∴f(x)=|x−2|+|x−5| ={x−2+5−x=3,2≤x≤5} Thus f(x)=3,2≤x≤5 f′(x)=0,2<x<5 f′(4)=0 ∵ Statement-1 is true
Since f(x)=3,2≤x≤5 is constant function. So, it continuous in 2,5 and differentiable in (2,5) ∵f(2)=0+|2−5|=3 and f(5)=|5−2|+0=3 statement- 2 is also true.