Let f x = log e s i n x , 0 < x < π and g x = sin - 1 ( e - x ) , ( x ≥ 0 ) . If α is a positive real number such that a = f o g ' ( α ) and b = f o g ( α ) , then

Step-by-step solution

Given f x = log e s i n x ,   0 < x < π and g x = sin - 1 ( e - x ) ,   ( x ≥ 0 ) , then we have f o g x = f g x = f sin - 1 e - x = log e sin sin - 1 e - x = log e e - x = - x ⇒ f o g x = - x ⇒ b = f g α = - α Now ( f g ( x ) ) ' = - 1 ⇒ a = f g α ' = - 1 ∴   a α 2 - b α - a = - 1 α 2 - - α α - - 1 = - α 2 + α 2 + 1 = 1 .