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If x=−2/3, which of the following inequalities properly lists its term

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If x=−2/3, which of the following inequalities properly lists its term  [#permalink]

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New post 16 Apr 2018, 05:20
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A
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E

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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. \(\frac{1}{x}<x<x^3<x^2\)

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Re: If x=−2/3, which of the following inequalities properly lists its term  [#permalink]

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New post 16 Apr 2018, 05:33
Bunuel wrote:
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. \(\frac{1}{x}<x<x^3<x^2\)


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."
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Re: If x=−2/3, which of the following inequalities properly lists its term  [#permalink]

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New post 17 Apr 2018, 16:15
Bunuel wrote:
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. \(\frac{1}{x}<x<x^3<x^2\)


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
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Re: If x=−2/3, which of the following inequalities properly lists its term  [#permalink]

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New post 24 Apr 2018, 06:24
Bunuel wrote:
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. \(\frac{1}{x}<x<x^3<x^2\)


We have x = 2/3 ( a negative fraction)

We know x^2 is always positive irrespective of the value of x.

x^3 will be negative, therefore highest value of x will be x^2.

We can eliminate A, C, and D.

From B and E, we have x = -2/3 and 1/x = -3/2 as first term in the order.

We know -3/2 < -2/3

therefore (E) is the correct answer.
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Re: If x=−2/3, which of the following inequalities properly lists its term &nbs [#permalink] 24 Apr 2018, 06:24
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