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Re M2522 [#permalink]
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16 Sep 2014, 01:23
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Official Solution:If function \(f(x)\) satisfies \(f(x) = f(x^2)\) for all \(x\), which of the following must be true? A. \(f(4) = f(2)f(2)\) B. \(f(16)  f(2) = 0\) C. \(f(2) + f(4) = 0\) D. \(f(3) = 3f(3)\) E. \(f(0) = 0\) We are told that some function \(f(x)\) has the following property \(f(x) = f(x^2)\) for all values of \(x\). Note that we don't know the actual function, just this one property of it. For example for this function \(f(3)=f(3^2)\) or \(f(3)=f(9)\), similarly: \(f(9)=f(81)\), so \(f(3)=f(9)=f(81)=...\). Now, the question asks: which of the following MUST be true? A. \(f(4)=f(2)*f(2)\): we know that \(f(2)=f(4)\), but it's not necessary \(f(2)=f(2)*f(2)\) to be true (it will be true if \(f(2)=1\) or \(f(2)=0\) but as we don't know the actual function we cannot say for sure); B. \(f(16)  f(2) = 0\): again \(f(2)=f(4) =f(16)=...\) so \(f(16)f(2)=f(16)f(16)=0\) and thus this option is always true; C. \(f(2) + f(4) = 0\): \(f(2)=f(4)\), but it's not necessary \(f(4) + f(4)=2f(4)=0\) to be true (it will be true only if \(f(4)=0\), but again we don't know that for sure); D. \(f(3)=3*f(3)\): is \(3*f(3)f(3)=0\)? is \(2*f(3)=0\)? is \(f(3)=0\)? As we don't know the actual function we cannot say for sure; E. \(f(0)=0\): And again as we don't know the actual function we cannot say for sure. Answer: B
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Re: M2522 [#permalink]
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23 Oct 2014, 05:37
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Bunuel wrote: Official Solution:
If function \(f(x)\) satisfies \(f(x) = f(x^2)\) for all \(x\), which of the following must be true?
A. \(f(4) = f(2)f(2)\) B. \(f(16)  f(2) = 0\) C. \(f(2) + f(4) = 0\) D. \(f(3) = 3f(3)\) E. \(f(0) = 0\)
We are told that some function \(f(x)\) has the following property \(f(x) = f(x^2)\) for all values of \(x\). Note that we don't know the actual function, just this one property of it. For example for this function \(f(3)=f(3^2)\) or \(f(3)=f(9)\), similarly: \(f(9)=f(81)\), so \(f(3)=f(9)=f(81)=...\). Now, the question asks: which of the following MUST be true? A. \(f(4)=f(2)*f(2)\): we know that \(f(2)=f(4)\), but it's not necessary \(f(2)=f(2)*f(2)\) to be true (it will be true if \(f(2)=1\) or \(f(2)=0\) but as we don't know the actual function we cannot say for sure); B. \(f(16)  f(2) = 0\): again \(f(2)=f(4) =f(16)=...\) so \(f(16)f(2)=f(16)f(16)=0\) and thus this option is always true; C. \(f(2) + f(4) = 0\): \(f(2)=f(4)\), but it's not necessary \(f(4) + f(4)=2f(4)=0\) to be true (it will be true only if \(f(4)=0\), but again we don't know that for sure); D. \(f(3)=3*f(3)\): is \(3*f(3)f(3)=0\)? is \(2*f(3)=0\)? is \(f(3)=0\)? As we don't know the actual function we cannot say for sure; E. \(f(0)=0\): And again as we don't know the actual function we cannot say for sure.
Answer: B Could you please explain as to why E is wrong



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Re: M2522 [#permalink]
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23 Oct 2014, 10:54
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Rahul2000 wrote: Could you please explain as to why E is wrong
Hi Rahul2000, in my opinion the reason for it to be wrong is that it doesn't say: f(0) = f(0). It says f(0) = 0 which is not the same. Do you see the difference?



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Re: M2522 [#permalink]
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23 Oct 2014, 11:02
mdacaret wrote: Rahul2000 wrote: Could you please explain as to why E is wrong
Hi Rahul2000, in my opinion the reason for it to be wrong is that it doesn't say: f(0) = f(0). It says f(0) = 0 which is not the same. Do you see the difference? Aaaah I see it now. So you mean F(X) could be say X+1 . Hence F(0) = F(0) would hold but not F(0) = 0.



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Re: M2522 [#permalink]
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07 Dec 2014, 12:34
Request you to please explain the answer with some other method or please explain more elaborately. Most of these kind of function questions I am unable to understand. Can you please suggest some study material for the same? I have already studied MNAHATTAN Books.



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Re: M2522 [#permalink]
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08 Dec 2014, 05:13



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Re: M2522 [#permalink]
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08 Dec 2014, 07:03
Sir,
I have gone through questions , you have given only answer keys but no explained solutions. can you please give me links where i can read more about functions.



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08 Dec 2014, 07:30



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04 Nov 2015, 16:07
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I think this is a highquality question and the explanation isn't clear enough, please elaborate. I don't get how f(2) = f(4) = f(16) are we squaring terms to get to this ?



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Re: M2522 [#permalink]
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20 Feb 2016, 21:13
This question was pretty tough. I guess since we don't know the actual function all we know is the F(x)=F(x^2) and this could go on infinitely. So F(x)=f(x^2)=F(x^2^2)=F(x^2^2^2) and so on. So basically any number equals the square of that number. So B is the only answer that works. I fell for the E trap though.



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Re: M2522 [#permalink]
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26 Feb 2016, 15:16
Hard Question, fell for the trap. So all you need to see is that F(2) can become 16 and F(16) is 16. Hope don't see this on test day.



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Re: M2522 [#permalink]
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28 Jul 2016, 06:35
We are given f(x) = f(x^2) If possible you can construct an equation that meets this condition ... e.g [ x1/1x] Simply plug in each answer choice and verify
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Re M2522 [#permalink]
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05 Aug 2016, 03:20
I think this is a highquality question and I agree with explanation.



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Re: M2522 [#permalink]
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05 Aug 2016, 03:52
Hi,
I resolved this question assuming that a function that satisfies F(x)=F(X^2) for all x is a constant function. Am I correct?



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Re: M2522 [#permalink]
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11 May 2018, 12:46
So in B, f(2) can be f(4) or f(16) but f(16) cannot be f(16^2). Why?



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11 May 2018, 13:21



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Re: M2522 [#permalink]
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11 May 2018, 13:32
Bunuel wrote: urvashis09 wrote: So in B, f(2) can be f(4) or f(16) but f(16) cannot be f(16^2). Why? f(x) = f(x^2), so f(2) = f(4) = f(16) = f(16^2) = f(256^2) = ... Thank you, now I get it!










