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Tough one

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Intern
Joined: 01 Nov 2010
Posts: 38

Kudos [?]: 4 [0], given: 3

Location: India
Schools: Tuck, Kellogg, Tepper

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17 Jun 2011, 09:56
if f(x)>0 and f(x+y)=f(x)f(y) for all x,y. Then f(2) =?

Kudos [?]: 4 [0], given: 3

Manager
Status: 700 (q47,v40); AWA 6.0
Joined: 16 Mar 2011
Posts: 82

Kudos [?]: 127 [0], given: 3

GMAT 1: 700 Q47 V40

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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?

Regards
Rahul
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Regards
Rahul

Kudos [?]: 127 [0], given: 3

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18 Jun 2011, 07:40
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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Kudos [?]: 17369 [0], given: 232

Intern
Joined: 01 Nov 2010
Posts: 38

Kudos [?]: 4 [0], given: 3

Location: India
Schools: Tuck, Kellogg, Tepper

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18 Jun 2011, 10:14

rt2
1
0
-1
-rt2

Kudos [?]: 4 [0], given: 3

Intern
Joined: 01 Nov 2010
Posts: 38

Kudos [?]: 4 [0], given: 3

Location: India
Schools: Tuck, Kellogg, Tepper

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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

Kudos [?]: 4 [0], given: 3

Re: Tough one   [#permalink] 18 Jun 2011, 10:15
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