Having trouble with this PS problem...

This topic has expert replies
Source: — Problem Solving |

Senior | Next Rank: 100 Posts
Posts: 97
Joined: Mon Jan 05, 2009 11:04 pm
Thanked: 2 times

by sjd00d » Wed Jan 21, 2009 9:55 pm
unfortunately no easy way but to try different options (start from the middle when confronted with a problem that requires trying all combinations).

So, let's take A for example to understand how we do this

f(x) = 1 - x
f(1-x) - 1 - (1-x) = x

thus, f(x) != f(1-x).

now f(x) = x^2(1-x)^2
f(1-x) = (1-x)^2 (1-(1-x))^2 = x^2(1-x)^2

that's why this is the correct option

User avatar
Master | Next Rank: 500 Posts
Posts: 138
Joined: Thu Jan 15, 2009 7:52 am
Location: Steamboat Springs, CO
Thanked: 15 times

by gaggleofgirls » Wed Jan 21, 2009 9:56 pm
The best way I can think to solve this is to plug in some numbers. I will chose X=4 just to stay away from 0 and not deal with 0^2 which might be a special case.


So f(4) = f(1-4)
f(4) = f(-3)

a) 1-x
f(4) = 1-4 = -3
f(-3) = 1- - 3 = 1+3 = 4 => not equal

b) 1 - x^2
f(4) = 1 - (4^2) = 1-16 = -15
f(-3) = 1 - (-3^2) = 1 - 9 = -8 => not equal

c) x^2 - (1-x)^2
f(4) = 4^2 - (1-4)^2) = 16 - (-3)^2 = 16 - 9 = 7
f(-3) = (-3)^2 - (1- -3)^2 = 9 - 4^2 = 9-16 = -7 = > not equal

d) x^2 (1-x)^2
f(4) = 4^2 (1-4)^2 = 16 (9)
f(-3) = (-3)^2 (1- -3)^2 = 9 (4)^2 = 9 (16) => equal so possible

e) x/1-x
f(4) = 4/(1-4) = 4/-3 = -4/3
f(-3) = -3/((1--3) = -3/4 = >not equal

Since all the others fail for this example, D is the only possible answer.

But my issue is getting this work done in 2 minutes.

-Carrie

Senior | Next Rank: 100 Posts
Posts: 45
Joined: Sun Nov 30, 2008 8:41 pm
Thanked: 4 times

by mrsmarthi » Wed Jan 21, 2009 9:57 pm
IMO D. Here is my explanation.

Go by the answer choice replacing the following.

for each f(x), check for f(1-x) by replacing x with 1-x on the RHS and still should lead the RHS of the answer choice.

For choice a, f(x) = 1 - x, f(1-x) = 1 - (1-x) = X ==> f(x) <> f(1-x)
For choice b, f(x) = 1 - x^2, f(1-x) = 1 - (1-x)^2 = 2x-x^2 ==> f(x) <> f(1-x)
For choice c, f(x) = x^2 - (1 - x)^2 = 2x - 1, f(1-x) = (1-x) ^ 2 - (1 - (1-x)) ^ 2 = 1 - 2x ==> f(x) <> f(1-x)
For choice d, f(x) = x^2 * (1-x)^2, f(1-x) = (1-x)^2 - (1-(1-x))^2 = (1-x)^2 * x^2 ==> f(x) = f(1-x).
...didn't try to solve choice e.

Hope you understand it

Junior | Next Rank: 30 Posts
Posts: 13
Joined: Sat Jul 26, 2008 6:32 pm

by coolguy55220 » Wed Jan 21, 2009 10:02 pm
awesome... thank you guys... i guess i could have done all that, i just didnt know what the question was asking for. First time i ran into a f(x) type of problem, been studying for the gmat for bout 6 months.

Master | Next Rank: 500 Posts
Posts: 160
Joined: Sat Dec 20, 2008 9:12 pm
Thanked: 11 times

by aroon7 » Wed Jan 21, 2009 10:51 pm
i prefer plugging-in for this problem.

find f(x) and f(1-x) for x=1
i chose 1 because we have (1-x^2) or 1-x. so it will be easy to solve
if more than one option satisfies the condition we can try another round of numbers...

Option f(x) f(1-x) OK?
A 0 1 No
B 0 1 No
C 1 -1 No
D 0 0 Hold
E undefined 0 No

so D is the answer...
--------------------------
i am back!