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 function \( f(x) \) is both continuous and differentiable at \( x = 1 \). The function is defined as follows:

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

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

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

\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) \]

Calculating the left-hand limit:

\[ \lim_{x \to 1^-} f(x) = \frac{1}{2} \]

Calculating the right-hand limit:

\[ \lim_{x \to 1^+} f(x) = a + \tan^{-1}(1 - b) = a + \frac{\pi}{4} - b \]

Setting these equal gives us the first equation:

\[ a + \frac{\pi}{4} - b = \frac{1}{2} \quad \text{(1)} \]

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

For the function to be differentiable at \( x = 1 \), the derivatives from both sides must also be equal.

Calculating the derivative for \( x \geq 1 \):

\[ f'(x) = \frac{d}{dx}(a + \tan^{-1}(x - b)) = \frac{1}{1 + (x - b)^2} \]

Calculating the derivative for \( x < 1 \):

\[ f'(x) = \frac{1}{2} \]

Evaluating the derivative at \( x = 1 \):

\[ f'(1^+) = \frac{1}{1 + (1 - b)^2} \]

Setting the derivatives equal gives us the second equation:

\[ \frac{1}{1 + (1 - b)^2} = \frac{1}{2} \quad \text{(2)} \]

### Step 3: Solve Equation (2)

Cross-multiplying gives:

\[ 2 = 1 + (1 - b)^2 \]

Rearranging:

\[ (1 - b)^2 = 1 \]

Taking the square root:

\[ 1 - b = 1 \quad \text{or} \quad 1 - b = -1 \]

From \( 1 - b = 1 \):

\[ b = 0 \]

From \( 1 - b = -1 \):

\[ b = 2 \]

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

**Case 1:** If \( b = 0 \):

Substituting \( b = 0 \) into Equation (1):

\[ a + \frac{\pi}{4} = \frac{1}{2} \]

Thus,

\[ a = \frac{1}{2} - \frac{\pi}{4} \]

Calculating \( 4a - b \):

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

**Case 2:** If \( b = 2 \):

Substituting \( b = 2 \) into Equation (1):

\[ a + \frac{\pi}{4} - 2 = \frac{1}{2} \]

Thus,

\[ a = \frac{1}{2} + 2 - \frac{\pi}{4} = \frac{5}{2} - \frac{\pi}{4} \]

Calculating \( 4a - b \):

\[ 4a - b = 4\left(\frac{5}{2} - \frac{\pi}{4}\right) - 2 = 10 - \pi - 2 = 8 - \pi \]

### Conclusion

The possible values for \( 4a - b \) are \( 2 - \pi \) and \( 8 - \pi \).