Mathematics - Continuity - Differentiability Question with Solution | TestHub

We have, $f\left(x\right)=\left\{\begin{array}{ll}{\int }_0^x\left(5+\left|1-t\right|\right)dt,& x>2\\ 5x+1,& x\le 2\end{array}\right.$

Therefore, $f\left(x\right)={\int }_0^1\left(5+\left(1-t\right)\right)dt+{\int }_1^x\left(5+\left(t-1\right)\right)dt$

$=6-\frac{1}{2}+\left(4t+\frac{t^2}{2}\right){|}_1^x$

$=\frac{11}{2}+4x+\frac{x^2}{2}-4-\frac{1}{2}$

$=\frac{x^2}{2}+4x+1$

Now, $f\left(2^{+}\right)=2+8+1=11$

and $f\left(2\right)=f\left(2^{-}\right)=5\times 2+1=11$

Here, $f\left(2^{+}\right)=f\left(2^{-}\right)$

So, $f\left(x\right)$ is continuous at $x=2$

Clearly, $f\left(x\right)$ is differentiable at $x=1$

Now, $\mathrm{LHD} f^{\text{'}}\left(2^{-}\right)=5$ and $\mathrm{RHD} f^{\text{'}}\left(2^{+}\right)=6$

Therefore, $f\left(x\right)$ is not differentiable at $x=2$