)(27) < 0 \Rightarrow$ No solution exists

$\therefore$ Option A is correct.

$f(x)=\frac{x}{\sqrt{1-x^{2}}}$ and g(x) =

$f(x)=\frac{x}{\sqrt{1+x^{2}}}$ , then (fog)(x) =

0%
$f(x)=\frac{x}{\sqrt{1-x^{2}}}$
0%
$f(x)=\frac{x}{\sqrt{1+x^{2}}}$
0%
$x^{2}$
0%
x

Explanation

Given:-

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

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

To find:-

$fog{\left( x \right)} = ?$

$fog{\left( x \right)}$

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

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

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

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

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

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

$= \cfrac{x}{\sqrt{1 + {x}^{2}}}$

$\Rightarrow fog{\left( x \right)} = \cfrac{x}{\sqrt{1 + {x}^{2}}}$

Hence the correct answer is

$\cfrac{x}{\sqrt{1 + {x}^{2}}}$ .

The domain of the function

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

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

Explanation

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

$for\,domain$

$x-\sqrt{1-{{x}^{2}}}\ge 0$

$\Rightarrow x\ge \sqrt{1-{{x}^{2}}}$

$squaring\,both\,sides$

${{x}^{2}}\ge 1-{{x}^{2}}$

$\Rightarrow 2{{x}^{2}}-1\ge 0$

$\Rightarrow 2{{x}^{2}}\ge 1$

$\Rightarrow {{x}^{2}}\ge \frac{1}{2}$

$\Rightarrow x\ge \pm \frac{1}{\sqrt{2}}$

$rejecting\,the\,negative\,term$

$Domain:\left[ \frac{1}{\sqrt{2}},1 \right]$

The domain of the functions

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

0%
$(0,2)$
0%
$[0,2]$
0%
$\left\{ 1 \right\}$
0%
none

Explanation

$2x-x^{2} > 0$

$x^{2}-2x < 0$

$x (x-2) < 0$

$x \in (0,2)$

also,

$\log (2x-x^{2}) \ge 0$

$2x-x^{2} \ge 1$

$x^{2} - 2x+ 1 \le 0$

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

$x \le 1$

$x 6 \left( 0,1 \right]$

The domain of the function

$f(x)=\cfrac { 1 }{ \sqrt { \left| x \right| -x } }$ is

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$(-\infty, \infty)$
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$(0,\infty)$
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$(-\infty, 0)$
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$(-\infty, \infty)-\left\{ 0 \right\}$

Explanation

$\dfrac{1}{\sqrt{|x|}-x}$

$|x|-x> 0$

$C_{1}\rightarrow x> 0$

$x-x> 0$

$0> 0$

$C_{2}\rightarrow x< 0$

$-x-x> 0$

$-2x> 0$

$x< 0$

$x\epsilon (-\infty ,0)$

1066139_1164988_ans_ebebb33fdf6749f989bccbce9a69b185.jpg

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

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$x$
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$1$
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$f(x)$
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$g(x)$

$f:c \to c$ is defined as

$f(x) = \dfrac{{ax + b}}{{cx + d}},bd \ne 0$ then

$f$ is a constant function when,

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a=c
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b=d
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ad=bc
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ab=cd

Explanation

$x$ )=

$\frac{ax+b}{cx+d}$ is a constant function,

then lets say it equal to same constant m.

$m(cx+d)=ax+b$

$a=mc$

$b=md$

$\frac{a}{c}$ =

$\frac{b}{d}=m$

$\frac{a}{b}$ =

$\frac{c}{d}$

$ad=bc$

C is correct.

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

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

$f(x)=\frac{1}{x},g(x)=\frac{1}{x^{2}}$ and h

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

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$fog (x)= x^{2}, x\neq \bar{0}, h(g(x))= \frac{1}{x^{2}}$
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$h(g(x))= \frac{1}{x^{2}}x\neq 0, fog(x)= x^{2}$
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$fog(x)=x^{2},x\neq 0,h(g(x))= (g(x))^{2},x\neq 0$
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None of these

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

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

$\left( {f(g(x)))} \right)$ is equal to

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$- f\left( x \right)$
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$3f\left( x \right)$
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${\left( {f\left( x \right)} \right)^3}$
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$f\left( {3x} \right)$

Explanation

$f(x)=\log\bigg(\dfrac{1+x}{1-x}\bigg)\ and\ g(x)=\dfrac{3x+x^{3}}{1+3x^{2}}$

$f(g(x))=\log\bigg(\dfrac{1+\dfrac{3x+x^{3}}{1+3x^{2}}}{1-\dfrac{3x+x^{3}}{1+3x^{2}}}\bigg)$

$f(g(x))=\log\bigg(\dfrac{\dfrac{1+3x^{2}+3x+x^{3}}{1+3x^{2}}}{\dfrac{1+3x^{2}-3x-x^{3}}{1+3x^{2}}}\bigg)$

$f(g(x))=\log\bigg(\dfrac{1+3x^{2}+3x+x^{3}}{1+3x^{2}-3x-x^{3}}\bigg)$

$f(g(x))=\log\bigg(\dfrac{(1+x)^3}{(1-x)^{3}}\bigg)$

$f(g(x))=\log\bigg(\dfrac{1+x}{1-x}\bigg)^{3}$

$f(g(x))=3\log\bigg(\dfrac{1+x}{1-x}\bigg)$

$f(g(x))=3f(x)$

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

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$21$
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$12$
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$11$
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$22$

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

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

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$R$
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$\left\{ {0} \right\}$
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$\left\{ {0,\,2} \right\}$
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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

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

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

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

Explanation

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

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

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

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

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

$4 a ^ { 2 } < 36$

$a \in ( - 3,3 )$

Domain of

$f\left( x \right) = \sin ^{ -1 }{ \left[ \log _{ 2 }{ \left( \frac { { x }^{ 2 } }{ 2 } \right) } \right] }$ , where

$\left[ . \right]$ denotes greatest integer functions, is

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$\left[ -\sqrt { 8, } \sqrt { 8 } \right]$
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$\left[ -\sqrt { 8, } -1 \right] \cup (1,\quad \sqrt { 8 } ]$
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$(-2,-1)\cup (1,2)$
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None of these

Explanation

$f(x)sin^{-1}[log_{2}(\frac{x^{2}}{2})]$

For

$sin^{1}x,x$ must lie strietly in the range [-1,1]

$\Rightarrow -1\leq log_{2}\frac{x^{2}}{2}\leq 1$

$\Rightarrow \frac{1}{2}\leq \frac{x^{2}}{2}\leq 2$

$\Rightarrow 1 \leq x^{2}\leq 4$

$\Rightarrow 1\leq x\leq 2$ or

$-2\leq x\leq -1$

$\therefore$ Domain of f is

$[-2,-1]\cup [1,2]$

$f(x)$ is defined on

$(0,1)$ then the domain of definition of

$f({e}^{x})+f(\ln{\left| x \right| })$ is

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$(-e,-1)$
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$(-e,-1)\cup (1,e)$
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$(-5,-1)\cup (1,\infty)$
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$(-e,e)$

Explanation

$g=f{\left( e \right)^x} + f\left( {In\left| x \right|} \right)$

Domain of g

$= = 0 < {e^x} < 1$

$x \in \left( { - \infty ,0} \right)$ -----------------------

$(i)$

And

$,$

$0\left( {In\left| x \right|} \right) < 1$

$\left| x \right| < e$

$x \in \left( { - e, - 1} \right) \cup \left( {1,e} \right)$ ------------

$(ii)$

from

$(i)$ and

$(ii),$ we have

$x \in \left( { - e, - 1} \right)$

Hence

$,$ option

$(A)$ is correct

$.$

The domain of

$\sqrt {|x - 2| - 1} + \sqrt {3 - |x - 2|} \,\,is\,$

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$[ - 1,3]\, \cup \,[5,\infty )$
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$[ - 1,5]$
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$[1,3]$
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$[ - 1,1] \cup [3,5]$

$f:R \rightarrow R$ 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 -

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

Explanation

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

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

$g(2)=?$

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

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

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

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

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

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

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

CBSE Questions for Class 11 Commerce Applied Mathematics Functions Quiz 7 - 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 7 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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