Mean Value Theorems — Maths STD 12 Science — Question

$\text{f}(\text{x})=\cos2{\text{x}}\text{ on }[0,\pi]$

We know that, cosine function is continuous and differentiable every where, so f(x) is continuous is $[0,\pi]$ and differentiable is $(0,\pi).$

Now,

$\text{f}(0)=\cos0=1$

$\text{f}(\pi)=\cos(2\pi)=1$

$\Rightarrow\text{f}(0)=\text{f}(\pi)$

So, Rolle's theorem is applicable, so there must exist a point $\text{c}\in(0,\pi)$ such that f'(c) = 0.

Now,

$\text{f}(\text{x})=\cos2\text{x}$

$\text{f}'(\text{x})=-2\sin2\text{x}$

So, $\text{f}'(\text{c})=0$

$\Rightarrow-2\sin2\text{c}=0$

$\Rightarrow\sin2\text{c}=0$

$\Rightarrow2\text{c}=0$ or $2\text{c}=\pi$

$\Rightarrow\text{c}=0$ or $\text{c}=\frac{\pi}{2}\in(0,\pi)$

Hence, Rolle's theorem is verified.