2025年考研数学二第18题

$f^{\prime}(0)=5$ ## 【解析】

$-3=\displaystyle\lim _{x \rightarrow 0} \displaystyle\frac{x f(x)-e^{2 \sin x}+1}{\ln (1+x)+\ln (1-x)}$ $=\displaystyle\lim _{x \rightarrow 0} \displaystyle\frac{x f(x)-e^{2 \sin x}+1}{\ln \left(1-x^{2}\right)}$

$$ \begin{equation*} =\lim _{x \rightarrow 0} \frac{x f(x)-e^{2 \sin x}+1}{-x^{2}} \tag{1} \end{equation*} $$

$=\displaystyle\lim _{x \rightarrow 0} \displaystyle\frac{f(x)+\displaystyle\frac{1-e^{2 \sin x}}{x}}{-x}$

因为 $\displaystyle\lim _{x \rightarrow 0} \displaystyle\frac{1-e^{2 \sin x}}{x}=-2$

由（1）知 $-3=\displaystyle\lim _{x \rightarrow 0} \displaystyle\frac{x[f(x)-f(0)]+2 x-e^{2 \sin x}+1}{-x^{2}}$ $=-\displaystyle\lim _{x \rightarrow 0} \displaystyle\frac{f(x)-f(0)}{x}+\displaystyle\lim _{x \rightarrow 0} \displaystyle\frac{2 x-e^{2 \sin x}+1}{-x^{2}}$ $=-\displaystyle\lim _{x \rightarrow 0} \displaystyle\frac{f(x)-f(0)}{x}+\displaystyle\lim _{x \rightarrow 0} \displaystyle\frac{2-2 \cos x e^{2 \sin x}}{-2 x}$ $=-\displaystyle\lim _{x \rightarrow 0} \displaystyle\frac{f(x)-f(0)}{x}+\displaystyle\lim _{x \rightarrow 0} \displaystyle\frac{1-\cos x+\cos x\left(1-e^{2 \sin x}\right)}{-x}$ $=-\displaystyle\lim _{x \rightarrow 0} \displaystyle\frac{f(x)-f(0)}{x}+2$

所以 $\displaystyle\lim _{x \rightarrow 0} \displaystyle\frac{f(x)-f(0)}{x}=5$ 即 $f(x)$ 在 $x=0$ 处可导且 $f^{\prime}(0)=5$ （ 19 ）（本题满分 12 分）设函数 $f(x, y)$ 可微 且 满 足 $d f(x, y)=-2 x e^{-y} d x+e^{-y}\left(x^{2}-y-1\right) d y, f(0,0)=2$ ，求 $f(x, y)$ ，并求 $f(x, y)$ 的极值

【答案】 $f(x, y)=-x^{2} e^{-y}+(y+2) e^{-y} ; f(0,-1)$ 为极大值

由题意知：$f_{x}^{\prime}(x, y)=-2 x e^{-y}, f_{y}^{\prime}(x, y)=e^{-y}\left(x^{2}-y-1\right)$ ， 故

$$ \begin{aligned} & f(x, y)=\int-2 x e^{-y} d x=-x^{2} e^{-y}+C(y) \Rightarrow f_{y}^{\prime}(x, y)=x^{2} e^{-y}+C^{\prime}(y)=e^{-y}\left(x^{2}-y-1\right) \\ & C^{\prime}(y)=-(y+1) e^{-y} \Rightarrow C(y)=(y+2) e^{-y}+C \Rightarrow f(x, y)=-x^{2} e^{-y}+(y+2) e^{-y}+C \end{aligned} $$

因为 $f(0,0)=2 \Rightarrow C=0 \Rightarrow f(x, y)=-x^{2} e^{-y}+(y+2) e^{-y}$ 令 $\left\{\begin{array}{l}f_{x}^{\prime}(x, y)=-2 x e^{-y}=0 \\ f_{y}^{\prime}(x, y)=e^{-y}\left(x^{2}-y-1\right)=0\end{array} \Rightarrow\left\{\begin{array}{l}x=0 \\ y=-1\end{array}\right.\right.$ $\left\{\begin{array}{l}f_{x x}^{\prime \prime}(x, y)=-2 e^{-y} \\ f_{x y}^{\prime \prime}(x, y)=2 x e^{-y} \\ f_{y y}^{\prime \prime}(x, y)=-e^{-y}\left(x^{2}-y-1\right)-e^{-y}\end{array} \Rightarrow\left\{\begin{array}{l}A=-2 e \\ B=0 \\ C=-e\end{array} \Rightarrow\left\{\begin{array}{l}A C-B^{2}>0 \\ A<0\end{array}\right.\right.\right.$ 故极大值为 $f(0,-1)=e$ ．