Question

The value of $c$ in Rolles theorem for the function $f\left(x\right)=\sin x-\sin 2x$ on $[0,\pi ]$ is

• A. ${\cos }^{-1}\left(\frac{1+\sqrt{33}}{8}\right)$
• B. ${\cos }^{-1}\left(\frac{1+\sqrt{35}}{8}\right)$
• C. ${\cos }^{-1}\left(\frac{1-\sqrt{38}}{5}\right)$
• D. does not exist

Answer 1

Answer: (A)

${\cos }^{-1}\left(\frac{1+\sqrt{33}}{8}\right)$

$f\left(x\right)$ is continous & differentiable in $[0,\pi ]$
$f\left(0\right)=f\left(\pi \right)$ $a=0;b=\pi$
We have to find $c$ in $[0,1]$,

such that, $f^{\prime }\left(c\right)=0$ ( in Rolle's Theorem)

$\Rightarrow \cos c-2\cos 2c=0$
$\Rightarrow \cos c-2(2{\cos }^{2}c-1)=0$
$\Rightarrow 4{\cos }^{2}c-\cos c-2=0$
$\Rightarrow \cos c=\frac{1\pm \sqrt{1+32}}{8}$
$\Rightarrow \cos c=\frac{1\pm \sqrt{33}}{8}$

$\Rightarrow c={\cos }^{-1}\left(\frac{1\pm \sqrt{33}}{8}\right)$