MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 11 Commerce Applied Mathematics Functions Quiz 2 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Functions
Quiz 2
Let $$f(x)=\dfrac{Kx}{x+1}(x\neq -1)$$ then the value of $$K$$ for which $$(fof)(x)=x$$ is
Report Question
0%
$$1$$
0%
$$-1$$
0%
$$2$$
0%
$$\sqrt{2}$$
Explanation
$$f\left ( f\left ( x \right ) \right )=f\left ( \dfrac{kx}{x+1} \right )$$
$$=\dfrac{k\left ( \dfrac{kx}{x+1} \right )}{\dfrac{kx}{x+1}=1}=\dfrac{k^{2}x}{\left ( k+1 \right )x+1}$$
$$\therefore$$ for $$f\left ( f\left ( x \right ) \right )=x$$
$$\Rightarrow k+1=0$$
$$k^{2}=1$$
$$\therefore k=-1$$
If $$n (A) = 4$$ and $$n(B) = 6$$, then the number of surjections from $$A$$ to $$B$$ is
Report Question
0%
$$4^{6}$$
0%
$$6^{4}$$
0%
$$0$$
0%
$$24$$
Explanation
$$n(A)<n(B)$$
$$\therefore$$ Co-domain can never be equal to range.
$$\therefore$$ No of surjections is zero
$$f:\left ( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right )\rightarrow \left ( -\infty ,\infty \right )$$ defined by $$f(x)=1+3x$$ is
Report Question
0%
one-one but not onto
0%
onto but not one-one
0%
neither one - one nor onto
0%
bijective
Explanation
$$f(x_{1})=f(x_{2}) \Rightarrow 1+3x_{1}=1+3x$$
$$x_{1}=x_{2}$$
$$\therefore f(x_{1})=f(x_{2})\Rightarrow x_{1}=x_{2}$$
$$\therefore$$ f is one-one.
$$f$$ lies b/w $$(1-\dfrac{3\pi }{2},\dfrac{1+3\pi }{2})$$
$$\therefore$$ co-domain $$(-\infty ,\infty )$$ is not equal to range $$(\dfrac{1-3\pi }{2},\dfrac{1+3\pi }{2})$$
$$\Rightarrow$$ f is not onto.
$$\therefore$$ f is one-one but not onto.
Let $$A=\{1,2,3\}, B =\{a, b, c\}$$ and If $$f=\{(1,a),(2,b),(3,c)\}, g=\{(1,b),(2,a),(3,b)\}, h=\{(1,b)(2,c),(3,a)\}$$ then
Report Question
0%
$$g$$ and $$h$$ are injections
0%
$$f$$ and $$h$$ are injections
0%
$$f$$ and $$g$$ injections
0%
$$f,g$$ and $$h$$ are injections
Explanation
$$g(1)=g(3)=b \Rightarrow$$ g is not one-one.
domain of $$f =$$ domain of $$h = A.$$
range of $$f =$$ range of $$h = B$$
$$n(A)=n(B)=3$$
$$\therefore f$$ and $$h$$ are injections.
If $$f:R\rightarrow R, g:R\rightarrow R$$ are defined by $$f(x)=4x-1,g(x)=x^{3}+2,$$ then $$(gof)\left(\dfrac{a+1}{4}\right)=$$
Report Question
0%
$$43$$
0%
$$4a^3-1$$
0%
$$a^{3}+2$$
0%
$$64a^3 - 8a^{2}-1$$
Explanation
$$g\left ( f\left ( \dfrac{a+1}{4} \right ) \right )=g\left ( 4\left ( \dfrac{a+1}{4} \right )-1 \right )$$
$$=g\left ( a \right )$$
$$=a^{3}+2$$
If $$f(x)=2x+1$$ and $$g(x)=x^{2}+1$$ then $$ (go(fof))(2)=$$
Report Question
0%
$$112$$
0%
$$122$$
0%
$$12$$
0%
$$124$$
Explanation
$$(go(fof))\left( x \right) =g\left( f\left( f\left( x \right) \right) \right) $$
$$=g\left ( f\left ( 2x+1 \right ) \right )$$
$$=g\left ( 2\left ( 2x+1 \right )+1 \right )$$
$$=g\left ( 4x+3 \right )$$
$$=\left ( 4x+3 \right )^{2}+1$$
$$(go(fof))\left ( 2 \right )=\left ( 11 \right )^{2}+1=122$$
If $$f(x)=\dfrac{1}{x}, g(x)=\sqrt{x}$$ and $$ (go\sqrt{f})(16)=$$
Report Question
0%
$$2$$
0%
$$1$$
0%
$$\dfrac{1}{2}$$
0%
$$4$$
Explanation
$$\sqrt{f\left ( x \right )}=\dfrac{1}{\sqrt{x}}$$
$$g\left ( \sqrt{f\left ( x \right )} \right )=g\left ( \dfrac{1}{\sqrt{x}} \right )=\dfrac{1}{\sqrt{\sqrt{x}}}=\dfrac{1}{x^{\frac{1}{4}}}$$
$$\left ( go\sqrt{f} \right )(16)=\dfrac{1}{\left ( 16 \right )^{\dfrac{1}{4}}}=\dfrac{1}{2}$$
If $$f(x)=x, g(x)=2x^{2}+1$$ and $$h(x)=x+1$$ then $$(hogof)(x)$$ is equal to
Report Question
0%
$$x^{2}+2$$
0%
$$2x^{2}+1$$
0%
$$x^{2}+1$$
0%
$$2(x^{2}+1)$$
Explanation
Given
$$f(x)=x, g(x)=2x^{2}+1$$ and $$h(x)=x+1$$
$$hogof(x)$$
$$=h(g(f(x)))$$
$$=h\left ( g\left ( x \right ) \right )\dots\dots \left[ \because f\left ( x \right )=x \right ]$$
$$=h\left ( 2x^{2}+1 \right )\dots\dots\left [ \because g\left ( x \right )=2x^{2}+1 \right ]$$
$$=2x^{2}+1+1$$
$$=2x^2+2$$
$$=2\left (x ^{2} +1\right )$$
$$\therefore hogof(x)=2(x^2+1)$$
The number of injections possible from $$A=\{1,3,5,6\}$$ to $$B =\{2,8,11\}$$ is
Report Question
0%
$$8$$
0%
$$64$$
0%
$$2^{12}$$
0%
$$0$$
Explanation
co-domain $$n(B) <$$ domain $$n(A)$$
There can't be any one-one function from $$A$$ to $$B.$$
The number of possible surjection from $$A=\{1,2,3,...n\}$$ to $$B = \{1,2\}$$ (where $$n \geq 2)$$ is $$62$$, then $$n=$$
Report Question
0%
$$5$$
0%
$$6$$
0%
$$7$$
0%
$$8$$
Explanation
Each element in $$A$$ can be mapped onto any of two elements of $$B$$
$$\therefore$$ Total possible functions are $$2^{n}$$
For the $$f^{{n}'s}$$ to be surjections , they shouldn't be mapped alone to any of the two elements.
$$\therefore$$ Total no of surjections $$= 2^{n}-2$$
$$2^{n}-2=62$$
$$\Rightarrow n=6$$
If $$f:R\rightarrow R, g:R\rightarrow R$$ are defined by $$f(x)=x^{2}, g(x)=\cos x$$ then $$(gof)(x)=$$
Report Question
0%
$$\cos 2x$$
0%
$$x^{2}\cos x$$
0%
$$\cos x^{2}$$
0%
$$\cos^{2} x^{2}$$
Explanation
$$gof\left ( x \right )=g\left ( f\left ( x \right ) \right )=g\left ( x^{2} \right )$$
$$=\cos x^{2}$$
If $$f(x)=(1-x)^{1/2}$$ and $$g(x)= \ln(x)$$ then the domain of $$(gof)(x)$$ is
Report Question
0%
$$(-\infty ,2)$$
0%
$$(-1,1)$$
0%
$$(-\infty ,1]$$
0%
$$(-\infty ,1)$$
Explanation
Given $$f(x)=(1-x)^{\frac{1}{2}}$$ and $$g(x)=ln(x)$$
$$gof(x)$$
$$=g(f(x))$$
$$=\ln\left ( 1-x \right )^{1/2}$$
$$=\dfrac{1}{2}\ln\left ( 1-x \right )$$
$$\therefore$$ For the composite function to be defined $$1-x>0$$
$$x<1$$
$$\therefore$$ Domain is $$\left ( -\infty ,1 \right )$$
If the function is $$f:R\rightarrow R, g:R\rightarrow R$$ are defined as $$f(x)=2x+3, g(x)=x^{2}+7$$ and $$f[g(x)]=25$$ then $$x=$$
Report Question
0%
$$f(x)$$
0%
$$\pm 2$$
0%
$$\pm 3$$
0%
$$\pm 4$$
Explanation
$$f(g(x))=f(x^{2}+7) =25$$
$$=2(x^{2}+7)+3$$
$$=2x^{2}+17$$
$$\Rightarrow 2x^{2}=8$$
$$x^{2}=4 \Rightarrow x=\pm 2$$
If $$f(x)=\dfrac{x+1}{x-1}(x\neq 1)$$ then $$fofofof(x)=$$
Report Question
0%
$$f(x)$$
0%
$$2\left ( \dfrac{x+1}{x-1} \right )$$
0%
$$\dfrac{x-1}{x+1}$$
0%
$$x$$
Explanation
$$fofofof\left(x\right)=fofof\left(\dfrac{x+1}{x-1}\right)$$
$$=fof\left(\dfrac{\dfrac{x+1}{x-1}+1}{\dfrac{x+1}{x-1}-1}\right)=fof\left(\dfrac{2x}{2}\right)$$
$$=fof\left(x\right)$$
$$=f\left(\dfrac{x+1}{x-1}\right)=x$$
If $$F(n)=(-1)^{k-1}(n-1), G(n)=n-F(n)$$ then $$ (GoG)(n)=$$ (where $$k$$ is odd)
Report Question
0%
$$1$$
0%
$$n$$
0%
$$2$$
0%
$$n-1$$
Explanation
$$G(n)=n-(-1)^{k-1}(n-1)$$
$$GoG(n)=G(n-(-1)^{k-1}(n-1))$$
$$=n-(-1)^{k-1}(n-1)-(-1)^{k-1}((n-1)-(-1)^{k-1}(n-1))$$
$$=n-(n-1)$$
$$=1$$
If $$f(x)=\dfrac{x}{\sqrt{1-x^{2}}}$$, then $$ (fof)(x)=$$
Report Question
0%
$$\dfrac{x}{\sqrt{1-x^{2}}}$$
0%
$$\dfrac{x}{\sqrt{1-2x^{2}}}$$
0%
$$\dfrac{x}{\sqrt{1-3x^{2}}}$$
0%
$$x$$
Explanation
$$fof\left ( x \right )=f\left ( f\left ( x \right ) \right )=\left ( x/\sqrt{1-x^{2}} \right )/\left ( \sqrt{1-\left ( x^{2}/\left ( 1-x^{2} \right ) \right )} \right )$$
$$=\left ( x/\sqrt{1-x^{2}} \right )/\sqrt{\left ( 1-x^{2}-x^{2} \right )/\left ( 1-x^{2} \right )}$$
$$=\left ( x/\sqrt{1-x^{2}} \right )/\left ( \sqrt{1-2 x^{2}} /\sqrt{1-x^{2}}\right)$$
$$=x/\sqrt{1-2x^{2}}$$
If $$f:[1,\infty )\rightarrow B$$ defined by the function $$ f(x)=x^{2}-2x+6$$ is a surjection, then $$B$$ is equals to
Report Question
0%
$$[1,\infty )$$
0%
$$[5,\infty )$$
0%
$$[6,\infty )$$
0%
$$[2,\infty )$$
Explanation
$$f(x)=x^2-2x+6$$ is a surjection.
So the range of $$f(x)$$ will be equal to its codomain.
$$f(x)=x^2-2x+6$$
$$f^1(x)=2x-2$$
$$=2(x-1)$$
$$f(x)$$ will be increasing when $$x\geqslant 1$$.
$$\therefore f(1)=1-2+6$$
$$=5$$
$$\therefore B=[5, \infty)$$
If $$f:R\rightarrow R,f(x)=3x-2$$ then $$ (fof)(x)+2=$$
Report Question
0%
$$f(x)$$
0%
$$2f(x)$$
0%
$$3f(x)$$
0%
$$-f(x)$$
Explanation
$$fof (x)=f(f(x))$$
$$=f(3x-2)$$
$$=3(3x-2)-2$$
$$=9x-8$$
$$fof(x)+2=9x-6=3(3x-2)$$
$$=3f(x)$$
If $$f(x)=\dfrac{x}{\sqrt{1+x^{2}}}$$ then $$fofof(x)=$$
Report Question
0%
$$\dfrac{x}{\sqrt{1+3x^{2}}}$$
0%
$$\dfrac{x}{\sqrt{1-x^{2}}}$$
0%
$$\dfrac{2x}{\sqrt{1+2x^{2}}}$$
0%
$$\dfrac{x}{\sqrt{1+x^{2}}}$$
Explanation
$$fofof\left(x\right)=fof\left(\dfrac{x}{\sqrt{1+x^{2}}}\right)$$
$$=f\left(\dfrac{\dfrac{x}{\sqrt{1+x^{2}}}}{\sqrt{1+\dfrac{x^{2}}{1+x^{2}}}}\right)$$
$$=f\left(\dfrac{x}{\sqrt{1+2x^{2}}}\right)$$
$$=\left(\dfrac{\dfrac{x}{\sqrt{1+2x^{2}}}}{\sqrt{1+\dfrac{x^{2}}{1+2x^{2}}}}\right)$$
$$=\dfrac{x}{\sqrt{1+3x^{2}}}$$
If $$f(n+1)=f(n)$$ for all $$n\in N, f(7)=5$$ then $$f(35)=$$
Report Question
0%
$$25$$
0%
$$49$$
0%
$$35$$
0%
$$5$$
If $$f(x)=\displaystyle \dfrac{x}{\sqrt{1-x^{2}}},g(x)=\displaystyle \dfrac{x}{\sqrt{1+x^{2}}} $$, then $$(fog)(x)=$$
Report Question
0%
$$x$$
0%
$$\dfrac{x}{\sqrt{1+x^{2}}}$$
0%
$$\sqrt{1+x^{2}}$$
0%
$$2x$$
Explanation
$$fog\left(x\right)=f\left(g\left(x\right)\right)$$
$$\displaystyle =f\left(\dfrac{x}{\sqrt{1+x^{2}}}\right)$$
$$\displaystyle =\dfrac{\dfrac{x}{\sqrt{1+x^{2}}}}{\sqrt{1-\dfrac{x^{2}}{1+x^{2}}}}=x$$
The domain of $$\sqrt{\log_{e/3}\mathrm{x}+1}$$ is
Report Question
0%
$$\left(0,\displaystyle \frac{3}{\mathrm{e}}\right]$$
0%
$$\left(-\displaystyle \infty\frac{3}{\mathrm{e}}\right]$$
0%
$$\left[\displaystyle \frac{3}{\mathrm{e}}\infty\right)$$
0%
$$R$$
Explanation
$$\log_{e/3}{x}+1\geq 0$$ and $${x}>0$$
$$\log_{e/3}x\geq-1$$
$$\Rightarrow x\leq \left(\dfrac{e}{3}\right)^{-1}= \dfrac{3}{e}$$
$$\Rightarrow x\in \left(0, \dfrac{3}{e}\right]$$
Hence, option 'A' is correct.
The domain of $${ f }({ x })=\tan ^{ -1 } (5x) $$ is
Report Question
0%
$$(-\infty,\infty)$$
0%
$$(0,\infty)$$
0%
$$(-\infty,0)$$
0%
$$\left(\displaystyle -\frac{1}{5},\frac{1}{5}\right)$$
Explanation
$$\tan^{-1}x$$ is defined for all $$x$$ as range of $$\tan x$$ is from $$(-\infty ,\infty )$$.
$$\therefore$$ domain of $$\tan^{-1}5x$$ is $$(-\infty ,\infty )$$.
Hence, option 'A' is correct.
The domain of $$ \displaystyle f(x)= { \sin }^{ -1 }\left(\dfrac { 2x-3 }{ 5 } \right)$$ is
Report Question
0%
$$[1,3]$$
0%
$$[1,4]$$
0%
$$[2, 14]$$
0%
$$[-1,4]$$
Explanation
$$\sin^{-1}x$$ is defined if $$x\in [-1,1]$$
$$\Rightarrow -1\leq \dfrac{2x-3}{5}\leq 1$$
$$-5\leq 2x-3\leq 5$$
$$-2\leq 2x\leq 8$$
$$-1\leq x\leq 4$$
If $$f:R\rightarrow R$$ is defined by $$f(x)=x^{2}-10x+21 $$ then $$ f^{-1}(-3)$$ is
Report Question
0%
$$\left \{ -4,6 \right \}$$
0%
$$\left \{ 4,6 \right \}$$
0%
$$\left \{ -4, 4, 6 \right \}$$
0%
Not Invertible
Explanation
Let $$f^{-1}(-3)=t$$
$$\Rightarrow f(t)=-3$$
$$t^{2}-10t+21=-3$$
$$t^{2}-10t+24=0$$
$$t^{2}-6t-4t+24=0$$
$$t(t-6)-4(t-6)=0$$
$$\Rightarrow t=6,4 = f^{-1}(-3)$$
$$f:[-2,2]\rightarrow R $$ is defined as $$f(x)=\left\{\begin{array}{l}-1,-2\leq x\leq 0\\x-1,0\leq x\leq 2\end{array}\right.$$ then
$$\{x\in[-2,2]:x\leq 0,\; f(|x|)=x\}=$$
Report Question
0%
$$\{-1\}$$
0%
$$\{0\}$$
0%
$$\displaystyle \left\{\frac{-1}{2}\right\}$$
0%
$$\phi$$
The domain of $${f}({x})=\log|x^{2}-9|$$ is
Report Question
0%
$$\mathrm{R}- \{-3,3\}$$
0%
$$(-\infty,-3)$$
0%
$$(3,\infty)$$
0%
$$(-\infty,\infty)$$
Explanation
$$f$$ is defined for all $$[x^{2}-9]>0 \Rightarrow x\neq \pm 3$$
$$\therefore$$ domain of $$f$$ is $$R-\left \{ -3,3 \right \}$$
Hence, option 'A' is correct.
The domain of $$\displaystyle \mathrm{f}(\mathrm{x})=\cos^{-1}\left(\frac{2}{2+\sin \mathrm{x}}\right)$$ contained in $$[0,2\pi]$$ is
Report Question
0%
$$\left[0,\displaystyle \frac{\pi}{2}\right]$$
0%
$$\left[\displaystyle \frac{\pi}{2},\pi \right]$$
0%
$$[0, \pi]$$
0%
$$\left[\displaystyle \frac{-\pi}{2},\frac{\pi}{2}\right]$$
Explanation
Domain of the function is value of $$x$$ in $$\displaystyle {f}({x})=\cos^{-1}\left(\frac{2}{2+\sin {x}}\right)$$
Since value of $$\cos x$$ is between $$-1$$ to $$1$$ thus
$$-1\displaystyle \leq\frac{2}{2+\sin{x}}\leq 1$$
$$\Rightarrow \sin x\geq0$$
$$\Rightarrow x\in[0, \pi]$$
Hence, option 'C' is correct.
The domain of $$f(x)=\sqrt{\log\left ( \dfrac{7x-x^{2}}{12} \right )}$$ is
Report Question
0%
$$(-\infty ,\infty )$$
0%
$$(-\infty ,4]$$
0%
$$[3 ,\infty )$$
0%
$$[3,4]$$
Explanation
Domain of a function is value of $$x$$ for which function is defined
$$f(x)=\sqrt{\log\left ( \dfrac{7x-x^{2}}{12} \right )} $$
For log to be always positive, value into log should be greater than $$1$$ so
$$\displaystyle \frac{7{x}-{x}^{2}}{12}\geq 1$$
$$x^2-7x+12=0$$
So solving the inequality, we get
$$x\in[3,4]$$
Hence, option 'D' is correct.
The domain of the function $$\displaystyle \mathrm{f}({x})=\frac{1}{\sqrt{x-2}}+\frac{1}{\sqrt{5-x}}$$ is
Report Question
0%
$$[2,5]$$
0%
$$(2,5)$$
0%
$$[2,5 )$$
0%
$$(2,5 ]$$
Explanation
Given,
$$\displaystyle \mathrm{f}({x})=\frac{1}{\sqrt{x-2}}+\frac{1}{\sqrt{5-x}}$$
For the $$f(x)$$ to be defined.
$$x-2>0 \Rightarrow x>2$$
$$5-x>0 \Rightarrow x<5$$
$$\Rightarrow x\in (2,5)$$
The domain of $$\mathrm{f}({x})=\mathrm{c}\mathrm{o}\mathrm{t}^{-1} \left(\displaystyle \frac{x}{3}\right)$$ is
Report Question
0%
$$(-\infty,\infty)$$
0%
$$(0,\infty)$$
0%
$$(1,\infty)$$
0%
$$\left(\displaystyle -\frac{1}{3},\frac{1}{3}\right)$$
Explanation
$$\cot^{-1}$$ is defined for all $$x$$
$$\therefore$$ Domain of $$f$$ is $$(-\infty ,\infty )$$
The domain of $$\displaystyle \mathrm{f}({x})=\frac{1}{\sqrt{9-x^{2}}}+\sqrt{x^{2}-4}$$ is
Report Question
0%
$$(-4,-2)\cup (2,4)$$
0%
$$(-3,-2]\cup [2,3)$$
0%
$$(-\infty,-3)\cup (2,\infty)$$
0%
$$(-\infty,\infty)$$
Explanation
$$f$$ is defined if both terms under square root exist.
$$\Rightarrow 9-x^{2}> 0$$ and $$ x^{2}-4\geq 0$$
$$\Rightarrow (3-x)(3+x)>0$$ and $$(x-2)(x+2)\geq 0$$
$$\Rightarrow (x-3)(3+x)<0$$ and $$(x-2)(x+2)\geq 0$$
$$\Rightarrow -3<x<3$$ and $$x<-2$$ and $$ x>2 $$
$$x\in (-3,3)$$ and $$ x\in [-\infty ,-2]\cup [2,\infty ]$$
Intersection of both gives the following solution
$$\Rightarrow x\in (-3,-2]\cup [2,3)$$
The domain of $$\mathrm{f}({x})=\mathrm{e}^{\sqrt{{x}}}+\cos x$$ is
Report Question
0%
$$(-\infty,\infty)$$
0%
$$[0, \infty)$$
0%
$$(0, 1)$$
0%
$$(1, \infty)$$
Explanation
$$\cos x$$ is defined for all $$x$$
$$\therefore$$ For $$f(x)$$ to be defined.
$$\sqrt{x}\geq 0$$
$$\Rightarrow x\geq 0 \Rightarrow x\in [0,\infty )$$
Let $$S$$ be set of all rational numbers. The functions $$f:R\rightarrow R,\ g:R\rightarrow R$$ are
defined as
$$f(x)=\begin{cases}
0, & x \in S \\
1, & x \notin S
\end{cases}$$
$$g(x)=\begin{cases}
-1 & x\in S \\
0 & x\notin S
\end{cases}$$
then, $$(fog) (\pi)+(gof)(e)=$$
Report Question
0%
$$-1$$
0%
$$0$$
0%
$$1$$
0%
$$2$$
Explanation
$$f(x)=\begin{cases}
0, & x \text{ is rational} \\
1, & x \text{ is irrational}
\end{cases}$$
$$g(x)=\begin{cases}
-1 & x\text { is rational} \\
0& x\text { is irrational }
\end{cases}$$
$$fog(\pi )=f(g(\pi ))=f(0)=0$$
so $$gof(e)=g(f(e))=g(1)=-1$$
$$\therefore fog(\pi )+gof(e)=-1$$
The domain of $$f(x)= \text{cosec} x-\cot x$$ is
Report Question
0%
$$(-\infty,\infty)$$
0%
$$ R-\{n\pi$$ : $$n\in Z\}$$
0%
$$R-\displaystyle \{(2n+1)\frac{\pi}{2}$$ : $$n\in Z\}$$
0%
$$R-\displaystyle \left\{\frac{n\pi}{2}:n\in Z\right\}$$
Explanation
$$\text{cosec} x $$ & $$\cot x$$ are not defined at $$x=n\pi \ \ \forall n\in Z$$
$$\therefore$$ Domain of $$f$$ is $$R- \{ n\pi\} , n\in Z $$
The domain of $$\mathrm{f}({x})=\mathrm{c}\mathrm{o}\mathrm{s}^{-1} (\sqrt{3x})$$ is
Report Question
0%
$$[-1, 1]$$
0%
$$\left[0,\displaystyle \frac{1}{3}\right]$$
0%
$$[0,1]$$
0%
$$[0,3 ]$$
Explanation
For $$f$$ to be defined $$\sqrt{3x}$$ has to exist.
$$\Rightarrow 3x\geq 0$$
$$x\geq 0$$
$$\cos^{-1}$$ is defined only if $$-1\leq \sqrt{3x}\leq 1$$
As $$x\geq 0$$
$$\Rightarrow \sqrt{3x}\leq 1$$
$$\Rightarrow x\leq \dfrac{1}{3}$$
$$\therefore x\in \left[0,\dfrac{1}{3}\right]$$
lf $$f:[-6,6]\rightarrow \mathbb{R}$$ is defined by $$f(x)=x^{2}-3$$ for $$x\in \mathbb{R}$$ then
$$(fofof)(-1)+(fofof)(0)+(fofof)(1)=$$
Report Question
0%
$$f(4\sqrt{2})$$
0%
$$f(3\sqrt{2})$$
0%
$$f(2\sqrt{2})$$
0%
$$f(\sqrt{2})$$
Explanation
$$fofof(-1)=fof(-2)=f(1)=-2$$
$$fofof(1)=fof(-2)=f(1)=-2$$
$$fofof(0)=fof(-3)=f(0)=33$$
$$\therefore fofof(-1)+fofof(1)+fofof(0)=29$$
$$=32-3$$
$$=(4\sqrt{2})^{2}-3$$
$$\Rightarrow f(4\sqrt{2})=(4\sqrt{2})^{2}-3$$
The domain of $$\mathrm{f}({x})=\sqrt{x^{2}-9x+18}$$ is
Report Question
0%
$$[3,6]$$
0%
$$(-\infty,3]\cup[6,\infty)$$
0%
$$[3,\infty)$$
0%
$$[6, \infty)$$
Explanation
For given function to be defined, $$x^2-9x+18\ge 0$$
$$\Rightarrow (x-6)(x-3)\ge 0\Rightarrow x\in (-\infty,3]\cup [6,\infty) =$$ required domain
Hence option 'B' is correct
The domain of $$\mathrm{f}({x})=\sqrt{1-2x}+\cos^{-1}(1-2x)$$
Report Question
0%
$$\left(-\displaystyle \infty,\frac{1}{2}\right)$$
0%
$$\left(\displaystyle \frac{1}{2},\infty\right)$$
0%
$$\left[0,\displaystyle \frac{1}{2}\right]$$
0%
$$(0,\infty)$$
Explanation
$$f$$ is defined if $$\sqrt{1-2x}$$ exist $$\sum cos^{-1}(1-2x)$$ is define.
$$\Rightarrow 1-2x\geq 0 -1\leq 1-2x\leq 1$$
$$\Rightarrow x\leq \dfrac{1}{2} -2\leq -2x\leq 0$$
$$0\leq 2x\leq 2$$
$$ \Rightarrow 0\leq x\leq 1$$
$$\therefore$$ For $$x\in [0,\dfrac{1}{2}],$$ $$f$$ is defined.
The domain of the function $$\displaystyle \mathrm{f}({x})=\dfrac{\sqrt{2+x}+\sqrt{2-x}}{x}$$ is
Report Question
0%
$$(-2,2)$$
0%
$$[-2,0)\cup (0,2]$$
0%
$$[-2,2]$$
0%
$$(-\infty,2)$$
Explanation
Given,
$$\displaystyle \mathrm{f}({x})=\dfrac{\sqrt{2+x}+\sqrt{2-x}}{x}$$
For $$f(x)$$ to be defined
$$x\neq 0$$
$$2+x\geq 0 \Rightarrow x\geq -2$$
$$2-x\geq 0 \Rightarrow x\leq 2$$
$$\therefore x\in [-2,0)\cup (0,2]$$
lf $${f}\left({x}\right)=\sin^{2}{x}+\sin^{2}\left({x}+\displaystyle \dfrac{\pi}{3}\right)+ \cos x \cos \left({x}+\displaystyle \dfrac{\pi}{3}\right)$$ and $${g}\left(\displaystyle\dfrac{5}{4}\right)=1$$, $$g\left(1\right) = 0 $$ then $$\left({g}{o}{f}\right)\left({x}\right)=$$
Report Question
0%
$$1$$
0%
$$0$$
0%
$$\sin x$$
0%
Data is insufficient
Explanation
$$\displaystyle {f}\left({x}\right)=\sin^{2}{x}+\left(\sin{x}\cos\frac{\pi}{3}+ \cos{x} \displaystyle \sin\frac{\pi}{3}\right)^{2}+ {c}{o}{s} {x}\left(\displaystyle \cos {x}\cos\frac{\pi}{3}-\sin {x}\sin\frac{\pi}{3}\right)$$
$$\displaystyle =\sin ^{ 2 }{ x } +{ \left[ \frac { \sin{ x } }{ 2 } +\frac { \sqrt { 3 } \cos{ x } }{ 2 } \right] }^{ 2 }+\frac { \cos^{ 2 }{ x } }{ 2 } -\frac { \sqrt { 3 } }{ 2 } \cos x \sin x$$
$$\displaystyle =\sin ^{ 2 }{ x } +\frac { \sin ^{ 2 }{ x } }{ 4 } +\frac { 3 }{ 4 } \cos ^{ 2 }{ x } +\frac { \sqrt { 3 } }{ 2 } \sin x\cos{ x }+\frac { \cos ^{ 2 }{ x } }{ 2 } -\frac { \sqrt { 3 } }{ 2 } \sin { x\cos { x } } $$
$$=\displaystyle \frac{5}{4}\left(\sin^{2}{x}+\cos^{2}{x}\right)=\frac{5}{4}$$
$$\displaystyle \therefore $$ $$[{g}{o}{f}]\left({x}\right)={g}[{f}\left({x}\right)]={g}\left(\displaystyle \frac{5}{4}\right)=1$$
If $$n\geq 1$$ is any integer, $$\mathrm{d}(n)$$ denotes the number of positive factors of $$n$$, then for any prime number $$\mathrm{p},\ \mathrm{d}(\mathrm{d}(\mathrm{d}(\mathrm{p}^{7})))=$$
Report Question
0%
$$1$$
0%
$$2$$
0%
$$3$$
0%
$$4$$
Explanation
For a number being $$p^{n}$$ the total number of positive factors excluding $$1$$ and the number itself will be $$n-1$$.
Therefore total number of positive factors will be $$n-1+2$$
$$=n+1$$
Hence for $$p^{7}$$ we will have $$8$$ factors.
Now for $$8=2^{3}$$.
Hence $$8$$ will have $$4$$ factors.
Now $$4=2^{2}$$, hence the number of factors will be $$3$$.
Set of values of $$x$$ for which the function $$\displaystyle \mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}}+2^{\sin^{-1}\mathrm{x}}+\frac{1}{\sqrt{\mathrm{x}-2}}$$ exists is
Report Question
0%
$$R$$
0%
$$\mathrm{R}-\{\mathrm{0}\}$$
0%
$$\phi$$
0%
$$\mathrm{R}-\{1\}$$
Explanation
$$f(x)=2 \sin^{-1}x+\dfrac{1}{\sqrt{x-2}}+\dfrac{1}{x}$$
$$x\neq 0$$ for $$\dfrac{1}{x}$$
$$-1\leqslant x\leqslant1$$ for $$\sin^{-1}x$$
$$\sqrt{x-2}\neq 0$$ and $$ x-2>0$$ for $$\dfrac{1}{\sqrt {x-2}}$$
$$x>2$$
n combining, we get domain of $$f(x)$$ is $$\phi $$.
The domain of the function $$\displaystyle \mathrm{f}(\mathrm{x})=\frac{1}{\sqrt{|\mathrm{x}|-\mathrm{x}}}$$ is
Report Question
0%
$$(-\infty, \infty)$$
0%
$$(0, \infty)$$
0%
$$(-\infty, 0)$$
0%
$$(-\infty, \infty)-\{0\}$$
Explanation
Given function is $$\displaystyle \mathrm{f}(\mathrm{x})=\frac{1}{\sqrt{|\mathrm{x}|-\mathrm{x}}}$$
$$\Rightarrow$$
$$\mathrm{f}(\mathrm{x})$$ is defined if $$ |x| > x $$
$$= |\mathrm{x}|>\mathrm{x}$$
$$= \mathrm{x}<0$$
So domain of $$\mathrm{f}(\mathrm{x})$$ is $$(-\infty, 0)$$.
lf $$f$$ : $$R\rightarrow R$$ is defined by
$$f(x)=\left\{\begin{array}{l}x+4 & x<-4\\3x+2 & -4\leq x<4\\x-4 & x\geq 4\end{array}\right.$$
then the correct matching of list I to List II is.
List - I
List - II
$$\mathrm{A}) f(-5)+f(-4)=$$
$$\mathrm{i}) 14$$
$$\mathrm{B}) f(|f(-8)|)=$$
ii $$) 4$$
$$\mathrm{C}) f(f(-7)+f(3))=$$
$$\mathrm{i}\mathrm{i}\mathrm{i})-11$$
$$\mathrm{D}) f(f(f(f(0)))+1=$$
$$\mathrm{i}\mathrm{v})-1$$
v) $$1$$
vi) $$0$$
Report Question
0%
A-iii , B-vi , C-ii , D- v
0%
A-iii , B-iv , C-ii , D- vi
0%
A-iv , B-iii , C-ii , D- i
0%
A-ii , B-vi , C-v , D- ii
Explanation
$$f(-5)=-5+4=-1 ; f(-4)=3(-4)+2=-10$$
$$\therefore f(-5)+f(4)=-11$$
$$f(|f(-8)|)=f((-8+4))=f(4)=4-4=0$$
$$f(f(-7)) tf(3) = f(-7+4)+3(3)+2$$
$$=3(-3)+2+3(3)+2$$
$$=4$$
$$f(f(f(f(0))))+1=f(f(f(2)))+1=f(f(8))+1$$
$$=f(4)+1=0+1=1$$
Set A has 3 elements and set B has 4 elements. The number of injections that can be defined from A into B is :
Report Question
0%
144
0%
12
0%
24
0%
64
Explanation
lf $$g(f(x)) =|\sin \mathrm{x}|,f(g(x)) =(\sin\sqrt{\mathrm{x}})^{2}$$, then
Report Question
0%
$${f}({x})=\sin^{2} {x},{g}({x})=\sqrt{{x}}$$
0%
$${f}({x})=\sin x,g({x})=|{x}|$$
0%
$${f}({x})={x}^{2},{g}({x})=\sin\sqrt{{x}}$$
0%
$${f}, {g}$$ cannot be determined
Explanation
$$g(f(x))=|\sin x|=\sqrt{\sin^{2}x}$$ .......(i)
$$f(g(x))=\sin^{2}\sqrt{x}$$ .......(ii)
Comparing $$i$$ and $$ii$$
$$\Rightarrow$$
$$g(f(x))=|\sin x|=\sqrt{\sin^{2}x}$$
$$\Rightarrow$$
$$g(\sin^{2}x)=\sqrt{\sin^{2}x}=g(f(x))$$
And
$$\Rightarrow$$
$$f(g(x))=\sin^{2}\sqrt{x}$$
$$\Rightarrow$$
$$f(\sqrt{x})=\sin^{2}\sqrt{x}=f(g(x))$$
Therefore
$$\Rightarrow$$
$$g(x)=\sqrt{x}$$
$$\Rightarrow$$
$$f(x)=\sin^{2}x$$
The domain of $$\sqrt{(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)}$$ is
Report Question
0%
$$(1,2)$$
0%
$$[1,2]$$
0%
$$(1,2)\cup (3, \infty)$$
0%
$$[1,2] \cup [3, \infty)$$
Domain of definition of the function $$\displaystyle{ f }({ x })=\sqrt { \sin ^{ -1 } (2{ x })+\frac { \pi }{ 6 } } $$ for real valued $$x$$, is
Report Question
0%
$$[-\displaystyle \frac{1}{4}, \displaystyle \frac{1}{2}]$$
0%
$$[-\displaystyle \frac{1}{2}, \displaystyle \frac{1}{2}]$$
0%
$$(-\displaystyle \frac{1}{2}, \displaystyle \frac{1}{9})$$
0%
$$[-\displaystyle \frac{1}{4}, \displaystyle \frac{1}{4}]$$
Explanation
Here $$\displaystyle f\left( x \right) =\sqrt { \sin ^{ -1 }{ \left( 2x \right) } +\frac { \pi }{ 6 } } $$ to find domain we must have
$$\displaystyle \sin ^{ -1 }{ \left( 2x \right) } +\frac { \pi }{ 6 } \ge 0$$ ( but $$\displaystyle -\frac { \pi }{ 2 } \le \sin ^{ -1 }{ \theta } \le \frac { \pi }{ 2 } $$)
As $$\displaystyle -\frac { \pi }{ 6 } \le \sin ^{ -1 }{ \left( 2x \right) } \le \frac { \pi }{ 2 } \\\Rightarrow \sin { \left( -\cfrac { \pi }{ 6 } \right) } \le 2x\le \sin { \left( \cfrac { \pi }{ 2 } \right) } $$
$$\displaystyle \Rightarrow -\cfrac { 1 }{ 2 } \le 2x\le 1\\\Rightarrow -\cfrac { 1 }{ 4 } \le x\le \cfrac { 1 }{ 2 } \\\Rightarrow x\in \left[ -\cfrac { 1 }{ 4 } ,\cfrac { 1 }{ 2 } \right] $$
Find the domain of $$ e^x$$.
Report Question
0%
N
0%
R
0%
Z
0%
All of the above
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 11 Commerce Applied Mathematics Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
