A real-valued function `f(x)` satisfies the functional equation `f(x-y)=f(x)f(y)-f(a-x)f(a+y),` where a is a given constant and `f(0)=1.` Then prove that `f(x)` is symmetrical about point (a, 0).

Explanation

We have `f(x-y)=f(x)f(y)-f(a-x)f(a+y)`
Putting `x=a " and " y=x-a,` we get
`f(a-(x-a))=f(a)f(x-a)-f(0)f(x) " (1)" `
Putting `x=0,y=0, ` we get
`f(0)=f(0)(f(0))-f(a) f(a)`
or ` f(0)=(f(0))^(2)-(f(a))^(2)`
or `1=(1)^(2)-(f(a))^(2)`
or `f(a)=0`
` :. f(2a-x)= -f(x) " (from (1))" `
Replacing x by `a-x,` we get
`f(2a-(a-x))= -f(a-x)`
or `f(a+x)= -f(a-x)`
Therefore, `f(x)` is symmetrical about point (a, 0).