MCQ Questions for Class 11 Commerce Applied Mathematics Functions Quiz 13

Let f(x)=

$y=\begin{cases} { x }^{ 2 }-3x+4\quad \quad :\quad X<3 \\ \quad \quad \quad x+7\quad \quad :\quad X\ge 3 \end{cases}\quad and\quad g(x)=\begin{cases} \quad \quad x+6\quad \quad \quad :\quad X<4 \\ { x }^{ 2 }+x+2\quad \quad :\quad X\ge 4 \end{cases}$
then which of the following is/ are true-

0%
$(f+g) (1)=9$
0%
$(f-g)(3.5)= 1$
0%
$(f+g) (0)=24$
0%
$\left( \dfrac { f }{ g } \right) (5)=\dfrac { 8 }{ 3 }$

If =

$f=\left\{ (-2,4),(0,6),(2,8) \right\}$ and

$g=\left\{ (-2,-1),(0,3),(2,5) \right\}$ , then

$\left( \frac { 2f }{ 3g } +\frac { 3g }{ 2f } \right) (0)=\quad$

0%
1/12
0%
25/12
0%
5/12
0%
13/12

If f(x)=x+tanx and g(x) is inverse of f(x) then g'(x) is equal to

0%
$\dfrac { 1 }{ 1+(g(x)-{ x) }^{ 2 } }$
0%
$\dfrac { 1 }{ 1-(g(x)-{ x) }^{ 2 } }$
0%
$\dfrac { 1 }{ 1+(g(x)-{ x) }^{ 2 } }$
0%
$\dfrac { 1 }{ 2-(g(x)-{ x) }^{ 2 } }$

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

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

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

For the function

$F(x)=\sqrt { { 4-x }^{ 2 } } +\sqrt { { x }^{ 2 }-1 }$

0%
Domain is (-2,2)
0%
Domain is (-2, -1) (1,2)
0%
Range is $\left( \sqrt { 3 } ,\sqrt { 5 } \right)$
0%
Range is $\left( \sqrt { 3 } ,\sqrt { 6 } \right)$

If f(x) = x + 4, g(x) = 5x and h(x) = 12/x. find the value of

$f^{ -1 }$ (g(h(6))).

0%
10
0%
14
0%
6
0%
0

Let

$f(x)=x+\cos x+2$ and g(x) be the inverse function of f(x) then

$g^1(3)=$

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

$f:A\rightarrow B$ given by

${ 3 }^{ f(x) }+{ 2 }^{ -x }=4$ is a bijection, then

0%
$A=\left\{ x\in R:-1< x< \infty \right\} ;B=\left\{ x\in R:2< x< 4 \right\}$
0%
$A=\left\{ x\in R:-3< x <\infty \right\} ;B=\left\{ x\in R:0 < x <4 \right\}$
0%
$A=\left\{ x\in R:-2< x <\infty \right\} ;B=\left\{ x\in R:0< x <4 \right\}$
0%
None of these

Explanation

$f:A\rightarrow B$

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

$\Rightarrow$

$3^{f(x)}=4-2^{-x}$

Taking log on both the sides,

$\Rightarrow$

$f(x)\log 3=\dfrac{\log(4-2^{-x})}{\log 3}$

$\Rightarrow$

$f(x)=\dfrac{\log(4-2^{-x})}{\log 3}$

Logarithmic function will only be defined if

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

$\Rightarrow$

$4>2^{-2}$

$\Rightarrow$

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

$\Rightarrow$

$2>-x$

$\Rightarrow$

$x\in(-2,\infty)$

That means

$A=\{x\in R:-2<x<\infty\}$

As we know that,

$f(x)=\dfrac{\log(4-2^{-x})}{\log 3}$

We take

$x=0\in (-2,\infty)$

$\Rightarrow$

$f(x)=1$ , which does not match to any of the options.

$f(x)=8x^3$ and

$g(x)=x^{1/3}$ then

$(g o f)(x)=?$

0%
$x$
0%
$2x$
0%
$\dfrac{x}{2}$
0%
$3x^2$

Explanation

$(g o f)(x)=g[f(x)]=g(8x^3)=(8x^3)^{1/3}=2x$ .

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

$(f o f o f)(x)=?$

0%
$\dfrac{1}{(1-3x)}$
0%
$\dfrac{x}{(1+3x)}$
0%
$x$
0%
None of these

Explanation

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

$\Rightarrow \{f o (f o f)\}(x)=f\{(f o f)(x)\}$

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

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

$f(x)=x^2, g(x)=\tan x$ and

$h(x)=log x$ then

$\{h o (g o f)\}\left(\sqrt{\dfrac{\pi}{4}}\right)=?$

0%
$0$
0%
$1$
0%
$\dfrac{1}{x}$
0%
$\dfrac{1}{2} \log \dfrac{\pi}{4}$

Explanation

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

$=h\{g(x^2)\}=h(\tan x^2)=log (\tan x^2)$ .

$\therefore \{h o (g o f)\}\sqrt{\dfrac{\pi}{4}}=\log\left(\tan\dfrac{\pi}{4}\right)=\log 1=0$ .

$f=\{(1, 2), (3, 5), (4, 1)\}$ and

$g=\{(2, 3), (5, 1), (1, 3)\}$ then

$(g o f)=?$

0%
$\{(3, 1), (1, 3), (3, 4)\}$
0%
$\{(1, 3), (3, 1), (4, 3)\}$
0%
$\{(3, 4), (4, 3), (1, 3)\}$
0%
$\{(2, 5), (5, 2), (1, 5)\}$

Explanation

$Dom(g o f)=dom(f)=\{1, 3, 4\}$ .

$(g o f)(1)=g\{f(1)\}=g(2)=3,$

$(g o f)(3)=g\{f(3)\}=g(5)=1$

$(g o f)(4)=g\{f(4)\}=g(1)=3$

$\therefore g o f=\{(1, 3), (3, 1), (4, 3)\}$ .

$f(x)=(x^2-1)$ and

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

$(g o f)(x)=?$

0%
$(2x^2+3)$
0%
$(3x^2+2)$
0%
$(2x^2+1)$
0%
None of these

Explanation

$(g o f)(x)=g[f(x)]=g(x^2-1)=2(x^2-1)+3=(2x^2+1)$ .

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

$(f o f)(x)=?$

0%
$x^4$
0%
$x^4-6x^3$
0%
$x^4-6x^3+10x^2$
0%
None of these

Which of the following is not true about

$h_2(x)$

0%
Domain R
0%
It periodic function with period $2\pi$
0%
Range is [0, 1]
0%
None of these

Let

$f: R \rightarrow R$ defined by

$f(x)= e\frac { { e }^{ { x }^{ 2 } }-{ e }^{ { -x }^{ 2 } } }{ { e }^{ { x }^{ 2 } }+{ e }^{ { -x }^{ 2 } } }$ then

0%
$f(x)$ is one-one but not onto
0%
$f(x)$ is neither one-one nor onto
0%
$f(x)$ is many one but onto
0%
$f(x)$ is one-one and onto

Explanation

$\textbf{Step-1: Apply the relevant concept of functions to get the required unknown}$

$\text{We have,}$

$f(x)= e\frac { { e }^{ { x }^{ 2 } }-{ e }^{ { -x }^{ 2 } } }{ { e }^{ { x }^{ 2 } }+{ e }^{ { -x }^{ 2 } } }$

$f(-x)= e\frac { { e }^{ { x }^{ 2 } }-{ e }^{ { -x }^{ 2 } } }{ { e }^{ { x }^{ 2 } }+{ e }^{ { -x }^{ 2 } } }$

$= f(x)$

$f(1) = f(-1)$

$\text{f is not one - one}$

$f(x)= e\frac { { e }^{ { x }^{ 2 } }-{ e }^{ { -x }^{ 2 } } }{ { e }^{ { x }^{ 2 } }+{ e }^{ { -x }^{ 2 } } }$

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

$e^{2x^2}\geq1$

$e^{2x}+1\geq2$

$0\leq\dfrac{2}{e^{2x^2}+1}\leq1$

$f(x) \in[0,1) \neq$

$\text{co-domain}$

$\text{Neither one -one nor onto}$

$\textbf{Hence, option is B}$

$f(x)=\dfrac{2}{x-3}, g(x)=\dfrac{x-3}{x+4}$ and

$h(x)=-\dfrac{2(2 x+1)}{x^{2}+x-12}$ then

$\lim _{x \rightarrow 3}[f(x)+g(x)+h(x)]$ is

0%
$-2$
0%
$-1$
0%
$-\dfrac{2}{7}$
0%
0

Explanation

c. We have

$f(x)+g(x)+h(x)=\dfrac{x^{2}-4 x+17-4 x-2}{x^{2}+x-12}$

$=\dfrac{x^{2}-8 x+15}{x^{2}+x-12}=\dfrac{(x-3)(x-5)}{(x-3)(x+4)}$

$\therefore \lim _{x \rightarrow 3}[f(x)+g(x)+h(x)]=\lim _{x \rightarrow 3} \dfrac{(x-3)(x-5)}{(x-3)(x+4)}=-\dfrac{2}{7}$ .

Which of thefollowing functions are indentical?

0%
f(x)= ln $x^{2}$ and g(x)= 2 In x
0%
f(x)= $log_x e$ and $g(x)= \dfrac{1}{log_e x}$
0%
f(x)= sin $(cos^{-1}x)$ and g(x)= $cos({sin^{-1}x})$
0%
none of these

For a real number y, let [y] denotes the greatest integer less than or equal to y. Then the function

$f(x)= \dfrac{tan(\pi \left[ x-\pi \right ])}{1+[x]^2}$ is

0%
discontinuous at some x
0%
continuous at all x, but the derivative $f'(x_{0})$ does not exist for some x
0%
$f'(x)$ exist for all x but the derivative $f'(x_{0})$ does not exist second for some x
0%
$f'(x)$ exists for all x

The domain of the function

$f(x) = \sqrt{\ln_{(|x| - 1)} \left (x^{2} + 4x +4 \right )}$ is

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

The number of roots of the equation g(x) = 1 is

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

$f : X \rightarrow Y$ , where

$X$ and

$Y$ are sets containing natural numbers,

$f(x) = \frac{x + 5} {x + 2}$ then the number of elements in the domain and range of $$f(x) are respectively

0%
1 and 1
0%
2 and 1
0%
2 and 2
0%
1 and 2

The domain of the

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

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

Explanation

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

For

$f(x)$ to be defined

$|x^{2} -10x + 9| < 4x$

$\Rightarrow x^{2} - 10x + 9 < 4x$ and

$x^{2} - 10x + 9 > -4x$

$\Rightarrow x^{2} -14x + 9 < 0$ and

$x^{2} - 6x + 9 >0$

$x \in \dfrac {14\pm \sqrt {196-36}}{2}$ &

$x \in \dfrac {6\pm \sqrt {36-36}}{2}$

$\Rightarrow x \space \epsilon \left ( 7 - \sqrt{40}, 7 + \sqrt{40} \right )$ and

$x \space \epsilon \left ( 7 - \sqrt{40}, -3 \right) \cup \left ( -3, 7 + \sqrt{40} \right)$

If A = { 1 ,2 , 3 , 4} , then which following is a function in A

0%
$f_{1} = {(x , y) : y = x + 1}$
0%
$f_{2} = {( x , y) : x + y > 4 }$
0%
$f_{3} = {( x , y) : y < x}$
0%
$f_{4} = { ( x , y) : x + y = 5}$

CBSE Questions for Class 11 Commerce Applied Mathematics Functions Quiz 13 - 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 13 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.