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The operation Θ is defined by x Θ y = 1/x + 1/y for all

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The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink] New post 28 Sep 2006, 02:17
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The operation Θ is defined by x Θ y = 1/x + 1/y for all nonzero numbers x and y. If z is a number greater than 1, which of the following must be true?

I. \hspace{10} z \hspace{2} \theta \hspace{2} (-z) = 0

II. \hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z-1)}= 1

III. \hspace{10} \frac{2}{z} \hspace{2} \theta \hspace{2} \frac{2}{z}= z

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III
[Reveal] Spoiler: OA

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Last edited by Bunuel on 08 Sep 2013, 05:45, edited 3 times in total.
Edited the question.
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Re: The operation theta is defined by [#permalink] New post 08 Sep 2013, 03:19
Can anyone verify the question? I think that first option should be z Θ (-z) =0 for this to be true!..
So either D or E! cheers~!!~
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Re: The operation theta is defined by [#permalink] New post 08 Sep 2013, 05:48
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rskanumuri wrote:
Can anyone verify the question? I think that first option should be z Θ (-z) =0 for this to be true!..
So either D or E! cheers~!!~


Edited the question.

The operation Θ is defined by x Θ y = 1/x + 1/y for all nonzero numbers x and y. If z is a number greater than 1, which of the following must be true?

I. \hspace{10} z \hspace{2} \theta \hspace{2} (-z) = 0

II. \hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z-1)}= 1

III. \hspace{10} \frac{2}{z} \hspace{2} \theta \hspace{2} \frac{2}{z}= z

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III

I. \hspace{10} z \hspace{2} \theta \hspace{2} (-z) =\frac{1}{z}-\frac{1}{z}=0 --> true.

II. \hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z-1)}=\frac{1}{z}+\frac{z-1}{z}=1 --> true.

III. \hspace{10} \frac{2}{z} \hspace{2} \theta \hspace{2} \frac{2}{z}= \frac{z}{2}+\frac{z}{2}=z --> true.

Answer: E.
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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink] New post 01 Nov 2014, 04:00
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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all   [#permalink] 01 Nov 2014, 04:00
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