Question

If the function \( f(x) = \begin{cases} \frac{1}{|x|}, & |x| \geq 2 \\ ax^2 + 2b, & |x| < 2 \end{cases} \) is differentiable on \( \mathbb{R} \), then \( 48 (a + b) \) is equal to \(\_\_\_\_\_\_\_\_\_\).

Answer 1

Correct Answer: 15

Continuity at \( x = \pm 2 \): For \( f(x) \) to be continuous at \( x = 2 \), we need:
\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^-} f(x) = f(2) \]
Since \( f(x) = \frac{1}{x} \) for \( |x| \geq 2 \), the limit as \( x \to 2 \) from the right is \( \frac{1}{2} \). For \( f(x) = ax^2 + 2b \) on \( |x| < 2 \), setting \( f(2) = \frac{1}{2} \):
\[ a \times 4 + 2b = \frac{1}{2} \Rightarrow 4a + 2b = \frac{1}{2} \]
Similarly, for \( x = -2 \), we get the same equation, ensuring continuity:
\[ 4a + 2b = \frac{1}{2} \]
Differentiability at \( x = \pm 2 \): For differentiability at \( x = 2 \), calculate \( f'(x) \) for both cases:
\[ f'(x) = -\frac{1}{x^2} \quad \text{for } |x| \geq 2 \]
Using \( f(x) = ax^2 + 2b \) for \( |x| < 2 \):
\[ f'(2) = -\frac{1}{4} = 2a \Rightarrow a = -\frac{1}{8} \]
Substitute \( a = -\frac{1}{8} \) into \( 4a + 2b = \frac{1}{2} \) to solve for \( b \):
\[ b = \frac{3}{8} \]
Calculate \( 48(a + b) \): Substitute \( a = -\frac{1}{8} \) and \( b = \frac{3}{8} \):
\[ 48(a + b) = 48 \left(-\frac{1}{8} + \frac{3}{8}\right) = 48 \times \frac{1}{4} = 15 \]