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 these 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
, 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)
, the value of f(x)
will be always positive and will give the following values: for x=-5
; for x=0
. 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
in the expression must give same result.
is the expression for f(x)
, we want to find whether the expression for f(1-x)
would be the same: plug 1-x
. As 1-x
are different, so f(x)
does not equal to f(1-x)
The same with the other options:
, so f(1-x)=1-(1-x)=x
: no match.
, so f(1-x)=1-(1-x)^2=1-1+2x-x^2=2x-x^2
: no match.
, so f(1-x)=2(1-x)-1=1-2x
: no match.
, so f(1-x)=(1-x)^2*(1-1+x)^2=(1-x)^2*x^2
. Bingo! if f(x)=x^2*(1-x)^2
also equals to x^2*(1-x)^2
Still let's check (E)
: 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)
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 question: functions-problem-need-help-93184.html?hilit=which%20following%20functions#p717196
Hope it helps.
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