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

The problem asks for the value of \( 48(a+b) \), given that the piecewise function \( f(x) \) is differentiable on the entire set of real numbers \( \mathbb{R} \).

Concept Used:

For a function to be differentiable at a point, it must first be continuous at that point. The conditions are as follows:

1. Continuity at a point \( c \): The left-hand limit (LHL), right-hand limit (RHL), and the function's value at the point must all be equal.

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

2. Differentiability at a point \( c \): The left-hand derivative (LHD) and the right-hand derivative (RHD) at the point must be equal.

\[ \lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h} = \lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h} \]

For piecewise functions, it is often easier to compute the derivatives of the pieces and then check if their limits are equal at the boundary points.

\[ \lim_{x \to c^-} f'(x) = \lim_{x \to c^+} f'(x) \]

The points where the function definition changes are at \( |x| = 2 \), which means we need to check continuity and differentiability at \( x = 2 \) and \( x = -2 \).

Step-by-Step Solution:

Step 1: Write the function explicitly without the absolute value notation.

\[ f(x) = \begin{cases} -\frac{1}{x}, & x \leq -2 \\ ax^2 + 2b, & -2 < x < 2 \\ \frac{1}{x}, & x \geq 2 \end{cases} \]

Step 2: Apply the continuity condition.

For continuity at \( x = 2 \):

\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) \] \[ \lim_{x \to 2^-} (ax^2 + 2b) = \lim_{x \to 2^+} \frac{1}{x} \] \[ a(2)^2 + 2b = \frac{1}{2} \implies 4a + 2b = \frac{1}{2} \quad \text{(Equation 1)} \]

For continuity at \( x = -2 \):

\[ \lim_{x \to -2^-} f(x) = \lim_{x \to -2^+} f(x) \] \[ \lim_{x \to -2^-} \left(-\frac{1}{x}\right) = \lim_{x \to -2^+} (ax^2 + 2b) \] \[ -\frac{1}{-2} = a(-2)^2 + 2b \implies \frac{1}{2} = 4a + 2b \]

Both points give the same equation for continuity.

Step 3: Find the derivatives of the function's pieces.

\[ f'(x) = \begin{cases} \frac{1}{x^2}, & x < -2 \\ 2ax, & -2 < x < 2 \\ -\frac{1}{x^2}, & x > 2 \end{cases} \]

Step 4: Apply the differentiability condition.

For differentiability at \( x = 2 \):

\[ \lim_{x \to 2^-} f'(x) = \lim_{x \to 2^+} f'(x) \] \[ \lim_{x \to 2^-} (2ax) = \lim_{x \to 2^+} \left(-\frac{1}{x^2}\right) \] \[ 2a(2) = -\frac{1}{2^2} \implies 4a = -\frac{1}{4} \implies a = -\frac{1}{16} \]

For differentiability at \( x = -2 \):

\[ \lim_{x \to -2^-} f'(x) = \lim_{x \to -2^+} f'(x) \] \[ \lim_{x \to -2^-} \left(\frac{1}{x^2}\right) = \lim_{x \to -2^+} (2ax) \] \[ \frac{1}{(-2)^2} = 2a(-2) \implies \frac{1}{4} = -4a \implies a = -\frac{1}{16} \]

Both points yield the same value for \(a\).

Step 5: Substitute the value of \(a\) into the continuity equation to find \(b\).

Using Equation 1: \( 4a + 2b = \frac{1}{2} \).

\[ 4\left(-\frac{1}{16}\right) + 2b = \frac{1}{2} \] \[ -\frac{1}{4} + 2b = \frac{1}{2} \] \[ 2b = \frac{1}{2} + \frac{1}{4} = \frac{3}{4} \] \[ b = \frac{3}{8} \]

Final Computation & Result:

Step 6: Calculate the value of the required expression \( 48(a+b) \).

First, find the sum \(a+b\):

\[ a + b = -\frac{1}{16} + \frac{3}{8} = -\frac{1}{16} + \frac{6}{16} = \frac{5}{16} \]

Now, multiply by 48:

\[ 48(a+b) = 48 \times \frac{5}{16} \] \[ = 3 \times 5 = 15 \]

The value of \( 48(a+b) \) is 15.