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 $f\left(0\right)=1,f(2a-x)$ is equal to

• A. f(a) + f(a - x)
• B. f(-x)
• C. -f(x)
• D. f(x)

Answer 1

The correct option is C -f(x)

We have, $f(x-y)=f\left(x\right)f\left(y\right)-f(a-x)f(a+y)$
Putting $x=a\&y=x-a$, we get
$f(a-(x-a\left)\right)=f\left(a\right)f(x-a)-f\left(0\right)f\left(x\right)\cdots \left(i\right)$
$\Rightarrow f(2a-x)=f\left(a\right)f(x-a)-f\left(x\right)\cdots [\because f(0)=1]$
Now putting $x=0,y=0$, we get
$f\left(0\right)=\left(f\right(0){)}^2-[f\left(a\right){]}^2\Rightarrow f\left(a\right)=0\cdots [\because f(0)=1]\cdots \left(ii\right)$
Using $\left(i\right)\&\left(ii\right)$, we get
$f(2a-x)=-f\left(x\right)$