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Question

If for all x,y the function f is defined by f(x)+f(y)+f(x).f(y)=1 and f(x)>0, then

A
f(x) does not exist
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B
f(x)=0 for all x
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C
f(0)<f(1)
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D
None of these
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Solution

The correct option is B f(x)=0 for all x
Putting x=0,y=0, we get
2f(0)+{f(0)}2=1f(0)=21(f(x)>0)
Putting y=x,2f(x)+{f(x)}2=1
Differentiating w.r.t. x, we get
2f(x)+2f(x).f(x)=0f(x){1+f(x)}=0
f(x)=0, because f(x)>0

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