MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 11 Commerce Applied Mathematics Functions Quiz 13 - MCQExams.com
CBSE
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-
Report Question
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
$$\left( \frac { 2f }{ 3g } +\frac { 3g }{ 2f } \right) (0)=\quad $$
Report Question
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
Report Question
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
Report Question
0%
$$0$$
0%
$$3$$
0%
$$2$$
0%
$$4$$
For the function $$F(x)=\sqrt { { 4-x }^{ 2 } } +\sqrt { { x }^{ 2 }-1 } $$
Report Question
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))).
Report Question
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 $$g^1(3)=$$
Report Question
0%
$$1$$
0%
$$2$$
0%
$$3$$
0%
$$0$$
If $$f:A\rightarrow B$$ given by $${ 3 }^{ f(x) }+{ 2 }^{ -x }=4$$ is a bijection, then
Report Question
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.
If $$f(x)=8x^3$$ and $$g(x)=x^{1/3}$$ then $$(g o f)(x)=?$$
Report Question
0%
$$x$$
0%
$$2x$$
0%
$$\dfrac{x}{2}$$
0%
$$3x^2$$
Explanation
$$(g o f)(x)=g[f(x)]=g(8x^3)=(8x^3)^{1/3}=2x$$.
If $$f(x)=\dfrac{1}{(1-x)}$$ then $$(f o f o f)(x)=?$$
Report Question
0%
$$\dfrac{1}{(1-3x)}$$
0%
$$\dfrac{x}{(1+3x)}$$
0%
$$x$$
0%
None of these
Explanation
$$(f o f)(x)=f\{f(x)\}=f\left(\dfrac{1}{1-x}\right)=\dfrac{1}{\left(1-\dfrac{1}{1-x}\right)}=\dfrac{1-x}{-x}=\dfrac{x-1}{x}$$
$$\Rightarrow \{f o (f o f)\}(x)=f\{(f o f)(x)\}$$
$$=f\left(\dfrac{x-1}{x}\right)$$
$$(f o f)(x)=\dfrac{1}{1-\dfrac{x-1}{x}}=x$$.
If $$f(x)=x^2, g(x)=\tan x$$ and $$h(x)=log x$$ then $$\{h o (g o f)\}\left(\sqrt{\dfrac{\pi}{4}}\right)=?$$
Report Question
0%
$$0$$
0%
$$1$$
0%
$$\dfrac{1}{x}$$
0%
$$\dfrac{1}{2} \log \dfrac{\pi}{4}$$
Explanation
$$\{h o ( g o f)\}(x)=(h o g)\{f(x)\}=(h o g)(x^2)$$
$$=h\{g(x^2)\}=h(\tan x^2)=log (\tan x^2)$$.
$$\therefore \{h o (g o f)\}\sqrt{\dfrac{\pi}{4}}=\log\left(\tan\dfrac{\pi}{4}\right)=\log 1=0$$.
If $$f=\{(1, 2), (3, 5), (4, 1)\}$$ and $$g=\{(2, 3), (5, 1), (1, 3)\}$$ then $$(g o f)=?$$
Report Question
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
$$Dom(g o f)=dom(f)=\{1, 3, 4\}$$.
$$(g o f)(1)=g\{f(1)\}=g(2)=3,$$
$$ (g o f)(3)=g\{f(3)\}=g(5)=1$$
$$(g o f)(4)=g\{f(4)\}=g(1)=3$$
$$\therefore g o f=\{(1, 3), (3, 1), (4, 3)\}$$.
If $$f(x)=(x^2-1)$$ and $$g(x)=(2x+3)$$ then $$(g o f)(x)=?$$
Report Question
0%
$$(2x^2+3)$$
0%
$$(3x^2+2)$$
0%
$$(2x^2+1)$$
0%
None of these
Explanation
$$(g o f)(x)=g[f(x)]=g(x^2-1)=2(x^2-1)+3=(2x^2+1)$$.
If $$f(x)=x^2-3x+2$$ then $$(f o f)(x)=?$$
Report Question
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 $$h_2(x)$$
Report Question
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
Report Question
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
$$\textbf{Step-1: Apply the relevant concept of functions to get the required unknown}$$
$$\text{We have,}$$
$$f(x)= e\frac { { e }^{ { x }^{ 2 } }-{ e }^{ { -x }^{ 2 } } }{ { e }^{ { x }^{ 2 } }+{ e }^{ { -x }^{ 2 } } } $$
$$f(-x)= e\frac { { e }^{ { x }^{ 2 } }-{ e }^{ { -x }^{ 2 } } }{ { e }^{ { x }^{ 2 } }+{ e }^{ { -x }^{ 2 } } } $$ $$= f(x)$$
$$f(1) = f(-1)$$
$$\text{f is not one - one}$$
$$f(x)= e\frac { { e }^{ { x }^{ 2 } }-{ e }^{ { -x }^{ 2 } } }{ { e }^{ { x }^{ 2 } }+{ e }^{ { -x }^{ 2 } } } $$ $$= \dfrac{e^{2x^2}-1}{e^{2x^2}+1}$$
$$e^{2x^2}\geq1$$
$$e^{2x}+1\geq2$$
$$0\leq\dfrac{2}{e^{2x^2}+1}\leq1$$
$$f(x) \in[0,1) \neq$$ $$\text{co-domain}$$
$$\text{Neither one -one nor onto}$$
$$\textbf{Hence, option is B}$$
If $$ f(x)=\dfrac{2}{x-3}, g(x)=\dfrac{x-3}{x+4} $$ and $$ h(x)=-\dfrac{2(2 x+1)}{x^{2}+x-12} $$ then $$ \lim _{x \rightarrow 3}[f(x)+g(x)+h(x)] $$ is
Report Question
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?
Report Question
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
Report Question
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
Report Question
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
Report Question
0%
2
0%
1
0%
3
0%
0
If $$f : X \rightarrow Y$$, 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 $$f(x) are respectively
Report Question
0%
1 and 1
0%
2 and 1
0%
2 and 2
0%
1 and 2
The domain of the $$f(x) = \frac{1} {\sqrt{4x - |x^{2} -10x + 9|}}$$ is
Report Question
0%
$$\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)$$
If A = { 1 ,2 , 3 , 4} , then which following is a function in A
Report Question
0%
$$ 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} $$
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page