A real-valued function f(x) satisfies the functional equation f(x−y)=f(x)f(y)−f(a−x)f(a+y)∀x,y∈R, where a is a given constant and f(0)=1. Then, f(2a−x) is equal to
A
f(x)
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B
−f(x)
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C
f(x)
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D
f(a)+f(a−x)
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Solution
The correct option is B−f(x) Given that : f(x−y)=f(x)f(y)−f(a−x)f(a+y)∀x,y∈R
Putting y=0, we get f(x)=f(x)−f(a−x)f(a) ⇒f(a−x)f(a)=0 ⇒f(a)=0(As f(a−x)≠0)