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(1) \(x^2+y^2=1\). Recall that \((x-y)^2 \ge 0\) (square of any number is more than or equal to zero). Expand: \(x^2-2xy+y^2 \ge 0\) and since \(x^2+y^2=1\) then: \(1-2xy \ge 0\). So, \(xy \le \frac{1}{2}\). Sufficient.

(2) \(x^2-y^2=0\). Re-arrange and take the square root from both sides: \(|x|=|y|\). Clearly insufficient.

The same applicable to (x+y)^2 ==> x^2 + y^2 + 2 x*y is greter than or equal to ZERO. 1 + 2 x*y is greter than or equal to ZERO. xy greter than or equal -1/2...This can prove A is insufficient, right ???

The same applicable to (x+y)^2 ==> x^2 + y^2 + 2 x*y is greter than or equal to ZERO. 1 + 2 x*y is greter than or equal to ZERO. xy greter than or equal -1/2...This can prove A is insufficient, right ???

You need to consider \((x-y)^2\geq{0}\) to get sufficiency.
_________________

The same applicable to (x+y)^2 ==> x^2 + y^2 + 2 x*y is greter than or equal to ZERO. 1 + 2 x*y is greter than or equal to ZERO. xy greter than or equal -1/2...This can prove A is insufficient, right ???

You need to consider \((x-y)^2\geq{0}\) to get sufficiency.

Please explain why do we need to consider only (x-y)^2 to get sufficiency, as also in (x+y)^2 we use x^2 + y^ 2=1 premise, hence its not in all cases that(A) results in single conclusion or sufficiency.

The same applicable to (x+y)^2 ==> x^2 + y^2 + 2 x*y is greter than or equal to ZERO. 1 + 2 x*y is greter than or equal to ZERO. xy greter than or equal -1/2...This can prove A is insufficient, right ???

You need to consider \((x-y)^2\geq{0}\) to get sufficiency.

Please explain why do we need to consider only (x-y)^2 to get sufficiency, as also in (x+y)^2 we use x^2 + y^ 2=1 premise, hence its not in all cases that(A) results in single conclusion or sufficiency.

shrutikasat wrote:

yeah even i have a same doubt why not (x+y)^2 as harshalnamdeo88...

Guys, the point is that if you consider \((x+y)^2 \ge 0\), you'll get \(xy\geq{-\frac{1}{2}}\), which is not helpful at all. Actually since both \(xy\geq{-\frac{1}{2}}\) (from \((x+y)^2 \ge 0\)) and \(xy \le \frac{1}{2}\) (from \((x-y)^2 \ge 0\)) are true, then we get that \(-\frac{1}{2} \leq{xy} \leq{\frac{1}{2}}\). But only the approach given in the solution above gives you the answer we are looking for, while another one gives you an inequality which is not helpful.

(1) \(x^2+y^2=1\). Recall that \((x-y)^2 \ge 0\) (square of any number is more than or equal to zero). Expand: \(x^2-2xy+y^2 \ge 0\) and since \(x^2+y^2=1\) then: \(1-2xy \ge 0\). So, \(xy \le \frac{1}{2}\). Sufficient.

(2) \(x^2-y^2=0\). Re-arrange and take the square root from both sides: \(|x|=|y|\). Clearly insufficient.

Answer: A

Hi Bunuel,

Could you help me if I am wrong?

What I did was first to convert question stem to 2xy - 1 < 0, then from (1) I concluded that in order to satisfy x^2+y^2=1, The only x and y possible were; x=1, y=0, x=-1, y=0, x=0, y=1, x=0, y=-1; so question stem 2xy - 1 < 0 is always negative... Is my thinking process wrong or is correct?

(1) \(x^2+y^2=1\). Recall that \((x-y)^2 \ge 0\) (square of any number is more than or equal to zero). Expand: \(x^2-2xy+y^2 \ge 0\) and since \(x^2+y^2=1\) then: \(1-2xy \ge 0\). So, \(xy \le \frac{1}{2}\). Sufficient.

(2) \(x^2-y^2=0\). Re-arrange and take the square root from both sides: \(|x|=|y|\). Clearly insufficient.

Answer: A

Hi Bunuel,

Could you help me if I am wrong?

What I did was first to convert question stem to 2xy - 1 < 0, then from (1) I concluded that in order to satisfy x^2+y^2=1, The only x and y possible were; x=1, y=0, x=-1, y=0, x=0, y=1, x=0, y=-1; so question stem 2xy - 1 < 0 is always negative... Is my thinking process wrong or is correct?

Thanks a lot.

Regards.

Luis Navarro Looking for 700

Your thought process is not precise. We are NOT told that x and y are integers, thus x^2 + y^2 = 1 will have infinitely many solutions, not just the ones you mentioned.
_________________

(1) \(x^2+y^2=1\). Recall that \((x-y)^2 \ge 0\) (square of any number is more than or equal to zero). Expand: \(x^2-2xy+y^2 \ge 0\) and since \(x^2+y^2=1\) then: \(1-2xy \ge 0\). So, \(xy \le \frac{1}{2}\). Sufficient.

(2) \(x^2-y^2=0\). Re-arrange and take the square root from both sides: \(|x|=|y|\). Clearly insufficient.

Answer: A

Hi Bunuel,

Could you help me if I am wrong?

What I did was first to convert question stem to 2xy - 1 < 0, then from (1) I concluded that in order to satisfy x^2+y^2=1, The only x and y possible were; x=1, y=0, x=-1, y=0, x=0, y=1, x=0, y=-1; so question stem 2xy - 1 < 0 is always negative... Is my thinking process wrong or is correct?

Thanks a lot.

Regards.

Luis Navarro Looking for 700

Your thought process is not precise. We are NOT told that x and y are integers, thus x^2 + y^2 = 1 will have infinitely many solutions, not just the ones you mentioned.

I used geometrical approach to solve this problem. Did I get lucky?

(X)^2 +(Y)^2=1

That would indicate a right angle triangle with x and y as height and base. Maximize the area of triangle would maximise the x*y product. Thus X=Y=1/(root2) (not sure if this step is right.... I don't have a reason for this)

Thus the product will always be less than or equals 1/2

I don't understand why do we need this: "Recall that (x−y)2≥0(x−y)2≥0 (square of any number is more than or equal to zero)" to solve the equation. What's the relationship with x2+y2?

For statement 1 I used a different approach: I considered X^2+Y^2=1 a circle with centre in (0,0) and radius 1. The product of xy is represented by a parallelogram whose area is maximized in the first quadrant when the parallelogram is a square. Hence xy reaches the max when x=y = 1/square root of 2. So the product can be max 1/2.

This is a hard question, in particular because knowing to use the square of a difference is a tricky detail.

You can also use the fact that the maximum product of \(ab\) given that \(a+b=c\) is the set of numbers where \(a\) and \(b\) are nearest each other and still sum to \(c\).

For example, if \(a+b=20\), the maximum product of \(ab=10*10=100\)

In this case the maximum product of \(ab\) where \(a+b=1\) is when \(a=b=\frac{1}{2}\)

In this case \(x^2=a\) and \(y^2=b\), so the the maximum product of \(xy\) is \(\sqrt{\frac{1}{2}}*\sqrt{\frac{1}{2}}=\frac{1}{2}\)