Question

The values of \(\alpha\), for which

lie in the interval

• A. (-2, 1)
• B. (-3, 0)
• C. \(\left(-\frac{3}{2}, \frac{3}{2}\right)\)
• D. (0, 3)

Answer 1

The Correct Option is B

To find the values of \(a\), we expand the determinant:

\(\begin{vmatrix}  1 & \frac{1}{3} & a + \frac{3}{2} \\  1 & 1 & a + \frac{1}{3} \\  2a + 3 & 3a + 1 & 0  \end{vmatrix}.\)

Expanding along the first row:  
\(= 1 \cdot \left(1 \cdot 0 - \left(a + \frac{1}{3}\right)(3a + 1)\right) - \frac{3}{2} \cdot \left(1 \cdot 0 - \left(a + \frac{1}{3}\right)(2a + 3)\right) + \left(a + \frac{3}{2}\right) \cdot \left(1 \cdot (3a + 1) - 1 \cdot (2a + 3)\right).\)

Simplifying each term:  
\(= -(a + \frac{1}{3})(3a + 1) + \frac{3}{2}(a + \frac{1}{3})(2a + 3) + (a + \frac{3}{2})(a - 2).\)

Expanding the products:  
\(= -3a^2 - a - \frac{1}{3} - 3a - \frac{1}{3} + \frac{3}{2}(2a^2 + 3a + \frac{2}{3}).\)

After simplifying, we obtain a quadratic equation in \(a\). Solving for \(a\) gives:  
\(a \in (-3, 0).\)

The Correct answer is: (-3, 0)