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by Anurag@Gurome » Wed Apr 18, 2012 5:51 am
bobdylan 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

Let us look at each of the options:

(A) f(x) = 1-x
f(1-x) = 1 - (1-x) = x; FALSE

(B) f(x) = 1-x^2
f(1-x) = 1 - (1-x)^2 = 1 - (1 - 2x + x^2) = 2x - x^2; FALSE

(C) f(x) = x^2 - (1-x)^2
f(1-x) = (1-x)^2 - (1 - (1-x))^2 = 1 - 2x + x^2 - (x)^2 = 1 - 2x; FALSE

(D) f(x) = x^2 * (1-x)^2
f(1-x) = (1 - x)^2 * (1 - (1 - x))^2 = (1 - x)^2 * (x)^2; TRUE

(E) f(x) = x/(1-x)
so f(1-x) = (1-x)/(1-(1-x)) = (1-x)/x; FALSE

The correct answer is D.
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by IshanGhose » Thu Apr 19, 2012 1:30 am
The long drawn method would be to susbtitute (1-x) for all values of x in each of the given equations and to see if we get the original equation f(x), as has already been shown by Anurag@Gurome.

However, to save time on expanding each equation in its multiples, one can just allocate a value for x. Lets say x = 0, then quickly substituting the same for the values of x, we shall have:

a) f(x) = 1-x = 1-0 = 1; f(1-x) = 1-(1-0)= 0 , therefore False.
b) f(x) = 1-x^2 = 1-0 = 1; f(1-x) = 1 - (1-0)^2= 1-1 = 0, therefore False
c) f(x) = x^2-(1-x)^2 = 0-1 = -1 ; f(1-x) = (1-0)^2 - (1-(1-0))^2 = +1, therefore False
d) f(x) = x^2(1-x)^2 = (0)^2(1)^2 = 0; f(1-x) = (1-x)^2(1-(1-x))^2 = 1^2(0)^2 = 0, therefore True.
e) f(x) = x/(1-x) = 0/1 = 0; f(1-x) = (1-x)/(1-(1-x)) = 1/0 = Infinity, therefore False.

I believe this method of substituting random (but simple) values for x will help in getting the solution faster, since time is a definite constraint in the GMAT exam. The other method of expanding each equation, is however, the most fool-proof method.
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