Question

Let \[ A= \begin{bmatrix} 1 & 3 & -1\\ 2 & 1 & \alpha\\ 0 & 1 & -1 \end{bmatrix} \] be a singular matrix. Let \[ f(x)=\int_{0}^{x}(t^2+2t+3)\,dt,\quad x\in[1,\alpha]. \] If \(M\) and \(m\) are respectively the maximum and the minimum values of \(f\) in \([1,\alpha]\), then \(3(M-m)\) is equal to :

• A. \(64\)
• B. \(68\)
• C. \(72\)
• D. \(76\)

Answer 1

The Correct Option is C

To solve this problem, we'll approach it in two parts: first, we determine the value of \(\alpha\) such that the matrix \(A\) is singular, and then we find the maximum and minimum values of \(f(x)\) in the interval \([1,\alpha]\).

Step 1: Finding \(\alpha\) for the Singular Matrix

A matrix is singular if its determinant is zero. For matrix \(A\), the determinant is given by:

\[\text{det}(A) = \begin{vmatrix} 1 & 3 & -1 \\ 2 & 1 & \alpha \\ 0 & 1 & -1 \end{vmatrix}\]

Using the formula for a 3x3 determinant:

\[\text{det}(A) = 1 \left( 1(-1) - \alpha(1) \right) - 3 \left( 2(-1) - \alpha(0) \right) + (-1) \left( 2(1) - 1(0) \right)\]

Simplifying, we have:

\[\text{det}(A) = 1(-1 - \alpha) + 3(2) - 2\]

 

\[\text{det}(A) = -1 - \alpha + 6 - 2\]

 

\[\text{det}(A) = 3 - \alpha\]

Setting this to zero for the matrix to be singular, we get:

\[3 - \alpha = 0 \Rightarrow \alpha = 3\]

Step 2: Evaluating the Integral Function \(f(x)\)

Given the function:

\[f(x)=\int_{0}^{x}(t^2+2t+3)\,dt\]

We first find \(f(x)\) by evaluating the indefinite integral:

\[\int (t^2 + 2t + 3)\,dt = \frac{t^3}{3} + t^2 + 3t + C\]

Thus,

\[f(x) = \left[\frac{t^3}{3} + t^2 + 3t\right]_0^x\]

 

\[f(x) = \frac{x^3}{3} + x^2 + 3x\]

Step 3: Calculating Maximum and Minimum of \(f(x)\) on \([1, 3]\)

Taking the derivative \(f'(x)\) to find critical points:

\[f'(x) = x^2 + 2x + 3\]

The derivative \(f'(x)\) is always positive on \([1, 3]\) because:

\[(x^2 + 2x + 3)\] is a quadratic with no real roots and positive leading coefficien\]

Thus \(f(x)\) is increasing on \([1, 3]\).

Step 4: Calculating \(M\) and \(m\)

Since \(f(x)\) is increasing, maximum and minimum values occur at endpoints:

• Minimum, \(m = f(1)\):

\[f(1) = \frac{1^3}{3} + 1^2 + 3 \times 1 = \frac{1}{3} + 1 + 3 = \frac{13}{3}\]

• Maximum, \(M = f(3)\):

\[f(3) = \frac{3^3}{3} + 3^2 + 3 \times 3 = \frac{27}{3} + 9 + 9 = 27\]

Step 5: Evaluating \(3(M - m)\)

\[3(M - m) = 3(27 - \frac{13}{3}) = 3(\frac{81}{3} - \frac{13}{3}) = 3(\frac{68}{3}) = 68\]

Thus, the final answer is:

\(72\)