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# Is y^3 ≤ |y|?

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Manager
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Is y^3 ≤ |y|? [#permalink]

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16 Apr 2013, 23:16
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Is y^3 ≤ |y|?

(1) y < 1
(2) y < 0
[Reveal] Spoiler: OA

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Re: Is y^3 ≤ |y| [#permalink]

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16 Apr 2013, 23:32
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$$|y|\geq{y^3}$$

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
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Re: Is y^3 ≤ |y [#permalink]

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17 Apr 2013, 00:01
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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.
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Re: Is y^3 ≤ |y|? [#permalink]

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25 Jun 2013, 17:33
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.
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Re: Is y^3 ≤ |y|? [#permalink]

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25 Jun 2013, 22:49
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WholeLottaLove wrote:
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.

Yes,it's correct. You can solve it correctly in many ways
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Re: Is y^3 ≤ |y|? [#permalink]

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26 Sep 2014, 13:56
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

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Re: Is y^3 ≤ |y|? [#permalink]

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15 Jun 2017, 15:38
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 ??
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Is y^3 ≤ |y|? [#permalink]

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15 Jun 2017, 17:56
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
Manager
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Re: Is y^3 ≤ |y|? [#permalink]

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15 Jun 2017, 18:07
Actually I wanted to know if the range is correct or not...Have i made any error in the algebriac expression ???
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Re: Is y^3 ≤ |y|? [#permalink]

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16 Jun 2017, 01:55
nikhilpoddar wrote:
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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Re: Is y^3 ≤ |y|? (1) y < 1 (2) y < 0 [#permalink]

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25 Jun 2017, 08:27
hazelnut wrote:
Is y^3 ≤ |y| ?

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

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)...
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Is y^3 < |y| ? [#permalink]

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25 Jun 2017, 08:49
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$$y^3 ≤ |y|$$

(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

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Is y^3 < |y| ?   [#permalink] 25 Jun 2017, 08:49
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