JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Given f(x+y) = f(x) + 2y^2 + y + xy Substituting x = 0 and replacing y with x, we get: f(x) = f(0) + 2x^2 + x + (0)x Since f(0) = -1, we have: f(x) = 2x^2 + x - 1 To find , substitute f(x) into the original functional equation: 2(x+y)^2 + (x+y) - 1 = (2x^2 + x - 1) + 2y^2 + y + xy 2x^2 + 4xy + 2y^2 + x + y - 1 = 2x^2 + x - 1 + 2y^2 + y + xy Comparing both sides, we get = 4. We need to find the value of _ n=1 ^ 5 ( +f(n)): _ n=1 ^ 5 (4 + 2n^2 + n - 1) = _ n=1 ^ 5 (2n^2 + n + 3) = 2 _ n=1 ^ 5 n^2 + _ n=1 ^ 5 n + _ n=1 ^ 5 3 = 2 ( 5 6 11 6 ) + 5 6 2 + 3 5 = 2(55) + 15 + 15 = 110 + 30 = 140 Answer: 140