JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

For the quadratic expression ax^2 + 2 2 bx + c > 0 for all x R , the leading coefficient must be positive and the discriminant must be negative. Since a \ 1, 2, 3, 4\ , the condition a > 0 is always satisfied. The discriminant condition is: D (2 2 b)^2 - 4ac 8b^2 - 4ac 2b^2 The total number of possible triplets (a, b, c) is 4 4 4 = 64. Now, we find the number of favorable outcomes by checking the possible values of b \ 1, 2, 3, 4\ : Case 1: b = 1 We need 2(1)^2 2. The total number of pairs (a, c) is 4 4 = 16. The pairs for which ac 2 are (1, 1), (1, 2), (2, 1), which are 3 in number. So, the number of pairs with ac > 2 is 16 - 3 = 13. Case 2: b = 2 We need 2(2)^2 8. The possible pairs (a, c) from the given set are (3, 3), (3, 4), (4, 3), (4, 4). So, there are 4 pairs. Case 3: b = 3 We need 2(3)^2 18. Since the maximum possible value of ac is 4 4 = 16, there are 0 pairs. Case 4: b = 4 We need 2(4)^2 32. Again, there are 0 pairs. Total number of favorable outcomes = 13 + 4 = 17. The required probability is m n = 17 64 . Since (17, 64) = 1, we have m = 17 and n = 64. Therefore, m + n = 17 + 64 = 81. Answer: 81