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M25-22

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Math Expert
Joined: 02 Sep 2009
Posts: 48067

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16 Sep 2014, 01:23
2
21
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Difficulty:

85% (hard)

Question Stats:

36% (01:01) correct 64% (00:54) wrong based on 435 sessions

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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. $$f(0) = 0$$

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16 Sep 2014, 01:23
2
8
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.

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23 Oct 2014, 05:37
1
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.

Could you please explain as to why E is wrong
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23 Oct 2014, 10:54
1
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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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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07 Dec 2014, 12:34
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.
Math Expert
Joined: 02 Sep 2009
Posts: 48067

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08 Dec 2014, 05:13
manojpandey80 wrote:
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.

Check function questions in our Special Questions Directory:

Operations/functions defining algebraic/arithmetic expressions
Symbols Representing Arithmetic Operation
Rounding Functions
Various Functions
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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.
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08 Dec 2014, 07:30
manojpandey80 wrote:
Sir,

I have gone through questions , you have given only answer keys but no explained solutions.

All questions there have solutions, no?
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04 Nov 2015, 16:07
1
I think this is a high-quality 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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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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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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Joined: 21 Sep 2015
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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 [ x-1/1-x]

Simply plug in each answer choice and verify
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05 Aug 2016, 03:20
I think this is a high-quality question and I agree with explanation.
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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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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
1
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) = ...
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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!
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28 Jun 2018, 01:34
So, what I understand from this question is that the answer is B because the function described in the question could be stretched to infinity as every number can raised to it's square and there is no given value of x per se. Right, Bunuel ?
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28 Jun 2018, 02:12
1
HinaTabassum wrote:
So, what I understand from this question is that the answer is B because the function described in the question could be stretched to infinity as every number can raised to it's square and there is no given value of x per se. Right, Bunuel ?

The answer is B because for any value of x, f(-2) = f(4) = f(16) = f(16^2) = f(256^2) = .
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