A real-valued function f(x) satisfies the equation f(x−y)=f(x)f(y)−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)
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B
f(x)
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C
f(a)+f(a-x)=0
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D
f(-x)
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Solution
The correct option is A -f(x) Giventhatf(x−y)=f(x)f(y)−f(a−x)f(a+y)Puttingx=y=0inaboveequation,⇒f(0)=(f(0)]2−(f(a)]2⇒f(a)=0f(2a−x)=f(a−(x−a)]=f(a)f(x−a)−f(0)f(x)=0−f(x)(usingthevaluesoff(a)andf(0)]=−f(x)