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Is y^3 ≤ |y|?

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Is y^3 ≤ |y|? [#permalink] New post 16 Apr 2013, 22:16
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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] New post 16 Apr 2013, 22:32
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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] New post 16 Apr 2013, 23:01
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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] New post 25 Jun 2013, 21:49
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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] New post 16 Apr 2013, 23:02
Thanks for the reply zarrollu and bunnel
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Re: Is y^3 ≤ |y|? [#permalink] New post 25 Jun 2013, 16:33
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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y^3 <= |y| [#permalink] New post 27 Jun 2013, 13:54
y^3 <= |y|

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

Is it possible to solve it algebraically?
How could we know whether we should use algebra or picking numbers?

Thanks!
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Re: y^3 <= |y| [#permalink] New post 27 Jun 2013, 14:10
Expert's post
danzig wrote:
y^3 <= |y|

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

Is it possible to solve it algebraically?
How could we know whether we should use algebra or picking numbers?

Thanks!


Merging topics.

Please refer to the solutions above.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Is y^3 ≤ |y|? [#permalink] New post 24 Sep 2014, 21:03
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Re: Is y^3 ≤ |y|? [#permalink] New post 26 Sep 2014, 12:56
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] 26 Sep 2014, 12:56
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