If the function `f(x) ={(1/|x|,|x|,ge,2),(zx^2+2b,|x|,<,2):}` differentiable on R then `48(a + b)` is equal to

Options

A. 19
B. 16
C. 15
D. 20

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous and differentiable at the points where its definition changes, specifically at \( x = 2 \) and \( x = -2 \).

### Step-by-Step Solution:

1. **Define the function**:
\[ f(x) = \begin{cases} \frac{1}{|x|} & \text{if } |x| \geq 2 \\ ax^2 + 2b & \text{if } |x| < 2 \end{cases} \]

2. **Check continuity at \( x = 2 \)**:
- The value of \( f(2) \) is:
\[ f(2) = \frac{1}{2} \]
- The limit from the left as \( x \) approaches 2:
\[ \lim_{x \to 2^-} f(x) = a(2^2) + 2b = 4a + 2b \]
- The limit from the right as \( x \) approaches 2:
\[ \lim_{x \to 2^+} f(x) = \frac{1}{2} \]
- Setting these equal for continuity:
\[ 4a + 2b = \frac{1}{2} \quad \text{(1)} \]

3. **Check continuity at \( x = -2 \)**:
- The value of \( f(-2) \) is:
\[ f(-2) = \frac{1}{2} \]
- The limit from the left as \( x \) approaches -2:
\[ \lim_{x \to -2^-} f(x) = a(-2)^2 + 2b = 4a + 2b \]
- The limit from the right as \( x \) approaches -2:
\[ \lim_{x \to -2^+} f(x) = \frac{1}{2} \]
- Setting these equal for continuity:
\[ 4a + 2b = \frac{1}{2} \quad \text{(2)} \]

4. **Differentiate the function**:
- Left-hand derivative at \( x = 2 \):
\[ f'(x) = 2ax \quad \text{for } |x| < 2 \quad \Rightarrow \quad f'(2^-) = 4a \]
- Right-hand derivative at \( x = 2 \):
\[ f'(x) = -\frac{1}{x^2} \quad \Rightarrow \quad f'(2^+) = -\frac{1}{4} \]
- Setting these equal for differentiability:
\[ 4a = -\frac{1}{4} \quad \Rightarrow \quad a = -\frac{1}{16} \quad \text{(3)} \]

5. **Differentiate at \( x = -2 \)**:
- Left-hand derivative at \( x = -2 \):
\[ f'(-2^-) = 4a \]
- Right-hand derivative at \( x = -2 \):
\[ f'(-2^+) = -\frac{1}{4} \]
- Setting these equal for differentiability:
\[ 4a = -\frac{1}{4} \quad \Rightarrow \quad a = -\frac{1}{16} \quad \text{(same as before)} \]

6. **Substituting \( a \) into equation (1)**:
- Substitute \( a = -\frac{1}{16} \) into equation (1):
\[ 4\left(-\frac{1}{16}\right) + 2b = \frac{1}{2} \]
\[ -\frac{1}{4} + 2b = \frac{1}{2} \]
\[ 2b = \frac{1}{2} + \frac{1}{4} = \frac{3}{4} \quad \Rightarrow \quad b = \frac{3}{8} \quad \text{(4)} \]

7. **Calculate \( 48(a + b) \)**:
- Substitute \( a \) and \( b \):
\[ a + b = -\frac{1}{16} + \frac{3}{8} = -\frac{1}{16} + \frac{6}{16} = \frac{5}{16} \]
- Now calculate:
\[ 48(a + b) = 48 \times \frac{5}{16} = 3 \times 5 = 15 \]

### Final Answer:
\[ 48(a + b) = 15 \]