3 and 4 .Determinants and Matrices
hard

माना $A =\left[\begin{array}{ccc}2 & b & 1 \\ b & b ^{2}+1 & b \\ 1 & b & 2\end{array}\right]$ जहाँ $b > 0$ है। तब $\frac{\operatorname{det}( A )}{ b }$ का न्यूनतम मान होगा 

A

$2\sqrt 3$

B

$-2\sqrt 3$

C

$-\sqrt 3$

D

$\sqrt 3$

(JEE MAIN-2019)

Solution

Det $A = {b^2} + 3$

$\frac{{\det \,A}}{b} = b + \frac{3}{b}$

$\therefore $ Least value $ = 2\sqrt 3 $

Standard 12
Mathematics

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