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

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For which of the following functions f is f(x)=f(1-x) for [#permalink] New post 25 Oct 2009, 01:34
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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)
[Reveal] Spoiler: OA
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Re: Function [#permalink] New post 28 Oct 2009, 03:18
Can someone please solve this. I know what the answer is. I am looking for an explanation on how to get to the answer.

Bunuel?

Thanks.
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Re: Function [#permalink] New post 28 Oct 2009, 15:17
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study 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)


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.

Similar questions:
functions-problem-need-help-93184.html
for-which-of-the-following-functions-is-f-a-b-f-b-f-a-124491.html
let-the-function-g-a-b-f-a-f-b-143311.html

Hope it helps.
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Re: Function [#permalink] New post 29 Oct 2009, 03:29
Helps very much

Thanks to Lagomez and Bunuel..
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Re: Function problem = Can someone please explain to me [#permalink] New post 03 Jul 2010, 22:15
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Okay, so you need to check if f(x) = f(1-x)

Let's do this one by one.

Case A: - Wrong
f(x) = 1-x
f(1-x) = 1-(1-x) = x

So we clearly see that f(x) is not f(1-x)

Case B: Wrong

f(x) = 1-x-x^2
f(1-x) = 1-(1-x)-(1-x)^2 = x - 1 - x^2 + 2x = 3x-1-x^2

Case C: Wrong

f(x) = x^2 - (1-x)^2 = x^2 - 1 +2x - x^2= 2x-1
f(1-x) = (1-x)^2 - (1-(1-x))^2 = (1-x)^2 - (x)^2= 1-2x

Case D: Right

f(x) = x^2 (1-x)^2
f(1-x) =(1-x)^2 (1-(1-x))^2= (1-x)^2(x)^2=f(x)

Let's check E just to be sure.

Case E: Wrong

f(x) = \frac{x}{1-x}
f(1-x) = \frac{1-x}{1-(1-x)}=\frac{1-x}{x}
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Re: Function [#permalink] New post 04 Jul 2010, 15:07
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Bunuel wrote:
Merging similar topics.

Your explanations are always lucid and very helpful;
Your contributions make the forum very resourceful.
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Re: Function [#permalink] New post 04 Jul 2010, 21:33
Bunuel:

sometimes, I almost die from the awesomeness of your posts.
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Re: Functions question [#permalink] New post 29 Dec 2010, 20:38
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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.
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Re: Function [#permalink] New post 16 Jan 2011, 09:48
Thanks guys, great help!
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Re: Please, someone explain [#permalink] New post 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.
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Re: Please, someone explain [#permalink] New post 23 Jan 2012, 00:40
Yes indeed simple after explanations, but was difficult to understand. Many thanks
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Re: Which of the following function.. [#permalink] New post 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.

Similar question to practice: functions-problem-need-help-93184.html#p717196
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For which of the following functions f is f(x)=f(1-x) for [#permalink] New post 12 Dec 2012, 23:40
A.
1-x ? 1-(1-x)
1-x ? 1-1+x
1-x ? xNOT EQUAL!

B.
1-x^2 ? 1-(1-x)^2
1-x^2 ? 2x + x^2 NOT EQUAL!

C.
x^2 - (1-x)^2 ? (1-x)^2 - (1-(1-x))^2
x^2 - (1-x)^2 ? (1-x)^2 - x^2 NOT EQUAL!

D.
x^2(1-x)^2 ? (1-x)^2(1-(1-x))^2
x^2(1-x)^2 ? (1-x)^2(x)^2 EQUAL!

Answer: D
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Re: For which of the following functions f is f(x)=f(1-x) for [#permalink] New post 27 Feb 2013, 02:42
Karishma....that is a useful tip....does it work in majority of the problems and thus safe to follow every time any such problem crops up???
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Re: For which of the following questions is f(x)=f(1-x) for all [#permalink] New post 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.

D is the right answer.

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For which of the following functions f is f(x) = f (1-x) [#permalink] New post 04 Aug 2013, 05:57
For which of the following fns 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)

Last edited by mau5 on 04 Aug 2013, 06:02, edited 1 time in total.
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Re: For which of the following functions f is f(x) = f (1-x) [#permalink] New post 04 Aug 2013, 07:25
smrutipattnaik wrote:
For which of the following fns 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)


Option A
f(x)=1-x
f(1-x) = 1- (1-x) =x
Not Equal

Option B
f(x)=1-x^2
f(1-x) = 1- (1-x)^2 =1- (1-2x +x^2) = 2x -x^2
Not Equal

Option C
f(x)=x^2-(1-x)^2
f(1-x) = (1-x)^2- (1-1+x)^2 =(1-x)^2- (x)^2
Not Equal

Option D
f(x)=x^2(1-x)^2
f(1-x) = (1-x)^2 (1-1+x)^2 =(1-x)^2 (x)^2
Equal

Answer is D
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Re: For which of the following functions f is f(x) = f (1-x) [#permalink] New post 11 Aug 2013, 05:16
Expert's post
smrutipattnaik wrote:
For which of the following fns 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)


Merging similar topics. Please refer to the solutions above.
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Re: For which of the following functions f is f(x)=f(1-x) for [#permalink] New post 25 Dec 2013, 17:31
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)
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Re: For which of the following functions f is f(x)=f(1-x) for [#permalink] New post 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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Re: For which of the following functions f is f(x)=f(1-x) for   [#permalink] 26 Dec 2013, 00:21
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