Question

A real valued function $f\left(x\right)$ satisfies the function equation $f(x-y)=f\left(x\right)f\left(y\right)-f(a-x)f(a+y)$ where a is a given constant and $f\left(0\right)=1,f(2a-x)$ is equal to.

• A. $f\left(1\right)+f(a-x)$
• B. $f(-x)$
• C. $-f\left(x\right)$
• D. $f\left(x\right)$

Answer 1

The correct option is C $-f\left(x\right)$

Given f=(0)=1.

and f(x-y)=f(x)f(y)-f(a-x)f(a+y)....(I)

Here if y=0 then (1)becomes.

f(x-0)=f(x)f(0)-f(a-x)f(a+0)

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

$\Rightarrow$f(x)f(x)-f(a-x)f(a)

$\Rightarrow$ f(a-x)f(a)=0. ...(II)

$\therefore$f(2a-x)=f(a+a-x)-f(a-a)f(a+x-a)

=f(a)f(x-a)-f(a-a)f(a+x-a)

=f(a)f(x-a)-f(0)f(x)

=0-1.f(x)[from (II)]

=-f(x)