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 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 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 question:
functions-problem-need-help-93184.html?hilit=which%20following%20functions#p717196Hope it helps.
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68% (02:12) correct
