Let _ x 2 ( (x-2))(rx^2 + (p-2)x - 2p) (x-2)^2 = 5 for some r, p R . If the set of all possible values of q , such that the roots of the equation rx^2 - px + q = 0 lie in (0, 2) , be the interval ( , ] , then 4( + ) equals :

Step-by-step solution

Given limit is _ x 2 (x-2) x-2 rx^2 + (p-2)x - 2p x-2 = 5 Since _ x 2 (x-2) x-2 = 1 , we have: _ x 2 rx^2 + (p-2)x - 2p x-2 = 5 For the limit to exist, the numerator must be zero at x = 2 : r(2)^2 + (p-2)(2) - 2p = 0 4r + 2p - 4 - 2p = 0 4r = 4 r = 1 Substituting r = 1 into the limit expression: _ x 2 x^2 + (p-2)x - 2p x-2 = 5 _ x 2 (x-2)(x+p) x-2 = 5 _ x 2 (x+p) = 5 2 + p = 5 p = 3 The quadratic equation is rx^2 - px + q = 0 , which becomes x^2 - 3x + q = 0 . For both roots of f(x) = x^2 - 3x + q = 0 to lie in the