If the function f(x) = begin{cases} -x, & x

Question

(A) For $f(x)$ to be differentiable at $x = 1$,it must be continuous at $x = 1$.$\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) = f(1)$$-1 = a + \cos^{

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

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(A) For $f(x)$ to be differentiable at $x = 1$,it must be continuous at $x = 1$.$\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) = f(1)$$-1 = a + \cos^{-1}(1 + b) \implies \cos^{-1}(1 + b) = -1 - a \quad \dots(1)$For differentiability,$LHD = RHD$ at $x = 1$.$LHD = \frac{d}{dx}(-x) = -1$.$RHD = \frac{d}{dx}(a + \cos^{-1}(x + b)) = \frac{-1}{\sqrt{1 - (x + b)^2}}$.At $x = 1$,$RHD = \frac{-1}{\sqrt{1 - (1 + b)^2}}$.Equating $LHD = RHD$: $-1 = \frac{-1}{\sqrt{1 - (1 + b)^2}} \implies \sqrt{1 - (1 + b)^2} = 1$.Squaring both sides: $1 - (1 + b)^2 = 1 \implies (1 + b)^2 = 0 \implies b = -1$.Substituting $b = -1$ into equation $(1)$: $\cos^{-1}(0) = -1 - a$.Since $\cos^{-1}(0) = \frac{\pi}{2}$,we have $\frac{\pi}{2} = -1 - a \implies a = -1 - \frac{\pi}{2} = \frac{-\pi - 2}{2}$.Therefore,$\frac{a}{b} = \frac{(-\pi - 2)/2}{-1} = \frac{\pi + 2}{2}$.

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

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