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Let A=[(2,b,1),(b,b^(2)+1,b),(1,b,2)] wh...

Let `A=[(2,b,1),(b,b^(2)+1,b),(1,b,2)]` where `b gt 0`. Then the minimum value of `("det.(A)")/(b)` is

A

`-sqrt(3)`

B

`2-sqrt(3)`

C

`2sqrt(3)`

D

`sqrt(3)`

Text Solution

Verified by Experts

The correct Answer is:
A
Given matrix, `A= [{:(2, " "b,1),(b, b^(2)+1,b),(1," " b,2):}], b gt 0`
So, det (A) ` = |A| = [{:(2, " "b,1),(b, b^(2)+1,b),(1," " b,2):}]`
` = 2[2(b^(2) + 1)-b^(2)]-b(2b-b)+1(b^(2)-b^(2)-1)`
`= 2[2b^(2) + 4-b^(2)]-1 = b^(2) + 3`
`rArr ("det(A)")/(b) = (b^(2) + 3)/(b) = b + (3)/(b)`
Now, by `"AM" ge"GM"`, we get
`(b+(3)/(b))/(2) ge (b xx (3)/(b))^(1//2) " " {because b gt 0}`
`rArr b+(3)/(b) ge 2sqrt(3)`
So, minimum value of `("det(A)")/(b) = 2sqrt(3)`
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