MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 12 Commerce Maths Relations And Functions Quiz 7 - MCQExams.com
CBSE
Class 12 Commerce Maths
Relations And Functions
Quiz 7
If $$f(x)=ax+b $$ and $$g(x)=cx+d$$, then $$f\left( g(x) \right) =g\left( f(x) \right) \Leftrightarrow$$
Report Question
0%
$$f(a)=g(c)$$
0%
$$f(b)=g(b)$$
0%
$$f(d)=g(b)$$
0%
$$f(x)=g(a)$$
Explanation
We have
$$f(x)=ax+b,g(x)=cx+d$$
Therefore, $$ f\left\{ g(x) \right\} =g\left\{ f(x) \right\} \Leftrightarrow f(cx+d)=g(ax+b)$$
$$\Leftrightarrow a(cx+d)+b=c(ax+b)+d$$
$$\Leftrightarrow ad+b=cb+d\Leftrightarrow f(d)=g(b)$$
The inverse of the function $$f(x) = \log (x^{2} + 3x + 1), x\epsilon [1, 3]$$, assuming it to be an onto function, is
Report Question
0%
$$\dfrac {-3 + \sqrt {5 + 4e^{x}}}{2}$$
0%
$$\dfrac {-3 \pm \sqrt {5 + 4e^{x}}}{2}$$
0%
$$\dfrac {-3 - \sqrt {5 + 4e^{x}}}{2}$$
0%
None of the above
Explanation
Given, $$f(x) = \log (x^{2} + 3x + 1)$$
Therefore, $$ f'(x) = \dfrac {2x + 3}{(x^{2} + 3x + 1)} > 0\forall x\epsilon [1, 3]$$
which is a strictly increasing function. Thus, $$f(x)$$ is injective, given that $$f(x)$$ is onto. Hence, the given function $$f(x)$$ is invertible.
Now, $$f\left \{f^{-1}(x)\right \} = x$$
$$\Rightarrow \log \left \{f^{-1}(x)\right \}^{2} + 3\left \{f^{-1}(x) + 1 \right \} = x$$
$$\Rightarrow \left \{f^{-1}(x)\right \}^{2} + 3\left \{f^{-1}(x)\right \} + 1 - e^{x} = 0$$
Thus $$ f^{-1}(x) = \dfrac {-3\pm \sqrt {9 - 4.1 (1 - e^{x})}}{2}$$
$$= \dfrac {-3\pm \sqrt {5 + 4e^{x}}}{2}$$
$$= f^{-1}(x) = \dfrac {-3\pm \sqrt {5 + 4e^{x}}}{2} [\because f^{-1}(x) \epsilon [1, 3]]$$
Hence, $$f^{-1}(x) = \dfrac {-3 + \sqrt {5 + 4e^{x}}}{2}$$
Let $$f(x)={ x }^{ 3 }-3x+1$$. The number of different real solutions of $$f(f(x))=0$$
Report Question
0%
$$2$$
0%
$$4$$
0%
$$5$$
0%
$$7$$
If $$f\left( x \right) $$ and $$g\left( x \right) $$ are two functions with $$g\left( x \right) =x-\dfrac { 1 }{ x } $$ and $$f\circ g\left( x \right) ={ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } $$, then $$f^{ ' }\left( x \right) $$ is equal to
Report Question
0%
$$3{ x }^{ 2 }+3$$
0%
$${ x }^{ 2 }-\dfrac { 1 }{ { x }^{ 2 } } $$
0%
$$1+\dfrac { 1 }{ { x }^{ 2 } } $$
0%
$$3{ x }^{ 2 }+\dfrac { 3 }{ { x }^{ 4 } } $$
Explanation
$$f\circ g\left( x \right) ={ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } $$
Writing $${ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } $$ using $${ \left( a-b \right) }^{ 3 }={ a }^{ 3 }-{ b }^{ 3 }-3ab\left( a-b \right) $$, we have
$${ \left( x-\dfrac { 1 }{ x } \right) }^{ 3 }={ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } -3x\cdot \dfrac { 1 }{ x } \left( x-\dfrac { 1 }{ x } \right) $$
$$\Rightarrow { x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } ={ \left( x-\dfrac { 1 }{ x } \right) }^{ 3 }+3\left( x-\dfrac { 1 }{ x } \right) $$
We have,
$$f\left( g\left( x \right) \right) ={ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } ={ \left( x-\dfrac { 1 }{ x } \right) }^{ 3 }+3\left( x-\dfrac { 1 }{ x } \right) $$
As $$g\left( x \right) =x-\dfrac { 1 }{ x } $$, this yields
$$f\left( x-\dfrac { 1 }{ x } \right) ={ \left( x-\dfrac { 1 }{ x } \right) }^{ 3 }+3\left( x-\dfrac { 1 }{ x } \right) $$
On putting $$x-\dfrac { 1 }{ x } =t$$, we get
$$f\left( t \right) ={ t }^{ 3 }+3t$$
Thus, $$f\left( x \right) ={ x }^{ 3 }+3x$$
and $$f^{ ' }\left( x \right) =3{ x }^{ 2 }+3$$
If $$f:R\rightarrow R$$, $$g:R\rightarrow R$$ are defined by$$ f(x)=5x-3$$, $$g(x)=x^{2}+3$$, then $$(gof^{-1})(3)$$=
Report Question
0%
$$\dfrac{25}{3}$$
0%
$$ \dfrac{111}{25}$$
0%
$$\dfrac{9}{25} $$
0%
$$\dfrac{25}{111} $$
Explanation
$$f:R\rightarrow R$$
$$ g:R\longrightarrow R$$
$$ f\left( x \right) =5x-3$$
$$g\left( x \right) ={ x }^{ 2 }+3$$
$$\left(gof^{-1}\right)\left(3\right)$$
$$f\left(x\right)={5}{x}-{3}$$
Let
$$f\left(x\right)=y$$
$$\Rightarrow{x}=f^{-1}\left(y\right)$$
$$f\left(x\right)=y={5}x-3$$
$$\Rightarrow x=\dfrac{y+3}{5}$$
$$\Rightarrow f^{-1}\left(y\right)=\dfrac{y+3}{5}$$
$$ f^{-1}\left(x\right)=\dfrac{x+3}{5}$$ and $$g\left(x\right)=x^{2}+{3}$$
$$\left(gof^{-1}\right)\left(3\right)=g\left[f^{-1}\left(3\right)\right]=g\left(\dfrac{3+3}{5}\right)=g\left(\dfrac{6}{5}\right)$$
$$=\left[\left(\dfrac{6}{5}\right)^{2}+3\right]=\dfrac{36}{25}+{3}=\dfrac{111}{25}$$
Hence, $$\dfrac{111}{25}$$ is the correct answer.
The inverse of the function $$y = 5^{\ln\ x}$$ is
Report Question
0%
$$x = y^{\frac {1}{\ln 5}}, y > 0$$
0%
$$x = y^{\ln 5}, y > 0$$
0%
$$x = y^{\frac {1}{\ln 5}}, y < 0$$
0%
$$x = 5\ln y, y > 0$$
Explanation
Given that
$$ y = 5^{\ln x} $$
$$ \ln y = \ln (5^{\ln x}) $$
$$ = (\ln x) (\ln 5)$$
$$ \therefore \ln x = \dfrac{\ln y}{\ln 5} $$
$$ x = e^{\frac{\ln y}{\ln 5} }$$
$$ = (e^{\ln y})^{\frac{1}{\ln 5}}$$
$$ = y^{\frac{1}{\ln 5}}$$
Since the domain of the natural log function ln() is the set of positive real numbers, it is required here that
$$ y > 0$$.
The inverse function:
$$ x = y^{\frac{1}{\ln 5}}, \; \; y > 0$$
If $$f(x) = 2x^{3} + 7x - 5$$ then $$f^{-1}(4)$$ is
Report Question
0%
Equal to $$1$$
0%
Equal to $$2$$
0%
Equal to $$1/3$$
0%
Non existent
Explanation
$$f\left( x \right) =2{ x }^{ 3 }+7x-5$$
$$x=2\left( { f^{ -1 }\left( x \right) }^{ 3 } \right) +7f^{ -1 }\left( x \right) -5$$
$$ f^{ -1 }\left( 4 \right) =?$$
$$ 4=2\left( { f^{ -1 }\left( 4 \right) }^{ 3 } \right) +7f^{ -1 }\left( 4 \right) -5$$
$$ 2\left( { f^{ -1 }\left( 4 \right) }^{ 3 } \right) +7f^{ -1 }\left( 4 \right) -9=0$$
Let $$f^{ -1 }\left( 4 \right) =t$$
$$ 2{ t }^{ 3 }+7t-9=0$$
$$ 2{ t }^{ 3 }-2+7t-7=0$$
$$ 2\left( { t }^{ 3 }-1 \right) +7\left( t-1 \right) =0$$
$$ 2\left( t-1 \right) \left( { t }^{ 2 }+t+1 \right) +7t-1=0$$
$$ \left( t-1 \right) \left( 2{ t }^{ 2 }+2t+2+7 \right) =0$$
$$ t=1\quad 2{ t }^{ 2 }+2t+9=0$$
So, $$f^{ -1 }\left( 4 \right) =1$$
Answer A
$$f,g:R\rightarrow R$$ are functions such that $$f(x)=3x-\sin { \left( \cfrac { \pi x }{ 2 } \right) } ,g(x)={ x }^{ 3 }+2x-\sin { \left( \cfrac { \pi x }{ 2 } \right) } $$
The value of $$\cfrac { d }{ dx } { f }^{ -1 }{ \left( { g }^{ -1 }(x) \right) }_{ x=12 }$$ is equal to
Report Question
0%
$$\cfrac { 2 }{ 30+x } $$
0%
$$\cfrac { 2 }{ 30-x } $$
0%
$$\cfrac { 2 }{ 3\left( 28-\pi \right) } $$
0%
$$\cfrac { 2 }{ 3\left( 28+\pi \right) } $$
Explanation
Given that,
$$g:R \to R$$
$$f\left( x \right) = 3x - \sin \left( {\dfrac{{\pi x}}{2}} \right)$$
$$g\left( x \right) = {x^3} + 2x - \sin \left( {\dfrac{{\pi x}}{2}} \right)$$
$$\therefore \dfrac{{d{f^{ - 1}}}}{{dx}}{\left( {g'\left( x \right)} \right)_{x = 12}} = \dfrac{2}{{30 + x}}$$
Hence, the option $$(A)$$ is correct.
If $$f: R\rightarrow R$$ and $$g: R\rightarrow R$$ are defined $$f(x) = x - [x]$$ and $$g(x) = [x]\forall x\epsilon R, f(g(x))$$.
Report Question
0%
$$x$$
0%
$$0$$
0%
$$f(x)$$
0%
$$g(x)$$
Explanation
We have, $$f(x)=x-[x]=\{x\}\dots (1)$$
where $$\{x\}$$ is the fractional part of $$x$$
If $$g(x)=[x]$$, where $$[x]$$ is the greatest integer part of $$x$$, then
The value of $$g(x)$$ will always be an integer.
And $$\therefore f(g(x))=f([x])=0$$
$$\because $$ the fractional part of $$g(x)$$ i.e $$[x]$$ is always $$0$$.
Let $$f : A \rightarrow B$$ be a function defined as $$f(x) = \dfrac {x - 1}{x - 2}$$, where $$A = R - \left \{2\right \}$$ and $$B = R - \left \{1\right \}$$. Then $$f$$ is :
Report Question
0%
invertible and $$f^{-1}(y) = \dfrac {2y + 1}{y - 1}$$
0%
invertible and $$f^{-1}(y) = \dfrac {3y - 1}{y - 1}$$
0%
not invertible
0%
invertible and $$f^{-1}(y) = \dfrac {2y - 1}{y - 1}$$
Explanation
Let $$y = f(x)$$
$$\Rightarrow y = \dfrac{x-1}{x-2}$$
$$\Rightarrow yx - 2y = x - 1$$
$$\Rightarrow (y-1)x = 2y-1 $$
$$\Rightarrow x = f^{-1} (y) = \dfrac{2y-1}{y-1}$$
So on the given domain the function is invertible and its inverse can be computed as shown above.
So, option D is the correct answer.
If $$f(x)$$ is a real valued function, then which of the following is one-one function?
Report Question
0%
$$f(x)=e^{|x|}$$
0%
$$f(x)=|e^x|$$
0%
$$f(x)=\sin x$$
0%
$$f(x)=|\sin x|$$
Explanation
$$(1)$$ $$f(x)=e^{|x|}$$
Since$$f(x)=f(-x)$$
So $$f(x)$$ is many one.
$$(2)$$ $$f(x)=|e^{x}|$$
so, $$f(x)=e^x$$
and as $$e^x$$ is one-one
So $$f(x)$$ is one-one.
$$(3)$$ $$f(x)=\sin x$$
Since $$\sin x$$ is many one
So $$f(x)$$ is also many-one.
$$(4)$$ $$f(x)=|\sin x|$$
Since $$\sin x$$ is many one
So $$f(x)$$ is also many-one.
If $$A =\{1, 2, 3\}$$ and $$ B = \{4, 5\}$$ then the number of function $$f : A \rightarrow B$$ which is not onto is ______
Report Question
0%
$$2$$
0%
$$6$$
0%
$$8$$
0%
$$4$$
Explanation
Total number of function $$f:A\rightarrow B$$ is the number of relations from $$A$$ to $$B$$ such that all the elements of $$A$$ are in the first elements of function $$f$$.
Hence, total number of functions are $$=2\times 2\times 2 =2^3=8$$
There are two functions for which the function is $$\{(1,4),(2,4),(3,4)\}$$ and $$\{(1,5),(2,5),(3,5)\}$$.
Hence there are two non-onto functions.
If $$f\,: R \rightarrow R, g: R \rightarrow R\,$$ are defined by $$f(x)= 5x -3,g(x)=x^2 + 3$$, then, $$(gof^{-1})(3) =$$
Report Question
0%
$$\displaystyle \frac{25}{3}$$
0%
$$\displaystyle \frac{111}{25}$$
0%
$$\displaystyle \frac{9}{25}$$
0%
$$\displaystyle \frac{25}{111}$$
Explanation
$$f(x)=5x-3,$$ $$g(x)=x^2+3$$
Let $$f(x)=y$$
$$\Rightarrow$$ $$5x-3=y$$
$$\Rightarrow$$ $$x=\dfrac{y+3}{5}$$ ----( 1 )
Now,
$$f(x)=y$$
$$f^{-1}(y)=x$$
$$f^{-1}(y)=\dfrac{y+3}{5}$$ [ From ( 1 ) ]
$$f^{-1}(x)=\dfrac{x+3}{5}$$
Next,
$$(gof^{-1})(3)$$
$$\Rightarrow$$ $$g[f^{-1}(3)]$$
$$\Rightarrow$$ $$g\left[\dfrac{3+3}{5}\right]$$
$$\Rightarrow$$ $$g\left[\dfrac{6}{5}\right]$$
$$\Rightarrow$$ $$\left(\dfrac{6}{5}\right)^2+3$$ [ Since, $$g(x)=x^2+3$$ ]
$$\Rightarrow$$ $$\dfrac{36}{25}+3$$
$$\Rightarrow$$ $$\dfrac{36+75}{25}$$
$$\Rightarrow$$ $$\dfrac{111}{25}$$
Let $$f:A \to b$$ be a function defined by f(x) =$$\sqrt {1 - {x^2}} $$
Report Question
0%
f(x) is one-one if A =[0,1]
0%
f(x) is onto if B = [0,1]
0%
f(x) is one-one if A =[-1 , 0]
0%
f(x) is onto if B = [-1,1]
Explanation
$$f:A\rightarrow B$$
$$ f(x)=\sqrt { 1-{ x }^{ 2 } } $$
When $$ x=0$$
$$ f(0)=\sqrt { 1-0 } $$
$$ =1$$
$$ f(1)=\sqrt { 1-{ 1 }^{ 2 } } $$
$$ =0$$
$$ f(x)$$ is one - one function when $$A=[0,1]$$
If $$f:R\rightarrow R,f(x)=\begin{cases} 1\quad \quad x>0 \\ 0\quad \quad x=0 \\-1\quad x<0 \end{cases}$$ and $$g:R\rightarrow R,g(x)=\left[ x \right] $$, then $$\left( f\circ g \right) \left( \pi \right)$$ is:
Report Question
0%
$$\pi$$
0%
$$0$$
0%
$$1$$
0%
$$-1$$
Explanation
Given that $$g(x)=[x]$$
$$\Rightarrow$$ $$g(\pi )=g(3.141....)$$
i.e $$g(\pi )=3$$
$$(f\circ g)(x)=f(g(x))$$
$$\Rightarrow$$
$$(f\circ g)(\pi ))=f(g(\pi ))$$
But $$g(\pi)=3$$
$$\Rightarrow f(3)=1$$ ...... $$[\because f(x)=1 \text{ for } x>0]$$
The inverse of the function $$y=\large{\frac{e^x-e^{-x}}{e^x+e^{-x}}}$$ is
Report Question
0%
$$\large{\frac{1}{2}}$$ $$\log \large{\frac{1+x}{1-x}}$$
0%
$$\large{\frac{1}{2}}$$ $$\log \large{\frac{2+x}{2-x}}$$
0%
$$\large{\frac{1}{2}}$$ $$\log \large{\frac{1-x}{1+x}}$$
0%
$$2 \log (1+x)$$
Explanation
Given $$y=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}$$
or, $$\dfrac{1+y}{1-y}=\dfrac{2e^x}{2e^{-x}}$$
or, $$e^{2x}=\dfrac{1+y}{1-y} $$
or, $$ x=\dfrac{1}{2}\log \dfrac{1+y}{1-y}$$
So the inverse is $$\dfrac{1}{2}\log \dfrac{1+x}{1-x}$$.
Let $$f\left( x \right) = {x^2}$$ and $$g\left( x \right) = \sqrt x $$ (where $$x > 0$$),then
Report Question
0%
$$f\left( {g\left( x \right)} \right) = x$$
0%
$$g\left( {f\left( x \right)} \right) = x$$
0%
The least value of $$f\left( {g\left( x \right)} \right) + {1 \over {g\left( {f\left( x \right)} \right)}}$$ is $$2$$
0%
The least value of $$g\left( {f\left( x \right)} \right) + {1 \over {f\left( {g\left( x \right)} \right)}}$$ is $$2$$
Explanation
$$f(x)=x^{2},g(x)=\sqrt{x}$$
$$f(g(x))=(\sqrt{x})^{2}=x$$
$$g(f(x))=\sqrt{x^{2}}=x$$
$$f(g(x))+\dfrac{1}{g(f(x))}=g(f(x))+\dfrac{1}{f(g(x))}=x+\dfrac{1}{x}\ge 2$$
Hence, all options are correct.
The solution of $$( 3 * 4) * 3$$, when $$*$$ is a binary operation on Z such that: $$a * b = a + b$$, is.
Report Question
0%
$$10$$
0%
$$-10$$
0%
$$-\dfrac16$$
0%
$$-6$$
Explanation
Given $$( 3 * 4) * 3$$
$$(3*4)*3=(3+4)+3$$ (as$$(a \ast b)=(a+b)$$)
$$=7+3$$
$$=10$$
If $$g\left( x \right) = {x^2} + x - 2$$ and $$\frac{1}{2}gof\left( x \right) = 2{x^2} + 5x + 2$$, then $$f\left( x \right)$$ is
Report Question
0%
$$2x-3$$
0%
$$2x+3$$
0%
$$2{x^2} + 3x + 1$$
0%
$$2{x^2} - 3x -1$$
Explanation
$$g(x)={ x }^{ 2 }+x-2\\ \cfrac { 1 }{ 2 } g(f(x))=2{ x }^{ 2 }+5x+2\\ \Rightarrow f^{ 2 }\left( x \right) +f(x)-2=4{ x }^{ 2 }-10x+4\\ \Rightarrow f^{ 2 }\left( x \right) +f(x)-(4{ x }^{ 2 }-10x+6)=0\\ f(x)=\cfrac { -1\pm \sqrt { 1+4(4{ x }^{ 2 }-10x+6) } }{ 2 } \\ f(x)=\cfrac { -1\pm \sqrt { 1+16{ x }^{ 2 }-40x+24 } }{ 2 } \\ f(x)=\cfrac { -1\pm (4x-5) }{ 2 } =(2x-3)$$
Let f : $$R \to R$$ and g : $$R \to R$$ be two one-one and onto functions such that they are the mirror images of each other about the line y =If h(x) = f(x) + g(x), then h(0) equal to
Report Question
0%
2
0%
4
0%
0
0%
1
Explanation
Given $$g(x)$$ and $$f(x)$$ are mirror images about $$y=2$$
$$\implies g(x)-2=2-f(x)\implies f(x)+g(x)=4\implies h(x)=4$$
$$h(0)=4$$
If $$f: A\rightarrow A$$ defined by $$f(x) =\dfrac{4x +3}{6x-4}$$ where $$A= R-\dfrac{2}{3}$$. Find $$f^{-1}$$
Report Question
0%
$$2x$$
0%
$$\dfrac {4x+3}{6x-4}$$
0%
$$\dfrac x2$$
0%
None of these
Explanation
$$f\left( x \right) =\cfrac { 4x+3 }{ 6x-4 } \\ let\quad y=\cfrac { 4x+3 }{ 6x-4 } \\ \Rightarrow 6xy-4y=4x+3\\ \Rightarrow 6xy-4x=3+4y\\ \Rightarrow x(6y-4)=3+4y\\ \Rightarrow x=\cfrac { 3+4y }{ 6y-4 } \\ Replace\quad x\quad by\quad y\quad \& \quad y\quad by\quad x.\\ y=\cfrac { 3+4x }{ 6x-4 } \\ f^{ ' }\left( x \right)=\cfrac { 4x+3 }{ 6x-4 } $$
If the binary operation $$*$$ is defined on a set of integers as $$a * b = a + 3 b ^{2} $$ , then the value of $$2 * 3$$ is
Report Question
0%
$$27$$
0%
$$29$$
0%
$$2$$
0%
None of these
Explanation
$$a\ast b =a+3b^{2}$$
$$\Rightarrow 2\ast 3=2+3(3)^{2}$$
$$=2+27$$
$$=29$$
Let $$f$$, $$g:R\rightarrow {R}$$ be two functions defined as $$ f\left( x \right) =\left| x \right| +x$$, $$ g\left( x \right) =\left| x \right| -x, \forall x\in R$$. Then, find $$fog(x)$$
Report Question
0%
$$||x|-x|-|x|-x$$
0%
$$||x|-x|+|x|-x$$
0%
$$||x|-x|-|x|+x$$
0%
None of thesse
Explanation
$$f\left( x \right) = \left| x \right| + x$$
$$g\left( x \right) = \left| x \right| - x$$
$$fog\left( x \right) = f\left( {g\left( x \right)} \right)$$
$$ = f\left( {\left| x \right| - x} \right)$$
$$ = \left| {\left| x \right| - x} \right| + \left| x \right| - x$$
Consider set $$A={1,2,3,4}$$ and set $$B={0,2,4,6,8}$$, then the
number of one-one function from set $$A$$ to set $$B$$ is ?
Report Question
0%
$$5$$
0%
$$24$$
0%
$$120$$
0%
None of these
Explanation
Number of one-one function from
$$\underset{(m)}{A}$$ to $$\underset{(n)}{B} = \left\{\begin{matrix} \, ^nP_m, & if \, n \ge m \\ 0, & if \ n < m \end{matrix}\right.$$
$$m = 4, \ n = 5$$
One-one function $$=\, ^5P_4 = \dfrac{5!}{1!} = 120$$
The function $$*$$ on $$N$$ as
$$a * b = ( a - b)^{2} $$ is a binary operator
Report Question
0%
True
0%
False
Explanation
$$\ast$$ is relation of $$N: a\ast b=(a-b)^{2}$$
Checking at $$a=b, a,b\in N,\Rightarrow a\ast b=(a-b)^{2}=0 \notin N$$
$$\Rightarrow \ast $$ is not a binary opeation.
If the binary operation $$*$$ is on set of integers $$Z$$ is defined as
$$a * b = a + 2b ^{2}$$ , then the value of $$(8 * 3) * 2$$
Report Question
0%
$$26$$
0%
$$22$$
0%
$$32$$
0%
$$34$$
Explanation
$$a\ast b=a+2b^{2}$$
$$\Rightarrow (8\ast 3)\ast 2=(8+2(3)^{2})\ast 2$$
$$=(26\ast 2)$$
$$=26+2(2)^{2}$$
$$=26+8$$
$$=34$$
If $$f(x)=2x+5$$ and $$g(x)=x^2+1$$ be two real function , then value of
$$fog$$ at x=1
Report Question
0%
$$9$$
0%
$$6$$
0%
$$5$$
0%
$$4$$
If $$g(f(x) ) = |\sin x |$$ and $$f(g(x))=(\sin\sqrt x)^2$$ , then
Report Question
0%
$$f(x) = \sin^2x. g(x) =\sqrt x$$
0%
$$f(x) = \sin x , g(x) =|x| $$
0%
$$f(x) = x^2, g(x) = \sin \sqrt x$$
0%
f and g can not be determined
Explanation
$$g(f(x))=|sinx|=\sqrt{(sinx)^2}\\f(g(x))=(sin\sqrt x )^2=sin^2\sqrt x\\\therefore f(x)=sin^2x \>and \>g(x)=\sqrt x$$
Let $$f : R \rightarrow R$$ be defined by $$f(x) = x^2 - 3x + 4 $$ for all $$x \epsilon R$$, then $$f^{-1}(2)$$ is
Report Question
0%
2
0%
1
0%
Not defined
0%
$$\dfrac{1}{2}$$
Explanation
$$F:R\to R$$
$$f(x)=x^2-3x+4$$
$$f(1)=1-3+4$$
$$f(1)=2$$
$$\boxed {f^{-1}(2)=1}$$
Let $$f(x+\dfrac{1}{x})=x^2+\dfrac{1}{x^2}(x\neq 0)$$, then $$f(x)=$$
Report Question
0%
$$x^2$$
0%
$$x^2-1$$
0%
$$x^2-2$$
0%
N.O.T
Explanation
We are given
$$f(x+ \dfrac{1}{2})= x^{2}+ \dfrac{1}{x^{2}}$$ (where $$x \neq 0$$)
$$= x^{2}+ \dfrac{1}{x^{2}}+2-2$$.
$$=x^{2}+ \dfrac{1}{x^{2}}+ 2 (x^{2}) \left( \dfrac{1}{x^{2}} \right)-2$$.
$$f\left(x+ \dfrac{1}{x} \right)= \left(x+ \dfrac{1}{x} \right)^{2}-2$$
So, simply put $$x+ \dfrac{1}{x} \rightarrow x$$
we get $$f(x) = x^{2}-2$$
The set onto which the derivative of the function $$f(x)=x(\log x-1)$$ maps the range $$[1,\infty )$$ is
Report Question
0%
$$\left[1,\infty \right)$$
0%
$$\left( e,\infty \right)$$
0%
$$\left[e,\infty \right)$$
0%
$$\left( 0,0 \right)$$
Explanation
Given $$f (x) = x [\log x-1] x \in [+, \infty]$$
$$f (x) = x \left[ \dfrac{d}{dx} (\log x-1) \right]+ (\log x-1) \dfrac{d}{dx} (x)$$
$$=x. \dfrac{1}{x}+ (\log x-1).1$$
$$=1 -1 \log x-1$$
$$f(x)= \log x$$
$$\therefore $$ The set of $$f(x)= \log x$$ that maps
$$[1, \infty )$$ is $$[e, \infty )$$
Let $$E=\{1, 2, 3, 4\}$$ and $$F=\{1, 2\}$$ then the number of onto functions from E to F is
Report Question
0%
$$14$$
0%
$$16$$
0%
$$12$$
0%
$$8$$
Explanation
Number of onto function from $$E$$ to $$F.$$
$$E=\{1,2,3,4\}$$ $$F=\{1,2\} $$
Number of function from $$E$$ to $$F=2\times 2\times 2\times 2=16$$
We have to exclude functions where $$f(x)=1$$ & $$ f(x)=2$$.
$$\therefore$$ Total number of onto function $$=16-2=14$$.
Let $$f\left( x \right) ={ x }^{ 2 },g\left( x \right) ={ 2 }^{ x }$$, then solution set of $$fog\left( x \right) =gof\left( x \right) $$ is
Report Question
0%
R
0%
$$\left\{ 0 \right\} $$
0%
$$\left\{ 0,2 \right\} $$
0%
None of these
If $$f\left( x \right) =\begin{cases} 2+x,\quad x\ge 0 \\ 2-x,\quad x<0 \end{cases}$$ then $$f\left( f\left( x \right) \right) $$ is given by
Report Question
0%
$$f\left( f\left( x \right) \right) =\begin{cases} 2+x,\quad x\ge 0 \\ 4-x,\quad x<0 \end{cases}$$
0%
$$f\left( f\left( x \right) \right) =\begin{cases} 2+x,\quad x\ge 0 \\ 2-x,\quad x<0 \end{cases}$$
0%
$$f\left( f\left( x \right) \right) =\begin{cases} 4+x,\quad x< 0 \\ x,\quad x\ge 0 \end{cases}$$
0%
$$f\left( f\left( x \right) \right) =\begin{cases} 4+x,\quad x\ge 0 \\ x,\quad x<0 \end{cases}$$
Let $$f\left[-1, \dfrac{-1}{2}\right] \to [-1, 1]$$ is defined by $$f(x) = 4x^3 - 3x$$, then $$f^{-1} (x) =$$ ____ .
Report Question
0%
$$\cos \left(\dfrac{1}{3}\cos^{-1} x\right)$$
0%
$$\cos \left(3\cos^{-1} x\right)$$
0%
$$\sin \left(\dfrac{1}{3}\sin^{-1} x\right)$$
0%
$$\cos \left(\dfrac{2\pi}{3}+\dfrac{1}{3} \cos^{-1} x\right)$$
Explanation
$$f(x)=4x^3-3x$$
$$f\left[-1,\cfrac{-1}{2}\right]\longrightarrow \left[-1,1\right]$$
let,$$x=\cos\theta$$
$$\because y=\cos 3\theta$$
$$f(\cos\theta)=4cos^3\theta-3\cos\theta$$
$$=\cos3\theta$$
$$\because 3\theta=\cos^{-1}y$$
$$\theta=\cfrac{1}{3}\cos^{-1}y$$
$$f^{-1}(\cos3\theta)=\cos\theta$$
$$\because y=f(x)$$
$$f^{-1}(y)=(x)$$
$$\because f^{-1}(\cos3\theta)=\cos\theta$$
$$f^{-1}(y)=\cos\left(\cfrac{1}{3}\cos^{-1}y\right)$$
$$f^{-1}(x)=\cos\left(\cfrac{1}{3}\cos^{-1}x\right)$$
If : $$f(x) = 5 {x}^{2}$$, $$g(x) = 3x^{4}$$, then : $$(fog) (-1) =$$
Report Question
0%
$$45$$
0%
$$-54$$
0%
$$-32$$
0%
$$-64$$
Let $$f:X \to \left[ {1,\,27} \right]$$ be a function by $$f\left( x \right) = 5\sin x + 12\cos x + 14$$. The set $$X$$ so that $$f$$ is one-one and onto is
Report Question
0%
$$\left[ { - \pi /2,\pi /2\,} \right]$$
0%
$$\left[ {0,\,\pi } \right]$$
0%
$$\left[ {0,\,\pi /2} \right]$$
0%
non of these
For $$a,\ b\ \in \ R-\left\{ 0 \right\}$$, let $$f(x)=ax^{2}+bx+a$$ satisfies $$f\left(x+\dfrac{7}{4}\right)=f\left(\dfrac{7}{4}-x\right) \forall \ x\ \in\ R$$.
Also the equation $$f(x)=7x+a$$ has only one real distinct solution. The minimum value of $$f(x)$$ in $$\left[0,\dfrac{3}{2}\right]$$ is equal to
Report Question
0%
$$\dfrac{-33}{8}$$
0%
$$0$$
0%
$$4$$
0%
$$-2$$
If $$f\left( x \right) = (1 - x)$$ , $$x \in \left[ { - 3,3} \right]$$ , then the domain of $$f\left( {f\left( x \right)} \right)$$ is
Report Question
0%
$$\left[ { - 2,3} \right]$$
0%
$$\left( { - 2,3} \right)$$
0%
$$\left[ { - 2,3]} \right.$$
0%
$$( - 2,3]$$
Explanation
$$f\left(x\right)=1-x$$
$$f\left(f\left(x\right)\right)=f\left(1-x\right)=1-1+x=x$$
And $$x\in\left[-3,3\right]$$
Domain of $$f\left(x\right)$$ is $$\left[-3,3\right]$$
Domain of $$f\left(f\left(x\right)\right)=$$Domain of $$f\left(1-x\right)$$
Domain of $$f\left(f\left(x\right)\right)=\left[1-3,3\right]=\left[-2,3\right]$$
If f(g(x))=5x+2 and g(x)=8x then f(x)=
Report Question
0%
$$ \frac{5}{8}x+2$$.
0%
$$ \frac{8}{5}x+2$$.
0%
$$ \frac{5}{8}x-2$$.
0%
8x-2
0%
5x-2
Let $$g\left( x \right) =1+x-\left[ x \right] $$ and $$f\left( x \right) =\begin{cases} -1,x<0 \\ 0,x=0 \\ 1,x>0 \end{cases}$$ Then for all $$x,f\left( g\left( x \right) \right) $$ is equal to (where $$\left[ . \right] $$ represents the greatest integer function)
Report Question
0%
$$x$$
0%
$$1$$
0%
$$f\left( x \right)$$
0%
$$g\left( x \right)$$
Let $$f:(2,3) \rightarrow (0,1)$$ be defined by $$f(x)=x-[x]$$ then $$f^{-1}(x)$$ equals
Report Question
0%
$$x-2$$
0%
$$x+1$$
0%
$$x-1$$
0%
$$x+2$$
If $$f(x)=\frac{x}{\sqrt{1-x^{2}}}$$ and g(x) = $$f(x)=\frac{x}{\sqrt{1+x^{2}}}$$ , then (fog)(x) =
Report Question
0%
$$f(x)=\frac{x}{\sqrt{1-x^{2}}}$$
0%
$$f(x)=\frac{x}{\sqrt{1+x^{2}}}$$
0%
$$x^{2}$$
0%
x
Explanation
Given:-
$$f{\left( x \right)} = \cfrac{x}{\sqrt{1 - {x}^{2}}}$$
$$g{\left( x \right)} = \cfrac{x}{\sqrt{1 + {x}^{2}}}$$
To find:-
$$fog{\left( x \right)} = ?$$
$$fog{\left( x \right)}$$
$$= f{\left( g{\left( x \right)} \right)}$$
$$= f{\left( \cfrac{x}{\sqrt{1 + {x}^{2}}} \right)}$$
$$= \cfrac{\left( \cfrac{x}{\sqrt{1 + {x}^{2}}} \right)}{\sqrt{1 - {\left( \cfrac{x}{\sqrt{1 + {x}^{2}}} \right)}^{2}}}$$
$$= \cfrac{\left( \cfrac{x}{\sqrt{1 + {x}^{2}}} \right)}{\sqrt{1 - \left( \cfrac{{x}^{2}}{1 + {x}^{2}} \right)}}$$
$$= \cfrac{\left( \cfrac{x}{\sqrt{1 + {x}^{2}}} \right)}{\sqrt{\left( \cfrac{1 + {x}^{2} - {x}^{2}}{1 + {x}^{2}} \right)}}$$
$$= \cfrac{\left( \cfrac{x}{\sqrt{1 + {x}^{2}}} \right)}{\left( \cfrac{1}{1 + {x}^{2}} \right)}$$
$$= \cfrac{x}{\sqrt{1 + {x}^{2}}}$$
$$\Rightarrow fog{\left( x \right)} = \cfrac{x}{\sqrt{1 + {x}^{2}}}$$
Hence the correct answer is $$\cfrac{x}{\sqrt{1 + {x}^{2}}}$$.
Let $$f:X\rightarrow Y$$ be an invertible function. Then f has unique inverse.
Report Question
0%
True
0%
False
If $$f:R \rightarrow R, f(x)=2x-1$$ and $$g; R \rightarrow R, g(x)=x^{2}+2$$, then $$(gof)(x)$$ equals-
Report Question
0%
$$2x^{2}-1$$
0%
$$(2x-1)^{2}$$
0%
$$2x^{2}+3$$
0%
$$4x^{2}-4x+3$$
Explanation
$$f(x)=2x-1$$
$$g(x)=x^2+2$$
$$g(f(x))=(2x-1)^2+2=4x^2-4x+3$$.
Let $$f(x)=\dfrac{1}{x^{2}}$$ for $$x\ge 1$$, and $$g(x)$$ is its reflection in the line mirror $$y=x$$, then function $$h(x)=\begin{cases} f\left( x \right) & x\ge 1 \\ g\left( x \right) & 0<x<1 \end{cases}$$, is
Report Question
0%
derivable at $$x=1$$
0%
continuous at $$x=1$$
0%
not derivable at $$x=1$$
0%
not continuous at $$x=1$$
Explanation
Let $$\left(t, \dfrac{1}{t^2} \right)$$ be a point on f(x)
$$\therefore $$ reflection in line mirror y = x
$$\therefore t = \gamma$$
$$\therefore \dfrac{\gamma - \dfrac{1}{t^2}}{1} = \dfrac{x - t}{-1} = -2 \dfrac{(-t + \dfrac{1}{t^2})}{2}$$
$$\therefore \gamma = \dfrac{1}{t^2} + t - \dfrac{1}{t^2}$$
$$\therefore \gamma = t$$
$$\therefore x = \dfrac{1}{t^2}$$
$$\therefore x = \dfrac{1}{y^2}$$
$$\therefore xy^2 = 1$$ $$=9(x)$$
$$\therefore h'(x) $$ at $$x = 1$$ for $$x \ge 1 = f'(x)$$
$$= \dfrac{-2}{x^3} = \dfrac{-2}{1} = -2$$
$$\therefore h'(x) $$ at x = 1 for $$x \le 1 = 9'(x)$$
$$= \dfrac{-2}{13} = -2$$
derivable at $$x = 1$$.
If $$f(x)=\begin{cases} x+1 & x\epsilon \left[ -1,0 \right] \\ { x }^{ 2 }+1 & x\epsilon \left( 0,1 \right) \end{cases}$$, then the value of $$\dfrac{f^{-1}(0)+f^{-1}(1)+f^{-1}(2)}{f(-1)+f(0)+f(1)}$$ is-
Report Question
0%
$$0$$
0%
$$1$$
0%
$$2$$
0%
$$\dfrac{1}{3}$$
Explanation
$$f(x) = \begin{cases} x +1 & x \in (-1, 0) \\ x^2 + 1 & x \in (0,1) \end{cases}$$
$$\therefore x = \begin{cases} f^{-1} (x + 1) & x \in [-1, 0] \\ f^{-1} (x^2 + 1) & x \in (0, 1)\end{cases}$$
$$f^{-1} (0) $$:- put $$x = -1$$
$$\therefore f^{-1} (0) = x = -1$$
$$\therefore f^{-1} (1) $$ put $$x = 0$$
$$f^{-1} = 0$$
$$f^{-1} (2)$$:- put $$x = 1$$
$$f^{-1} (2) = 1$$
$$\therefore f(0) = 1$$
$$\therefore f(1) = 1 + 1 = 2$$
$$\therefore f(-1) = -1 + 1 = 0$$
$$\therefore \dfrac{f^{-1} (0) + f^{-1} (1) + f^{-1} (2)}{f(-1) + f(0) + f(1)} = \dfrac{-1 + 0 + 1}{3}$$
$$= \dfrac{0}{3}$$
$$= 0$$.
The last three digits, if $$(12345956)_{10}$$ is expressed in binary system.
Report Question
0%
$$110$$
0%
$$210$$
0%
$$100$$
0%
$$010$$
Explanation
undefined
If $$ f ( x ) = \left( a - x ^ { n } \right) ^ { 1 / n }$$ where $$ a > 0$$ and } $$n$$ is a positive integer then$$( f o f ) ( x )$$ is
Report Question
0%
$$f ( x )$$
0%
x
0%
0
0%
1
Explanation
$$f(x)=(a-x^n)^{1/n}$$
$$\therefore f(f(x))=(a-(f(x))^n))^{1/n}$$
$$\therefore f(f(x))=(a-((a-x^n)^{1/n)^n)^{1/n}}$$
$$\therefore f(f(x))=(a-((a-x^n)))^{1/n}$$
$$\therefore f(f(x))=(a-a+x^n)^{1/n}$$
$$\therefore f(f(x))=(x^n)^{1/n}$$
$$\therefore(f(x))=x$$
if $$f\left( x \right) = \log \left( {\dfrac{{1 +x}}{{1 - x}}} \right)$$ and $$g\left( x \right) = \dfrac{{3x + {x^3}}}{{1 + 3{x^2}}}$$ then $$\left( {f(g(x)))} \right)$$ is equal to
Report Question
0%
$$ - f\left( x \right)$$
0%
$$3f\left( x \right)$$
0%
$${\left( {f\left( x \right)} \right)^3}$$
0%
$$f\left( {3x} \right)$$
Explanation
$$f(x)=\log\bigg(\dfrac{1+x}{1-x}\bigg)\ and\ g(x)=\dfrac{3x+x^{3}}{1+3x^{2}}$$
$$f(g(x))=\log\bigg(\dfrac{1+\dfrac{3x+x^{3}}{1+3x^{2}}}{1-\dfrac{3x+x^{3}}{1+3x^{2}}}\bigg)$$
$$f(g(x))=\log\bigg(\dfrac{\dfrac{1+3x^{2}+3x+x^{3}}{1+3x^{2}}}{\dfrac{1+3x^{2}-3x-x^{3}}{1+3x^{2}}}\bigg)$$
$$f(g(x))=\log\bigg(\dfrac{1+3x^{2}+3x+x^{3}}{1+3x^{2}-3x-x^{3}}\bigg)$$
$$f(g(x))=\log\bigg(\dfrac{(1+x)^3}{(1-x)^{3}}\bigg)$$
$$f(g(x))=\log\bigg(\dfrac{1+x}{1-x}\bigg)^{3}$$
$$f(g(x))=3\log\bigg(\dfrac{1+x}{1-x}\bigg)$$
$$f(g(x))=3f(x)$$
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 12 Commerce Maths Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
