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Applications of Derivative — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsApplications of Derivative4 Marks
Question
Prove that function $\sin ^2 x(1+\cos x)$, at $\cos x$ $=\frac{1}{3}$ is maximum.
✓

Answer

Suppose $\quad y=\sin ^2 x(1+\cos x)$
So,$
\begin{aligned}
\frac{d y}{d x}=\sin ^2 x(-\sin x)+ & (1+\cos x) \\
& \times 2 \sin x \cos x
\end{aligned}
$or $\quad \frac{d y}{d x}=-\sin ^3 x+2 \sin x \cos x+2 \sin x \cos ^2 x$$
\begin{aligned}
\frac{d y}{d x} & =\sin x\left(-\sin ^2 x+2 \cos x+2 \cos ^2 x\right) \\
& =\sin x\left(-1+\cos ^2 x+2 \cos x+2 \cos ^2 x\right) \\
& =\sin x\left(3 \cos ^2 x+2 \cos x-1\right)
\end{aligned}
$
For maxima and minima $\frac{d y}{d x}=0$$
\begin{array}{l}
\quad \quad \quad 0=\sin x\left(3 \cos ^2 x+2 \cos x-1\right)=0 \\
\Rightarrow \quad \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}
\end{array}
$$x=0$ and $\cos x=\frac{1}{3},-1$
now, $\frac{d^2 y}{d x^2}=\sin x[-6 \cos x \sin x-2 \sin x]+$$
\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+$$
2 \cos ^2 x-\cos x
$now at $\cos x=\frac{1}{3}, \frac{d^2 y}{d x^2}=-6 \times \frac{1}{3} \times \frac{2}{9}-2 \times \frac{2}{9}$$
\begin{array}{l}
\quad+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
\end{array}
$
Hence value of function will be maximum at cos $x=\frac{1}{3}$.

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