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Senior Manager
Senior Manager
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Joined: 05 Oct 2008
Posts: 270

Kudos [?]: 538 [0], given: 22

Another function [#permalink]

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New post 08 Oct 2008, 20:55
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Can someone please post an explanation to the foll. qs

with reference to the foll question, I cannot get myself to understand how f(16) - f(-2) = 0
If f(x) = f(x^2)
then f(16) = 16^2 ie 256
and f(-2) = -2^2 = 4

or f(16) = f(4^2)
but then f(2) = f(1.4^2) right?


Function f(x) satisfies f(x) = f(x^2) for all x . Which of the following must be true?

* f(4) = f(2)f(2)
* f(16) - f(-2) = 0
* f(-2) + f(4) = 0
* f(3) = 3f(3)
* f(0) = 0

Kudos [?]: 538 [0], given: 22

Intern
Intern
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Joined: 11 Jun 2008
Posts: 1

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

Re: Another function [#permalink]

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New post 09 Oct 2008, 01:29
I think it supposed to be like this:

f(2)=f(4)
f(4)=f(16) so f(2) or f(-2)=f(16)

study wrote:
Can someone please post an explanation to the foll. qs

with reference to the foll question, I cannot get myself to understand how f(16) - f(-2) = 0
If f(x) = f(x^2)
then f(16) = 16^2 ie 256
and f(-2) = -2^2 = 4

or f(16) = f(4^2)
but then f(2) = f(1.4^2) right?


Function f(x) satisfies f(x) = f(x^2) for all x . Which of the following must be true?

* f(4) = f(2)f(2)
* f(16) - f(-2) = 0
* f(-2) + f(4) = 0
* f(3) = 3f(3)
* f(0) = 0

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

8 KUDOS received
Intern
Intern
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Joined: 24 Sep 2008
Posts: 18

Kudos [?]: 53 [8], given: 0

Re: Ya its a tricky one. [#permalink]

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New post 09 Oct 2008, 02:06
8
This post received
KUDOS
Answer choice f(16) - f(-2) = 0 is the correct one. I solved this as follows:

f(16) - f(-2) = 0
Is f(16) = f(-2) ?

Taking LHS f(16) could be f(16x16) or alternatively (f4) as f(4x4) = f(16)
so f(16) is either f(4) or f(256)

Now RHS f(-2) could be f(-2 x -2) = f(4) but watchout it cannot be (1.4 x 1.4) as it is a negative number.(square of any number is always positive)
so f(-2) is f(4)

Try same approach on all other options and it will fail.

Hope that helps.

Kudos [?]: 53 [8], given: 0

Re: Ya its a tricky one.   [#permalink] 09 Oct 2008, 02:06
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