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

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

It is given that f(x) is differentiable at x = 1. So, it is continuous at x = 1

∴  $\lim\limits_{x \rightarrow 1^{-}} f(x)=f(1)=\lim\limits_{x \rightarrow 1^{+}} f(x)$

$\Rightarrow \lim\limits_{x \rightarrow 1^{-}}-x=a+\cos ^{-1}(1+b)=\lim\limits_{x \rightarrow 1} a+\cos ^{-1}(x+b)$

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

$\Rightarrow \cos ^{-1}(1+b)=-1-a$                      .......(i)

f(x) is differentiable at x = 1

∴  (LHD at x = 1) = (RHD at x = 1)

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

$\Rightarrow \sqrt{1-(1+b)^2}=1 \Rightarrow(1+b)^2=0 \Rightarrow b=-1$

Putting b = -1 in (i), we get:

Hence, $\frac{a}{b}=\frac{\frac{-\pi}{2}-1}{-1}=\frac{\pi}{2}+1-1-a=\cos ^{-1}(0) \Rightarrow-1-a=\frac{\pi}{2} \Rightarrow a=-\frac{\pi}{2}-1$