Question

If $c$ is a value of $x$ for which Rolle's theorem holds for the function $f\left(x\right)=\sin x-\sin 2x$ on the interval $[0,\pi ],$ then the value of $\cos c$ is

• A. $\frac{1+\sqrt{33}}{16}$
• B. $\frac{1+\sqrt{31}}{8}$
• C. $\frac{1+\sqrt{33}}{8}$
• D. $\frac{1+\sqrt{31}}{16}$

Answer 1

The correct option is C $\frac{1+\sqrt{33}}{8}$

Given, $f\left(x\right)=\sin x-\sin 2x,x\in [0,\pi ]$
$f\left(0\right)=0,f\left(\pi \right)=0$ and $f$ is continuous in $[0,\pi ]$,differentiable in $(0,\pi )$
$\therefore$ from rolle's theorem, $f^{\prime }\left(c\right)=0,$ for $c\in (0,\pi )$
$\Rightarrow \cos c-2\cos 2c=0$
$\Rightarrow \cos c-2(2{\cos }^{2}c-1)=0$
$\Rightarrow 4{\cos }^{2}c-\cos c-2=0$
So, $\cos c=\frac{1\pm \sqrt{33}}{8}$