As   f(x) is differentiable at x = 1

$$ \therefore $$   $$\mathop {\lim }\limits_{x \to {1^ - }} \left( { - x} \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {a + {{\cos }^{ - 1}}\left( {x + b} \right)} \right) = f(1)$$

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

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

As   f(x) is differentiable, so,

2 . H . D $$=$$ R . H . D

Here, L . H . D $$=$$ $$\mathop {\lim }\limits_{h \to 0} $$ $${{f\left( {1 - h} \right) - f\left( 1 \right)} \over { - h}}$$

$$ = \mathop {\lim }\limits_{h \to 0} {{ - \left( {1 - h} \right) - \left( { - 1} \right)} \over { - h}}$$

$$ = \mathop {\lim }\limits_{x \to 0} {{ - 1 + h + 1} \over { - h}}$$

$$=$$ $$ = \mathop {\lim }\limits_{x \to 0} {h \over { - h}}$$

$$=$$ $$-$$ 1

R. H. D $$ = \mathop {\lim }\limits_{x \to 0} {{f\left( {1 + h} \right) - f\left( 1 \right)} \over h}$$

$$ = \mathop {\lim }\limits_{h \to 0} {{a + {{\cos }^{ - 1}}\left( {1 + h + b} \right) - \left[ {a + {{\cos }^{ - 1}}\left( {1 + b} \right)} \right]} \over h}$$

$$ = \mathop {\lim }\limits_{h \to 0} {{{{\cos }^{ - 1}}\left( {1 + h + b} \right) - {{\cos }^{ - 1}}\left( {1 + b} \right)} \over h}\left[ {{0 \over 0}\,\,form\left. \, \right]} \right.$$

$$ = \mathop {\lim }\limits_{h \to 0} {{ - 1} \over {\sqrt {1 - {{\left( {1 + h + b} \right)}^2}} }}$$ [ Using L' Hospital Rule]

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

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

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

$$ \Rightarrow $$  $${\left( {1 + b} \right)^2} = 0$$

$$ \Rightarrow $$   $$b = - 1$$

putting value of b in equation (1), we get,

$${\cos ^{ - 1}}\left( {1 - 1} \right) = - 1 - a$$

$$ \Rightarrow $$   $${\pi \over 2} = - 1 - a$$

$$ \Rightarrow $$   $$a = - 1 - {\pi \over 2}$$

$$ \therefore $$   $${a \over b} = {{ - 1 - {\pi \over 2}} \over { - 1}} = 1 + {\pi \over 2} = {{\pi + 2} \over 2}$$