Key Concepts and Formulas

• Continuity at a point: A function $f(x)$ is continuous at $x=a$ if $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$. For a piecewise function, this means the left-hand limit, right-hand limit, and the function value at the point must all be equal.
• Differentiability at a point: A function $f(x)$ is differentiable at $x=a$ if the left-hand derivative equals the right-hand derivative at $x=a$.
• Fundamental Theorem of Calculus (Part 1): If $F(x) = \int_c^x g(t) dt$, then $F'(x) = g(x)$.
• Properties of Modulus Function: $|u| = u$ if $u \ge 0$, and $|u| = -u$ if $u < 0$.

Step-by-Step Solution

Step 1: Understand the function definition and the condition for continuity. The function $f(x)$ is defined piecewise: $f(x)= \begin{cases}\int\limits_{0}^{x}(5-|t-3|) d t, & x>4 \\ x^{2}+b x & , x \leq 4\end{cases}$ We are given that $f$ is continuous at $x=4$. This means the left-hand limit, the right-hand limit, and the function value at $x=4$ must be equal. $\lim_{x \to 4^-} f(x) = \lim_{x \to 4^+} f(x) = f(4)$.

Step 2: Calculate the left-hand limit and the function value at x=4. For $x \leq 4$, $f(x) = x^2 + bx$. So, $f(4) = 4^2 + b(4) = 16 + 4b$. The left-hand limit is $\lim_{x \to 4^-} f(x) = \lim_{x \to 4^-} (x^2 + bx) = 4^2 + b(4) = 16 + 4b$.

Step 3: Calculate the right-hand limit at x=4. For $x > 4$, $f(x) = \int_{0}^{x}(5-|t-3|) dt$. The right-hand limit is $\lim_{x \to 4^+} f(x) = \lim_{x \to 4^+} \int_{0}^{x}(5-|t-3|) dt$. Since the integrand is continuous, we can evaluate this limit by substituting $x=4$ into the integral: $\lim_{x \to 4^+} f(x) = \int_{0}^{4}(5-|t-3|) dt$.

Step 4: Evaluate the definite integral $\int_{0}^{4}(5-|t-3|) dt$. We need to split the integral based on the definition of $|t-3|$. $|t-3| = t-3$ if $t-3 \ge 0 \implies t \ge 3$. $|t-3| = -(t-3) = 3-t$ if $t-3 < 0 \implies t < 3$. So, we split the integral at $t=3$: $\int_{0}^{4}(5-|t-3|) dt = \int_{0}^{3}(5-(3-t)) dt + \int_{3}^{4}(5-(t-3)) dt$ $= \int_{0}^{3}(5-3+t) dt + \int_{3}^{4}(5-t+3) dt$ $= \int_{0}^{3}(2+t) dt + \int_{3}^{4}(8-t) dt$

Now, evaluate each integral: $\int_{0}^{3}(2+t) dt = [2t + \frac{t^2}{2}]_{0}^{3} = (2(3) + \frac{3^2}{2}) - (0) = 6 + \frac{9}{2} = \frac{12+9}{2} = \frac{21}{2}$. $\int_{3}^{4}(8-t) dt = [8t - \frac{t^2}{2}]_{3}^{4} = (8(4) - \frac{4^2}{2}) - (8(3) - \frac{3^2}{2}) = (32 - \frac{16}{2}) - (24 - \frac{9}{2})$ $= (32 - 8) - (24 - \frac{9}{2}) = 24 - 24 + \frac{9}{2} = \frac{9}{2}$.

So, $\int_{0}^{4}(5-|t-3|) dt = \frac{21}{2} + \frac{9}{2} = \frac{30}{2} = 15$. Thus, $\lim_{x \to 4^+} f(x) = 15$.

Step 5: Use the continuity condition to find the value of b. Since $f$ is continuous at $x=4$, we have: $\lim_{x \to 4^-} f(x) = \lim_{x \to 4^+} f(x)$ $16 + 4b = 15$ $4b = 15 - 16$ $4b = -1$ $b = -\frac{1}{4}$.

Now the function is: $f(x)= \begin{cases}\int\limits_{0}^{x}(5-|t-3|) d t, & x>4 \\ x^{2}-\frac{1}{4} x & , x \leq 4\end{cases}$

Step 6: Analyze the differentiability of f at x=4. To check differentiability at $x=4$, we need to compute the left-hand and right-hand derivatives. For $x < 4$, $f(x) = x^2 - \frac{1}{4}x$. $f'(x) = 2x - \frac{1}{4}$. The left-hand derivative at $x=4$ is $f'_{-}(4) = \lim_{x \to 4^-} (2x - \frac{1}{4}) = 2(4) - \frac{1}{4} = 8 - \frac{1}{4} = \frac{32-1}{4} = \frac{31}{4}$.

For $x > 4$, $f(x) = \int_{0}^{x}(5-|t-3|) dt$. By the Fundamental Theorem of Calculus, $f'(x) = 5 - |x-3|$. For $x > 4$, $x-3 > 0$, so $|x-3| = x-3$. Therefore, for $x > 4$, $f'(x) = 5 - (x-3) = 5 - x + 3 = 8 - x$. The right-hand derivative at $x=4$ is $f'_{+}(4) = \lim_{x \to 4^+} (8 - x) = 8 - 4 = 4$.

Since $f'_{-}(4) = \frac{31}{4}$ and $f'_{+}(4) = 4$, and $\frac{31}{4} \neq 4$, the function $f$ is not differentiable at $x=4$. This means statement (A) "f is not differentiable at x=4" is TRUE. The question asks for the statement that is NOT true.

Step 7: Evaluate statement (B) $f^{\prime}(3)+f^{\prime}(5)=\frac{35}{4}$. We need to find $f'(3)$ and $f'(5)$. For $x \leq 4$, $f'(x) = 2x - \frac{1}{4}$. So, $f'(3) = 2(3) - \frac{1}{4} = 6 - \frac{1}{4} = \frac{24-1}{4} = \frac{23}{4}$.

For $x > 4$, $f'(x) = 8 - x$. So, $f'(5) = 8 - 5 = 3$.

Now, check the sum: $f'(3) + f'(5) = \frac{23}{4} + 3 = \frac{23}{4} + \frac{12}{4} = \frac{35}{4}$. Statement (B) is TRUE.

Step 8: Evaluate statement (C) $f$ is increasing in $\left(-\infty, \frac{1}{8}\right) \cup(8, \infty)$. To determine where $f$ is increasing, we need to analyze the sign of $f'(x)$. For $x \leq 4$, $f'(x) = 2x - \frac{1}{4}$. $f'(x) > 0$ when $2x - \frac{1}{4} > 0 \implies 2x > \frac{1}{4} \implies x > \frac{1}{8}$. So, for $x \leq 4$, $f$ is increasing in $(\frac{1}{8}, 4]$.

For $x > 4$, $f'(x) = 8 - x$. $f'(x) > 0$ when $8 - x > 0 \implies x < 8$. So, for $x > 4$, $f$ is increasing in $(4, 8)$.

Combining these intervals, $f$ is increasing in $(\frac{1}{8}, 4] \cup (4, 8) = (\frac{1}{8}, 8)$. Statement (C) claims $f$ is increasing in $\left(-\infty, \frac{1}{8}\right) \cup(8, \infty)$. Let's check the intervals: In $(-\infty, \frac{1}{8})$, for $x \leq 4$, $f'(x) = 2x - \frac{1}{4} < 0$, so $f$ is decreasing. In $(8, \infty)$, for $x > 4$, $f'(x) = 8 - x < 0$, so $f$ is decreasing. Therefore, statement (C) is FALSE.

Step 9: Evaluate statement (D) $f$ has a local minima at $x=\frac{1}{8}$. We look at the sign change of $f'(x)$ around $x=\frac{1}{8}$. For $x < \frac{1}{8}$ (and $x \leq 4$), $f'(x) = 2x - \frac{1}{4} < 0$, so $f$ is decreasing. For $x > \frac{1}{8}$ (and $x \leq 4$), $f'(x) = 2x - \frac{1}{4} > 0$, so $f$ is increasing. Since $f'(x)$ changes from negative to positive at $x=\frac{1}{8}$, $f$ has a local minimum at $x=\frac{1}{8}$. Statement (D) is TRUE.

Step 10: Identify the statement that is NOT true. We found that statements (A), (B), and (D) are true, and statement (C) is false. The question asks for the statement that is NOT true.

The final answer is $\boxed{A}$.