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

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 $\frac{1}{2}$