The maximum value of $\sin x \cdot \cos x$ is

$\frac{1}{2}$

The correct answer is Option (2) → $\frac{1}{2}$ ##

We have, $f(x) = \sin x \cdot \cos x = \frac{1}{2} \sin 2x$

$∴f'(x) = \frac{1}{2} \cdot \cos 2x \cdot 2 = \cos 2x$

Now, $f'(x) = 0 \Rightarrow \cos 2x = 0$

$\Rightarrow \cos 2x = \cos \frac{\pi}{2} \Rightarrow x = \frac{\pi}{4}$

Also $f''(x) = \frac{d}{dx} \cos 2x = -\sin 2x \cdot 2 = -2 \sin 2x$

$∴[f''(x)]_{\text{at } x = \pi/4} = -2 \cdot \sin 2 \cdot \frac{\pi}{4} = -2 \sin \frac{\pi}{2} = -2 < 0$

At $\frac{\pi}{4}$, $f(x)$ is maximum and $\frac{\pi}{4}$ is point of maxima.

$∴f\left(\frac{\pi}{4}\right) = \frac{1}{2} \cdot \sin 2 \cdot \frac{\pi}{4} = \frac{1}{2}$