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x and y be two variables such that xgt0 ...

x and y be two variables such that `xgt0` and `xy=1`. Then the minimum value of `x+y` is

A

2

B

3

C

4

D

0

Text Solution

Verified by Experts

The correct Answer is:
A
`xy=1impliesy=(1)/(x)` and let `z=x+y`
`z=x+(1)/(x)implies(dz)/(dx)=1-(1)/(x^(2))`
Now `(dz)/(dx)=0implies1-(1)/(x^(2))=0`
`x=-1,1 and (d^(2)z)/(dx^(2))=(2)/(x^(3))`
`((d^(2)z)/(dx^(2)))_(x=1)=(2)/(1)=2=+ive`
`because` Hence `x=1` is point of minima and `x=1` and `y=1` Minimum value `x+y=2`.
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