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

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Re: If x > y > 0, which of the following must be negative? [#permalink]
Bunuel wrote:
If x > y > 0, which of the following must be negative?

I. $$y - x^3$$
II. $$x^2y - x^3$$
III. $$x^2y - xy^2$$

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

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

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Re: If x > y > 0, which of the following must be negative? [#permalink]
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Bunuel wrote:
If x > y > 0, which of the following must be negative?

I. $$y - x^3$$
II. $$x^2y - x^3$$
III. $$x^2y - xy^2$$

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

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.

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.
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Re: If x > y > 0, which of the following must be negative? [#permalink]
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
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Re: If x > y > 0, which of the following must be negative? [#permalink]
Bunuel wrote:
If x > y > 0, which of the following must be negative?

I. $$y - x^3$$
II. $$x^2y - x^3$$
III. $$x^2y - xy^2$$

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

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 $$x^3=\frac{1}{8}<\frac{1}{3}=y$$

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
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Re: If x > y > 0, which of the following must be negative? [#permalink]

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

I. $$y - x^3$$
II. $$x^2y - x^3$$
III. $$x^2y - xy^2$$

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

­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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Re: If x > y > 0, which of the following must be negative? [#permalink]

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

I. $$y - x^3$$
II. $$x^2y - x^3$$
III. $$x^2y - xy^2$$

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

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

Attachment:

Screenshot 2024-04-08 at 9.05.50 PM.png [ 22.17 KiB | Viewed 2893 times ]

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.