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metallicafan
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20 Jun 2012, 10:44
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
47% (02:10) correct
53%
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wrong
based on 90
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Is x^2 greater than x ?
(1) x^2 > x^3.
(2) x^2 > x^4.
(1) x^2 > x^3.
(2) x^2 > x^4.
I already solved it, but I tried to solve it by also the following method, and I don't know in which part I am wrong.
(1) \(x^2 > x^3\).
\(x^2 - x^3 > 0\)
\(x^2 * (1-x) > 0\)
So, we get: x= 0 and x = 1
-----(+)------0----(-)-------1------(+)------
Therefore, the intervals are x<0 and x>1
However, if we solve it in this way:
\(x^2 > x^3\)
We divide both sides by \(x^2\):
\(1 > x\)
The interval is not the same.
Please, tell me in which part of my first way I am wrong.
Thanks!
(1) \(x^2 > x^3\).
\(x^2 - x^3 > 0\)
\(x^2 * (1-x) > 0\)
So, we get: x= 0 and x = 1
-----(+)------0----(-)-------1------(+)------
Therefore, the intervals are x<0 and x>1
However, if we solve it in this way:
\(x^2 > x^3\)
We divide both sides by \(x^2\):
\(1 > x\)
The interval is not the same.
Please, tell me in which part of my first way I am wrong.
Thanks!
20 Jun 2012, 12:32
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Bookmarks
metallicafan
Hi,
When you say,
\(x^2 (1-x) > 0\)
if \(x\neq 0\) then,\(x^2 >0\) for any real value of x.
thus, inequality reduces to \(1-x>0\)
or \(x < 1\)
Also, to use the number line method, you should change the coefficient of x as positive (for simplicity)
\(x^2 (x-1) < 0\) (multiply by -1 and reverse the sign)
now, use the number line;
-----(-)-----1---(+)---------
Please go through the following post:
Solving inequalities
Regards,
20 Jun 2012, 12:41
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Hi,
Is \(x^2>x\)?
or \(x^2-x>0\)?
or \(x(x-1)>0\)?
or x<0 or x>1?
Detailed solution:
Using (1),
\(x^2>x^3\)
or \(x^2(1-x)>0\)
or x < 1, for x = 0.5
\(x^2=0.25<x\)
but for x=-1,
\(x^2=1>x\). Thus, Insufficient.
Using (2),
\(x^2>x^4\)
or \(x^2(1-x^2)>0\)
or \(x^2(1-x)(1+x)>0\)
or \(x^2(x-1)(1+x)<0\) (multiplied by -1)
-(+)-------(-1)---(-)-----(+1)-----(+)------
or -1<x<1
Again, for x = 0.5
\(x^2=0.25<x\)
but for x=-0.5,
\(x^2=0.25>x\). Thus, Insufficient.
Even after using (1) & (2), we have -1 < x <1, Insufficient.
Answer is (E),
Regards,
Is \(x^2>x\)?
or \(x^2-x>0\)?
or \(x(x-1)>0\)?
or x<0 or x>1?
Detailed solution:
Using (1),
\(x^2>x^3\)
or \(x^2(1-x)>0\)
or x < 1, for x = 0.5
\(x^2=0.25<x\)
but for x=-1,
\(x^2=1>x\). Thus, Insufficient.
Using (2),
\(x^2>x^4\)
or \(x^2(1-x^2)>0\)
or \(x^2(1-x)(1+x)>0\)
or \(x^2(x-1)(1+x)<0\) (multiplied by -1)
-(+)-------(-1)---(-)-----(+1)-----(+)------
or -1<x<1
Again, for x = 0.5
\(x^2=0.25<x\)
but for x=-0.5,
\(x^2=0.25>x\). Thus, Insufficient.
Even after using (1) & (2), we have -1 < x <1, Insufficient.
Answer is (E),
Regards,
