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

$$f(x) = \begin{cases} \int_0^x (5 - |t - 3|) \, dt, & x > 4 \\ x^2 + bx, & x \leq 4 \end{cases}$$

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

For $$x \gt 4$$ the function is defined as $$f(x)=\displaystyle\int_{0}^{x}\bigl(5-\lvert t-3\rvert\bigr)\,dt$$ and for $$x\le 4$$ it is $$f(x)=x^{2}+bx$$.

Step 1: Determine $$b$$ from continuity at $$x=4$$
Continuity requires $$\displaystyle\lim_{x\to4^{-}}f(x)=\lim_{x\to4^{+}}f(x)=f(4)$$.

Left-hand value (polynomial part):
$$f(4^{-})=4^{2}+4b=16+4b$$

Right-hand value (integral part): split the integral at $$t=3$$ where the absolute value changes.

For $$0\le t\le3$$, $$\lvert t-3\rvert=3-t\;\Rightarrow\;5-\lvert t-3\rvert=5-(3-t)=2+t$$.
For $$3\le t\le4$$, $$\lvert t-3\rvert=t-3\;\Rightarrow\;5-\lvert t-3\rvert=5-(t-3)=8-t$$.

Hence
$$\begin{aligned} f(4^{+})&=\int_{0}^{3}(2+t)\,dt+\int_{3}^{4}(8-t)\,dt\\[2mm] &=\Bigl(2t+\tfrac{t^{2}}{2}\Bigr)_{0}^{3}+\Bigl(8t-\tfrac{t^{2}}{2}\Bigr)_{3}^{4}\\[2mm] &=\bigl(6+4.5\bigr)+\bigl(24-19.5\bigr)=10.5+4.5=15. \end{aligned}$$

Set the two one-sided limits equal:
$$16+4b=15 \;\Longrightarrow\; 4b=-1 \;\Longrightarrow\; b=-\tfrac14.$$ Thus $$b=-\dfrac14$$ keeps $$f$$ continuous at $$x=4$$.

Step 2: Find the derivatives where they exist

• For $$x\le4$$,
$$f(x)=x^{2}+bx \;\Longrightarrow\; f'(x)=2x+b=2x-\tfrac14.$$
• For $$x\gt4$$, by the Fundamental Theorem of Calculus,
$$f'(x)=5-\lvert x-3\rvert.$$ Since $$x\gt4\gt3$$, here $$\lvert x-3\rvert=x-3$$, so
$$f'(x)=5-(x-3)=8-x.$$

Therefore
$$f'(3)=2(3)-\tfrac14=\tfrac{23}{4},\qquad f'(5)=8-5=3.$$

Step 3: Check each statement

Option A: At $$x=4$$ the left derivative is $$2(4)-\tfrac14=\tfrac{31}{4}$$ while the right derivative is $$8-4=4$$. They are unequal, so $$f$$ is not differentiable at $$x=4$$. Option A is true.

Option B: $$f'(3)+f'(5)=\tfrac{23}{4}+3=\tfrac{23}{4}+\tfrac{12}{4}=\tfrac{35}{4}.$$ Hence Option B is true.

Option C: To decide monotonicity, use the sign of $$f'(x).$$
• For $$x\lt\frac18$$ (still inside $$x\le4$$ part), $$f'(x)=2x-\tfrac14\lt0,$$ so $$f$$ is decreasing, not increasing.
• For $$x\gt8$$ (inside $$x\gt4$$ part), $$f'(x)=8-x\lt0,$$ so $$f$$ is again decreasing, not increasing.
Thus the claim that $$f$$ is increasing on $$\bigl(-\infty,\tfrac18\bigr)\cup(8,\infty)$$ is false. Option C is NOT true.

Option D: In the region $$x\le4$$, $$f'(x)=0$$ gives $$2x-\tfrac14=0\Rightarrow x=\tfrac18$$. Here $$f''(x)=2\gt0$$, so $$x=\tfrac18$$ is indeed a point of local minimum. Option D is true.

Exactly one statement is not true, namely Option C. Hence the answer (third option) matches the given key.