MCQMediumJEE Main 2024 · 27 January, Shift 2Determinants Basics

Mathematics Question from JEE Main 2024 · 27 January, Shift 2

The values of α\alpha, for which 132α+32113α+132α+33α+10=0\begin{vmatrix} 1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2\alpha+3 & 3\alpha+1 & 0 \end{vmatrix} = 0, lie in the interval:

  • A

    (2,1)(-2,1)

  • B

    (3,0)(-3,0)

  • C

    (32,32)\left(-\frac{3}{2},\frac{3}{2}\right)

  • D

    (0,3)(0,3)

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