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Mean Value Theorems — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMean Value Theorems5 Marks
Question
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\text{e}^{\text{x}}\cos{\text{x}}\text{ on }\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big]$
✓

Answer

The given function is $\text{f}(\text{x})=\text{e}^{\text{x}}\cos{\text{x}}$
Since $\cos\text{x}\ \&\ \text{e}^{\text{x}}$ are everywhere continuous and differentiable.
Therefore, f(x) being a product of these two, is continuous on $\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big]$ and differentiable on $\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)$
Also,
$\text{f}\Big(\frac{-\pi}{2}\Big)=\text{f}\Big(\frac{\pi}{2}\Big)=0$
Thus, f(x) satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists $\text{c}\in\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)$ such that f'(c) = 0.
We have
$\text{f}(\text{x})=\text{e}^{\text{x}}\cos{\text{x}}$
$\Rightarrow\text{f}'(\text{x})=\text{e}^{\text{x}}(\cos\text{x}+\sin\text{x})$
$\therefore\ \text{f}'(\text{x})=0$
$\Rightarrow\text{e}^{\text{x}}(\cos\text{x}+\sin\text{x})=0$
$\Rightarrow\sin\text{x}+\cos\text{x}=0$
$\Rightarrow\tan\text{x}=-1$
$\Rightarrow\text{x}=\frac{\pi}{4}$
Since, $\text{c}=\frac{\pi}{4}\in\Big(\frac{\pi}{4},\frac{\pi}{2}\Big)$ such that f'(c) = 0
Hence, Rolle's theorem verified.

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