Is xy

Question

Is xy < 1?

(1) x + y = 1
(2) x^2 + y^2 = 1

Bunuel · Answer

Is xy < 1?

(1) x + y = 1 --> the question becomes: is \(x(1-x)<1\)? --> is \(x^2-x+1>0\)? The curve of \(f(x)=x^2-x+1\) is a parabola which is entirely above the x-axes, thus \(x^2-x+1>0\) is true for any x. Sufficient.

(2) x^2 + y^2 = 1. We know that \((x-y)^2\geq{0}\) (square of any number is more than or equal to zero) --> \(x^2-2xy+y^2\geq{0}\). Since given that \(x^2 + y^2 = 1\), then \(1-2xy\geq{0}\) --> \(2xy\leq{1}\) --> \(xy\leq{\frac{1}{2}}\). Sufficient.

Answer: D.

Zarrolou · Answer

Is xy < 1?

(1) x + y = 1 
The greatest value it can assume is the case in which  x,y=0.5  and MAX  xy=\frac{1}{4}<1 .
Sufficient

(2) x^2 + y^2 = 1 
Same thing here, the gratest value xy can have is if the terms are equal  (\frac{1}{2}) , we take their positive root (to max the multiplication) so  x,y=\sqrt{\frac{1}{2}} ; also in this case  xy  is less than 1 ( xy=\frac{1}{2}  at most).
Sufficient

Hi Bunuel, can you please confirm the rule I've used? I remember it from old books, but I am not 100% sure of it
Thanks in advance

Bunuel · Answer

Yes. The property you are referring to is:  for given sum of two numbers, their product is maximized when they are equal .

Hope it helps.

Bunuel · Answer

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mau5 · Answer

Maybe another approach which is a bit visual: F.S 1:Imagine the straight line on the x-y axis : Omitting the x/y intercepts where anyways the value of xy = 0 ,any point on the line in the 2nd quadrant will always have the sign of (x,y) as (-,+)--> The product xy<0 and hence xy<1. Similarly for any point on the line in the 4th quadrant, where (x,y) will always be (+,-) and xy<1. For any point on the line in the first quadrant except (1,0) or (0,1), 0 0

bumpbot · Answer

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Is xy < 1? (1) x + y = 1 (2) x^2 + y^2 = 1

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