If f(x) = \ cases ∫ _ 0 ^ x (5 + |1 - t|)dt, & x > 2 5x + 1, & x ≤...

We have, f\left(x\right) = \left\{\begin{cases}{∫}_{0}^{x}\left(5 + \left|1 - t\right|\right)dt, & x > 2 \\ 5x + 1, & x \le 2\end{cases}\right)</p><p>Therefore,f\left(x\right) = {∫}_{0}^{1}\left(5 + \left(1 - t\right)\right)dt + {∫}_{1}^{x}\left(5 + \left(t - 1\right)\right)dt$</p><p>$= 6 - \dfrac{1}{2} + \left(4t + \dfrac{{t}^{2}}{2}\right){|}_{1}^{x}$</p><p>$= \dfrac{11}{2} + 4x + \dfrac{{x}^{2}}{2} - 4 - \dfrac{1}{2}$</p><p>$= \dfrac{{x}^{2}}{2} + 4x + 1$</p><p>Now,$f\left({2}^{+}\right) = 2 + 8 + 1 = 11$</p><p>and$f\left(2\right) = f\left({2}^{-}\right) = 5 \times 2 + 1 = 11$</p><p>Here,$f\left({2}^{+}\right) = f\left({2}^{-}\right)$</p><p>So,$f\left(x\right)$is continuous at$x = 2$</p><p>Clearly,$f$\left(x\right)$ is differentiable at $x = 1$

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

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