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yellowho
- Master | Next Rank: 500 Posts
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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.
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.
