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Let $R$ be the set of all real numbers and $f(x)=\sin ^{10} x\left(\cos ^8 x+\cos ^4 x+\cos ^2 x+1\right)$ $x \in R$. Let $S=\{\lambda \in R$ there exists a point $c \in(0,2 \pi)$ with $\left.f^{\prime}(c)=\lambda f(c)\right\}$ Then,
$S=R$
$S=\{0\}$
$S=[0,2 \pi]$
$S$ is a finite set having more than one element
Solution
(a)
Given, $\ln f(x)=10 \ln \sin x+$ $\ln \left(1+\cos ^2 x+\cos ^4 x+\cos ^8 x\right)$
$\frac{f^{\prime}(x)}{f(x)}=10 \cot x-\sin 2 x$
${\left[\frac{1+2 \cos ^2 x+4 \cos ^6 x}{1+\cos ^2 x+\cos ^4 x+\cos ^8 x}\right] }$
$\left[\frac{1+2 \cos ^2 x+4 \cos ^6 x}{1+\cos ^2 x+\cos ^4 x+\cos ^8 x}\right]$
$\because f(x)$ is periodic with period $\pi$ and $\frac{f^{\prime}(x)}{f(x)}$ is continuous in $(0, \pi) .$
$\frac{f^{\prime}(x)}{f(x)}=\underbrace{10 \cot x}_{\begin{array}{c}\text { Range is }(-\infty, \infty) \\ \text { for } x \in(0, \pi)\end{array}}$
$\underbrace{-\sin 2 x\left[\frac{1+2 \cos ^2 x+4 \cos ^6 x}{1+\cos ^2 x+\cos ^4 x+\cos ^8 x}\right]}_{\text {finite number for } x \in(0, \pi)}$
So, $\frac{f^{\prime}(x)}{f(x)}$ has range $(-\infty, \infty)$ Hence, $\lambda \in R$.
