If $$fog = |\sin x|$$ and $$gof = \sin^{2}\sqrt {x}$$, then $$f(x)$$ and $$g(x)$$ are

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

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

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

If $$f(x)=\left| x \right| ,x\in R$$, then

$$f(x)=\left( f\circ f \right) \left( x \right) $$

$$f(x)=\left( f\times f \right) \left( x \right) $$

$$f(x)=\left( f\times f \right) \left( x^2 \right) $$

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

Range of f(g(x)) is $$(0,\frac{7}{2\sqrt{3}})$$

MCQ Questions for Class 11 Commerce Applied Mathematics Functions Quiz 6

Let

$f : R \rightarrow R$ be defined by

$f(x) = x^4$ then

0%
f may be one-one and onto
0%
f is one-one and onto
0%
f is one-one but not onto
0%
f is neither one-one nor onto

Explanation

Given

$f:R\rightarrow R$ defined by

$f\left( x \right) ={ x }^{ 4 }$
Since

$-2$ and

$2$ are in the domain and

$f\left( -2 \right) =4=f\left( 2 \right)$
Thus,

$-2$ and

$2$ have same images under

$f$

Hence,

$f$ is not a one-one function.

Also, there is no real number which maps to

$-2$ or any other negative real number in co-domain because if

${ x }^{ 4 }=-2\Rightarrow x=\pm \sqrt { -2 },$ which is not a real number

Hence,

$f$ is not onto function

$\therefore f\left( x \right)$ is neither one-one nor onto.

$f:[0, \infty)\to [0,\infty)$ and

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

$f$ is

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

Explanation

Here,

$f:[0, \infty) \rightarrow [0, \infty)$ i.e, domain is

$[0, \infty)$ and codomain is

$[0, \infty)$

For one-one

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

$f'(x) = \dfrac{1}{(1+x)^2} > 0, \forall x \in [0, \infty)$

$\therefore f(x)$ is increasing in this domain. Thus

$f(x)$ is one-one in its domain

For onto (we find range)

$f(x) = \dfrac{x}{1+x}$ i.e.

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

$\Rightarrow y+yx=x$

$\Rightarrow x=\dfrac{y}{1-y}$

$\Rightarrow x=\dfrac{y}{1-y} \ge 0$ as

$x \ge 0$

$\therefore 0 \le y \neq 1$ and

$y<1$

$\therefore$ Range

$\neq$ Codomain

$\therefore f(x)$ is one-one but not onto

The function

$f:A\rightarrow B$ given by

$f(x) = x ,x\in A$ , is one to one but not onto. Then;

0%
$B\subset A$
0%
$A=B$
0%
$A'\subset B'\quad$
0%
$A\subset B$
0%
$A'\cap B'=\phi$

Explanation

$f(x)=x$ is identity function
It is given that

$f(x)$ is one to one but not on to.
It means some element in

$B$ has no pre image in

$A$

$A\subset B$

The domain of definition of function

$f(x) = \dfrac {1 + 2(x + 4)^{-0.5}}{2 - (x + 4)^{0.5}} + (x + 4)^{0.5} + 4(x + 4)^{0.5}$ is

0%
$R$
0%
$(-4, 4)$
0%
$R^{+}$
0%
$(-4, 0)\cup (0, \infty)$

Explanation

We have,

$f(x) = \dfrac {1 + 2(x + 4)^{-0.5}}{2 - (x + 4)^{0.5}} + (x + 4)^{0.5} + 4(x + 4)^{0.5}$

$\Rightarrow f(x) = \dfrac {1 + \dfrac {2}{\sqrt {x + 4}}}{2 - \sqrt {x + 4}} + \sqrt {x + 4} + 4\sqrt {x + 4}$
Clearly,

$f(x)$ is defined for

$x + 4 > 0$ and

$x\neq 0$ . So, domain of

$f(x)$ is

$(-4, 0)\cup (0, \infty)$ .

$fog = |\sin x|$ and

$gof = \sin^{2}\sqrt {x}$ , then

$f(x)$ and

$g(x)$ are

0%
$f(x) = \sqrt {\sin x}, g(x) = x^{2}$
0%
$f(x) = |x|, g(x) = \sin x$
0%
$f(x) = \sqrt {x}, g(x) = \sin^{2}x$
0%
$f(x) = \sin \sqrt {x}, g(x) = x^{2}$

Explanation

$fog = f\left \{g(x)\right \} = |\sin x| = \sqrt {\sin^{2}x}$

Also,

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

Obviously,

$\sqrt {\sin^{2}x} = \sqrt {g(x)}$

and

$\sin^{2} \sqrt {x} = \sin^{2} \left \{f(x)\right \}$
i.e.,

$g (x) = \sin^{2}x$

and

$f(x) = \sqrt {x}$ .

The domain of the function

$f(x) = \sin^{-1} \left( \dfrac {x+5}{2} \right)$ is :

0%
$[-1.1]$
0%
$[2,3]$
0%
$[3,7]$
0%
$[-7,-3]$
0%
$(- \infty , \infty )$

Explanation

Given,

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

$\because$ Domain of

$sin^{-1} x$ is

$\epsilon [-1,1]$ ]
For

$f(x)$ to be defined

$-1 \le \dfrac {x+5}{2} \le 1$

$\Rightarrow -2 \le x + 5 \le 2$

$\Rightarrow -2 -5 \le x \le 2 - 5$

$\Rightarrow -7 \le x \le - 3$
Thus

$x \epsilon [-7, -3]$

Let

$f(x) = |x - 2|$ , where

$x$ is a real number. Which one of the following is true?

0%
$f$ is periodic
0%
$f(x + y) = f(x) + f(y)$
0%
$f$ is an odd function
0%
$f$ is not one-one function
0%
$f$ is an even function

Explanation

From the graph of the function it can be observed clearly that

$f$ is not one-one function.

So option

$D$ is correct.

$A = \left \{ 1 , 3 , 5 , 7 \right \}$ and

$B = \left \{ 1 , 2 , 3, 4 , 5 , 6 , 7 , 8 \right \}$ then the number of one-to-one functions from

$A$ into

$B$ is

0%
1340
0%
1860
0%
1430
0%
1880
0%
1680

Explanation

Given,

$A = \{1, 3, 5, 7\}$
and

$B =\{1, 2, 3, 4, 5, 6, 7, 8\}$
Here,

$n(A) = 4$ and

$n(B) = 8$
Thus number of one-one function from

$A$ into

$B$

$= \,^8 P_4 = 8.7.6.5 = 1680$

$f(x)=3x+5$ and

$g(x)={ x }^{ 2 }-1$ , then

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

$({ x }^{ 2 }-1)$ is equal to

0%
$3{ x }^{ 4 }-3x+5$
0%
$3{ x }^{ 4 }-6{ x }^{ 2 }+5$
0%
$6{ x }^{ 4 }+3{ x }^{ 2 }+5$
0%
$6{ x }^{ 4 }-6x+5$
0%
$3{ x }^{ 2 }+6x+4$

Explanation

$f(x)=3x+5$
and

$g(x)={x}^{2}-1$

$\therefore f\circ g\left( { x }^{ 2 }-1 \right) =f[g\left( { x }^{ 2 }-1 \right) ]=f\left[ { \left( { x }^{ 2 }-1 \right) }^{ 2 }-1 \right]$

$=3\left( { x }^{ 4 }-2{ x }^{ 2 } \right) +5=3{ x }^{ 4 }-6{ x }^{ 2 }+5$

$g(x)=1+\sqrt{x}$ and

$f\{g(x)\}=3+2\sqrt{x}+x$ , then

$f(x)$ is equal to

0%
$1+2x^2$
0%
$2+x^2$
0%
$1+x$
0%
$2+x$

Explanation

Given,

$g(x)=1+\sqrt{x}$
and

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

$(i)$

$\Rightarrow f(1+\sqrt{x})=3+2\sqrt{x}+x$
Put

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

$\therefore f(y)=3+2(y-1)+(y-1)^2=3+2y-2+y^{2}-2y+1=2+y^2$

$\therefore f(x)=2+x^2$

$f\left( x \right)$ and

$g\left( x \right)$ are two functions with

$g\left( x \right) =x-\dfrac { 1 }{ x }$ and

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

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

0%
$3{ x }^{ 2 }+3$
0%
${ x }^{ 2 }-\dfrac { 1 }{ { x }^{ 2 } }$
0%
$1+\dfrac { 1 }{ { x }^{ 2 } }$
0%
$3{ x }^{ 2 }+\dfrac { 3 }{ { x }^{ 4 } }$

Explanation

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

${ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } }$ using

${ \left( a-b \right) }^{ 3 }={ a }^{ 3 }-{ b }^{ 3 }-3ab\left( a-b \right)$ , we have

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

$\Rightarrow { x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } ={ \left( x-\dfrac { 1 }{ x } \right) }^{ 3 }+3\left( x-\dfrac { 1 }{ x } \right)$
We have,

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

$g\left( x \right) =x-\dfrac { 1 }{ x }$ , this yields

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

$x-\dfrac { 1 }{ x } =t$ , we get

$f\left( t \right) ={ t }^{ 3 }+3t$
Thus,

$f\left( x \right) ={ x }^{ 3 }+3x$
and

$f^{ ' }\left( x \right) =3{ x }^{ 2 }+3$

$f : R - \left \{-1, k\right \} \rightarrow R - \left \{\alpha, \beta \right \}$ is a bijective function defined by

$f(x) = \dfrac {(2x - 1)(2x^{2} - 4px + p^{3})}{(x + 1)(x^{2} - p^{2}x + p^{2})}$ (where

$p\geq 0$ ), then identify which of the following statement(s) is (are) correct?

0%
If $k\epsilon (-1, 1)$ then $\alpha + \beta = 2$
0%
If $k\epsilon (1, 3)$ then $\alpha + \beta = 6$
0%
If $k\epsilon (1, 3)$ then $\alpha + \beta = 4$
0%
If $k\epsilon (-1, 1)$ then $\alpha + \beta = 6$

$f(x)=\left| x \right| ,x\in R$ , then

0%
$f(x)=\left( f\times f \right) \left( x \right)$
0%
$f(x)=x$
0%
$f(x)=\left( f\times f \right) \left( x^2 \right)$
0%
$f(x)=\left( f\circ f \right) \left( x \right)$

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

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

$f(x)$ is

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

Explanation

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

$\Rightarrow \cfrac { 1 }{ 2 } g\left[ f(x) \right] =2{ x }^{ 2 }-5x+2\quad$

$\left[ { \left\{ f(x) \right\} }^{ 2 }+\left\{ f(x) \right\} -2 \right] =2\left[ 2{ x }^{ 2 }-5x+2 \right] \quad \left[ \because g(x)={ x }^{ 2 }+x-2 \right]$

$\Rightarrow f{ \left( x \right) }^{ 2 }+f(x)-\left( 4{ x }^{ 2 }-10x+6 \right) =0$ ....represents quadratic equation

We have a formula for solving quadratic equation

$ax^2+bx+c=0$ is

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$\therefore f(x)=\cfrac { -1\pm \sqrt { 1+4\left( 4{ x }^{ 2 }-10x+6 \right) } }{ 2 }$

$=\cfrac { -1\pm(4x-5) }{ 2 } =2x-3$ or

$2-2x$

$(ax^2 + bx + c)y +a'x^2+b'x+c=0$ , then the condition that x may be a rational function of y is

0%
$(ac'-a'c)^2=(ab'-a'b)(bc'-b'c)$
0%
$(ab'-a'b)^2=(ab'-a' c)(bc'-b'c)$
0%
$(bc'-b'c)^2=(ab'-a'b)(ac'-a'c)$
0%
None of these

$f(x)=\sin ^{ 2 }{ x } +\sin ^{ 2 }{ \left( x+\cfrac { \pi }{ 3 } \right) } +\cos { x } \cos { \left( x+\cfrac { \pi }{ 3 } \right) }$ and

$g\left( \cfrac { 5 }{ 4 } \right) =1$ , then

$g\circ f(x)$ is equal to

0%
$0$
0%
$1$
0%
$\sin { { 1 }^{ o } }$
0%
None of these

Explanation

Given,

$f(x)=\sin ^{ 2 }{ x } +\sin ^{ 2 }{ \left( x+\cfrac { \pi }{ 3 } \right) } +\cos { x } \cos { \left( x+\cfrac { \pi }{ 3 } \right) }$

$=\sin ^{ 2 }{ x } +\left( \sin { x } \cos { \cfrac { \pi }{ 3 } } +\cos { x } \sin { \cfrac { \pi }{ 3 } } \right) +\cos { x } \left( \cos { \cfrac { \pi }{ 3 } } -\sin { x } \sin { \cfrac { \pi }{ 3 } } \right)$

$=\sin ^{ 2 }{ x } +\cfrac { \sin ^{ 2 }{ x } }{ 4 } +\cfrac { 3\cos ^{ 2 }{ x } }{ 4 } +\cfrac { 2\sqrt { 3 } }{ 2.2 } \sin { x } \cos { x } +\cfrac { \cos ^{ 2 }{ x } }{ 2 } -\cos { x } \sin { x } \cfrac { \sqrt { 3 } }{ 2 }$

$=\cfrac { 5 }{ 4 } \left( \sin ^{ 2 }{ x } +\cos ^{ 2 }{ x } \right) =\cfrac { 5 }{ 4 } \quad$
Therefore,

$g\circ f(x)=g(f(x))=g\left( \cfrac { 5 }{ 4 } \right) =1\quad \quad$

$f(x)=ax+b$ and

$g(x)=cx+d$ , then

$f\left( g(x) \right) =g\left( f(x) \right) \Leftrightarrow$

0%
$f(a)=g(c)$
0%
$f(b)=g(b)$
0%
$f(d)=g(b)$
0%
$f(x)=g(a)$

Explanation

We have

$f(x)=ax+b,g(x)=cx+d$
Therefore,

$f\left\{ g(x) \right\} =g\left\{ f(x) \right\} \Leftrightarrow f(cx+d)=g(ax+b)$

$\Leftrightarrow a(cx+d)+b=c(ax+b)+d$

$\Leftrightarrow ad+b=cb+d\Leftrightarrow f(d)=g(b)$

Number of solution of the equation

$f ( x ) = g ( x )$ are same as number of point of intersection of the curves

$y = f ( x )$ and

$y = g ( x )$ hence answer the following question.
Number of the solution of the equation

$2 ^ { x } = | x - 1 | + | x + 1 |$ is

0%
$0$
0%
$1$
0%
$2$
0%
$\infty$

The domain of definition of the function y(x) given by the equation

$a^x + a^y = a(a>1)$ is

0%
$0 < x \leq 1$
0%
$0$ $\leq$ $x < 1$
0%
$- \infty < x < 1$
0%
$- \infty < x \leq 0$ .

Explanation

$a^x+a^y=a\Rightarrow a^y=a-a^x\\\Rightarrow y=\log_a{(a-a^x)}$

$\Rightarrow a-a^x>0\quad$ (Since logarithm of a negative number do not exist)

$\Rightarrow a^x<a$

This happen only if the power of x is less than

$1\quad$ (Since a>1)

$\Rightarrow x\in(-\infty,1)$

$\Rightarrow -\infty<x<1\quad[C]$

Let

$f(x)={ x }^{ 3 }-3x+1$ . The number of different real solutions of

$f(f(x))=0$

0%
$2$
0%
$4$
0%
$5$
0%
$7$

The domain of the function

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

0%
$\left( -\infty ,-3 \right) \cup \left( 3,\infty \right)$
0%
$\left[ -3,3 \right]$
0%
$\quad (-\infty ,-3]\cup [3,\infty )$
0%
$\phi$

If $f:R\rightarrow R$ , $g:R\rightarrow R$ are defined by $f(x)=5x-3$ , $g(x)=x^{2}+3$ , then $(gof^{-1})(3)$ =

0%
$\dfrac{25}{3}$
0%
$\dfrac{111}{25}$
0%
$\dfrac{9}{25}$
0%
$\dfrac{25}{111}$

Explanation

$f:R\rightarrow R$

$g:R\longrightarrow R$

$f\left( x \right) =5x-3$

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

$\left(gof^{-1}\right)\left(3\right)$

$f\left(x\right)={5}{x}-{3}$

Let

$f\left(x\right)=y$

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

$f\left(x\right)=y={5}x-3$

$\Rightarrow x=\dfrac{y+3}{5}$

$\Rightarrow f^{-1}\left(y\right)=\dfrac{y+3}{5}$

$f^{-1}\left(x\right)=\dfrac{x+3}{5}$ and

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

$\left(gof^{-1}\right)\left(3\right)=g\left[f^{-1}\left(3\right)\right]=g\left(\dfrac{3+3}{5}\right)=g\left(\dfrac{6}{5}\right)$

$=\left[\left(\dfrac{6}{5}\right)^{2}+3\right]=\dfrac{36}{25}+{3}=\dfrac{111}{25}$

Hence,

$\dfrac{111}{25}$ is the correct answer.

If the real-valued function

$f(x)=px+\sin x$ is a bijective function, then the set of all possible values of

$p\epsilon R$ is?

0%
$R-\{0\}$
0%
$R$
0%
$(0, \infty)$
0%
None of these

Explanation

Given, the function

$f(x)=px \ + \ \sin{x}$ is a bijective function. Which implies, there exists a function

$g(x)$ such that

$f(g(x))=g(f(x))=x$ for,

$x \epsilon R$ .

This argument doesn't work if

$p=0$ because it would involve division by zero.

Hence, the set of all possible values of

$p \epsilon R$ is

$R- \left \{0 \right \}$ .

$f,g:R\rightarrow R$ are functions such that

$f(x)=3x-\sin { \left( \cfrac { \pi x }{ 2 } \right) } ,g(x)={ x }^{ 3 }+2x-\sin { \left( \cfrac { \pi x }{ 2 } \right) }$
The value of

$\cfrac { d }{ dx } { f }^{ -1 }{ \left( { g }^{ -1 }(x) \right) }_{ x=12 }$ is equal to

0%
$\cfrac { 2 }{ 30+x }$
0%
$\cfrac { 2 }{ 30-x }$
0%
$\cfrac { 2 }{ 3\left( 28-\pi \right) }$
0%
$\cfrac { 2 }{ 3\left( 28+\pi \right) }$

Explanation

Given that,

$g:R \to R$

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

$g\left( x \right) = {x^3} + 2x - \sin \left( {\dfrac{{\pi x}}{2}} \right)$

$\therefore \dfrac{{d{f^{ - 1}}}}{{dx}}{\left( {g'\left( x \right)} \right)_{x = 12}} = \dfrac{2}{{30 + x}}$

Hence, the option

$(A)$ is correct.

The graph of a constant function

$f(x)=k$ is?

0%
A straight line parallel to $X$ -axis
0%
A straight line parallel to $Y$ -axis
0%
A straight line passing through orgin
0%
None

Explanation

A graph of a constant function is always a horizontal line.
Horizontal line is a line which is parallel to

$x$ -axis.

$f,g,h$ are three functions from a set of positive real numbers into itself satisfying the condition,

$f(x) \cdot g(x)=h \sqrt{x^2 + y^2}$ such that

$x,y \epsilon (0,\infty)$ .then,

$\dfrac{f(x)}{g(x)}$ is a?

0%
Constant function
0%
Identity function
0%
Zero function
0%
Signum function

A constant function is a periodic function.

0%
True
0%
False

Explanation

$f(x)=f(x+a)$
A constant function is a period with any value of

$'a'$ as a period

$f: R\rightarrow R$ and

$g: R\rightarrow R$ are defined

$f(x) = x - [x]$ and

$g(x) = [x]\forall x\epsilon R, f(g(x))$ .

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

Explanation

We have,

$f(x)=x-[x]=\{x\}\dots (1)$

where

$\{x\}$ is the fractional part of

$x$

$g(x)=[x]$ , where

$[x]$ is the greatest integer part of

$x$ , then

The value of

$g(x)$ will always be an integer.

And

$\therefore f(g(x))=f([x])=0$

$\because$ the fractional part of

$g(x)$ i.e

$[x]$ is always

$0$ .

Find number of all such function

$y = f(x)$ which are onto?

0%
$243$
0%
$93$
0%
$150$
0%
None of these

Explanation

1. For all possible functions from A to B

${ 3 }^{ 5 }= 243$ (as each value in A can get matched to any one of the elements in B)

Now we remove functions in which all match to one elements only.

3. Number of such elements $= 1 + 1 + 1= 3$

(One when all match to 6, one when all match to 7 and one when all match to 8)

For functions in which 2 elements are only matched.

4. $3[\binom {5} {1} + \binom {5} {2} + \binom {5} {3} + \binom {5} {4}]=90$

If we select any two numbers from $6,7$ and $8.$ Possible ways are $3$ .

Now we distribute these $5$ numbers from $1$ to $5$ in different possible ways between the two numbers.

This is done by $[\binom {5} {1} + \binom {5} {2} + \binom {5} {3} + \binom {5} {4}]$ .

5. Hence the total onto functions are $= 243 - ( 3 + 90 ) = 150$

On differentiating an identity function, we get?

0%
Signum function
0%
Sinc function
0%
Constant function
0%
None

Explanation

Derivative of an Identity function gives a Constant function

$\dfrac{d(cx)}{dx} = c$

Where

$c$ is a constant.

Find the domain of definition of

$f(x) = \dfrac {\log_{2}(x + 3)}{x^{2} + 3x + 2}$ .

0%
$(-3, \infty)$
0%
$\left \{-1, -2\right \}$
0%
$(-3, \infty) - \left \{-1, -2\right \}$
0%
$(-\infty, \infty)$

Explanation

We know

$\log _{ 2 }{ P }$

$P>0$

and hence here

$x+3>0\\ \Rightarrow x>-3$

Also,

$x^2+3x+2\neq 0 \Rightarrow x^{2}+x+2x+2\neq0\\ \Rightarrow x(x+1)+2(x+1)\neq0\\ \Rightarrow (x+1)(x+2)\neq0\\ \Rightarrow x\neq -1, -2\\ \therefore x\sum { \left( -3,\infty \right) } -\left\{ -1,-2 \right\}$

Answer C

$f(x_{1})=f(x_{2})\Rightarrow x_{1}=x_{2}\forall x_{1},x_{2}\epsilon A$ , then what type of a function is

$f:A\rightarrow B?$

0%
One-one
0%
Constant
0%
Onto
0%
Many one

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

0%
Domain of f(g(x)) ids[0,1]
0%
Range of f(g(x)) is $(0,\frac{7}{2\sqrt{3}})$
0%
f (g(x)) is many -one function
0%
f(g(x)) is unbounded function

The domain of the function

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

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

Explanation

For domain of the given function:-

$(|x|-x)^\dfrac{1}{2} >0$

$|x|-x >0$

therefore,

$|x|-x$ , wil be positive only if

$x<0$

hence,

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

If A = {1, 3, 5, 7} , B = {2, 4, 6, 8, 10} and let R = {(1,8), (3,6), (5,2), (1,4)} be a relation from A to B. Then,

Domain (R) = ?

0%
$\{1, 3, 5\}$
0%
$\{8, 6, 2, 4\}$
0%
$\{1, 2, 3, 4\}$
0%
None of these

Explanation

The set of all first components or coordinates of the ordered pairs belonging to

$R$ is called the domain of

$R.$

Thus, Domain

$(R) = \{1, 3, 5\}$

Consider the following functions are odd function in their default domains
(i)

$\cfrac { { 2 }^{ x }-1 }{ { 2 }^{ x }+1 }$
(ii)

$\cfrac { { x }^{ 2 }+1 }{ x\sin { x } }$
(iii)

$\ln { \left( \cfrac { 1+x }{ 1-x } \right) }$
(iv)

$x{ e }^{ \left| x \right| +\cos { x } }\quad$
Which of these is/are odd

0%
(i) and (iii)
0%
(i) and (iv)
0%
all four
0%
(i), (iii) and (iv)

Explanation

$(i)$

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

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

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

Hence is an odd function.

$(ii)$

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

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

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

Hence is an even function.

$(iii)$

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

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

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

Hence is an odd function.

$(iv)$

$f(x)=xe^{|x|+\cos{x}}$

$f(-x)=(-x)e^{|-x|+\cos{(-x)}}$

$f(-x)=-xe^{|x|+\cos{x}}=-f(x)$

Hence is an odd function.

Hence (i), (iii) and (iv) are odd functions.

Answer-(D)

Let

$f$ be an injective map with domain {x, y, z} and range {1, 2, 3} such that exactly one of the following statements is correct and the remaining are false :

$f (x) = 1, f (y) \sqrt 1, f (z) \sqrt 2$ . The value of

$f^{-1} (1)$ is

0%
x
0%
y
0%
z
0%
none of these

Explanation

$f(x)=1,\quad f(y)\neq 1,\quad f(z)\neq 1$

Case 1:

$f(x)=1\\ f(z)=2\\ f(y)=1$

$\therefore f$ is not injective

Case 2:

$f(y)\neq 1,\quad f(z)=2,\quad f(x)=1$

Case 3:

$f(z)\neq 2\quad \quad \quad f(z)=3\\ f(x)\neq 1\quad \quad \quad f(x)=2\\ f(y)=1\quad \quad \quad f(y)=1\\ f(x)=2,f(y)=1,f(z)=3\\ f^{ -1 }\left( 2 \right) =x,f^{ -1 }\left( 1 \right) =y,f^{ -1 }\left( 3 \right) =z$

Let

$f(x)=\dfrac{[x]} {[x+2]}$ . Find the domain of

$f(x)$

0%
$x\,\epsilon R, x$ is not an integer
0%
$(-\infty,-2)\cup[-1,\infty)$
0%
$x\,\epsilon R, x\neq 1$
0%
$(-\infty,-1]$

Explanation

As from the definition of

$[x]$ we get,

$[x]=\begin{cases}.\\.\\ -1 &\mbox{for}& -1\le x\lt 0 \\ 0 &\mbox{for}&\ 0\le x\lt 1\\ 1 &\mbox{for}& \ 1\le x\lt 2\\.\\.\end {cases}$ .

Now the given function

$f(x)$ will be defined for

$[x+2]\ne 0$ .

And this occurs of

$x \in (-\infty,-2)\cup [-1,\infty)$ .

So option (B) is correct.

$f(x)$ is a real valued function, then which of the following is one-one function?

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

Explanation

$(1)$

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

Since

$f(x)=f(-x)$

$f(x)$ is many one.

$(2)$

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

so,

$f(x)=e^x$

and as

$e^x$ is one-one

$f(x)$ is one-one.

$(3)$

$f(x)=\sin x$

Since

$\sin x$ is many one

$f(x)$ is also many-one.

$(4)$

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

Since

$\sin x$ is many one

$f(x)$ is also many-one.

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

$B = \{4, 5\}$ then the number of function

$f : A \rightarrow B$ which is not onto is ______

0%
$2$
0%
$6$
0%
$8$
0%
$4$

Explanation

Total number of function

$f:A\rightarrow B$ is the number of relations from

$A$ to

$B$ such that all the elements of

$A$ are in the first elements of function

$f$ .

Hence, total number of functions are

$=2\times 2\times 2 =2^3=8$

There are two functions for which the function is

$\{(1,4),(2,4),(3,4)\}$ and

$\{(1,5),(2,5),(3,5)\}$ .

Hence there are two non-onto functions.

$f\,: R \rightarrow R, g: R \rightarrow R\,$ are defined by

$f(x)= 5x -3,g(x)=x^2 + 3$ , then,

$(gof^{-1})(3) =$

0%
$\displaystyle \frac{25}{3}$
0%
$\displaystyle \frac{111}{25}$
0%
$\displaystyle \frac{9}{25}$
0%
$\displaystyle \frac{25}{111}$

Explanation

$f(x)=5x-3,$

$g(x)=x^2+3$

Let

$f(x)=y$

$\Rightarrow$

$5x-3=y$

$\Rightarrow$

$x=\dfrac{y+3}{5}$ ----( 1 )

Now,

$f(x)=y$

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

$f^{-1}(y)=\dfrac{y+3}{5}$ [ From ( 1 ) ]

$f^{-1}(x)=\dfrac{x+3}{5}$

Next,

$(gof^{-1})(3)$

$\Rightarrow$

$g[f^{-1}(3)]$

$\Rightarrow$

$g\left[\dfrac{3+3}{5}\right]$

$\Rightarrow$

$g\left[\dfrac{6}{5}\right]$

$\Rightarrow$

$\left(\dfrac{6}{5}\right)^2+3$ [ Since,

$g(x)=x^2+3$ ]

$\Rightarrow$

$\dfrac{36}{25}+3$

$\Rightarrow$

$\dfrac{36+75}{25}$

$\Rightarrow$

$\dfrac{111}{25}$

A function R on the set N of natural numbers is defined as R ={

$(2n, 2n+1):n\in N$ }
The domain of R= {2, 4, 6, 8,............}

0%
True
0%
False

Explanation

$R=\left\{(x,y); x, y\ \in R\right\}$

Domain of

$R$ is all values of

$x$

hence here

$R=\left\{(2n, 2m+1); n\ \in N\right\}$

$\therefore \$ Domain of

$'R'$ is

$2n; n\in N$

$\Rightarrow \ \boxed {R=\left\{2, 4, 6,....\right\}}$

Let

$f:A \to b$ be a function defined by f(x) =

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

0%
f(x) is one-one if A =[0,1]
0%
f(x) is onto if B = [0,1]
0%
f(x) is one-one if A =[-1 , 0]
0%
f(x) is onto if B = [-1,1]

Explanation

$f:A\rightarrow B$

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

When

$x=0$

$f(0)=\sqrt { 1-0 }$

$=1$

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

$=0$

$f(x)$ is one - one function when

$A=[0,1]$

$f:R\rightarrow R,f(x)=\begin{cases} 1\quad \quad x>0 \\ 0\quad \quad x=0 \\-1\quad x<0 \end{cases}$ and

$g:R\rightarrow R,g(x)=\left[ x \right]$ , then

$\left( f\circ g \right) \left( \pi \right)$ is:

0%
$\pi$
0%
$0$
0%
$1$
0%
$-1$

Explanation

Given that

$g(x)=[x]$

$\Rightarrow$

$g(\pi )=g(3.141....)$

i.e

$g(\pi )=3$

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

$\Rightarrow$

$(f\circ g)(\pi ))=f(g(\pi ))$

But

$g(\pi)=3$

$\Rightarrow f(3)=1$ ......

$[\because f(x)=1 \text{ for } x>0]$

$f: ( 0,\infty )\rightarrow [0,\infty )$ defined by

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

0%
one - one but not onto
0%
onto but not one - one
0%
bijective
0%
neither one - one nor onto

Let

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

$g\left( x \right) = \sqrt x$ (where

$x > 0$ ),then

0%
$f\left( {g\left( x \right)} \right) = x$
0%
$g\left( {f\left( x \right)} \right) = x$
0%
The least value of $f\left( {g\left( x \right)} \right) + {1 \over {g\left( {f\left( x \right)} \right)}}$ is $2$
0%
The least value of $g\left( {f\left( x \right)} \right) + {1 \over {f\left( {g\left( x \right)} \right)}}$ is $2$

Explanation

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

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

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

$f(g(x))+\dfrac{1}{g(f(x))}=g(f(x))+\dfrac{1}{f(g(x))}=x+\dfrac{1}{x}\ge 2$

Hence, all options are correct.

Let

$A = \{ 1,2,3,4,5,6\} .$ The number of onto functions from

$A$ to

$A$ such that.

$f\left( x \right) \ne x$ for all

$x \in A$ is

0%
$720$
0%
$240$
0%
$245$
0%
$265$

Explanation

$A=\{1,2,3,4,5,6\}$

$f:A\rightarrow A$

$\therefore \$ Number of onto function

$=n!$

$=6!$

$=6\times 5\times 4\times 3\times 2\times 1$

$\therefore \$ Number of onto function

$=720$

The domain of definition of the function y(x) given by equation

${2^x} + {2^y} = 2$ is

0%
$0 < x \le 1$
0%
$0 \le x \le 1$
0%
$- \infty < x \le 0$
0%
$- \infty < x < 1$

Explanation

${ 2 }^{ x }+{ 2 }^{ y }=2\\ { 2 }^{ y }=2-{ 2 }^{ x }\\ \because { 2 }^{ y }>0\quad [exponential\quad function]\\ \therefore 2-{ 2 }^{ x }>0\\ { 2 }^{ x }<2\\ x<1\\ \therefore x\epsilon (-\infty ,1)$

The domain of the function

$f(x) = sin^{-1} (log_2(\dfrac{x^2}{2}))$ is

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

Explanation

Given

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

We know that in

$\sin^{-1}t-1\le t\le 1$

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

$\Rightarrow \dfrac{1}{2}\le \dfrac{1}{2}x^{2}\le 2$

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

Domain of

$x=(-2, -1]\cup [1,2]$

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

$\frac{1}{2}gof\left( x \right) = 2{x^2} + 5x + 2$ , then

$f\left( x \right)$ is

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

Explanation

$g(x)={ x }^{ 2 }+x-2\\ \cfrac { 1 }{ 2 } g(f(x))=2{ x }^{ 2 }+5x+2\\ \Rightarrow f^{ 2 }\left( x \right) +f(x)-2=4{ x }^{ 2 }-10x+4\\ \Rightarrow f^{ 2 }\left( x \right) +f(x)-(4{ x }^{ 2 }-10x+6)=0\\ f(x)=\cfrac { -1\pm \sqrt { 1+4(4{ x }^{ 2 }-10x+6) } }{ 2 } \\ f(x)=\cfrac { -1\pm \sqrt { 1+16{ x }^{ 2 }-40x+24 } }{ 2 } \\ f(x)=\cfrac { -1\pm (4x-5) }{ 2 } =(2x-3)$

CBSE Questions for Class 11 Commerce Applied Mathematics Functions Quiz 6 - 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 6 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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