Vavali wrote:
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
Let's look at this question logically. There will be some key takeaways here so don't focus on the question and the (long) solution. Focus on the logic.
First of all, we are just dealing with positives so life is simpler.
To compare two terms e.g. \(x^2\) and \(2x\), we should focus on the points where they are equal. \(x^2 = 2x\) holds when \(x = 2\).
When \(x < 2, x^2 < 2x\)
When \(x > 2, x^2 > 2x\)
Similarly \(1/x = x^2\) when \(x = 1\)
When \(x < 1, 1/x > x^2\).
When \(x > 1, 1/x < x^2\)
Going on, \(1/x = 2x\) when \(x = 1/\sqrt{2}\)
When \(x < 1/\sqrt{2}, 1/x > 2x\)
When \(x > 1/\sqrt{2}, 1/x < 2x\)
So now you know that:
If \(x < 1/\sqrt{2}\),
\(1/x > 2x, 1/x > x^2\) and \(x^2 < 2x\)
So \(x^2 < 2x < 1/x\) is possible.
If \(1/\sqrt{2} < x < 1\)
\(1/x < 2x, 1/x > x^2\)
So \(x^2 < 1/x < 2x\) is possible.
If \(1 < x < 2\)
\(1/x < 2x, 1/x < x^2, x^2 < 2x\)
So \(1/x < x^2 < 2x\) is possible.
If \(x > 2\)
\(1/x < 2x, 1/x < x^2, x^2 > 2x\)
So \(1/x < 2x < x^2\) is possible.
For no positive values of x is the third relation possible.
*Edited
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Karishma
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