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Q.
The function $\sin x(1+\cos x), 0 \leq x \leq \frac{\pi}{2}$ has maximum value, when $x$ is equal to
ManipalManipal 2018
Solution:
Let $z=\sin x(1+\cos x)$
or $z=\sin x+\frac{1}{2} \sin 2 x$
On differentiating w.r.t. x, we get
$\frac{d z}{d x}=\cos x+\cos 2 x$
For maximum or minimum, put $\frac{d z}{d x}=0$
$\Rightarrow \cos x+\cos 2 x=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 $0 \leq x \leq \frac{\pi}{2}$,
therefore we take only $x=\frac{\pi}{3}$
$\therefore \frac{d^{2} z}{d x^{2}}=-\sin x-2 \sin 2 x=-$ ve, for $x=\frac{\pi}{3}$
$\therefore $ It is maximum at $x=\frac{\pi}{3}$