Question

The function $\sin x(1+\cos x),0\le x\le \frac{\pi }{2}$ has maximum value, when $x$ is equal to

• A. $0$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{6}$
• D. None of these

Answer 1

Answer: (D)

Let $z=\sin x(1+\cos x)$
or $z=\sin x+\frac{1}{2}\sin 2x$
On differentiating w.r.t. x, we get
$\frac{dz}{dx}=\cos x+\cos 2x$
For maximum or minimum, put $\frac{dz}{dx}=0$
$\Rightarrow \cos x+\cos 2x=0$
$\Rightarrow 2{\cos }^{2}x-1+\cos x=0$
$\Rightarrow (\cos x+1)(2\cos x-1)=0$
$\Rightarrow \cos x=-1,\frac{1}{2}$
$\Rightarrow x=\frac{\pi }{3},\pi$
But $0\le x\le \frac{\pi }{2}$,
therefore we take only $x=\frac{\pi }{3}$
$\therefore \frac{d^{2}z}{d{x}^2}=-\sin x-2\sin 2x=-$ ve, for $x=\frac{\pi }{3}$
$\therefore$ It is maximum at $x=\frac{\pi }{3}$