Question need help . thanks.

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by kishore » Thu Jun 05, 2008 8:58 pm
Answer:
D. f(x) = x^2(1-x)^2

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by kishore » Thu Jun 05, 2008 9:01 pm
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)


Go by answers.

a. f(x) = 1-x

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

f(x) = 1-x and f(1-x) = x which are not equal.

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by shrikantkamble » Fri Jun 06, 2008 4:11 am
Hi ,

I can see this problem as substitution.
As mention F(x) = F(1-x). Let us take any value of X and evaluate it.
let x = 1.
ie 1-x = 0.

A. f(X) = 1-X

f(x)=f(1) = 1 - 1 = 0

f(1-x) = f(1-1) = f(0) = 1-0 = 1
This is not a correct anser.

B. f(X) = 1-X^2

f(x)=f(1) = 1 - 1^2 = 1 - 1 = 0 .

f(1-x) = f(1-1)= f(0) = 1 - 0^2 = 1 - 0 = 1 .
This is not a correct anser.

C. f(x) = x^2-(1-x)^2

f(x)=f(1) = 1^2 - (1 - 1)^2 = 1 - 0 = 1 .

f(1-x) = f(1-1)= f(0) = 0^2 - (1 - 0)^2 = 0 - 1 = -1 .
This is not a correct anser.

D. f(x) = x^2(1-x)^2
f(x)=f(1) = 1^2 *(1 - 1)^2 = 1 * 0 = 0 .
f(1-x) = f(1-1)= f(0) = 0^2 *(1 - 0)^2 = 0 * 1 = 0 .

This may be correct answer. Let us evaluate last option.

E. f(x) = x/(1-x)

f(x)=f(1) = 1 /(1 - 1) = 1 / 0 = infinity .
f(1-x) = f(1-1)= f(0) = 0/ (1 - 0) = 0 / 1 = 0 .

This is not a correct answer.
We only left with Option D.
Thanks & Regards,
Shrikant