Question
Prove that function $\sin ^2 x(1+\cos x)$, at $\cos x =\frac{1}{3}$ is maximum.
✓
Answer
Suppose $ y=\sin ^2 x(1+\cos x)$
So,$\frac{d y}{d x}=\sin ^2 x(-\sin x)+ (1+\cos x) \times 2 \sin x \cos x$
or $ \frac{d y}{d x}=-\sin ^3 x+2 \sin x \cos x+2 \sin x \cos ^2 x$
$\frac{d y}{d x} =\sin x\left(-\sin ^2 x+2 \cos x+2 \cos ^2 x\right)$
$ =\sin x\left(-1+\cos ^2 x+2 \cos x+2 \cos ^2 x\right)$
$ =\sin x\left(3 \cos ^2 x+2 \cos x-1\right)$
For maxima and minima $\frac{d y}{d x}=0$$ 0=\sin x\left(3 \cos ^2 x+2 \cos x-1\right)=0$
$\Rightarrow \sin x=0,3 \cos ^2 x+2 \cos x-1=0$
$\therefore x=\sin ^{-1}(0)=0 \text { and } \cos x=\frac{-2 \pm \sqrt{4-4 \times 3(-1)}}{2 \times 3}$
$\therefore x=0 \text { and } \cos x=\frac{-2 \pm 4}{6}=\frac{2}{6}, \frac{-6}{6}$$x=0$ and $\cos x=\frac{1}{3},-1$
now, $\frac{d^2 y}{d x^2}=\sin x[-6 \cos x \sin x-2 \sin x]+\left(3 \cos ^2 x+2 \cos x-1\right) \times \cos x$
or $\frac{d^2 y}{d x^2}=-6 \cos x \sin ^2 x-2 \sin ^2 x+3 \cos ^3 x+ 2 \cos ^2 x-\cos x$
now at $\cos x=\frac{1}{3}, \frac{d^2 y}{d x^2}$
$=-6 \times \frac{1}{3} \times \frac{2}{9}-2 \times \frac{2}{9} +3 \times \frac{1}{27}+2 \times \frac{1}{9}-\frac{1}{3}$
$=-\frac{12}{27}-\frac{4}{9}+\frac{3}{27}+\frac{2}{9}-\frac{1}{3}$
$=\frac{-12-12+3+6-9}{27}=\frac{-24}{27}<0$
Hence value of function will be maximum at $\cos x=\frac{1}{3}$.
So,$\frac{d y}{d x}=\sin ^2 x(-\sin x)+ (1+\cos x) \times 2 \sin x \cos x$
or $ \frac{d y}{d x}=-\sin ^3 x+2 \sin x \cos x+2 \sin x \cos ^2 x$
$\frac{d y}{d x} =\sin x\left(-\sin ^2 x+2 \cos x+2 \cos ^2 x\right)$
$ =\sin x\left(-1+\cos ^2 x+2 \cos x+2 \cos ^2 x\right)$
$ =\sin x\left(3 \cos ^2 x+2 \cos x-1\right)$
For maxima and minima $\frac{d y}{d x}=0$$ 0=\sin x\left(3 \cos ^2 x+2 \cos x-1\right)=0$
$\Rightarrow \sin x=0,3 \cos ^2 x+2 \cos x-1=0$
$\therefore x=\sin ^{-1}(0)=0 \text { and } \cos x=\frac{-2 \pm \sqrt{4-4 \times 3(-1)}}{2 \times 3}$
$\therefore x=0 \text { and } \cos x=\frac{-2 \pm 4}{6}=\frac{2}{6}, \frac{-6}{6}$$x=0$ and $\cos x=\frac{1}{3},-1$
now, $\frac{d^2 y}{d x^2}=\sin x[-6 \cos x \sin x-2 \sin x]+\left(3 \cos ^2 x+2 \cos x-1\right) \times \cos x$
or $\frac{d^2 y}{d x^2}=-6 \cos x \sin ^2 x-2 \sin ^2 x+3 \cos ^3 x+ 2 \cos ^2 x-\cos x$
now at $\cos x=\frac{1}{3}, \frac{d^2 y}{d x^2}$
$=-6 \times \frac{1}{3} \times \frac{2}{9}-2 \times \frac{2}{9} +3 \times \frac{1}{27}+2 \times \frac{1}{9}-\frac{1}{3}$
$=-\frac{12}{27}-\frac{4}{9}+\frac{3}{27}+\frac{2}{9}-\frac{1}{3}$
$=\frac{-12-12+3+6-9}{27}=\frac{-24}{27}<0$
Hence value of function will be maximum at $\cos x=\frac{1}{3}$.
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