Let a function $\mathrm{f}:\mathrm{ℝ}\to \mathrm{ℝ}$ be defined as :

$\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}{\int }_0^{\mathrm{x}} (5-|\mathrm{t}-3\left|\right)\mathrm{dt},\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}>4\\ {\mathrm{x}}^2+\mathrm{bx} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}\le 4\end{array}\right.$

where $\mathrm{b}\in \mathrm{ℝ}$. If f is continuous at x = 4, then which of the following statements is NOT true ?

Given $\mathrm{f}\left(\mathrm{x}\right)\left\{\begin{array}{l}{\int }_0^{\mathrm{x}} (5-|\mathrm{t}-3\left|\right)\mathrm{dt},\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}>4\\ {\mathrm{x}}^2+\mathrm{bx}, \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}\le 4\end{array}\right.$

f(x) is continuous at x = 4

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

So $16+4\mathrm{b}={\int }_0^3 (2-\mathrm{t})\mathrm{dt}+{\int }_3^4 (8-\mathrm{t})\mathrm{dt}$

$\Rightarrow 16+4\mathrm{b}=15$

So  $\mathrm{b}=\frac{-1}{4}$

At x = 4

LHD$=2\mathrm{x}+\mathrm{b}=\frac{31}{4}$

RHD$=5-|\mathrm{x}-3|=4$

LHD $\ne$ RHD

Option (A) is true

$\text{ and }{\mathrm{f}}^{\mathrm{\prime }}\left(3\right)+{\mathrm{f}}^{\mathrm{\prime }}\left(5\right)=\frac{23}{4}+3=\frac{35}{4}$

Option (B) is true

$\because \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2-\frac{\mathrm{x}}{4}\text{ at }\mathrm{x}\le 4$

${\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=2\mathrm{x}-\frac{1}{4}$

This function is not increasing.

In the interval in $\mathrm{x}\in \left(-\mathrm{\infty },\frac{1}{8}\right)$

Option (C) is NOT TRUE.

This function f(x) is also local minima at $\mathrm{x}=\frac{1}{8}$