f(x) + f(y) + f(x)f(y) = 1

For all values of x,yR

f'(x) = ?

Putting y = x we have,

2f(x)+[f(x)]2=1

2f(x)+[f(x)]2+1=1+1

[f(x)+1]2=2

f(x)=1±2

Since f(x) is a constant function, f(x)=0.

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