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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

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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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I was checking my phone all day. I wasn’t sure when I would receive the admission decision from Tepper. I received an acceptance from Goizueta in the early morning...

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