November International SAT MATH Discussion.

Math112358
Sun, 04 Nov 2012 13:12:10 +0000

Hey guys, was looking over that last question again, here’s what i came up with.
If we assume that the interior shaded region does not contain its boundary, which we can infer from the wording of the question, then II and III are correct; here’s why:

I is obviously incorrect because x can take negative values

II is correct because the region we are looking at rests above y = x^2 which means that all y values in the region are greater than the value of x squared. Remember that the values of x and y are dependent on one another, that is, y = f(x) therefore we can take any point on the curve as (x, f(x)); we cannot just take arbitrary points of y and x and then compare. This condition holds as long as the boundary is not contained in the shaded region, because on the boundary y = x which breaks the condition.
For example, let (x,y) be a point on the curve y = x^2 and let (x,y) be a point in the region bounded by y = x^2 and y = 4 we can take the point x = 0.5. On the boundary, y = x^2 = (0.5)^2 = 0.25. Now this point is on the curve y = x^2 but since the shaded region lies above this curve, the values of y for x = 0.5 are 0.25 < y* < 4.

If the interior does not include the boundary, then x can never reach 2 but is infinitely close to 2 which satisfies condition III.
The answer is II and III unless the boundary is contained in the shaded region in which case the answer is none, but this is far less likely.