Đáp án đúng: D Giải chi tiết:Sử dụng Lopitan ta có: \(\mathop {\lim }\limits_{x \to t} \dfrac{{xf\left( t \right) - tf\left( x \right)}}{{{x^2} - {t^2}}} = \mathop {\lim }\limits_{x \to t} \dfrac{{f\left( t \right) - tf'\left( x \right)}}{{2x}} = \dfrac{{f\left( t \right) - tf'\left( t \right)}}{{2t}} = 1\)\(\begin{array}{l} \Rightarrow \dfrac{{f\left( x \right) - xf'\left( x \right)}}{{2x}} = 1 \Rightarrow \dfrac{{f\left( x \right) - xf'\left( x \right)}}{{2{x^2}}} = \dfrac{1}{x}\\ \Rightarrow \dfrac{{f\left( x \right) - xf'\left( x \right)}}{{{x^2}}} = \dfrac{2}{x} \Rightarrow \dfrac{{f'\left( x \right).x - f\left( x \right)}}{{{x^2}}} = \dfrac{{ - 2}}{x}\end{array}\)\(f\left( 1 \right) = 1 \Rightarrow \dfrac{1}{1} = - 2\ln 1 + C \Rightarrow C = 1\)\( \Rightarrow f\left( x \right) = - 2x\ln x + x\)\(f\left( e \right) = - 2e\ln e + e = - e\)Chọn D
