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Tough one

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Tough one [#permalink]

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New post 17 Jun 2011, 09:56
if f(x)>0 and f(x+y)=f(x)f(y) for all x,y. Then f(2) =?
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Status: 700 (q47,v40); AWA 6.0
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Re: Tough one [#permalink]

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New post 17 Jun 2011, 10:34
Quite a vague question but the best extent of information I can draw from this one is that f(0) = 1 and f(n) = f(1)^n

What is the source by the way?

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Re: Tough one [#permalink]

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New post 18 Jun 2011, 07:40
Expert's post
passion4ivymba wrote:
if f(x)>0 and f(x+y)=f(x)f(y) for all x,y. Then f(2) =?


Some information is missing here.
You get \(f(0+0) = f(0)^2\)
\(f(0)[f(0) - 1] = 0\)
\(f(0) = 1\)

Then, you get infinite GPs each having a different value for f(2) depending on what value you choose for f(1).
Say f(1) = 2
... f(-3) = 1/8, f(-2) = 1/4, f(-1) = 1/2, f(0) = 1, f(1) = 2, f(2) = 4, f(3) = 8 and so on

Say f(1) = 3
...f(-3) = 1/27, f(-2) = 1/9, f(-1) = 1/3, f(0) = 1, f(1) = 3, f(2) = 9, f(3) = 27 and so on
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Re: Tough one [#permalink]

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New post 18 Jun 2011, 10:14
Following are the answer options...

rt2
1
0
-1
-rt2
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Re: Tough one [#permalink]

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New post 18 Jun 2011, 10:15
Thanks for the explaination.....


VeritasPrepKarishma wrote:
passion4ivymba wrote:
if f(x)>0 and f(x+y)=f(x)f(y) for all x,y. Then f(2) =?


Some information is missing here.
You get \(f(0+0) = f(0)^2\)
\(f(0)[f(0) - 1] = 0\)
\(f(0) = 1\)

Then, you get infinite GPs each having a different value for f(2) depending on what value you choose for f(1).
Say f(1) = 2
... f(-3) = 1/8, f(-2) = 1/4, f(-1) = 1/2, f(0) = 1, f(1) = 2, f(2) = 4, f(3) = 8 and so on

Say f(1) = 3
...f(-3) = 1/27, f(-2) = 1/9, f(-1) = 1/3, f(0) = 1, f(1) = 3, f(2) = 9, f(3) = 27 and so on
Re: Tough one   [#permalink] 18 Jun 2011, 10:15
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