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

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