JEE Advanced Mathematics Question (2026) — Solution

Step-by-step solution

The given differential equation is x dy dx = y - x^3. Dividing by x, we get: dy dx - 1 x y = -x^2 This is a linear differential equation of the form dy dx + P(x)y = Q(x), where P(x) = - 1 x and Q(x) = -x^2. Integrating Factor (IF) = e^ - 1 x dx = e^ - x = 1 x . Multiplying the differential equation by the IF: d dx ( y 1 x ) = -x Integrating both sides with respect to x: y x = - x^2 2 + C y = - x^3 2 + Cx Given y(1) = 0: 0 = - 1 2 + C C = 1 2 Thus, f(x) = x 2 - x^3 2 . To find local extrema, we find f'(x): f'(x) = 1 2 - 3x^2 2 Setting f'(x) = 0 gives 3x^2 = 1 x = 1 3 (since x > 0). Now, f''(x) = -3x. At x = 1 3 , f'' ( 1 3 ) = - 3 So, f(x) has a local maximum at x = 1 3 . For x (1, 2), x^2 > 1, so f'(x) = 1 - 3x^2 2 For the intersection of f(x) and g(x): x 2 - x^3 2 = 4x^3 - 5x^2 + 3 2 x x - x^3 = 8x^3 - 10x^2 + 3x 9x^3 - 10x^2 + 2x = 0 x(9x^2 - 10x + 2) = 0 Since x (0, ), x 0. We solve 9x^2 - 10x + 2 = 0. Discriminant = (-10)^2 - 4(9)(2) = 100 - 72 = 28 > 0. Sum of roots = 10 9 > 0 and Product of roots = 2 9 > 0. Both roots are real and positive. Therefore, there are exactly 2 elements in the set. Answer: The function f has a local maximum at x = 1 3 ; If g(x) = 4x^3 - 5x^2 + 3 2 x for x > 0, then the number of elements in the set \ x (0, ) : f(x) = g(x)\ is 2