Question

If \(f(x) = \begin{cases} ax^2 + bx - \frac{13}{8}, & x \le 1 \\ 3x - 3, & 1<x \le 2 \\ bx^3 + 1, & x>2 \end{cases}\)is differentiable \(\forall x \in \mathbb{R}\), then \( a - b = \)

• A. \( \frac{9}{8} \)
• B. \( \frac{5}{4} \)
• C. \( \frac{11}{8} \)
• D. \( \frac{1}{4} \)

Hint:

If \(f(x)\) is differentiable at \(x = c\), then \(f(x)\) is continuous at \(x = c\).

Answer 1

The Correct Option is A

Since \( f(x) \) is differentiable for all \( x \in \mathbb{R} \), it must be continuous for all \( x \in \mathbb{R} \). In particular, \( f(x) \) must be continuous at \( x = 1 \) and \( x = 2 \). For continuity at \( x = 1 \), we must have:

\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1). \] 

This gives us:

\[ a(1)^2 + b(1) - \frac{13}{8} = 3(1) - 3 \] \[ a + b - \frac{13}{8} = 0 \] \[ a + b = \frac{13}{8} \]

For continuity at \( x = 2 \), we must have:

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

This gives us:

\[ 3(2) - 3 = b(2)^3 + 1 \] \[ 3 = 8b + 1 \] \[ 8b = 2 \] \[ b = \frac{1}{4} \]

Then,

\[ a = \frac{13}{8} - b = \frac{13}{8} - \frac{1}{4} = \frac{13}{8} - \frac{2}{8} = \frac{11}{8} \]

Thus,

\[ a - b = \frac{11}{8} - \frac{1}{4} = \frac{11}{8} - \frac{2}{8} = \frac{9}{8} \]

Since \( f(x) \) is differentiable for all \( x \in \mathbb{R} \), \( f'(x) \) must be continuous for all \( x \in \mathbb{R} \). We have:

\[ f'(x) = \begin{cases} 2ax + b, & x \leq 1 \\[8pt] 3, & 1 < x \leq 2 \\[8pt] 3bx^2, & x > 2 \end{cases} \]

For continuity of \( f'(x) \) at \( x = 1 \), we must have:

\[ 2a(1) + b = 3 \] \[ 2a + b = 3 \]

For continuity of \( f'(x) \) at \( x = 2 \), we must have:

\[ 3 = 3b(2)^2 \] \[ 3 = 12b \] \[ b = \frac{1}{4} \]

Then,

\[ 2a + \frac{1}{4} = 3 \] \[ 2a = \frac{11}{4} \] \[ a = \frac{11}{8} \]

Thus,

\[ a - b = \frac{11}{8} - \frac{1}{4} = \frac{11}{8} - \frac{2}{8} = \frac{9}{8} \]