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Manager  Joined: 05 Oct 2008
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For which of the following functions f is f(x) = f(1-x) for all x?  [#permalink]

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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. $$f(x) = \frac{x}{(1 - x)}$$
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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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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:
for-which-of-the-following-functions-is-f-a-b-f-a-f-b-for-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: For which of the following functions f is f(x) = f(1-x) for all x?  [#permalink]

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

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Bunuel:

sometimes, I almost die from the awesomeness of your posts.
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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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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: For which of the following functions f is f(x) = f(1-x) for all x?  [#permalink]

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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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Similar question to practice: http://gmatclub.com/forum/functions-pro ... ml#p717196
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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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A.
$$1-x ? 1-(1-x)$$
$$1-x ? 1-1+x$$
$$1-x ? x$$NOT 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!

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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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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.

Please give a kudo if you like my explanation.
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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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Thank you Bunuel.

I didn't understand till the time i read about the option of picking numbers.

Even then it took a while for me to understand that i have to check both the values (-1 and 2) and check if i get the same answer.

Please do let me know if there are more similar questions besides the ones you mentioned.

Thank you

Rajat
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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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rajatsp wrote:
Please do let me know if there are more similar questions besides the ones you mentioned.

Thank you

Rajat

Here are several more:
for-which-of-the-following-functions-does-f-x-f-2-x-155813.html
for-which-of-the-following-does-f-a-f-b-f-a-b-164979.html
for-which-of-the-following-functions-f-is-f-x-f-1-x-for-85751.html

Hope this helps.
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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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Given: f(x)= f(1-x)
This means if we replace x by 1 - x in the function, still the result should be same.

Checking each option:

A f(x)=1-x
f(1 - x) = 1 - (1-x) = x. Not equal to f(x)
INCORRECT

B f(x)=1-x^2
f(1-x) = 1- (1-x)^2 = 1 - (1 +x^2 - 2x). Not equal to f(x)
INCORRECT

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 to f(x)
INCORRECT

D f(x)=x^2(1-x)^2
f(1-x) = (1-x)^2(1 - 1 + x)^2 = (1-x)^2*x^2.
This is equal to f(x)
CORRECT

E f(x)= x/(1-x)
f(1-x) = 1-x/1-x + x = (1-x)/x. Not equal to f(x)
INCORRECT

Correct Option: D
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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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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)

Since we are not given any restrictions on the value of x, let’s let x = 1. Thus, we are determining for which of the following functions is f(1) = f(1-1), i.e., f(1) = f(0). Next, we can test each answer choice using our value x = 1.

A. f(x) = 1 - x

f(1) = 1 - 1 = 0

f(0) = 1 - 0 = 1

Since 0 does not equal 1, A is not correct.

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

f(1) = 1 - 1^2 = 1 - 1 = 0

f(0) = 1 - 0^2 = 1 - 0 = 1

Since 0 does not equal 1, B is not correct.

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

f(1) = 1^2 - (1 - 1)^2 = 1 - 0 = 1

f(0) = 0^2 - (1 - 0)^2 = 0 - 1 = -1

Since 1 does not equal -1, C is not correct.

D. f(x) = x^2*(1 - x)^2

f(1) = 1^2*(1 - 1)^2 = 1(0)= 0

f(0) = 0^2*(1 - 0)^2 = 0(2) = 0

Since 0 equals 0, D is correct.

Alternate Solution:

Let’s test each answer choice using x and 1 - x.

A. f(x) = 1 - x

f(x) = 1 - x

f(1 - x) = 1 - (1 - x) = x

Since 1 - x does not equal x, A is not correct.

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

f(x) = 1 - x^2

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

Since 1 - x^2 does not equal 2x - x^2, B is not correct.

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

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

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

Since 2x - 1 does not equal 1 - 2x, C is not correct.

D. f(x) = x^2*(1 - x)^2

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

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

Since x^2*(1 - x)^2 equals (1 - x)^2*x^2, D is correct.

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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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Most people seem to suck at explaining these function related questions. Here is a good video explaining some function skills that helps on other questions. https://www.youtube.com/watch?v=T6-Zdr5w_bE

The key to these questions is understanding that f(x) is the function. Meaning that f(3) would mean everytime you see an x you sub in a 3. Or in this case we sub in an X-3. So for instance the first question is F(x) = 1-x............. Imagine the X is a blank or a blank parentheses waiting to be filled in by a number. so F(3) = 1-3 or F(x) = 1 - X or F (blank)= 1-(blank) or F(1-x) = 1-(1-x)
Attachments Function question 6.png [ 417.34 KiB | Viewed 11580 times ]

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

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Think of functions in terms of inputs and outputs. We want a function for which the input of x will lead to exactly the same result as the input of (1-x). Below is a video explanation

For a function question, it is almost always best to pick numbers, and to choose numbers that are small and manageable. Lets chose 1 for x. The quantity (1-x) would therefore equal 0.

If you put each of those inputs into the function in answer choice A you'll see very quicky that the two outputs are not equal to each other. You'll see very quickly that answer B is also incorrect. Answer C takes a bit more time. The correct answer is D.

The question if easiest to answer if you create separate columns for f(x) and f(1-x). Once again, a video explanation can be found here:
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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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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. $$f(x) = \frac{x}{(1 - x)}$$

Let's try plugging in an easy value for x. How about x = 0.
So, we can reword the question as, For which of the following functions is f(0)=f(1-0)
In other words, we're looking for a function such that f(0) = f(1)

A) f(x)=1-x
f(0)=1-0 = 1
f(1)=1-1 = 0
Since f(0) doesn't equal f(1), eliminate A

B) f(x) = 1 - x^2
f(0) = 1 - 0^2 = 1
f(1) = 1 - 1^2 = 0
Since f(0) doesn't equal f(1), eliminate B

C) f(x) = x^2 - (1-x)^2
f(0) = 0^2 - (1-0)^2 = -1
f(1) = 1^2 - (1-1)^2 = 1
Since f(0) doesn't equal f(1), eliminate C

D) f(x) = x^2(1-x)^2
f(0) = 0^2(1-0)^2 = 0
f(1) = 1^2(1-1)^2 = 0
Since f(0) equals f(1), keep D for now

E) f(x) = x/(1-x)
f(0) = 0/(1-0) = 0
f(1) = 1/(1-1) = undefined
Since f(0) doesn't equal f(1), eliminate E

Since only D satisfies the condition that f(x)=f(1-x) when x=0, the correct answer is D

Cheers,
Brent
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