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CBSE Questions for Class 11 Commerce Applied Mathematics Functions Quiz 3 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Functions
Quiz 3
If $$ f : R \rightarrow R$$ is defined by $$f(x)=2x-2,$$ then $$(f\circ f) (x) + 2 =$$
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$$f(x)$$
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$$2f(x)$$
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$$3f(x)$$
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$$-f(x)$$
Explanation
Let $$f : R \rightarrow R$$ is defined by $$f(x)=2x-2$$.
Then
$$(f\circ f)(x)+2= f[f(x)]+2$$
$$=f(2x-2)+2$$
$$=2(2x-2)-2+2$$
$$=2f(x)$$
If $$f(x) =\displaystyle \frac{x}{\sqrt{1-x^2}}, g(x)=\frac{x}{\sqrt{1+x^2}}$$ then $$(f\circ g)(x) =$$
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$$\displaystyle \frac{x}{\sqrt{1-x^2}}$$
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$$\displaystyle \frac{x}{\sqrt{1+x^2}}$$
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$$\displaystyle \frac{1-x^2}{\sqrt{1-x^2}}$$
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$$x$$
Explanation
Let $$x= \tan \theta$$
$$(f\circ g)(x)=f[g(x)] = f[g(\tan \theta)]$$.
Since $$\displaystyle g(x) = \frac{x}{\sqrt{1+x^2}}$$, we have
$$f[g(\tan \theta)]=\displaystyle f \left [ \frac{\tan \theta}{\sqrt{1+\tan^2 \theta}} \right ] = f \left [ \frac{\tan \theta}{\sec \theta} \right ]=f (\sin \theta)$$
Also since $$f(x)=\cfrac{x}{\sqrt{1-x^2}}$$ , we have
$$\displaystyle f (\sin \theta) = \frac{\sin \theta}{\sqrt{1-\sin^2 \theta}}$$
$$=\tan \theta =x$$
If $$f(x)=\log x, g(x) = x^3$$ then $$f[g(a)]+f[g(b)]= $$
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$$f[g(a)+g(b)]$$
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$$f[g(ab)]$$
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$$g[f(ab)]$$
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$$g[f(a)+f(b)]$$
Explanation
Given that $$g(x)=x^3$$. Therefore,
$$f[g(a)]+f[g(b)]=f(a^3)+f(b^3)$$.
Since $$f(x)=\log x$$, we have
$$f(a^3)+fb^3=\log a^3+\log b^3$$
$$=\log (a^3b^3)=\log ((ab)^{3})$$
$$=f[g(ab)]$$
Find the domain of $$x^2+2$$
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R
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N
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Z
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All of the above
Explanation
The function is a polynomial function. Hence its domain includes all real numbers which includes the set of all integers, Z and the set of all natural numbers, N. So correct option is D.
The domain of $$f(x) = \displaystyle \frac{3}{4-x^2}+\log_{10} (x^3-x)$$ is:
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$$(1, 2)$$
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$$[-1, 1) \cup (1, 2)$$
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$$(1, 2) \cup (2, \infty)$$
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$$(-1, 0) \cup (1,2) \cup (2, \infty)$$
Explanation
$$ f(x)$$ is defined if
either
(1) $$4-{ x }^{ 2 }\neq 0$$
$$ \Rightarrow x\neq \pm 2$$
or (2) $$ { x }^{ 3 }-x>0$$
$$ \Rightarrow x\left( { x }^{ 2 }-1 \right) >0$$
$$ \Rightarrow \quad x(x+1)(x-1)>0$$
$$ \Rightarrow x>0, x>1, x>-1$$
Therefore, we have $$x>-1, x>0, x>1$$ and $$x\neq 2, x\neq -2$$.
$$ \therefore x$$ lies between $$(-1,0), (1,2), (2,\infty )$$.
$$ \therefore$$ Domain is $$(-1,0)\cup (1,2)\cup (2,\infty )$$.
If $$f:R \rightarrow R$$ and $$g : R \rightarrow R$$ are defined by $$f(x)=2x+3$$ and $$g(x)=x^2+7$$, then the values of $$x$$ such that $$g(f(x)) =8$$ are:
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$$1, 2$$
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$$-1, 2$$
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$$-1, -2$$
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$$1, -2$$
Explanation
Since $$f(x)=2x+3$$,
$$g[f(x)]=8 \Rightarrow g(2x+3)=8$$.
Also since $$g(x)=x^2+7$$,
$$g(2+3x)=8\Rightarrow (2x+3)^2+7=8$$
$$\Rightarrow 2x+3 = \pm 1$$
$$\Rightarrow x=-1$$ or $$-2$$.
If $$f(x) = \dfrac{2x+5}{x^{2} + x + 5}$$, then $$f\left [ f(- 1 ) \right ]$$ is equal to
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$$\dfrac{149}{155}$$
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$$\dfrac{155}{147}$$
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$$\dfrac{155}{149}$$
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$$\dfrac{147}{155}$$
Explanation
Given expression is $$ f(x)=\dfrac { 2x+5 }{ { x }^{ 2 }+x+5 } $$
$$ \therefore f(-1)=\dfrac { 2\times (-1)+5 }{ (-1)^{ 2 }+(-1)+5 } =\dfrac { 3 }{ 5 } $$
$$ \therefore f\left( f\left( -1 \right) \right) =\dfrac { 2\times \dfrac { 3 }{ 5 } +5 }{ \left( \dfrac { 3 }{ 5 } \right) ^{ 2 }+\dfrac { 3 }{ 5 } +5 } =\dfrac { 155 }{ 149 }$$
Which one of the following relation is a function
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All of these
Explanation
As in option C every element in the domain has a unique image.
If $$f : R \rightarrow R$$ and $$g :R \rightarrow R$$ are defined by $$f(x) = x -[x]$$ and $$g(x) = [x]$$ for $$x \in R$$, where $$[x]$$ is the greatest integer not exceeding $$x$$, then for every $$x \in R, f(g(x)) =$$
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$$x$$
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$$0$$
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$$f(x)$$
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$$g(x)$$
Explanation
Since $$f(x)=x-[x]$$, we have
$$f[g(x)] =g(x)-[g(x)]$$
Also since $$g(x)=[x]$$,we have
$$g(x)-[g(x)]=[x] - [[x]]$$.
Here, $$[x]$$ is the greatest integer not exceeding $$x$$. Therefore,
$$[[x]]=[x]$$ and hence,
$$[x]-[[x]]=[x]-[x]=0$$
If $$y=f(x) = \dfrac{2x-1}{x-2}$$, then $$f(y)=$$
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$$x$$
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$$y$$
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$$2y-1$$
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$$y-2$$
Explanation
$$\displaystyle f(y) = \frac{2y-1}{y-2} = \frac{2 \left ( \dfrac{2x-1}{x-2} \right ) -1}{\left ( \dfrac{2x-1}{x-2} \right ) -2}$$
$$=\displaystyle \frac{4x-2-x+2}{2x-1-2x+4} = \frac{3x}{3} = x$$
If $$f(g(x))$$ is one-one function, then
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g(x) must be one-one
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f(x) must be one-one
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f(x) may not be one-one
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g(x) may not be one-one
Explanation
It is fundamental concept that, If $$f(g(x))$$ is one-one function then,
$$f(x)$$ must be one-one function and $$g(x)$$ may be one-one or many one
Which of the following functions are one-one?
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$$f:R\rightarrow R$$ given by $$ f(x)={ 2x }^{ 2 }+1$$ for all $$\quad x\in R$$
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$$g:Z\rightarrow Z$$ given by $$ g(x)={ x }^{ 4 }$$ for all $$\quad x\in R$$
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$$h:R\rightarrow R$$ given by $$ h(x)={ x }^{ 3 }+4$$ for all $$\quad x\in R$$
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$$\phi :C\rightarrow C$$ given $$ \phi (z)={ 2z }^{ 6 }+4$$ for all $$\quad x\in R$$
Explanation
For all even powered function,
$$f(x_{1})=f(x_{2})$$ where $$x_{1}=\pm(x_{2})$$
Hence, the only possible one-one function is $$f(x)=x^3+4$$
$$f(x)$$ is a strictly increasing function. So, it is one-one.
A mapping function $$f:X\rightarrow Y$$ is one-one, if
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$$f({ x }_{ 1 })\neq f({ x }_{ 2 })\ $$for all $$ { x }_{ 1 },{ x }_{ 2 }\in X$$
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$$f({ x }_{ 1 })=f({ x }_{ 2 })\Rightarrow { x }_{ 1 }={ x }_{ 2 }$$ for all $${ x }_{ 1 },{ x }_{ 2 }\in X$$
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$${ x }_{ 1 }={ x }_{ 2 }\Rightarrow f({ x }_{ 1 })=f({ x }_{ 2 })$$ for all $${ x }_{ 1 },{ x }_{ 2 }\in X$$
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none of these
Explanation
For a function to be one one
$$f(x_{1})=f(x_{2})$$
Implies that
$$x_{1}=x_{2}$$
Hence the answer is
option B
Find the domain of $$x$$ if $$f(x)=\sqrt {x^2-|x|-2}$$
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$$x\in R-(-2, 2)$$
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$$x\in R$$
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$$x\in R-(0, 2)$$
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None of these
Explanation
Case I $$x<0$$
Hence $$f(x)=\sqrt{x^2-(-x)-2}$$
$$=\sqrt{x^2+x-2}$$
$$=\sqrt{(x+2)(x-1)}$$
Therefore the domain will be $$(-\infty,-2]$$....$$(D_{1})$$
Case II $$x>0$$
Hence $$f(x)=\sqrt{x^2-(x)-2}$$
$$=\sqrt{x^2-x-2}$$
$$=\sqrt{(x-2)(x+1)}$$
Therefore the domain will be $$[2,\infty)$$....$$(D_{2})$$
Hence the domain of the function $$f(x)$$ from $$D_{1}$$ and $$ D_{2}$$ will be
$$=(-\infty,-2]\cup[2,\infty)$$
If $$f:R\rightarrow R$$ given by $$f(x)={ x }^{ 3 }+({ a+2)x }^{ 2 }+3ax+5$$ is one-one, then $$a$$ belongs to the interval
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$$(-\infty ,1)$$
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$$(1 ,\infty)$$
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$$(1 ,4)$$
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$$(4 ,\infty)$$
Explanation
If $$ f(x) $$ is one one then $$ f'\left( x \right) $$ should be either positive or negative for all $$ x $$.
$$f'\left( x \right) = 3{ x }^{ 2 }+2(a+2)x+3a$$
$$\Rightarrow D = 4{ (a+2) }^{ 2 }-4\times3\times3a<0$$
$$\Rightarrow$$ $${ a }^{ 2 }-5a+4<0$$
$$\Rightarrow$$ $$(a-1)(a-4)<0$$
Hence, $$a\in \left( 1,4 \right) $$
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Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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Assertion is correct but Reason is incorrect
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Assertion is incorrect but Reason is correct
Explanation
Domain of $$g(x)$$:
$$g(x)$$ is defined if $$3-x\ge 0$$ and $$(x-1)(x-2)(x-2)\neq 0$$
$$\Rightarrow x\le3$$ and $$x\neq 1,2,3$$
$$\therefore $$ Domain of $$g(x)=(\infty,3)-{1,2,3}$$
Domain of $$h(x):$$
$$\displaystyle h\left( x \right) =\sin ^{ -1 }{ \left[ \frac { 3x-2 }{ 2 } \right] } \Rightarrow -1\le \left[ \frac { 3x-2 }{ 2 } \right] \le 1$$
Case I:
If $$\displaystyle \left[ \frac { 3x-2 }{ 2 } \right] =-1\Rightarrow -1\le \frac { 3x-2 }{ 2 } <0\Rightarrow -2\le 3x-2<0\Rightarrow 0\le x<\frac { 2 }{ 3 } $$ ...(1)
Case II:
$$\displaystyle \left[ \frac { 3x-2 }{ 2 } \right] =0\Rightarrow 0\le \frac { 3x-2 }{ 2 } <1\Rightarrow 0\le 3x-2<2$$
$$\displaystyle \Rightarrow 2\le 3x<4\Rightarrow \frac { 2 }{ 3 } \le x<\frac { 4 }{ 3 } $$ ...(2)
Case III:
If $$\displaystyle \left[ \frac { 3x-2 }{ 2 } \right] =1\Rightarrow 1\le \frac { 3x-2 }{ 2 } <2\Rightarrow 2\le 3x-2<4\Rightarrow \frac { 4 }{ 3 } \le x<2$$ ...(3)
Thus, from (1),(2) and (3), we have
Domain of $$h(x)=[0,2)$$
$$\therefore $$ Domain of $$f=[2,0)-{1}$$
Which of the following function is one-one?
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$$f:R\rightarrow R$$ given by$$ f(x)=|x-1|$$ for all $$x\in R$$
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$$g:\left[ -\dfrac{\pi }{ 2 },\dfrac{ \pi }{ 2 } \right] \rightarrow R$$ given by $$g(x)=|sinx|$$ for all $$ x\in \left[ \dfrac{ -\pi }{ 2 },\dfrac{ \pi }{ 2 } \right] $$
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$$h:\left[ \dfrac{ -\pi }{ 2 },\dfrac{ \pi }{ 2 } \right] \in R$$ given by $$ h(x)=sinx$$ for all $$ x\in \left[ \dfrac{ -\pi }{ 2 },\dfrac{ \pi }{ 2 } \right] $$
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$$\phi :R\rightarrow R$$ given by $$f(x)={ x }^{ 2 }-4$$ for all $$ x\in R$$
Explanation
Option A - Consider $$x = 2 $$ and $$ x =0 $$ , the value of $$f(x)$$ is same. Hence it is not one-one
Option B - If we replace $$x$$ by $$-x$$, then the value of $$g(x)$$ remains the same. Hence it is not one-one
Option C - $$h(x)$$ is an increasing function for the given values of $$x$$. Hence it is one-one function
Option D -$$f(x)$$ is an even function. So it is not one-one for the given values of $$x$$
If $$f$$ and $$g$$ are one-one functions from $$R\to R$$, then
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$$f+g$$ is one-one
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$$fg$$ is one-one
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$$fog$$ is one-one
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none of these
Explanation
Let $${x_1},{x_2} \in R$$ be two distinct elements, then $$g\left( {{x_1}} \right) \ne g\left( {{x_2}} \right)$$, as $$g$$ is one-one function. Similarly $$f\left( {g\left( {{x_1}} \right)} \right) \ne f\left( {g\left( {{x_2}} \right)} \right)$$ as $$f$$ is also one-one function.
Hence, $$f\circ g$$ is one-one function.
Note that $$f+g$$ and $$f\cdot g$$ may not one-one functions even if $$f$$ and $$g$$ are one one. For example consider $$f\left( x \right) = x$$ and $$g\left( x \right) =- x$$, then $$f+g$$ and $$f\cdot g$$ are not one-one.
Let $$\displaystyle f:R\rightarrow A=\left \{ y: 0\leq y< \dfrac{\pi}{2} \right \}$$ be a function such that $$\displaystyle f(x)=\tan^{-1}(x^{2}+x+k),$$ where $$k$$ is a constant. The value of $$k$$ for which $$f$$ is an onto function is
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$$1$$
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$$0$$
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$$\displaystyle \frac{1}{4}$$
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none of these
Explanation
For $$f$$ to be an onto function, Range and co-domain of $$f$$ should be equal
$$\Rightarrow f(x)\ge 0 \forall x\in R$$
$$\Rightarrow \tan^{-1}(x^2+x+k)\ge 0$$
For the above equation to be valid for all $$x$$
We must have, the discriminant of $$x^2+x+k=0$$, is zero
$$ b^2-4ac=0$$
$$\Rightarrow 1-4k=0\Rightarrow k =\dfrac{1}{4}$$
If $$f(x) = \sqrt{| x-1|}$$ and $$g(x) = \sin x$$, then $$(fog) (x)$$ equals
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$$\sin \sqrt{| x-1|}$$
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$$\left|\sin\dfrac{x}{2} - \cos\dfrac{x}{2}\right|$$
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$$\left|\sin x + \cos x\right|$$
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$$\left|\sin\dfrac{x}{2} + \cos\dfrac{x}{2}\right|$$
Explanation
$$\displaystyle fog\left( x \right) =f\left( g\left( x \right) \right) =\sqrt { \left| \sin { x } -1 \right| } $$
$$=\sqrt { \left| 1-\cos { \left( \dfrac { \pi }{ 2 } -x \right) } \right| } =\sqrt { 2 } \left| \sin { \left( \dfrac { \pi }{ 4 } -\dfrac { x }{ 2 } \right) } \right| =\left| \sin { \dfrac { x }{ 2 } } -\cos { \dfrac { x }{ 2 } } \right| $$
The domain of the function $$\displaystyle f(x)=\sqrt{x^{2}-[x]^{2}},$$ where $$[x]=$$ the greatest integer less than or equal to $$x$$, is
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$$R$$
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$$[0,+\infty)$$
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$$(-\infty,0]$$
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none of these
Explanation
If $$x \ge 0$$, then $$\left[x\right]$$ is an integer part of $$x$$.
Hence $${x^2} \ge {\left[ x \right]^2}$$ and $$\displaystyle f\left( x \right) = \sqrt {{x^2} - {{\left[ x \right]}^2}}$$ is well-defined.
If $$x < 0$$ say $$x=-a$$ where $$a$$ is a positive number,
then $$\displaystyle \left[ x \right] =- \left( {\left[ a \right] + 1} \right)$$.
Hence, $${x^2} < {\left[ x \right]^2}$$ and the function $$\displaystyle f\left( x \right) = \sqrt {{x^2} - {{\left[ x \right]}^2}}$$ is not defined.
Therefore the domain of the function is $$\left[ {0,\infty } \right)$$.
If $$f(x)=ax+b$$ and $$g(x)=cx+d$$, then $$f(g(x))=g(f(x))$$ implies
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$$f(a)=g(c)$$
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$$f(b)=g(b)$$
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$$f(d)=g(b)$$
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$$f(c)=g(a)$$
Explanation
$$f(g(x))=a(cx+d)+b$$
$$=acx+ad+b$$...(i)
$$g(f(x))=c(ax+b)+d$$
$$=ac(x)+bc+d$$ ...(ii)
$$f(g(x))=g(f(x))$$
From (i) and (ii)
$$ac(x)+bc+d=acx+ad+b$$
$$cb+d=ad+b$$
$$g(b)=f(d)$$
If $$\displaystyle f(x)=\frac{1}{1-x},x\neq 0,1$$ then the graph of the function $$\displaystyle y=f\left \{ f(f(x)) \right \},x> 1,$$ is
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a circle
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an ellipse
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a straight line
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a pair of straight lines
Explanation
$$f(f(x))=\dfrac{1}{1-\dfrac{1}{1-x}}$$
$$=-\dfrac{1-x}{x} =g(x)$$
$$f(g(x))=\dfrac{1}{1+\dfrac{1-x}{x}}$$
$$=\dfrac{x}{x+1-x}$$
$$=x$$
$$y=x$$ is an equation of straight line.
Hence, 'C' is correct.
Let $$\displaystyle f:\left \{ x,y,z \right \}\rightarrow \left \{ a,b,c \right \}$$ be a one-one function and only one of the conditions $$(i)f(x)\neq b, (ii)f(y)=b,(iii)f(z)\neq a$$ is true then the function $$f$$ is given by the set
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$$\displaystyle \left \{ (x,a),(y,b),(z,c)\right \}$$
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$$\displaystyle \left \{ (x,a),(y,c),(z,b)\right \}$$
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$$\displaystyle \left \{ (x,b),(y,a),(z,c)\right \}$$
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$$\displaystyle \left \{ (x,c),(y,b),(z,a)\right \}$$
Explanation
$$f:\left\{ x,y,z \right\} \rightarrow \left\{ a,b,c \right\} $$ is a one-one function
$$ \Rightarrow$$ each element in $$ \left\{ x,y,z \right\} $$ will have exactly one image in $$\left\{ a,b,c \right\}$$
and no two elements of $$ \left\{ x,y,z \right\}$$ will have same image in $$\left\{ a,b,c \right\} $$
Coming to the given 3 conditions, only one is true.
1) if $$f\left( x \right) \neq b$$ is true then $$f\left( y \right) =b$$ is false which makes $$f\left( z \right) \neq a$$ true $$\Longrightarrow f\left( x \right) \neq b$$ is false.
2) if $$f\left( y \right) =b$$ is true then $$f\left( x \right) \neq b$$ will also be true $$\Longrightarrow f\left( y \right) =b$$ is false
$$ \therefore f\left( z \right) \neq a$$ is the true condition and remainig two are false conditions.
$$\therefore f\left( x \right) =b, f\left( y \right) =a, f\left( z \right) =c$$
hence $$f=\left\{ \left( x,b \right) ,\left( y,a \right) ,\left( z,c \right) \right\} $$
Let $$ f:R \rightarrow R$$ and $$g:R \rightarrow R$$ be defined by $$f(x)=x^2+2x-3,g(x)=3x-4$$ then $$(gof) (x)=$$
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$$3x^2+6x-13$$
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$$3x^2-6x-13$$
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$$3x^2+6x+13$$
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$$-3x^2+6x-13$$
Explanation
Given that, $$f(x)=x^2+2x-3$$ and $$g(x)=3x-4$$
$$(gof) (x)=g\left(f(x)\right)$$
$$=3(x^2+2x-3)-4$$
$$(gof) (x)=3x^2+6x-13$$
Hence, option A.
The domain of the function $$\displaystyle f(x)=\log_{10}\log_{10}(1+x^{3})$$ is
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$$(-1,+\infty)$$
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$$(0,+\infty)$$
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$$[0,+\infty)$$
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$$(-1,0)$$
Explanation
Given,
$$\displaystyle f(x)=\log_{10}\log_{10}(1+x^{3})$$
The value inside a log function must not be negative.
Hence
$$\log_{10}(1+x^3)>0$$
$$1+x^{3}>10^{0}$$ ....(by taking anti-log).
$$x^{3}>0$$
$$x>0$$
Hence the domain for $$f(x)$$ is $$x>0$$.
If $$f$$ and $$g$$ are two functions such that $$\displaystyle \left ( fg \right )\left ( x \right )=\left ( gf \right )\left ( x \right )$$ for all $$x$$. Then $$f $$ and $$g$$ may be defined as
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$$\displaystyle f\left ( x \right )=\sqrt{x}, g\left ( x \right )=\cos x$$
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$$\displaystyle f\left ( x \right )=x^{3}, g\left ( x \right )=x+1$$
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$$\displaystyle f\left ( x \right )=x-1, g\left ( x \right )=x^{2}+1$$
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$$\displaystyle f\left ( x \right )=x^{m}, g\left ( x \right )=x^{n}$$ where $$m, n$$ are unequal integers
Explanation
A$$)f(x)=$$$$\sqrt{x},g(x)=cosx then (fg)(x) = \sqrt{\cos x} and (gf)(x) = \cos \sqrt{x}$$.They are not equal.
B)$$f(x)=x^{3},g(x)=x+1 then (fg)(x) = (x+1)^{3} and (gf)(x)= x^{3}+1$$.They are not equal.
C)$$f(x)=x1,g(x)=x^{2}+1 then (fg)(x)=x^{2} and (gf)(x)=x^{2}-2x$$. They are not equal.
D)$$f(x)=x^{m}, g(x)=x^{n} then (fg)(x)=x^{n+m} and (gf)(x)=x^{n+m}$$.Hence $$(fg)(x)= (gf)(x)$$
If $$\displaystyle f(x)=x^{n},n\in N$$ and $$(gof)(x)=ng(x)$$ then $$g(x)$$ can be
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$$n\:|x|$$
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$$3.\sqrt[3]{x}$$
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$$e^{x}$$
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$$\log\:|x|$$
Explanation
$$f(x)=x^{n}$$
$$g(f(x))=ng(x)$$ ...$$(i)$$
$$\log(f(x))=n\log(x)$$ ...$$(ii)$$
Taking $$\log(|x|)$$ as $$g(x)$$ the above expression is reduced to eq $$i$$.
Hence
$$g(x)=\log(|x|)$$.
The domain of $$\displaystyle f(x)=\sqrt { \log_{ x^{ 2 }-1 }(x) } $$ is
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$$(\sqrt{2},+\infty)$$
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$$(0,+\infty)$$
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$$(1,+\infty)$$
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none of these
Explanation
$$\log_{x^{2}-1}(x)\geq 0$$
Hence,
$$x\geq 1 \quad \dots (1)$$
Now, $$x^{2}-1>1$$
$$x^{2}>2$$
$$x<-\sqrt{2}$$ and $$x>\sqrt{2} \quad \dots (2)$$
Hence from $$(1)$$ and $$(2)$$,
$$x\in(\sqrt{2},\infty)$$
The composite mapping $$fog$$ of the map $$f: R\rightarrow R,f(x)=\sin x$$ and $$g: R\rightarrow R, g(x)=x^2$$ is
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$$x^2 \sin x$$
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$$(\sin x)^2$$
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$$\sin x^2$$
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$$\dfrac{ \sin x}{x^2}$$
Explanation
Composite mapping will be $$f(g(x))$$
$$=f(x^2)$$
$$=\sin(x^2)$$
Hence option $$'C'$$ is the answer.
If $$\displaystyle f\left ( x \right )=\left\{\begin{matrix}
x^{2} x \geq 0\\
x x < 0
\end{matrix}\right.$$
then $$\displaystyle (f o f)(x)$$ is given by
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$$x^{2}$$ for $$x\geq 0$$ and $$x$$ for $$ x< 0$$
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$$\displaystyle x^{4}$$ for $$\displaystyle x\geq 0$$ and $$x^{2}$$ for $$x< 0$$
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$$ \displaystyle x^{4}$$ for $$ \displaystyle x\geq 0$$ and $$-x^{2} $$ for $$x < 0$$
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$$\displaystyle x^{4}$$ for $$ x\geq 0$$ and $$x $$ for $$ x< 0$$
Explanation
For $$x\ge0$$,we have $$\displaystyle f \circ f\left( x \right)= {\left( {{x^2}} \right)^2}=\left( {{x^4}} \right)$$
For $$x<0$$,we have $$\displaystyle f \circ f\left( x \right)=x$$
Let $$\displaystyle g(x)=1+x-[x]$$ and $$\displaystyle f(x)=\left\{\begin{matrix}{-1}\quad {x< 0} \\ {0} \quad {x=0}\\{1} \quad {x> 0} \end{matrix}\right.$$ Then for all $$\displaystyle x, f\left \{ g\left ( x \right ) \right \}$$ is equal to
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$$x$$
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$$1$$
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$$\displaystyle f(x)$$
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$$\displaystyle g(x)$$
Explanation
Let $$g(x) = 1+x-[x] = 1+\{x\}> 0$$
since $$\{x\}\in [0,1) \forall x\in R$$
Hence $$f\{g(x)\} = 1$$
Let $$\displaystyle f(x)=\frac{ax}{x+1}$$, where $$\displaystyle x\neq -1$$. Then for what value of $$\displaystyle a$$ is $$\displaystyle f( f(x))=x$$ always true
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$$\displaystyle \sqrt{2}$$
0%
$$\displaystyle -\sqrt{2}$$
0%
$$1$$
0%
$$-1$$
Explanation
$$\displaystyle f\left( {f\left( x \right)} \right) = \dfrac{{a\dfrac{{ax}}{{x + 1}}}}{{\dfrac{{ax}}{{x + 1}} + 1}} = \dfrac{{\dfrac{{{a^2}x}}{{x + 1}}}}{{\dfrac{{ax + x + 1}}{{x + 1}}}} = \dfrac{{{a^2}x}}{{ax + x + 1}}$$
Since, $$\displaystyle f\left( {f\left( x \right)} \right) =x$$, we have,
$$\displaystyle\dfrac{{{a^2}x}}{{ax + x + 1}}=x$$.
Simplifying the equation we get,
$${a^2}x = \left( {a + 1} \right){x^2} + x$$
$$\therefore \left( {a + 1} \right){x^2} + \left( {1 - {a^2}} \right)x = 0$$
or $$\left( {a + 1} \right)x\left( {x + 1 - a} \right) = 0$$
Hence the only possible value is $$a=-1$$
If $$\displaystyle f(y)=\frac{y}{\sqrt{1-y^2}}$$; $$\displaystyle g(y)=\frac{y}{\sqrt{1+y^2}}$$ then $$(fog)y$$ is equal to
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0%
$$\displaystyle \frac{y}{\sqrt{1-y^2}}$$
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$$\displaystyle \frac{y}{\sqrt{1+y^2}}$$
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$$y$$
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$$2f(x)$$
Explanation
Given $$\displaystyle f(y)=\frac{y}{\sqrt{1-y^2}}$$ and $$\displaystyle g(y)=\frac{y}{\sqrt{1+y^2}}$$
$$\therefore \displaystyle (fog)y = f(g(y))=\dfrac{\dfrac{y}{\sqrt{1+y^2}}}{\sqrt{1-\dfrac{y^2}{1+y^2}}}=y$$
If $$\displaystyle f(x)= (x-1)+(x+1)$$ and
$$\displaystyle g(x)= f\left \{ f(x) \right \}$$ then $$\displaystyle {g}'(3)$$
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equals $$1$$
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equals $$0$$
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equals $$3$$
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equals $$4$$
Explanation
Simplifying,$$f(x)$$ we get
$$f(x)=2x$$
Hence
$$f(f(x))=2(f(x))$$
$$=2(2x)$$
$$=4x$$
Hence
$$g(x)=f(f(x))$$
$$=4x$$
Thus
$$g'(x)=4$$
Hence $$g'(3)=4$$ .
Set $$A$$ has $$3$$ elements and set $$B$$ has $$4$$ elements. The number of injections that can be defined from $$A$$ to $$B$$ is
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0%
$$144$$
0%
$$12$$
0%
$$24$$
0%
$$64$$
Explanation
The total number of injective mappings from a set with $$m$$ elements to a set with $$n$$ elements,$$\displaystyle m\leq n,$$ is
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$$\displaystyle m^{n}$$
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$$\displaystyle n^{m}$$
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$$\displaystyle \frac{n!}{\left ( n-m \right )!}$$
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$$\displaystyle n!$$
Explanation
Let $$A=\{a_1,a_2, a_3.....a_m\}$$
and $$B=\{b_1,b_2, b_3.....b_n\}$$ where $$m \le n$$
Given $$f:A\rightarrow B$$ be an injective mapping.
So, for $$a_1 \in A$$, there are $$n$$ possible choices for $$f(a_1)\in B$$.
For $$a_2 \in A$$, there are $$(n-1)$$ possible choices for $$f(a_2)\in B$$.
Similarly for $$a_m \in A$$, there are $$(n-m-1)$$ choices for $$f(a_m)\in B$$
So, there are $$n(n-1)(n-2).....(n-m-1)=\dfrac{n!}{(n-m)!}$$ injective mapping from $$A$$ to $$B.$$
Let f(x)=tan x, x$$\displaystyle \epsilon \left [ -\frac{\pi }{2},\frac{\pi }{2} \right ]$$ and $$\displaystyle g\left (x \right )=\sqrt{1-x^{2}}$$ Determine $$g o f(1)$$.
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0%
1
0%
0
0%
-1
0%
not defined
Find $$\displaystyle \phi \left [ \Psi \left ( x \right ) \right ]$$ and $$\displaystyle \Psi \left [ \phi \left ( x \right ) \right ]$$ if $$\displaystyle \phi \left ( x \right )=x^{2}+1$$ and $$\displaystyle \Psi \left ( x \right )=3^{x}.$$
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$$\displaystyle \Psi \left [ \phi \left ( x \right ) \right ]=3^{x^{2}+1}.$$
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$$\displaystyle \phi \left [ \Psi \left [ x \right ] \right ]=3^{2x}+1$$
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$$\displaystyle \Psi \left [ \phi \left ( x \right ) \right ]=3^{x^{3}+1}.$$
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$$\displaystyle \phi \left [ \Psi \left [ x \right ] \right ]=3^{x}+1$$
Explanation
$$\displaystyle \phi \left [ \Psi \left [ x \right ] \right ]=\phi \left
( 3^{x} \right )=\left ( 3^{x} \right )^{2}+1=3^{2x}+1$$
and
$$\displaystyle \Psi \left [ \phi \left ( x \right ) \right ]=\Psi \left
( x^{2} +1\right )=3^{x^{2}+1}.$$
If $$\displaystyle f\left ( x \right )=\frac{ax+b}{cx+d}$$ and $$\displaystyle \left ( fof \right )x=x,$$ then d=?
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0%
$$a$$
0%
$$-a$$
0%
$$b$$
0%
$$-b$$
Explanation
$$\displaystyle \left ( fof \right )x=x\Rightarrow f\left [ f\left ( x \right ) \right ]=x$$ or $$\displaystyle f\left [ \frac{ax+b}{cx+d} \right ]=x$$ or $$\displaystyle\frac{a\left [ \frac{ax+b}{cx+d}\right ]+b}{c\left [\frac{ax+b}{cx+d} \right ]+d}=x$$ $$\displaystyle \therefore \frac{x\left ( a^{2}+bc \right )+b\left ( a+d \right )}{cx\left ( a+d \right )+\left ( bc+d^{2} \right )}=x$$ Clearly if $$\displaystyle a+d=0$$ or $$\displaystyle d=-a$$ then in that case $$\displaystyle L.H.S.=\frac{x\left ( a^{2}+bc \right )+0}{0+\left ( bc+a^{2} \right )}=x$$ $$\displaystyle \because d=-a$$
Given $$\displaystyle f\left ( x \right )=\log \left ( \frac{1+x}{1-x} \right )$$ and $$\displaystyle g\left ( x \right )=\frac{3x+x^{3}}{1+3x^{2}}, fog (x)$$ equals
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0%
$$-f(x)$$
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$$3f(x)$$
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$$\displaystyle \left [ f\left ( x \right ) \right ]^{3}$$
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none of these
Explanation
Given $$f(x)\, =\, \log\, \left( \displaystyle \dfrac{1\, +\, x}{1\, -\,
x} \right)$$ and $$g(x)\, =\, \displaystyle \dfrac{3x\, +\, x^{3}}{1\,
+\, 3x^{2}}$$
$$\therefore fog (x)\, =\, \log \left( \displaystyle \dfrac{1\, +\, \dfrac{3x\, +\,
x^{3}}{1\, +\, 3x^{2}}}{1\, -\,\displaystyle \dfrac{3x\, +\, x^{3}}{1\,
+\, 3x^{2}}} \right)$$
$$=\, \log\, \left(\displaystyle \dfrac{(1\, +\,
x)^{3}}{(1\, -\, x)^{3}} \right)\, =\, 3\, \log\, \left(\displaystyle
\dfrac{1\, +\, x}{1\, -\, x} \right)\, =\, 3\, f(x)$$
Let $$f(x)=x^{2}-2x$$ and $$g(x)=f(f(x)-1)+f(5-f(x)),$$ then
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$$g(x)<0,\forall x\in R$$
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$$g(x)<0$$ for some $$x\in R$$
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$$g(x)\leq 0$$ for some $$x\in R$$
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$$g(x)\geq 0,\forall x\in R$$
Explanation
Given, $$ f(x) = x^2 - 2x $$
Now, $$ f(f(x) - 1) \\ = f(x^2 - 2x - 1 ) \\ = (x^2 - 2x - 1 )^2 - 2 (x^2 - 2x - 1 ) $$
And,
$$ f(5 - f(x) ) \\ = f(5 - x^2 + 2x ) \\ = (5 - x^2 + 2x )^2 - 2 (5 - x^2 + 2x ) $$
Hence, $$ g(x) = f(f(x) - 1) + f (5 - f(x)) \\ \therefore g(x) = (x^2 - 2x - 1)^2 - 2 (x^2 - 2x -1 ) + (5 - x^2 + 2x )^2 - 2(5 - x^2 + 2x ) \\ \Rightarrow g(x) = (x^2 - 2x - 1)^2 + (5- x^2 + 2x)^2 - 2(x^2 - 2x - 1 + 5 - x^2 + 2x) \\ \therefore g(x) = (x^2 + 2x - 1 )^2 + ( 5 - 2x ^2 + 2 x^2)^2 - 8$$
Since the first two terms are in square,
$$\therefore $$ it can not be negative and if x = 9 then we also get positive value.
$$\therefore g(x) \geq 0 \, \, \forall x \in R $$
$$ \therefore $$ option D is
correct
If $$g(x)=1+\sqrt { x } $$ and $$f(g(x))=3+2\sqrt { x } +x$$, then $$f(x)=$$
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$$1+2{ x }^{ 2 }$$
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$$2+{ x }^{ 2 }$$
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$$1+x$$
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$$2+x$$
Explanation
We have $$g\left( x \right)=1+\sqrt { x } $$ and $$f(g(x))=3+2\sqrt { x } +x$$ ...(1)
Also, $$f\left( g\left( x \right) \right) =f\left( 1+\sqrt { x } \right) $$ ...(2)
From (1) and (2), we get
$$f\left( 1+\sqrt { x } \right) =3+2\sqrt { x } +x$$
Let $$1+\sqrt { x } =y\Rightarrow x={ \left( y-1 \right) }^{ 2 }$$
$$\therefore f\left( y \right) =3+2\left( y-1 \right) +{ \left( y-1 \right) }^{ 2 }\\ =3+3y-2+{ y }^{ 2 }-2y+1=2+{ y }^{ 2 }\\ \therefore f\left( x \right) =2+{ x }^{ 2 }$$
Are the following sets of ordered pairs functions? If so, examine whether the mapping is surjective or injective :
{(x, y): x is a person, y is the mother of x}
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injective (one- one ) and surjective (into)
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injective (one- one ) and not surjective (into)
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not injective (one- one ) and surjective (into)
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not injective (one- one ) and not surjective (into)
Explanation
We have {(x, y) : x is a person, y is the mother of x}. Clearly each person 'x' has only one biological mother. So above set of ordered pair is a function. Now more than one person may have same mother. So function is many-one and surjective.
If $$f:R\rightarrow R$$ and $$g:R\rightarrow R$$ are functions defined by $$f(x)=3x-1; g(x)=\sqrt{x+6}$$, then the value of $$(g\circ f^{-1})(2009)$$ is
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$$26$$
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$$29$$
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$$16$$
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$$15$$
Explanation
Given $$f(x)=3x-1, g(x)=\sqrt{x+6}$$
Let $$f(x)=y$$
$$ \Rightarrow y=3x-1\Rightarrow y+1-3x\Rightarrow x=\cfrac{y+1}{3}$$
Now, $$ f^{-1}(y)=x=\cfrac{y+1}{3}$$
$$\Rightarrow f^{-1}(x)=\cfrac{x+1}{3}$$ and $$g(x)=\sqrt {x+6}$$
Consider, $$(g\circ f^{-1})(2009)=g[{f^{-1}(2009)}]$$
$$ =g\left(\cfrac{2009+1}{3}\right)=g\left(\cfrac{2010}{3}\right)$$
$$=\sqrt{\cfrac{2010}{3}+6}=\sqrt{\cfrac{2028}{3}}=\sqrt{676}=26$$
Hence, option $$A$$ is correct.
Let f : {x,y,z} $$\rightarrow$$ {a,b,c} be a one-one function. It is known that only one of the following statment is true, and only one such function exists :
find the function f (as ordered pair).
(i) f(x) $$\neq$$ b
(i) f(y) = b
(ii) f(z) $$\neq$$ a
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{(x,b), (y,a), (z,c)}
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{(x,a), (y,b), (z,c)}
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{(x,b), (y,c), (z,a)}
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{(x,c), (y,a), (z,b)}
Explanation
When (i) is true, then $$f(x) \neq b, f(y) \neq b , f(z) = a,b,c $$
$$\Rightarrow$$ Two ordered pair function is possible $$ f(x) = a, f(y) = c, f(z) = b$$ or $$f(x) = c, f(y) = a, f(z) = b$$
But given $$f$$ is one-one and only one such function is possible. Hence (i) can't be true.
When (ii) is true, then $$f(y) = b, f(z) =c , f(x) = a\Rightarrow f(x) = a, f(y) = b, f(z) = c$$
Clearly if (ii) is true then it is satisfying every condition.
Hence ordered pair of $$f$$ is $$\{(x,a), (y,b), (z,c)\}$$
If $$f_{0}(x)\, =\, \dfrac{x}{(x\, +\, 1)}$$ and $$f_{n\, +\, 1}\, =\, f_{0}\circ f_{n}(x)$$ for $$n = 0, 1, 2,\cdots$$ then $$f_{n}(x)$$ is
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$$\displaystyle \frac{x}{(n\, +\, 1) x\, +\, 1}$$
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$$f_{0}(x)$$
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$$\displaystyle \frac{nx}{nx\, +\, 1}$$
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$$\displaystyle \frac{x}{nx\, +\, 1}$$
Explanation
Given $$f_{0}\left(x\right)=\dfrac{x}{ \left(x\, +\, 1\right)}\cdots\cdots\left(1\right)$$
Also given $$f_{n\, +\, 1}\, =\, f_{0}\, o\, f_{n}$$ for $$(n = 0, 1, 2,\cdots)\ldots\left(2\right)$$
Put $$n=0$$ in eqn (2)
$$f_1=f_{0} o f_0$$
$$f_1\left(x\right)=f_{0}[f_{0}\left(x\right)]$$
$$\Rightarrow f_{1}\left(x\right)=f_{0}\left(\dfrac{x}{x+1}\right)$$
$$=\dfrac{\dfrac{x}{x+1}}{\dfrac{x}{x+1}+1}$$
$$=\dfrac{x}{2x+1}$$
$$\Rightarrow f_1\left(x\right)=\dfrac{x}{2x+1}$$
Put $$n=1$$ in eqn $$(2)$$
$$f_2=f_{0} o f_1$$
$$f_2\left(x\right)=f_{0}[f_{1}\left(x\right)]$$
$$\Rightarrow f_{2}\left(x\right)=f_{0}\left(\dfrac{x}{2x+1}\right)$$
$$=\dfrac{\dfrac{x}{2x+1}}{\dfrac{x}{2x+1}+1}$$
$$=\dfrac{x}{3x+1}$$
$$\Rightarrow f_2\left(x\right)=\dfrac{x}{3x+1}$$
Hence, $$f_n\left(x\right)=\dfrac{x}{\left(n+1\right)x+1}$$
Suppose f and g both are linear function with $$\displaystyle f(x)=-2x+1$$ and $$\displaystyle f \left ( g\left ( x \right ) \right )=6x-7$$ then slope of line $$y=g(x)$$ is
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$$3$$
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$$-3$$
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$$6$$
0%
$$-2$$
Explanation
Given, $$f(x)=-2x+1$$
$$\therefore f(g(x))=-2g(x)+1=6x-7$$
$$\Rightarrow g(x)=-3x+4$$
$$\Rightarrow g'(x)=-3$$
Hence slope of $$g(x)$$ is $$-3$$
If the function $$f : R \rightarrow$$ A given by $$f(x)\, =\, \displaystyle \frac{x^{2}}{x^{2}\, +\, 1}$$ is a surjection, then A is
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$$R$$
0%
$$[0, 1]$$
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$$(0, 1]$$
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$$[0, 1)$$
Explanation
As 'f' is surjective,
Range of
f
= co-domain of '
f
'
$$\implies$$ A = range of 'f'
Since, $$f(x)=\dfrac{x^2}{x^2+1}$$
$$\implies$$
$$y=\dfrac{x^2}{x^2+1}$$
$$\implies$$ $$y(x^2+1)=x^2$$
$$\implies$$ $$(y-1)x^2+y=0$$
$$\implies$$ $$x^2=\dfrac{-y}{y-1}$$
$$\implies$$ $$x=\sqrt{\dfrac{y}{1-y}}$$
$$\implies$$ $$\dfrac{y}{1-y} \geq 0$$
$$\implies$$ $$y \epsilon [0,1)$$
$$\implies$$ $$A=[0,1)$$
If $$f(x)=\begin{cases} 2x+3\quad \quad x\le 1 \\ a^{ 2 }x+1\quad x>1 \end{cases}$$, then the values of $$a$$ for which $$f(x)$$ is injective.
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$$-3$$
0%
$$1$$
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$$0$$
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none of these
Explanation
A function is called injective (one-one) if it is monontonic.
Clearly for $$x<1$$ f is increasing and it's maximum values is $$2(1)+3=5$$
Hence for f to be monotonic $$(x>1)$$, $$f(1) \geq 5$$
$$\Rightarrow a^2(1)+1\geq 5 \Rightarrow a^2\geq 2 \Rightarrow a\in R-(-2,2)$$
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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