JEE Main 2026 — Determinants Question with 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:
$\Delta =\left|\begin{array}{ccc}1& 2& t\\ 6& 1& 5t\\ 3& t^2& f(t)\end{array}\right|$

Expanding the determinant along the first row, we get:
$\Delta =1(1\cdot f(t)-5t\cdot t^2)-2(6\cdot f(t)-3\cdot 5t)+t(6\cdot t^2-3\cdot 1)$

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

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

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

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

$f(t)=\frac{t^3+27t}{11}$

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

$f^{\prime }(t)=\frac{3t^2+27}{11}$

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

Therefore, $f^{\prime }(t)>0$ for all $t\in \mathbb{R}$, which implies that $f(t)$ is a strictly increasing function on $\mathbb{R}$.

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