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

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

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

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

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

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

If $$f(x) = -{ \frac { x\left| x \right| }{ 1+x^{ 2 } } }$$ then $$f^{-1} (x) $$ equals

$$(Sgn x) \sqrt { \frac { x\left| x \right| }{ 1-\left| x \right| } } $$

$$\sqrt { \frac { x\left| x \right| }{ 1-\left| x \right| } } $$

MCQ Questions for Class 12 Commerce Maths Relations And Functions Quiz 12

Let A=(5,6); how many binary operations can be defined on this set?

0%
8
0%
10
0%
16
0%
20

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

$f:A\rightarrow B?$

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

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

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

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

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%
$4$

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

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

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

$f(x)= sin^{2}x$ and the composite functions g{f(x)}=|sin x|, then the function g(x)=

0%
$\sqrt{x-1}$
0%
$\sqrt{x}$
0%
$\sqrt{x+1}$
0%
$-\sqrt{x}$

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%
$\infty$

$f:A\rightarrow B$ then for

$f^{-1}$ =

0%
$I_{A}$
0%
$I_{B}$
0%
$A\rightarrow B$
0%
none of these

$f:N\rightarrow N,f(x)=x+3$ , then

$\quad { f }^{ -1 }(x)=.....$

0%
$x+3$
0%
does not exist
0%
$x-3$
0%
$3-x$

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

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

$f(x)=8x^3$ 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

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

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

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

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

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

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

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

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

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

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

$\text{We have,}$

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

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

$= f(x)$

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

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

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

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

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

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

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

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

$\text{co-domain}$

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

$\textbf{Hence, option is B}$

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

$f^{-1} (x)$ equals

0%
$\sqrt { \frac { x\left| x \right| }{ 1-\left| x \right| } }$
0%
$(Sgn x) \sqrt { \frac { x\left| x \right| }{ 1-\left| x \right| } }$
0%
$- \sqrt { \frac {x} {1-x}}$
0%
None of these

Explanation

$\textbf{Step-1: Apply the concept of function & simplify}$

$\text{Let y = f(x)}$

$= \dfrac{-x^2}{1+x^2},x\geq0$

$= \dfrac{x^2}{1+x^2}, x<0$

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

$\sqrt{\dfrac{-y}{1+y}},y<0$

$= -\sqrt{\dfrac{y}{1-y}},y>0$

$f'^{-1}(x) = \sqrt{\dfrac{-x}{1+x}}, x<0$

$=\sqrt{\dfrac{-x}{1-x}},x>0$

$\text{= (sgn x)} \sqrt{\dfrac{|x|}{1-|x|}}$

$\textbf{Hence, option B}$

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

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

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

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

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

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

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

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

CBSE Questions for Class 12 Commerce Maths Relations And Functions Quiz 12 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Relations And Functions Quiz 12 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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