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For which of the following functions f is f(x) = f(1-x) for all x?

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For which of the following functions f is f(x) = f(1-x) for all x? [#permalink] New post 20 Nov 2005, 05:33
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For which of the following functions f is f(x) = f(1-x) for all x?

A. f (x) = 1 - x
B. f (x) = 1 - x^2
C. f (x) = x^2 - (1 - x)^2
D. f (x) = x^2*(1 - x)^2
E. f (x) = x/(1 - x)

OPEN DISCUSSION OF THIS QUESTION IS HERE: for-which-of-the-following-functions-f-is-f-x-f-1-x-for-85751.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 25 Sep 2014, 22:48, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink] New post 20 Nov 2005, 06:42
Should be D. F(x)=(x^2) (1-x^2)

D. is the only function which is the greatest at 1/2.
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink] New post 20 Nov 2005, 10:12
njss750 wrote:
For which of the following functions f is f(x)= f(1-x) for all x
F(x)=1-x
F(x)=1-x^2
F(x)=x^2-(1-x)^2
F(x)=(x^2) (1-x^2)
F(x)-x/ 1-x


I remember seeing this problem in a GMATprep test. I think choice D should be F(x)=(x^2) (1-x)^2. If that is true then substituting (1-x) for will give us the same function back and the answer choice is D, where the function is multiplicative
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink] New post 25 Nov 2005, 07:09
Agreed, none of the original answers made sense.
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink] New post 25 Sep 2014, 22:06
Can someone please update the OA / answers?
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink] New post 25 Sep 2014, 22:49
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njss750 wrote:
For which of the following functions f is f(x) = f(1-x) for all x?

A. f (x) = 1 - x
B. f (x) = 1 - x^2
C. f (x) = x^2 - (1 - x)^2
D. f (x) = x^2*(1 - x)^2
E. f (x) = x/(1 - x)


\(f(x)="some \ expression \ with \ variable \ x"\), means that the value of \(f(x)\) can be found by calculating the expression for the particular \(x\).

For example: if \(f(x)=3x+2\), what is the value of \(f(3)\)? Just plug \(3\) for \(x\), \(f(3)=3*3+2=11\), so if the function is \(f(x)=3x+2\), then \(f(3)=11\).

There are some functions for which \(f(x)=f(-x)\). For example: if we define \(f(x)\) as \(f(x)=3*x^2+2\), the value of \(f(x)\) will be always positive and will give the following values: for \(x=-5\), \(f(x)=3*(-5)^2+2=77\); for \(x=0\), \(f(0)=3*0^2+2=2\). Please note that \(f(x)\) in this case is equal to \(f(-x)\), meaning that for positive values of \(x\) you'll get the same values of \(f(x)\) as for the negative values of \(x\).

So, basically in original question we are told to define the expression, for which \(f(x)=f(1-x)\), which means that plugging \(x\) and \(1-x\) in the expression must give same result.

A. \(f(x)=1-x\) --> \(1-x\) is the expression for \(f(x)\), we want to find whether the expression for \(f(1-x)\) would be the same: plug \(1-x\) --> \(f(1-x)=1-(1-x)=x\). As \(1-x\) and \(x\) are different, so \(f(x)\) does not equal to \(f(1-x)\).

The same with the other options:

(A) \(f(x)=1-x\), so \(f(1-x)=1-(1-x)=x\) --> \(1-x\) and \(x\): no match.

(B) \(f(x)=1-x^2\), so \(f(1-x)=1-(1-x)^2=1-1+2x-x^2=2x-x^2\) --> \(1-x^2\) and \(2x-x^2\): no match.

(C) \(f(x)=x^2-(1-x)^2=x^2-1+2x+x^2=2x-1\), so \(f(1-x)=2(1-x)-1=1-2x\) --> \(2x-1\) and \(1-2x\): no match.

(D) \(f(x)=x^2*(1-x)^2\), so \(f(1-x)=(1-x)^2*(1-1+x)^2=(1-x)^2*x^2\) --> \(x^2*(1-x)^2\) and \((1-x)^2*x^2\). Bingo! if \(f(x)=x^2*(1-x)^2\) then \(f(1-x)\) also equals to \(x^2*(1-x)^2\).

Still let's check (E)

(E) \(f(x)=\frac{x}{1-x}\) --> \(f(1-x)=\frac{1-x}{1-1+x}=\frac{1-x}{x}\). \(\frac{x}{1-x}\) and \(\frac{1-x}{x}\): no match.

But this problem can be solved by simple number picking: plug in numbers.

As stem says that "following functions f is f(x) = f (1-x) for all x", so it should work for all choices of \(x\).

Now let \(x\) be 2 (note that: -1, 0, and 1 generally are not good choices for number picking), then \(1-x=1-2=-1\). So we should check whether \(f(2)=f(-1)\).

(A) \(f(2)=1-x=1-2=-1\) and \(f(-1)=1-(-1)=2\) --> \(-1\neq{2}\);

(B) \(f(2)=1-x^2=1-4=-3\) and \(f(-1)=1-1=0\) --> \(-3\neq{0}\);

(C) \(f(2)=x^2-(1-x)^2=x^2-1+2x+x^2=2x-1=2*2-1=3\) and \(f(-1)=2*(-1)-1=-3\) --> \(3\neq{-3}\);

(D) \(f(2)=x^2*(1-x)^2=(-2)^2*(-1)^2=4\) and \(f(-1)=(-1)^2*2^2=4\) --> \(4=4\), correct;

(E) \(f(2)=\frac{x}{1-x}=\frac{2}{1-2}=-2\) and \(f(-1)=\frac{-1}{1-(-1)}=-\frac{1}{2}\) --> \(-2\neq{-\frac{1}{2}}\).

It might happen that for some choices of \(x\) other options may be "correct" as well. If this happens, just pick some other number for \(x\) and check again these "correct" options only.

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OPEN DISCUSSION OF THIS QUESTION IS HERE: for-which-of-the-following-functions-f-is-f-x-f-1-x-for-85751.html
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Re: For which of the following functions f is f(x) = f(1-x) for all x?   [#permalink] 25 Sep 2014, 22:49
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