JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Given f(xy) = f(x)f(y) for all x, y R and f(0) 0. Substituting y = 0, we get f(0) = f(x)f(0). Since f(0) 0, dividing by f(0) gives f(x) = 1 for all x R . The given integral equation is: x^2 g(x) = _1^x (t^2 f(t) - tg(t))\,dt Substituting f(t) = 1: x^2 g(x) = _1^x (t^2 - tg(t))\,dt Differentiating both sides with respect to x using Leibniz's rule: 2x g(x) + x^2 g'(x) = x^2 - xg(x) x^2 g'(x) + 3xg(x) = x^2 Dividing by x (since x 1): x g'(x) + 3g(x) = x g'(x) + 3 x g(x) = 1 This is a linear differential equation. The integrating factor is: IF = e^ 3 x \,dx = e^ 3 x = x^3 Multiplying the differential equation by x^3: x^3 g'(x) + 3x^2 g(x) = x^3 d dx (x^3 g(x)) = x^3 Integrating both sides with respect to x: x^3 g(x) = x^4 4 + C From the integral equation, substituting x = 1 gives 1^2 g(1) = _1^1 (t^2 - tg(t))\,dt = 0, so g(1) = 0. Substituting x = 1 and g(1) = 0 into the integrated equation: 1^3(0) = 1^4 4 + C C = - 1 4 Thus, the function g(x) is given by: x^3 g(x) = x^4 - 1 4 g(x) = x^4 - 1 4x^3 Substituting x = 2: g(2) = 2^4 - 1 4(2^3) = 16 - 1 32 = 15 32 Answer: 15 32