JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Given f(x) = _1^x f(t)\,dt + (1-x)( _e x - 1) + e Substituting x = 1 in the given equation: f(1) = _1^1 f(t)\,dt + (1-1)( _e 1 - 1) + e f(1) = 0 + 0 + e = e Differentiating the given equation with respect to x using the Leibniz rule: f'(x) = f(x) + d dx [(1-x)( _e x - 1)] f'(x) = f(x) + (-1)( _e x - 1) + (1-x) ( 1 x ) f'(x) = f(x) - _e x + 1 + 1 x - 1 f'(x) - f(x) = 1 x - _e x This is a linear differential equation of the form dy dx + Py = Q, where P = -1 and Q = 1 x - _e x. Integrating Factor (IF) = e^ -1 \,dx = e^ -x Multiplying the differential equation by the IF: e^ -x f'(x) - e^ -x f(x) = e^ -x ( 1 x - _e x ) d dx [e^ -x f(x)] = e^ -x 1 x - e^ -x _e x Notice that d dx [e^ -x _e x] = -e^ -x _e x + e^ -x 1 x . Thus, d dx [e^ -x f(x)] = d dx [e^ -x _e x] Integrating both sides with respect to x: e^ -x f(x) = e^ -x _e x + C f(x) = _e x + C e^x Using the initial condition f(1) = e: e = _e 1 + C e^1 e = 0 + C e C = 1 Therefore, the function is f(x) = _e x + e^x. We need to find f(f(1)): Since f(1) = e, we evaluate f(e): f(e) = _e e + e^e = 1 + e^e Answer: (1 + e^e)