Question

Consider the system of linear equations in x, y, z:
x+2y+tz = 0,
6x + y + 5t z = 0,
3x + y + f(t) z = 0,
where f: R$\rightarrow$ R is a differentiable function. If this system has infinitely many solutions for all t $\in$ R, then f

• A. is a constant function
• B. is strictly increasing on R
• C. is strictly decreasing on R
• D. has two critical points

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We have a system of homogeneous linear equations. A homogeneous system always has the trivial solution \((x=0, y=0, z=0)\).
The condition that the system has infinitely many solutions means it must have non-trivial solutions.
This occurs if and only if the determinant of the coefficient matrix is zero. This condition must hold for all \(t \in \mathbb{R}\).

Step 2: Key Formula or Approach:
For the system of equations \(A\mathbf{x} = \mathbf{0}\) to have a non-trivial solution, we must have \(\det(A) = 0\).
We will set up the coefficient matrix, calculate its determinant, set it to zero for all \(t\), and solve for the function \(f(t)\). Then we will analyze the properties of \(f(t)\).

Step 3: Detailed Explanation:
The given system of equations is:
1. \(x + 2y + tz = 0\)
2. \(6x + y + 5tz = 0\)
3. \(3x + y + f(t)z = 0\)

The coefficient matrix \(A\) is:
\[ A = \begin{pmatrix} 1 & 2 & t \\ 6 & 1 & 5t \\ 3 & 1 & f(t) \end{pmatrix} \] For the system to have infinitely many solutions, the determinant of \(A\) must be zero for all \(t \in \mathbb{R}\):
\[ \det(A) = 0 \] Now calculate the determinant:
\[ \det(A) = 1 \cdot \begin{vmatrix} 1 & 5t \\ 1 & f(t) \end{vmatrix} - 2 \cdot \begin{vmatrix} 6 & 5t \\ 3 & f(t) \end{vmatrix} + t \cdot \begin{vmatrix} 6 & 1 \\ 3 & 1 \end{vmatrix} = 0 \] \[ = 1(1 \cdot f(t) - 5t \cdot 1) - 2(6 \cdot f(t) - 5t \cdot 3) + t(6 \cdot 1 - 1 \cdot 3) \] \[ = (f(t) - 5t) - 2(6f(t) - 15t) + t(6 - 3) \] \[ = f(t) - 5t - 12f(t) + 30t + 3t \] \[ = (1 - 12)f(t) + (-5 + 30 + 3)t \] \[ = -11f(t) + 28t = 0 \] Solving for \(f(t)\):
\[ 11f(t) = 28t \] \[ f(t) = \frac{28}{11}t \] Now analyze the function:
\[ f'(t) = \frac{d}{dt} \left( \frac{28}{11}t \right) = \frac{28}{11} \] Since \(\frac{28}{11} > 0\), the function is strictly increasing on \(\mathbb{R}\).

Other observations:
- It is not constant
- It is not decreasing
- It has no critical points

Step 4: Final Answer:
The function \(f(t)\) is strictly increasing on \(\mathbb{R}\).