|y|\geq{y^3}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 SufficientLet me know if it's clear

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