The values of, for which | 1 3 2 + 3 2 1 1 3 + 1 3 2 + 3 3 + 1 0 | = 0 , lie in the interval [2024]

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Solution

Q.
The values of $α,$ for which $| \begin{matrix} 1 & \frac{3}{2} & α + \frac{3}{2} \\ 1 & \frac{1}{3} & α + \frac{1}{3} \\ 2 α + 3 & 3 α + 1 & 0 \end{matrix} |$ $= 0,$ lie in the interval [2024]

1 $(- \frac{3}{2}, \frac{3}{2})$

2 $(- 2, 1)$

3 $(- 3, 0)$

4 $(0, 3)$

Ans.
(3)

Given, $| \begin{matrix} 1 & \frac{3}{2} & α + \frac{3}{2} \\ 1 & \frac{1}{3} & α + \frac{1}{3} \\ 2 α + 3 & 3 α + 1 & 0 \end{matrix} |$ $= 0$

$\Rightarrow (2 α + 3) (\frac{7 α}{6}) - (3 α + 1) (\frac{- 7}{6}) = 0$

$\Rightarrow 14 α^{2} + 42 α + 7 = 0 \Rightarrow 2 α^{2} + 6 α + 1 = 0$

$Roots of this equation are α = \frac{- 3 \pm \sqrt{7}}{2}$

$\frac{- 3 + \sqrt{7}}{2} \in (- 3, 0) and \frac{- 3 - \sqrt{7}}{2} \in (- 3, 0)$

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