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  • Please solve RD Sharma class 12 Chapter Maxima and Minima exercise Multiple choice question, question 32 maths textbook solution.

Please solve RD Sharma class 12 Chapter Maxima and Minima exercise Multiple choice question, question 32 maths textbook solution.

Answers (1)

Answer: option(b) \frac{1}{2}

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

Given: f(x)=\sin x \cos x

Solution:

We have,

f(x)=\sin x \cos x      

We know that,

\begin{aligned} &\sin 2 x=2 \sin x \cos x \\ &f(x)=\frac{1}{2} \sin 2 x \\ &\Rightarrow f^{\prime}(x)=\frac{1}{2} \frac{d}{d x}(\sin 2 x) \end{aligned}   

\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{1}{2}(\cos 2 x) \times 2 \\ &\Rightarrow f^{\prime}(x)=\cos 2 x \end{aligned}

For maxima and minima f'(x)=0

\Rightarrow \cos 2x=0

\Rightarrow \cos 2x=\cos \frac{\pi}{2}

\Rightarrow 2x= \frac{\pi}{2}

\Rightarrow x= \frac{\pi}{4}

\begin{aligned} &f^{\prime \prime}(x)=-2 \sin 2 x \\ &f^{\prime \prime}\left(\frac{\pi}{4}\right)=-2 \sin 2\left(\frac{\pi}{4}\right)=-2 \sin \left(\frac{\pi}{2}\right)=-2<0 \end{aligned}

=\therefore x= \frac{\pi}{4}  is local maxima.

f\left(\frac{\pi}{4}\right)=\sin \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{4}\right)=\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)=\frac{1}{2}

Hence, maximum value of f (x) is \frac{1}{2}

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