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Mathematics Question on Relations and functions

A real valued function f(x)f (x) .Satisfies the functional equation f(x−y)=f(x)f(y)−f(a−x)f(a+y)f(x- y) =f (x) f(y)-f(a-x) f(a+y) where a is a given constant and f(0)=1f(0) = 1. Then f(2a−x)f(2a - x) is equal to

A

f(−x)f(-x)

B

f(a)+f(a−x)f(a)+f(a-x)

C

f(x)f(x)

D

−f(x)-f(x)

Answer

−f(x)-f(x)

Explanation

Solution

f(x−y)=f(x)f(y)−f(a−x)f(a+y)f(x - y) = f(x) f (y) - f (a - x) f (a + y) Put x=0=yx = 0 = y. ∴  f(0)=f(0)f(0)−f(a)f(a)=(f(0))2−(f(a))2\therefore \:\: f(0) = f(0) f(0) -f(a) f(a) = (f(0))^2 - (f (a))^2 ⇒1=1−(f(a))2⇒f(a)=0\Rightarrow 1= 1-(f(a))^2 \Rightarrow f(a) = 0 Now f(2a−x)=f(a−(x−a))f(2a-x) = f(a - (x - a)) =f(a)f(x−a)−f(a−a)f(a+x−a)= f(a)f(x - a)-f(a- a) f(a + x - a) 0⋅f(x−a)−f(0)f(x)0\cdot f(x-a)-f(0) f(x) =0−1.f(x)=−f(x)= 0 - 1. f(x) = - f(x) .