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# M10: Q21 - PS

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Manager
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22 Aug 2008, 14:37
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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) $$F(3) = 3*F(3)$$
(E) $$F(0) = 0$$

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

Between B and E, which one is correct and why?

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Re: M10: Q21 - PS [#permalink]

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23 Aug 2008, 02:32
balboa wrote:
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) = 3*F(3)
e) F(0) = 0

Given : F(x) = F (x^2)
F(-2) = F (-2^2) = F (4)
F(4) = F(4^2) = F(16)

Therefore, F(16) - F(-2) = F (16) - F(16)=0

Ans is B

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Re: M10: Q21 - PS [#permalink]

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23 Aug 2008, 22:58
You cant come to the conclusion that F(0) = 0 There is not information given about the vlaue of F(x).

Whats the OA?

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Re: M10: Q21 - PS [#permalink]

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30 Aug 2008, 11:35
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LoyalWater wrote:
balboa wrote:
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) = 3*F(3)
e) F(0) = 0

Given : F(x) = F (x^2)
F(-2) = F (-2^2) = F (4)
F(4) = F(4^2) = F(16)

Therefore, F(16) - F(-2) = F (16) - F(16)=0

Ans is B

I don't understand why F(-2) is taken to the 4th power....when the original function states that is to be taken to the 2nd power.

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Re: M10: Q21 - PS [#permalink]

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01 Sep 2008, 00:15
Refer to the second post in this thread. The trick is here:

F(-2)=F(4)
and also
F(4)=F(16)
therefore
F(-2)=F(16)

I hope this helps.
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Re: M10: Q21 - PS [#permalink]

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02 Sep 2008, 13:04
It does not provide information about the value of F(x) or F(x^2) .. so we can not perform any operation with F(x) or F(x^2).
The leaves us only F(16) - F(-2) = 0 and F(-2)+F(4) = 0 to be examined..

F(x) = F(x^2) => f(-2) = f(4) =f(16) .. so B is the answer.

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Re: M10: Q21 - PS [#permalink]

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12 Apr 2010, 19:51
hear F(x)=F(x^2) but not any specific operation so,

F(x) = F(x^2) => f(-2) = f(4) =f(16) .. so B is the answer.
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Re: M10: Q21 - PS [#permalink]

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14 Apr 2011, 05:50
F(-2) = F((-2)^2) = F(4)

F(4) = F(4^2) = F(16)

=> F(16) - F(-2) = 0

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Re: M10: Q21 - PS [#permalink]

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16 Apr 2011, 13:50
Just by going the same logic why "C" cant be the answer?

As (-2)^2 = 4
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Re: M10: Q21 - PS [#permalink]

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19 Apr 2011, 01:18
vijayahir wrote:
Just by going the same logic why "C" cant be the answer?

As (-2)^2 = 4

Because F(4) + f(4) can not be equal to zero unless F(4) = 0 which we have no information about, so the answer can not be C.
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Re: M10: Q21 - PS [#permalink]

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21 Apr 2011, 16:57
balboa 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) $$F(3) = 3*F(3)$$
(E) $$F(0) = 0$$

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

Between B and E, which one is correct and why?

If for some reason the answer is B, why not A?
We are saying F(-2) = F(4) = F(16) and therefore F(16) - F(-2) => F(16) - F(16) =0 must be true.
So in the above example we are doing a simple arithmetic of 16 - 16 for F(16) - F(16).

Why not do 4 = 2 * 2 for F(4) = F(2)*F(2)? Can someone please help explain?

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Re: M10: Q21 - PS [#permalink]

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21 Apr 2011, 23:36
mniyer wrote:
balboa 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) $$F(3) = 3*F(3)$$
(E) $$F(0) = 0$$

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

Between B and E, which one is correct and why?

If for some reason the answer is B, why not A?
We are saying F(-2) = F(4) = F(16) and therefore F(16) - F(-2) => F(16) - F(16) =0 must be true.
So in the above example we are doing a simple arithmetic of 16 - 16 for F(16) - F(16).

Why not do 4 = 2 * 2 for F(4) = F(2)*F(2)? Can someone please help explain?

We don't know what the function actually returns; we just know that the function returns the same value for a number and its square.

e.g..

f(2^{1/4})=100
f(2^{1/2})=100
f(2)=100
f(4)=100
f(16)=100

OR

f(2^{1/4})=2
f(2^{1/2})=2
f(2)=2
f(4)=2
f(16)=2

AND
f(3^{1/4})=200
f(3^{1/2})=200
f(3)=200
f(9)=200
f(81)=200

Now,
f(4) = f(2)*f(2)

If;
f(2)=0
f(4)=0
f(16)=0

f(2)*f(2)=0
f(4) = f(2)*f(2) as 0 = 0.

If,
f(2)=2
f(4)=2
f(16)=2
f(2)*f(2)=2*2=4

f(4) != f(2)*f(2) as 2 != 4

So, this function may be true but not MUST be true.
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Re: M10: Q21 - PS [#permalink]

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08 May 2011, 11:48
What if one of the option is f(-2)-f(4)=0

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Re: M10: Q21 - PS [#permalink]

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08 May 2011, 11:50
agdimple333 wrote:
What if one of the option is f(-2)-f(4)=0

This will also be correct.
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Re: M10: Q21 - PS [#permalink]

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18 Apr 2012, 05:04
balboa wrote:
Do you mind telling me how E is incorrect?

Because there are many possibilities other than one given what if F(x) = 1 then F(0)=1

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Re: M10: Q21 - PS [#permalink]

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19 Apr 2012, 13:31
the key to solving this problem is that if not given a restriction as in b... the function can go on forever right?

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Re: M10: Q21 - PS [#permalink]

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19 Apr 2012, 13:37
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snkrhed wrote:
the key to solving this problem is that if not given a restriction as in b... the function can go on forever right?

Below is a detailed solution of this problem. Hope it helps.

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.

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Re: M10: Q21 - PS [#permalink]

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19 Apr 2013, 05:27
Bunuel wrote:
snkrhed wrote:
the key to solving this problem is that if not given a restriction as in b... the function can go on forever right?

Below is a detailed solution of this problem. Hope it helps.

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.

Great explanation! I missed out on the initial part and got lost (f'x is the same as f'x2 the same as f'x4) - Great question!

Thanks & Regards,
Vishnu
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Re: M10: Q21 - PS [#permalink]

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21 Apr 2013, 09:15
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When F(x) = F(x²) for all x, then F must be a constant function. That means F(0)=F(1)=F(...) whatever.
That is why B must be correct.
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Re: M10: Q21 - PS [#permalink]

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23 Apr 2013, 09:03
Bunuel wrote:
snkrhed wrote:
the key to solving this problem is that if not given a restriction as in b... the function can go on forever right?

Below is a detailed solution of this problem. Hope it helps.

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 Bunuel,

thanks for the explanation. I found this question to be very tough and I seem to be the only one not being able to understand this even after your explanation

I could understand why B is the answer and why not C and D. But I am still confused with A and E. I have tried to explain my thought process below -

We know that f(x) = f(x^2) means f(2) = f(4) = f(16) and so on.....
A. f(4)=f(2)*f(2)
LHS is f(4) which means (16)
RHS is f(2)*f(2) means 4*4 means (16)
Hence LHS = RHS?
As I am writing this, it occurred to me that here we are multiplying to functions {f(2)*f(2)} and we don't really know if multiplying of two functions will actually result in the multiplication of those two numbers/integers? it could result in some other function as well? Am I thinking in the right direction?

Coming to option E - it says f(0)=0
we know that f(x) = f(x^2)
if x is 0...its square (infact any exponent) or function of its square should always result in 0. I can't think through what's wrong with E?

Kind regards

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Re: M10: Q21 - PS   [#permalink] 23 Apr 2013, 09:03

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# M10: Q21 - PS

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