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CBSE Questions for Class 12 Commerce Maths Relations And Functions Quiz 4 - MCQExams.com
CBSE
Class 12 Commerce Maths
Relations And Functions
Quiz 4
If $$f:\left( 3,6 \right) \rightarrow \left( 1,3 \right) $$ is a function defined by $$\displaystyle f\left( x \right)=x-\left[ \frac { x }{ 3 } \right] $$ $$($$ where $$\left[ . \right] $$ denotes the greatest integer function $$),$$ then $$\displaystyle f^{ -1 }\left( x \right)=$$
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$$x-1$$
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$$x+1$$
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$$x$$
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none of these
Explanation
Given that $$\displaystyle f\left( x \right) =x-\left[ \frac { x }{ 3 } \right] $$
where $$\displaystyle f:\left( 3,6 \right) \rightarrow \left( 1,3 \right) $$
Now, inside the given domain, we will always have $$\displaystyle \left[ \frac { x }{ 3 } \right] =1$$
$$\displaystyle \therefore f\left( x \right) =x-1$$ throughout its domain
$$\displaystyle \Rightarrow y=x-1$$
$$\displaystyle \Rightarrow x=y+1={ f }^{ -1 }\left( y \right) $$
$$\displaystyle \therefore { f }^{ -1 }\left( x \right) =x+1$$
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 $$f\{g(x)\} = 1$$
The inverse of the function $$\displaystyle f(x) = \log_{2}(x+\sqrt{x^{2}+1}) $$ is
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$$\displaystyle 2^{x}+2^{-x} $$
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$$\displaystyle \frac{2^{x}+2^{-x}}{2}$$
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$$\displaystyle \frac{2^{-x}-2^{x}}{2}$$
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$$\displaystyle \frac{2^{x}-2^{-x}}{2}$$
Explanation
We have to find the inverse of $$f(x)$$ which means $$f$$ is one one and onto function
Let $$\displaystyle f\left ( x \right )= y= \log _{2}\left (
x+\sqrt{x^{2}+1} \right )$$
$$\displaystyle \therefore 2^{y}=
x+\sqrt{x^{2}+1}$$ ...(i)
$$\displaystyle 2^{-y}=
\frac{1}{x+\sqrt{x^{2}+1}}= \frac{\sqrt{x^{2}+1}-x}{1}$$ ...(ii)
By subtracting (i) and (ii) we have
$$ \therefore 2^{y}-2^{-y}=
2x$$
or $$\displaystyle x= \frac{1}{2}\left ( 2^{y}-2^{-y} \right )$$
$$\displaystyle
\therefore f^{-1}\left ( y \right )= \frac{1}{2}\left ( 2^{y}-2^{-y}
\right )$$
Replace $$\displaystyle y\rightarrow x$$
$$\displaystyle \therefore f^{-1}\left ( x \right )= \frac{1}{2}\left ( 2^{x}-2^{-x} \right )$$
If $$f : \{1,2,3,...\} \rightarrow \{0, \pm 1, \pm 2,...\}$$ is defined by
$$\displaystyle y=f(x)=\begin{cases} \displaystyle \frac { x }{ 2 } \quad \quad \text{ if x is even } \\ -\displaystyle \frac { (x-1) }{ 2 } \quad ,\text{ if x is odd } \end{cases}$$, then $$f^{-1}(100)$$ is
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Function is not invertible as it is not onto
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$$199$$
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$$201$$
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$$200$$
Explanation
Since $$100$$ is even $$f^{-1}(x)=2x$$
Hence $$f^{-1}(100)=200$$
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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$$\displaystyle \frac{y}{\sqrt{1-y^2}}$$
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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 $$g'(3)=4$$ .
If $$\displaystyle f\left ( x \right )=x+\tan x$$ and $$\displaystyle g^{-1}=f$$ then $$\displaystyle g{}'\left ( x \right )$$ equals
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$$\displaystyle \frac{1}{2+\left [ g\left ( x \right )+x \right ]^{2}}$$
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$$\displaystyle \frac{1}{1+\left [ g\left ( x \right )-x \right ]^{2}}$$
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$$\displaystyle \frac{1}{2+\left [ g\left ( x \right )-x \right ]^{2}}$$
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$$\displaystyle \frac{1}{2-\left [ g\left ( x \right )-x \right ]^{2}}$$
Explanation
$$f\left( x \right) =x+\tan { x } \dashrightarrow \left( i \right) \\ { g }^{ -1 }(x)=f(x)\dashrightarrow \left( ii \right) \\ f\left( g(x) \right) =g(x)+\tan { g(x) } \quad \left( from\quad equation\quad \left( i \right) \quad \right) \\ { g }^{ -1 }\left( g(x) \right) =g(x)+\tan { g(x) } \left( from\quad equation\quad \left( ii \right) \quad \right) \\ x=g(x)+\tan { g(x) } \quad \dashrightarrow \left( iii \right) \quad \left( { \because \quad g }^{ -1 }\left( g(x) \right) =x \right) \\ $$
Differentiating the above equation w.r.t x,
$$\\ 1=g^{ ' }\left( x \right) +(\sec ^{ 2 }{ g(x) } )g^{ ' }\left( x \right) \\ g^{ ' }\left( x \right) =\dfrac { 1 }{ 1+\sec ^{ 2 }{ g(x) } } =\dfrac { 1 }{ 2+\tan ^{ 2 }{ g\left( x \right) } } \quad \left( \because \sec ^{ 2 }{ g(x) } =1+\tan ^{ 2 }{ g\left( x \right) } \quad \right) \\ from\quad eqn\quad \left( iii \right) ,\quad \tan { g(x) } =x-g\left( x \right) \\ \therefore g^{ ' }\left( x \right) =\dfrac { 1 }{ 2+{ (x-g\left( x \right) ) }^{ 2 } } $$
Let $$f:[4,\infty )\rightarrow [4,\infty )$$ be a function defined by $$f\left( x \right)={ 5 }^{ x\left( x-4 \right) }$$, then $$f^{ -1 }\left( x \right)$$ is
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$$2-\sqrt { 4+\log _{ 5 }{ x } } $$
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$$2+\sqrt { 4+\log _{ 5 }{ x } } $$
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$$\displaystyle { \left( \frac { 1 }{ 5 } \right) }^{ x\left( x-4 \right) }$$
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None of these
Explanation
Let $$y={ 5 }^{ x\left( x-4 \right) }\Rightarrow x\left( x-4 \right) =\log _{ 5 }{ y } \Rightarrow { x }^{ 2 }-4x-\log _{ 5 }{ y } =0$$
$$\displaystyle \Rightarrow x=\frac { 4\pm \sqrt { 16+4\log _{ 5 }{ y } } }{ 2 } =\left( 2\pm \sqrt { 4+\log _{ 5 }{ y } } \right) $$
But $$x\ge 4$$, so $$x=\left( 2+\sqrt { 4+\log _{ 5 }{ y } } \right) $$
$$\therefore f^{ -1 }\left( y \right) =2+\sqrt { 4+\log _{ 5 }{ y } } $$
$$\therefore f^{ -1 }\left( x \right) =2+\sqrt { 4+\log _{ 5 }{ x } } $$
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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
For Onto function
codomain $$\displaystyle =\left [ 0, \frac{\pi}{2} \right )=$$ Range
Which is possiable where $$\displaystyle x^{2}+x+a\geq 0$$
Now using fact $$ \displaystyle f(x)\geq 0 \Rightarrow D\leq 0$$
$$\displaystyle \left ( A=1=b \right )\ and \ C=a$$
$$\displaystyle \Rightarrow 1-4a\leq 0$$
$$\displaystyle \Rightarrow a\geq \frac{1}{4}$$
$$\displaystyle \therefore $$ Assertion is false, Reason is true
Let $$\displaystyle f:N\rightarrow Y$$ be a function defined as $$f(x)=4x+3$$ where $$\displaystyle Y=\left \{ y \in N:y=4x+3 \right \}$$ for some $$\displaystyle x\in N$$ such that $$f$$ is invertible then its inverse is
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$$\displaystyle g\left ( y \right )=4+\frac{y+3}{4}$$
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$$\displaystyle g\left ( y \right )=\frac{2y-3}{4}$$
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$$\displaystyle g\left ( y \right )=\frac{3y+4}{3}$$
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$$\displaystyle g\left ( y \right )=\frac{y-3}{4}$$
Explanation
Let $$\displaystyle y=f(x)\Rightarrow x= f^{-1}(y)$$ ........(A)
$$\displaystyle \therefore y=4x+3$$
$$\displaystyle \Rightarrow 4x=y-3 \Rightarrow x=\frac{y-3}{4}$$
$$\displaystyle \Rightarrow f^{-1}(y)=\frac{y-3}{4}$$ .......(using A)
$$\displaystyle \Rightarrow g(y)=\frac{y-3}{4}$$
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 $$\displaystyle f \circ f\left( x \right)=x$$
If $$\displaystyle f(x)= \frac{3x+2}{5x-3}$$ then
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$$\displaystyle f^{-1}(x)= -f(x)$$
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$$\displaystyle f^{-1}(x)= f(x)$$
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$$\displaystyle fo(f(x))= -x $$
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$$\displaystyle f^{-1}(x)= -\frac{1}{19}f(x)$$
Explanation
Let $$y=\displaystyle \frac{3x+2}{5x-3}$$
$$\Rightarrow \displaystyle x=\frac{3y+2}{5y-3}=f^{-1}(y)$$
$$\displaystyle \therefore f^{-1}(x)=\frac{3x+2}{5x-3}=f(x)$$
Let $$\displaystyle f:R \rightarrow R $$ be defined as $$\displaystyle f(x)= x^{2}+5x+9$$ then $$\displaystyle f^{-1}(8) $$ equals to
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$$\displaystyle \left \{ \frac{-5+\sqrt{20}}{2},\frac{-5-\sqrt{21}}{2} \right \}$$
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$$\displaystyle \left \{ \frac{-5+\sqrt{21}}{2} ,\frac{-5-\sqrt{21}}{2}\right \}$$
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$$\displaystyle \left \{ \frac{5-\sqrt{21}}{2} ,\frac{21-\sqrt{5}}{2}\right \}$$
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Does not exist.
Explanation
Let $$\displaystyle f^{-1}(8)= x, $$ then $$\displaystyle f(x)= 8$$
$$\implies x^{2}+5x+9= 8 $$
$$\implies x^{2}+5x+1= 0 $$
$$\displaystyle
\Rightarrow x= \frac{-5\pm \sqrt{25-4}}{2} $$
$$\displaystyle x=
\frac{-5+\sqrt{21}}{2},\frac{-5-\sqrt{21}}{2}\in R $$
Hence
$$\displaystyle f^{-1}(8)= \left \{ \frac{-5+\sqrt{21}}{2}
,\frac{-5-\sqrt{21}}{2}\right \}$$
Which of the following functions have inverse defined on the ranges
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$$\displaystyle f\left ( x \right )=x^{2}, x $$ $$\displaystyle \in $$ R
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$$\displaystyle f\left ( x \right )=x^{3},x$$ $$\displaystyle \in $$ R
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$$\displaystyle f\left ( x \right )=e^{x},x$$ $$\displaystyle \in $$ R
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$$\displaystyle f(x)=\sin x,$$ $$\displaystyle 0< x< 2\pi$$
Explanation
We know that only one-one onto functions are invertible.
In (A) it is not one-one and hence not invertible.
In (B) and (C)both are one-one onto and both are invertible.
In (D) Since$$\displaystyle f\left ( \frac{\pi}{4}
\right )=f\left ( \frac{3\pi }{4} \right )=\frac{1}{\sqrt{2}}$$
$$\displaystyle \therefore $$ (D) is not invertible.
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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1
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0
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-1
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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
$$\displaystyle \Psi \left [ \phi \left ( x \right ) \right ]=\Psi \left
( x^{2} +1\right )=3^{x^{2}+1}.$$
The inverse of the function$$\displaystyle f(x)=(1-(x-5)^{3})^{1/5} $$is
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$$5-(1-x^{5})^{1/3}$$
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$$5+(1-x^{5})^{1/3}$$
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$$5+(1+x^{5})^{1/3}$$
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$$5-(1+x^{5})^{1/3}$$
Explanation
Let $$y=(1-(x-5)^{3})^{\frac{1}{5}}$$
$$y^{5}=1-(x-5)^{3}$$
$$(x-5)^{3}=1-y^{5}$$
$$x-5=(1-y^{5})^{\frac{1}{3}}$$
$$x=5+(1-y^{5})^{\frac{1}{3}}$$
Replacing $$x$$ with $$y$$ gives us
$$f^{-1}(x)=5+(1-x^{5})^{\frac{1}{3}}$$
If $$\displaystyle f\left ( x \right )=\frac{ax+b}{cx+d}$$ and $$\displaystyle \left ( fof \right )x=x,$$ then d=?
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$$a$$
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$$-a$$
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$$b$$
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$$-b$$
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 $$\displaystyle L.H.S.=\frac{x\left ( a^{2}+bc \right )+0}{0+\left ( bc+a^{2} \right )}=x$$ $$\displaystyle \because d=-a$$
The inverse of the function $$\log_{e}x$$ is
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$$10^{x}$$.
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$$e ^{x}$$
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$$10^{e}$$.
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$$x^{e}$$.
Explanation
$$y=f^{ -1 }\left( x \right) $$
$$ \Rightarrow x=f\left( y \right) =\log _{ e }{ y } $$
$$\Rightarrow y={ e }^{ x }$$
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 $$B.$$
If $$\displaystyle A= \left \{ a,b,c,d \right \}, B= \left \{ 1,2,3 \right \}$$ find whether or not the following sets of ordered pairs are relations from $$A$$ to $$B$$ or not.
$$\displaystyle R_{1}= \left \{ \left ( a,1 \right ), \left ( a,3 \right ) \right \}$$
$$\displaystyle R_{2}= \left \{ \left ( a,1 \right ), \left ( c,2 \right ), \left ( d,1 \right ) \right \}$$
$$\displaystyle R_{3}= \left \{ \left ( a,1 \right ), \left ( b,2 \right ), \left ( 3,c \right ) \right \}.$$
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$$R_{1}$$ $$R_{2}$$ are relations but $$R_{3}$$ is not a relation.
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$$R_{1}$$ $$R_{3}$$ are relations but $$R_{2}$$ is not a relation.
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All are relations
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none of these
Explanation
Given, $$ A= \left \{ a,b,c,d \right \}, B= \left \{ 1,2,3 \right \}$$ and
$$ R_{1}= \left \{ \left ( a,1 \right ), \left ( a,3 \right ) \right \} $$
$$ R_{2}= \left \{ \left ( a,1 \right ), \left ( c,2 \right ), \left ( d,1 \right ) \right \} $$
$$ R_{3}= \left \{ \left ( a,1 \right ), \left ( b,2 \right ), \left ( 3,c \right ) \right \}.$$
We know that every relation from $$A$$ to $$B$$ will be a subset of $$A \times B$$.
Thus, both $$ R_{1} $$ and $$R_{2} $$ are subsets of $$ A\times B$$ and hence are relations but $$ R_{3}$$ is not a relation because $$R_{3}$$ is not a subset of $$A\times B$$ because the element $$ (3,c) \notin (A\times B).$$ It belongs to $$B\times A.$$
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.
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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$$-f(x)$$
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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 $$R$$ be a relation from a set $$A$$ to a set $$B$$,then
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$$\displaystyle R=A\cup B$$
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$$\displaystyle R=A\cap B$$
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$$\displaystyle R\subseteq A\times B$$
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$$\displaystyle R\subseteq B\times A$$
Explanation
If $$R$$ is a relation from $$A$$ to $$B$$ then $$R$$ is the subset of $$A \times B$$ because it contains all the possible mappings from $$A$$ to $$B$$.
If $$A=\{a,b,c,d\}, B=\{p,q,r,s\}$$, then which of the following are relations from $$A$$ to $$B$$?
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$$\displaystyle R_{1}= \left \{ \left ( a,p \right ), \left ( b,r \right ), \left ( c,s \right ) \right \}$$
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$$\displaystyle R_{2}= \left \{ \left ( q,b \right ), \left ( c,s \right ), \left ( d,r \right ) \right \}$$
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$$\displaystyle R_{3}= \left \{ \left ( a,p \right ), \left ( a,q \right ), \left ( d,p \right ), \left ( c,r \right ), \left ( b,r \right ) \right \}$$
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$$\displaystyle R_{4}= \left \{ \left ( a,p \right ), \left ( q,a \right ), \left ( b,s \right ), \left ( s,b \right ) \right \}$$
Explanation
From then definiation of Relation
Let $$R$$ be $$a$$ relation from $$A$$ to $$B$$
i.e., $$R\subseteq A\times B,$$ then
Domain of {$$a:a\in A,\left( a,b \right) \in R$$ for some $$b\in B$$}
Range of $$R=$${ $$b:b\in B,\left( a,b \right) \in R$$ for some $$a\in A$$}.
Therefore,
(A) $$\displaystyle R,=\left\{ \left( a,p \right) ,\left( b,s \right) ,\left( c,s \right) \right\} $$ is a relation as $$\left\{ a,b,c \right\} \subseteq A$$ and $$\left\{ p,r,s \right\} \subseteq B.$$
(B) $${ R }_{ 2 }=\left\{ \left( q,b \right) ,\left( c,s \right) ,\left( d,r \right) \right\} $$ is not a relation as $$q$$ is not from set $$A.$$
(C) $${ R }_{ 3 }=\left\{ \left( a,p \right) ,\left( a,q \right) ,\left( d,p \right) ,\left( c,r \right) ,\left( b,r \right) \right\} $$ is a relation as $$\displaystyle \left\{ a,d,c,b \right\} \subseteq A$$ and $$\displaystyle \left\{ p,q,r \right\} \subseteq B.$$
(D)$${ R }_{ 4 }=\left\{ \left( a,p \right) ,\left( q,a \right) ,\left( b,s \right) ,\left( s,b \right) \right\} $$ is not a relation as $$q$$ and $$s$$ does not belong to set $$A.$$
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.
If $$\displaystyle X= \left \{ 1,2,3,4,5 \right \}, Y= \left \{ 1,3,5,7,9 \right \}$$ determine which of the following sets are mappings, relations or neither from A to B:
(i)$$\displaystyle F= \left \{ \left ( x,y \right ) \because y= x+2, x \in X, y \in Y \right \}$$
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It is clearly a one-one onto mapping i.e. a bijection. It is also a relation.
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It is clearly a many-one onto mapping. It is also a relation.
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It is clearly a one-one but not onto mapping. It is also a relation.
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It is not a mapping but a relation
Explanation
Given def of F as
$$\displaystyle F= \left \{ \left ( x,y \right ) \because y= x+2, x \epsilon X, y \epsilon Y \right \}$$
where $$\displaystyle X= \left \{ 1,2,3,4,5 \right \}, Y= \left \{ 1,3,5,7,9 \right \}$$
So, we get $$\left \{ \left ( 1,3 \right ), \left ( 2,4 \right ), \left ( 3,5 \right ), \left ( 4,6 \right ), \left ( 5,7 \right ) \right \} $$
Since, $$\displaystyle y \in Y$$ and in the above 4,6 do not belong to Y.
$$\displaystyle \therefore $$ we exclude the ordered pair (2,4) and (4,6)
$$\displaystyle \therefore \displaystyle F=\left \{ \left ( 1,3 \right ), \left ( 3,5 \right ), \left ( 5,7 \right ) \right \}$$
F is not a mapping because the elements 2,4 $$\displaystyle \epsilon X$$ do not have any image. $$\displaystyle F\subset X\times Y$$ and hence F is a relation from X to Y.
Let $$f:[2,\infty)\rightarrow [1,\infty)$$defined by $$f(x)=2^{x^{4}-4x^{2}}$$ and $$\displaystyle g:\left[ \frac{\pi}{2},\pi \right] \rightarrow A$$ defined by $$ \displaystyle g(x)=\frac {\sin x+4}{\sin x-2}$$ be two invertible functions, then
$$f^{-1}(x)$$ is equal to
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$$\sqrt{2+\sqrt{4-\log_{2}x}}$$
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$$\sqrt{2+\sqrt{4+\log_{2}x}}$$
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$$\sqrt{2-\sqrt{4+\log_{2}x}}$$
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None of these
Explanation
$$\because ff^{-1}(x)=x$$
$$2^{(f^{-1}(x))^{4}}-4^{(f^{-1}(x))^{2}}=x$$
$$\Rightarrow (f^{-1}(x))^{4}-4(f^{-1}(x))^{2}=\log _{2}x=0$$
$$\therefore (f^{-1}(x))^{2}=2+\sqrt{4+\log _{2}x}$$
$$\therefore $$ Range of $$f^{-1}(x)$$ is $$(2, \infty )$$.
$$\therefore f^{-1}(x)=\sqrt{2+\sqrt{4+\log _{2}x}}$$
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, $$f_n\left(x\right)=\dfrac{x}{\left(n+1\right)x+1}$$
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 $$f(x)=\begin{cases} 2x+3\quad \quad x\le 1 \\ a^{ 2 }x+1\quad x>1 \end{cases}$$, then the values of $$a$$ for which $$f(x)$$ is injective.
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$$-3$$
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$$1$$
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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 $$(x>1)$$, $$f(1) \geq 5$$
$$\Rightarrow a^2(1)+1\geq 5 \Rightarrow a^2\geq 2 \Rightarrow a\in R-(-2,2)$$
Which of the functions defined below are NOT one-one function(s)
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$$f(x)\, =\, 5(x^{2}\, +\, 4),\, (x\, \in\, R)$$
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$$g(x)\, =\, 2x\, +\, \dfrac1x$$
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$$h(x)\, =\, ln(x^{2}\, +\, x\, +\, 1)\,, (x\, \in\, R)$$
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$$f(x)\, =\, e^{-x}$$
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 $$1+\sqrt { x } =y\Rightarrow x={ \left( y-1 \right) }^{ 2 }$$
$$\therefore f\left( y \right) =3+2\left( y-1 \right) +{ \left( y-1 \right) }^{ 2 }\\ =3+3y-2+{ y }^{ 2 }-2y+1=2+{ y }^{ 2 }\\ \therefore f\left( x \right) =2+{ x }^{ 2 }$$
Let $$X=\left\{ 1,2,3,4 \right\} $$ and $$Y=\left\{ 1,2,3,4 \right\} $$. Which of the following is a relation from $$X$$ to $$Y$$.
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$${R}_{1}=\left\{ (x,y)| y=2+x, x\in X, y\in Y \right\} $$
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$${R}_{2}=\left\{ (1,1),(2,1),(3,3),(4,3),(5,5) \right\} $$
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$${R}_{3}=\left\{ (1,1),(1,3),(3,5),(3,7),(5,7) \right\} $$
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$${R}_{4}=\left\{ (1,3),(2,5),(2,4),(7,9) \right\} $$
Explanation
Given $$X=\left\{ 1,2,3,4 \right\} $$ and $$Y=\left\{ 1,2,3,4 \right\} $$
$$X\times Y=\{ (1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)$$
$$(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(4,4)\} $$
Now, $$R_1=\{(1,3),(2,2)\}$$
Since, $$R_1\subset X\times Y$$
Hence, $$R_1$$ is a relation from $$X$$ to $$Y$$
$${R}_{2}=\left\{ (1,1),(2,1),(3,3),(4,3),(5,5) \right\} $$
Since, $$(5,5) \in R_2$$ but $$\notin X\times Y$$
So, $$R_2$$ is not a relation from $$X$$ to$$Y$$
$${R}_{3}=\left\{ (1,1),(1,3),(3,5),(3,7),(5,7) \right\} $$
Since, $$(3,5),(3,7),(5,7) \in R_3$$ but $$\notin X\times Y$$
So, $$R_3$$ is not a relation from $$X$$ to $$Y$$
$${R}_{4}=\left\{ (1,3),(2,5),(2,4),(7,9) \right\} $$
Since, $$(2,5),(7,9) \in R_4$$ but $$\notin X\times Y$$
So, $$R_4$$ is not a relation from $$X$$ to $$Y$$
Find inverse $$f(x)=\log_{e}(x+\sqrt{x^{2}+1})$$
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$$\sinh(x) $$
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$$\cosh(x)$$
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$$\tanh(x) $$
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$$\coth(x)$$
Explanation
$$y=\log(x+\sqrt{x^{2}+1})$$
$$\Rightarrow e^{y}=x+\sqrt{x^{2}+1}$$
$$\Rightarrow e^{y}-x=\sqrt{x^{2}+1}$$
$$\Rightarrow x^{2}-2xe^{y}+e^{2y}=x^{2}+1$$
$$\Rightarrow e^{2y}=1+2xe^{y}$$
$$\Rightarrow 2x=\dfrac{e^{2y}-1}{e^{y}}$$
$$\Rightarrow x=\dfrac{e^{y}-e^{-y}}{2}$$
Replacing x with y
$$f^{-1}(x)=\dfrac{e^{x}-e^{-x}}{2}$$
$$=\sinh(x)$$
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 $$\{(x,a), (y,b), (z,c)\}$$
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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$$6$$
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$$-2$$
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 $$-3$$
from the given statement $$N$$ denotes the natural number and $$W$$ denotes the whole number, so which statement in the following is correct
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N=W
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N $$\subset$$ W
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W $$\subset$$ N
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N $$\cong$$ W
Explanation
Natural numbers are naturally occurring counts of an object $$1, 2, 3, 4,… $$forming an infinite list in which the first number is $$1$$ and every next number is equal to the preceding number $$+1.$$
Whole number include the set of all Natural numbers and also $$0.$$
Let $$N$$ and $$W$$ elements of natural and whole numbers.
$$N=\{1,2,3,4,....\infty\}$$
$$W=\{0,1,2,3,4,...\infty\}$$
So, here we can say $$N$$ is subset of $$W.$$
$$\therefore$$ $$N\subset W$$ is correct statement.
If
$$f(x)=\begin{cases} x+1,\quad \quad if\quad x\, \leq \, 1 \\ 5-x^{ 2 }\quad \quad if\quad x>1 \end{cases},g(x)=\begin{cases} x\quad \quad if\quad x\leq 1 \\ 2-x\quad if\quad x>1 \end{cases}$$
and $$x\, \in\, (1, 2)$$, then $$g(f(x))$$ is equal to
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$$x^{2}\, +\, 3$$
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$$x^{2}\, -\, 3$$
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$$5\, -\, x^{2}$$
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$$1 - x$$
Explanation
$$x\, \epsilon\, (1, 2)$$
Hence for x >1, $$f(x) = 5-x^2$$and $$f(x) \epsilon (1,4) $$
$$f(x) >1$$
$$g(f(x)) = 2 - f(x)= 2-5+x^2 = x^2-3$$
If $$g(x) = 2x + 1$$ and $$h(x) = 4x^{2} + 4x + 7$$, find a function $$f$$ such that $$f o g = h$$
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$$f(x) = x^{3} - 6$$
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$$f(x) = x^{2} + 6$$
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$$f(x) = x^{2} - 6$$
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$$f(x) = (2x+1)^2 + 6$$
Explanation
$$f(g(x)) = h = 4x^2+4x+7=(2x+1)^2+6=(g(x))^2+6$$
$$\Rightarrow f(x) =x^2+6$$
Let $$X = \left\{1,2,3,4\right\}$$ and $$Y = \left\{1,3,5,7,9\right\}$$. Which of the following is relations from $$X$$ to $$Y$$
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$$R_1 = \left\{(x,y) | y = 2x+1, x \in X, y \in Y\right\}$$
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$$R_2 = \left\{(1,1),(2,1),(3,3),(4,3),(5,5)\right\}$$
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$$R_3 = \left\{(1,1),(1,3),(3,5),(3,7),(5,7)\right\}$$
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$$R_4 = \left\{(1,3),(2,5), (2,4), (7,9)\right\}$$
Explanation
We have
$$X = \left\{1,2,3,4\right\}$$ and $$Y = \left\{1,3,5,7,9\right\}$$.
$$X \times Y=\{(1,1),(1,3),(1,5),(1,7),(1,9), (2,1),(2,3),(2,5),$$
$$(2,7),(2,9), (3,1),(3,3),(3,5),(3,7),(3,9),(4,1),(4,3),(4,5),(4,7),(4,9)\}$$
Let's take option A,
$$R_1 = \left\{(x,y) | y = 2x+1, x \in X, y \in Y\right\}$$
$$R_{1}= \{(1,3),(2,5),(3,7), (4,9)\}$$
Since, $$R_1 \subseteq X\times Y$$
So, $$R_1$$ is a relation from $$X$$ to $$Y$$
Clearly, $$R_2$$ is not a relation from $$X$$ to $$Y$$ as $$(5,5) \notin X\times Y$$
Similarly, $$R_3$$ is not a relation from $$X$$ to $$Y$$ as $$(5,7) \notin X\times Y$$
In the same manner, $$R_4$$ is also not a relation from $$X$$ to $$Y$$ as $$(7,9) \notin X\times Y$$
Which of the following are two distinct linear functions which map the interval $$[-1, 1]$$ onto $$[0, 2]$$
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$$f(x) = 1 + x$$ or $$1 - x$$
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$$f(x) = 1 + 2x$$ or $$1 - x$$
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$$f(x) = 1 + x$$ or $$1 - 2x$$
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$$f(x) = 1 + x$$ or $$2 - x$$
Explanation
Out of two linear functions one will be increasing and other will be decreasing.
Let $$f(x) = ax+b$$
For increasing: $$f(-1)=0=> b-a=0\Rightarrow b=a$$ and $$f(1) = 2\Rightarrow a+b=2\Rightarrow a=b=1\therefore f(x) = 1+x$$
For decreasing: $$f(-1)=2=> b-a=2$$ and $$f(1) = 0\Rightarrow a+b=2\Rightarrow b=1, a=-1\therefore f(x) = 1-x$$
If $$f(x) =\ln {\displaystyle \frac { 1+x }{ 1-x } } $$ and $$g(x)=\displaystyle \frac {3x+x^3}{1+3x^2}$$, then $$f[g(x)]$$ equals.
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$$f(x)$$
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$$[f(x)]^3$$
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$$3f(x)$$
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$${f(x)}^2$$
Explanation
Given $$f(x)=\ln {\displaystyle \frac { 1+x }{ 1-x } } $$ and $$g(x)=\displaystyle \frac {3x+x^3}{1+3x^2}$$
then $$f[g(x)]=\ln {\displaystyle \frac { 1+\displaystyle\frac { 3x+x^{ 3 } }{ 1+3x^{ 2 } } }{ 1-\displaystyle\frac { 3x+x^{ 3 } }{ 1+3x^{ 2 } } } } $$
$$=\ln {\displaystyle \frac { 1+3x^{ 2 }+3x+x^{ 3 } }{ 1+3x^{ 2 }-3x-x^{ 3 } } } $$
$$=\ln\left[ {\displaystyle \frac { 1+x }{ 1-x } }\right]^3 $$
$$=3\ln {\displaystyle \frac { 1+x }{ 1-x } } $$
$$\Rightarrow f[g(x)]=3f(x)$$
Hence, option C.
Let $$f(x) = e^{3x}, g(x) = \log_ex, x > 0$$, then $$fog (x)$$ is
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$$3x$$
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$$x^3$$
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$$\log_{10}3x$$
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$$\log3x$$
Explanation
Given $$f(x) = e^{3x}, g(x) = \log_ex, x > 0$$
$$fog(x)=f(\log_ex)=e^{3\log_ex}=x^3$$
Hence, option B.
$$f(x)\, >\, x;\, \forall\, x\, \epsilon\, R.$$ The equation $$f (f(x)) -x = 0$$ has
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Atleast one real root
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More than one real root
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No real root if f(x) is a polynomial & one real root if f(x) is not a polynomial
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No real root at all
Explanation
Given $$f(x) > x $$ $$\forall x \epsilon R$$
Hence, $$ f(x)-x$$ has no real roots, because $$f(x)-x >0$$
Since, $$x \epsilon R \implies f(x) \epsilon R$$
$$f(f(x))>x$$
$$\implies f(f(x))-x>0$$
Hence it will have no real root.
If $$ f : R \rightarrow R, f(x) = (x + 1)^2$$ and $$g : R \rightarrow R, g(x) = x^2 + 1 $$ then $$(fog)(3)$$ is equal to
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$$121$$
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$$144$$
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$$112$$
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$$11$$
Explanation
Given,
$$ f : R \rightarrow R, f(x) = (x + 1)^2$$ and $$g : R \rightarrow R,g(x) = x^2 + 1 $$
$$g(3)=3^2+1=10$$
$$f(10)=(10+1)^2=121$$
$$\therefore fog(3)=f\left(g(3)\right)=f(10)=121$$
Hence, option A.
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
$$\displaystyle fog\left( x \right) =f\left( g\left( x \right) \right) =\sqrt { \left| \sin { x } -1 \right| } $$
$$=\sqrt { \left| 1-\cos { \left( \dfrac { \pi }{ 2 } -x \right) } \right| } =\sqrt { 2 } \left| \sin { \left( \dfrac { \pi }{ 4 } -\dfrac { x }{ 2 } \right) } \right| =\left| \sin { \dfrac { x }{ 2 } } -\cos { \dfrac { x }{ 2 } } \right| $$
If $$f(x) = \log x$$, $$g(x) = x^3$$, then $$f[g(a)] + f[g(b)]$$ equals
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$$f[g(a) + g(b)]$$
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$$3f(ab)$$
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$$g[f(ab)]$$
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$$g[f(a) + f(b)]$$
Explanation
We have $$f(x) = \log x$$
$$g(x) = x^3$$
$$g(a) = a^3 $$
$$f(g(a)) = \log (a^3) $$
$$f(g(a)) =3 \log a $$
Similarly,
$$f(g(b)) = 3 \log b $$
$$\Rightarrow f(g(a)) + f(g(b)) = 3 \log a + 3 \log b $$
$$\Rightarrow f(g(a)) + f(g(b)) = 3 \log ab $$
$$\Rightarrow f(g(a)) + f(g(b)) = 3 f(ab)$$
If $$f(x) = x^3 $$ and $$g(x) = sin2x$$, then
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$$g[f(1)] = 1$$
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$$f(g(\pi/12) = 1/8$$
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$$g{f(2)} = \sin 2$$
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none of these
Explanation
Given,
$$ f(x)=x^3 $$ and $$g(x)=\sin 2x$$
Clearly,
$$ f(1)=1 \Rightarrow g[f(1)]=\sin 2 $$
$$ g(\cfrac{\pi}{12}) =\dfrac{1} {2} \Rightarrow f[g(\dfrac{\pi}{12})]=\dfrac{1}{8}$$
and,$$ f(2)=8 \Rightarrow g[f(2)]= \sin 16 $$
So, the correct option is (B).
If $$f(x) = (a x^n)^{1/n},$$ where $$\ n \in N$$, then $$f\{f(x)\}$$ equals
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$$0$$
0%
$$x$$
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$$x^n$$
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none of these
Explanation
We have, $$f(x) = (a x^n)^{1/n}=a^{1/n}x=kx$$ where $$k = a^{1/n}$$
$$\therefore f\{f(x)\} = kf(x) = k^2x =a^{2/n} x$$
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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