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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

04 Jan 2011, 15:21

4

This post received KUDOS

7

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

75% (hard)

Question Stats:

38% (01:47) correct
62% (00:50) wrong based on 168 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.

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.

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

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

Please clarify

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.
_________________

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

Please clarify

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.

Last edited by Joy111 on 24 May 2012, 03:13, edited 3 times in total.

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

Please clarify

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 \(f(16)=f(16^2)\).
_________________

\(f(16)-f(-2)\) 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 .

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".
_________________

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

Please clarify

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.

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.

Answer: B.

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.

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.

Answer: B.

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.

Re: If function f(x) satisfies f(x) = f(x^2) for all x [#permalink]

Show Tags

18 Oct 2012, 13:50

1

This post received KUDOS

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.

Re: If function f(x) satisfies f(x) = f(x^2) for all x [#permalink]

Show Tags

16 Jul 2014, 21:48

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: If function f(x) satisfies f(x) = f(x^2) for all x [#permalink]

Show Tags

03 Jan 2016, 02:28

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

There’s something in Pacific North West that you cannot find anywhere else. The atmosphere and scenic nature are next to none, with mountains on one side and ocean on...

This month I got selected by Stanford GSB to be included in “Best & Brightest, Class of 2017” by Poets & Quants. Besides feeling honored for being part of...

Joe Navarro is an ex FBI agent who was a founding member of the FBI’s Behavioural Analysis Program. He was a body language expert who he used his ability to successfully...