cornell2005 wrote:I get C also, but am not sure how to prove that 1 and 2 together are sufficient. Anyone have any ideas? Thanks.
the presence of x^(1/2), also known as √x, in the problem guarantees that x is nonnegative. therefore, there are four cases:
(1) x = 0
(2) 0 < x < 1
(3) x = 1
(4) x > 1
cases (1) and (3): in these cases, √x = x^3 = x. therefore, saying that
any one of these three quantities is greater than y is equivalent to saying that any other one of them is equivalent to y.
**case (2): if 0 < x < 1, then x^3 < x < √x. if both statements are true, then y must be smaller than
both x^3 and √x, so that we have y < x^3 < x < √x.
***case (4): if x > 1, then √x < x < x^3 (the "normal" order of these three quantities). if both statements are true, then y must be smaller than
both x^3 and √x, so that we have y < √x < x < x^3.
in all four cases, statements 1 and 2 together are sufficient, so answer = c.
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**note: this is the case that proves statement 1 insufficient, as y could be less than √x but more than x.
***note: this is the case that proves statement 2 insufficient, as y could be less than x^3 but more than x.
Ron has been teaching various standardized tests for 20 years.
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