JEE Main 10 Jan 2019 Shift 2 Solved Paper

Section: Mathematics

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Question : 39 of 90

Marks: +1, -0

Let A = [

2	b	1
b	b2+1	b
1	b	2

] where b > 0. Then the minimum value of

det(A)

b

is :

[10 Jan 2019 Shift 2]

√3
- √3
- 2 √3
2 √3

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