Let x = 2 be a root of the equation x 2 + p x + q = 0 and f x = 1 - cos x 2 - 4 p x + q 2 + 8 q + 16 x - 2 p 4 , x ≠ 2 p 0 , x = 2 p . Then lim x → 2 p + f x where · denotes greatest integer function, is

Step-by-step solution

Given, x = 2 be the root of the given equation x 2 + p x + q = 0 , Putting x = 2 in given equation we get, ⇒ 4 + 2 p + q = 0 ⇒ q + 4 = - 2 p ∵   x 2 - 4 p x + q 2 + 8 q + 16 = x 2 - 4 p x + q + 4 2 = x 2 - 4 p x + 4 p 2                 ( ∵ q + 4 = - 2 p ) = ( x - 2 p ) 2 Now, solving the limit lim x → 2 p + f x = lim x → 2 p + 1 - cos x - 2 p 2 x - 2 p 4 Let x - 2 p = θ ⇒ lim θ → 0 + f x = lim θ → 0 + 1