Bunuel wrote:
Is \(xy \le \frac{1}{2}\)?
(1) \(x^2+y^2=1\)
(2) \(x^2-y^2=0\)
Hello, everyone. If it helps, I thought I would add a graphical solution. Notice that the equation in Statement (1) conforms to that for a circle:
\((x-h)^2+(y-k)^2=r^2\)
in which (h, k) is the center of the circle and
r is the radius. This circle would fit the equation thus:
\((x - 0)^2 + (y-0)^2=1^2\)
\(x^2 + y^2=1\)
This is the very unit circle that many trigonometry and pre-calculus students (at least, in the US) are taught, and recognizing it here can save a lot of time and effort.
Attachment:
Screen Shot 2022-04-18 at 14.26.33.png [ 29.03 KiB | Viewed 1151 times ]
Where it touches the circle, the
green segment represents the maximum value for
xy, since, in quadrants I and III, the two unknowns will have the same sign. The maximum value for the product will be found at 45 (or 225) degrees, equidistant from the
x- and
y- axes. (You can use a 45-45-90 special right triangle if you like. You know that in the corresponding side ratios
x,
x, and
x√2, respectively, 1, the radius of the circle, is equivalent to
x√2. This leads to a leg length of 1/√2 or, when the denominator is rationalized, √2/2.)
Thus, the maximum value of
xy is
\(\frac{1}{\sqrt{2}}*\frac{1}{\sqrt{2}} = \frac{1}{2}\)
Since 1/2 is
equal to 1/2, but we know the product cannot be any greater,
Statement (1) is SUFFICIENT.
Statement (2) should take little more than a glance to see that the equation will prove insufficient on its own. Just as
Bunuel has done above, start by rearranging.
\(x^2-y^2=0\)
\(x^2 = y^2\)
\(|x| = |y|\)
On its own, the statement tells us nothing about the nature of
xy, relative to 1/2.
The answer must be (A).Although knowledge of the circle equation and the unit circle is not required for the GMAT™, it can come in handy in unexpected ways, sometimes on tougher questions such as this one.
Good luck with your studies.
- Andrew