Inequality

[yellowho]

If x is positive which of the following could be correct ordering of 1/x, 2x, x^2.
1) x^2 < 2x < 1/x

2) x^2 < 1/x < 2x

3) 2x < x^2 < 1/x

Choices are:

1) None
2) 1 Only
3) 3 Only
4) I and 2 only
5) 1, 2, 3

1) How do you know what numbers to pick? (I know the standard numbers are frations, negative etc..) Depending what you pick here, you might miss some items.

2) My approach: Please critique

a) I picked number. Pretty straight forward fraction will work.

b) x^2<1/x<2x, Since X is positive per the stem. I multiply=> x^3<1<2x^2

For X^3<1 to be true X<1
For 2x^2>1 simplify => x^2>1/2 => square-root |X|>sqroot(1/2) [Can someone comment on this step; I think you can skip the absolute value part since X is defined as positive)
So B can happen at X>sroot(1/2) and X<1.

C) Same procedure for 3.

2x<x^2 => divide by x since x is positive 2<x
x^2<1/x=> x^3<1 meaning x<1
Since X cannot be greater than 2 and less than 1. error.

[Anurag@Gurome]

Since x is positive, so we can do it straightaway without picking numbers also.
1) x^2 < 2x < 1/x implies x^2 < 2x or x < 2 (considering the 1st 2 inequalities).
Consider the 2nd and 3rd inequalities, 2x < 1/x or x^2 < 1/2 or x < 1/√2
So, x < 2 and x < 1/√2 can be true. So, ordering in (1) can be true.

(2) x^2 < 1/x < 2x implies x^2 < 1/x or x^3 < 1 or x < 1
1/x < 2x implies 1/2 < x^2 or x > 1/√2. So, 1/√2 < x < 1 can be true again. So, ordering in (2) can be true.

(3) 2x < x^2 < 1/x implies 2x < x^2 or x > 2
x^2 < 1/x implies x^3 < 1 or x < 1
Here x > 2 and x < 1 cannot be true at one time. So, ordering in (3) cannot be true.

Hence, the correct answer is [spoiler]4)[/spoiler].

[yellowho]

Anurag,

1) How would one do this with plugging in? How do you realize that x<1/root(2) would work if you didn't do the algebra

2) For x^2<1/2, you have to do |x|<1/root(2) if you don't know that x>0?



[quote="Anurag@Gurome"]Since x is positive, so we can do it straightaway without picking numbers also.
1) x^2 < 2x < 1/x implies x^2 < 2x or x < 2 (considering the 1st 2 inequalities).
Consider the 2nd and 3rd inequalities, 2x < 1/x or x^2 < 1/2 or x < 1/√2
So, x < 2 and x < 1/√2 can be true. So, ordering in (1) can be true.

(2) x^2 < 1/x < 2x implies x^2 < 1/x or x^3 < 1 or x < 1
1/x < 2x implies 1/2 < x^2 or x > 1/√2. So, 1/√2 < x < 1 can be true again. So, ordering in (2) can be true.

(3) 2x < x^2 < 1/x implies 2x < x^2 or x > 2
x^2 < 1/x implies x^3 < 1 or x < 1
Here x > 2 and x < 1 cannot be true at one time. So, ordering in (3) cannot be true.

Hence, the correct answer is [spoiler]4)[/spoiler].[/quote]

[santosh_surathkal]

Thanks Anurag