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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