Function : Quant Question Archive [LOCKED]
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# Function

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Manager
Joined: 12 Feb 2008
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23 Sep 2008, 06:10
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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23 Sep 2008, 06:21
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..
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23 Sep 2008, 07:13
f(-2) = f(4) = f(16)
Hence, f(-2) = f(16)
f(16) - f(-2) = 0
Ans. is B
Manager
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24 Sep 2008, 04:28
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)...........
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25 Sep 2008, 11:41
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).
Re: Function   [#permalink] 25 Sep 2008, 11:41
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