GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 Oct 2019, 20:22 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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  M25-22

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 58445

Show Tags

2
22 00:00

Difficulty:   85% (hard)

Question Stats: 37% (01:21) correct 63% (01:22) wrong based on 312 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. $$f(0) = 0$$

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58445

Show Tags

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.

_________________
Intern  Joined: 15 Jun 2013
Posts: 4

Show Tags

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
Intern  Joined: 02 Oct 2014
Posts: 1

Show Tags

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?
Intern  Joined: 09 Mar 2014
Posts: 3
GPA: 3.01
WE: General Management (Energy and Utilities)

Show Tags

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.
Math Expert V
Joined: 02 Sep 2009
Posts: 58445

Show Tags

manojpandey80 wrote:
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.

Check function questions in our Special Questions Directory:

Operations/functions defining algebraic/arithmetic expressions
Symbols Representing Arithmetic Operation
Rounding Functions
Various Functions
_________________
Senior Manager  Joined: 31 Mar 2016
Posts: 375
Location: India
Concentration: Operations, Finance
GMAT 1: 670 Q48 V34 GPA: 3.8
WE: Operations (Commercial Banking)

Show Tags

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 Show Tags

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?
Senior Manager  P
Joined: 15 Oct 2017
Posts: 295
GMAT 1: 560 Q42 V25 GMAT 2: 570 Q43 V27 GMAT 3: 710 Q49 V39 Show Tags

So in B, f(-2) can be f(4) or f(16) but f(16) cannot be f(16^2). Why?
Math Expert V
Joined: 02 Sep 2009
Posts: 58445

Show Tags

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) = ...
_________________
BSchool Forum Moderator G
Joined: 23 May 2018
Posts: 546
Location: Pakistan

Show Tags

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 ?
_________________
If you can dream it, you can do it.

Practice makes you perfect.

Kudos are appreciated.
Math Expert V
Joined: 02 Sep 2009
Posts: 58445

Show Tags

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) = .
_________________
Senior Manager  P
Status: Gathering chakra
Joined: 05 Feb 2018
Posts: 434

Show Tags

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
Display posts from previous: Sort by

M25-22

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  