Functions Question 557
Question: A real valued function $ f(x) $ satisfies the function equation $ f(x-y)=f(x)f(y)-f(a-x)f(a+y) $ where a is a given constant and $ f(0)=1 $ , $ f(2a-x) $ is equal to
[AIEEE 2005]
Options:
A) $ f(a)+f(a-x) $
B) $ f(-x) $
C) $ -f(x) $
D) $ f(x) $
Show Answer
Answer:
Correct Answer: C
Solution:
$ f(a-(x-a))=f(a)f(x-a)-f(0)f(x) $ …..(i) Put $ x=0,y=0 $ ; $ f(0)={{(f(0))}^{2}}-{{[f(a)]}^{2}}\Rightarrow f(a)=0 $ $ [\because f(0)=1]. $ From (i), $ f(2a-x)=-f(x) $ .
BETA
