Question
Consider the system of linear equations in $x, y, z$:
$x + 2y + tz = 0$,
$6x + y + 5tz = 0$,
$3x + t^2y + f(t)z = 0$,
where $f: \mathbb{R} \to \mathbb{R}$ is a differentiable function. If this system has infinitely many solutions for all $t \in \mathbb{R}$, then $f$
The correct answer is
is strictly increasing on $\mathbb{R}$
We are tasked with analyzing a system of linear equations in three variables \(x, y, z\), depending on a parameter \(t\), given by:
\(x + 2y + tz = 0, \\ 6x + y + 5tz = 0, \\ 3x + t^2y + f(t)z = 0.\)
The goal is to determine the nature of the function \(f(t)\) if the system has infinitely many solutions for all \(t \in \mathbb{R}\).
- To have infinitely many solutions in a system of linear equations, the coefficient matrix and the augmented matrix must be rank-deficient, that is, their ranks must be less than the number of variables (which is 3 here). This implies the determinant of the matrix of coefficients must be zero for all \(t\).
- Construct the coefficient matrix \(A\) of the system:
- The determinant of \(A\), denoted as \(\text{det}(A)\), should be zero for all \(t\) for the system to have infinitely many solutions:
- Simplify the expression:
- Further simplification yields:
- Set the determinant to zero:
- The function \(f(t)\) must be such that the determinant is zero for all real numbers \(t\). Notice this derived function:
- Now, analyze the nature of \(f(t)\). Compute the derivative \(f'(t)\) to determine the behavior:
- Since the quadratic \(3t^2 + 27 > 0\) for all \(t\), \(f'(t) > 0\) everywhere.
- Thus, \(f(t)\) is strictly increasing for all \(t\).
Based on this analysis, the correct option is that \(f(t)\) is strictly increasing on \(\mathbb{R}\).
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