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Mean Value Theorems — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMean Value Theorems4 Marks
Question
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\cos2\text{x}\text{ on }\Big[\frac{-\pi}{4},\frac{\pi}{4}\Big]$
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Answer

Here

$\text{f}(\text{x})=\cos2\text{x}\text{ on }\Big[\frac{-\pi}{4},\frac{\pi}{4}\Big]$

We know that $\cos\text{x}$ is continuous and differentiable everywhere. So, f(x) is continuous in $\Big[\frac{-\pi}{4},\frac{\pi}{4}\Big]$ and differentiable is $\Big(\frac{-\pi}{4},\frac{\pi}{4}\Big)$.

Now,

$\text{f}\Big(-\frac{\pi}{4}\Big)=\cos2\Big(-\frac{\pi}{4}\Big)=\cos\Big(-\frac{\pi}{2}\Big)=0$

$\text{f}\Big(\frac{\pi}{4}\Big)=\cos2\Big(\frac{\pi}{4}\Big)=\cos\Big(\frac{\pi}{2}\Big)=0$

$\Rightarrow\text{f}\Big(-\frac{\pi}{4}\Big)=\text{f}\Big(\frac{\pi}{4}\Big)$

So, Rolle's theorem is applicable, so, there must exist a $\text{c}\in\Big(0,\frac{\pi}{2}\Big)$ such that f'(c) = 0

Now,

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

$\text{f}'(\text{c})=2\sin2\text{c}=0$

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

$\Rightarrow2\text{c}=0$

$\Rightarrow\text{c}=0\in\Big(\frac{-\pi}{4},\frac{\pi}{4}\Big)$

Thus, Rolle's theorem verified.

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