If the function f(x) = $$\left\{ {\matrix{ { - x} & {x < 1} \cr {a + {{\cos }^{ - 1}}\left( {x + b} \right),} & {1 \le x \le 2} \cr } } \right.$$ is differentiable at x = 1, then ${a \over b}$ is equal to :

Question

If the function

f(x) = $$\left\{ {\matrix{ { - x} & {x < 1} \cr {a + {{\cos }^{ - 1}}\left( {x + b} \right),} & {1 \le x \le 2} \cr } } \right.$$

is differentiable at x = 1, then ${a \over b}$ is equal to :

${{\pi - 2} \over 2}$

${{ - \pi - 2} \over 2}$

${{\pi + 2} \over 2}$

$ - 1 - {\cos ^{ - 1}}\left( 2 \right)$

Solution

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