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M27#12 [#permalink] New post 13 Sep 2012, 18:04
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Question Stats:

40% (01:58) correct 60% (00:46) wrong based on 0 sessions
Is xy≤1/2?

(1) x^2+y^2=1
(2) x^2−y^2=0

OFFICIAL EXPLANATION
(1) x^2+y^2=1. Recall that (x−y)2≥0 (square of any number is more than or equal to zero). Expand: x^2−2xy+y^2≥0 and since x2+y2=1 then: 1−2xy≥0. So, xy≤1/2. Sufficient.
(2) x^2−y^2=0. Re-arrange and take the square root from both sides: |x|=|y|. Clearly insufficient.

I am fine with (2). I have trouble with (1).

What if, instead of using x^2-2xy+y^2≥0, I decided to use x^2+2xy+y^2≥0 (note the positive). That would result in xy≥-1/2 instead of the xy≤1/2 that is sufficient. In the end, I would have to use x^2-2xy+y^2≥0?

Anyone care to elaborate on this please? :)
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Re: M27#12 [#permalink] New post 14 Sep 2012, 04:15
lesliehh wrote:
Is xy≤1/2?

(1) x^2+y^2=1
(2) x^2−y^2=0

OFFICIAL EXPLANATION
(1) x^2+y^2=1. Recall that (x−y)2≥0 (square of any number is more than or equal to zero). Expand: x^2−2xy+y^2≥0 and since x2+y2=1 then: 1−2xy≥0. So, xy≤1/2. Sufficient.
(2) x^2−y^2=0. Re-arrange and take the square root from both sides: |x|=|y|. Clearly insufficient.

I am fine with (2). I have trouble with (1).

What if, instead of using x^2-2xy+y^2≥0, I decided to use x^2+2xy+y^2≥0 (note the positive). That would result in xy≥-1/2 instead of the xy≤1/2 that is sufficient. In the end, I would have to use x^2-2xy+y^2≥0?

Anyone care to elaborate on this please? :)


Yes, to solve this question you should use (x−y)^2≥0 not (x+y)^2≥0.
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Re: M27#12 [#permalink] New post 18 Feb 2013, 15:38
Because square of any number is more than or equal to zero). so x2−2xy+y2≥0 is used instead of x2+2xy+y2≥0?????????????????????????????

Bunuel wrote:
lesliehh wrote:
Is xy≤1/2?

(1) x^2+y^2=1
(2) x^2−y^2=0

OFFICIAL EXPLANATION
(1) x^2+y^2=1. Recall that (x−y)2≥0 (square of any number is more than or equal to zero). Expand: x^2−2xy+y^2≥0 and since x2+y2=1 then: 1−2xy≥0. So, xy≤1/2. Sufficient.
(2) x^2−y^2=0. Re-arrange and take the square root from both sides: |x|=|y|. Clearly insufficient.

I am fine with (2). I have trouble with (1).

What if, instead of using x^2-2xy+y^2≥0, I decided to use x^2+2xy+y^2≥0 (note the positive). That would result in xy≥-1/2 instead of the xy≤1/2 that is sufficient. In the end, I would have to use x^2-2xy+y^2≥0?

Anyone care to elaborate on this please? :)


Yes, to solve this question you should use (x−y)^2≥0 not (x+y)^2≥0.
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Re: M27#12 [#permalink] New post 21 Mar 2013, 03:03
Bunuel,

I am not able to understand this explanation. What triggered you to choose this method to find the solution?

OFFICIAL EXPLANATION
(1) x^2+y^2=1. Recall that (x−y)2≥0 (square of any number is more than or equal to zero). Expand: x^2−2xy+y^2≥0 and since x2+y2=1 then: 1−2xy≥0. So, xy≤1/2. Sufficient.
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Re: M27#12   [#permalink] 21 Mar 2013, 03:03
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