DeeptiM wrote:
fluke wrote:
Baten80 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)
For
x=1;
1-x=0
a) f(1)=1-1=0; f(0)=1-0=1
b) f(1)=1-1=0; f(0)=1-0=1
c) f(1)=1-(1-1)^2=1-0=1; f(0)=0-(1-0)^2=-1
d) f(1)=1*(1-1)^2=0; f(0)=0---- Correct. f(1)=f(0)
e) f(1)=1/0=undefined; f(0)=0
Ans: "D"
Thanks for directing me here fluke..need your help with point c and d..
we took x=1 and 1-x=0..if so then c should be
f(x) = x^2 - (1-x)^2 -------- Original Statement
1 = 1 - (0)^2 {:Note: Here LHS should be f(1) not 1. You need to consider both case one by one}
1=1
and d should be
f(x) = x^2 * (1-x)^2 -------- Original Statement
1 = 1 * (0)^2
1=0
I am sure there is a miss in my understanding for which i need your help.
If there is a function;
f(x)=x-1
f(1)=1-1=0
f(2)=2-1=0
f(z)=z-1
f(1+x)=1+x-1=x
You see how we replace x with whatever is in the bracket of f().
Likewise;
f(x) = x^2 - (1-x)^2
f(1)=(1)^2-(1-1)^2=1-0=1 {:Note: Replace x with 1}
f(0)=0^2-(1-0)^2=0-1=-1 {:Note: Replace x with 0}
You see:
f(1) NOT EQUAL f(0). Thus, it is incorrect.
We can't do partial replacement.
Or algebraically:
C:
f(x) = x^2 - (1-x)^2= x^2-(1+x^2-2x)=x^2-1-x^2+2x=2x-1
f(1-x)=(1-x)^2 - (1-(1-x))^2=1+x^2-2x-x^2=1-2x {:Note: I replaced x with 1-x everywhere}
Try the same with D yourself.
_________________
~fluke
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