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 determine the value of \( b - a \) given the function:

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

We know that for \( f(x) \) to be differentiable at \( x = 1 \), it must be continuous at that point. This means that the left-hand limit (LHL) and the right-hand limit (RHL) at \( x = 1 \) must be equal.

### Step 1: Find the left-hand limit as \( x \) approaches 1

For \( x < 1 \), the function is defined as \( f(x) = -x \). Therefore, the left-hand limit as \( x \) approaches 1 is:

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

### Step 2: Find the right-hand limit as \( x \) approaches 1

For \( x \geq 1 \), the function is defined as \( f(x) = a + \cos^{-1}(x + b) \). Therefore, the right-hand limit as \( x \) approaches 1 is:

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

### Step 3: Set the left-hand limit equal to the right-hand limit

Since \( f(x) \) is continuous at \( x = 1 \), we set the two limits equal to each other:

\[ -1 = a + \cos^{-1}(1 + b) \]

### Step 4: Find the derivatives

Next, we need to find the derivatives from both sides at \( x = 1 \).

- For \( x < 1 \), \( f'(x) = -1 \).
- For \( x \geq 1 \), we differentiate \( f(x) = a + \cos^{-1}(x + b) \):

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

### Step 5: Evaluate the derivative at \( x = 1 \)

We need to find the right-hand derivative at \( x = 1 \):

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

### Step 6: Set the derivatives equal

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

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

This simplifies to:

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

Squaring both sides gives:

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

### Step 7: Solve for \( b \)

Rearranging gives:

\[ (1 + b)^2 = 0 \implies 1 + b = 0 \implies b = -1 \]

### Step 8: Substitute \( b \) back to find \( a \)

Now substitute \( b = -1 \) into the continuity equation:

\[ -1 = a + \cos^{-1}(1 - 1) = a + \cos^{-1}(0) \]

Since \( \cos^{-1}(0) = \frac{\pi}{2} \):

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

### Step 9: Find \( b - a \)

Now we calculate \( b - a \):

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

### Final Answer

Thus, the value of \( b - a \) is:

\[ \boxed{\frac{\pi}{2}} \]