Let ƒ(x) =  $\{\begin{array}{cc}\underset{0}{\overset{x}{\int }}\{5+|1-t\left|\right\}dt,& ifx<2\\ 5x+1& ifx\ge 2\end{array}$ then -

If x < 2, 

$f\left(x\right)= \underset{0}{\overset{x}{\int }}\{5+|1-t\left|\right\}dt=\underset{0}{\overset{1}{\int }}(5+1-t)dt+\underset{1}{\overset{x}{\int }}(5+t-1)dt$

$={[6t-\frac{t^2}{2}]}_0^1+{[4+\frac{t^2}{2}]}_1^x= 1 + 4x +\frac{x^2}{2}$

So,  $f\left(x\right)=\{\begin{array}{cc}1+4x+\frac{x^2}{2},& ifx<2\\ 5x+1,& ifx\ge 2\end{array}$

Clearly f(x) is continuous everywhere including x = 2
Again  $f^{\text{'}}$(x) =  $\{\begin{array}{cc}4+x& ifx<2\\ 5& ifx>2\end{array}$

 $f^{\text{'}}$ (x) is not continuous at x = 2
  $\therefore$  f(x) is not differentiable only at x = 2.