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Explain solution RD Sharma class 12 chapter Maxima and Minima exercise Multiple choice question, question 24 maths

Answers (1)

Answer: option (a) Minimum at x=\frac{\pi}{2}

Hint: For local maxima or minima, we must have f'(x)=0.

Given: f(x)=1+2\sin x+3\cos^2x


We have,

f(x)=1+2\sin x+3\cos^2x

f'(x)=2\cos x-6\cos x\sin x

f'(x)=2\cos x(1-3 \sin x)              (i)

For maxima and minima f'(x)=0

\begin{aligned} &\Rightarrow 2 \cos x(1-3 \sin x)=0 \\ &\Rightarrow 2 \cos x=0 \underset{\text { or }} \Rightarrow(1-3 \sin x)=0 \end{aligned}

\Rightarrow \cos x=0\Rightarrow \sin x =\frac{1}{3}

\Rightarrow x=\frac{\pi}{2}                                    \Rightarrow x=\sin ^{-1}\left ( \frac{1}{3} \right )


\begin{aligned} &f^{\prime}(x)=2 \cos x(-3 \cos x)+(1-3 \sin x)(-2 \sin x) \\ &=-6 \cos ^{2} x-2 \sin x+6 \sin ^{2} x \\ &=-6\left(\cos ^{2} x-\sin ^{2} x\right)-2 \sin x \\ &-6 \cos 2 x-2 \sin x \quad\left[\because \cos ^{2} x-\sin ^{2} x=\cos 2 x\right] \end{aligned}

\begin{aligned} &\text { at } x=\frac{\pi}{2}, f^{\prime \prime}\left(\frac{\pi}{2}\right) \\ &=-6 \cos \pi-2 \sin \frac{\pi}{2} \\ &=-6(-1)-2 \times 1 \quad\left[\because \cos \pi=-1, \sin \frac{\pi}{2}=1\right] \\ &-2+6=4>0 \end{aligned}

So, x=\frac{\pi}{2} is local minima

\begin{aligned} &\text { at } x=\sin ^{-1}\left(\frac{1}{3}\right) \\ &f^{\prime \prime}(x)=f^{\prime \prime}\left(\sin ^{-1}\left(\frac{1}{3}\right)\right) \\ &=-2 \sin \left(\sin ^{-1}\left(\frac{1}{3}\right)\right)-6\left(1-2 \sin ^{2} x\right) \quad\left[\because \cos 2 x=1-2 \sin ^{2} x\right] \end{aligned}

\begin{aligned} &=-2 \times \frac{1}{3}-6+12(\sin x)^{2} \\ &=\frac{-2}{3}-6+12\left(\sin \left(\sin ^{-1}\left(\frac{1}{3}\right)\right)\right)^{2} \\ &=\frac{-2}{3}-6+12 \times\left(\frac{1}{3}\right)^{2}=\frac{-2}{3}-6+12 \times \frac{1}{9} \end{aligned}

\begin{aligned} &=\frac{-2}{3}-6-\frac{4}{3}=\frac{-2-18-4}{3} \\ &=\frac{-24}{3}=-8 \\ &f^{\prime \prime}(x)<0 \end{aligned}

So, x=\sin ^{-1} \left (\frac{1}{3} \right ) is local maxima

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