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- Rolle's Theorem: Let f be a real valued function that satisfies the following conditions:
- (i) f is defined and continuous on the closed interval $[a,b]$
- (ii) f is differentiable in the open interval $(a,b)$
- (iii) $f(a)=f(b)$
- Then there exists at least one value $c \in (a,b)$ such that $f'(c)=0$

$f(x)=\sin x \; 0 \leq x \leq \pi$

Step 1:

$f(x)$ is continous on $[0,\pi]$, differentiable in $(0,\pi)$ and $f(0)=f(\pi)=0$

The conditions for Rolle's Theorem are satisfied.

To find $c \in (0, \pi)$ such that $f'(c) =0$

$f'(x)=\cos x$

$f'(c)=0=> \cos c =0$ and $c=\large\frac{\pi}{2} $$\in (0, \pi)$ with $\cos {\large\frac{\pi}{2}}$$=0$

The theorem is verified

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