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If 0 < x < 1, which of the following is the greatest?

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V
Joined: 02 Sep 2009
Posts: 55274
If 0 < x < 1, which of the following is the greatest?  [#permalink]

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New post 10 Sep 2017, 05:10
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

62% (01:09) correct 38% (00:55) wrong based on 65 sessions

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Joined: 12 Feb 2017
Posts: 70
If 0 < x < 1, which of the following is the greatest?  [#permalink]

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New post 10 Sep 2017, 09:58
1
0 < x < 1
lets assume x=0.16

A)x^(-1/2)
=1/{x^(1/2)}
=1/(0.16)^1/2
=1/0.4
=2.5

B)x^0=1

c)x^1/2
=0.16^1/2
=0.4

D)x^1
=0.16^1
=0.16

E)x^2
=0.16^2
=0.0256

Hence Answer is option A

Kudos if it helps.
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Joined: 22 May 2016
Posts: 2766
If 0 < x < 1, which of the following is the greatest?  [#permalink]

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New post 10 Sep 2017, 14:09
Bunuel wrote:
If 0 < x < 1, which of the following is the greatest?


A. \(x^{(-\frac{1}{2})}\)

B. \(x^0\)

C. \(x^{(\frac{1}{2})}\)

D. \(x^1\)

E. \(x^2\)

If 0 < x < 1, it is a proper fraction, (e.g., \(\frac{1}{2}\)).

Number properties
If very fluent in properties of these fractions, finding the answer is quick and no calculations are needed:

The negative exponent with the smallest value will yield the greatest answer. Only choice A has a negative exponent.

Given their equal absolute values, you could compare Answer A and Answer C (see below for calculation) to be sure.

After that or at a glance, the greatest result is

Answer A

Test a value

Because the answer choices contain square roots, and to keep things simple, choose a unit fraction whose denominator is a perfect square, e.g. \(\frac{1}{9}\)

A. \((\frac{1}{9})^{(-\frac{1}{2})}\) =\(9^{(\frac{1}{2})}\) = \(\sqrt{9} =
3\)

B. \((\frac{1}{9})^0\) = 1

C.\((\frac{1}{9})^{(\frac{1}{2})}\) =\(\sqrt{\frac{1}{9}}\) = \(\frac{1}{3}\)

D. \((\frac{1}{9})^1\) = \(\frac{1}{9}\)

E. \((\frac{1}{9})^2\) = \(\frac{1}{81}\)

Greatest of these results?

ANSWER A
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If 0 < x < 1, which of the following is the greatest?   [#permalink] 10 Sep 2017, 14:09
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