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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 _________________

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|.

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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...