JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Given the limit _ t x t^2 y(x) - x^2 y(t) x - t = 3. The limit is in 0 0 form. Applying L'Hopital's Rule with respect to t: _ t x d dt (t^2 y(x) - x^2 y(t)) d dt (x - t) = 3 _ t x 2t y(x) - x^2 y'(t) -1 = 3 Substituting t = x: 2x y(x) - x^2 y'(x) -1 = 3 x^2 y'(x) - 2x y(x) = 3 Dividing by x^4 to make it a linear differential equation or recognizing the quotient rule form: x^2 y'(x) - 2x y(x) x^4 = 3 x^4 d dx ( y(x) x^2 ) = 3 x^4 Integrating both sides with respect to x: y(x) x^2 = 3x^ -4 dx = 3x^ -3 -3 + C = - 1 x^3 + C Given y(1) = 2: 2 1^2 = - 1 1^3 + C 2 = -1 + C C = 3 So, y(x) x^2 = - 1 x^3 + 3 y(x) = - 1 x + 3x^2 We need to find 2y(2): y(2) = - 1 2 + 3(2^2) = - 1 2 + 12 = 23 2 2y(2) = 2 23 2 = 23