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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