Google
Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm liên tục trên khoảng $\left( 0;+\infty \right)$ và thỏa mãn $\underset{x\to t}{\mathop{\lim }} \dfrac{xf\left( t \right)-tf\left( x \right)}{{{x}^{2}}-{{t}^{2}}}=1$ với mọi $t>0.$ Biết rằng $f\left( 1 \right)=1,$ tính $f\left( e \right).$
A. $\dfrac{3e+1}{2}$
B. $3e$
C. $2e$
D. $-e$
A. $\dfrac{3e+1}{2}$
B. $3e$
C. $2e$
D. $-e$
Cách giải:
Sử dụng Lopitan ta có: $\underset{x\to t}{\mathop{\lim }} \dfrac{xf\left( t \right)-tf\left( x \right)}{{{x}^{2}}-{{t}^{2}}}=\underset{x\to t}{\mathop{\lim }} \dfrac{f\left( t \right)-tf'\left( x \right)}{2x}=\dfrac{f\left( t \right)-tf'\left( t \right)}{2t}=1$
$\Rightarrow \dfrac{f\left( x \right)-xf'\left( x \right)}{2x}=1\Rightarrow \dfrac{f\left( x \right)-xf'\left( x \right)}{2{{x}^{2}}}=\dfrac{1}{x}$
$\Rightarrow \dfrac{f\left( x \right)-xf'\left( x \right)}{{{x}^{2}}}=\dfrac{2}{x}\Rightarrow \dfrac{f'\left( x \right).x-f\left( x \right)}{{{x}^{2}}}=\dfrac{-2}{x}$
$\Rightarrow \int\limits_{{}}^{{}}{\dfrac{f'\left( x \right).x-f\left( x \right)}{{{x}^{2}}}dx}=-2\ln x+C$
$\Rightarrow \dfrac{f\left( x \right)}{x}=-2\ln x+C$
$f\left( 1 \right)=1\Rightarrow \dfrac{1}{1}=-2\ln 1+C\Rightarrow C=1$
$\Rightarrow f\left( x \right)=2x\ln x+x$
$f\left( e \right)=-2e\ln e+e=-e$
Sử dụng Lopitan ta có: $\underset{x\to t}{\mathop{\lim }} \dfrac{xf\left( t \right)-tf\left( x \right)}{{{x}^{2}}-{{t}^{2}}}=\underset{x\to t}{\mathop{\lim }} \dfrac{f\left( t \right)-tf'\left( x \right)}{2x}=\dfrac{f\left( t \right)-tf'\left( t \right)}{2t}=1$
$\Rightarrow \dfrac{f\left( x \right)-xf'\left( x \right)}{2x}=1\Rightarrow \dfrac{f\left( x \right)-xf'\left( x \right)}{2{{x}^{2}}}=\dfrac{1}{x}$
$\Rightarrow \dfrac{f\left( x \right)-xf'\left( x \right)}{{{x}^{2}}}=\dfrac{2}{x}\Rightarrow \dfrac{f'\left( x \right).x-f\left( x \right)}{{{x}^{2}}}=\dfrac{-2}{x}$
$\Rightarrow \int\limits_{{}}^{{}}{\dfrac{f'\left( x \right).x-f\left( x \right)}{{{x}^{2}}}dx}=-2\ln x+C$
$\Rightarrow \dfrac{f\left( x \right)}{x}=-2\ln x+C$
$f\left( 1 \right)=1\Rightarrow \dfrac{1}{1}=-2\ln 1+C\Rightarrow C=1$
$\Rightarrow f\left( x \right)=2x\ln x+x$
$f\left( e \right)=-2e\ln e+e=-e$
Đáp án D.