If `f(x)={{:(a+cos^(-1)(x+b),":",xge1),(-x,":",xlt1):}` is differentiable at x = 1, then the value of `b-a` is equal to

Options

A. 0
B. 1
C. `-1`
D. `(pi)/(2)`

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:

\[ f(x) = \begin{cases} a + \cos^{-1}(x + b) & \text{if } x \geq 1 \\ -x & \text{if } x < 1 \end{cases} \]

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

For \( f(x) \) to be continuous at \( x = 1 \), we need the left-hand limit (LHL) and right-hand limit (RHL) to be equal at that point.

1. **Left-hand limit as \( x \) approaches 1:**
\[ \text{LHL} = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (-x) = -1 \]

2. **Right-hand limit as \( x \) approaches 1:**
\[ \text{RHL} = \lim_{x \to 1^+} f(x) = a + \cos^{-1}(1 + b) \]
Since \( \cos^{-1}(1) = 0 \), we have:
\[ \text{RHL} = a + \cos^{-1}(1 + b) = a + 0 = a \]

Setting LHL equal to RHL for continuity:
\[ -1 = a \]

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

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

1. **Left-hand derivative:**
\[ f'(x) = -1 \quad \text{for } x < 1 \]
Therefore,
\[ \text{LHD} = -1 \]

2. **Right-hand derivative:**
We need to differentiate \( f(x) = a + \cos^{-1}(x + b) \):
\[ f'(x) = -\frac{1}{\sqrt{1 - (x + b)^2}} \quad \text{for } x \geq 1 \]
Evaluating at \( x = 1 \):
\[ \text{RHD} = -\frac{1}{\sqrt{1 - (1 + b)^2}} = -\frac{1}{\sqrt{1 - (1 + b)^2}} \]

Setting LHD equal to RHD:
\[ -1 = -\frac{1}{\sqrt{1 - (1 + b)^2}} \]
This simplifies to:
\[ 1 = \frac{1}{\sqrt{1 - (1 + b)^2}} \]
Squaring both sides gives:
\[ 1 = 1 - (1 + b)^2 \]
Rearranging:
\[ (1 + b)^2 = 0 \]
Thus:
\[ 1 + b = 0 \implies b = -1 \]

### Step 3: Find \( b - a \)

Now we have:
- \( a = -1 \)
- \( b = -1 \)

Calculating \( b - a \):
\[ b - a = -1 - (-1) = -1 + 1 = 0 \]

### Final Answer
The value of \( b - a \) is \( 0 \).