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# m25 #22

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m25 #22 [#permalink]  20 Oct 2008, 14:27
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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) $$f(0) = 0$$

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

Can someone please explain the correct answer to me? This one has me completely puzzled.

Correct answer is b. Explanation f(-2) = f(2) = f(4) = f(16). Thus, B must be true. The other choices are not necessarily true. Consider $$f(x) = 3$$ for all x.
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Re: m25, 22 [#permalink]  16 Feb 2009, 01:11
5
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Hi jairus,

Here goes my explanation:
--------------------------------
Let f(x) = f(x^2) be equation 1

A.) If we consider f(2) from the R.H.S part, we have
f(2) = f(2^2) = f(4) ---> using equation 1

Now since f(2) = f(4), we cannot have f(2)f(2) = f(4). Got it?

B.) f(-2) = f(-2^2) = f(4) ---> using equation 1
On applying equation 1 again on f(4), we get
f(4) = f(4^2) = f(16)

From above calculations, since f(-2) = f(16), rearranging it, we get, f(16) - f(-2) = 0

C.) As we have proved above in B.), we have
f(-2) = f(4),
so f(-2) - f(4) = 0, which is NOT stated in the option.

D.) f(3) = 3f(3)
Not possible

E.) f(0) = 0
Can't say as we don't have the value of f(0) with us.
Moreover, equation 1 just shows relation between two functions. There is no way we can find the values.
--------------------------------

Hope that helps.

Regards,
Technext
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+++ Please tell me why other options are wrong. +++

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Re: m25 #22 [#permalink]  26 Mar 2012, 07:35
5
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Expert's post
duuuma 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) $$f(0) = 0$$

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

Can someone please explain the correct answer to me? This one has me completely puzzled.

Correct answer is b. Explanation f(-2) = f(2) = f(4) = f(16). Thus, B must be true. The other choices are not necessarily true. Consider $$f(x) = 3$$ for all x.

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 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 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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Re: m25 #22 [#permalink]  30 Jun 2010, 07:53
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Which of the options satisfies f(x) = f(x^2) for all values of x.
A. f(4) = f(2)f(2)...
L.H.S: f(4);
R.H.S: f(2)f(2)=> f(4)f(4) not equal to f(4)
from L.H.S f(4) can be reduced to f(2) - since f(x) = f(x^2)
WRONG

B. f(16) - f(-2) = 0 => f(16) = f(-2)
L.H.S: f(16)
R.H.S: f(-2); also f(-2) = f(4) = f(16)
thus f(16) - f(-2) = 0 satisfies the stem
CORRECT

C. f(-2) + f(4) = 0 => f(4) = -f(-2)
L.H.S: f(4)
R.H.S: -f(-2) = -f(4)
but because we don't know what f(4) is, we cannot certainly say
f(4) = -f(4) -except f(4)=0...WRONG

D. f(3) = 3f(3)
L.H.S: f(3)
R.H.S: 3f(3)
f(3) = f(9) but we cannot say f(9) = 3f(3)...WRONG

E. f(0) = 0
L.H.S: F(0)
R.H.S: 0
Because we don't know the value of f(0), it's not certain to conclude that
f(0^2) - which is also f(0) - is equal to zero
WRONG

Hope am correct anyway
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Re: m25 #22 [#permalink]  22 Oct 2008, 01:56
2
KUDOS
Function f(x) satisfies f(x) = f(x^2) for all x. So f(2 ) or f(-2) = f( square of 2 or -2) = f(4) = f( square of 4) = f(16) etc....

Notice that only choice b has a minus between them , and hence the result will be zero.
a. f(4) = f(2)f(2) -> f(4) * f(4) which cannot be equal to f(4)
b. f(16) - f(-2) = 0 , This has to be true because you are subtracting same function from the same one.
c. f(-2) + f(4) = 0 -> there is a plus sign to its not going to be zero anyways ....
d. f(3) = 3f(3) -> it cannot be 3 * f(3) , the qs stem does not agree with this
e. f(0) = 0 -> we can't say f(x) = x , so this is not true as well

Hence only b is true.
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Re: m25 #22 [#permalink]  20 Nov 2010, 04:55
1
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greatchap wrote:
I am a little confused. Can someone provide a background as to how to solve questions with functions ?

Thanks.

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

Now we are given f(x) = f(x^2) , hence
f(-2) = f(4) .....Subst. x=-2 in the given function
Also, f(4) = f(16)......Subst. x=4 in the given function
Hence we get the following relation: f(-2)=f(4)=f(16).
Option B: f(16) - f(-2)=0 holds true
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Re: m25 #22 [#permalink]  06 Jul 2011, 17:49
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$$f(256) - f(-2)$$ is also equal to 0.
if $$f(x) = f(x^2)$$ for all $$x$$,
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m25, 22 [#permalink]  03 Feb 2009, 10:18
OA says: f(-2) = f(2) = f(4) = f(16) . Thus, B must be true. The other choices are not necessarily true. Consider for all x.
I do not get the logic; can somebody also explain why A is not good?
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Re: m25, 22 [#permalink]  30 Sep 2009, 12:24
[quote="Technext"]Hi jairus,

Here goes my explanation:
--------------------------------
Let f(x) = f(x^2) be equation 1

A.) If we consider f(2) from the R.H.S part, we have
f(2) = f(2^2) = f(4) ---> using equation 1
COOL

Now since f(2) = f(4), we cannot have f(2)f(2) = f(4). Got it?
TOTALLY STILL WITH YOU HERE

B.) f(-2) = f(-2^2) = f(4) ---> using equation 1
On applying equation 1 again on f(4), we get
f(4) = f(4^2) = f(16)
THIS IS WHERE I GET LOST? WHY ARE YOU PERFORMING THE FUNCTION TWICE??
From above calculations, since f(-2) = f(16), rearranging it, we get, f(16) - f(-2) = 0

I REALIZE THERES SOMETHING OBVIOUS IM MISSING...CAN YOU SPELL IT OUT, PLEASE? THANKS.
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Re: m25 #22 [#permalink]  03 Mar 2010, 01:48
Hey,
I will give it a try;
I think crucial is the + sign in answer (3). Although it is possible that the two equations are the same f(-2) = f(2) = f(4) we never end up at 0 by adding the same two terms. The first term has to be negativ, but this is not possible since the two terms are EXACTLY the same.

Answer (2) shows us, how it would be possible. We have the same two terms, but due to the - sign we arrive exactly at 0!

Hope its clear now and I didnt screw up my first serious contribution to this forum.
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Re: m25, 22 [#permalink]  24 Oct 2010, 14:50
I know this is a super-late response, but anyways:
f(-2) = f(-2^2)=f(4), now since question stem has a f(16) lets see if we can fit that in the equation: f(4) = f(4^2) = f(16)
Hence f(-2)=f(2^2)=f(2^4) Ans: B
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Re: m25 #22 [#permalink]  04 Jul 2011, 05:51
Straightforward "functions" question. Just substitute the value of "x" and solve -- only B holds up.

Choice A, D and E are all trap answers.
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Re: m25 #22 [#permalink]  04 Jul 2011, 07:51
greatchap wrote:
I am a little confused. Can someone provide a background as to how to solve questions with functions ?

Thanks.

Ans B

The best way to solve such questions is to substitute values in place of variables. Generally u'd get questions/options wherein integers can be used. You just have to tweak around a bit to get the desired solution...Like in this case..
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Re: m25 #22 [#permalink]  06 Jul 2012, 19:13
Yes. Anwser B obviously is always True at all time
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Re: m25, 22 [#permalink]  23 Jul 2012, 10:30
quantjock wrote:
Technext wrote:
Hi jairus,

Here goes my explanation:
--------------------------------
Let f(x) = f(x^2) be equation 1

A.) If we consider f(2) from the R.H.S part, we have
f(2) = f(2^2) = f(4) ---> using equation 1
COOL

Now since f(2) = f(4), we cannot have f(2)f(2) = f(4). Got it?
TOTALLY STILL WITH YOU HERE

B.) f(-2) = f(-2^2) = f(4) ---> using equation 1
On applying equation 1 again on f(4), we get
f(4) = f(4^2) = f(16)
THIS IS WHERE I GET LOST? WHY ARE YOU PERFORMING THE FUNCTION TWICE??
From above calculations, since f(-2) = f(16), rearranging it, we get, f(16) - f(-2) = 0

I REALIZE THERES SOMETHING OBVIOUS IM MISSING...CAN YOU SPELL IT OUT, PLEASE? THANKS.

Hi Quantjack,

[b]THIS IS WHERE I GET LOST? WHY ARE YOU PERFORMING THE FUNCTION TWICE??[/b]
From above calculations, since f(-2) = f(16), rearranging it, we get, f(16) - f(-2) = 0

here its been mentioned from the function f(x)=f(x^2) , where x is a common integer, so we need to choose the right integer for x from the option given below...

as per the options given below, only option B holds good,

It should be rewritten as f(4)=f(16) i.e {f(x)=f(x^2)}...

Hope it helps....

Cheers
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Re: m25 #22 [#permalink]  31 Aug 2012, 19:59
I was actually stumped by this question initially and failed to understand the given statement. But I was able to quickly move on and then picked up on the function and the co-relation.

However was not able to solve it correctly and guessed on the answer. The explanations above have been particularly helpful in realizing my mistake.

I was thinking that f(x) = f(x^2) would also mean that the converse is true i.e. for eg. f(2) could be f(x^2) for f(x) as f($$\sqrt{2}$$) . This, I thought us possible since the function is true for all values of x and there is no restriction on x.
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Re: m25 #22 [#permalink]  23 Dec 2012, 14:00
Sorry to revive an old thread. I just wanted to add another way to look at this question.

I thought about it as f(x) defining a line in a coordinate plane. The only way for f(x) = f(x^2) is if f(x) defines a horizontal line (e.g., y = 5, y = 0, y = -2, etc.). Then, I tested each answer choice:

Suppose f(x) defines the line y = 2 (you can repeat this with any y = n)

A) f(4) = 2 and f(2) = 2. This choice is false
B) f(16) and f(-2) are the same so this choice is true

etc.

Thanks everyone

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Re: m25 #22 [#permalink]  08 Jul 2013, 08:28
Another simple way to do this (related to dlpoht's method) is to define f(x). We shouldn't do this all the time, but we are able to do so here because we are being asked for something that must hold true under all conditions, regardless of what f(x) actually is. For example:

Say f(x) = (0)(x) + 3

Thus, f(x) always equals 3. Additionally, f(x^2) = (0)(x^2) + 3 = 3, so this formula works because f(x) = f(x^2). Trying out the answers with this equation, we see the following:

(A) f(4) = f(2)f(2)
f(4) = (0)(4) + 3 = 3
f(2) = (0)(2) + 3 = 3
3 != (3)(3)
WRONG

(B) f(16) - f(-2) = 0
f(16) = (0)(16) + 3 = 3
f(-2) = (0)(-2) + 3 = 3
3 - 3 = 0
WORKS

(C) f(-2) + f(4) = 0
f(-2) = (0)(-2) + 3 = 3
f(4) = (0)(4) + 3 = 3
3 + 3 != 0
WRONG

(D) f(3) = 3f(3)
f(3) = (0)(3) + 3 = 3
3 != (3)(3)
WRONG

(E) f(0) = 0
f(0) = (0)(0) + 3 = 3
3 != 0
WRONG

Is it the most elegant solution? Absolutely not. However, you have to be versatile in the GMAT because there may be times when you are unable to see the elegant solution. Creating specific examples can help you to understand a hard question when you are stuck, especially if it asks what "must be true"
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Re: m25 #22 [#permalink]  28 May 2014, 00:55
Bunuel wrote:
duuuma 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) $$f(0) = 0$$

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

Can someone please explain the correct answer to me? This one has me completely puzzled.

Correct answer is b. Explanation f(-2) = f(2) = f(4) = f(16). Thus, B must be true. The other choices are not necessarily true. Consider $$f(x) = 3$$ for all x.

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 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.

HI Bunnel,

I have a doubt here .

it is given that f(x) = f(x^2)

so for 0 it will be same. what yiu mean by we don't know the actual function we can not say for sure
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Re: m25 #22 [#permalink]  28 May 2014, 03:55
Expert's post
PathFinder007 wrote:
Bunuel wrote:
duuuma 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) $$f(0) = 0$$

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

Can someone please explain the correct answer to me? This one has me completely puzzled.

Correct answer is b. Explanation f(-2) = f(2) = f(4) = f(16). Thus, B must be true. The other choices are not necessarily true. Consider $$f(x) = 3$$ for all x.

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 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.

HI Bunnel,

I have a doubt here .

it is given that f(x) = f(x^2)

so for 0 it will be same. what yiu mean by we don't know the actual function we can not say for sure

I mean that we don't know the actual function f(x). Is it f(x) = 0, f(x) = 1, ...?
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