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# M25-22

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Math Expert
Joined: 02 Sep 2009
Posts: 58445

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16 Sep 2014, 01:23
2
22
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Difficulty:

85% (hard)

Question Stats:

37% (01:21) correct 63% (01:22) wrong based on 312 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
9
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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Joined: 02 Oct 2014
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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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Joined: 09 Mar 2014
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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: 58445

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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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Posts: 375
Location: India
Concentration: Operations, Finance
GMAT 1: 670 Q48 V34
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05 Aug 2016, 03:20
I think this is a high-quality question and I agree with explanation.
Manager
Joined: 04 Mar 2016
Posts: 51
Location: United Kingdom
GMAT 1: 710 Q49 V38

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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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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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26 Aug 2019, 11:31
fredurst89 wrote:
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?

This is was my first thought too (thinking about constant 0,1 output) but it's not necessarily the case that it's a constant. It could be some expression, but it's easier to conceptualize with a constant.

If we test x=0,1 we can see that several answers work, if we test x=10 we can see only B) works:

A) 10 = 10 * 10
B) 10 - 10 = 0
C) 10 + 10 = 0
D) 10 = 3 * 10
E) 10 = 0

The point is to understand that given the square in 2nd function, f(-2) = f(4) = f(16)

Bunuel chetan2u is the above reasoning correct?

Takeaway: when given functions without expressions, just plug (and re-plug) numbers to see what works
M25-22   [#permalink] 26 Aug 2019, 11:31
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# M25-22

Moderators: chetan2u, Bunuel