Let a function $f: \mathbb{R} \rightarrow \mathbb{R}$ be defined as :

$$f(x)= \begin{cases}\int\limits_{0}^{x}(5-|t-3|) d t, & x>4 \\ x^{2}+b x & , x \leq 4\end{cases}$$

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

<p>$\because$ f(x) is continuous at x = 4</p> <p>$\Rightarrow f({4^ - }) = f({4^ + })$</p> <p>$\Rightarrow 16 + 4b = \int\limits_0^4 {(5 - |t - 3|)dt}$</p> <p>$= \int\limits_0^3 {(2 + t)dt + \int\limits_3^4 {(8 - t)dt} }$</p> <p>$$ = \left. {2t + {{{t^2}} \over 2}} \right)_0^3 + \left. {8t - {{{t^2}} \over 3}} \right]_3^4$$</p> <p>$= 6 + {9 \over 2} - 0 + (32 - 8) - \left( {24 - {9 \over 2}} \right)$</p> <p>$16 + 4b = 15$</p> <p>$\Rightarrow b = {{ - 1} \over 4}$</p> <p>$$ \Rightarrow f(x) = \left\{ {\matrix{ {\int\limits_0^x {5 - |t - 3|\,dt} } & {x > 4} \cr {{x^2} - {x \over 4}} & {x \le 4} \cr } } \right.$$</p> <p>$$ \Rightarrow f'(x) = \left\{ {\matrix{ {5 - |x - 3|} & {x > 4} \cr {2x - {1 \over 4}} & {x \le 4} \cr } } \right.$$</p> <p>$$ \Rightarrow f'(x) = \left\{ {\matrix{ {8 - x} & {x > 4} \cr {2x - {1 \over 4}} & {x \le 4} \cr } } \right.$$</p> <p>$f'(x) < 0 = x \in \left( { - \infty ,{1 \over 8}} \right) \cup (8,\infty )$</p> <p>$f'(3) + f'(5) = 6 - {1 \over 4} = {{35} \over 4}$</p> <p>$f'(x) = 0 \Rightarrow x = {1 \over 8}$ have local minima</p> <p>$\therefore$ (C) is only incorrect option.</p>