JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

f(x) = e^x + _0^1 (y + xe^x)f(y)\, dy Splitting the integral: f(x) = e^x + _0^1 yf(y)\, dy + xe^x _0^1 f(y)\, dy Let A = _0^1 yf(y)\, dy and B = _0^1 f(y)\, dy f(x) = (1 + Bx)e^x + A At x = 0: f(0) = 1 + A Now compute B using the expression for f(y): B = _0^1 [(1 + By)e^y + A ] dy B = _0^1 e^y\, dy + B _0^1 ye^y\, dy + _0^1 A\, dy Evaluating each integral: _0^1 e^y\, dy = e - 1 _0^1 ye^y\, dy = [ye^y]_0^1 - _0^1 e^y\, dy = e - (e - 1) = 1 _0^1 A\, dy = A B = (e - 1) + B(1) + A 0 = e - 1 + A A = 1 - e f(0) = 1 + A = 1 + (1 - e) = 2 - e e + f(0) = e + (2 - e) = 2 Answer: 2