JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Let the polynomial of degree 5 be f(x) = ax^5 + bx^4 + cx^3 + dx^2 + ex + k. Given _ x 0 ( f(x) x^3 ) = -5, the terms of degree less than 3 must be zero, and the coefficient of x^3 must be -5. Thus, k = 0, e = 0, d = 0, and c = -5. The polynomial becomes f(x) = ax^5 + bx^4 - 5x^3. Differentiating with respect to x, we get: f'(x) = 5ax^4 + 4bx^3 - 15x^2 Since f(x) has extrema at x = 1 and x = -1, we have f'(1) = 0 and f'(-1) = 0. f'(1) = 5a + 4b - 15 = 0 f'(-1) = 5a - 4b - 15 = 0 Adding both equations, we get 10a - 30 = 0 a = 3. Subtracting the equations, we get 8b = 0 b = 0. So, the polynomial is f(x) = 3x^5 - 5x^3. Now, we find f(2) and f(-2): f(2) = 3(2)^5 - 5(2)^3 = 3(32) - 5(8) = 96 - 40 = 56 f(-2) = 3(-2)^5 - 5(-2)^3 = 3(-32) - 5(-8) = -96 + 40 = -56 Therefore, f(2) - f(-2) = 56 - (-56) = 112. Answer: 112