If the function f(x) = \ cases - x, & x < 1 a + ^ - 1 (x + b), & 1...

f\left(x\right) = \left\{\begin{cases}- x & x < 1 \\ a + \cos^{- 1} \left(x + b\right) & 1 \le x \le 2\end{cases}\right)<br /><br />f\left(x\right)$is continuous at$x = 1$as$f\left(x\right)$is differentiable at$x = 1$.<br /><br />$⇒ \lim_{x \to 1} f( x ) = \lim_{x \to {1}^{+}} ( a + \cos^{- 1} ( x + b ) ) = f( 1 )$<br /><br />$⇒ - 1 = a + \cos^{- 1} \left(1 +b\right)$<br /><br />$\cos^{- 1} \left(1 + b\right) = - 1 - a$........\left(i\right)$

$f\left(x\right)$ is differentiable at $x = 1$.

$⇒$ $LHD = RHD$

$⇒$ $- 1 = \dfrac{- 1}{\sqrt{1 - {\left(1 + b\right)}^{2}}}$

$⇒$ $1 - {\left(1 + b\right)}^{2} = 1$

$⇒$ $b = - 1$ ......\left(ii\right)$<br /><br />From \left(i\right)$⇒ \cos^{- 1} ( 0 ) = - 1 - a$<br /><br />$∴ - 1 - a = \dfrac{\pi}{2}$<br /><br />$a = - 1 - \dfrac{\pi}{2}$<br /><br />$a = \dfrac{- \pi - 2}{2}$......\left(iii\right)$

$∴$ $\dfrac{a}{b} = \dfrac{\pi + 2}{2}$