JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Let g(x) = x^2 - x - 1 2 . Differentiating with respect to x, we get g'(x) = 2x - 1. For x [2, 4], g'(x) > 0, which implies that g(x) is strictly increasing in the interval [2, 4]. The values of g(x) at the endpoints are: g(2) = 2^2 - 2 - 1 2 = 1.5 g(4) = 4^2 - 4 - 1 2 = 11.5 The function f(x) = [g(x)] is discontinuous at all points where g(x) is an integer, as g(x) is strictly monotonic and crosses these integer values. The integers between 1.5 and 11.5 are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. Since g(x) is strictly increasing, it attains each of these 10 integer values exactly once in the interval (2, 4). At the endpoints x = 2 and x = 4, g(x) is not an integer, so f(x) is continuous from the right at x = 2 and continuous from the left at x = 4. Thus, the number of points of discontinuity in the interval [2, 4] is 10. Answer: 10