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# M10: Q21 - PS

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Math Expert
Joined: 02 Sep 2009
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Re: M10: Q21 - PS [#permalink]

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24 Apr 2013, 05:14
tirbah wrote:
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 [color=#ff0000]f(2)*f(2) means 4*4 means (16)[/color]
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

f(4)=f(16), not 16 and f(2)*f(2)=f(4)*f(4), not 4*4.

As for E: f(x)=f(x^2) does not necessarily means that f(0)=0.
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Re: M10: Q21 - PS [#permalink]

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24 Apr 2013, 05:30
Bunuel wrote:
tirbah wrote:
Bunuel 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 [color=#ff0000]f(2)*f(2) means 4*4 means (16)[/color]
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

f(4)=f(16), not 16 and f(2)*f(2)=f(4)*f(4), not 4*4.

As for E: f(x)=f(x^2) does not necessarily means that f(0)=0.[/quote][/quote]

Oh yeah...silly me...I got that now....we are not given the value of the function so we cant presume what that function might be equal to.....thanks a ton for your reply

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Re: M10: Q21 - PS [#permalink]

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11 Apr 2014, 10:42
I agree with vishnuns, Great explanation! made it look like a cakewalk! :D
thanks Bunuel
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Kudos [?]: 37 [0], given: 47

Re: M10: Q21 - PS   [#permalink] 11 Apr 2014, 10:42

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# M10: Q21 - PS

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