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CBSE Questions for Class 11 Commerce Applied Mathematics Functions Quiz 8 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Functions
Quiz 8
If $$f(x) = \sqrt {4 - {x^2}} + \dfrac{1}{{\sqrt {\left| {\sin x} \right| - \sin x} }}$$, then the domain of f(x) is
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0%
[-2,0]
0%
(0,2]
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[-2,2}
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[$$\dfrac{\pi}{2} $$, 2]
Explanation
undefined
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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$$-\sqrt{x}$$
the equation $$|x+1$^{log}(x+1)^{3+2x-x^{2}}= (x-3)|x|$$ has
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Uniques solution
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Two solution
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No solutions
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more thatn two solution
if f(x)=$$\sqrt {4 - {x^2}} + \dfrac{1}{{\sqrt {|\sin x| - \sin x} }}$$, then the domain of f(x) is
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$$[-2,0)$$
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$$[0,2]$$
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$$[-2,2]$$
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$$None\ of\ these$$
Explanation
$$f ( x ) = \sqrt { 4 - x ^ { 2 } } + \dfrac { 1 } { \sqrt { | \sin x | } - \sin x }$$
$$\Rightarrow 4 - x ^ { 2 } \geqslant 0\ and\ \sqrt{(|\sin x|)} - \sin x > 0$$
$$\Rightarrow x ^ { 2 } \leq 4 \quad 4 \quad \sin x < | \sin x |$$
$$\Rightarrow \quad x \in [ - 2,2 ] \quad \& \quad \sin x < 0$$
$$\Rightarrow \quad x \in [ 2,2 ] + x \in [ - 2,0 )$$
$$\Rightarrow \quad x \in [ - 2,0 )$$
The domain of the function
$$f ( x ) = \log _ { 1 / 2 } \left( - \log _ { 2 } \left( 1 + \dfrac { 1 } { \sqrt [ 4 ] { x } } \right) - 1 \right)$$ is
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$$0 < x < 1$$
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$$0 < x \leq 1$$
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$$x \geq 1$$
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null set
Explanation
$$ f\left( x \right)=\log _{2}^{{}}\left( -\log _{1/2}^{{}}\left( 1+\frac{1}{{{x}^{1/4}}} \right)-1 \right) $$
$$ \,it\,is\,defined\,for $$
$$ \,\left( -\log _{1/2}^{{}}\left( 1+\frac{1}{{{x}^{1/4}}} \right)-1 \right)>0 $$
$$ \Rightarrow \left( \log _{1/2}^{{}}\left( 1+\frac{1}{{{x}^{1/4}}} \right)+1 \right)<0 $$
$$ \Rightarrow \left( \log _{1/2}^{{}}\left( 1+\frac{1}{{{x}^{1/4}}} \right) \right)<-1 $$
$$ \Rightarrow \left( 1+\frac{1}{{{x}^{1/4}}} \right)>{{\left( \frac{1}{2} \right)}^{-1}} $$
$$ \Rightarrow 1+\frac{1}{{{x}^{1/4}}}>2 $$
$$ \Rightarrow \frac{1}{{{x}^{1/4}}}>1 $$
$$ \Rightarrow 0<x<1 $$
Find the domain of function $$\sin^{-1}\left[\dfrac {1+x^{2}}{2x}\right].$$
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$$[-1,1]$$
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$$(-1,1)$$
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$$\left\{-1,1\right\}$$
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$$\left\{0\right\}$$
Explanation
$$ {{\sin }^{-1}}\left( \frac{1+{{x}^{2}}}{2x} \right) $$
$$ -1\le \frac{1+{{x}^{2}}}{2x}\le 1 $$
$$ =-2{{x}^{2}}\le x\left( {{x}^{2}}+1 \right)\le 2{{x}^{2}} $$
$$ -2{{x}^{2}}\le x\left( {{x}^{2}}+1 \right) $$
$$ \Rightarrow x\left( {{x}^{2}}+1+2x \right)\ge 0 $$
$$ \Rightarrow x{{\left( x+1 \right)}^{2}}\ge 0 $$
$$ which\,is\,true\,for\,x\ge 0 $$
$$ x=-1 $$
$$ x\left( {{x}^{2}}+1 \right)\le 2{{x}^{2}} $$
$$ \Rightarrow x\left( {{x}^{2}}+1-2x \right)\le 0 $$
$$ \Rightarrow x{{\left( x-1 \right)}^{2}}\le 0 $$
$$ which\,is\,true\,for\,x\le 0,x=1 $$
$$ Hence\,domain:\left\{ \left. -1,1 \right\} \right. $$
$$ $$
The domain of $$ f(x)= e^{sin (x-|x|)}+|x|cos (\dfrac{x}{|x+1|})$$ , where [.] represents greatest integer function , is
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R
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R-[-1, 0]
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R-[0,1]
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R-[-1)
The equation $$|x-1|+|a|=4$$ can have real solution for $$x$$ if a belongs to the interval
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$$(-\infty,4]$$
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$$(4,\infty)$$
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$$(-4,4)$$
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$$(\infty,-4)\cup (4,\infty)$$
If $$f ( x ) = \sin ^ { - 1 } ( \sin x ) + \cos ^ { - 1 } ( \sin x ) \text { and } \phi ( x ) = f ( f ( f ( x ) ) )$$ then $$\phi ^ { \prime } ( x )$$
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1
0%
$$\sin x$$
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0
0%
none of these
Explanation
$$f(x)=\sin ^{ -1 }{ \left( \sin { x } \right) } +\cos ^{ -1 }{ \left( \sin { x } \right) } =\cfrac { \pi }{ 2 } \left[ \because \sin ^{ -1 }{ \theta } +\cos ^{ -1 }{ \theta } =\cfrac { \pi }{ 2 } \right] $$
$$f(f(x))=f\left(\cfrac { \pi }{ 2 }\right )=\cfrac { \pi }{ 2 } $$
$$\phi (x)=f(f(f(x)))=f\left(\cfrac { \pi }{ 2 }\right )=\cfrac { \pi }{ 2 } $$
$$\phi (x)=\cfrac { \pi }{ 2 } $$
$$\phi '(x)=0$$
if $$f\left( x \right) = 3x + 2$$ , $$g\left( x \right) = {x^2} + 1$$,then the values of $$\left( {f_og} \right)\left( {{x^2} - 1} \right)$$
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$$3{x^4} - 6{x^2} + 8$$
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$$3{x^4} + 3x + 4$$
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$$6{x^4} + 3{x^2} - 2$$
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$$6{x^4} + 3{x^2} + 2$$
Explanation
$$f\left(x\right)=3x+2$$
$$g\left(x\right)={x}^{2}+1$$
$$f.g\left(x\right)=f\left({x}^{2}+1\right)$$
$$f.g\left(x\right)=3\left({x}^{2}+1\right)+2=3{x}^{2}+5$$
$$f.g\left(x\right)=3{x}^{2}+5$$
$$f.g\left({x}^{2}-1\right)=3{\left({x}^{2}-1\right)}^{2}+5$$
$$=3\left({x}^{4}-2{x}^{2}+1\right)+5$$
$$=3{x}^{4}-6{x}^{2}+3+5$$
$$=3{x}^{4}-6{x}^{2}+8$$
Let A = {1,2,3,4,5} and B={1,2,3,4,5}. If $$f:A\rightarrow B$$ is an one-one function and $$f(x)=x$$ holds only for one value of $$x\epsilon \{ 1,2,3,4,5\} ,$$ then the number of such possible function is
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$$120$$
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$$36$$
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$$45$$
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$$44$$
Number of positive integers in the domain of the function $$f(x)=\sqrt {\log_{0.5}\log_{6}\left(\dfrac {x^{2}+x}{x+4}\right)}$$ is
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$$5$$
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$$6$$
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$$7$$
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$$8$$
Difference between the greatest and the least values of the function
$$f(x) = x(ln x - 2)$$ on $$[1, e^{2}]$$ is
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$$2$$
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$$e$$
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$$e^{2}$$
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$$1$$
The function $$f :\left[-\dfrac{1}{2}, \dfrac{1}{2}\right]\rightarrow \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$$ defined by $$f(x)=\sin^{-1}(3x-4x^{3})$$ is
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both one-one and onto
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onto but not one-one
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one-one but not onto
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neither one-one nor onto
Explanation
$$f:[-\cfrac{1}{2},\cfrac{1}{2}]\rightarrow[-\cfrac{\pi}{2},\cfrac{\pi}{2}]\\f(x)=\sin^{-1}(3x-4x^3)$$
Range of $$\sin^{-1}(3x-4x^3)$$ is $$
[-\cfrac{\pi}{2},\cfrac{\pi}{2}]$$
And $$-\cfrac{\pi}{2}\le\sin^{-1}(3x-4x^3)\le\cfrac{\pi}{2}$$
$$\Rightarrow \sin(\cfrac{-\pi}{2})\le3x-4x^3\le\sin\cfrac{\pi}{2}$$
$$\Rightarrow-1\le3x-4x^3\le1$$
$$3x-4x^3\le-1$$
$$\Rightarrow 4x^3-3x-1\le0$$
$$\Rightarrow x\le\cfrac{1}{2}$$
and$$3x-4x^3\le1\Rightarrow 4x^3-3x+1\le0\\ \Rightarrow x\le-\cfrac{1}{2}\\ \therefore x\epsilon [-\cfrac{1}{2},\cfrac{1}{2}]\\f{x}=\sin^{-1}(3x-4x^3)\\ \Rightarrow f^1(x)=\cfrac{1}{\sqrt{1-(3x-4x^3)^2}}\times(3-12x^2)\\ \Rightarrow f^1(x)=3-12x^2\\ \quad=-(12x^2-3)\\x^2\le0\Rightarrow 12x^2\le0\Rightarrow 12x^2-3\le-3\\-3(12x^2-3)\le3\\ \therefore f^1(x)\le0$$
$$\therefore f^1(x)$$ is a decreasing function.
Hence $$f(x)$$ is one-one and range of function $$=$$ its co-domain.
Hence it is both one-one and onto.
The domain of the function $$f\left(x\right)=\sin^{-1}{\left(1+{e}^{x}\right)^{-1}}$$ is
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$$R$$
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$$[-1,0]$$
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$$[0,1]$$
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$$[-1,1]$$
The interval on which the function $$f(x)=2x^3+9x^2+12x-1$$ is decreasing is?
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$$[-1, \infty)$$
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$$[-2, -1]$$
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$$(-\infty, -2]$$
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$$[-1, 1]$$
Explanation
$$ f(x) = 2x^{3} + 9x^{2} + 12x - 1 $$
$$ f'(x) = 6x^{2} + 18x +12 $$
$$ = 6(x^{2} + 3x + 2)$$
$$ = 6 (x+2)(x+1)$$
$$ \therefore f(x)$$ decreases in interval $$xt [-2,-1]$$
Let g be the inverse function of differentiable function f and $$G\left( x \right) =\frac { 1 }{ g\left( x \right) } if\quad f\left( 4=2 \right) $$ and $$f'\left( 4 \right) =\frac { 1 }{ 16 } $$, then the value of $${ \left( G'\left( 2 \right) \right) }^{ 2 }$$ equals to:
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1
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4
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16
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64
The domain of the function $$f(x) = \frac {1} {\sqrt {^{10}C_{x - 1} - 3 \times ^{10} C_x}}$$ contains the points
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9, 10, 11
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9, 10, 12
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all natural numbers
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None of these
Explanation
Given function is defined if $$^{10} C_{x + 1} > 3 ^{10}C_x$$
$$\Rightarrow \dfrac{1} {11 - x} > \dfrac {3} {4} \Rightarrow 4x > 33$$
$$\Rightarrow x \geqslant 9 \ but \ x \leq 10 \Rightarrow x = 9, 10$$
If $$f:( - 1,1) \to B$$ , is a function defined by $$f(x) = {\tan ^{ - 1}}\dfrac{{2x}}{{1 - {x^2}}}$$, then find $$B$$ when $$f(x)$$ is both one-one and onto function.
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$$\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$$
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$$\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$$
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$$\left( {0,\frac{\pi }{2}} \right)$$
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$$\left[ {0,\frac{\pi }{2}} \right)$$
Explanation
For $$x \epsilon (-1,1)$$, we have
$$f(x)=\tan^{-1}\left[\dfrac {2x}{1-x^2}\right]$$
Substituting $$x=\tan\theta$$ in above equation.
Therefore, $$f(\tan \theta)=\tan^{-1}\left[\dfrac {2\tan \theta}{1-\tan^2\theta}\right]$$
$$=\tan^{-1}\tan (2\theta)=2\theta$$
$$=2\tan ^{-1}x$$
Thus $$-\dfrac {\pi}{2}<\tan ^{-1}\left[\dfrac {2x}{1-x^2}\right]<\dfrac {\pi}{2}$$
Thus option B is correct.
The sum of all real values of $$x$$ satisfying the equation
$${\left( {{{\rm{x}}^{{\rm{2 - }}}}{\rm{5x + 5}}} \right)^{{{\rm{x}}{{\rm{^2 + 4x - 60}}}}}}{\rm{ = 1}}$$ is
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$$-4$$
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$$6$$
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$$5$$
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$$3$$
Explanation
$$(x^2 - 5x + 5)^{x^2 + 4x - 60} = 1$$
$$\Rightarrow (x^2 - 5x + 5)^0$$
$$\Rightarrow x^2 + 4x - 60 = 0$$
Solving for x.
$$x = -10, 6$$
And, $$x^2 - 5x + 5$$
$$1^a = 1 $$ or $$b^o = 1$$ or $$(-1)^x = 1$$
$$\therefore x^2 - 5x + 5 = 1$$
$$x^2 - 5x + 4 = 0$$
$$(x - 4) (x - 1) = 0$$
$$x = 4 $$ or $$1$$
Or
$$x^2 - 5x + 5 = -1$$
$$x^2 - 5x + 6 = 0$$
$$(x - 3) (x - 2) = 0$$
$$x = 3, 2$$
2 will satisfy not 3 because power will be odd
Hence $$x = \{-10, 1, 2, 4, 6\}$$
$$\therefore $$ sum of solutions $$=3$$
If $$f \left( \dfrac { x + y } { 2 } \right) = \dfrac { f ( x ) + f ( y ) } { 2 }$$ for all $$x , y \in R$$ and $$f ^ { \prime } ( o ) = - 1 , f ( o ) = 1$$ then $$f(2)=$$
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$$\dfrac { 1 } { 2 }$$
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$$1$$
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$$-1$$
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$$\dfrac { -1 } { 2 }$$
Explanation
let $$f(x)=ax+b$$
$$f(0)=1\implies b=1$$
$$f'(0)=-1 \implies a=-1$$
$$\implies f(x)=1-x$$
$$\implies f(2)=-1$$
The domain of $$f(x) = \sqrt{2 - log_3 (x - 1)}$$ is
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$$(2, 12]$$
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$$(\infty, 10]$$
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$$(3, 12]$$
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$$(1, 10]$$
Explanation
$$\sqrt{2-log_3(x-1)}$$
$$x-1 > 0$$
$$\Rightarrow x > 1$$ …………$$(1)$$
Also,
$$2-log_3(x-1)\geq 0$$
$$log_3(x-1)\leq 2$$
$$(x-1)\leq 3^2$$
$$\Rightarrow x-1\leq 9$$
$$x\leq 10$$ …………..$$(2)$$
Intersection of $$(1)$$ & $$(2)$$
$$x\in (1, 10]$$.
If $$f(x)=x^{3}+x^{2}f'(1)+xf''(2)+f'''(3)\ \forall x\ \epsilon \ R$$, then $$f(x)$$ is
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one-one and onto
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one-one and into
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many-one and onto
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non-invertible
Let $$E=\left\{1,2,3,4\right\}$$ and $$F=\left\{1,2\right\}$$. Then the number of onto functions from $$E$$ to $$F$$ is
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$$14$$
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$$16$$
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$$12$$
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$$8$$
Explanation
The total number of ways of distributing $$(1,2)$$ considering them as 2 objects among $$(1,2,3,4)$$ considering them as 4 people is equal to total no. of Onto functions from E to F.
There can be two ways of distributing $$(1,2),$$
Either we give both to one people
Or give $$1$$ to one people and $$2$$ to another.
So, Total no. of ways $$=4×1+^4C_2×2=16$$
The domain of the function, $$f(x)=\dfrac{|x|-2}{|x|-3}$$ is
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$$R$$
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$$R-\{2,3\}$$
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$$R-\{2,-2\}$$
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$$R-\{-3,3\}$$
Explanation
$$f(x)=\dfrac{|x|-2}{|x|-3}$$
Here, $$x$$ cannot be $$\pm 3$$.
(f is not defined when $$x=\pm 3$$)
Therefore,
Domain $$= R-\{-3,3\}$$
If $$f\left( x \right) =\sqrt { { x }^{ 2 }-4 } $$ and $$g\left( x \right) =\dfrac { x-1 }{ x-3 } $$ then number of integer elements, which are not in the domain of the function $$(f.g)(x)$$ equals
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0%
3
0%
4
0%
5
0%
None of these
Let $$N$$ be the set of natural numbers and two functions $$f$$ and $$g$$ be defined as $$f,g : N\to N$$ such that :
$$f (n)= \begin{cases}\dfrac{n+1}{2}& \text{if n is odd}\\ \dfrac{n}{2} & \text{in n is even} \end{cases}$$
and $$g(n) = n - (-1)^n$$. The fog is:
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0%
Both one-one and onto
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One-one but not onto
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Neither one-one nor onto
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onto but not one-one
Explanation
$$fx = \begin{cases} \dfrac{n+1}{2}& \text{n is odd} \\\dfrac{n}{2}& \text{n is even}\end{cases}$$
$$g(x) = n -(-1)^n \begin{cases}n+1; \text{n is odd} \\n-1; \text{n is even} \end{cases}$$
$$f(g(n)) = \begin{cases}\dfrac{n}{2}; & \text{n is even}\\ \dfrac{n+1}{2}; &\text{n is odd} \end{cases}$$
$$\therefore$$ onto but not one-one
If $$f(x)=\dfrac {4^{x}}{4^{x}+2}$$, then the value of $$f(x)+f(1-x)$$ is
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0%
$$0$$
0%
$$-1$$
0%
$$1$$
0%
$$can't\ be\ determined$$
Explanation
Given $$f{\left( x \right)} = \cfrac{{4}^{x}}{{4}^{x} + 2}$$
$$\therefore f{\left( 1 - x \right)} = \cfrac{{4}^{1-x}}{{4}^{1-x} + 2}$$
$$\Rightarrow f{\left( 1 - x \right)} = \cfrac{\left( \cfrac{4}{{4}^{x}} \right)}{\left( \cfrac{4}{{4}^{x}} \right) + 2}$$
$$\Rightarrow f{\left( 1 - x \right)} = \cfrac{4}{4 + {4}^{x} \cdot 2} = \cfrac{2}{2 + {4}^{x}}$$
Therefore,
$$f{\left( x \right)} + f{\left( 1 - x \right)} = \cfrac{{4}^{x}}{2 + {4}^{x}} + \cfrac{2}{2 + {4}^{x}}$$
$$\Rightarrow f{\left( x \right)} + f{\left( 1 - x \right)} = \cfrac{{4}^{x} + 2}{2 + {4}^{x}} = 1$$
Domain of the function $$f\left( x \right) =\sqrt { 2-2x-{ x }^{ 2 } } $$ is
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$$-\sqrt { 3 } \le x\le \sqrt { 3 } $$
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$$-1-\sqrt { 3 } \le x\le -1+\sqrt { 3 } $$
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$$-2\le x\le 2$$
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$$-2+\sqrt { 3 } \le x\le -2-\sqrt { 3 } $$
Explanation
Given,
$$f(x)=\sqrt{2-2x-x^2}$$
$$\sqrt{f\left(x\right)}\quad \Rightarrow \quad \:f\left(x\right)\ge 0$$
$$2-2x-x^2\ge \:0$$
$$-\left(x+1\right)^2+3\ge \:0$$
$$-\left(x+1\right)^2\ge \:-3$$
$$\left(x+1\right)^2\le \:3$$
$$-\sqrt{3}\le \:x+1\le \sqrt{3}$$
$$-\sqrt{3}\le \:x+1\quad \mathrm{and}\quad \:x+1\le \sqrt{3}$$
$$-\sqrt{3}\le \:x+1\quad :\quad x\ge \:-\sqrt{3}-1$$
$$x+1\le \sqrt{3}\quad :\quad x\le \sqrt{3}-1$$
$$\Rightarrow -\sqrt{3}-1\le \:x\le \sqrt{3}-1$$
Let $$f(x)=x^ {135}+x^ {125}-x^ {115}+x^ {5}+1$$. If $$f(x)$$ divided by $$x^ {3}-x$$, then the remainder is some function of $$x$$ say $$g(x)$$. Then $$g(x)$$ is an:-
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one-one function
0%
many one function
0%
into function
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onto function
Explanation
The domain of the function $$\sin^{-1} (log_2(\frac{x}{3}))$$ is-
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[$$\frac{1}{2},3$$]
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[$$\frac{1}{2},3$$]
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[$$\frac{3}{2},6$$]
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[$$\frac{1}{2},2$$]
Domain of function $$f(x)=\dfrac{|x|-x}{2x}$$ is
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$$\mathbb{R}$$
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$$\mathbb{R}-\{0\}$$
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$$\mathbb{Z}$$
0%
$$\mathbb{N}$$
The domain of $$f(x)= e^{\sqrt{x}}+cos x $$ is
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$$(-\infty ,\infty )$$
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$$[0,\infty )$$
0%
(0,1)
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$$(1,\infty )$$
The domain of function $$f ( x ) = \dfrac { x ^ { 2 } - 10 x + 26 } { x ^ { 4 } \left( x ^ { 2 } - 9 \right) \left( 1 + 27 x ^ { 2 } \right) }$$
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$$\mathbf { x } \in \mathbf { R } - \{ 0,\pm3 \}$$
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$$\mathbf { x } \in \mathrm { R } - \{ 0,3 \}$$
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$$x \in R$$
0%
none
Explanation
$$f(x)=\dfrac{x^2-10x+26}{x^4(x^2-9)(1+27x^2)}$$
The function is defined only when
$$x^4(x^2-9)(1+27x^2)\neq 0\\\implies x\neq 0,x^2-9\neq 0\\x^2\neq 9\\x\neq \pm 3$$
The domain is $$R-\{0,\pm3\}$$
The domain of the function $$f(x)=\sqrt {\dfrac{x^{2}-1}{x-2}}$$ is
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0%
$$(2,\infty )$$
0%
$$(1,\infty )$$
0%
$$[-1,1] \cup (2,\infty)$$
0%
none of these
Explanation
Given,
$$f\left(x\right)=\sqrt{\dfrac{x^2-1}{x-2}}$$
$$\dfrac{x^2-1}{x-2}\ge \:0:\quad -1\le \:x\le \:1\quad \mathrm{or}\quad \:x>2$$
undefined singularity point: $$x=2$$
combined domain of function,
$$-1\le \:x\le \:1\quad \mathrm{or}\quad \:x>2$$
$$\left[-1,\:1\right]\cup \left(2,\:\infty \:\right)$$
The domain of the function $$f ( x ) = \log _ { 2 } x^2$$ is
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0%
$$\mathbf { x } \in \mathbf { R }$$
0%
$$x \in [ 0 , \infty )$$
0%
$$x \in (- \infty ,0)\cup( 0 , \infty )$$
0%
$$x \in R - \{ x | x \in 1 \}$$
Explanation
option C is correct
Domain of the function $$f(x) = \dfrac{x^2-3x+2}{x^2+x-6}$$ is
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{$$x:x \epsilon R, x \neq -3$$}
0%
{$$x:x \epsilon R, x \neq 2$$}
0%
{$$x:x \epsilon R$$}
0%
{$$x:x \epsilon R, x \neq 2, x \neq -3$$}
Explanation
Given,
$$f(x)=\dfrac{x^2-3x+2}{x^2+x-6}$$
$$x^2+x-6=0$$
roots of above equation are:
$$x=-3,2$$
So, $$x$$ can't be equal to $$-3 \, or \, 2$$
The above points are undefined
The function domain is
$$x<-3,-3<x<2,x>2$$
$$\left \{ x:x\in R,x\neq 2,x\neq -3 \right \}$$
The function $$y=\dfrac { x }{ 1+{ x }^{ 2 } } $$ has its domain as
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$$x \in R$$
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$$x\in R-(-1,1)$$
0%
$$x \in \left( 0,\infty \right) $$
0%
$$x \in \left( -\infty ,-1 \right) $$
Explanation
The function $$y=\dfrac x{1+x^2}$$
is defined only when
$$1+x^2\neq 0$$
$$x^2\neq -1$$
$$\implies x\in R$$
The domain of the function, $$f(x)=\sqrt{2-x}$$$$-\dfrac{1}{\sqrt{9-x^{2}}}$$ is
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(-3,1)
0%
[-3,1]
0%
(-3,2)
0%
(-3,2]
Explanation
Given,
$$f\left(x\right)=\sqrt{2-x}-\dfrac{1}{\sqrt{9-x^2}}$$
$$\sqrt{f\left(x\right)}\quad \Rightarrow \quad \:f\left(x\right)\ge 0\:$$
$$2-x\ge \:0:\quad x\le \:2$$
$$9-x^2\ge \:0:\quad -3\le \:x\le \:3$$
Combined interval:
$$-3\le \:x\le \:2$$
Undefined singularity points: $$x=3,\:x=-3$$
Combining real regions and undefined singularity points:
$$-3<x\le \:2$$
$$(-3,\:2]$$
$$f : R \rightarrow R , f ( x ) = e ^ { | x | } - e ^ { - x }$$ is many-one into function.
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0%
True
0%
False
Explanation
$$f(x)=e^{(x)}-e^{-x}$$
For every $$x$$, there will be different $$f(x) $$
$$\therefore $$ It is a one - one function
$$\therefore False $$
Number of one-one functions from A to B where $$n(A)=4, n(B)=5$$.
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0%
$$4$$
0%
$$5$$
0%
$$120$$
0%
$$90$$
Explanation
$$n(A)=4$$ and $$n(B)=5$$
For one-one mapping
4 elements can be selected out of 5 elements of set B in $${}^5C_4$$ ways
and then those 4 selected elements can be mapped with 4 elements of set A in $$4!$$ ways.
Number of one-one mapping from $$A$$ to $$B$$ $$={}^5C_4\times4!={}^5P_4=\dfrac{5!}{(5-4)!}=5!=120$$
Consider $$f(x) = \dfrac{x^2}{1 + x^3}$$ ; $$g(t) = \displaystyle \int f(t) dt$$ . If $$g(1) = 0$$ then $$g(x)$$ equals
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0%
$$\dfrac{1}{3} ln(1 + x^3)$$
0%
$$\dfrac{1}{3} ln\left ( \dfrac{1 + x^3}{2} \right )$$
0%
$$\dfrac{1}{2} ln\left ( \dfrac{1 + x^3}{3} \right )$$
0%
$$\dfrac{1}{3}l n\left ( \dfrac{1 + x^3}{3} \right )$$
Domain of the function $$f(x)=$$ $$\dfrac { x-3 }{ (x-1)\sqrt { { x }^{ 2 }-4 } } $$
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0%
$$(1,2)$$
0%
$$(-\infty,-2)\cup(2,\infty)$$
0%
$$(-\infty,-2)\cup(1,\infty)$$
0%
$$(-x,x)-{(t\pm2)}$$
Explanation
Given,
$$f(x)=\dfrac{x-3}{\left(x-1\right)\sqrt{x^2-4}}$$
$$\sqrt{f\left(x\right)}\quad \Rightarrow \quad \:f\left(x\right)\ge 0$$
$$x^2-4\ge \:0:\quad x\le \:-2\quad \mathrm{or}\quad \:x\ge \:2$$
Undefined singularity points $$x<-2\quad \mathrm{or}\quad \:x>2$$
Domain $$x<-2\quad \mathrm{or}\quad \:x>2$$
So, the domain is $$(-\infty,-2)\cup(2,\infty)$$
In a set $$A=\left\{1,2,3,4\right\}$$, the relation R is defined as $$x\quad R\quad y\quad \Longleftrightarrow \quad x\le y$$, then the domain of the inverse relation is
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0%
$$\left\{1,2,3\right\}$$
0%
$$\left\{3,4,5,6\right\}$$
0%
$$\left\{1,2,3,4\right\}$$
0%
$$\left\{4,5,6\right\}$$
$$f : R \rightarrow R , f ( x ) = 2 x + | \sin x |$$ is one-one onto.
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0%
True
0%
False
Explanation
$$ f(x) = 2x + \left | sin x \right |$$
for query x,there will be a defferent f(x)
$$ \therefore one-one $$ & domain $$\rightarrow$$ R
$$ \therefore onto $$
$$ \therefore True. $$
If $$f:R\rightarrow R,f\left( x \right) =\dfrac { { ax }^{ 2 }+6x-8 }{ a+6x-{ 8x }^{ 2 } } $$ is onto, then $$\alpha \in $$
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0%
$$\left( 1,\infty \right) $$
0%
$$\left( 0,\infty \right) $$
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$$\left( 2,12 \right) $$
0%
$$\left[ 2,14 \right] $$
If f : $$R\rightarrow S$$, defined by f(x) =sin x -$$\sqrt{3}$$ cos x +1, is onto, then the interval of S is
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0%
[0, 3]
0%
[-1, 1]
0%
[0, 1]
0%
[-1, 3]
If $$ f : R \rightarrow R $$ be given by $$ f(x)=\left(3-x^{3}\right)^{\dfrac{1}{3}}, $$ then $$fof(x)$$ is
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0%
$$
x^{\dfrac{1}{3}}
$$
0%
$$
1^{3}
$$
0%
x
0%
$$
\left(3-x^{3}\right)
$$
Explanation
Given,
$$ f(x)=\left(3-x^{3}\right)^{\dfrac{1}{3}} $$.
Now,
$$fof(x)$$
$$=f[f(x)]$$
$$ =\left(3-[f(x)]^{3}\right)^{\dfrac{1}{3}}, $$
$$ =\left(3-[(3-x^3)^{\dfrac{1}{3}}]^{3}\right)^{\dfrac{1}{3}}, $$
$$ =\left(3-[(3-x^3)]\right)^{\dfrac{1}{3}}, $$
$$=[x^3]^{\dfrac{1}{3}}$$
$$=x$$.
Let : $$R\rightarrow R$$ defined as $$f\left( x \right) =\dfrac { x\left( x+1 \right) \left( { x }^{ 4 }+1 \right) +{ 2x }^{ 4 }+{ x }^{ 2 }+2 }{ { x }^{ 2 }+x+1 } $$
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0%
odd and one-one
0%
even and one-one
0%
many to one and even
0%
many to one and neither even nor odd
The domain of the function $$f(x) =$$$$\dfrac{\sin^{-1}(x-3)}{\sqrt{9-x^2}}$$ is
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0%
$$[1, 2]$$
0%
$$[2, 3)$$
0%
$$[2, 3]$$
0%
$$[1, 2)$$
Explanation
Given the function is
$$f(x) =$$$$\dfrac{\sin^{-1}(x-3)}{\sqrt{9-x^2}}$$.
Now, $$f(x)$$ will be defined if
$$|x-3|\le 1$$
or, $$-1\le x-3\le 1$$
or, $$2\le x\le 4$$......(1),
and $${{9-x^2}}\gt 0$$
or, $$x^2<9$$
or, $$-3<x<3$$.....(2).
So $$f(x)$$ will be defined in the common portion of (1) and (2).
And the common protion is $$2\le x\lt 3$$ or, $$x\in [2,3)$$.
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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