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CBSE Questions for Class 12 Commerce Maths Relations And Functions Quiz 12 - MCQExams.com
CBSE
Class 12 Commerce Maths
Relations And Functions
Quiz 12
Let A=(5,6); how many binary operations can be defined on this set?
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8
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10
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16
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20
If $$f(x_{1})=f(x_{2})\Rightarrow x_{1}=x_{2}\forall x_{1},x_{2}\epsilon A$$, then what type of a function is $$f:A\rightarrow B?$$
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One-one
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Constant
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Onto
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Many one
If =$$f=\left\{ (-2,4),(0,6),(2,8) \right\} $$ and
$$g=\left\{ (-2,-1),(0,3),(2,5) \right\} $$, then
$$\left( \frac { 2f }{ 3g } +\frac { 3g }{ 2f } \right) (0)=\quad $$
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1/12
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25/12
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5/12
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13/12
If $$g(x)-x^{2}-x+1 and f(x)=\sqrt{\frac{1}{x}-x}$$, then-
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Domain of f(g(x)) ids[0,1]
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Range of f(g(x)) is $$(0,\frac{7}{2\sqrt{3}})$$
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f (g(x)) is many -one function
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f(g(x)) is unbounded function
$$f: ( 0,\infty )\rightarrow [0,\infty )$$ defined by $$f(x)=x^{2}$$ is
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one - one but not onto
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onto but not one - one
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bijective
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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
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$$0$$
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$$3$$
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$$2$$
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$$4$$
If $$f(x)=\frac{1}{x},g(x)=\frac{1}{x^{2}} $$ and h $$(x)= x^{2},$$ then
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$$fog (x)= x^{2}, x\neq \bar{0}, h(g(x))= \frac{1}{x^{2}}$$
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$$h(g(x))= \frac{1}{x^{2}}x\neq 0, fog(x)= x^{2}$$
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$$fog(x)=x^{2},x\neq 0,h(g(x))= (g(x))^{2},x\neq 0$$
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None of these
If $$f(x)= sin^{2}x$$ and the composite functions g{f(x)}=|sin x|, then the function g(x)=
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$$\sqrt{x-1}$$
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$$\sqrt{x}$$
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$$\sqrt{x+1}$$
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$$-\sqrt{x}$$
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
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0%
$$0$$
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$$1$$
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$$2$$
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$$\infty$$
If $$f:A\rightarrow B$$ then for $$f^{-1}$$=
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$$I_{A}$$
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$$I_{B}$$
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$$A\rightarrow B$$
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none of these
If $$f:N\rightarrow N,f(x)=x+3$$, then $$\quad { f }^{ -1 }(x)=.....$$
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$$x+3$$
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does not exist
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$$x-3$$
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$$3-x$$
If $$f(x)=(x^2-1)$$ and $$g(x)=(2x+3)$$ then $$(g o f)(x)=?$$
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$$(2x^2+3)$$
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$$(3x^2+2)$$
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$$(2x^2+1)$$
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None of these
Explanation
$$(g o f)(x)=g[f(x)]=g(x^2-1)=2(x^2-1)+3=(2x^2+1)$$.
If $$f(x)=8x^3$$ and $$g(x)=x^{1/3}$$ then $$(g o f)(x)=?$$
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$$x$$
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$$2x$$
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$$\dfrac{x}{2}$$
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$$3x^2$$
Explanation
$$(g o f)(x)=g[f(x)]=g(8x^3)=(8x^3)^{1/3}=2x$$.
If $$f(x)=\dfrac{1}{(1-x)}$$ then $$(f o f o f)(x)=?$$
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$$\dfrac{1}{(1-3x)}$$
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$$\dfrac{x}{(1+3x)}$$
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$$x$$
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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 $$\{h o (g o f)\}\left(\sqrt{\dfrac{\pi}{4}}\right)=?$$
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$$0$$
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$$1$$
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$$\dfrac{1}{x}$$
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$$\dfrac{1}{2} \log \dfrac{\pi}{4}$$
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 $$(g o f)=?$$
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$$\{(3, 1), (1, 3), (3, 4)\}$$
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$$\{(1, 3), (3, 1), (4, 3)\}$$
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$$\{(3, 4), (4, 3), (1, 3)\}$$
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$$\{(2, 5), (5, 2), (1, 5)\}$$
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 $$f(x)=x^2-3x+2$$ then $$(f o f)(x)=?$$
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$$x^4$$
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$$x^4-6x^3$$
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$$x^4-6x^3+10x^2$$
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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
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$$f(x)$$ is one-one but not onto
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$$f(x)$$ is neither one-one nor onto
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$$f(x)$$ is many one but onto
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$$f(x)$$ 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 $$f(x) = -{ \frac { x\left| x \right| }{ 1+x^{ 2 } } }$$ then $$f^{-1} (x) $$ equals
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$$\sqrt { \frac { x\left| x \right| }{ 1-\left| x \right| } } $$
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$$(Sgn x) \sqrt { \frac { x\left| x \right| }{ 1-\left| x \right| } } $$
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$$- \sqrt { \frac {x} {1-x}}$$
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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 $$h_2(x)$$
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Domain R
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It periodic function with period $$2\pi$$
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Range is [0, 1]
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None of these
Which of thefollowing functions are indentical?
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f(x)= ln $$x^{2}$$ and g(x)= 2 In x
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f(x)= $$log_x e$$ and $$g(x)= \dfrac{1}{log_e x}$$
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f(x)= sin $$(cos^{-1}x)$$ and g(x)= $$cos({sin^{-1}x})$$
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none of these
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
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$$ -2 $$
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$$ -1 $$
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$$ -\dfrac{2}{7} $$
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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
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2
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1
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3
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0
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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