MathematicsHard179×since 2002Q4181If 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 ab{a \over b}ba is equal to :Aπ−22{{\pi - 2} \over 2}2π−2B−π−22{{ - \pi - 2} \over 2}2−π−2Cπ+22{{\pi + 2} \over 2}2π+2D−1−cos−1(2)- 1 - {\cos ^{ - 1}}\left( 2 \right)−1−cos−1(2)Check answerSkip