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CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 15 - MCQExams.com
CBSE
Class 12 Commerce Maths
Continuity And Differentiability
Quiz 15
if the function $$f(x)=\begin{cases} a+{\sin}^{-1}(x+b),\,x\geq 1 \\ x,\,x<1 \end{cases}$$ is differentiable at $$x=1$$, then $$\displaystyle \frac{a}{b}$$ is equal to
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$$1$$
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$$0$$
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$$-1$$
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$$2$$
The function $$f(x) = \dfrac {x}{1 + |x|}$$ is differentiable at which of the following?
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Every where
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Everywhere except at $$x = 1$$
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Everywhere except at $$x = 0$$
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Everywhere except at $$x = 0$$ or $$1$$
If $$y=e^{\sin^{-1}(t^{2}-1)}$$ & $$x=e^{\sec^{-1}\left (\frac{1}{t^{2}-1}\right)}$$, then $$\dfrac{dy}{dx}$$ is equal to
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$$\frac{x}{y}$$
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$$\dfrac{-y}{x}$$
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$$\dfrac{y}{x}$$
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$$\dfrac{-x}{y}$$
If $$f(x)={ sin }^{ -1 }\left[ \dfrac { 2x }{ 1+{ x }^{ 2 } } \right] $$,then $$f(x)$$ is differentiable on
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[-1,1]
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R-{-1,1}
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R-(-1,1)
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None of these
f(X)=|x|+|x-1| is continuous at
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'0' only
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0,1 only
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Every where
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No where
Explanation
$$\textbf{Step-1: Finding values of}$$ $$\mathbf{f(x)}$$
$$f(x)=|x|+|x-1|$$
$$f(x)=-x-|x-1|$$ $$\text{at}$$ $$x\leq{0}$$
$$=x-(x-1)$$ $$\text{at}$$ $$0\leq{x}<1$$
$$=x+(x-1)$$ $$\text{at}$$ $$x\geq{1}$$
$$\text{Hence, we get}$$
$$f(x)=1-2x$$ $$\text{at}$$ $$x\leq{0}$$
$$=1$$ $$\text{at}$$ $$0\leq{x}<1$$
$$=2x-1$$ $$\text{at}$$ $$x\geq{1}$$
$$\textbf{Step-2: Finding limits of the function at constraint values}$$
$$\text{At}$$ $$x=0,$$
$$\lim_{x\to 0} f(x)=1$$
$$\text{Hence,}$$ $$f(x)$$ $$\text{is continuous at}$$ $$x=0$$
$$\text{At}$$ $$x=1,$$
$$\lim_{x\to 1} f(x)=1$$
$$\text{Hence,}$$ $$f(x)$$ $$\text{is continuous at}$$ $$x=1$$
$$\text{Hence,}$$ $$f(x)$$ $$\text{is continuous everywhere}$$
$$\textbf{Hence,Correct option is (C)}$$
If $$\displaystyle f_{n}(x) = e^{f_{n-1}(x)} $$ for all $$\displaystyle n \,\epsilon \,N $$ and $$ f_{0} (x) = x $$ then $$ \dfrac{d}{dx} \left \{f_{n} (x)\right \} $$ is equal to
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$$\displaystyle f_{n}(x) \,.\dfrac{d}{dx}\left \{f_{n - 1} (x)\right \} $$
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$$\displaystyle f_{n}(x) \,.f_{n - 1}(x) $$
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$$\displaystyle f_{n}(x) \,.f_{n - 1}(x) ..... f_{2}(x)\,.f_{1} (x) $$
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$$\displaystyle \prod_{i = 1}^{n} f_{i}(x) $$
Explanation
Hence, Option (A,C,D) is the correct answer.
The order of the differential equation of all circles whose radius is $$4$$, is?
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$$1$$
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$$2$$
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$$3$$
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$$4$$
Explanation
Circle with radius $$4$$ is represented as follows:-
$$(x - h)^2 + (y - k)^2 = 4^2$$ ..... $$(1)$$
Differentiating w.r.t $$x$$
$$2(x - h) + 2(y - k) \dfrac{dy}{dx} = 0$$
$$\Rightarrow\ (x - h) + (y - k) \dfrac{dy}{dx} = 0$$ .....$$(2)$$
Again differentiating w.r.t $$x$$
$$1 + (y - k)\dfrac{d^2y}{dx^2} + \left (\dfrac{dy}{dx} \right )^2 = 0$$
$$\Rightarrow\ (y - k) = \dfrac{-\left (\dfrac{dy}{dx} \right )^2 - 1}{\dfrac{d^2y}{dx^2}} $$ ....$$(3)$$
From $$(2)$$ we get
$$\Rightarrow\ (x - h) = - (y - k) \dfrac{dy}{dx} $$
$$\Rightarrow\ (x - h) = - \left \{ \dfrac{\left (\dfrac{dy}{dx} \right )^2 - 1 }{\dfrac{d^2y}{dx^2}} \right \} \dfrac{dy}{dx} $$
$$\therefore\ (x - h) = \dfrac{\left (\dfrac{dy}{dx} \right )^3 + \left (\dfrac{dy}{dx} \right )}{\dfrac{d^2 y}{dx^2}} $$ .... $$(4)$$
Substituting $$(4)$$ and $$(3)$$ in eq. $$(1)$$
$$\left (\dfrac{\left(\dfrac{dy}{dx} \right )^3 + \dfrac{dy}{dx}}{\dfrac{d^2y}{dx^2}} \right )^2 + \left (\dfrac{-\left (\dfrac{dy}{dx} \right )^2 - 1}{\dfrac{d^2y}{dx^2}} \right )^2 = 4^2$$
$$ \left [\left (\dfrac{dy}{dx} \right )^3 + \dfrac{dy}{dx} \right ]^2 + \left [\left (\dfrac{dy}{dx} \right )^2 + 1 \right ]^2 = 16 \dfrac{d^2y}{dx^2} $$
$$\therefore$$ Order of D.E of circle whose radius is $$4$$ is $$2$$.
If $$f\left( x \right) = {\left| x \right|^{\left| {\sin x} \right|}}$$, then $${f'}\left( { - \dfrac{\pi }{4}} \right)$$ is equals
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$${\left( {\dfrac{\pi }{4}} \right)^{1/\sqrt 2 }}\left( { - \dfrac{{\sqrt 2 }}{2}{\text{ln}}\dfrac{4}{\pi } - \dfrac{{2\sqrt 2 }}{\pi }} \right)$$
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$${\left( {\dfrac{\pi }{4}} \right)^{1/\sqrt 2 }}\left( {\dfrac{{\sqrt 2 }}{2}{\text{ln}}\dfrac{4}{\pi } + \dfrac{{2\sqrt 2 }}{\pi }} \right)$$
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$${\left( {\dfrac{\pi }{4}} \right)^{1/\sqrt 2 }}\left( {\dfrac{{\sqrt 2 }}{2}{\text{ln}}\dfrac{\pi }{4} + \dfrac{{2\sqrt 2 }}{\pi }} \right)$$
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None of these.
Let $$f:(-1,1)\rightarrow R$$ be a differentiable function satisfying
$$(f'(x))^4=16(f(x))^2$$ for all $$x\in (-1,1)$$
$$f(0)=0$$
The number of such functions is
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$$2$$
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$$3$$
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$$4$$
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more than $$4$$
Explanation
Given:$${\left({f}^{\prime}{\left(x\right)}\right)}^{4}=16{\left(f{\left(x\right)}\right)}^{2}$$ for all $$x\in\,\left(–1 , 1\right)$$
$$f\left(0\right)=0$$
$$\Rightarrow\, {\left({f}^{\prime}{\left(x\right)}\right)}^{2}=\pm\,4f\left(x\right)$$
$$\Rightarrow\, {f}^{\prime}{\left(x\right)}=\pm\,2\sqrt{\pm\,f\left(x\right)}$$
$$(i)$$Case $$1$$
$${f}^{\prime}{\left(x\right)}=2\sqrt{\pm\,f\left(x\right)}$$
$$\displaystyle\int{\dfrac{d\left(f\left(x\right)\right)}{f\left(x\right)}}=\displaystyle\int{2\,dx}$$
$$\Rightarrow\,2\sqrt{f\left(x\right)}=2x+c$$
$$\Rightarrow\,f\left(0\right)=0\Rightarrow\,c=0$$
$$\Rightarrow\,\sqrt{f\left(x\right)}=x$$
$$\Rightarrow\,x\ge 0$$
$$f\left(x\right)={x}^{2},\,\,0\le x<1$$
$$(ii)$$Case $$2$$
$${f}^{\prime}{\left(x\right)}=-2\sqrt{f\left(x\right)}$$
$$\Rightarrow\,\sqrt{f\left(x\right)}=-x\Rightarrow\,x\le 0$$
$$f\left(x\right)={x}^{2};\,\,\,-1<x\le 0$$
$$(iii)$$Case $$3$$
$${f}^{\prime}{\left(x\right)}=2\sqrt{-f\left(x\right)}$$
$$\sqrt{-f\left(x\right)}=x$$
$$-f\left(x\right)={x}^{2}$$
$$f\left(x\right)=-{x}^{2};\,\,\,0\le x< 1$$
$$(iv)$$Case $$4$$
$${f}^{\prime}{\left(x\right)}=-2\sqrt{-f\left(x\right)}$$
$$\sqrt{-f\left(x\right)}=-x$$
$$f\left(x\right)=-{x}^{2};\,\,\,-1< x\le 0$$
$$(v)$$Also, one singular solution of given differential equation is
$$f\left(x\right)=0,\,\,-1<x<1$$
Hence, there are more than $$4$$ function possible
$${f}_{1}\left( x \right) =\begin{cases} {x}^{2};0\le x< 1 \\ {-x}^{2}; -1<x<0 \end{cases}$$
$${f}_{2}\left( x \right) =\begin{cases} -{x}^{2};0\le x< 1 \\ {x}^{2}; -1<x<0 \end{cases}$$
$${f}_{3}\left( x \right) = {x}^{2},\,-1<x<1$$
$${f}_{4}\left( x \right) =-{x}^{2},\,-1<x<1$$
$${f}_{5}\left( x \right) =0,\,-1<x<1$$...
Hence there are more than $$4$$ solutions
If $$\displaystyle y = \dfrac{\sqrt{(1 + t^{2})} - \sqrt{(1 - t^{2})}}{\sqrt{(1 + t^{2})} +\sqrt{(1 - t^{2})}} $$ and $$\displaystyle x = \sqrt {(1 - t^{4})} $$ , then $$ \dfrac{dy}{dx} $$
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$$\displaystyle \dfrac{-1}{t^{2}\left \{1 +\sqrt{1 - t^{4}} \right \}} $$
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$$\displaystyle \dfrac{\left \{\sqrt{(1 - t^{4})} - 1\right \}}{t^{6}} $$
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$$\displaystyle \dfrac{1}{t^{2}\left \{1 + \sqrt{(1 - t^{4})}\right \}} $$
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$$\displaystyle \dfrac{1 - \sqrt{( 1 - t^{4})}}{t^{6}} $$
f(x) is diffrentiable function and (f(x). g(x)) is differentiable a x=a , then
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g(x) must be differentiable at x=a
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if g(x) is discontinuous , then f(a) =0
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f(a) $$\neq $$ 0 then g(x) must be differentiable
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nothing can be said
Give that f(x) =xg(x) /$$ \left | x \right | $$ , g(0) = 0 and f(x) is continous at x=Then the value of f' (0)
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Does not exist
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is -1
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is 1
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is 0
Let f(x)=
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f' is differentiable
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f is differentiable
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f' is continuous
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f is continuous
If $$f(x) = \left\{\begin{matrix} \dfrac{x\log \cos x}{\log(1+x^2)}, & x \neq 0\\ 0, & x=0\end{matrix}\right.$$ then
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f(x) is not continuous at x=0.
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f(x) is continuous at x=0.
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f(x) is continuous at x=0 but not differentiable at x=0.
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f(x) is differentiable at x=0.
Let f: $$ R\rightarrow R $$ be a function such that f(x+y)= f(x)+f(y),$$ \forall $$ x,y$$ \epsilon R $$. If f(x) is differentiable at x=0, then
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f(x) is differentiable only in a finite interval containing zero
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f(x) is continuous $$ \forall x\epsilon R $$
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f(x) is constant $$ \forall x\epsilon R $$
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f(x) is differentiable except at finitely many points
Let $$f(x)$$ be a function satisfying $$f(x+y) = f(x)+f(y)$$ and $$f(x) = xg(x)$$ $$\forall x, ~y \in$$
R, where $$g(x)$$ is a continuous function then, which of the following is true?
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$$f'(x)=g'(0)$$
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$$f'(x)=g (0)$$
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$$f(x)=g (0)$$
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$$f(x)=g'(0)$$
Which of the following is differentiable at x= 0
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cos $$ \left ( \left | x \right | \right )+\left | x \right | $$
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cos $$ \left ( \left | x \right | \right )-\left | x \right | $$
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sin$$ \left ( \left | x \right | \right )+\left | x \right | $$
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$$sin \left ( \left | x \right | \right )-\left | x \right |$$
$$\cos |x|$$ is differentiable everywhere.
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True
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False
If the function $$f:[0,8] \rightarrow R$$ is differentiable, then for$$0<a, b<2, \int_{0}^{8} f(t) d t$$ is equal to
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$$3\left[\alpha^{3} f\left(\alpha^{2}\right)+\beta^{2} f\left(\beta^{2}\right)\right]$$
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$$3\left[\alpha^{3} f(\alpha)+\beta^{3} f(\beta)\right]$$
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$${3}\left[\alpha^{2} f\left(\alpha^{3}\right)+\beta^{2} f\left(\beta^{3}\right)\right]$$
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$$3\left[\alpha^{2} f\left(\alpha^{2}\right)+\beta^{2} f\left(\beta^{2}\right)\right]$$
Explanation
$$\operatorname{Let} g(x)=\int_{0}^{x^{3}} f(t) d t\\$$
$$ \text { Now } \int_{0}^{8} f(t) d t=g(2) =\dfrac{g(2)-g(1)}{2-1}+\dfrac{g(1)-g(0)}{1-0} \\$$
$$=g^{\prime}(\alpha)+g^{\prime}(\beta) \\$$
$$=3\left[\alpha^{2} f\left(\alpha^{3}\right)+\beta^{2} f\left(\beta^{3}\right)\right] $$
Let f(x) and g(x) be differentiable for $$0\leq x\leq 1$$, such that f(0) such that f'(c)=2g'(c), then the value of g(1) must be
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1
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3
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-2
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-1
$$ f(x)=x^{2}+x g^{\prime}(1)+g^{\prime \prime}(2) $$ and $$ g(x)=f(1) x^{2}+x f^{\prime}(x)+f^{\prime \prime}(x) $$
The value of $$ g(0) $$ is
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0
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-3
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2
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None of these
Explanation
$$ f(x)=x^{2}+x g^{\prime}(1)+g^{\prime \prime}(2) $$ and $$ g(x)=f(1) x^{2}+x f^{\prime}(x)+f^{\prime \prime}(x) $$
$$\text { Here put } g^{\prime}(1)=a, g^{\prime \prime}(2)=b \quad\quad\quad(1)$$
$$\text { Then } f(x)=x^{2}+a x+b, f(1)=1+a+b \Rightarrow f^{\prime}(x)=2 x+a$$
$$\quad f^{\prime \prime}(x)=2 $$
$$\therefore g(x)=(1+a+b) x^{2}+(2 x+a) x+2=x^{2}(3+a+b)+a x\quad+2$$
$$\Rightarrow g^{\prime}(x)=2 x(3+a+b)+a \text { and } g^{\prime \prime}(x)=2(3+a+b)$$
$$\text { Hence, } g^{\prime}(1)=2(3+a+b)+a \quad\quad\quad(2)$$
$$g^{\prime \prime} (2)=2(3+a+b)\quad\quad\quad(3)$$
From (1),(2) and (3) we have
$$ a=2(3+a+b)+a $$
$$\text { and } b=2(3+a+b) $$
$$\Rightarrow 3+a+b=0 \text { and } b+2 a+6=0$$
Hence, $$ b=0 $$ and $$ a=-3 . $$ So, $$ f(x)=x^{2}-3 x $$ and $$ g(x)=-3 x+2 $$
$$ g(x)=-3 x+2 \Rightarrow g(0)=2 $$
$$f(x)$$ is not invertible for
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$$x \in\left[-\dfrac{\pi}{2}-\tan ^{-1} 2, \dfrac{\pi}{2}-\tan ^{-1} 2\right]$$
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$$x \in\left[\tan ^{-1} \dfrac{1}{2}, \pi+\tan ^{-1} \dfrac{1}{2}\right]$$
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$$x \in\left[\pi+\cot ^{-1} 2,2 \pi+\cot ^{-1} 2\right]$$
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None of these
Explanation
$$\begin{array}{l}f(x)=\sin x+\sin x \int_{-\pi / 2}^{\pi / 2} f(t) d t+\cos x \int_{-\pi / 2}^{\pi / 2} t f(t) d t \\ \qquad \begin{aligned}=\sin x\left(1+\int_{-\pi / 2}^{\pi / 2} f(t) d t\right)+\cos x \int_{-\pi / 2}^{\pi / 2} t(t) d t & \\=& A \sin x+B \cos x \\ \text { Thus, } A=& 1+\int_{-\pi / 2}^{\pi / 2} f(t) d t \\=& 1+\int_{-\pi / 2}^{\pi / 2}(A \sin t+B \cos t) d t \\=& 1+2 B \int_{0}^{\pi / 2} \cos t d t \\ &=\int_{-\pi / 2}^{\pi / 2} t(A \sin t+B \cos t) d t \\ &=1+2 B \\ = & 2 A \int_{0}^{\pi / 2} t f(t) d t \end{aligned}\end{array} \\ $$
$$\begin{aligned}&=2 A[-t \cos t+\sin t]_{0}^{\pi / 2} \\\Rightarrow \quad B &=2 A\end{aligned}\\$$
From equations (1) and $$(2),$$ we get
$$A=-1 / 3, B=-2 / 3 \\$$
$$\Rightarrow f(x)=-\dfrac{1}{3}(\sin x+2 \cos x)\\$$
Thus, the range of $$f(x)$$ is $$\left[-\dfrac{\sqrt{5}}{3}, \dfrac{\sqrt{5}}{3}\right]\\$$
$$\begin{aligned}f(x) &=-\dfrac{1}{3}(\sin x+2 \cos x) \\&=-\dfrac{\sqrt{5}}{3} \sin \left(x+\tan ^{-1} 2\right) \\&=-\dfrac{\sqrt{5}}{3} \cos \left(x-\tan ^{-1} \dfrac{1}{2}\right)\end{aligned}\\$$
$$f(x)$$ is invertible if $$-\dfrac{\pi}{2} \leq x+\tan ^{-1} 2 \leq \dfrac{\pi}{2}\\$$
$$\Rightarrow-\dfrac{\pi}{2}-\tan ^{-1} 2 \leq x \leq \dfrac{\pi}{2}-\tan ^{-1} 2\\$$
or $$0 \leq x-\tan ^{-1} \dfrac{1}{2} \leq \pi\\$$
$$\Rightarrow \tan ^{-1} \dfrac{1}{2} \leq x \leq \pi+\tan ^{-1} \dfrac{1}{2}\\$$
or $$\pi \leq x-\tan ^{-1} \dfrac{1}{2} \leq 2 \pi\\$$
$$\Rightarrow x \in\left[\pi+\cot ^{-1} 2,2 \pi+\cot ^{-1} 2\right]\\$$
Let $$f(0,\infty)\rightarrow R$$ be a differentiable function such that $$f'(x)=2-\dfrac{f(x)}{x}$$ for all $$x\epsilon (0,\infty)$$ and $$f(1)\neq 1$$ Then
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$$\underset{x\rightarrow 0+}{lim}f'(\dfrac{1}{x})=1$$
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$$\underset{x\rightarrow 0+}{lim}xf'(\dfrac{1}{x})=2$$
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$$\underset{x\rightarrow 0+}{lim}x^2f'(\dfrac{1}{x})=0$$
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$$|f(x)|\le 2$$ for all $$x\epsilon(0,2)$$
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