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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
$\text{f}(\text{x})=\frac{\text{x}}{2}-\sin\frac{\pi\text{x}}{6}\text{ on }[-1,0]$
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Answer

The given function is $\text{f}(\text{x})=\frac{\text{x}}{2}-\sin\frac{\pi\text{x}}{6}$

Since $\sin\text{x}\ \&\ \frac{\text{x}}{2}$ are everywhere continuous and differentiable, f(x) is continuous on [-1, 0] and differentiable on (-1, 0).

Also,

f(-1) - f(0) = 0

Thus, f(x) satisfies all the conditions of Rolle's theorem.

Now, we have to show that there must exist a point $\text{c}\in(-1,0)$ such that f'(c) = 0.

We have

$\text{f}(\text{x})=\frac{\text{x}}{2}-\sin\frac{\pi\text{x}}{6}$

$\Rightarrow\text{f}'(\text{x})=\frac{1}{2}-\frac{\pi}{6}\cos\frac{\pi\text{x}}{6}$

$\therefore\ \text{f}'(\text{x})=0$

$\Rightarrow\frac{1}{2}-\frac{\text{x}}{6}\cos\frac{\pi\text{x}}{6}=0$

$\Rightarrow\cos\frac{\pi\text{x}}{6}=\frac{3}{\pi}$

$\Rightarrow\text{x}=\frac{-6}{\pi}\cos^{-1}\Big(\frac{3}{\pi}\Big)$

Thus, $\text{c}=\frac{-6}{\pi}\cos^{-1}\Big(\frac{3}{\pi}\Big)\in(-1,0)$ such that f'(c) = 0.

Hence, Rolle's theorem is verified.

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