The function $f(x)=sinx(1+cosx); x \in \left[0, \frac{3\pi}{2}\right]$ wil have :

Maximum value $\frac{3\sqrt{3}}{4}$ at x$=\frac{\pi}{3}$

The correct answer is Option (1) → Maximum value $\frac{3\sqrt{3}}{4}$ at x$=\frac{\pi}{3}$

$f(x)=\sin x(1+\cos x)$

$f'(x)=\cos x(1+\cos x)-\sin x\sin x$

$=\cos x+\cos^2x-\sin^2x$

$=\cos x+\cos 2x$

$=\cos x+2\cos^2x-1=0$

so $(2\cos x-1)(\cos x+1)=0$

$\cos x=-1,\cos x = \frac{1}{2}$

Max occurs at $x=\frac{π}{3}$

$f(\frac{π}{3})=\frac{3\sqrt{3}}{4}$