JEE Main · 2012 · 29 Apr 2012 · Offline · Continuity and Differentiability

On the interval $[2,5]$ both $x-2\ge 0$ and $x-5\le 0$, so $f(x)=(x-2)+(5-x)=3$, a constant. Being constant, $f$ is continuous on $[2,5]$, differentiable on $(2,5)$, and satisfies $f(2)=f(5)=3$, so the second statement holds in full. A constant also has derivative zero throughout, in particular $f^{\prime }(4)=0$, so the first statement holds — and it follows from the second by Rolle's theorem, which the second statement states exactly. The second statement therefore correctly explains the first.