JEE Main Mathematics Question (2026) — Solution

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 interval (0, 2), the following conditions must hold: 1) Discriminant D 0 (-3)^2 - 4(1)(q) 0 9 - 4q 0 q 9 4 2) f(0) > 0 q > 0 3) f(2) > 0 2^2 - 3(2) + q > 0 q - 2 > 0 q > 2 4) The x-coordinate of the vertex must lie in (0, 2) 0 Taking the intersection of all conditions, we get q (2, 9 4 ]. Comparing this with ( , ], we have = 2 and = 9 4 . Therefore, 4( + ) = 4(2 + 9 4 ) = 8 + 9 = 17. Answer: 17