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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)

I've already seen quite a few GMAT questions of this type. They are quite easy to solve, since you understand the concept.

\(f(x)="some \ expression \ with \ variable \ x"\), 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.

Is there a faster way to solve this question rather than replacing each "x" by (1-x)? Thanks!

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. \(f(x) = x / (1-x)\)

Tip: Try to first focus on the options where terms are added/multiplied rather than subtracted/divided. They are more symmetrical and a substitution may not change the expression. I will intuitively check D first since it involves multiplication of the terms.
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]

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22 Jan 2012, 23:40

2

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This one is quite simple. We have to substitute (1 - x) for x in each of the options and see which option yields the same result for both f(x) and f(1 - x).

However, careful observation of the options will easily tell you that only D will fulfill this condition because in D, the x now becomes (1 - x) and (1 - x) now becomes x. The overall outcome will remain the same. You don't even need to expand or do any calculations whatsoever.
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]

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23 Mar 2013, 21:01

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You need to check for every option. Either you can replace x in every option with 1-x or you can take x = 1 and then check for which option f(1) = f(1-1) = f(0). In A f(1) = 1-1 = 0. f(0) = 1-0 = 1. In B f(1) = 1 - 1^2 = 0. f(0) = 1 - 0 = 1 In C f(1) = 1^2 - (1 - 1 )^2 = 1. f(0) = 0 - (1-0)^2 = 0 - 1 = -1 in D f(1) = 1(1-1)^2 = 0. f(0) = 0.(1-0)^2 = 0. ( Right answer ) And for option E you can take x = 2 because if you take x = 1 in denominator the denominator becomes zero, which makes the f(x) undefined. f(2) = 2/(1 - 2) = -2. f(1-2) = f(-1) = -1/(1-(-1)) = -1/2.

Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]

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26 Dec 2013, 00:21

Policarpa wrote:

Replacing x=4 and 1-x = -3 I get 16(9) - why is that wrong 4^2 (1-4)^2 = 16 (9), not at all f(4)

I'm sorry but I don't get your point. You mean if we plug in x = 4, the answer is incorrect? \(f(4) = 4^2*(1-4)^2 = 16*9\) \(f(1-4) = (1-4)^2*[1-(1-4)]^2 = 9*16\)

D is correct.
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Please +1 KUDO if my post helps. Thank you.

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