# Let A = 2 2b1 bb 1b 1b2 é ù ê ú ê ú + ê ú ë û where b > 0. Then the minimum value of det (A) b is:

Question:

Let $\mathrm{A}=\left[\begin{array}{ccc}2 & \mathrm{~b} & 1 \\ \mathrm{~b} & \mathrm{~b}^{2}+1 & \mathrm{~b} \\ 1 & \mathrm{~b} & 2\end{array}\right]$ where $\mathrm{b}>0$. Then the minimum

value of $\frac{\operatorname{det}(\mathrm{A})}{\mathrm{b}}$ is:

1. $2 \sqrt{3}$

2. $-2 \sqrt{3}$

3. $-\sqrt{3}$

4. $\sqrt{3}$

Correct Option: 1

JEE Main Previous Year 1 Question of JEE Main from Mathematics Matrices and Determinants chapter.
JEE Main Previous Year Jan. 10, 2019 (II)

Solution:

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