Is y^3 ≤ |y|? (1) y

Question

Is y^3 ≤ |y|?

(1) y < 1
(2) y < 0

Zarrolou · Accepted Answer

|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 Sufficient Let me know if it's clear

Bunuel · Answer

Is y^3 ≤ |y|?

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|.

So, we have that y^3 ≤ |y| holds true when y ≤ 1.

(1) y < 1. Sufficient.
(2) y <0. Sufficient.

Answer: D.

Hope it's clear.

WholeLottaLove · Answer

Is y^3 ≤ |y|?

(1) y < 1
(2) y < 0

Is y^3 ≤ |y|?

If y≥0 then: 
y^3 ≤ |y|
y^3 ≤ y
y^3 - y ≤ 0

If y≤0 then:
y^3 ≤ |y|
y^3 ≤ -y
y^3 + y ≤ 0

1.) y < 1

For either case, positive or negative it holds true. For example:

for the positive case y<1. 
If y=1/2 then 1/2^3 - 1/2 ≤ 0. 
we only take cases between 0 and 1

for the negative case y<1
if y = -1/2 then -1/2^3 + -1/2 ≤ 0
we only take cases less than zero

2.) y < 0

We don't consider the positive case because we are only considering the negative case of y.

for the negative case y<0
if y = -1/2 then -1/2^3 + -1/2 ≤ 0
we only take cases less than zero
(Same as above)

Is this a correct way to solve the problem? It seems like others solved it in a different fashion.

Zarrolou · Answer

Yes,it's correct. You can solve it correctly in many ways

kalaits · Answer

stmt 1 : 
y<1

test numbers:

y=1/2 , 1/8<1/2 - yes
y=0 0=0 , yes
y= any negative value

-ve <= +ve , always yes

sufficient.

stmt 2: 
y<0

for any negative value, y3 will always be negative
-ve <= +ve ( always yes)

sufficient 
Ans D

please give kudos if you like the solution

nikhilpoddar · Answer

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 ??

Mo2men · Answer

Is y^3 ≤ |y|?

(1) y < 1

Test Values with integers and fraction with positive and negative sign.

Let y = 1/2........Answer to question is yes

Let y = 0........Answer to question is yes

Let y = -1/2........Answer to question is yes

Let y =-2........Answer to question is yes

Sufficient

(2) y < 0

Use the same third and fourth examples above....Answer is Yes

Sufficient

Answer: D

nikhilpoddar · Answer

Actually I wanted to know if the range is correct or not...Have i made any error in the algebriac expression ???

abhijitatprepitt · Answer

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.

sashiim20 · Answer

y is negative integer or negative fraction from both statement (1) and (2).

|y|  will give only the positive value of y.

y ^3  Cube of any negative integer or negative fraction will be negative and less than the positive value of y.

I is Sufficient. II is Sufficient. Answer (D)...

ydmuley · Answer

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 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 Did you like it? 1 Kudos Please :-D

MBAHOUSE · Answer

Is y^3 ≤ |y|?

(1) y < 1
(2) y < 0

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Is y^3 ≤ |y|? (1) y < 1 (2) y < 0

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