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16 Apr 2018, 05:20
Kudos
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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:
15%
(low)
Question Stats:
79% (01:34) correct
21%
(01:30)
wrong
based on 223
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If \(x = −\frac{2}{3}\), which of the following inequalities properly lists its terms in ascending order?
A. \(\frac{1}{x}<x<x^2<x^3\)
B. \(x<x^3<\frac{1}{x}<x^2\)
C. \(x^3<x^2<x<\frac{1}{x}\)
D. \(x<x^2<x^3<\frac{1}{x}\)
E. \(\frac{1}{x}<x<x^3<x^2\)
A. \(\frac{1}{x}<x<x^2<x^3\)
B. \(x<x^3<\frac{1}{x}<x^2\)
C. \(x^3<x^2<x<\frac{1}{x}\)
D. \(x<x^2<x^3<\frac{1}{x}\)
E. \(\frac{1}{x}<x<x^3<x^2\)
16 Apr 2018, 05:33
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Bunuel
E
Courtesy: Ron
"if you ever see a problem like this -- on which you're comparing different powers of the same variable -- then, immediately, you should think about the following number properties:
* signs
* ""fractions"" (between 0 and 1, or between -1 and 0) vs. ""non-fractions"" (greater than 1 or less than -1)
for each of these ranges of numbers -- less than -1; between -1 and 0; between 0 and 1; greater than 1 -- the powers have different properties, and so the ordering of the powers is different.
it's rather pointless (and possibly confusing) to memorize what happens to the powers. instead, just throw in a few numbers of each type and watch what they do.
if the inequalities are not strict inequalities (i.e., if they are the < or > type), then you should also mess around with the numbers 0, 1, and -1."
17 Apr 2018, 16:15
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Bunuel
Let’s plug -2/3 in for x:
1/x = -3/2
x = -2/3
x^2 = 4/9
x^3 = -8/27
So, in ascending order we have:
1/x , x, x^3, x^2
Alternate Solution:
We notice that x^2 is positive, whereas x, 1/x and x^3 are negative. Thus, among the given quantities, x^2 will be the greatest, which eliminates A, C and D.
To decide between B and E, we only need to compare x and 1/x. Since -3/2 < -2/3, 1/x is less than x and thus, E is the correct answer.
Answer: E
