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If f :RR and g:RR  are one to one, real valued functions, then the value of the integral ππ[f(x)+f(x)][g(x)g(x)]dx is 

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detailed solution

Correct option is A

Let,ϕ(x)=[f(x)+f(−x)][g(x)−g(−x)]then ϕ(−x)=[f(−x)+f(x)][g(−x)−g(x)]ϕ(−x)=−ϕ(x)⇒it odd function.∴    ∫−ππ ϕ(x)dx=0⇒    ∫−ππ [f(x)+f(−x)][g(x)−g(−x)]dx=0


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