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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.

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

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Answer

We have,

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

Since, $\sin\text{x},\sin2\text{x}\ \&\ \text{x}$ are everywhere continuous and differentiable.

Therefore, f(x) is continuous on $[0,\pi]$ and differentiable on $(0,\pi)$

Concequently, there exist some $\text{c}\in(0,\pi)$such that

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

Now, $\text{f}(\text{x})=\sin\text{x}-\sin2\text{x}-\text{x}$

$\text{f}'(\text{x})=\cos\text{x}-2\cos2\text{x}-1,\text{f}(\pi)=-\pi,\text{f}(0)=0$

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

$\Rightarrow\cos\text{x}-2\cos2\text{x}-1=-1$

$\Rightarrow\cos\text{x}-2\cos2\text{x}=0$

$\Rightarrow\cos\text{x}-4\cos^2\text{x}=-2$

$\Rightarrow4\cos^2\text{x}-\cos\text{x}-2=0$

$\Rightarrow\cos\text{x}=\frac{1}{8}\big(1\pm\sqrt{33}\big)$

$\Rightarrow\text{x}=\cos^{-1}\Big[\frac{1}{8}\big(1\pm\sqrt{33}\big)\Big]$

Thus, $\text{c}=\cos^{-1}\Big(\frac{1\pm\sqrt{33}}{8}\Big)\in(0,\pi)$ such that $\text{f}'(\text{c})=\frac{\text{f}(\pi)-\text{f}(0)}{\pi-0}.$

Hence, Lagrange's mean value theorem is verified.

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