If `f(x)={{:(,ax^(2)-b,|x|lt 1),(,(1)/(|x|),|x| ge1):}` is differentiable at x=1, then

Options

A. `a=(1)/(2),b=-(1)/(2)`
B. `a=-(1)/(2),b=-(3)/(2)`
C. `a=b=(1)/(2)`
D. `a=b=-(1)/(2)`

Answer

To determine the values of \( a \) and \( b \) such that the function

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

is differentiable at \( x = 1 \), we need to ensure that the function is both continuous and differentiable at that point.

### Step 1: Check 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 at \( x = 1 \) and also equal \( f(1) \).

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

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

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

Setting the left-hand limit equal to the right-hand limit gives:
\[ a - b = 1 \quad \text{(Equation 1)} \]

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

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

1. **Calculate the left-hand derivative**:
\[ f'(x) = \frac{d}{dx}(ax^2 - b) = 2ax \quad \text{for } |x| < 1 \]
Thus,
\[ f'(1) = 2a(1) = 2a \]

2. **Calculate the right-hand derivative**:
\[ f'(x) = \frac{d}{dx}\left(\frac{1}{|x|}\right) = -\frac{1}{x^2} \quad \text{for } |x| \geq 1 \]
Thus,
\[ f'(1) = -\frac{1}{1^2} = -1 \]

Setting the left-hand derivative equal to the right-hand derivative gives:
\[ 2a = -1 \quad \text{(Equation 2)} \]

### Step 3: Solve the Equations

From Equation 2:
\[ a = -\frac{1}{2} \]

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

### Final Result

The values of \( a \) and \( b \) that make \( f(x) \) differentiable at \( x = 1 \) are:
\[ a = -\frac{1}{2}, \quad b = -\frac{3}{2} \]