JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The given function is f(x) = _ y 0 (1 - (xy)) (xy) y^3 . Multiplying and dividing by x^3, we get: f(x) = _ y 0 1 - (xy) (xy)^2 (xy) xy x^3 Using the standard limits _ t 0 1 - t t^2 = 1 2 and _ t 0 t t = 1, we obtain: f(x) = 1 2 1 x^3 = x^3 2 We need to find the number of solutions for the equation f(x) = x, which is x^3 2 = x. Let g(x) = x^3 2 - x. Differentiating with respect to x: g'(x) = 3x^2 2 - x g''(x) = 3x + x For x > 0, g''(x) > 0, which implies that g'(x) is strictly increasing on (0, ). We have g'(0) = -1 0. Thus, g'(x) = 0 has exactly one root in (0, ), say at x = x_0. This means g(x) decreases on (0, x_0) and increases on (x_0, ). Since g(0) = 0 and g(x) initially decreases, g(x_0) Since g(x) is an odd function (g(-x) = -g(x)), it must also have exactly one negative root. Including the trivial root x = 0, the equation g(x) = 0 has exactly 3 solutions. Answer: 3