If the function
`f(x)={{:(,-x,x lt 1),(,a+cos^(-1)(x+b),1 le xle 2):}` is differentiable at x=1, then `(a)/(b)` is equal to

Options

A. `(-pi-2)/(2)`
B. `-1-cos^(-1)`
C. `(pi)/(2)+1`
D. `(pi)/(2)-1`

Explanation

To solve the problem, we need to ensure that the function \( f(x) \) is continuous and differentiable at \( x = 1 \). The function is defined as follows:

\[ f(x) = \begin{cases} -x & \text{if } x < 1 \\ a + \cos^{-1}(x + b) & \text{if } 1 \leq x \leq 2 \end{cases} \]

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

For \( f(x) \) to be continuous at \( x = 1 \), we need:

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

Calculating the left-hand limit:

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

Calculating the right-hand limit:

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

Setting these equal for continuity:

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

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

For \( f(x) \) to be differentiable at \( x = 1 \), we need:

\[ f'(1^-) = f'(1^+) \]

Calculating the left-hand derivative:

\[ f'(x) = -1 \quad \text{for } x < 1 \quad \Rightarrow \quad f'(1^-) = -1 \]

Calculating the right-hand derivative:

Using the derivative of \( \cos^{-1}(x) \):

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

Thus,

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

Setting these equal for differentiability:

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

### Step 3: Solve the Differentiability Equation

From the equation:

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

Squaring both sides:

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

This implies:

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

Thus:

\[ 1 + b = 0 \quad \Rightarrow \quad b = -1 \]

### Step 4: Substitute \( b \) into the Continuity Equation

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

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

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

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

### Step 5: Find \( \frac{a}{b} \)

Now we have:

\[ a = -1 - \frac{\pi}{2}, \quad b = -1 \]

Calculating \( \frac{a}{b} \):

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

Thus, the final answer is:

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

### Final Answer:

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

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