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Let A=[ … ], where b>0. Then the minimum value of det(A) b is:

JEE Main 2019 — Mathematics Algebra

2019mcqmedium

Let A=[2b1bb2+1b1b2],A=[\begin{matrix}2 & b & 1 \\ b & {b}^{2}+1 & b \\ 1 & b & 2\end{matrix}],A=[2b1​bb2+1b​1b2​], where b>0b>0b>0. Then the minimum value of det(A)b\frac{\mathrm{det}(A)}{b}bdet(A)​ is:

Official previous-year question

Held on 10 Jan 2019 · Verified 6 Jul 2026.

Options

  1. A

    232\sqrt{3}23​

  2. B

    −23-2\sqrt{3}−23​

  3. C

    3\sqrt{3}3​

  4. D

    −3-\sqrt{3}−3​

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