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f is a real-valued function not identically zero, satisfying f(x+y)+f(xy)=2f(x)f(y)x,yRf(x)  is definitely 

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a
odd
b
even
c
neither even nor odd
d
none of these

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detailed solution

Correct option is B

Putting x=0,y=0 in the functional equation we get 2f(0)=2f2(0)⇒f(0)=0,1. But if f(0) is equal to zero then f(x)=0∀x∈R. Thus, f(0)=1. Now putting x=0 in  the functional equation we get, f(y)+f(−y)=2f(0)⋅f(y)⇒f(y)=f(−y). Thus, f(x) is even.


Similar Questions

 Let G(x)=1ax1+12F(x), where a is a positive real  number not equal to 1 and F(x) is an odd function. Which  of the following statements is true? 


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