Determinants Question 4
Question 4 - 2024 (27 Jan Shift 2)
The values of $\alpha$, for which
$\left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}\right|=0$,
lie in the interval
(1) $(-2,1)$
(2) $(-3,0)$
(3) $\left(-\frac{3}{2}, \frac{3}{2}\right)$
(4) $(0,3)$
Show Answer
Answer (2)
Solution
$$ \begin{aligned} & \left|\begin{array}{ccc} 1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 \text { nathon } \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0 \end{array}\right|=0 \\ & \Rightarrow(2 \alpha+3){\frac{7 \alpha}{6} }-(3 \alpha+1){\frac{-7}{6} }=0 \\ & \Rightarrow(2 \alpha+3) \cdot \frac{7 \alpha}{6}+(3 \alpha+1) \cdot \frac{7}{6}=0 \\ & \Rightarrow 2 \alpha^{2}+3 \alpha+3 \alpha+1=0 \\ & \Rightarrow 2 \alpha^{2}+6 \alpha+1=0 \\ & \Rightarrow \alpha=\frac{-3+\sqrt{7}}{2}, \frac{-3-\sqrt{7}}{2} \end{aligned} $$
Hence option (2) is correct.
