AIEEE 2012 (Offline) 29th April Code AQuestion 1
Consider the function, $f(x)=|x-2|+|x-5|$, $x\in\mathbf{R}$.
Statement 1 : $f'(4)=0$
Statement 2 : $f$ is continuous in $[2,5]$, differentiable in $(2,5)$ and $f(2)=f(5)$.
Limits & Continuity
Solution
On $(2,5)$, $f(x)=(x-2)+(5-x)=3$, so $f\prime(4)=0$ (true). $f$ is continuous on $[2,5]$, differentiable on $(2,5)$ and $f(2)=f(5)=3$ (true), but Rolle only gives some $c$ with $f\prime(c)=0$, it does not explain the value at $x=4$.