JEE Advanced Mathematics Question (2026) — Solution

Step-by-step solution

Let f(x) = g(x) + h(x), where g(x) = (|x| + |x - 1|) x and h(x) = [x x]. First, we analyze h(x) = [x x] on the interval (- 2 , 2 ). The function y = x x is an even function. For x [0, 2 ), x x is strictly increasing. At x = 0, x x = 0. As x 2 , x x 2 1.57. Thus, x x [0, 1.57) for x (- 2 , 2 ). The value of [x x] will be 0 when 0 x x Let x_1 (0, 2 ) be the unique point where x_1 x_1 = 1. By symmetry, -x_1 is the unique point in (- 2 , 0 ) where (-x_1) (-x_1) = 1. Therefore, h(x) has jump discontinuities at exactly two points: x = x_1 and x = -x_1. At x = 0, x x = 0 and is non-negative in its neighborhood, so [x x] = 0 around x = 0, making h(x) continuous and differentiable at x = 0. Next, we analyze g(x) = (|x| + |x - 1|) x. The absolute value functions have critical points at x = 0 and x = 1. For x in a small neighborhood of 0 (specifically x (-1, 1)): If x [0, 1), g(x) = (x - (x - 1)) x = x g'(0^+) = 0 = 1. If x (-1, 0), g(x) = (-x - (x - 1)) x = (1 - 2x) x g'(0^-) = -2 0 + 1 0 = 1. Since g'(0^+) = g'(0^-), g(x) is differentiable at x = 0. For x in a small neighborhood of 1: If x (0, 1], g(x) = x g'(1^-) = 1. If x (1, 2 ), g(x) = (x + (x - 1)) x = (2x - 1) x g'(1^+) = 2 1 + 1 1. Since g'(1^-) g'(1^+), g(x) is not differentiable at x = 1. Now, combining the results for f(x) = g(x) + h(x): 1. Discontinuities: g(x) is continuous everywhere. h(x) is discontinuous at x_1 and -x_1. Thus, f(x) is discontinuous at 2 points. = 2. 2. Non-differentiability: f(x) is not differentiable where it is discontinuous (x_1 and -x_1). Additionally, at x = 1, h(x) is constant (0) because 1 1 0.84 This gives 3 points of non-differentiability: -x_1, 1, x_1. = 3. Finally, + = 2 + 3 = 5. Answer: 5