Question

Let $y=\sin x(1+\cos x)$
$x\in [0,\frac{\pi }{2}]$. The maxima or minima point for the above curve:

• A. Maxima at $\frac{\pi }{6}$
• B. Minima at $\frac{\pi }{6}$
• C. Maxima at $\frac{\pi }{3}$
• D. Minima at $\frac{\pi }{4}$

Answer 1

Answer: (C)

Maxima at $\frac{\pi }{3}$

$\Rightarrow y^{{}^{\prime }}=\sin x(-\sin x)+\cos x(1+\cos x)$

For maxima or minima, $y^{{}^{\prime }}=0$

$\Rightarrow -{\sin }^{2}x+\cos x+{\cos }^{2}x=0$

$\Rightarrow {\cos }^{2}x-1+\cos x+{\cos }^{2}x=0$

$\Rightarrow 2{\cos }^{2}x+\cos x-1=0$

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

$\Rightarrow \cos x=\frac{1}{2}$ or $\cos x=-1$

$\Rightarrow x=\frac{\pi }{3}$ or $x=\pi$

But given $x\in [0,\frac{\pi }{2}]$

$\therefore x=\frac{\pi }{3}$

$\Rightarrow y^{{}^{\prime \prime }}\left(x\right)=-2\sin x\cos x-\sin x-2\sin x\cos x$

$\Rightarrow y^{{}^{\prime \prime }}\left(\frac{\pi }{3}\right)=-\left(\frac{\sqrt{3}}{2}\right)(1+1+1)$

$\Rightarrow y^{{}^{\prime \prime }}\left(\frac{\pi }{3}\right)<0$, $y\left(x\right)$ has maxima at $x=\frac{\pi }{3}$