JEE Main 2026 — Mathematics Algebra

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:

$$\Delta = \begin{vmatrix} 1 & 2 & t \\ 6 & 1 & 5t \\ 3 & t^2 & f(t) \end{vmatrix}$$Δ=​163​21t2​t5tf(t)​​

Expanding the determinant along the first row, we get:

$$\Delta = 1(1 \cdot f(t) - 5t \cdot t^2) - 2(6 \cdot f(t) - 3 \cdot 5t) + t(6 \cdot t^2 - 3 \cdot 1)$$Δ=1(1⋅f(t)−5t⋅t2)−2(6⋅f(t)−3⋅5t)+t(6⋅t2−3⋅1)

$$\Delta = f(t) - 5t^3 - 12f(t) + 30t + 6t^3 - 3t$$Δ=f(t)−5t3−12f(t)+30t+6t3−3t

$$\Delta = -11f(t) + t^3 + 27t$$Δ=−11f(t)+t3+27t

Since the system has infinitely many solutions for all $t \in \mathbb{R}$t∈R, we must have $\Delta = 0$Δ=0 for all $t \in \mathbb{R}$t∈R.

$$-11f(t) + t^3 + 27t = 0$$−11f(t)+t3+27t=0

$$f(t) = \dfrac{t^3 + 27t}{11}$$f(t)=11t3+27t​

To determine the nature of the function $f(t)$f(t), we find its derivative with respect to $t$t:

$$f'(t) = \dfrac{3t^2 + 27}{11}$$f′(t)=113t2+27​

Since $t^2 \ge 0$t2≥0 for all $t \in \mathbb{R}$t∈R, we have $3t^2 + 27 \ge 27 > 0$3t2+27≥27>0.

Therefore, $f'(t) > 0$f′(t)>0 for all $t \in \mathbb{R}$t∈R, which implies that $f(t)$f(t) is a strictly increasing function on $\mathbb{R}$R.

Answer: is strictly increasing on $\mathbb{R}$R