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GRE Prep Club Team Member
Joined: 20 Feb 2017
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If x is positive, which of the following could be correct ordering of
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27 Nov 2021, 04:22
27 Nov 2021, 04:22
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If x is positive, which of the following could be correct ordering of \(\frac{1}{x}\), \(2x\), and \(x^2\)?
I. \(x^2 < 2x < \frac{1}{x}\)
II. \(x^2 < \frac{1}{x} < 2x\)
III. \(2x < x^2 < \frac{1}{x}\)
(A) none
(B) I only
(C) III only
(D) I and II
(E) I, II and III
I. \(x^2 < 2x < \frac{1}{x}\)
II. \(x^2 < \frac{1}{x} < 2x\)
III. \(2x < x^2 < \frac{1}{x}\)
(A) none
(B) I only
(C) III only
(D) I and II
(E) I, II and III
Re: If x is positive, which of the following could be correct ordering of
[#permalink]
28 Nov 2021, 09:56
28 Nov 2021, 09:56
1
Expert Reply
First note that we are asked "which of the following COULD be the correct ordering" not MUST be.
Basically we should determine relationship between \(x\), \(\frac{1}{x}\) and \(x^2\) in three areas: \(0<1<2<\).
\(x>2\)
\(1<x<2\)
\(0<x<1\)
When \(x>2\) --> \(x^2\) is the greatest and no option is offering this, so we know that x<2.
If \(1<x<2\) --> \(2x\) is greatest then comes \(x^2\) and no option is offering this.
So, we are left with \(0<x<1\):
In this case \(x^2\) is least value, so we are left with:
I. \(x^2<2x<\frac{1}{x}\) --> can \(2x<\frac{1}{x}\)? Can \(\frac{2x^2-1}{x}<0\), the expression \(2x^2-1\) can be negative or positive for \(0<x<1\). (You can check it either algebraically or by picking numbers)
II. \(x^2<\frac{1}{x}<2x\) --> can \(\frac{1}{x}<2x\)? The same here \(\frac{2x^2-1}{x}>0\), the expression \(2x^2-1\) can be negative or positive for \(0<x<1\). (You can check it either algebraically or by picking numbers)
Answer: D.
Second condition: \(x^2<\frac{1}{x}<2x\)
The question is which of the following COULD be the correct ordering not MUST be.
Put \(0.9\) --> \(x^2=0.81\), \(\frac{1}{x}=1.11\), \(2x=1.8\) --> \(0.81<1.11<1.8\). Hence this COULD be the correct ordering.
Basically we should determine relationship between \(x\), \(\frac{1}{x}\) and \(x^2\) in three areas: \(0<1<2<\).
\(x>2\)
\(1<x<2\)
\(0<x<1\)
When \(x>2\) --> \(x^2\) is the greatest and no option is offering this, so we know that x<2.
If \(1<x<2\) --> \(2x\) is greatest then comes \(x^2\) and no option is offering this.
So, we are left with \(0<x<1\):
In this case \(x^2\) is least value, so we are left with:
I. \(x^2<2x<\frac{1}{x}\) --> can \(2x<\frac{1}{x}\)? Can \(\frac{2x^2-1}{x}<0\), the expression \(2x^2-1\) can be negative or positive for \(0<x<1\). (You can check it either algebraically or by picking numbers)
II. \(x^2<\frac{1}{x}<2x\) --> can \(\frac{1}{x}<2x\)? The same here \(\frac{2x^2-1}{x}>0\), the expression \(2x^2-1\) can be negative or positive for \(0<x<1\). (You can check it either algebraically or by picking numbers)
Answer: D.
Second condition: \(x^2<\frac{1}{x}<2x\)
The question is which of the following COULD be the correct ordering not MUST be.
Put \(0.9\) --> \(x^2=0.81\), \(\frac{1}{x}=1.11\), \(2x=1.8\) --> \(0.81<1.11<1.8\). Hence this COULD be the correct ordering.
gmatclubot
Re: If x is positive, which of the following could be correct ordering of [#permalink]
28 Nov 2021, 09:56
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