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 f(0)=1,f(2a−x) is equal to
We have, f(x−y)=f(x)f(y)−f(a−x)f(a+y)
Putting x=a & y=x−a, we get
f(a−(x−a))=f(a)f(x−a)−f(0)f(x)⋯(i)
⇒f(2a−x)=f(a)f(x−a)−f(x)⋯[∵f(0)=1]
Now putting x=0,y=0, we get
f(0)=(f(0))2−[f(a)]2⇒f(a)=0⋯[∵f(0)=1]⋯(ii)
Using (i) & (ii), we get
f(2a−x)=−f(x)