The maximum value of sin x cos x is

Solution:

Let $f(x)=\sin x \cos x$

$\Rightarrow f(x)=\frac{1}{2} \sin 2 x$

Differentiating both sides with respect to $x$, we get

$f^{\prime}(x)=\frac{1}{2} \cos 2 x \times 2=\cos 2 x$

For maxima or minima,

$f^{\prime}(x)=0$

$\Rightarrow \cos 2 x=0$

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

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

Now,

$f^{\prime \prime}(x)=-2 \sin 2 x$

$\Rightarrow f^{\prime \prime}\left(\frac{\pi}{4}\right)=-2 \sin \left(2 \times \frac{\pi}{4}\right)=-2 \sin \frac{\pi}{2}=-2<0$

So, $x=\frac{\pi}{4}$ is the point of local maximum.

∴ Maximum value of f(x)

$=f\left(\frac{\pi}{4}\right)$

$=\frac{1}{2} \sin \left(2 \times \frac{\pi}{4}\right)$

$=\frac{1}{2} \sin \frac{\pi}{2}$

$=\frac{1}{2} \times 1$

$=\frac{1}{2}$

Thus, the maximum value of $\sin x \cos x$ is $\frac{1}{2}$.

Hence, the correct answer is option (b).