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Question

If the function f satisfies the relation f(x+y)+f(xy)=2f(x)f(y) x,yR and f(0)0, then

A
f(x) is an even function
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B
f(x) is an odd function
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C
If f(2)=a, then f(2)=a
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D
If f(4)=b, then f(4)=b
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Solution

The correct option is C If f(2)=a, then f(2)=a
f(x+y)+f(xy)=2f(x)f(y) ....[1]
Putting x=0 in [1], we get
f(y)+f(y)=2f(0)f(y) ....[2]
Putting x=0,y=0 in [1], we get
f(0)+f(0)=2f(0)f(0)f(0)=1
(As f(0)0)
f(y)=f(y) (From[2])
Hence, the function is even.
Then, f(2)=f(2)=a

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