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]?
A 7
B 9
C 8
D 13
find the minimum integer value
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- karthikgmat
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- Vemuri
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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?
IMO answer should be A. What is the OA?
Last edited by Vemuri on Sat May 02, 2009 4:05 am, edited 1 time in total.
- Vemuri
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DamnVemuri 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?
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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]?
================================
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)
================================
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)
- dumb.doofus
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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.
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.
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- karthikgmat
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It said the answer is 13 ..
I want to know the solution
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.
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.