JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The possible points of discontinuity for the composite function g(f(x)) are the points where f(x) is discontinuous and the points where f(x) is equal to a point of discontinuity of g(x). First, we find the points of discontinuity of f(x) and g(x). For f(x), the only possible point of discontinuity is at x = 0. _ x 0^- f(x) = _ x 0^- (x^3 + 8) = 8 _ x 0^+ f(x) = _ x 0^+ (x^2 - 4) = -4 Since _ x 0^- f(x) _ x 0^+ f(x), f(x) is discontinuous at x = 0. For g(x), the only possible point of discontinuity is at x = 0. _ x 0^- g(x) = _ x 0^- (x - 8)^ 1/3 = -2 _ x 0^+ g(x) = _ x 0^+ (x + 4)^ 1/2 = 2 Since _ x 0^- g(x) _ x 0^+ g(x), g(x) is discontinuous at x = 0. Next, we find the points where f(x) equals the point of discontinuity of g(x), which is f(x) = 0. For x For x 0, x^2 - 4 = 0 x = 2. Thus, the possible points of discontinuity for g(f(x)) are x = -2, x = 0, and x = 2. We check the continuity at each of these points. At x = -2: As x -2^-, f(x) 0^-, so _ x -2^- g(f(x)) = _ y 0^- g(y) = -2 As x -2^+, f(x) 0^+, so _ x -2^+ g(f(x)) = _ y 0^+ g(y) = 2 Since the left-hand limit and right-hand limit are not equal, g(f(x)) is discontinuous at x = -2. At x = 0: _ x 0^- g(f(x)) = g(8) = (8 + 4)^ 1/2 = 2 3 _ x 0^+ g(f(x)) = g(-4) = (-4 - 8)^ 1/3 = (-12)^ 1/3 Since the limits are not equal, g(f(x)) is discontinuous at x = 0. At x = 2: As x 2^-, f(x) 0^-, so _ x 2^- g(f(x)) = _ y 0^- g(y) = -2 As x 2^+, f(x) 0^+, so _ x 2^+ g(f(x)) = _ y 0^+ g(y) = 2 Since the limits are not equal, g(f(x)) is discontinuous at x = 2. Therefore, there are 3 points of discontinuity. Answer: 3