QJEEM19Mac390: JEE Main 2019 Mathematics - Class 12 JEEM19Mac390 Solution: Let A = [ 2 b 1 b b 2 + 1 b 1 b 2 ] where b > 0. Then the…

Question

Verified Solution • 2025–26 Session moderate

Let A =

[\begin{matrix} 2 & b & 1 \\ b & b^{2} + 1 & b \\ 1 & b & 2 \end{matrix}]

where b > 0.

Then the minimum value of

{{\det \left( A \right)} \over b}

is -

Accepted Answer

A =

[\begin{matrix} 2 & b & 1 \\ b & b^{2} + 1 & b \\ 1 & b & 2 \end{matrix}]

(b > 0)

\left| A \right|

= 2(2b² + 2

-

b²)

-

b(2b

-

b) + 1(b₂

-

b₂

-

1)

\left| A \right|

= 2(b² + 2)

-

b²

-

1

\left| A \right|

= b² + 3

{{\left| A \right|} \over b} = b + {3 \over b} \Rightarrow {{b + {3 \over b}} \over 2} \ge \sqrt 3

b + {3 \over b} \ge 2\sqrt 3

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