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

Answer

$\left|\begin{array}{ccc}
1 & \frac{3}{2} & \alpha+\frac{3}{2} \\
1 & \frac{1}{3} & \alpha+\frac{1}{3} \\
2 \alpha+3 & 3 \alpha+1 & 0
\end{array}\right|=0 \\$
$\Rightarrow(2 \alpha+3)\left\{\frac{7 \alpha}{6}\right\}-(3 \alpha+1)\left\{\frac{-7}{6}\right\}=0$
$\Rightarrow(2 \alpha+3) \cdot \frac{7 \alpha}{6}+(3 \alpha+1) \cdot \frac{7}{6}=0$
$\Rightarrow 2 \alpha^2+3 \alpha+3 \alpha+1=0$
$\Rightarrow 2 \alpha^2+6 \alpha+1=0$
$\Rightarrow \alpha=\frac{-3+\sqrt{7}}{2}, \frac{-3-\sqrt{7}}{2}$
Hence option $(2)$ is correct.

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