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Case y>0 \(y\geq{y^3}=y-y^3\geq{0}=y(1-y^2)\geq{0}\) this is +ve if \(y\geq{0}\) and \(1-y^2\geq{0}\) (\(-1\leq{x}\leq{1}\)) so the only interval in which this is positive is \(1\geq{y}\geq{0}\) ( you have to intersect \(y\geq{0}\) with \(-1\leq{x}\leq{1}\) AND with the rage we are considering y>0 and take the common part)

Case y<0 \(-y\geq{y^3}=-y-y^3\geq{0}=-y(1+y^2)\geq{0}\) this is +ve if \(-y\geq{0}=y\leq{0}\) and \(1+y^2\geq{0}=y^2\geq{-1}=\) always true => always positive. Intersect those ranges and find out that if y<0 the function is always +ve.

Take case 1 and 2: interval \(1\geq{y}\geq{0}+y\leq{0} = y\leq{1}\)

The question now:is \(y\leq{1}\)? (1) y < 1 Sufficient (2) y < 0 Sufficient

Let me know if it's clear
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It is beyond a doubt that all our knowledge that begins with experience.

You can use simple reasoning instead of algebra. Let's see for which range(s) y^3 ≤ |y| holds true. When y ≤ 0, then LHS = y^3 ≤ 0 and RHS = |y| >= 0, thus in this case y^3 ≤ |y|. When 0 < y ≤ 1, then also y^3 ≤ |y|. When y>1, then obviously y^3 > |y|.

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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Hey, guys, So, I’ve decided to run a contest in hopes of getting the word about the site out to as many applicants as possible this application season...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...