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# Function

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Manager
Joined: 12 Feb 2008
Posts: 175

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23 Sep 2008, 07:10

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

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

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Current Student
Joined: 28 Dec 2004
Posts: 3280
Location: New York City
Schools: Wharton'11 HBS'12

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23 Sep 2008, 07:21
if f(x)=f(x^2) then f(-2)=f(4)...if f(4) then assume if x=4 then f(4)=f(4^2) i.e f(16)..

I will go with B..
Manager
Joined: 22 Jul 2008
Posts: 139

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23 Sep 2008, 08:13
f(-2) = f(4) = f(16)
Hence, f(-2) = f(16)
f(16) - f(-2) = 0
Ans. is B
Manager
Joined: 12 Feb 2008
Posts: 175

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24 Sep 2008, 05:28
KASSALMD wrote:
f(-2) = f(4) = f(16)
Hence, f(-2) = f(16)
f(16) - f(-2) = 0
Ans. is B

Why do you square twice.

f(-2)=f(4) and thats it.
where does f(16) come from. i understand that f(4)=f(16) but the question says only that f(x)=f(x^2)...........
VP
Joined: 17 Jun 2008
Posts: 1474

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25 Sep 2008, 12:41
elmagnifico wrote:
KASSALMD wrote:
f(-2) = f(4) = f(16)
Hence, f(-2) = f(16)
f(16) - f(-2) = 0
Ans. is B

Why do you square twice.

f(-2)=f(4) and thats it.
where does f(16) come from. i understand that f(4)=f(16) but the question says only that f(x)=f(x^2)...........

What f(x) = f(x^2) means that f(2) = f(4) = f(16) = f(256) = ........

Hence, one can use f(16) in place of f(4).

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This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

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Re: Function &nbs [#permalink] 25 Sep 2008, 12:41
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# Function

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