If `f(x)={{:(a+tan^(-1)(x-b),,,x ge1),((x)/(2),,,xlt1):}` is differentiable at x = 1, then `4a-b` can be

Options

A. 0
B. 1
C. `-1`
D. `pi`

Answer

To solve the problem, we need to ensure that the piecewise function \( f(x) \) is differentiable at \( x = 1 \). This means the function must be continuous at that point, and the derivatives from both sides must also match.

Given the function:
\[ f(x) = \begin{cases} a + \tan^{-1}(x - b) & \text{if } x \geq 1 \\ \frac{x}{2} & \text{if } x < 1 \end{cases} \]

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

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

1. **Evaluate \( f(1) \)** from the right:
\[ f(1) = a + \tan^{-1}(1 - b) = a + \tan^{-1}(1 - b) \]

2. **Evaluate \( f(1) \)** from the left:
\[ f(1) = \frac{1}{2} \]

Setting these equal for continuity:
\[ a + \tan^{-1}(1 - b) = \frac{1}{2} \quad \text{(Equation 1)} \]

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

Next, we need to ensure that the derivatives from both sides are equal at \( x = 1 \).

1. **Derivative for \( x \geq 1 \)**:
\[ f'(x) = \frac{1}{1 + (x - b)^2} \]
Evaluating at \( x = 1 \):
\[ f'(1) = \frac{1}{1 + (1 - b)^2} \]

2. **Derivative for \( x < 1 \)**:
\[ f'(x) = \frac{1}{2} \]

Setting these equal for differentiability:
\[ \frac{1}{1 + (1 - b)^2} = \frac{1}{2} \quad \text{(Equation 2)} \]

### Step 3: Solve Equation 2

From Equation 2:
\[ 1 + (1 - b)^2 = 2 \]
\[ (1 - b)^2 = 1 \]
Taking the square root:
\[ 1 - b = 1 \quad \text{or} \quad 1 - b = -1 \]
This gives us two cases:
1. \( b = 0 \)
2. \( b = 2 \)

### Step 4: Solve for \( a \) using Equation 1

#### Case 1: \( b = 0 \)
Substituting \( b = 0 \) into Equation 1:
\[ a + \tan^{-1}(1 - 0) = \frac{1}{2} \]
\[ a + \frac{\pi}{4} = \frac{1}{2} \]
\[ a = \frac{1}{2} - \frac{\pi}{4} \]

#### Case 2: \( b = 2 \)
Substituting \( b = 2 \) into Equation 1:
\[ a + \tan^{-1}(1 - 2) = \frac{1}{2} \]
\[ a + \tan^{-1}(-1) = \frac{1}{2} \]
\[ a - \frac{\pi}{4} = \frac{1}{2} \]
\[ a = \frac{1}{2} + \frac{\pi}{4} \]

### Step 5: Calculate \( 4a - b \)

#### For \( b = 0 \):
\[ 4a - b = 4\left(\frac{1}{2} - \frac{\pi}{4}\right) - 0 = 2 - \pi \]

#### For \( b = 2 \):
\[ 4a - b = 4\left(\frac{1}{2} + \frac{\pi}{4}\right) - 2 = 2 + \pi - 2 = \pi \]

### Conclusion

The possible values for \( 4a - b \) are \( 2 - \pi \) and \( \pi \). Among the options provided, the only valid answer is:
\[ \boxed{\pi} \]