find the minimum integer value

I have to admit that this one was tricky for me. But, after thinking for a while, it occured to me that if I assume x & y to be equal, I am getting an answer of 6. So, if x & y are slightly different (maybe by a few tenths), then the minimum value of the expression will be 7.

IMO answer should be A. What is the OA?

Vemuri wrote: I have to admit that this one was tricky for me. But, after thinking for a while, it occured to me that if I assume x & y to be equal, I am getting an answer of 6. So, if x & y are slightly different (maybe by a few tenths), then the minimum value of the expression will be 7.

IMO answer should be A. What is the OA?

Damn ....if x & y are equal, I am getting 18 as the answer. I am missing something here.

If 0<x<1 and 0<y<1 and x+y=1, find the minimum integer value of [(x+1/x)pow 2]+[(y+1/y)pow 2]?
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x^2 + 1/x^2 + 2 + y^2 + 1/y^2 + 2
=4 + x^2+y^2+ (x^2+y^2)/x^2Y^2
=4 + (x^2+y^2)(x^2y^2 + 1)/(x^2y^2)
=4 + ((x+y)^2 - 2xy) + ((x+y)^2 - 2xy)/(x^2y^2)
=4 + (1 - 2xy)(1+1/x^2y^2)

Well, here is the thing..

The lowest value will occur only when x = y. As any of the value increases and so the other one decreases, the more the difference, the higher the value.

A very quick check was to take x = 0.1 this gives 121 as the answer for the equation (x+1)^2/x^2

so I think there should be an option E with value as 18 and that would be the minimum integer value.

It said the answer is 13 ..
I want to know the solution

i got answer 13

since 0<x<1 and 0<y<1

we will get the minimum value when x=y

also given is x+y=1

thus the minimum value when x=y and x+y=1 is 0.5

subsituting these values in the given equation you get answer as 12.50

thus the only possible option given is 13 so choose 13.