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JEE Main 202427 JanuaryEvening Shift
Exam
JEE Main
Year
2024
Date
27 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Determinants

JEE Main 2024 · 27 January · Evening Shift · Mathematics · Determinants

The values of α , for which | cases 1 & (3)/(2) & α + (3)/(2) 1 &...

The values of α\alpha, for which \left|{132α+32113α+132α+33α+10\begin{cases}1 & \dfrac{3}{2} & \alpha + \dfrac{3}{2} \\ 1 & \dfrac{1}{3} & \alpha + \dfrac{1}{3} \\ 2\alpha + 3 & 3\alpha + 1 & 0\end{cases}\right| = 0$, lie in the interval

Options

  1. A(−2,1)( - 2, 1 )
  2. B(−3,0)( - 3, 0 )
  3. C(−32,32)\left(- \dfrac{3}{2}, \dfrac{3}{2}\right)
  4. D(0,3)( 0, 3 )

Correct answer

B. (−3,0)( - 3, 0 )

Step-by-step solution

Given: \left|\begin{cases}1 & \dfrac{3}{2} & \alpha + \dfrac{3}{2} \\ 1 & \dfrac{1}{3} & \alpha + \dfrac{1}{3} \\ 2\alpha + 3 & 3\alpha + 1 & 0\end{cases}\right| = 0

Applying, R1→R1−R2{R}_{1} \to {R}_{1} - {R}_{2}

⇒ \left|\begin{cases}0 & \dfrac{7}{6} & \dfrac{7}{6} \\ 1 & \dfrac{1}{3} & \alpha + \dfrac{1}{3} \\ 2\alpha + 3 & 3\alpha + 1 & 0\end{cases}\right| = 0

Applying, C2→C2−C3{C}_{2} \to {C}_{2} - {C}_{3}

⇒ \left|\begin{cases}0 & 0 & \dfrac{7}{6} \\ 1 & - \alpha & \alpha + \dfrac{1}{3} \\ 2\alpha + 3 & 3\alpha + 1 & 0\end{cases}\right| = 0

⇒0−0+76(3α+1+2α2+3α)=0⇒ 0 - 0 + \dfrac{7}{6}\left(3\alpha + 1 + 2{\alpha}^{2} + 3\alpha\right) = 0

⇒2α2+6α+1=0⇒ 2{\alpha}^{2} + 6\alpha + 1 = 0

⇒α=−6±36−4(2)(1)2×2⇒ \alpha = \dfrac{- 6 \pm \sqrt{36 - 4\left(2\right)\left(1\right)}}{2 \times 2}

⇒α=−6±274⇒ \alpha = \dfrac{- 6 \pm 2\sqrt{7}}{4}

⇒α=−3±72⇒ \alpha = \dfrac{- 3 \pm \sqrt{7}}{2}

⇒α≈−3±2.62⇒ \alpha \approx \dfrac{- 3 \pm 2 . 6}{2}

⇒α≈−0.42,−5.62⇒ \alpha \approx \dfrac{- 0 . 4}{2}, \dfrac{- 5 . 6}{2}

⇒α≈−0.2,−2.8⇒ \alpha \approx - 0 . 2, - 2 . 8

⇒α∈(−3,0)⇒ \alpha \in \left(- 3, 0\right)

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