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by nasir » Mon Sep 27, 2010 8:10 am
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

OA= B

If someone can explain the answer.
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by euro » Mon Sep 27, 2010 8:32 am
I am not too sure but here is my explanation for choice (B)...

Given: f(x) = f(x^2)

For option (B),
f(16) = f(4^2) = f(4)
f(-2) = f(-2^2) = f(4)
Hence, f(16) - f(-2) = f(4) - f(4) = 0

Therefore, LHS = 0 = RHS.

However, I would like to know a better approach to solve this - one that is more generic.
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by narik11 » Mon Sep 27, 2010 10:02 am
euro wrote:I am not too sure but here is my explanation for choice (B)...

Given: f(x) = f(x^2)

For option (B),
f(16) = f(4^2) = f(4)
f(-2) = f(-2^2) = f(4)
Hence, f(16) - f(-2) = f(4) - f(4) = 0

Therefore, LHS = 0 = RHS.

However, I would like to know a better approach to solve this - one that is more generic.
I didn't get you..

For f(16) it has to be f(16^2) but it has been done as f(4^2), whereas for f(-2) it is done as f(-2^2)

I think it is Option (A)

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

=> 16=16

Also i have a doubt with Option (E)
As f(0)= 0^2=0

Please correct me if am wrong.
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by GMATGuruNY » Mon Sep 27, 2010 11:51 am
nasir 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

OA= B

If someone can explain the answer.
The problem stipulates that f(x) = f(x^2) for all values of x.

If x=4:
f(4) = f(4^2)
f(4) = f(16)
So f(16) = f(4).

If x= -2:
f(-2) = f((-2)^2)
So f(-2) = f(4).

Substituting into answer choice B:
f(16) - f(-2) = f(4) - f(4)
f(16) - f(-2) = 0.

Does this help?
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by euro » Mon Sep 27, 2010 12:02 pm
GMATGuruNY wrote:
The problem stipulates that f(x) = f(x^2) for all values of x.

If x=4:
f(4) = f(4^2)
f(4) = f(16)
So f(16) = f(4).

If x= -2:
f(-2) = f((-2)^2)
So f(-2) = f(4).

Substituting into answer choice B:
f(16) - f(-2) = f(4) - f(4)
f(16) - f(-2) = 0.

Does this help?
Thanks for the explanation Mitch. I had used the same logic but you could explain it better.

I was wondering if theres any another way of attacking such problems (other than back solving from answer options)??
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by GMATGuruNY » Mon Sep 27, 2010 12:09 pm
euro wrote:
GMATGuruNY wrote:
The problem stipulates that f(x) = f(x^2) for all values of x.

If x=4:
f(4) = f(4^2)
f(4) = f(16)
So f(16) = f(4).

If x= -2:
f(-2) = f((-2)^2)
So f(-2) = f(4).

Substituting into answer choice B:
f(16) - f(-2) = f(4) - f(4)
f(16) - f(-2) = 0.

Does this help?
Thanks for the explanation Mitch. I had used the same logic but you could explain it better.

I was wondering if theres any another way of attacking such problems (other than back solving from answer options)??
I would scan the answers, looking for an answer choice that's easy to prove. I saw relatively quickly that, according to the function, f(-2) = f(4) = f(16), so it must be true that f(16) - f(-2) = 0.
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by narik11 » Mon Sep 27, 2010 7:49 pm
GMATGuruNY wrote:
nasir 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

OA= B

If someone can explain the answer.
The problem stipulates that f(x) = f(x^2) for all values of x.

If x=4:
f(4) = f(4^2)
f(4) = f(16)
So f(16) = f(4).

If x= -2:
f(-2) = f((-2)^2)
So f(-2) = f(4).

Substituting into answer choice B:
f(16) - f(-2) = f(4) - f(4)
f(16) - f(-2) = 0.

Does this help?
Thanks a lot GuruNY.. :D
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