MCQ Questions for Class 11 Commerce Applied Mathematics Functions Quiz 11

Let

$f(x) = \begin{cases} 1+x, & 0\leq x\leq 2 \\ 3-x, & 2<x\leq 3 \end{cases}$ , then find

$(fof)(x)$

0%
$\left\{\begin{matrix}2 + x, &0 \leq x \leq 1 \\ 2 - x, &1 < x \leq 2 \\ 4 - x, &2 < x \leq 3 \end{matrix}\right.$
0%
$\left\{\begin{matrix}2 - x, &0 \leq x \leq 1 \\ 2 + x, &1 < x \leq 2 \\ 4 - x, &2 < x \leq 3 \end{matrix}\right.$
0%
$\left\{\begin{matrix}2 + x, &0 \leq x \leq 1 \\ 2 - x, &1 < x \leq 2 \\ 4 + x, &2 < x \leq 3 \end{matrix}\right.$
0%
None of these

Explanation

Given

$f(x) = \begin{cases} 1+x, & 0\leq x\leq 2 \\ 3-x, & 2<x\leq 3 \end{cases}$

Case I: When

$0\leq x\leq 2$

$(fof)(x)=f[f(x)]=f(1+x)$
Since,

$0\leq x\leq 2$

$\Rightarrow 1\leq 1+x\leq 3$

$(fof)(x)=1+1+x=2+x$ when

$1\leq 1+x\leq 2$

$\Rightarrow (fof)(x)=2+x$ when

$0\leq x\leq 1$
And

$(fof)(x)=3-(1+x)=2-x$ when

$2<1+x\leq 3$

$\Rightarrow (fof)(x)=2-x$ when

$1<x \le 2$

Case I: When

$2< x\leq 3$

$(fof)(x)=f[f(x)]=f(3-x)$
Since,

$0\leq x\leq 2$

$\Rightarrow -2< -x \leq 0$

$\Rightarrow 1< 3-x \le 3$

$(fof)(x)=1+3-x=4-x$ when

$1\le 3-x\leq 2$

$\Rightarrow (fof)(x)=4-x$ when

$1\le x\leq 2$
And

$(fof)(x)=3-(3-x)=x$ when

$2<3-x\leq 3$

$\Rightarrow (fof)(x)=x$ when

$0<x \le 1$

$f(x)\, =\, -1\, + |x\, -\, 2|, \, 0\, \leq\, x\, \leq\, 4$

$g(x)\, =\, 2\, -\, |x|,\, -1\, \leq\, x\, \leq\, 3$
Which of the following is true

0%
$fog(x)\, =\, \begin{cases} -(1\, +\, x), & & -1\, \leq\, x\, \leq\, 0\\ x + 1 & & 0\, <\, x\, \leq\, 2 \end{cases}$
0%
$gof(x)\, =\, \begin{cases}x\,+\,1, & 0\, \leq\, x\, <\, 1\\ 3\, -\, x, & 1\, \leq\, x\, \leq\, 2 \\ x\, -\, 1, & 2\, <\, x\, \leq\, 3\\ 5\, -\, x, & 3\, <\, x\, \leq\, 4\end{cases}$
0%
$fog(x)\, =\, \begin{cases} -(1\, +\, 2x), & & -1\, \leq\, x\, \leq\, 0\\ x - 1 & & 0\, <\, x\, \leq\, 2 \end{cases}$
0%
$gof(x)\, =\, \begin{cases}x\,+\,1, & 0\, \leq\, x\, <\, 1\\ 3\, -\, x, & 1\, \leq\, x\, \leq\, 2 \\ x\, +\, 1, & 2\, <\, x\, \leq\, 3\\ 5\, -\, x, & 3\, <\, x\, \leq\, 4\end{cases}$

Explanation

$f(x)\, =\, -1\, + |x\, -\, 2|, \, 0\, \leq\, x\, \leq\, 4$

$g(x)\, =\, 2\, -\, |x|,\, -1\, \leq\, x\, \leq\, 3$

$g(f(x))=2-|-1+|x-2||,$

$gof(x)=\begin{cases} 2-[2-x-1] & 0\, \leq \, x\, <\, 1 \\ 2+[2-x-1], & 1\, \leq \, x\, \leq \, 2 \\ 2+[x-2-1], & 2\, <\, x\, \leq \, 3 \\ 2-[x-2-1], & 3\, <\, x\, \leq \, 4 \end{cases}\\ gof(x)\, =\, \begin{cases} x\, +\, 1, & 0\, \leq \, x\, <\, 1 \\ 3\, -\, x, & 1\, \leq \, x\, \leq \, 2 \\ x\, -\, 1, & 2\, <\, x\, \leq \, 3 \\ 5\, -\, x, & 3\, <\, x\, \leq \, 4 \end{cases}\\$

If f(x)=x+5 and g(x)=

$\displaystyle \sqrt{x^{2}-9}$ then find the domain of gof(x)

0%
(-8,-2)
0%
$\displaystyle \left ( -\infty ,-8 \right )\cup \left ( -2,\infty \right )$
0%
$\displaystyle (-\infty ,-8]\cup [-2,\infty )$
0%
$\displaystyle (-(-\infty ,-8]\cup [-2,\infty )$

Explanation

$gof(x) = g(f(x)) = \sqrt {{(x+5)}^{2} -9} = \sqrt {x^2 + 25 + 10x -9}$

$= \sqrt {x^2 + 16 + 10x} = \sqrt {(x+8)(x+2)}$

Now, for the square root to have a real value, both

$(x+8)\ and\ (x+2)$ should be positive or equal to zero or both should be negative or equal to zero.

Case 1: When both

$x+8 \ge 0$ and

$x + 2 \ge 0$

$=> x \ge -8$ and

$x\ge -2$

$=> x \ge -2$ is the solution for this case

Case 2: When both

$x+8 \le 0$ and

$x + 2 \le 0$

$=> x \le -8$ and

$x\le -2$

$=> x \le -8$ is the solution for this case.

So, combining these both, we get the domain,

$x \le -8\ and\ x \ge -2$

Let

$f(x)=\ln x$ and

$g(x)\, =\, \left (\displaystyle \frac{x^{4}\, -\, x^{3}\, +\, 3x^{2}\, -\, 2x\, +\, 2}{2x^{2}\, -\, 2x\, +\, 3 )}\right )$ . The domain of

$f(g(x))$ is

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

Explanation

$f(x)=\ln x$ and

$g(x)\, =\, \displaystyle \frac{x^{4}\, -\, x^{3}\, +\, 3x^{2}\, -\, 2x\, +\, 2}{2x^{2}\, -\, 2x\, +\, 3}$

Now,

$f(g(x))=f \left( \displaystyle \frac{x^{4}\, -\, x^{3}\, +\, 3x^{2}\, -\, 2x\, +\, 2}{2x^{2}\, -\, 2x\, +\, 3} \right)$

$\Rightarrow f(g(x))=\ln\left( \displaystyle \frac{x^{4}\, -\, x^{3}\, +\, 3x^{2}\, -\, 2x\, +\, 2}{2x^{2}\, -\, 2x\, +\, 3} \right)$

For log to be defined

$\displaystyle \frac{x^{4}\, -\, x^{3}\, +\, 3x^{2}\, -\, 2x\, +\, 2}{2x^{2}\, -\, 2x\, +\, 3} >0$
The above inequality holds for all

$x\in R$

$f(x)=\begin{cases} x+1,\quad \quad if\quad x\, \leq \, 1 \\ 5-x^{ 2 }\quad \quad if\quad x>1 \end{cases},g(x)=\begin{cases} x\quad \quad if\quad x\leq 1 \\ 2-x\quad if\quad x>1 \end{cases}$

Number of negative integral solutions of

$g(f(x)) + 2 = 0$ are

0%
$0$
0%
$3$
0%
$1$
0%
$2$

Explanation

$f(x)=\begin{cases} x+1,\quad \quad if\quad x\, \leq \, 1 \\ 5-x^{ 2 }\quad \quad if\quad x>1 \end{cases},$

$g(x)=\begin{cases} x\quad \quad if\quad x\leq 1 \\ 2-x\quad if\quad x>1 \end{cases}$

$g(f(x))+2=0,\\ g(f(x))=-2,\\ g(f(x))=\begin{cases} x+1\quad \quad if\quad x\leq 0 \\ 1-x\quad if\quad 1\ge x>0 \\ x^{ 2 }-3\quad if\quad 2>x>1 \\5-x^2\quad if \quad x\geq 2\end{cases}\\ For\quad x<0\quad to\quad be\quad a\quad solution,\\ x+1=-2\longrightarrow 1\quad solution\\$

Hence, option C.

Given two functions

$f(x)$ and

$g(x)$ such that

$f(x) = \sin (arctan x), g(x) =\tan (arc\sin x)$ , and

$0\leq x < \dfrac {\pi}{2}$ . The value of the composite function

$f\left (g\left (\dfrac {\pi}{10}\right )\right )$ is:

0%
$0.314$
0%
$0.354$
0%
$0.577$
0%
$0.707$
0%
$0.866$

Explanation

$g(x)=\tan { (\sin ^{ -1 }{ x } ) } =\tan { \theta }$

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

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

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

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

$f(g(\dfrac { \pi }{ 10 } ))=0.314$

Find

$g(x)$ , if

$f(x) = 5x^{2} + 4$ and

$f(g(3)) = 84$

0%
$3x - 10$
0%
$4x - 7$
0%
$6x - 17$
0%
$x^{2} - 5$
0%
$x^{2} - 3$

Explanation

Given,

$f(x)=5x^2+4$ ,

$f(g(3))=84$

$\Rightarrow f(g(3)) = 5{g(3)}^{2} +4 = 84$

$\Rightarrow 5{g(3)}^{2}=80$

$\Rightarrow g(3) = \sqrt{16} = 4$

$\Rightarrow g(x) = {x}^{2}-5$ satisfies the relation

$g(3) = 4$
Therefore, the correct option is

$D$ .

Let

$f:\{x, y , z\} \rightarrow \{1, 2, 3\}$ be a one-one mapping such that only one of the following three statements and remaining two are false :

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

0%
$f(x) > f(y) > f(z)$
0%
$f(x) < f(y) < f(z)$
0%
$f(y) < f(y) < f(z)$
0%
$f(y) < f(z) < f(x)$

Explanation

$f:\left\{ x,y,z \right\} \rightarrow \left\{ 1,2,3 \right\}$

$I-f\left( x \right) \neq 2$

$II-f\left( y \right) =2$

$III-f\left( z \right) \neq 1$
1. Let statement I is true,

$\therefore$ other two are false

$\therefore f\left( x \right) \neq 2,f\left( y \right) \neq 2$

$\therefore f\left( z \right) =1$

$f\left( x \right) \& f\left( y \right)$ can be

$2$ or

$3$
We don't know exact value of

$f\left( x \right) \& f\left( y \right)$
2. Let statement II is true

$f\left( y \right) =2\quad \therefore f\left( z \right) =1\quad$ [

$\because f\left( z \right) \neq 1$ is false]

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

$f\left( x \right) >f\left( y \right) >f\left( z \right)$

$h(x) = x^2, g(x)= x^2 -3$ and f(x)= x -2, what can you say about ho(gof) and (hog)of?

0%
(hog)of $\neq$ ho(gof)
0%
ho(gof) = (hog)of
0%
(hog)of = 4 ho(gof)
0%
ho(gof)= $(x^2-4x-1)^2$

Which of the following functions are not identical?

0%
$f(x)=\frac{x}{x^2}$ and $g(x) = \frac{1}{x}$
0%
$f(x)=\frac{x^2}{x}$ and $g(x) =$
0%
$f(x)=In \,x^4$ and g(x)= 4 In Xx
0%
f(x) = In {(x-1)(x-2)} and g(x) = In (x-2)+In (x-3)

The function

$f(x)={x}^{2}+bx+c$ , where

$b$ and

$c$ real constants, describes

0%
one-to-one mapping
0%
onto mapping
0%
not one-to-one onto mapping
0%
neither one-to-one nor onto mapping

Explanation

$f\left( x \right) =x^{2}+bx+c$ when

$f\left( x_{1} \right) =f\left( x_2 \right)$

Since this a quadratic function and quadratic function is neither one-one nor onto.

so option D is correct.

If f(x)=x+2and

$g(x)=x^2-3$ , then which is true?

0%
fog $\neq$ gog
0%
2fog = gof
0%
fog = gof
0%
fog = 2 gof

$f(x) = 4x^{2} - 1$ and

$g(x) = 8x + 7, g\circ f(2) =$

0%
$15$
0%
$23$
0%
$127$
0%
$345$
0%
$2115$

Domain of definition of the function

$f\left( x \right) =\sqrt { 2\sin ^{ -1 }{ \left( 2x \right) +\dfrac { \pi }{ 3 } } }$ , for real value x, is

0%
$\left[ -\dfrac { 1 }{ 4 } ,\dfrac { 1 }{ 2 } \right]$
0%
$\left[ -\dfrac { 1 }{ 2 } ,\dfrac { 1 }{ 2 } \right]$
0%
$\left( -\dfrac { 1 }{ 2 } ,\dfrac { 1 }{ 9 } \right)$
0%
$\left[ -\dfrac { 1 }{ 4 } ,\dfrac { 1 }{ 4 } \right]$

The domain of definition of the function

$f(x) = \left \{x\right \}^{\left \{x\right \}} + [x]^{[x]}$ is (where

$\left \{\cdot \right \}$ represents fractional part and

$[\cdot ]$ represents greatest integral function).

0%
$R - I$
0%
$R - [0, 1)$
0%
$R - \left \{I\cup (0, 1)\right \}$
0%
$I\cup (0, 1)$

Let

$f$ be real valued function defined by

$f(x)=\sin ^{ -1 }{ \left( \cfrac { 1-\left| x \right| }{ 3 } \right) } +\cos ^{ -1 }{ \left( \cfrac { \left| x \right| -3 }{ 5 } \right) }$ . Then domain of

$f(x)$ is given by

0%
$\left[ -4,4 \right]$
0%
$\left[ 0,4 \right]$
0%
$\left[ -3,3 \right]$
0%
$\left[ -5,5 \right]$

$f:\left( 0,\infty \right) \rightarrow \left( 0,\infty \right)$ is defined by

$f(x)=\begin{cases} { 2 }^{ x },\quad x\in \left( 0,1 \right) \\ { 5 }^{ x },\quad x\in [1,\infty ) \end{cases}$ is

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

Let

$A=\left\{ { a }_{ 1 },{ a }_{ 2 },{ a }_{ 3 },{ a }_{ 4 }{ a }_{ 5 },{ a }_{ 6 } \right\}$ and

$B=\left\{ { b }_{ 1 },{ b }_{ 2 },{ b }_{ 3 } \right\}$ . The number of functions of

$f:A\rightarrow B$ such that it is onto and there are exactly three elements in

$A$ such that

$f(A)=b$ , is

0%
$75$
0%
$90$
0%
$100$
0%
$120$

Explanation

Since there should be exactly 3 elements in

$A$ such that

$f(A) = b$

the function

$f$ is a bijection from a subset

$A_1$ of

$A$ to

$B$ ., each having 3 elements.

Number of ways to select 3 elements from

$A$

$=\; ^6C_3$

Number of bijection from

$A_1$ to

$B$

$= 3 !$

Hence, the number of functions

$=\; ^6C_3 \; 3!$

$= 120$

Given that

$f'(x) > g'(x)$ for all real x, and

$f(0) = g(0)$ . Then

$f(x) < g(x)$ for all x belongs to

0%
$(0, \infty)$
0%
$(- \infty, 0)$
0%
$(- \infty, \infty)$
0%
none of these

The number of solutions of the equation

$9x^2 - 18 |x| + 5 = 0$ belonging to the domain of definition of

$log_e \{(x + 1) (x + 2)\}$ , is

0%
1
0%
2
0%
3
0%
4

Explanation

$f(x)=\log _{ e }{ \{ (x+1)(x+2)\} }$

For the function to be defined

$(x+1)(x+2)>0\\ \Rightarrow -2>x>-1\\ \Rightarrow x\in \left( -\infty ,-2 \right) \cup \left( -1,\infty \right)$

$9{ x }^{ 2 }-18|x|+5=0\\ 9{ |x| }^{ 2 }-18|x|+5=0\\ \Rightarrow 9|{ x| }^{ 2 }-18|x|+5=0\quad for(x>0)\\ \Rightarrow 9|{ x| }^{ 2 }-3|x|-15|x|+5=0\\ \Rightarrow 3|x|(3|x|-1)-5(3|x|-1)=0\\ \Rightarrow (3|x|-1)(3|x|-5)=0\\ \Rightarrow |x|=\dfrac { 1 }{ 3 } ,\dfrac { 5 }{ 3 } \\ \Rightarrow x=\dfrac { 1 }{ 3 } ,\dfrac { 5 }{ 3 } ,-\dfrac { 1 }{ 3 } ,-\dfrac { 5 }{ 3 }$

Clearly $-\dfrac { 5 }{ 3 }$ do not belong to the domain of the function $f(x)$

So there are $3$ solutions belonging to the domain of $f(x)$

Option $C$ is ccorrect.

Let

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

$\alpha$ be a real number. If

$x_{0} = \alpha, x_{1} = f(x_{0}), x_{2} = f(x_{1}), ....$ and

$x_{2011} = - \dfrac {1}{2012}$ then the value of

$\alpha$ is

0%
$\dfrac {2011}{2012}$
0%
$1$
0%
$2011$
0%
$-1$

Explanation

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

$x_0=\alpha$

$x_1 =f(x_0)=f(\alpha)=\dfrac {\alpha}{1-\alpha}$

$x_2 =f(x_1)=f\left (\dfrac {\alpha}{1-\alpha}\right)=\dfrac {\dfrac {\alpha}{1-\alpha}}{1-\dfrac {\alpha}{1-\alpha}}=\dfrac {\alpha}{1-2\alpha}$

$x_3 =f(x_2)=f\left (\dfrac {\alpha}{1-2\alpha}\right)=\dfrac {\dfrac {\alpha}{1-2\alpha}}{1-\dfrac {\alpha}{1-2\alpha}}=\dfrac {\alpha}{1-3\alpha}$

Analysis the values

$x_1, x_2, x_3$ , we can say that

$x_{2011}=\dfrac {\alpha}{1-20011\alpha}=-\dfrac {1}{2012}$

$\Rightarrow \ 2012\alpha =-1+2011\alpha$

$\Rightarrow \ 2012\alpha -2011\alpha =-1$

$\Rightarrow \ \boxed {\alpha =-1}$

Domain of

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

$\left\{ . \right\}$ denotes the fractional part, in

$\left\{ . \right\}$ is

0%
$\left[ -1,1 \right] \sim \left( \dfrac { 1 }{ 2 } ,1 \right)$
0%
$\left[ -1,-\dfrac { 1 }{ 2 } \right] \bigcup \left[ 0,\dfrac { 1 }{ 2 } \right] \bigcup \left\{ 1 \right\}$
0%
$\left[ -1,\dfrac { 1 }{ 2 } \right]$
0%
$\left[ -\dfrac { 1 }{ 2 } ,1 \right]$

Consider set

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

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

$A$ to set

$B$ in which

$f(i)\neq i$ is,

0%
$84$
0%
$78$
0%
$42$
0%
$24$

$f:R\rightarrow S$ defined by

$f(x)=4\sin { x } -3\cos { x } +1$ is onto, then

$S$ is equal to

0%
$[-5,5]$
0%
$(-5,5)$
0%
$(-4,6)$
0%
$[-4,6]$

Explanation

We have

$f(x)=4sin\:x-3cos\:x+1$

Therefore

maximum of

$f=\sqrt{4^2+(-3)^2}+1=\sqrt{16+9}+1=\sqrt{25}+1=5+1=6$ and

minimum of

$f=-\sqrt{4^2+(-3)^2}+1=-\sqrt{16+9}+1=-\sqrt{25}+1=-5+1=-4$

$S=$ Range of

$f$ =

$[minimum\:f,maximum\:f]=[-4,6]$

Let for

$a \neq a_{1} \neq 0,\ f(x)=ax^{2}+bx+c,\ g(x)=a_{1}x^{2}+b_{1}x+c_{1}$ and

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

$p(x)=0$ only for

$x=-1$ and

$p(-2)=2$ , then the value of

$p(2)$ is

0%
$6$
0%
$18$
0%
$3$
0%
$9$

Explanation

Given,

$f\left(x\right)=ax^{2}+bx+c,g\left(x\right)=ax^{2}+bx+c,$

and

$p\left(x\right)=f\left(x\right)-g\left(x\right)$

$p\left(x\right)=0$ for only

$x=-1,p\left(-2\right)=2$ and

$a\neq a_{1}\neq 0$

$\therefore p\left(x\right)=(a-a_{1})x^{2}+(b-b_{1})x+(c-c_{1})$

The standard quadratic equation is

$ax^2+bx+c=0$

For roots to be equal

$\displaystyle \Rightarrow b^{2}-4ac= 0$

Then Sum of roots =

$-\dfrac {b}{a}$

Here, Only root is

$-1 \Rightarrow D=0 \Rightarrow (b-b_{1})^{2}=4(a-a_{1})(c-c_{1})\rightarrow(1)$

Sum of roots

$\Rightarrow (-1)+(-1)=\dfrac{-(b-b_{1})}{(a-a_{1})} \Rightarrow (b-b_{1})=2(a-a_{1}) \rightarrow(2)$

From (1) and (2),

$c-c_{1}=(a-a_{1})\rightarrow(3)$

$\Rightarrow p\left(x\right)=(a-a_{1})x^{2}+2(a-a_{1})x+(a-a_{1})$

Now,

$p\left(-2\right)=2$

$\Rightarrow 2=(a-a_{1})(-2)^{2}-2(a-a_{1})(-2)+(a-a_{1})$

$\Rightarrow (a-a_{1})=2$

Now

$p(2)=2(2)^{2}+2(2)(2)+2=18$

Let

$f(-2, 2)\rightarrow(-2, 2)$ be a continuous function given

$f(x)=f{(x}^{2})$ . Given

$f(0)=\dfrac{1}{2}$ then the

$4f(\dfrac{1}{2})$

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

The domain of the function

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

$[x]$ denotes the greatest integer not greater than

$x$ , is :

0%
$R$
0%
$R - \{ 0\}$
0%
$R - \left\{ { \pm \sqrt n :n \in {I^ + } \cup \{ 0\} } \right\}$
0%
$R - \{ n:n \in I\}$

What percent of the domain of the function

$\displaystyle f\left( x \right) = \frac{{\sqrt {9 - {x^2}} }}{{\sqrt[4]{{9 - \left| {2x + 5} \right|}}}}$ consists of non-negative integers.

0%
$40\%$
0%
$50\%$
0%
$30\%$
0%
$65\%$

$\phi (x) = 3 f(\frac{x^2}{3} ) + f(3-x^2) \forall x \in (3,4)$ where

$f(x) >0 \forall x (-3,4)$ then

$\phi (x)$ is ____________.

0%
(a) increasing in $( \frac{3}{2} ,4)$
0%
(b) decreasing in $( 3, \frac{3}{2} )$
0%
(c) increasing $( -\frac{3}{2} , 0)$
0%
decreasing in $( 0, \frac{3}{2})$

Let the function

$f:D \to R$ ;

$f\left( x \right){\log _5}\left( {{{\log }_{{1 \over 3}}}\left( {{{\log }_8}\left( {2x + 1} \right)} \right)} \right)$ where

$D$ is the maximum domain of

$f\left( x \right)$ . If

$S$ represents the sum of the absolute values of all integers form

$D$ . Then the value of

$S$ , is

0%
$15$
0%
$10$
0%
$6$
0%
$3$

The domain of

$f(x)=\dfrac{1}{\sqrt{1[\vert x \vert -1]\vert-5}}$ (where [.] G.I.F) is

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

Let f(x) and g(x) be the differentiable functions for

$1\le x\le 3$ such that f(1)=2=g(1) and f(3)=Let there exist exactly one real number

$cE (1,3)$ such that 3f'(c)=g'(c), then the value of g(3) must be

0%
12
0%
13
0%
16
0%
26

Let

$f\left( x \right)={2^{10}}x + 1$ and

$g\left( x \right) = {3^{10}}x - 1$ , If

$\left( {fog} \right)\left( x \right) = x$ , then

$x$ is equal to

0%
$\dfrac{{{3^{10}} - 1}}{{{3^{10}} - {2^{ - 10}}}}$
0%
$\dfrac{{{2^{10}} - 1}}{{{2^{10}} - {3^{ - 10}}}}$
0%
$\dfrac{{1 - {3^{10}}}}{{{2^{10}} - {3^{ - 10}}}}$
0%
$\dfrac{{1 - {2^{-10}}}}{{{3^{10}} - {2^{ - 10}}}}$

Explanation

Given:

$f\left(x\right)={2}^{10}x+1$ and

$g\left(x\right)={3}^{10}x-1$

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

$\Rightarrow\,f\left({3}^{10}x-1\right)=x$

$\Rightarrow\,{2}^{10}\left({3}^{10}x-1\right)+1=x$

$\Rightarrow\,{2}^{10}{3}^{10}x-{2}^{10}+1=x$

$\Rightarrow\,-{2}^{10}+1=x-{2}^{10}{3}^{10}x$

$\Rightarrow\,-{2}^{10}+1=x\left(1-{2}^{10}{3}^{10}\right)$

$\Rightarrow\,x=\dfrac{-{2}^{10}+1}{1-{2}^{10}{3}^{10}}$

$\Rightarrow\,x=\dfrac{{2}^{10}-1}{{2}^{10}{3}^{10}-1}$

$\Rightarrow\,x=\dfrac{{2}^{10}-1}{{2}^{10}\left({3}^{10}-{2}^{-10}\right)}$

$\Rightarrow\,x=\dfrac{1-{2}^{-10}}{\left({3}^{10}-{2}^{-10}\right)}$

$f : R^+ \rightarrow R$ defined by

$f(x) = 2^x , \, x \in (0, 1), \, f(x) = 3^x , \, x \in [1, \, \infty)$ is

0%
onto
0%
one-one
0%
neither one-one nor onto
0%
one one onto

A real valued function

$f(x)$ satisfies the function equation

$f(x-y)=f(x)f(y)-f(a-x)f(a+y)$ where a is a given constant and

$f(0)=1, f(2a-x)$ is equal to?

0%
$f(a)+f(a-x)$
0%
$f(-x)$
0%
$-f(x)$
0%
$f(x)$

The domain of the function

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

0%
$x \in ( - \infty , - 1) \cup (1,\infty )$
0%
$x \in ( - \infty , - 2) \cup (2,\infty )$
0%
$x \in ( - 2 , - 1) \cup (1,2 )$
0%
none of these

$f(x)=|x|$ and

$g(x)=[x]$ , then value of

$fog \left(-\dfrac {1}{4}\right)+gof \left(-\dfrac {1}{4}\right)$ is ?

0%
$0$
0%
$1$
0%
$-1$
0%
$1/4$

The function

$f : R\rightarrow R$ given by, then

$f(x)=3-2\sin x$ is

0%
one-one
0%
onto
0%
bijective
0%
None of these

The domain of the function

$f\left( x \right) = {\sin ^{ - 1}}\left( {{{\log }_2}\left( {\dfrac{{{x^2}}}{2}} \right)} \right)$ is

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

Explanation

Given :

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

we know that it

$y = sin^{-1}x$

then 'x' should belongs to

$[-1,1]$

So,

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

take antilog So-

$2^{-1}\leq \dfrac{x^{2}}{2}\leq 2^{1}$

$\dfrac{1}{2}\leq \dfrac{x^{2}}{2}\leq 2$

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

So,

$x\epsilon [-2,-1]\cup [1,2]\rightarrow$ option 'D'

Let

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

$|x|>2$ , then the function

$f:(-\infty, -2)\cup [2,\infty)\rightarrow (-1,1)$ is

0%
One-one into
0%
One-one onto
0%
Many one into
0%
Many one onto

Explanation

$\begin{array}{l} f\left( x \right) =\frac { { { x^{ 2 } }-4 } }{ { { x^{ 2 } }+4 } } \\ let\, \, f\left( x \right) =y \\ \Rightarrow y=\frac { { { x^{ 2 } }-4 } }{ { { x^{ 2 } }+4 } } \\ \Rightarrow { x^{ 2 } }y+4y-{ x^{ 2 } }+4=0 \\ \Rightarrow { x^{ 2 } }\left( { y-1 } \right) +4\left( { y+1 } \right) =0 \\ \Rightarrow { x^{ 2 } }=\frac { { -4\left( { y+1 } \right) } }{ { y-1 } } \\ \Rightarrow x=\frac { { \sqrt { 4\left( { y+1 } \right) } } }{ { \left( { 1-y } \right) } } \\ \Rightarrow \frac { { 4\left( { y+1 } \right) } }{ { 1-y } } 0 \\ \Rightarrow y+,0 \\ \Rightarrow y-1 \\ \Rightarrow 1-y0 \\ \Rightarrow y1 \\ Hence,\, \, range\, \, of\, \, { f^{ b } }\, \, is\left( { -1,1\, \, which\, \, is\, \, equal\, \, to\, \, Codomain } \right) \\ \therefore \, \, { f^{ h } }\, \, is\, \, one-one-onto{ 1mu } \end{array}$

$f(x)=ax+b$ and

$f(f(f(x)))=27x+13$ where a and b are real numbers, then-

0%
a+b=3
0%
a+b=4
0%
f'(x)=3
0%
f'(x)=-3

Explanation

$\begin{array}{l} f\left( x \right) =ax+b \\ \Rightarrow f\left( { f\left( { f\left( x \right) } \right) } \right) =27x+13 \\ \Rightarrow f\left( { a\left( { ax+b } \right) +b } \right) =27x+13 \\ \Rightarrow \left[ { a\left\{ { a\left( { ax+b } \right) +b } \right\} +b } \right] =27x+13 \\ \Rightarrow \left[ { a\left\{ { { a^{ 2 } }x+ab+b } \right\} +b } \right] =27x+13 \\ \Rightarrow { a^{ 3 } }x+{ a^{ 2 } }b+ab+b=27x+13 \\ \Rightarrow { a^{ 3 } }x=27x \\ \Rightarrow { a^{ 3 } }=27 \\ \therefore a=3 \\ and\, \, { a^{ 2 } }b+ab+b=13 \\ \Rightarrow ab+3b+b=13 \\ 13b=13 \\ \therefore b=1 \end{array}$

$\therefore a+b=3+1=4$

Hence

$f\left(x\right)=3x+4\Rightarrow {f}^{\prime}{\left(x\right)}=3$

$f\left( x \right) = \sin ^{ 2 }{ x } + \sin ^{ 2 }({ { x }+\frac { \pi }{ 3 }) + \cos { x\cos { \left( { x } + \frac { \pi }{ 3 } \right), ~g(\frac { 5 }{ 4 }) = 1, \text{then} \left( gof \right)\left( x \right) } }\ \text{ is}\ \text{equal}\ \text{to} }$

0%
$1$
0%
$0$
0%
$\frac{1}{4}$
0%
$\frac{1}{2}$

$f:A \rightarrow A,A=\left\{a_{1},a_{2},a_{3},a_{4},a_{5}\right\}$ , then the number of one one function so that

$f(x_{i})\neq x_{i},x_{i}\ \in\ A$ is

0%
$44$
0%
$88$
0%
$22$
0%
$20$

$f(x)=x-\cfrac{1}{x}$ then number of solutions of

$f(f(f(x)))=1$ is

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Domain of the functon

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

0%
(- $6$ , $6$ )
0%
(- $6$ , $6$ )
0%
$(-\infty , - 6) \cup [6,\infty )$
0%
none

Show that the function

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

$f(x)=\dfrac{2x}{1+2x}$ is?

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

$f(x)=1+|x-1|,-1 \le x \le 3$ and

$g(x)=2-|x+1|,-2 \le x \le 2$ then choose the appropriate option.

0%
$fog(x)=x-1$ for $x\ \in\ (0,1)$
0%
$fog(x)=x$ for $x\ \in\ (-1,1)$
0%
$gog(x)=x$ for $x\ \in\ (-1,2)$
0%
$all\ of\ these$

The domain of

$f(x)=|x-2|-|x-5|$ is

0%
$R-(2, 5)$
0%
$R-\left\{0\right\}$
0%
$(0, \infty)$
0%
$R$

The domain of

$f(x)=\dfrac{1}{\sqrt{(x-1)(x-2)(x-3)}}$ is

0%
$(1, 2)$
0%
$(3, \infty)$
0%
$(2, 3)$
0%
$(-\infty, 1)$

Let

$f:R\rightarrow R$ be defined by

$f(x)=\dfrac {x|x|}{2}+\cos x+1$ then

$f(x)$ is

0%
One-one only
0%
Onto only
0%
Neither one-one nor onto
0%
Bijection

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

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.