Question

If the system of equations (in variables \(x, y, z\)): \(x - 2y + tz = 0\), \(3x + 5y + t^2 z = 0\), and \(6x + ty + f(t)z = 0\) has infinitely many solutions (where \(f(t)\) represents a real function), then:

• A. \( y = f(t) \) is strictly increasing
• B. \( y = f(t) \) is strictly decreasing
• C. \( y = f(t) \) is decreasing
• D. \( y = f(t) \) is increasing

Hint:

A quadratic \( ax^2 + bx + c \) is always positive if \( a > 0 \) and \( D > 0 \). This is a very common way to prove that a function is strictly increasing in calculus problems.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a homogeneous system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix must be equal to zero.
Step 2: Key Formula or Approach:
Set the determinant \( \Delta = 0 \):
\[ \begin{vmatrix} 1 & -2 & t \\ 3 & 5 & t^2 \\ 6 & t & f(t) \end{vmatrix} = 0 \] Step 3: Detailed Explanation:
1. Expand the determinant:
\[ 1(5f(t) - t^3) - (-2)(3f(t) - 6t^2) + t(3t - 30) = 0 \] \[ 5f(t) - t^3 + 6f(t) - 12t^2 + 3t^2 - 30t = 0 \] 2. Group the terms:
\[ 11f(t) = t^3 + 9t^2 + 30t \] 3. To check monotonicity, find \( f'(t) \):
\[ 11f'(t) = 3t^2 + 18t + 30 \] \[ f'(t) = \frac{3}{11}(t^2 + 6t + 10) \] 4. Analyze the quadratic \( t^2 + 6t + 10 \):
The discriminant \( D = 6^2 - 4(1)(10) = 36 - 40 = -4 < 0 \).
Since the leading coefficient is positive and \( D < 0 \), the expression is always positive for all real \( t \).
5. Since \( f'(t) > 0 \), the function is strictly increasing.
Step 4: Final Answer:
The function \( y = f(t) \) is strictly increasing.