Question

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

• A. -f(x)
• B. f(x)
• C. f(a)+f(a-x)=0
• D. f(-x)

Answer 1

The correct option is A -f(x)

$\mathrm{Given}\mathrm{that}f\left(x-y\right)=f\left(x\right)f\left(y\right)-f\left(a-x\right)f\left(a+y\right)\mathrm{Putting}x=y=0\mathrm{in}\mathrm{above}\mathrm{equation},\Rightarrow f\left(0\right)={\left(f\left(0\right)\right]}^2-{\left(f\left(a\right)\right]}^2\Rightarrow f\left(a\right)=0f\left(2a-x\right)=f\left(a-\left(x-a\right)\right]=f\left(a\right)f\left(x-a\right)-f\left(0\right)f\left(x\right)=0-f\left(x\right)\left(\mathrm{using}\mathrm{the}\mathrm{values}\mathrm{of}f\left(a\right)\mathrm{and}f\left(0\right)\right]=-f\left(x\right)$