\(y^3 ≤ |y|\)

(1) \(y < 1\)

Consider \(y = \frac{1}{2}\)

\({\frac{1}{2}}^3 ≤ {|\frac{1}{2}|}\)

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

Lets check for ZERO as well

\({0}^3 ≤ |0| = 0 = 0\) ==> TRUE

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 SUFFICIENT2) \(y < 0\)

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 SUFFICIENTAs both (1) and (2) are SUFFICIENT

Answer is DDid you like it? 1 Kudos Please
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