If x > y > 0, which of the following must be negative?

x & y both positive
x>y

I. \(y - x^3\)
x=3, y=1 , negative
x=1/2, y=1/4, positive

II. \(x^2y - x^3\)
\(x^2(y - x)\)
\(x^2\) positive
\((y - x)\) negative (as x>y)

So, \(x^2y - x^3\) Always Negative


III. \(x^2y - xy^2\)
\(xy(x-y)\)
x & y both positive , so xy :- positive
x>y, so (x-y):- positive.
So,\(x^2y - xy^2\) Always Positive

IMO Answer: B

I. If x = 1/2 and y = 1/3, we have:

y – x^3 = 1/3 – (1/2)^3 = 1/3 – 1/8 > 0, so we can cross out statement I.

II. (x^2)y – x^3 = (x^2)(y – x)

Since x ≠ 0 (x > 0), the expression x^2 must be positive. Since x > y, it must be true that 0 > y – x.

(x^2)(y – x) = (positive)(negative) = negative, so we can give a checkmark to statement II.

III. If we look at the answer choices, we can realize that this statement cannot be true, so we can cross out statement III.

Answer: B

Note: Let’s check statement III.

(x^2)y – x(y^2) = xy(x – y)

Since x > 0, y > 0, and x – y > 0 [because x > y], we have:

xy(x – y) = (positive)(positive)(positive) = positive, so we can indeed cross out statement III.

Asked: If x > y > 0, which of the following must be negative?

x^3 > x if x>1
x^3 < x if 0<x<1

I. \(y - x^3<0 or >0 depending on value of x\)
II. \(x^2y - x^3 = x^2(y-x) < 0 since y-x<0\)
III. \(x^2y - xy^2 = xy(x-y) > 0 since xy>0 & x-y>0\)

A. I only
B. II only
C. III only
D. I and II
E. I and III

IMO B

We have to see if the expression is positive for any value.
I. \(y - x^3\)
We should know that a number between 0 and 1 will reduce as we increase power. So surely x may be greater than y but increasing powers of x can lead to a value lesser than y.
Let x and y be between 0 and 1. Say \(x=\frac{1}{2}\) and \(y=\frac{1}{3}\)=> Here x>y
But \(x^3=\frac{1}{8}<\frac{1}{3}=y\)
Discard

II. \(x^2y - x^3\)
No need to check values.
\(x^2y - x^3=x^2(y-x)\)
Now, \(x^2\) is positive, while y-x is negative. So their product will always be negative.
True

III. \(x^2y - xy^2=xy(x-y)\)
Here, both xy and x-y are positive. So, their product will always be positive.


Only II


B

 

­To eliminate answer choices, try to make I, II and III greater than or equal to 0.

Option I:
\(y - x^3 ≥ 0\)
\(y ≥ x^3\)
This is possible if y=1/4 and x=1/2, with the result that \(y > x^3\).
Since option I does not have to be negative, the correct answer cannot include option I.
Eliminate A, D and E.

Option II:
\(x^2y - x^3 ≥ 0\)
\(x^2y ≥ x^3\)
Dividing both sides by \(x^2\), we get:
\(y ≥ x\)
Not possible, since the prompt requires that \(x > y\).
Since option II cannot be greater than or equal to 0 -- implying that it must be negative -- the correct answer must include option II.
Eliminate C.


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The most important point to remember here is that numbers behave differently in the four sections of the number line­.



0 < y < x

I. \(y - x^3\)
-- \(x < x^3\) if x > 1 which means
\(0 < y < x < x^3\) in this case so the given statement is negative. 
but 
-- \(x > x^3\) if x < 1
x is greater than y but x^3 may not be. Hence the given statement may not be negative in this case.
e.g. when y = 1/4 and x = 1/2, x^3 = 1/8 which is less than y so y - x^3 is positive. 
Not necessarily negative

II. \(x^2y - x^3\)
\( = x^2(y - x)\)
Since x > y, this is definitely negative. 

III. \(x^2y - xy^2\)
\(= xy (x - y)\)
This is positive since x > y and x and y both are positive so xy is positive too.
Not negative. 

Answer (B) ­

You can use number plugging here but you must understand what the transition points are and how numbers behave differently on different sections of the number line. 
Here is a post on the same: 
https://anaprep.com/algebra-importance- ... -plugging/