Let the function f x = 2 x 2 - log e x , x > 0 , be decreasing in 0 , a and increasing in a , 4 . A tangent to the parabola y 2 = 4 a x at a point P on it passes through the point 8 a , 8 a - 1 but does not pass through the point - 1 a , 0 . If the equation of the normal at P is x α + y β = 1 , then α + β is equal to

Step-by-step solution

Given, f x = 2 x 2 - log e x ⇒ f ' x = 4 x - 1 x ⇒ f ' x = 4 x 2 - 1 x ⇒ f ' x = 0 ⇒ 4 x 2 - 1 = 0 ⇒ x = ± 1 2 But given x > 0   so   x = 1 2 So function is decreasing in 0 , 1 2 and increasing in the interval 1 2 , ∞ So, a = 1 2 Now equation of parabola will be y 2 = 2 x Now tangent to y 2 = 2 x will be given by, y = m x + 1 2 m , given this tangent passes through 8 a , 8 a - 1 ≡ 4 , 3 , So 3 = 4 m + 1 2 m ⇒ m = 1 2   or  1 4