Let A = 2 b 1 b b 2 + 1 b 1 b 2 , where b > 0 . Then the minimum value of det A b is:

Let A=2b1bb2+1b1b2, where b>0. Then the minimum value of detAb is:
  1. 23
  2. -23
  3. 3
  4. -3

Solution

A=2b1bb2+1b1b2

=22b2+2-b2-b2b-b+1b2-b2-1

=2b2+2-b2-1

=2b2+4-b2-1

=b2+3

Ab=b+3b=b-3b2+23

Hence, the minimum value of Ab is 23.  ∵b-3b2≥0

Asked in: JEE Main 2019 (10 Jan Shift 2)

Practice more Determinants questions on Aicharya