JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Given equation: 3 ^2 x + 12 x - 3 = p Substituting ^2 x = 1 - ^2 x: 3(1 - ^2 x) + 12 x - 3 = p -3 ^2 x + 12 x = p Let x = t. Since x R , t [-1, 1]. The equation becomes p = -3t^2 + 12t. Let f(t) = -3t^2 + 12t. Differentiating with respect to t: f'(t) = -6t + 12 = 6(2 - t) For t [-1, 1], f'(t) > 0, which means f(t) is strictly increasing in the interval [-1, 1]. The minimum value of f(t) occurs at t = -1: f(-1) = -3(-1)^2 + 12(-1) = -15 The maximum value of f(t) occurs at t = 1: f(1) = -3(1)^2 + 12(1) = 9 Thus, the range of p for which the equation has at least one solution is [-15, 9]. The integral values of p are -15, -14, -13, , 9. The sum of these integral values is: _ k=-15 ^ 9 k = (-15) + (-14) + (-13) + (-12) + (-11) + (-10) + _ k=-9 ^ 9 k Since _ k=-9 ^ 9 k = 0, the sum simplifies to: -15 - 14 - 13 - 12 - 11 - 10 = -75 Answer: -75