Let `f(x) = {{:(-1/|x|, "for " |x| ge 1),(ax^(2)-b, "for " |x| lt 1):}`, If f(x) is continuous and differentiable at any point, then:

Options

A. `a=1/2, b=-3/2`
B. `a=1/2, b=3/2`
C. `a=1, b=-1`
D. none of these

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous and differentiable at \( x = 1 \). The function is defined as follows:

\[ f(x) = \begin{cases} -\frac{1}{|x|} & \text{for } |x| \geq 1 \\ ax^2 - b & \text{for } |x| < 1 \end{cases} \]

### Step 1: Ensure Continuity at \( x = 1 \)

For \( f(x) \) to be continuous at \( x = 1 \), the left-hand limit as \( x \) approaches 1 must equal the right-hand limit and the function value at that point.

1. **Evaluate \( f(1) \)**:
\[ f(1) = -\frac{1}{|1|} = -1 \]

2. **Evaluate the left-hand limit as \( x \) approaches 1**:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (ax^2 - b) = a(1^2) - b = a - b \]

3. **Evaluate the right-hand limit as \( x \) approaches 1**:
\[ \lim_{x \to 1^+} f(x) = -\frac{1}{|1|} = -1 \]

Setting these equal for continuity:
\[ a - b = -1 \quad \text{(Equation 1)} \]

### Step 2: Ensure Differentiability at \( x = 1 \)

For \( f(x) \) to be differentiable at \( x = 1 \), the left-hand derivative must equal the right-hand derivative.

1. **Evaluate the left-hand derivative**:
\[ f'(x) = \frac{d}{dx}(ax^2 - b) = 2ax \]
\[ \lim_{x \to 1^-} f'(x) = 2a(1) = 2a \]

2. **Evaluate the right-hand derivative**:
\[ f'(x) = \frac{d}{dx}\left(-\frac{1}{|x|}\right) = \frac{1}{x^2} \]
\[ \lim_{x \to 1^+} f'(x) = \frac{1}{1^2} = 1 \]

Setting these equal for differentiability:
\[ 2a = 1 \quad \text{(Equation 2)} \]

### Step 3: Solve the System of Equations

Now we have two equations:
1. \( a - b = -1 \)
2. \( 2a = 1 \)

From Equation 2, we can solve for \( a \):
\[ a = \frac{1}{2} \]

Substituting \( a \) into Equation 1:
\[ \frac{1}{2} - b = -1 \]
\[ -b = -1 - \frac{1}{2} = -\frac{3}{2} \]
\[ b = \frac{3}{2} \]

### Final Values

Thus, the values of \( a \) and \( b \) are:
\[ a = \frac{1}{2}, \quad b = \frac{3}{2} \]