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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.

Re: Functions question [#permalink]
29 Dec 2010, 20:38

5

This post received KUDOS

Expert's post

metallicafan wrote:

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.
_________________

Re: Please, someone explain [#permalink]
22 Jan 2012, 23:40

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.
_________________

Re: Which of the following function.. [#permalink]
12 Oct 2012, 02:41

Expert's post

IanSolo 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)=\frac{x}{1-x}

I find this question in gmat prep software, I have tried to solve it with picking number ( I have sostitute x with 1 and -1) but for all the answer the result was different from 1 and -1. How I can solve quickly this question? What is the level of this question?

Merging similar topics. Please ask if anything remains unclear.

Re: For which of the following questions is f(x)=f(1-x) for all [#permalink]
23 Mar 2013, 21:01

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 [#permalink]
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.
_________________

Please +1 KUDO if my post helps. Thank you.

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Re: For which of the following functions f is f(x)=f(1-x) for
[#permalink]
26 Dec 2013, 00:21