The values of α , for which begin array l 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3/2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,α + 3/2\\1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1/3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,α + 1/3\\2α + 3\,\,\,\,\,\,\,\,\,\,3α + 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0end array = 0 lie in the interval

Mathematics · Determinants

JEE Main 2024 — 27 January, Shift 2 — Question 12

The values of $\alpha$ , for which $\left| \begin{array}{l}1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{3}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha + \frac{3}{2}\\1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha + \frac{1}{3}\\2\alpha + 3\,\,\,\,\,\,\,\,\,\,3\alpha + 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\end{array} \right| = 0$ lie in the interval

Option A:
$(-2,1)$
Option B:
$(-3,0)$
Correct
Option C:
$\left(-\frac{3}{2}, \frac{3}{2}\right)$
Option D:
$(0,3)$

Answer: B

Step-by-step solution

\begin{array}{ccc}\mid & \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 \end{array}

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

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

$\Rightarrow 2 \alpha^{2}+3 \alpha+3 \alpha+1=0$

$\Rightarrow 2 \alpha^{2}+6 \alpha+1=0$

$\Rightarrow \alpha=\frac{-3+\sqrt{7}}{2}, \frac{-3-\sqrt{7}}{2}$

Hence option (2) is correct.

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 27 January, Shift 2
Subject: Mathematics
Chapter: Determinants
Topic: Determinants