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

Answer

Correct option: B.
 $(-3,0)$
b
$\left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}\right|=0$

$ \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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