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:
https://gmatclub.com/forum/for-which-of- ... 93184.htmlhttps://gmatclub.com/forum/for-which-of- ... 24491.htmlhttps://gmatclub.com/forum/let-the-funct ... 43311.htmlHope it helps.
\(f(x)=x^2* (1-x)^2 = x^2 * ( 1-2x+x^2)\) = now plugged in (1-x) for x --> \((1-x)^2 * (1-2(1-x)+(1-x)^2) \)
And why in Option C you didn`t plug in 1-x to get 2x-1 whereas in D you plugged in x-1 straight away without expanding (1-x)^2 ?