JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

For a homogeneous system of linear equations to have infinitely many solutions, the determinant of its coefficient matrix must be zero. The coefficient matrix is: = vmatrix 1 & 2 & t \\ 6 & 1 & 5t \\ 3 & t^2 & f(t) vmatrix Expanding the determinant along the first row, we get: = 1(1 f(t) - 5t t^2) - 2(6 f(t) - 3 5t) + t(6 t^2 - 3 1) = f(t) - 5t^3 - 12f(t) + 30t + 6t^3 - 3t = -11f(t) + t^3 + 27t Since the system has infinitely many solutions for all t R , we must have = 0 for all t R . -11f(t) + t^3 + 27t = 0 f(t) = t^3 + 27t 11 To determine the nature of the function f(t), we find its derivative with respect to t: f'(t) = 3t^2 + 27 11 Since t^2 0 for all t R , we have 3t^2 + 27 27 > 0. Therefore, f'(t) > 0 for all t R , which implies that f(t) is a strictly increasing function on R . Answer: is strictly increasing on R