Discuss the applicability of Rolles theorem for `f(x)=tanx` on `[0,\ pi]`

Explanation

`{f}({x})=\tan {x} \text { in } 0 \leq {x} \leq \pi`


At ` x=\frac{\pi}{2} `



`{f}(\frac{\pi}{2}+) \equiv {f}(\frac{\pi}{2}-) \equiv {f}(\frac{\pi}{2})`


`\Rightarrow f(x)` is not continuous at `x=\frac{\pi}{2}` and not differentiable, therefore Rolle's first two conditions are not satisfied. Hence, Rolle's theorem is not applicable.