Let a function f: ℝ → ℝ be defined as: f(x) = \ cases ∫ _ 0 ^ x (5...

Given $f\left(x\right) =$\left\{$\begin{cases}{∫}_{0}^{x}\left(5 - \left|t - 3\right|\right)dt, & x > 4 \\ {x}^{2} + bx, & x \le 4\end{cases}$\right)

Also given, $f\left(x\right)$ is continuous at $x = 4$

So $\lim_{x \to {4}^{-}}f\left(x\right) = \lim_{x \to {4}^{+}}f\left(x\right) = f\left(4\right)$

So $16 + 4b = {∫}_{0}^{3}\left(2 - t\right)dt + {∫}_{3}^{4}\left(8 - t\right)dt$

$⇒ 16 + 4b = {\left(2t - \dfrac{{t}^{2}}{2}\right)}_{0}^{3} + {\left(8t - \dfrac{{t}^{2}}{2}\right)}$_{3}^{4}$</p><p>$⇒ 16 + 4b = 15$</p><p>So$b = \dfrac{- 1}{4}$</p><p>Now differentiating$f\left(x\right)$we get,</p><p>${f}^{'}\left(x\right) = \left\{$\begin{cases}\left(5 - \left|x - 3\right|\right), & x > 4 \\ 2x + b, & x < 4\end{cases}$\right)

Now at $x = 4$

$LHD = 2x + b = 2 \times 4 - \dfrac{1}{4} = \dfrac{31}{4}$

$RHD = 5 - \left|x - 3\right| = 4$

$LHD \ne RHD$

So, option \left(A\right)$is true</p><p>And${f}^{'}\left(3\right) + {f}^{'}$\left(5\right) = \dfrac{23}{4} + 3 = \dfrac{35}{4}$

So, option \left(B\right)$is true</p><p>Now$f\left(x\right) = {x}^{2} - \dfrac{x}{4}$at$x \le 4$</p><p>${f}^{'}\left(x\right) = 2x - \dfrac{1}{4}$</p><p>Thus the function is not increasing in the interval in$x \in \left(- \infty , \dfrac{1}{8}\right)$</p><p>So, option \left(C\right)$ is NOT TRUE.

And function $f\left(x\right)$ is also have local minima at $x = \dfrac{1}{8}$ so option \left(D\right) is also true.