Question

If the matrix \[ \begin{bmatrix} 1 & 3 & 1 \\ 2 & 1 & \alpha \\ 0 & 1 & -1 \end{bmatrix} \] is singular. Given a function \( f(x) = \int_{0}^{x} (t^2 + 2t + 3)\, dt \), \( \forall x \in [1, \alpha] \). If \( m \) and \( n \) are the maximum and minimum values of the function \( f(x) \), then the value of \( 3(m - n) \) is:

• A. 550
• B. 510
• C. 490
• D. 540

Hint:

If the derivative of an integral function \(f'(x) = g(x)\) is always positive, you don't need to evaluate the integral twice. The difference \(m-n\) is simply the definite integral over the given interval.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A matrix is singular if its determinant is zero. We find \(\alpha\) first. Then, since the integrand \(t^2 + 2t + 3\) is always positive for real \(t\), the function \(f(x)\) is strictly increasing. Therefore, the minimum value \(n\) occurs at the lower bound of the interval and the maximum value \(m\) occurs at the upper bound.

Step 2: Key Formula or Approach:
1. Singular matrix: \(\det(A) = 0\).
2. Fundamental Theorem of Calculus: \(f(x) = \frac{x^3}{3} + x^2 + 3x\).
3. \(m - n = f(\alpha) - f(1) = \int_{1}^{\alpha} (t^2 + 2t + 3) \, dt\).

Step 3: Detailed Explanation:
1. Solve for \(\alpha\): \[ \begin{vmatrix} 1 & 3 & 1
2 & 1 & \alpha
0 & 1 & -1 \end{vmatrix} = 1(-1 - \alpha) - 3(-2 - 0) + 1(2 - 0) = 0 \] \[ -1 - \alpha + 6 + 2 = 0 \implies \alpha = 7 \] 2. Calculate \(m - n\): Since \(f(x)\) is increasing on \([1, 7]\), \(m = f(7)\) and \(n = f(1)\). \[ m - n = \int_{1}^{7} (t^2 + 2t + 3) \, dt = \left[ \frac{t^3}{3} + t^2 + 3t \right]_1^7 \] \[ m - n = \left( \frac{343}{3} + 49 + 21 \right) - \left( \frac{1}{3} + 1 + 3 \right) \] \[ m - n = \frac{342}{3} + 70 - 4 = 114 + 66 = 180 \] 3. Final Value: \(3(m - n) = 3 \times 180 = 540\).

Step 4: Final Answer:
The value of \(3(m - n)\) is 540.