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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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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|.

If we square it on both the sides then how do we get y<=1 ?

y^6 <= y^2 y^2 (y^4-1) <=0 As per + - + - method, since y^2, we do not consider it. Y^4 - 1 gives us +1 and -1.

-1 +1 + - +

So -1<= y <= 1. How did we get y<=1 ??

Nikhil - squaring both sides is not a good approach here, because there is no guarantee that both sides are positive. If y is negative, the squaring could flip the inequality.

Example take y=-2

(-2)^3 <= |-2| --> correct?

squaring both sides (without flipping inequality) gives (-2)^6 <= (-2)^2 ==> This is obviously wrong!

I hope this helps and you see the point.
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GMAT Math Workshops ($90/session) | GMAT Math Private Tutoring ($70/hr)

Now, lets check for Negative values as well, as we know that mod/absolute function will always give us positive values and cube of negative will always give us negative values, our L.H.S. Should always be < R.H.S. Lets test

\({\frac{-1}{2}}^3 ≤ {|\frac{-1}{2}|}\)

\({\frac{-1}{8}} ≤ {\frac{1}{2}}\) ==> TRUE

Hence, Eq. (1) is SUFFICIENT

2) \(y < 0\)

As y is negative, we know for Negative values as well, as we know that mod/absolute function will always give us positive values and cube of negative will always give us negative values, our L.H.S. Should always be < R.H.S. Lets test

\({\frac{-1}{2}}^3 ≤ {|\frac{-1}{2}|}\)

\({\frac{-1}{8}} ≤ {\frac{1}{2}}\) ==> TRUE

Hence, Eq. (2) is SUFFICIENT

As both (1) and (2) are SUFFICIENT

Answer is D

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