Mean Value Theorems — Maths STD 12 Science — Question

$\text{f}(\text{x})=2\sin\text{x}+\sin2\text{x}\text{ on }[0,\pi]$

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

Now,

$\text{f}(0)=2\sin0+\sin0=0$

$\text{f}(\pi)=2\sin\pi+\sin2\pi=0$

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

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

Now,

$\text{f}(\text{x})=2\sin\text{x}+\sin2\text{x}$

$\text{f}'(\text{x})=2\cos\text{x}+2\cos2\text{x}$

Now,

$\text{f}'(\text{c})=0$

$2\cos\text{c}+2\cos2\text{c}=0$

$\Rightarrow2(\cos\text{c}+2\cos^2\text{c}-1)=0$

$\Rightarrow(2\cos^2+2\cos\text{c}-\cos\text{c}-1)=0$

$\Rightarrow(2\cos\text{c}-1)(\cos\text{c}+1)=0$

$\Rightarrow\cos\text{c}=\frac{1}{2},\cos\text{c}=-1$

$\Rightarrow\tan\text{c}=1$

$\text{c}=\frac{\pi}{3}\in(0,\pi),\text{c}=\pi$

Hence, Rolle's theorem is verified.