Let  A=2b1bb2+1b1b2 where b>0. Then the minimum value of det⁡(A)b is

Let  A=2b1bb2+1b1b2 where b>0. Then the minimum value of det(A)b is

  1. A

    3

  2. B

    23

  3. C

    3

  4. D

    23

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    Solution:

    We have,  A=2b1bb2+1b1b2, where b>0

    Now, det(A)=22b2+2b2b(2bb)+1b2b21

    =2b2+2b21=b2+3

    Now, det(A)b=b2+3b=b+3b

    Since,  A.M.  G.M.

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