GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 31 May 2020, 05:49

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# If function f(x) satisfies f(x) = f(x^2) for all x

Author Message
TAGS:

### Hide Tags

Manager
Status: Current MBA Student
Joined: 19 Nov 2009
Posts: 89
Concentration: Finance, General Management
GMAT 1: 720 Q49 V40
If function f(x) satisfies f(x) = f(x^2) for all x  [#permalink]

### Show Tags

04 Jan 2011, 14:21
4
26
00:00

Difficulty:

85% (hard)

Question Stats:

41% (01:32) correct 59% (01:33) wrong based on 218 sessions

### HideShow timer Statistics

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. $$f(0) = 0$$

I reviewed the previous function posts that Bunuel referred me to and seem to understand those problems decently. But I don't clearly understand the process of substitution and POE on this problem. I would appreciate if someone could help me with understand how to break the answer choices down.
Math Expert
Joined: 02 Sep 2009
Posts: 64249

### Show Tags

04 Jan 2011, 16:37
4
1
tonebeeze wrote:
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. $$f(0) = 0$$

I reviewed the previous function posts that Bunuel referred me to and seem to understand those problems decently. But I don't clearly understand the process of substitution and POE on this problem. I would appreciate if someone could help me with understand how to break the answer choices down.

This function question is different from the problems you are referring to.

We are told that some function $$f(x)$$ has 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 $$f(3)=f(3^2)$$ --> $$f(3)=f(9)$$, 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 can not 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 can not say for sure;

E. $$f(0)=0$$: And again as we don't know the actual function we can not say for sure.

Hope it's clear.
_________________
##### General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 64249

### Show Tags

05 Jan 2011, 02:39
ionutulescu wrote:
While C can be true, we need to find the option that MUST be true. The answer will be true in all cases.

As shown above each and every option COULD be true but only option B MUST be true.
_________________
Manager
Joined: 17 Oct 2010
Posts: 69

### Show Tags

24 May 2012, 01:15
f(16) =f(16^2) = f( 16^4) .....and f(-2)= f(4) =f(16)=f(256)=f( 256^2)..

as f(x) = f(x^2)

so we can write f(16)-f(-2) as= f(16^2) -f(256^2) != 0

so f(16) = f( 16^2) = f(256)
if f(-2) can be written as f(16)
then f(-2) can also be written as f( 256^2) as shown above then when we have

f(256)-f(256^2) then for this particular condition I believe that f(x) cannot be true , so how are we confirming that f(16)-f(-2) MUST always be equal to 0

Math Expert
Joined: 02 Sep 2009
Posts: 64249

### Show Tags

24 May 2012, 01:20
Joy111 wrote:
f(16) =f(16^2) = f( 16^4) .....and f(-2)= f(4) =f(16)=f(256)=f( 256^2)..

as f(x) = f(x^2)

so we can write f(16)-f(-2) as= f(16^2) -f(256^2) != 0

so f(16) = f( 16^2) = f(256)
if f(-2) can be written as f(16)
then f(-2) can also be written as f( 256^2) as shown above then when we have

f(256)-f(256^2) then for this particular condition I believe that f(x) cannot be true , so how are we confirming that f(16)-f(-2) MUST always be equal to 0

Can you please tell me what do you mean by the red part?

Anyway: we are told that $$f(x) = f(x^2)$$, so $$f(-2)=f(4) =f(16)=...$$ --> $$f(16)-f(-2)=f(16)-f(16)=0$$. Thus option D is always true.
_________________
Manager
Joined: 17 Oct 2010
Posts: 69

### Show Tags

Updated on: 24 May 2012, 02:13
Bunuel wrote:
Joy111 wrote:
f(16) =f(16^2) = f( 16^4) .....and f(-2)= f(4) =f(16)=f(256)=f( 256^2)..

as f(x) = f(x^2)

so we can write f(16)-f(-2) as= f(16^2) -f(256^2) != 0

so f(16) = f( 16^2) = f(256)
if f(-2) can be written as f(16)
then f(-2) can also be written as f( 256^2) as shown above then when we have

f(256)-f(256^2) then for this particular condition I believe that f(x) cannot be true , so how are we confirming that f(16)-f(-2) MUST always be equal to 0

Can you please tell me what do you mean by the red part?

Anyway: we are told that $$f(x) = f(x^2)$$, so $$f(-2)=f(4) =f(16)=...$$ --> $$f(16)-f(-2)=f(16)-f(16)=0$$. Thus option D is always true.

given statement = f(16)-f(-2)= 0

which can be written as

1) f(16)-f(4) lets say we stop here , how do we know that this equation will be 0
2) f(16)-f(16) this as we can see will of course yield a 0
3)f(16)-f(256) this is yet another way of writing f(16) - f(-2) , again how can we be sure that this will always be 0 without knowing the function .

we are manipulating f(16)-f(-2) to become f(16)-f(16) this of course as we can see will be zero

but what if we write f(16)-f(-2) as f(-2)-f(256 ) and stop here , without the function it is difficult to see how this will yield a 0
if (ii) is the answer then f(16)-f(-2) = 0 must be true for all functions for the value's of x=16 and x= -2
but since f(x) = f(x^2) then f(-2)- f( 256) = 0 must also be true for all functions for the value's of x=-2 and x= 256

I think we are only considering one condition, the condition when f(-2) = f(16) then of course it is 0, but what if f(-2) is not equal to f(16) , in that case how can we say that f(16)- f( -2) will always be zero.

is it possible to clearly prove this by taking the example of one actual function rather then dealing ambiguously. thank you.

Originally posted by Joy111 on 24 May 2012, 01:37.
Last edited by Joy111 on 24 May 2012, 02:13, edited 3 times in total.
Math Expert
Joined: 02 Sep 2009
Posts: 64249

### Show Tags

24 May 2012, 01:43
Joy111 wrote:
Bunuel wrote:
Joy111 wrote:
f(16) =f(16^2) = f( 16^4) .....and f(-2)= f(4) =f(16)=f(256)=f( 256^2)..

as f(x) = f(x^2)

so we can write f(16)-f(-2) as= f(16^2) -f(256^2) != 0

so f(16) = f( 16^2) = f(256)
if f(-2) can be written as f(16)
then f(-2) can also be written as f( 256^2) as shown above then when we have

f(256)-f(256^2) then for this particular condition I believe that f(x) cannot be true , so how are we confirming that f(16)-f(-2) MUST always be equal to 0

Can you please tell me what do you mean by the red part?

Anyway: we are told that $$f(x) = f(x^2)$$, so $$f(-2)=f(4) =f(16)=...$$ --> $$f(16)-f(-2)=f(16)-f(16)=0$$. Thus option D is always true.

since f(x) = f(x^2) so we can write f(16) = f( 256)
similarly f(-2)= f(4) = f(16)= f(256) =f( 256^2)

hence f(16) - f(-2) can also be written as f( 256) - f( 256^2)

so how is this statement f(16) -f(-2) = 0 = f( 256) - f( 256^2) always valid

I don't understand your question at all.

$$f(16)-f(-2)$$ and $$f(16)-f(16^2)$$ both equal to zero since $$f(x) = f(x^2)$$ for all $$x$$, from which we have that $$f(16)=f(-2)$$ and $$f(16)=f(16^2)$$.
_________________
Manager
Joined: 17 Oct 2010
Posts: 69

### Show Tags

24 May 2012, 02:26
I don't understand your question at all.

$$f(16)-f(-2)$$ and $$f(16)-f(16^2)$$ both equal to zero since $$f(x) = f(x^2)$$ for all $$x$$, from which we have that $$f(16)=f(-2)$$ and $$f(16)=f(16^2)$$.[/quote]

I think we are only considering one condition, the condition when f(-2) = f(16) then of course it is 0, but what if f(-2) is not equal to f(16) , in that case how can we say that f(16)- f( -2) will always be zero. f(-2) could equal = f(256) = f(256^2) etc

so f(16)- f( 256^2) , is this equal to 0 as well.

.

is it possible to clearly prove this by taking the example of one actual function .Thank you

every thing will fall into place , if this question was COULD be true rather than MUST be true , could you please check this once more .
Math Expert
Joined: 02 Sep 2009
Posts: 64249

### Show Tags

24 May 2012, 02:32
Joy111 wrote:
I think we are only considering one condition, the condition when f(-2) = f(16) then of course it is 0, but what if f(-2) is not equal to f(16) , in that case how can we say that f(16)- f( -2) will always be zero. f(-2) could equal = f(256) = f(256^2) etc

so f(16)- f( 256^2) , is this equal to 0 as well.

.

is it possible to clearly prove this by taking the example of one actual function .Thank you

every thing will fall into place , if this question was COULD be true rather than MUST be true , could you please check this once more .

I think that you just have some problems understanding the question.

See the red part "what if f(-2) is not equal to f(16) ". That's the point: since $$f(x) = f(x^2)$$ for all $$x$$ then $$f(16)=f(-2)$$ without any "what if".
_________________
Intern
Joined: 28 Feb 2012
Posts: 19
GMAT 1: 700 Q48 V39
WE: Information Technology (Computer Software)

### Show Tags

24 May 2012, 09:05
Joy111 wrote:
f(16) =f(16^2) = f( 16^4) .....and f(-2)= f(4) =f(16)=f(256)=f( 256^2)..

as f(x) = f(x^2)

so we can write f(16)-f(-2) as= f(16^2) -f(256^2) != 0

so f(16) = f( 16^2) = f(256)
if f(-2) can be written as f(16)
then f(-2) can also be written as f( 256^2) as shown above then when we have

f(256)-f(256^2) then for this particular condition I believe that f(x) cannot be true , so how are we confirming that f(16)-f(-2) MUST always be equal to 0

Let us assume f(x) = y.
f(x) = f(x^2) means that for all values of x, f(x) takes the exact same value as f(x^2) takes.

And vice versa f(x^2) is always equal to f(x).

so, f(x^2) also equals y.
Now irrespective of how far you go on squaring x, the value f(x) or f(x^2) or f(x^4) etc will always be y.
Intern
Joined: 26 May 2012
Posts: 3

### Show Tags

26 May 2012, 11:55
Bunuel wrote:
tonebeeze wrote:
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. $$f(0) = 0$$

I reviewed the previous function posts that Bunuel referred me to and seem to understand those problems decently. But I don't clearly understand the process of substitution and POE on this problem. I would appreciate if someone could help me with understand how to break the answer choices down.

This function question is different from the problems you are referring to.

We are told that some function $$f(x)$$ has 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 $$f(3)=f(3^2)$$ --> $$f(3)=f(9)$$, 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 can not 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 can not say for sure;

E. $$f(0)=0$$: And again as we don't know the actual function we can not say for sure.

Hope it's clear.

who created this question, the solution is trivial has no end, how can you stop f(16) just as it is, f(16) can be f(16 sqaured) so on and so forth, there is no end to it unless some boundary conditions mentioned.
Math Expert
Joined: 02 Sep 2009
Posts: 64249

### Show Tags

27 May 2012, 02:09
koro12 wrote:
Bunuel wrote:
tonebeeze wrote:
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. $$f(0) = 0$$

I reviewed the previous function posts that Bunuel referred me to and seem to understand those problems decently. But I don't clearly understand the process of substitution and POE on this problem. I would appreciate if someone could help me with understand how to break the answer choices down.

This function question is different from the problems you are referring to.

We are told that some function $$f(x)$$ has 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 $$f(3)=f(3^2)$$ --> $$f(3)=f(9)$$, 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 can not 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 can not say for sure;

E. $$f(0)=0$$: And again as we don't know the actual function we can not say for sure.

Hope it's clear.

who created this question, the solution is trivial has no end, how can you stop f(16) just as it is, f(16) can be f(16 sqaured) so on and so forth, there is no end to it unless some boundary conditions mentioned.

The question asks which of the options presented must be true.

Now, since $$f(x) = f(x^2)$$ then $$f(-2)=f(16)$$ so option B. $$f(16) - f(-2) = 0$$ is always true.

Next, the fact that $$f(16)=f(16^2)$$ does not make option B any less correct. So there is nothing wrong with the question above.

Hope it's clear.
_________________
Intern
Joined: 18 Oct 2012
Posts: 1

### Show Tags

18 Oct 2012, 08:17
Please can someone the solution for the appended problem. Dint quite get the explaination provided. -Source - Gmat Club Test

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)=f(0)

Ans is B

Thank you.
Math Expert
Joined: 02 Sep 2009
Posts: 64249

### Show Tags

18 Oct 2012, 09:00
carpediem1984sm wrote:
Please can someone the solution for the appended problem. Dint quite get the explaination provided. -Source - Gmat Club Test

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)=f(0)

Ans is B

Thank you.

Merging similar topics. Please refer to the discussion above and ask if anything remains unclear.

_________________
Director
Joined: 22 Mar 2011
Posts: 577
WE: Science (Education)
Re: If function f(x) satisfies f(x) = f(x^2) for all x  [#permalink]

### Show Tags

18 Oct 2012, 12:50
1
tonebeeze wrote:
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. $$f(0) = 0$$

I reviewed the previous function posts that Bunuel referred me to and seem to understand those problems decently. But I don't clearly understand the process of substitution and POE on this problem. I would appreciate if someone could help me with understand how to break the answer choices down.

Rather than analyzing each answer, I would like to point out how the correct answer should look like.
From the equation $$f(x) = f(x^2)$$ we can get a chain of equalities between the values of the function $$f$$ at different points.
So, we will be able to deduce different equalities of the type $$f(a)=f(b)$$, but there is no way to find explicit values of the function in any specific point.
The correct answer should be of this form, or its equivalent $$f(a)-f(b)=0$$.
Only answer B is of this type.

For any specific value of $$x$$, except $$0$$ and $$1$$, we can start an infinite chain of equalities. For example, start with $$x=2$$:

$$f(2)=f(4)=f((-2)^2)=f(-2)=f(4^2)=f(16)=f((-4)^2)=f(-4)=f(16^2)=f(256)=f((-16)^2)=f(-16)=...$$.
We can see that for a given $$x$$, the function $$f$$ will have the same value at all the points $$x, x^2,x^4,x^8,..., -x,-x^2,-x^4,-x^8,...$$

For $$0$$, we just get $$f(0)=f(0^2)$$, while for $$x=1$$, we have $$f(1)=f((-1)^2)=f(-1)$$.
_________________
PhD in Applied Mathematics
Love GMAT Quant questions and running.
Manager
Joined: 20 Dec 2013
Posts: 114
Re: If function f(x) satisfies f(x) = f(x^2) for all x  [#permalink]

### Show Tags

16 Jul 2014, 20:58
[quote="tonebeeze"]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. $$f(0) = 0$$

A- f(4) = f(2)

Now is it always true that f(2)f(2) = f(2) OR f(2) = 1
It might be or might not be true.

B - f(16) = f(-2)
f(16) = f(4) = f(2) = f(-2)
This must be true

C. f(-2) = -f(4)
f(-2) = f(4). It might or might not be equal to negative of it.

D. f(3) = 3f(3) Again this might not be true always

E. f(0) = 0 We don't know what will be the y value of the function as expressed in option C, D and E.

_________________
76000 Subscribers, 7 million minutes of learning delivered and 5.6 million video views

Perfect Scores
http://perfectscores.org
CEO
Joined: 03 Jun 2019
Posts: 2911
Location: India
GMAT 1: 690 Q50 V34
WE: Engineering (Transportation)
Re: If function f(x) satisfies f(x) = f(x^2) for all x  [#permalink]

### Show Tags

08 Oct 2019, 01:08
tonebeeze wrote:
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. $$f(0) = 0$$

I reviewed the previous function posts that Bunuel referred me to and seem to understand those problems decently. But I don't clearly understand the process of substitution and POE on this problem. I would appreciate if someone could help me with understand how to break the answer choices down.

f(16) = f(4) = f(-2)
f(16) - f(-2) = 0

IMO B
_________________
Kinshook Chaturvedi
Email: kinshook.chaturvedi@gmail.com
Manager
Joined: 23 Sep 2019
Posts: 92
Location: India
GMAT 1: 700 Q50 V35
Re: If function f(x) satisfies f(x) = f(x^2) for all x  [#permalink]

### Show Tags

20 Feb 2020, 23:23
Bunuel - Quick question

C. f(−2)+f(4)=0f(−2)+f(4)=0: f(−2)=f(4)f(−2)=f(4), but it's not necessary f(4)+f(4)=2f(4)=0f(4)+f(4)=2f(4)=0 to be true (it will be true only if f(4)=0f(4)=0, but again we don't know that for sure);

I disregarded this as the right option because f(-2) = -f(4) and we don't have any information on -f(x). Is that correct?
If not, can you explain why this option is incorrect again?
Re: If function f(x) satisfies f(x) = f(x^2) for all x   [#permalink] 20 Feb 2020, 23:23