Let g x = f x + f 1 - x and f " x > 0 , x ∈ 0 , 1 . If g is decreasing in the interval 0 , α and increasing in the interval α , 1 , then tan - 1 2 α + tan - 1 1 α + tan - 1 α + 1 α is equal to

Step-by-step solution

Given, g x = f x + f 1 - x and f " x > 0 , x ∈ 0 , 1 . If g is decreasing in the interval 0 , α and increasing in the interval α , 1 , Now solving, g x = f x + f 1 - x Now differentiating both side we get, g ' x = f ' x - f ' 1 - x Differentiating again we get, g " x = f " x + f " 1 - x > 0 So, g ' x is increasing as given f " x > 0 ⇒ g ' 0 < g ' 1   ⇒ f ' 0 - f ' 1 < f ' 1 - f ' 0 ⇒ f ' 0 &#60