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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\sin\text{x}-\sin2\text{x}\text{ on }[0,\pi]$
✓

Answer

Since trignometric functions are differentiable and continuous, the given function, $\text{f}(\text{x})=\sin\text{x}-\sin2\text{x}$ is also continuous and differentiable.
Now $\text{f}(0)=\sin0-2\times0=0$
and
$\text{f}(\pi)=\sin\pi-\sin2\times\pi=0$
$\Rightarrow\text{f}(0)=\text{f}(\pi)$
Thus, f(x) satisfies conditions of the Rolle's Theorem on $[0,\pi].$
Therefore there exist $\text{c}\in[0,\pi]$ such that f'(c)=0
Now $\text{f}(\text{x})=\sin\text{x}-\sin2\text{x}$
$\Rightarrow\text{f}'(\text{x})\cos\text{x}-2\cos2\text{x}=0$
$\Rightarrow\cos\text{x}=2\cos2\text{x}$
$\Rightarrow\cos\text{x}=2(2\cos^2\text{x}-1)$
$\Rightarrow\cos\text{x}=4\cos^2\text{x}-2$
$\Rightarrow4\cos^2\text{x}-\cos\text{x}-2=0$
$\Rightarrow\cos\text{x}=\frac{1\pm\sqrt{33}}{8}=0.8431\text{ or }-0.5931$
$\Rightarrow\text{x}=\cos^{-1}(0.8431)\text{ or }\cos^{-1}(-0.5931)$
$\Rightarrow\text{x}=\cos^{-1}(0.8431)\text{ or }180^{\circ}-\cos^{-1}(0.5931)$ $\big[\because\ \cos^{-1}(-\text{x})=\pi-\cos^{-1}(\text{x})\big]$
$\Rightarrow\text{x}=32^\circ32'\text{ or }\text{x}=126^\circ23'$
Both $32^\circ32'$ and $126^\circ23'\in[0,\pi]$ such that f'(c) = 0.
Hence Rolle's theorem is verified.

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