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
I would typically start here by seeing if I can spot any obvious x-values that satisfy any of the statements.
For statements both I and II, \(x^2\) is the smallest value.
This typically occurs when \(0<x<1\).
So let's test a value in that range.
If \(x = 0.1\), we get:
Statement I. \((0.1)^2 < 2(0.1) < \frac{1}{(0.1)}\), which simplifies to: \(0.01 < 0.2 < 10\) WORKS!!
So, statement I could be true, which means we can eliminate answer choices A and C, since they incorrectly state that statement I can't be true
Aside: Since \(x = 0.1\) makes statement I true, I know that it won't make the other two statements true since they have different orderings of \(x^2\), \(2x\) and \(\frac{1}{x}\)Now let's analyze statement II: \(x^2 < \frac{1}{x} < 2x\)
Since we know \(x\) is positive, we can safely multiply all parts by \(x\) to get: \(x^3 < 1 < 2x^2\)
If \(x^3 < 1\), then we know \(x<1\).
What about this part of the inequality: \(1 < 2x^2\)
Divide both sides by \(2\) to get: \(\frac{1}{2} < x^2\)
Find the square root of both sides: \(\sqrt{\frac{1}{2}} < x\)
We can rewrite this as follows: \(\frac{1}{\sqrt{2}}<x\)
ASIDE: Before test day, be sure to memorize the following approximations:
√2 ≈ 1.4
√3 ≈ 1.7
√5 ≈ 2.2Substitute 1.4 for √2 to get: \(\frac{1}{1.4}<x\)
Approximate: \(0.7 < x\)
When we combine our two conclusions we see that statement II is true when \(0.7 < x < 1\)
To confirm, let's plug \(x = 0.8\) into statement II to get: \((0.8)^2 < \frac{1}{0.8} < 2(0.8)\), which simplifies (approximately) to \(0.64 < 1.25 < 1.6\) WORKS!
So, statement II could be true, which means we can eliminate answer choice B, since it incorrectly states that statement II can't be true
Now onto statement III: \(2x < x^2 < \frac{1}{x}\)
Take this part of the inequality: \(2x < x^2\)
Subtract \(2x\) from both sides: \(0 < x^2 - 2x\)
Factor: \(0 < x(x- 2)\)
Since we already know \(x\) is positive, the only way \(x(x- 2)\) can be positive is if \(x > 2\)
However, if \(x > 2\), then there's no way that \(\frac{1}{x}\) can have the greatest value.
Therefore, statement III can never be true.
Answer: D