JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

We are given the function: f(x) = cases e^ x-1 , & x We need to analyze the continuity and differentiability of g(x) = f(|x|) + |f(x)|. For x 0. Thus, f(|x|) = (-x)^2 - 5(-x) + 6 = x^2 + 5x + 6. Also, for x 0, so |f(x)| = e^ x-1 . Therefore, for x For x 0, |x| = x. Thus, f(|x|) = f(x) = x^2 - 5x + 6. Therefore, for x 0, g(x) = x^2 - 5x + 6 + |x^2 - 5x + 6|. Let us check the continuity of g(x) at x = 0: _ x 0^- g(x) = _ x 0^- (x^2 + 5x + 6 + e^ x-1 ) = 6 + 1 e _ x 0^+ g(x) = _ x 0^+ (x^2 - 5x + 6 + |x^2 - 5x + 6|) = 6 + 6 = 12 g(0) = 12 Since _ x 0^- g(x) _ x 0^+ g(x), g(x) is discontinuous at x = 0. For all other x, g(x) is a sum of continuous functions and is therefore continuous. Thus, the number of points of discontinuity is = 1. Now, let us check the differentiability of g(x). Since g(x) is discontinuous at x = 0, it is not differentiable at x = 0. For x For x > 0, we can rewrite g(x) by analyzing the sign of x^2 - 5x + 6 = (x-2)(x-3): g(x) = cases 2(x^2 - 5x + 6), & x (0, 2] [3, ) \\ 0, & x (2, 3) cases Differentiating g(x) for x > 0, x 2, 3: g'(x) = cases 4x - 10, & x (0, 2) (3, ) \\ 0, & x (2, 3) cases Checking differentiability at x = 2: g'(2^-) = 4(2) - 10 = -2 g'(2^+) = 0 Since g'(2^-) g'(2^+), g(x) is not differentiable at x = 2. Checking differentiability at x = 3: g'(3^-) = 0 g'(3^+) = 4(3) - 10 = 2 Since g'(3^-) g'(3^+), g(x) is not differentiable at x = 3. Thus, g(x) is not differentiable at exactly three points: x = 0, 2, 3. So, the number of points of non-differentiability is = 3. Finally, + = 1 + 3 = 4. Answer: 4