If the function $$f(x) = \begin{cases} -x, & x < 1 \\ a + \cos^{-1}(x+b), & 1 \leq x \leq 2 \end{cases}$$ is differentiable at $$x = 1$$, then $$\frac{a}{b}$$ is equal to

To solve the problem, we need to find the value of $$\frac{a}{b}$$ such that the function $$f(x)$$ is differentiable at $$x = 1$$. The function is defined as:

$$ f(x) = \begin{cases} -x, & x < 1 \\ a + \cos^{-1}(x+b), & 1 \leq x \leq 2 \end{cases} $$

For a function to be differentiable at a point, it must be continuous at that point, and the left-hand derivative must equal the right-hand derivative at that point.

Step 1: Ensure continuity at $$x = 1$$

For continuity 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, and both must equal $$f(1)$$.

The left-hand limit ($$x \to 1^-$$) uses the first piece of the function:

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

The right-hand limit ($$x \to 1^+$$) and $$f(1)$$ use the second piece:

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

Setting these equal for continuity:

$$ -1 = a + \cos^{-1}(1 + b) \quad \text{(Equation 1)} $$

Step 2: Ensure differentiability at $$x = 1$$

The derivative from the left at $$x = 1$$ is found using the first piece. The derivative of $$-x$$ is $$-1$$, so:

$$ f'_{-}(1) = -1 $$

The derivative from the right at $$x = 1$$ is found using the second piece. The derivative of $$a + \cos^{-1}(x + b)$$ with respect to $$x$$ is computed as follows. Let $$u = x + b$$, so $$\frac{du}{dx} = 1$$. The derivative of $$\cos^{-1}(u)$$ is $$-\frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}$$. Thus:

$$ \frac{d}{dx} \left[ a + \cos^{-1}(x + b) \right] = -\frac{1}{\sqrt{1 - (x + b)^2}} \cdot 1 = -\frac{1}{\sqrt{1 - (x + b)^2}} $$

Evaluating at $$x = 1$$:

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

For differentiability, set the left-hand derivative equal to the right-hand derivative:

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

Simplify this equation. Multiply both sides by $$-1$$:

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

Take reciprocals on both sides (since both sides are positive):

$$ \sqrt{1 - (1 + b)^2} = 1 $$

Square both sides to eliminate the square root:

$$ 1 - (1 + b)^2 = 1 $$

Subtract 1 from both sides:

$$ - (1 + b)^2 = 0 $$

Thus:

$$ (1 + b)^2 = 0 $$

Taking square roots:

$$ 1 + b = 0 $$

Solving for $$b$$:

$$ b = -1 $$

Step 3: Find $$a$$ using continuity

Substitute $$b = -1$$ into Equation 1:

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

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

$$ -1 = a + \frac{\pi}{2} $$

Solving for $$a$$:

$$ a = -1 - \frac{\pi}{2} $$

Step 4: Compute $$\frac{a}{b}$$

Now substitute $$a = -1 - \frac{\pi}{2}$$ and $$b = -1$$:

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

Rewrite as a single fraction:

$$ 1 + \frac{\pi}{2} = \frac{2}{2} + \frac{\pi}{2} = \frac{2 + \pi}{2} = \frac{\pi + 2}{2} $$

Step 5: Verify the solution

Check if the derivative expression is valid at $$x = 1$$ with $$b = -1$$. The denominator in the derivative was $$\sqrt{1 - (x + b)^2}$$. At $$x = 1$$, $$x + b = 1 + (-1) = 0$$, so:

$$ \sqrt{1 - (0)^2} = \sqrt{1} = 1 \neq 0 $$

This is defined and non-zero, so the derivative exists. Also, the square root was valid before squaring because we had $$\sqrt{1 - (1+b)^2} = \sqrt{1 - 0} = 1 > 0$$.

Comparing with the options:

A. $$\frac{\pi + 2}{2}$$

B. $$\frac{\pi - 2}{2}$$

C. $$\frac{-\pi - 2}{2}$$

D. $$-1 - \cos^{-1}(2)$$

Our result matches option A.

Hence, the correct answer is Option A.