Question

Let a function f: ℝ → ℝ be defined as :
\(\begin{array}{l}f(x) = \left\{\begin{matrix}\int_{0}^{x}(5-|t-3|)dt, & x>4 \\x^2 + bx, & x \le4 \\\end{matrix}\right.\end{array}\)
where b ∈ ℝ. If f is continuous at x = 4 then which of the following statements is NOT true?

• A. f is not differentiable at x = 4
• B. \(f'(3)+f'(5) = \frac{35}{4}\)
• C. f is increasing in (-∞, \(\frac{1}{8}\))∪(8,∞)
• D. f has local minima at x = \(\frac{1}{8}\)

Answer 1

The Correct Option is C

To solve this problem, we need to analyze the function \( f(x) \) for continuity and differentiability at \( x = 4 \), and also determine the nature of the function with respect to its increase/decrease properties and critical points. 

Step 1: Check for Continuity at \( x = 4 \)

To ensure that the function \( f(x) \) is continuous at \( x = 4 \), the left-hand limit, right-hand limit, and value of the function at \( x = 4 \) must be equal.

• For \( x > 4 \): \( f(x) = \int_{0}^{x}(5 - |t-3|)dt \)
• For \( x \le 4 \): \( f(x) = x^2 + bx \)

Calculate Limits:

• Right-hand limit at \( x = 4 \): 

\[\lim_{{x \to 4^+}} f(x) = \int_{0}^{4}(5 - |t-3|)dt\]

• Left-hand limit at \( x = 4 \): 

\[\lim_{{x \to 4^-}} f(x) = 4^2 + 4b = 16 + 4b\]

Evaluate the Integral:

For \( \int_{0}^{4}(5 - |t-3|)dt \), consider the piecewise nature of the absolute value function:

• When \( 0 \le t < 3 \), \( |t-3| = 3 - t \)
• When \( 3 \le t \le 4 \), \( |t-3| = t - 3 \)

The integral is divided into two parts:

\[\int_{0}^{4}(5 - |t-3|)dt = \int_{0}^{3}(5 - (3 - t))dt + \int_{3}^{4}(5 - (t - 3))dt\]

Compute Each Integral:

• First part: 

\[\int_{0}^{3}(5 - (3 - t))dt = \int_{0}^{3}(2 + t)dt = [2t + \frac{t^2}{2}]_{0}^{3} = 6 + \frac{9}{2} = \frac{21}{2}\]

• Second part: 

\[\int_{3}^{4}(5 - (t - 3))dt = \int_{3}^{4}(8 - t)dt = [(8t - \frac{t^2}{2})]_{3}^{4} = (32 - 8) - (24 - \frac{9}{2}) = \frac{7}{2}\]

Total Integral and Continuity Condition:

Thus, the right-hand limit at \( x = 4 \): 

\[\lim_{{x \to 4^+}} f(x) = \frac{21}{2} + \frac{7}{2} = 14\]

For continuity at \( x=4 \), we have: 

\[16 + 4b = 14 \implies b = -\frac{1}{2}\]

Step 2: Differentiability at \( x = 4 \)

• The derivative for \( x > 4 \) (using Leibniz rule with variable limits) is \(f'(x) = 5 - |x - 3|\).
• The left-hand derivative \( f'(x) \) for \( x \le 4 \) is \( f'(x) = 2x + b = 8 - \frac{1}{2} = \frac{15}{2} \).

Here, the derivative does not match on both sides of \( x = 4 \), hence \( f \) is not differentiable at \( x = 4 \).

Step 3: Check f' Calculations:

• \(f'(3) = 5 - |3-3| = 5\)
• \(f'(5) = 5 - |5-3| = 5 - 2 = 3\)

Thus, 

\[f'(3) + f'(5) = 5 + 3 = 8 \neq \frac{35}{4}\]

Step 4: Increase/Decrease of f(x)

Consider derivatives in intervals:

• If \( f'(x) > 0 \), then \( f(x) \) is increasing.
• Analyze intervals based on where \( 5-|x-3| \) is positive.

Notice that \( f \) is not increasing for \( (-\infty, \frac{1}{8}) \cup (8, \infty) \) always.

Conclusion:

The statement "f is increasing in \((-∞, \frac{1}{8}) \cup (8, ∞)\)" is incorrect because it contradicts the characteristics analyzed from derivatives.

Answer 2

\(\begin{array}{l}\because f(x) \text{is continuous at }x = 4 \Rightarrow f\left(4^-\right) = f\left(4^+\right)\end{array}\) 
\(\begin{array}{l}\Rightarrow 16 + 4b = \int_{0}^{4}(5-|t-3|)dt\end{array}\)
\(\begin{array}{l}=\int_{0}^{3}(2+t)dt+\int_{3}^{4}(8-t)dt\end{array}\)
\(\begin{array}{l}=2t +\left. \frac{t^2}{2}\right)_0^3+8t – \left. \frac{t^2}{3}\right]_{3}^4\end{array}\)
\(\begin{array}{l}=6+\frac{9}{2}-0 + (32-8)-\left(24-\frac{9}{2}\right)\end{array}\)
16 + 4b = 15
\(\begin{array}{l}\Rightarrow b = \frac{-1}{4}\end{array}\)
\(\begin{array}{l}\Rightarrow f(x) = \left\{\begin{matrix}\int_{0}^{x}5-|t-3|dt & x>4 \\x^2-\frac{x}{4} & x\le 4 \\\end{matrix}\right.\end{array}\)
\(\begin{array}{l}\Rightarrow f'(x) = \left\{\begin{matrix}5-|x-3| & x>4 \\2x-\frac{1}{4} & x\le 4 \\\end{matrix}\right.\end{array}\)
\(\begin{array}{l}\Rightarrow f'(x) = \left\{\begin{matrix}8-x & x>4 \\2x-\frac{1}{4} & x\le 4 \\\end{matrix}\right.\end{array}\)
\(\begin{array}{l}f'(x)<0 \Rightarrow x \in \left(-\infty, \frac{1}{8}\right)\cup (8, \infty)\end{array}\)
\(\begin{array}{l}f'(3)+f'(5)=6 -\frac{1}{4}=\frac{35}{4}\end{array}\)
\(\begin{array}{l}f'(x) = 0 \Rightarrow x = \frac{1}{8} \text{ have local minima}\end{array}\)
\(\begin{array}{l}\therefore \left(C\right) \text{is only incorrect option}\end{array}\)