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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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The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink]
07 Mar 2015, 07:39
Bunuel wrote:
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.
I don't understand the algebraic form for option "II". I mean, lets say z = 3. Then I'd get \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z-1)}=\frac{1}{3}+\frac{3}{3-1}=/=1\)
Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink]
07 Mar 2015, 07:48
Expert's post
erikvm wrote:
Bunuel wrote:
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.
I don't understand the algebraic form for option "II". I mean, lets say z = 3. Then I'd get \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z-1)}=\frac{1}{3}+\frac{3}{3-1}=/=1\)
hi option ll says we are adding reciprocals of x and y..... it will not be \(\frac{1}{z}+\frac{z}{(z-1)}\) but \(\frac{1}{z}+\frac{(z-1)}{z}\) =1/3+2/3=1 hope it helped _________________
in the attached fig...x Θ y = 1/x + 1/y here x is z and y is\(\frac{z}{(z-1)}\)... you have taken z as 3... so x=z=3..and y=\(\frac{z}{(z-1)}\)=3/(3-1)=3/2... 1/x + 1/y=1/3+1/(3/2)=1/3+2/3=1 you are taking z as 3 for x and z as 1/3 for y.. that is why u are going wrong _________________
in the attached fig...x Θ y = 1/x + 1/y here x is z and y is\(\frac{z}{(z-1)}\)... you have taken z as 3... so x=z=3..and y=\(\frac{z}{(z-1)}\)=3/(3-1)=3/2... 1/x + 1/y=1/3+1/(3/2)=1/3+2/3=1 you are taking z as 3 for x and z as 1/3 for y.. that is why u are going wrong
Im sorry man, I'm sure you are really annoyed, but I really dont understand.. What values am I suppose to take?
in the attached fig...x Θ y = 1/x + 1/y here x is z and y is\(\frac{z}{(z-1)}\)... you have taken z as 3... so x=z=3..and y=\(\frac{z}{(z-1)}\)=3/(3-1)=3/2... 1/x + 1/y=1/3+1/(3/2)=1/3+2/3=1 you are taking z as 3 for x and z as 1/3 for y.. that is why u are going wrong
Im sorry man, I'm sure you are really annoyed, but I really dont understand.. What values am I suppose to take?
does not matter... look dont take any values and work with z itself.. z Θ \(\frac{z}{(z-1)}\).... since x Θ y = 1/x + 1/y... z Θ \(\frac{z}{(z-1)}\)=\(\frac{1}{z}\)+1/\(\frac{z}{(z-1)}\)= \(\frac{1}{z}\)+\(\frac{(z-1)}{z)}\).... \(\frac{(1+z-1)}{z}\)=z/z=1 _________________
Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink]
04 Jul 2015, 17:18
Bunuel wrote:
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.
Hi Bunuel,
I don't know if that is the ideal place to ask my question, but I haven't found any post gathering the broader topic, i.e Functions / Expressions.
In such exercise, can we consider the assumption right in any case if it did work using 1 number (respecting the constraint given by the wording, obviously). I used 2 for that one, but lost time double checking with 8 and 9, probably a bad reflex acquired after thoroughly testing numbers in DS questions!
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