If f(x⋅y)=f(x)+f(y)∀x,y∈R+, f(e2)=2, then the number of solution – InterviewSolution

1.	If f(x⋅y)=f(x)+f(y)∀x,y∈R+, f(e2)=2, then the number of solution of the equation ef([x])+ef({x})=x2+x is :(where [x],{x} represents greatest integer function and fractional part function resepectively)
Answer» If $f (x \cdot y) = f (x) + f (y) \forall x, y \in R^{+}$ , $f (e^{2}) = 2$ , then the number of solution of the equation $e^{f ([x])} + e^{f ({x})} = x^{2} + x$ is : (where $[x], {x}$ represents greatest integer function and fractional part function resepectively)

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