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

To analyze the given system of linear equations and determine the nature of the function \( f(t) \), we need to understand the conditions under which the system has infinitely many solutions.

The given system of equations is: 

1. \(x + 2y + tz = 0\)
2. \(6x + y + 5tz = 0\)
3. \(3x + y + f(t)z = 0\)

For a system of equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero. Let's construct the coefficient matrix:

	x	y	z
Equation 1	1	2	t
Equation 2	6	1	5t
Equation 3	3	1	f(t)

To find when the determinant is zero:

\[\begin{vmatrix} 1 & 2 & t \\ 6 & 1 & 5t \\ 3 & 1 & f(t) \end{vmatrix} = 0\]

Calculating the determinant, we obtain:

\[1 \left(1 \cdot f(t) - 5t \cdot 1\right) - 2 \left(6 \cdot f(t) - 15t\right) + t \left(6 - 3\right) = 0\]

So, simplifying further:

\[f(t) - 5t - 12f(t) + 30t + 3t = 0\]

Which simplifies to:

\[-11f(t) + 28t = 0 \Rightarrow f(t) = \frac{28}{11} t\]

The derived function \(f(t) = \frac{28}{11} t\) is a linear function with a positive slope, indicating it is strictly increasing for all \(t \in \mathbb{R}\).

Thus, the correct answer is:

is strictly increasing on R

Therefore, the function \( f(t) \) must be strictly increasing on \( \mathbb{R} \) for the given system of equations to have infinitely many solutions for all \( t \).