inequality dissection

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inequality dissection

by topspin360 » Thu Sep 20, 2012 3:57 am
hi all,

quick question:

I understand that the statement - is m^3 > m^2? - is really asking if m is positive and greater than 1.

but when i try to solve algebraically, i end up with the following:
m^3 - m^2 > 0?
m^2(m-1) > 0?
m>0 or m<0 or m>1?

What's wrong with the derivation above? Why am I getting m>0 and m<0?

Thanks.

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by GMATGuruNY » Thu Sep 20, 2012 5:17 am
topspin360 wrote:hi all,

quick question:

I understand that the statement - is m^3 > m^2? - is really asking if m is positive and greater than 1.

but when i try to solve algebraically, i end up with the following:
m^3 - m^2 > 0?
m^2(m-1) > 0?
m>0 or m<0 or m>1?

What's wrong with the derivation above? Why am I getting m>0 and m<0?

Thanks.
mÂ²(m-1) > 0.
The CRITICAL POINTS are m=0 and m=1.
These are the only values where the lefthand side is EQUAL to 0.
To determine the range(s) where the lefthand side is GREATER than 0, test only value to the left and right of each critical point.

Plug m = -1 into mÂ³ > mÂ²:
(-1)Â³ > (-1)Â²
- 1 > 1.
Doesn't work.
Thus, m<0 is not a viable range here.

Plug m = 1/2 into mÂ³ > mÂ²:
(1/2)Â³ > (1/2)Â²
1/8 > 1/4.
Doesn't work.
Thus, 0<m<1 is not a viable range here.

Plug m = 2 into mÂ³ > mÂ²:
2Â³ > 2Â²
8 > 4.
This works.
Thus, m>1 is a viable range here.

Only one range satisfies mÂ³ > mÂ²:
m>1.

An easier approach:
mÂ³ > mÂ² implies that m does not equal 0.
mÂ² cannot be negative, since the square of a value cannot be negative.
Thus, we can safely divide by mÂ², without having to change the direction of the inequality:
mÂ³/mÂ² > mÂ²/mÂ²
m > 1.
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by neelgandham » Sat Sep 22, 2012 8:25 am
My \$0.02

m^3 > m^2
m^3 - m^2 > 0
m^2 * (m-1) > 0

If a product of two numbers a,b is positive then
Case 1:Both a and b are positive.
m^2 > 0 ? Yes if the value m is not equal to 0. So, m>0 is the solution set.
m-1 > 0 ? Yes if the value of m is greater than 1. So. m>1 is the solution set.
The intersection of these two sets is the solution set. i.e m>1

Case 2:Both a and b are negative.
m^2 < 0 ? Sqaure of a number is never negative. So Case 2 is an invalid case
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by topspin360 » Sat Sep 22, 2012 4:19 pm
I have another very similar problem. Can you please explain what's wrong with my reasoning?

Q: y^3 <= abs(y)?

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

Before even getting to answer choices, I have two paths: y^3 <= y and y^3 >= -y.
The second option becomes invalid b/c it involves square root of -1.

The first option: y^3 - y <= 0
becomes: y(y-1)(y+1) <= 0.
using the number line, that means: either y <= -1 or 0 <= y <= 1 (that's when the equation is negative).

The answer is D. Fundamentally, the answer makes sense: y^3 must be smaller than or equal to abs(y) as long as y is less than 1. But I don't understand what I'm doing wrong in the derivation above.

Thanks.

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by gmatdriller » Sun Sep 23, 2012 3:49 am
topspin360 wrote:
Q: y^3 <= abs(y)?

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

Before even getting to answer choices, I have two paths: y^3 <= y and y^3 >= -y.
The second option becomes invalid b/c it involves square root of -1.

The first option: y^3 - y <= 0
becomes: y(y-1)(y+1) <= 0.
using the number line, that means: either y <= -1 or 0 <= y <= 1 (that's when the equation is negative).

The answer is D. Fundamentally, the answer makes sense: y^3 must be smaller than or equal to abs(y) as long as y is less than 1. But I don't understand what I'm doing wrong in the derivation above.

Thanks.
-1, 0, and +1 are not the values to be tested.

The values in bold only provide you with the boundaries for testing: -1, 0, and +1
You use the boundary points to test the validity of the solution subject to the constraints
provided in options I and II.

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by topspin360 » Sun Sep 23, 2012 8:28 am
gmatdriller, thanks for your explanation but didn't quite understand what you mean. Can you clarify?

Thanks.

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