If f(x) = \ cases a + cos^ - 1 (x + b) & : & x ≥ 1 - x & : & x < 1...

Since, $f ( x )$ is differentiable, it must be continuous at $x = 1$

$∴ f\left({1}^{-}\right) = f\left({1}^{+}\right) = f\left(1\right)<br /><br />$- 1 = a + co{s}^{- 1}\left(1 + b\right)$... \left(1\right)$

{f}^{'}\left(x\right) = \left\{\begin{cases}\dfrac{- 1}{\sqrt{1 - {\left(x + b\right)}^{2}}} & : & x \ge 1 \\ - 1 & : & x < 1\end{cases}\right)<br /><br />Now, forf\left(x\right) to be differentiable at $x = 1$

${f}^{′} \left({1}^{-}\right)$ = {f}^{′} \left({1}^{+}\right)$<br /><br />$\dfrac{- 1}{\sqrt{1 - {\left(1 + b\right)}^{2}}} = - 1 ⇒ 1 - {\left(1 + b\right)}^{2} = 1 ⇒ b = - 1$... \left(2\right)$

From equation \left(1\right)& \left(2\right)

- 1 = a + \dfrac{\pi}{2} ⇒ a = - \dfrac{\pi}{2} - 1