Question

The maximum value of sin x cos x is
$\left(\mathrm{a}\right)\frac{1}{4}\left(\mathrm{b}\right)\frac{1}{2}\left(\mathrm{c}\right)\sqrt{2}\left(\mathrm{d}\right)2\sqrt{2}$

Answer 1

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

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

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=\frac{1}{2}\cos 2x\times 2=\cos 2x$

For maxima or minima,

$f\text{'}\left(x\right)=0$

$\Rightarrow \cos 2x=0$

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

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

Now,

$f\text{'}\text{'}\left(x\right)=-2\sin 2x$

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

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

∴ Maximum value of f(x)

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

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

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

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

$=\frac{1}{2}$

Thus, the maximum value of sinx cosx is $\frac{1}{2}$.

Hence, the correct answer is option (b).