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CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 5 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Differentiation
Quiz 5
If P(x) is a polynomial such that $$P\left ( x^{2}+1 \right )=\left \{ P\left ( x^{2} \right ) \right \}^{2}+1$$ and $$P(0)=0$$ then $$P^{'}(0)$$ is equal to
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$$1$$
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$$0$$
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$$-$$1
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none of these
Explanation
Given, $$P\left ( x^{2}+1 \right )=\left \{ P\left ( x^{2} \right ) \right \}^{2}+1$$ ...(1)
Given $$P(0)=0$$
Put $$x=0$$ in (1)
$$\Rightarrow P(1)={P(0)}^{2}+1$$
$$\Rightarrow P(1)=1$$
Also, put $$x=1$$ in (1)
$$\Rightarrow P(2)=2$$
Also, since, P(x) is a polynomial so we have P(x)=x
$$\Rightarrow P'(x)=1$$
$$P'(0)=1$$
If $$\displaystyle y^{x}=x^{\sin y} $$, find $$\cfrac{dy}{dx}$$.
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$$\displaystyle \frac{y}{x}\left [ \frac{x \log y-\sin y}{y \:\log x\: \cos y-x} \right ]$$
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$$\displaystyle \frac{y}{x}\left [ \frac{x \log y+\sin y}{y \:\log x\: \cos y+x} \right ]$$
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$$\displaystyle \frac{-y}{x}\left [ \frac{x \log y-\sin y}{y \:\log x\: \cos y-x} \right ]$$
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$$\displaystyle \frac{y}{x}\left [ \frac{x \log y-\sin y}{y \:\log x\: \cos y+x} \right ]$$
Explanation
Given $$\displaystyle y^{x}\:=x^{\sin \:y}$$.
Take log both sides.
$$\displaystyle x \log y = \sin y \: \log \:x.$$
Differentiate w.r.t. x
$$\displaystyle \log y+x.\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \sin y+ (\log x)\cos y .\frac{dy}{dx}$$
$$\displaystyle \left ( \log y - \frac{\sin y}{x} \right )=\frac{dy}{dx}\left [ \cos y \log x-\frac{x}{y} \right ]$$
$$\displaystyle \therefore \frac{dy}{dx}=\frac{y}{x}\left [ \frac{x \log y-\sin y}{y \log x \cos y-x}\right ]$$
$$\displaystyle y=(\cot x)^{\sin x}+(\tan x)^{\cos x}$$.Find dy/dx
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$$\sin x(\cot x)^{ \sin x-1 }(-cosec ^{ 2 } x)+(\cot x)^{ \sin x }(log\cot x)\cos x+\\\cos { x } { (\tan { x } ) }^{ \cos { x-1 } }\sec ^{ 2 }{ x } +{ (\tan { x } ) }^{ cosx }(\log { tanx)(-sinx) } $$
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$$\sin x(\cot x)^{ \sin x-1 }(-co\sec ^{ 2 } x)+\cos { x } { (\tan { x } ) }^{ \cos { x-1 } }\sec ^{ 2 }{ x } $$
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$$\sin x(\cot x)^{ \sin x-1 }(-sec ^{ 2 } x)+(\cot x)^{ \sin x }(log\cot x)\cos x+\\\cos { x } { (\tan { x } ) }^{ \cos { x+1 } }co\sec ^{ 2 }{ x } +{ (\tan { x } ) }^{ cosx }(\log { tanx)(sinx) } $$
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None of these
Explanation
Given $$\displaystyle y=(\cot x)^{\sin x}+(\tan x)^{\cos x}$$
Let $$u=(\cot x)^{\sin x}$$
Taking log on both sides
$$\log u=\sin x \log(\cot x)$$
Differentiating w.r.t. x, we get,
$$\dfrac{1}{u}\dfrac{du}{dx}=\dfrac{\sin x}{\cot x}(-cosec^2 x)+\cos x \log \cot x$$
$$\Rightarrow \dfrac{du}{dx}=\sin x(\cot x)^{\sin x -1}(-cosec^2 x)+(\cot x)^{\sin x} \cos x \log \cot x$$
Now, $$v=(\tan x)^{\cos x}$$
Taking log on both sides
$$\log v=\cos x \log(\tan x)$$
Differentiating w.r.t. x, we get,
$$\dfrac{1}{v}\dfrac{dv}{dx}=\dfrac{\cos x}{\tan x}(\sec^2 x)-\sin x \log \tan x$$
$$\Rightarrow \dfrac{dv}{dx}=\cos x(\tan x)^{\cos x -1}(\sec^2 x)+(\tan x)^{\cos x}(-\sin x) \log \tan x$$
Since, $$y=u+v$$
$$\Rightarrow \dfrac{dy}{dx}=\dfrac{du}{dx}+\dfrac{dv}{dx}$$
$$\Rightarrow \dfrac{dy}{dx}=\sin x(\cot x)^{\sin x -1}(-cosec^2 x)+(\cot x)^{\sin x} \cos x \log \cot x+$$
$$ \cos x(\tan x)^{\cos x -1}(\sec^2 x)+(\tan x)^{\cos x}(-\sin x) \log \tan x$$
If $$\displaystyle x^{2}+y^{2}= 1$$ then $$\displaystyle \left ( where\ y'= \frac{dy}{dx}, y''= \frac{d^{2}y}{dx^{2}}\right )$$
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$$\displaystyle yy''-2y'^{2}+1= 0$$
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$$\displaystyle yy''+y'^{2}+1= 0$$
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$$\displaystyle yy''-y'^{2}-1= 0$$
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$$\displaystyle yy''+2y'^{2}+1= 0$$
Explanation
Given, $$\displaystyle x^{2}+y^{2}= 1$$
Differentiating w.r.t x,
$$2x+2yy'=0$$
$$\Rightarrow x+yy'=0$$
Again differentiating w.r.t. x, we get
$$1+yy''+y'y'=0$$
$$1+yy''+y'^{2}=0$$
$$\displaystyle \frac{d}{dx}(\log_{e}\left ( \frac{1+x}{1-x} \right )^{1/4}-\frac{1}{2}\tan^{-1}x.)$$
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$$\displaystyle \frac{x^{2}}{1-x^{4}}.$$
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$$\displaystyle \frac{x^{3}}{1-x^{4}}.$$
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$$\displaystyle \frac{x^{4}}{1-x^{4}}.$$
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$$-\displaystyle \frac{x^{2}}{1-x^{4}}.$$
Explanation
Let $$\displaystyle y=\frac{1}{4}\left [ \log\left ( 1+x \right )-\log\left ( 1-x \right ) \right ]-\frac{1}{2}\tan^{-1}x.$$
$$\Rightarrow \displaystyle
\frac{dy}{dx}=\frac{1}{4}\left [ \frac{1}{1+x}-\frac{1}{1-x}\left ( -1
\right ) \right ]-\frac{1}{2}.\frac{1}{1+x^{2}}$$
$$\displaystyle =\frac{1}{2\left ( 1-x^{2} \right )}-\frac{1}{2\left ( 1+x^{2} \right )}=\frac{x^{2}}{1-x^{4}}.$$
If $$\displaystyle x={e^{\displaystyle y+e^{\displaystyle y+e^{\displaystyle y+...\infty }}}}$$, $$\forall x> 0$$, then $$\displaystyle \frac{dy}{dx}=$$
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$$\displaystyle \frac{1-x}{x}$$
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$$\displaystyle \frac{1}{x}$$
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$$\displaystyle \frac{x}{1+x}$$
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$$\displaystyle \frac{1+x}{x}$$
Explanation
$$\displaystyle x=e^{y\displaystyle+e^{y\displaystyle+e^{y\displaystyle+...}}}$$
$$\Rightarrow \displaystyle x=e^{y+x}$$
Take logarithm on both sides and differentiating with respect to $$x$$
$$\therefore \displaystyle \frac{1}{x}=1+\frac{dy}{dx}$$ $$\Rightarrow $$ $$\displaystyle \frac{dy}{dx}=\frac{1-x}{x}$$
If $$f$$ is a real-valued differentiable function satisfying $$\displaystyle \left | f(x)-f(y) \right |\leq (x-y)^{2}$$ for all $$x,y\:\epsilon\:R$$ and $$f(0)=0$$ then $$f(1)$$ equals
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$$0$$
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$$-1$$
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$$1$$
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$$2$$
Explanation
Given $$\displaystyle \left| f\left( x \right) -f\left( y \right) \right| \le { \left| x-y \right| }^{ 2 }\quad x\neq y$$
$$\displaystyle \therefore \left| \frac { f\left( x \right) -f\left( y \right) }{ x-y } \right| \le \left| x-y \right| $$
Taking limit as $$y\rightarrow x$$, we get
$$\displaystyle \lim _{ y\rightarrow x }{ \left| \frac { f\left( x \right) -f\left( y \right) }{ x-y } \right| } \le \lim _{ y\rightarrow x }{ \left| x-y \right| } $$
$$\displaystyle\Rightarrow\left|\lim _{ y\rightarrow x }{ \frac { f\left( x \right) -f\left( y \right) }{ x-y } } \right| \le \left| \lim _{ y\rightarrow x }{ \left( x-y \right) } \right| $$
$$\Rightarrow \left| f'\left( x \right) \right| \le 0\Rightarrow \left| f'\left( x \right) \right| =0\left[ \because \left| f'\left( x \right) \right| \ge 0 \right] $$
$$ \therefore f'\left( x \right) =0\Rightarrow f\left( x \right) =c$$
$$ \therefore h\left( x \right) =\int { f\left( x \right) } dx=\int { cdx } =cx+d$$
where $$d$$ is constant of integraion
Therefore $$h\left( x \right) $$ is a linear function of $$x$$
which is constant for all $$x.$$
$$f(0)=0$$ ........given
$$\therefore f(1)=0$$
$$\displaystyle \frac{d}{dx}(\tan ^{-1}\frac{\sin x+\cos x}{\cos x-\sin x})$$
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$$-1$$
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$$-2$$
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$$1$$
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$$2$$
Explanation
Let $$\displaystyle y = \tan ^{-1}\frac{\cos x+\sin x}{\cos x-\sin x}=\tan ^{-1}\frac{1+\tan x}{1-\tan x}$$
$$\displaystyle \Rightarrow y =\tan ^{-1}\frac{\tan\frac{\pi}{4}+\tan x}{1-\tan\frac{\pi}{4}\tan x}=\tan^{-1}\tan(\frac{\pi}{4}+x)=\frac{\pi}{4}+x$$
$$\therefore \cfrac{dy}{dx}=1$$
$$\displaystyle \frac{d}{dx}\tan ^{-1}\left(\frac{a \cos x-b\sin x}{b\cos x+a\sin x}\right)$$
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$$-1$$
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$$-2$$
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$$1$$
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$$2$$
Explanation
Put $$\displaystyle a= r\cos \alpha , b= r\sin \alpha $$
$$\displaystyle \therefore r= \sqrt{a^{2}+b^{2}},\tan \alpha = \dfrac{b}{a}$$
$$\displaystyle \therefore y= \tan ^{-1}\frac{r\cos\left ( x+\alpha \right )}{r\sin\left ( x+\alpha \right )}=\tan^{-1}\cot \left ( x+\alpha \right )$$
$$\displaystyle =\tan^{-1}\tan\left [ \dfrac{\pi}{2} -\left ( x+\alpha \right )\right ]$$
$$\displaystyle \therefore y=\dfrac{\pi} {2}-x-\alpha $$
$$\therefore \dfrac{dy}{dx}=-1$$
If $$\displaystyle f\left( x \right) =\sqrt { 1+\sqrt { x } } , x > 0,$$ then $$\displaystyle f\left ( x \right )\cdot f'\left ( x \right )$$ is equal to
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$$\displaystyle \frac{1}{2\sqrt{x}}$$
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$$\displaystyle \frac{1}{2}$$
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$$\displaystyle \frac{1}{4\sqrt{x}}$$
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$$\displaystyle \frac{2\sqrt{x}+1}{4\sqrt{x}}$$
Explanation
Let $$y=f\left( x \right) =\sqrt { 1+\sqrt { x } } $$
Then, $${ y }^{ 2 }=1+\sqrt { x } $$
Differentiating w.r.t. x, we get,
$$\displaystyle 2y\frac{dy}{dx}=\frac{1}{2\sqrt{x}}$$
$$\Rightarrow f(x)f'(x)=\displaystyle \frac{1}{4\sqrt{x}}$$
$$\displaystyle \dfrac{d}{dx}\tan ^{-1}\left(\dfrac{\cos x}{1+\sin x}\right)$$
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$$-\displaystyle \dfrac{1}{2}$$
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$$-\displaystyle \dfrac{1}{4}$$
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$$-\displaystyle \dfrac{1}{8}$$
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$$\displaystyle \dfrac{1}{2}$$
Explanation
Let $$\displaystyle y = \tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)$$
$$\Rightarrow \displaystyle y = \tan ^{-1}\left(\dfrac{(\cos^2 \dfrac{x}{2}-\sin^2\dfrac{x}{2})}{\sin^2\dfrac{x}{2}+\cos^2\dfrac{x}{2}+2\sin \dfrac{x}{2}.\cos\dfrac{x}{2}}\right)$$
$$\displaystyle y=\tan ^{-1}\left(\dfrac{\cos\left ( x/2 \right )-\sin \left ( x/2 \right )}{\cos \left ( x/2 \right )+\sin \left ( x/2 \right )}\right)$$
$$\displaystyle =\tan^{-1}\left(\frac{1-\tan\left ( x/2 \right )}{1+\tan \left ( x/2 \right )}\right)$$
$$\displaystyle =\tan^{-1}\left(\tan \left ( \pi /4-x/2 \right )\right)=\pi/4-x/2$$
$$\displaystyle \therefore \dfrac{dy}{dx}=-\dfrac{1}{2}$$
Differentiate $$\displaystyle x^{\sin^{-1}x}$$ w.r.t. $$\displaystyle \sin ^{-1}x.$$
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$$\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x.\frac{\sqrt{\left ( 1-x^{2} \right )}}{x} \right ]$$
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$$-\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x\frac{\sqrt{\left ( 1-x^{2} \right )}}{x} \right ]$$
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$$\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x\frac{\sqrt{\left ( 1+x^{2} \right )}}{x} \right ]$$
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$$-\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x\frac{\sqrt{\left ( 1+x^{2} \right )}}{x} \right ]$$
Explanation
Let $$\displaystyle y = x^{\sin^{-1}x}$$ and $$z = \displaystyle \sin ^{-1}x\Rightarrow \sin z = x$$
we have to find $$\cfrac{dy}{dz}$$
$$\Rightarrow y = (\sin z)^z$$ taking $$\log$$ both side $$\log y = z\log \sin z$$
Differentiating w.r.t $$z$$
$$\displaystyle \frac{1}{y}\frac{dy}{dz}=\log\sin z+z\frac{\cos z}{\sin z}$$
$$\therefore \displaystyle \frac{dy}{dz}=y\left(\log\sin z+z\frac{\cos z}{\sin z}\right)=\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x.\frac{\sqrt{\left ( 1-x^{2} \right )}}{x} \right ]$$
If $$y = \displaystyle (\tan x)^{\log x}$$, then $$\cfrac{dy}{dx} = $$
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$$(\tan x)^{\log x} \left[\cfrac{\log \tan x}{x}+\cfrac{\log x}{\tan x}(\sec^2x) \right]$$
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$$\frac { 1 }{ x } { tanx }^{ logx }log(tanx)+\frac { 1 }{ tanx } \sec ^{ 2 }{ x } logx$$
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$$\frac { 1 }{ x } { tanx }^{ logx }log(tanx)+\frac { 1 }{ tanx } \sec ^{ 2 }{ x } $$
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none of these
Explanation
Let $$y = \displaystyle (\tan x)^{\log x}$$
$$\log y = \log x .\log \tan x$$
Differentiating both side w.r.t $$x$$
$$\cfrac{1}{y}.\cfrac{dy}{dx} = \cfrac{\log \tan x}{x}+\cfrac{\log x}{\tan x}(\sec^2x)$$
$$\Rightarrow \cfrac{dy}{dx} = (\tan x)^{\log x} \left[\cfrac{\log \tan x}{x}+\cfrac{\log x}{\tan x}(\sec^2x) \right]$$
Differentiate $$\displaystyle \tan x^{n}+\tan ^{n}x-\tan ^{-1}\frac{a+x^{n}}{1-ax^{n}}.$$
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$$ \left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n-1}x.\sec^{2}x-\left [\dfrac{ 1}{\left ( 1-x^{2n} \right ) }\right ]nx^{n-1}$$
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$$ \left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n-1}x.\sec^{2}x-\left [ \dfrac{1}{\left ( 1+x^{2n} \right )} \right ]nx^{n}$$
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$$ \left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n}x.\sec^{2}x-\left [\dfrac{ 1}{\left ( 1+x^{2n} \right )} \right ]nx^{n-1}$$
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$$ \left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n-1}x.\sec^{2}x-\left [ \dfrac{1}{\left ( 1+x^{2n} \right )} \right ]nx^{n-1}$$
Explanation
Let $$y = \displaystyle \tan x^{n}+\tan ^{n}x-\tan ^{-1}\frac{a+x^{n}}{1-ax^{n}}.$$
$$\Rightarrow \displaystyle y= \tan x^{n}+\tan^{n}x-\left ( \tan^{-1}a+\tan^{-1}x^{n} \right )$$
$$\displaystyle\therefore \frac{dy}{dx}= \left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n-1}x.\sec^{2}x-0-\left [\dfrac{ 1}{\left ( 1+x^{2n} \right ) }
\right ]nx^{n-1}$$
$$= \left ( \sec^{2}x^{n} \right
).nx^{n-1}+n\tan^{n-1}x.\sec^{2}x-\left [\dfrac{ 1}{\left ( 1+x^{2n} \right ) }
\right ]nx^{n-1}$$
If $$\displaystyle y=x^{n} log x+x(log x)^{n}$$, find $$dy/dx.$$
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$$x^{n-1}(1+n log x)+(log x)^{n-1}[n+log x]$$
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$$x^{n}(1+n log x)+(log x)^{n}[n+log x]$$
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$$x^{n-1}(1+(n-1) log x)+(log x)^{n-1}[n-1+log x]$$
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none of these
Explanation
$$\displaystyle y=x^{n} \log x+x (\log x)^{n}$$
$$\Rightarrow \cfrac{dy}{dx}=nx^{n-1} \log x+x^{n}
.(1/x)+x.n(\log x)^{n-1}.(1.x)+1(\log x)^{n}$$
$$\Rightarrow \cfrac{dy}{dx} =x^{n-1}(1+n \log x)+(\log
x)^{n-1}[n+\log x]$$
If $$\displaystyle x\sqrt{(1+y)}+y\sqrt{(1+x)}=0$$, then $$\displaystyle \frac{dy}{dx}$$=
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$$\displaystyle \frac{1}{(1+x)^{2}}$$
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$$\displaystyle -\frac{1}{(1+x)^{2}}$$
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$$\displaystyle -\frac{1}{(1-x)^{2}}$$
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$$\displaystyle \frac{1}{(1-x)^{2}}$$
Explanation
Given, $$\displaystyle x\sqrt{(1+y)}+y\sqrt{(1+x)}=0$$
$$\Rightarrow x\sqrt{(1+y)}=-y\sqrt{(1+x)}$$
Squaring both side we get, $$x^{2}(1+y)=y^{2}(1+x)$$
$$\displaystyle x+y+xy=0 $$ or $$y=x$$ (rejected)
$$\displaystyle y=-\frac{x}{1+x}=\frac{1}{1+x}-1$$
$$\displaystyle \therefore \frac{dx}{dy}=-\frac{1}{(1+x)^2}$$
If $$x^{m}.y^{n}=\left ( x+y \right )^{m+n}$$, then $$\dfrac{dy}{dx}=$$
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$$\displaystyle \frac{y}{x}$$
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$$\displaystyle \frac{-y}{x}$$
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$$\dfrac {my}{x}$$
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$$\dfrac {ny}{x}$$
Explanation
$$x^{m}\times y^{n}=\left ( x+y \right )^{m+n}$$
Taking log both sides we get
$$m\log x+n\log y=\left ( m+n \right )\log \left ( x+y \right )$$
Differentiating w.r.t. x we get
$$\displaystyle \dfrac{m}{x}+\dfrac{n}{y}\dfrac{dy}{dx}=\dfrac{m+n}{x+y}\left ( 1+\dfrac{dy}{dx} \right )$$
$$\Rightarrow $$ $$\displaystyle \dfrac{dy}{dx}\left ( \dfrac{n}{y}-\dfrac{m+n}{x+y} \right )=\dfrac{m+n}{x+y}-\dfrac{m}{x}$$
$$\Rightarrow $$ $$\displaystyle \dfrac{dy}{dx}\left ( \dfrac{nx+ny-my-ny}{y\left ( x+y \right )} \right )=\dfrac{mx+nx-mx-my}{x\left ( x+y \right )}$$
$$\Rightarrow $$ $$\displaystyle \dfrac{dy}{dx}=\left ( \dfrac{nx-my}{nx-my} \right )\dfrac{y}{x}=\dfrac{y}{x}$$ $$\Rightarrow $$ $$\displaystyle \dfrac{dy}{dx}=\dfrac{y}{x}$$
Differentiate the following: $$\displaystyle \cot ^{-1}\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{\left ( 1+\sin x \right )}-\sqrt{\left ( 1-\sin x \right )}}$$
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$$ \displaystyle \frac{1}{2}.$$
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$$\displaystyle \frac{-1}{2}.$$
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$$\displaystyle \frac{1}{4}.$$
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$$\displaystyle \frac{-1}{4}.$$
Explanation
Let $$y = \displaystyle \cot ^{-1}\frac{\sqrt{1+\sin x}+\sqrt{1-\sin
x}}{\sqrt{\left ( 1+\sin x \right )}-\sqrt{\left ( 1-\sin x \right )}}$$
$$\displaystyle y= \cot^{-1}\frac{\left (
\cos\frac{x}{2}+\sin\frac{x}{2} \right )+\left (
\cos\frac{x}{2}-\sin\frac{x}{2} \right )}{\left ( \cos\frac{x}{2}+\sin
\frac{x}{2} \right )-\left ( \cos \frac{x}{2}-\sin \frac{x}{2} \right
)}=\cot ^{-1}\frac{2\cos \left ( x/2 \right )}{2\sin \left ( x/2 \right
)}$$
or $$\displaystyle y=\cot ^{-1}\cot \frac{x}{2}= \frac{x}{2}, $$
$$\therefore \dfrac{dy}{dx}= \dfrac{1}{2}.$$
Note: Here assumption taken that $$\tan\frac{x}{2}\geq 1$$
Find the differential equation of the family of curves whose equations are
$$\displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}+\lambda }=1,$$ where $$\displaystyle \lambda $$ is parameter.
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$$\displaystyle -\frac{xy}{a^{2}{y}'}=\frac{a^{2}-x^2}{a^{2}}$$
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$$\displaystyle -\frac{xy}{a^{2}{y}'}=\frac{a^{2}+x^2}{a^{2}}$$
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$$\displaystyle -\frac{xy}{a^{2}{y}'}=\frac{a^{4}-x^2}{a^{2}}$$
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$$\displaystyle -\frac{xy}{a^{2}{y}'}=\frac{a^{4}+x^2}{a^{2}}$$
Explanation
Differentiating given equation w.r.t $$x$$ we get
$$\displaystyle \frac{2x}{a^{2}}+\frac{2y}{a^{2}+\lambda }\frac{dy}{dx}=0$$
$$\displaystyle \therefore \left ( a^{2}+\lambda \right )=-\frac{a^{2}y{y}'}{x}$$
Putting for $$\displaystyle a^{2}+\lambda $$ in the given equation , we get
$$\displaystyle\Rightarrow \frac{x^{2}}{a^{2}}-\frac{y^{2}x}{a^{2}y{y}'}=1$$
$$\displaystyle\Rightarrow \frac{x^{2}}{a^{2}}-\frac{y^{2}x}{a^{2}y{y}'}=1$$
$$\Rightarrow \displaystyle -\frac{xy}{a^{2}{y}'}=1-\frac{x^{2}}{a^{2}}=\frac{a^{2}-x^2}{a^{2}}$$
Let $$ \displaystyle f\left ( x \right ) $$ be defined by $$ \displaystyle f\left ( x \right )=\left\{\begin{matrix}\sin 2x & \text{if } 0< x\leq \dfrac{\pi}6\\ ax+b& \text{if } \dfrac{\pi}6< x\leq 1\end{matrix}\right. $$. The values of $$a$$ and $$b$$ such that $$ \displaystyle f $$ and $$ \displaystyle {f}' $$ are continuous, are
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$$ \displaystyle a=1,b=\dfrac1{\sqrt{2}}+\dfrac{\pi}6 $$
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$$ \displaystyle a=\dfrac1{\sqrt{2}},b=\dfrac1{\sqrt{2}} $$
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$$ \displaystyle a=1,b=\dfrac{\sqrt{3}}2-\dfrac{\pi}6 $$
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None of these
Explanation
For $$f$$ to be continuous
$$\displaystyle \frac { \sqrt { 3 } }{ 2 } =f\left( \frac { \pi }{ 6 } \right) =\displaystyle \lim_{ x\rightarrow \tfrac { \pi }{ 6 }^+ } f\left( x \right) =\displaystyle\lim_{ x\rightarrow \tfrac { \pi }{ 6 } + } \left( ax+b \right) =\frac { a\pi }{ 6 } +b$$
$$\displaystyle f'\left( x \right) =\begin{cases} 2\cos { 2x } \quad \quad \text{if }0<x<\dfrac { \pi }{ 6 } \\ a\quad \quad \quad \quad \quad \text{if }\dfrac { \pi }{ 6 } <x<1 \end{cases}$$
$$\displaystyle f'\left( \frac { \pi }{ 6 } + \right) =a$$ and $$\displaystyle f'\left( \frac { \pi }{ 6 } - \right) =1$$
Thus $$\displaystyle a=1,b=\frac { \sqrt { 3 } }{ 2 } -\frac { \pi }{ 6 } $$
Find the solution of $$\displaystyle \frac{dy}{dx}= \frac{2x+2y-2}{3x+y-5}.$$
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$$\displaystyle \left ( 2x+y-3 \right )= k\left ( x-y-3 \right )^{4}$$
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$$\displaystyle \left ( 2x-y-3 \right )= k\left ( x-y-3 \right )^{4}$$
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$$\displaystyle \left ( 2x+y+3 \right )= k\left ( x-y-3 \right )^{4}$$
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$$\displaystyle \left ( 2x+y-3 \right )= k\left ( 2x-y-3 \right )^{4}$$
If $$f\left( x \right) $$ is a polynomial of degree $$n(>2)$$ and $$f\left( x \right) =f\left( k-x \right) ,($$ where $$k$$ is a fixed real number$$),$$ then degree of $$f'(x)$$ is
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$$n$$
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$$n-1$$
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$$n-2$$
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None of these
Explanation
Clearly, $$f(x)$$ must be of the from
$$f\left( x \right) ={ a }_{ 0 }\left[ { x }^{ n }+{ \left( k-x \right) }^{ n } \right] +{ a }_{ 1 }\left[ { x }^{ n-1 }+{ \left( k-x \right) }^{ n-1 } \right] +...+{ a }_{ n-1 }\left[ x+\left( k-x \right) \right] +{ a }_{ n. }$$
It may be noted that $$n$$ must be even for otherwise $$f(x)$$ will become a polynomial of degree $$n-1.$$
Clearly, $$f'(x)$$ is a polynomial of degree $$n-1.$$
If $$2f\left( \sin { x } \right) +f\left( \cos { x } \right) =x$$, then $$\displaystyle \frac { d }{ dx } f\left( x \right)$$ is
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$$\sin{x}+\cos{x}$$
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$$2$$
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$$\displaystyle \frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } $$
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none of these
Explanation
$$\displaystyle 2f\left( \sin { x } \right) +f\left( \cos { x } \right) =x$$ ...(1)
Replace $$x$$ by $$\displaystyle \frac { \pi }{ 2 } -x$$
$$\displaystyle 2f\left( \cos { x } \right) +f\left( \sin { x } \right) =\frac { \pi }{ 2 } -x$$ ...(2)
Solving we get , $$\displaystyle 3f\left( \sin { x } \right) =\frac { \pi }{ 2 } +3x$$
$$\displaystyle \therefore f\left( x \right) =\frac { \pi }{ 6 } +\sin ^{ -1 }{ x } \quad \\\displaystyle\therefore \frac { d }{ dx } f\left( x \right) =\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } $$
Obtain the differential equation whose solutions are
$$\displaystyle y=A\cos \left ( x+3 \right ),$$ A being constant.
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$$\displaystyle \frac{dy}{dx}+y\tan \left ( x+3 \right )=0$$
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$$\displaystyle \frac{dy}{dx}+y\tan \left ( x-3 \right )=0$$
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$$\displaystyle -\frac{dy}{dx}+y\tan \left ( x-3 \right )=0$$
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$$\displaystyle -\frac{dy}{dx}+y\tan \left ( x+3 \right )=0$$
Explanation
$$y = A\cos(x+3)$$ (i)
$$\displaystyle \frac{dy}{dx}=-A\sin \left ( x+3 \right )$$ ..(ii)
by (ii)$$\div$$ (i)
$$\displaystyle
\frac{\frac{dy}{dx}}{y}=-\tan \left ( x+3 \right )$$
$$\displaystyle
\frac{dy}{dx}+y\tan \left ( x+3 \right )=0$$
If $$ \displaystyle {f}'\left ( x \right )=g\left ( x \right ) $$ and $$ \displaystyle {g}'\left ( x \right )=-f\left ( x \right ) $$ and $$ \displaystyle f\left ( 2 \right )=4={f}'\left ( 2 \right ) $$ then $$ \displaystyle f^{2}\left ( 16 \right )+g^{2}\left ( 16 \right ) $$ is
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16
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32
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64
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None of these
Explanation
$$\displaystyle \frac { d }{ dx } \left( { f }^{ 2 }\left( x \right) +{ g }^{ 2 }\left( x \right) \right) \\ =2\left[ f\left( x \right) f'\left( x \right) +g\left( x \right) g'\left( x \right) \right] \\ =2\left[ f\left( x \right) -g\left( x \right) f\left( x \right) \right] =0$$
Thus $${ f }^{ 2 }\left( x \right) +{ g }^{ 2 }\left( x \right) $$ is constant.
Therefore $${ f }^{ 2 }\left( 16 \right) +{ g }^{ 2 }\left( 16 \right) ={ f }^{ 2 }\left( 2 \right) +{ g }^{ 2 }\left( 2 \right) \\ ={ f }^{ 2 }\left( 2 \right) +{ \left( f'\left( 2 \right) \right) }^{ 2 }=16+16=32$$
Let $$f\left( x \right)=\sqrt { x-1 } +\sqrt { x+24-10\sqrt { x-1 } } ;1<x<26$$ be a real valued function. Then $$f'(x)$$ for $$1<x<26$$ is
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$$0$$
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$$\displaystyle \frac { 1 }{ \sqrt { x-1 } } $$
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$$2\sqrt { x-1 } -5$$
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none of these
Explanation
Let, $$ f(x) = \sqrt{x -1} + \sqrt{x + 24 -10\sqrt{x-1}}$$
$$ 1< x< 26$$ be a real valued function ,
We have,
$$ f(x) = \sqrt{x -1} + \sqrt{x + 24 -10\sqrt{x-1}}$$
This can be written as
$$ f(x) = \sqrt{x -1} + \sqrt{\left ( 5 - \sqrt{x-1} \right )^{2} }$$
$$ \because 1< x< 26$$
$$ f(x) = \sqrt{x-1} + 5 - \sqrt{x-1} $$
$$ f(x) = 5 $$
On differentiating wrt x , we get
$$ f{}'(x) = 0 $$
Hence ,Option A
A curve passing through the point $$(1,1)$$ is such that the intercept made by a tangent to it on x-axis is three times the x co-ordinate of the point of tangency, then the equation of the curve is:
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$$\displaystyle y=\frac{1}{x^{2}}$$
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$$\displaystyle y=\sqrt{x}$$
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$$\displaystyle y=\frac{1}{\sqrt{x}}$$
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none
Explanation
Equation of tangent on any point on the curve is,
$$\displaystyle Y-y=\frac{dy}{dx}\left ( X-x \right )$$
Thus intercept on the x-axis is,$$\displaystyle
I_{x}=x-\frac{y}{dy/dx}$$
Now using given condition,
$$\displaystyle
I_{x}=x-\frac{y}{dy/dx}=3x$$
$$\displaystyle \therefore
y\frac{dx}{dy}+2x=0$$
or $$\displaystyle
\frac{dx}{2x}+\frac{dy}{y}=0$$ or $$\displaystyle \frac{1}{2}\log x+\log
y=k$$ or $$\displaystyle \log y\sqrt{x}=e^{k}=constant$$
Now given this curve passing through (1,1) $$\Rightarrow k=0$$
$$\therefore $$ Hence required curve is, $$y=\cfrac{1}{\sqrt{x}}$$
Let $$ \displaystyle f $$ be a function satisfying $$ \displaystyle f\left ( x+y \right )=f\left ( x \right )f\left ( y \right ) $$ for all $$x$$ and $$y$$ and $$ \displaystyle f\left ( 0 \right )={f}'\left ( 0 \right )=1 $$ then
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$$ \displaystyle f $$ is differentiable for all x
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$$ \displaystyle {f}'\left ( x \right )=f\left ( x \right ) $$
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$$ \displaystyle f\left ( x \right )=e^{x} $$
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$$ \displaystyle f $$ is continuous for alI x
Explanation
We have, $$f(x+y)=f(x)f(y) ...........(1)$$
Now $$f'(x) =\displaystyle\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\dfrac{f(x)f(h)-f(x)}{h}$$ using (1)
$$\Rightarrow f'(x) =\displaystyle\lim_{h\to 0}\dfrac{f(x)[f(h)-1)}{h}=f(x)\lim_{h\to 0}\dfrac{f(0+h)-f(0)}{h}=f(x)f'(0)=f(x)$$
Integrating we get, $$\log(f(x)) = x+c$$
Since at x$$=0, f(x) =1\Rightarrow $$ we have, $$c=0$$
Hence $$f(x) =e^x$$, which is continuous and differentiable for all $$x$$
If $$ \displaystyle f\left ( 1 \right )=3 $$ and $$ \displaystyle {f}'\left ( 1 \right )=-\dfrac13 $$ then the derivative of $$ \displaystyle \left ( x^{11} +f\left ( x \right )\right )^{-2} $$ at $$ \displaystyle x=1 $$ is
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$$-\dfrac12$$
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$$-1$$
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$$1$$
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$$ \displaystyle {f}'\left ( 1 \right ) $$
Explanation
The given equation is:
$$y = (x^{11} + f(x))^{-2}$$
Differentiating the equation w.r.t to x once we get,
$$\Rightarrow y' = -2(x^{11} + f(x))^{-3}.(11x^{10} + f'(x))$$
Value of the derivative at $$x = 1$$ is given as,
$$\Rightarrow y' = \dfrac{-2}{(1+3)^3}.(11 - \dfrac{1}{3})$$
$$\Rightarrow y' = \dfrac{-2}{64}.\dfrac{32}{3}$$
$$\Rightarrow y' = \dfrac{-1}{3}$$ .....Answer
$$\Rightarrow y' = f'(1)$$
A polynomial $$f(x)$$ leaves remainder $$15$$ when divided by $$(x-3)$$ and $$(2x+1)$$ when divided by $$(x-1)^2$$. When $$f$$ is divided by $$(x-3)(x-1)^2,$$ the remainder is
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$$2x^2+2x+3$$
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$$2x^2-2x-3$$
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$$2x^2-2x+3$$
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none of these
Explanation
Since function $$f(x)$$ leaves remainder $$15$$ when divided by $$x-3$$, therefore $$f(x)$$ can be written as
$$f(x)=(x-3)l(x)+15$$ ...(1)
Also, $$f(x)$$ leaves remainder $$2x+1$$ when divided by $$(x-1)^2.$$
Thus, $$f(x)$$ can also be written as
$$f\left( x \right)={ \left( x-1 \right) }^{ 2 }m\left( x \right) +2x+1$$ ...(2)
If $$R(x)$$ be the remainder when $$f(x)$$ is divided by $$(x-3)(x-1)^2,$$ then we may write
$$f\left( x \right)=\left( x-3 \right) { \left( x-1 \right) }^{ 2 }n\left( x \right) +R\left( x \right) $$ ...(3)
Since $$(x-3)(x-1)^2$$ is a polynomial of degree three, the remainder has to be a polynomial of degree less than or equal to two.
Thus let
$$R(x)=ax^2+bx+c$$
From (1) and (3), we have
$$f(3)=15=R(3)\Rightarrow 9a+3b+c=15$$ ...(4)
From (2) and (3), we have
$$f(1)=3=R(1)\Rightarrow a+b+c=3$$ ...(5)
From (2) and (3), we have
$$f'(1)=2=R'(1)\Rightarrow 2a+b=2$$ ...(6)
Solving equation (4),(5) and (6), we get
$$a=2,b=-2,c=3$$
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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