For every x,y\in \Re ^{\plus{}}, the function f: \Re ^{\plus{}} \to \Re satisfies the condition f\left(x\right)\plus{}f\left(y\right)\equal{}f\left(x\right)f\left(y\right)\plus{}1\minus{}\frac{1}{xy}. If f(2)<1, then f(3) will be<spanclass=′latex−bold′>(A)</span>2/3<spanclass=′latex−bold′>(B)</span>4/3<spanclass=′latex−bold′>(C)</span>1<spanclass=′latex−bold′>(D)</span>More information needed<spanclass=′latex−bold′>(E)</span>There is no f satisfying the condition above.