JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Given the function: f(x) = _ 0 ^ x (t-x) dt - _ 0 ^ x f(t) t \, dt First, we simplify the first integral. Let u = x - t, then du = -dt. When t = 0, u = x; when t = x, u = 0. _ 0 ^ x (t-x) dt = _ x ^ 0 (-u) (-du) = _ x ^ 0 u \, du = - _ 0 ^ x u \, du = -[ ( u)]_ 0 ^ x = - ( x) = ( x) So, the equation becomes: f(x) = ( x) - _ 0 ^ x f(t) t \, dt Differentiating both sides with respect to x using the Leibniz rule: f'(x) = - x x - f(x) x f'(x) + f(x) x = - x This is a linear first-order differential equation. The integrating factor (IF) is: IF = e^ x \, dx = e^ ( x) = x Multiplying the differential equation by x: f'(x) x + f(x) x x = - x x d dx (f(x) x) = - x x Integrating both sides with respect to x: f(x) x = - x + C To find C, we use the initial condition. From f(x) = ( x) - _ 0 ^ x f(t) t \, dt, substituting x = 0 gives f(0) = (1) - 0 = 0. 0 0 = - 0 + C 0 = -1 + C C = 1 Thus, f(x) x = 1 - x f(x) = x - 1 Now, we find the required derivatives: f'(x) = - x f''(x) = - x Evaluating these at the given points: f ( 6 ) = ( 6 ) - 1 = 3 2 - 1 f' (- 6 ) = - (- 6 ) = ( 6 ) = 1 2 f'' ( 6 ) = - ( 6 ) = - 3 2 Finally, substituting these values into the required expression: f'' ( 6 ) + 12f' (- 6 ) + f ( 6 ) = - 3 2 + 12 ( 1 2 ) + ( 3 2 - 1 ) = - 3 2 + 6 + 3 2 - 1 = 5 Answer: 5