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Verify mean value theorem for each of the function $f(x)=\sin x-\sin 2x\;in\;[0,\pi].$

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Toolbox:
  • Let $f(x)$ be a function defined on $[a,b]$ such that
  • (i) it is continuous on $[a,b]$
  • (ii) it is differentiable on $(a,b)$
  • Then,there exists $c\in (a,b)$ such that $f'(c)=\large\frac{f(b)-f(a)}{b-a}$
Step 1:
$f(x)=\sin x-\sin 2x$ in $[0,\pi]$
Since $\sin x$ and $\sin 2x$ is continuous and differentiable everywhere,it is continuous on $[0,\pi]$ and differentiable on $(0,\pi)$.
Thus $f(x)$ satisfies both the conditions of Lagrange's Mean value theorem.
Hence there lies at least one $C\in (0,\pi)$,such that
$f'(c)=\large\frac{f(\pi)-f(0)}{\pi-0}$
Given : $f(x)=\sin x-\sin 2x$
Step 2:
$f'(x)=\cos x-2\cos 2x$
$f(0)=\sin 0-\sin 0=0$
$f(\pi)=\sin \pi-\sin \pi=0$
Therefore $\cos x-2\cos 2x=\large\frac{0-0}{\pi-0}=$$0$
$\Rightarrow \cos x-2\cos 2x=0$
$\cos x=2\cos 2x$
$\Rightarrow 2x=x$
$\Rightarrow x=0$
(i.e)$c=0$
Hence Lagrange's Mean value theorem is verified.
answered Jul 3, 2013 by sreemathi.v
 

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