JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Given f ( x+y 3 ) = f(x) + f(y) 3 Substituting x = 0 and y = 0, we get: f(0) = 2f(0) 3 f(0) = 0 Differentiating the given equation partially with respect to x, treating y as a constant: f' ( x+y 3 ) 1 3 = f'(x) 3 f' ( x+y 3 ) = f'(x) Substituting x = 0, we get: f' ( y 3 ) = f'(0) = 3 Since this is true for all y R , f'(x) = 3 for all x R . Integrating both sides with respect to x: f(x) = 3x + C Using f(0) = 0, we get C = 0. Thus, f(x) = 3x. Now, the function g(x) is given by: g(x) = 3 + e^x f(x) = 3 + 3x e^x To find the minimum value, we differentiate g(x) with respect to x: g'(x) = 3(e^x + x e^x) = 3e^x(1 + x) Setting g'(x) = 0 gives x = -1. For x -1, g'(x) > 0. Therefore, x = -1 is a point of global minimum. The minimum value of g(x) is: g(-1) = 3 + 3(-1)e^ -1 = 3 - 3 e = 3 ( e-1 e ) Answer: 3 ( e-1 e )