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CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\frac{6\text{x}}{\pi}-4\sin^{2}\text{x}\text{ on }\Big[0,\frac{\pi}{6}\Big]$
✓

Answer

The given function is $\text{f}(\text{x})=\frac{6\text{x}}{\pi}-4\sin^{2}\text{x}$ Since $\sin\text{x}\ \&\ \text{x}$ are everywhere continuous and differentiable, f(x) is continuous on $\Big[0,\frac{\pi}{6}\Big]$ and differentiable on $\Big(0,\frac{\pi}{6}\Big)$ Also, $\text{f}\Big(\frac{\pi}{6}\Big)=\text{f}(0)=0$ Thus, f(x) satisfies all the conditions of Rolle's theorem. Now, we have to show that there must exist a point $\text{c}\in\Big(0,\frac{\pi}{6}\Big)$ such that f'(c) = 0.We have
$\text{f}(\text{x})=\frac{6\text{x}}{\pi}-4\sin^{2}\text{x}$ $\Rightarrow\text{f}'(\text{x})=\frac{6}{\pi}-8\sin\text{x}\cos\text{x}$ $\therefore\ \text{f}'(\text{x})=0$ $\Rightarrow\frac{6}{\pi}-8\sin\text{x}\cos\text{x}=0$ $\Rightarrow\sin2\text{x}=\frac{3}{2\pi}$ $\Rightarrow\text{x}=\frac{1}{2}\sin^{-1}\Big(\frac{3}{2\pi}\Big)$ Thus, $\text{c}=\frac{1}{2}\sin^{-1}\Big(\frac{3}{2\pi}\Big)\in\Big(0,\frac{\pi}{6}\Big)$ such that f'(c) = 0. Hence, Rolle's theorem is verified.

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