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

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

Solution:

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 )

Now,

\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

Posted by

infoexpert24

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