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

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Manager
Joined: 21 Sep 2006
Posts: 111
Location: Where you mind is
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Kudos [?]: 17 [0], given: 0

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

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23 Dec 2007, 22:17
For which of the following function f is f(x) = f(1-x) for all x?

A. f(x) = 1-x
B. f(x) = 1-xsquared
C. f(x) = xsquared - (1-x)squared
D. f(x) = xsquared * (1-x)squared
E. f(x) = x / (1-x)

OA is D.
Director
Joined: 03 Sep 2006
Posts: 879
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Kudos [?]: 622 [0], given: 33

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24 Dec 2007, 00:58
Hang Tuah wrote:
For which of the following function f is f(x) = f(1-x) for all x?

A. f(x) = 1-x
B. f(x) = 1-xsquared
C. f(x) = xsquared - (1-x)squared
D. f(x) = xsquared * (1-x)squared
E. f(x) = x / (1-x)

OA is D.

Yeah it is D, none other, you have to check...I dont think any fast method is there other than checking.

x^2*(1-x)^2... this will be result whether it is f(x) or f(1-x)
Manager
Status: GMAT BATTLE - WIN OR DIE
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Concentration: General Management, Entrepreneurship
GMAT Date: 12-22-2011
GPA: 3.81
WE: General Management (Hospitality and Tourism)
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Kudos [?]: 67 [0], given: 13

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29 Jul 2011, 06:59
sorry for comebach thic question but i completele miss this question.

can expert explane how to do this type of questions

kudos for detail explanation
Intern
Joined: 18 Jul 2011
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Kudos [?]: 17 [0], given: 2

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29 Jul 2011, 12:20
I'll start by cleaning up the question:
For which of the following is $$f(x) = f(1 - x)$$?
(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) $$f(x) = \frac{x}{1-x}$$

There's really only one practical way to solve this:
(1) We can check all the answers - computing $$f(1 - x)$$ for each one and seeing if it is equal to the given function (if you're not familiar with this concept, think if it as a substitution of (1 - x) for all the x's that appear in the given function... I'll try to make it clear by showing all the work):
(A) $$f(1 - x) = 1 - (1 - x) = 1 - 1 + x = x \neq f(x)$$
(B) $$f(1 - x) = 1 - (1 - x)^2 = 1 - (1 - 2x + x^2) = 1 - 1 + 2x + x^2 = 2x + x^2 \neq f(x)$$
(C) $$f(1 - x) = (1 - x)^2 - (1 - (1 - x))^2 = 1 - 2x + x^2 - (1 - 1 + x)^2 = 1 - 2x + x^2 - x^2 = 1 - 2x \neq f(x)$$
(D) $$f(1 - x) = (1 - x)^2 (1 - (1 - x))^2 = ( 1 - x)^2(1 - 1 + x)^2 = (1 - x)^2(x^2) = x^2(1 - x)^2 = f(x)$$

So, the answer is D. You can also see that this would be the answer by just looking carefully at your choices - of course this is dependent on your comfort with the underlying concept of composition of functions.

WARNING - THE FOLLOWING CONTAINS SOME SOMEWHAT ADVANCED MATH THAT YOU DON'T NEED TO KNOW TO ROCK THE GMAT
An impractical way to approach the question is to consider the rules for transformations of functions
$$f(1 - x) = f(-x + 1)$$. Adding 1 moves the function to the left, and multiplying the x by negative 1 rotates the function around the y axis. A linear function could not be transformed this way without changing it's slope - this eliminates A and C. The rational function in E would not have the same asymptote after it was moved 1 unit to the left. This leaves us with B and D. B has roots at 1 and -1. Therefore it is symmetric about the y axis and the transformation would disturb that symmetry. So it must be D
Re: Some more Algebra   [#permalink] 29 Jul 2011, 12:20
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