\documentclass[../Main.tex]{subfiles}
\begin{document}
\chapter{2023-2024学年微积分（一）（上）期中考试}

\section{基本计算题(每小题 6 分，共 60 分)}
\begin{enumerate}
    \item 求极限$l=\lim_{n\to\infty}(-1)^{n}\sin\left(\sqrt{n^{2}+1}\pi\right)$.
          \vspace{11em}

    \item 求极限$l=\lim_{x\to0}\frac{\mathrm{e}^{\tan x}-\mathrm{e}^{x}}{x-\sin
                  x}.$
          \vspace{11em}

    \item 当$x\to 0$时，设$u=\sqrt{1+\tan x}-\sqrt{1+\sin x}$的主部为$cx^{k}$，试确定常数$c
              ,k$的值.
          \vspace{11em}

    \item 求极限$l=\lim_{x\to0}\left(\frac{\arcsin x}{x}\right)^{\frac{1}{x^{2}}}$.
          \vspace{11em}

    \item 设函数$f(x)=\frac{x-x^{2}}{\sin \pi x}$，指出$f(x)$的间断点并判断其类型.
          \vspace{9em}

    \item 设$y=\sqrt{x\sin x\sqrt{\mathrm{e}^{x}-1}}(0<x<\pi)$，求导数$y'$.
          \vspace{10em}

    \item 设$f(x)=
              \begin{cases}
                  a+x+\sqrt{1-x}, & x<0    \\
                  1+b\arctan x,   & x\geq0
              \end{cases}$，求常数$a,b$，使$f(x)$在$x=0$处可导.
          \vspace{10em}

    \item 设函数$y=y(x)$由方程$y=2\mathrm{e}^{y}\sin x-7x$确定的可微函数，求$\left
              .\mathrm{d}y\right|_{x=0}$.
          \vspace{10em}

    \item 设$y=\frac{1}{2-3x-2x^{2}}$，求$y^{(n)}(0)$.
          \vspace{8em}

    \item 设$y=\sin^{4} x-\cos^{4} x$，求$y'$及$y''$.
          \vspace{8em}
\end{enumerate}

\section{综合题(每小题 6 分，共 30 分)}
\begin{enumerate}
    \setcounter{enumi}{10}
    \item 设函数$y=y(x)$由参数方程$\begin{cases}
                  x=t^{2}+1, \\
                  y=4t-t^{2}
              \end{cases},t>0$确定，过点$(-1,0)$作$y=y(x)$的切线，求切点$(x_{0},y_{0})$，并写出该切线的方程.
          \vspace{10em}

    \item 设$y=g(x)$是$y=f(x)$的反函数，$f(x)=x+(1+x)^{x}$，求$y=g(2+x^{2})$在$x
              =1$处的导数.
          \vspace{10em}

    \item 设$b_{i}\geq0(i=1,2,\cdots,n,\cdots)$，$x_{n}=\frac{b_{1}}{1+b_{1}}
              +\frac{b_{2}}{(1+b_{1})(1+b_{2})}+\cdots+\frac{b_{n}}{(1+b_{1})(1+b_{2})\cdots(1+b_{n})}$，研究数列$\{
              x_{n}\}$的极限的存在性.
          \vspace{8em}

    \item 设函数$f(x)=
              \begin{cases}
                  \frac{g(x)-\mathrm{e}^{-x}}{x}, & x\neq0, \\
                  0,                              & x=0
              \end{cases}$，其中$g(x)$在$(-\infty,+\infty)$内具有一阶连续的导数，且$g''(0
              )$存在，$g(0)=1$，$g'(0)=-1$，求$f'(x)$，并讨论$f'(x)$在$x=0$处的连续性.
          \vspace{8em}

    \item 一观察者站在地面上用望远镜观察一架飞机，该飞机的高度为$20\mathrm{km}$，正以$1
              6\mathrm{km/min}$的速度向观察者水平地飞过来，试问当飞机离观察者的水平距离为$4
              0\mathrm{km}$时，望远镜视角改变的速率是多少？
          \vspace{8em}
\end{enumerate}

\section{证明题(每小题 5 分，共 10 分)}
\begin{enumerate}
    \setcounter{enumi}{15}
    \item 设函数$f(x)$在$(a,b)$内连续，且$x_{i}\in(a,b),i=1,2,\cdots,n$，证明至少存在一点$\xi
              \in(a,b)$，使
          \[
              f(\xi)=\frac{1}{n}\left(f(x_{1})+f(x_{2})+\cdots+f(x_{n})\right).
          \]
          \vspace{10em}

    \item 设函数$f(x)$在$[0,1]$上连续，在$(0,1)$内可导，并且$f(0)=0,f(1)=\frac{1}{2}$，证明存在$\xi
              ,\eta\in(0,1)$，且$\xi\neq\eta$，使得$f'(\xi)+f'(\eta)=\xi+\eta$.
          \vspace{8em}
\end{enumerate}

\chapter{2023-2024学年微积分（一）（上）期中考试参考答案}
\section{基本计算题(每小题 6 分，共 60 分)}
\begin{enumerate}
    \item \textbf{Solution}.
          \[
              \begin{aligned}
                  l= & \lim_{n\infty}(-1)^{n} n\sin\left((\sqrt{n^2+1}-n)\pi+n\pi\right)                                                    \\
                  =  & \lim_{n\to\infty}n\sin\left((\sqrt{n^2+1}-n)\pi\right)=\lim_{n\to\infty}n\sin\left(\frac{\pi}{\sqrt{n^2+1}+n}\right) \\
                  =  & \lim_{n\to\infty}\frac{n}{\sqrt{n^2+1}+n}\pi=\frac{\pi}{2}.
              \end{aligned}
          \]

    \item \textbf{Solution}.
          \[
              \begin{aligned}
                  l= & \lim_{x\to 0}\frac{\mathrm{e}^{x}(\mathrm{e}^{\tan x -x}-1)}{x-\sin x}=\lim_{x\to0}\frac{\tan x-x}{x-\sin x} \\
                  =  & \lim_{x\to0}\frac{\sec ^2 x-1}{1-\cos x}=\lim_{x\to0}\frac{\tan^2 x}{\frac{1}{2}x^2}=2.
              \end{aligned}
          \]

          或
          \[
              \begin{aligned}
                  l= & \lim_{x\to 0}\mathrm{e}^{\xi}\frac{\tan x -x}{x-\sin x}\qquad (\xi\,\text{介于}\,x\,\text{与}\tan x\,\text{之间}) \\
                  =  & \lim_{x\to 0}\frac{\tan x -x}{x-\sin x}                                                                      \\
                  =  & \lim_{x\to0}\frac{\sec ^2 x-1}{1-\cos x}=\lim_{x\to0}\frac{\tan^2 x}{\frac{1}{2}x^2}=2.
              \end{aligned}
          \]

    \item \textbf{Solution}.

          \textbf{法一.}
          \[
              \begin{aligned}
                  u=   & \frac{1}{\sqrt{1+\tan x}+\sqrt{1+\sin x}}(\tan x-\sin x)                    \\
                  \sim & \frac{1}{2}(\tan x-\sin x)=\frac{1}{2}\tan x(1-\cos x)\sim\frac{1}{4}x^{3},
              \end{aligned}
          \]

          即$k=3,c=\frac{1}{4}$.

          \textbf{法二.}由题意得$\lim_{x\to0}\frac{u}{cx^{k}}=1$. 而
          \[
              \begin{aligned}
                  \lim_{x\to0}\frac{u}{cx^k}= & \frac{\tan x-\sin x}{cx^k(\sqrt{1+\tan x}+\sqrt{1+\sin x})}              \\
                  =                           & \lim_{x\to0}\frac{\tan x(1-\cos x)}{2cx^k}                               \\
                  =                           & \lim_{x\to0}\frac{\frac{1}{2}x^3}{2cx^k}=\lim_{x\to0}\frac{x^{3-k}}{4c},
              \end{aligned}
          \]

          故$3-k=0,4c=1$，即$k=3,c=\frac{1}{4}$.

    \item \textbf{Solution}.
          \[
              \begin{aligned}
                  l= & \lim_{x\to0}\mathrm{e}^{\frac{1}{x^2}\ln\left(\frac{\arcsin x}{x}\right)}=\mathrm{e}^{\lim\limits_{x\to0}\frac{1}{x^2}\ln\left(\frac{\arcsin x}{x}\right)}                                                                                                                            \\
                  \
                  =  & \mathrm{e}^{\lim\limits_{x\to0}\frac{1}{x^2}\left(\frac{\arcsin x}{x}-1\right)}=\mathrm{e}^{\lim\limits_{x\to0}\frac{\arcsin x-x}{x^3}}=\mathrm{e}^{\lim\limits_{x\to0}\frac{\frac{1}{\sqrt{1-x^2}}-1}{3x^2}}=\mathrm{e}^{\lim\limits_{x\to0}\frac{1-\sqrt{1-x^2}}{3x^2\sqrt{1-x^2}}} \\
                  =  & \mathrm{e}^{\lim\limits_{x\to0}\frac{\frac{1}{2}x^2}{3x^2\sqrt{1-x^2}}}=\mathrm{e}^{\frac{1}{6}}.
              \end{aligned}
          \]

    \item \textbf{Solution}.

          当$\sin\pi x=0$即$x=k,k=0,\pm1,\pm2,\cdots$时，$f(x)$无定义，故$x=k,k=0,\pm
              1,\pm2\cdots$是$f(x)$的间断点.

          因
          \[
              \begin{aligned}
                  \lim_{x\to0}f(x)= & \lim_{x\to0}\frac{x-x^2}{\sin\pi x}=\lim_{x\to0}\frac{\pi x}{\sin\pi x}\cdot\frac{1-x}{\pi}=\frac{1}{\pi}; \\
                  \lim_{x\to1}f(x)= & \lim_{x\to1}\frac{\pi(1-x)}{\sin\pi(1-x)}\cdot\frac{x}{\pi}=\frac{1}{\pi},
              \end{aligned}
          \]

          所以$x=0,1$是可去间断点；

          当$k=-1,\pm2,\pm3,\cdots$时，$\lim_{x\to k}f(x)=\infty$，所以$x=-1,\pm2,\pm
              3,\cdots$是无穷间断点.

    \item \textbf{Solution}.因$\ln y=\frac{1}{2}\left(\ln x+\ln\sin x+\frac{1}{2}
              \ln\left(\mathrm{e}^{x}-1\right)\right)$，

          所以$y'=\frac{1}{2}\sqrt{x\sin x\sqrt{\mathrm{e}^{x}-1}}\left(\frac{1}{x}+\cot
              x+\frac{\mathrm{e}^{x}}{2(\mathrm{e}^{x}-1)}\right)$.

    \item \textbf{Solution}.要$f(x)$在$x=0$处可导，必须$f(x)$在$x=0$处连续，即
          \[
              \lim_{x\to0^+}f(x)=\lim_{x\to0^-}f(x)=f(0).
          \]

          因$f(0)=1$，$\lim_{x\to0^+}f(x)=\lim_{x\to0^+}(1+b\arctan x)=1=f(0)$，

          $\lim_{x\to0^-}f(x)=\lim_{x\to0^-}(a+x+\sqrt{1-x})=a+1$，所以$a+1=1$，即$a=
              0$.

          又$f'_{-}(0)=\lim_{x\to0}\frac{(x+\sqrt{1-x})-1}{x}=\lim_{x\to0}\frac{\sqrt{1-x}(1-\sqrt{1-x})}{x}
              =\lim_{x\to0}\frac{\sqrt{1-x}\cdot\frac{1}{2}x}{x}=\frac{1}{2}$，

          $f'_{+}(0)=\lim_{x\to0}\frac{1+b\arctan x-1}{x}=\lim_{x\to0}\frac{b\arctan x}{x}
              =b$，

          要$f(x)$在$x=0$处可导，必须$f'_{-}(0)=f'_{+}(0)$，所以$b=\frac{1}{2}$.

    \item \textbf{Solution}.

          \textbf{法一.} 当$x=0$时$y=0$.

          方程两边求导
          \[
              y'=2\mathrm{e}^{y}y'\sin x+2\mathrm{e}^{y}\cos x-7,
          \]

          即$y'=\frac{2\mathrm{e}^{y}\cos x-7}{1-2\mathrm{e}^{y}\sin x}$，

          所以$\left.y'\right|_{x=0}=-5$，$\left.\mathrm{d}y\right|_{x=0}=\left.y'\right
              |_{x=0}\,\mathrm{d}x=-5\,\mathrm{d}x$.

          \textbf{法二.} 当$x=0$时$y=0$.

          方程两边求微分，得
          \[
              \mathrm{d}y=2\mathrm{e}^{y}\sin x\,\mathrm{d}y+2\mathrm{e}^{y}\cos x\,\mathrm{d}
              x-7\,\mathrm{d}x,
          \]

          $\mathrm{d}y=\frac{2\mathrm{e}^{y}\cos x-7}{1-2\mathrm{e}^{y}\sin x}\,\mathrm{d}
              x$，$\left.\mathrm{d}y\right|_{x=0}=\left.\frac{2\mathrm{e}^{y}\cos x-7}{1-2\mathrm{e}^{y}\sin
                  x}\right|_{x=0}\,\mathrm{d}x=-5\,\mathrm{d}x$.

    \item \textbf{Solution}.$y=\frac{1}{(1-2x)(x+2)}=\frac{1}{5}\cdot\frac{1}{x+2}
              -\frac{2}{5}\cdot\frac{1}{2x-1}$.

          由公式$\left(\frac{1}{(1+x)}\right)^{(n)}=\frac{(-1)^{n}\,n!}{(1+x)^{n+1}}$，得
          \[
              \begin{aligned}
                  y^{(n)}(x)= & \frac{1}{5}\left(\frac{1}{x+2}\right)^{(n)}-\frac{2}{5}\left(\frac{1}{2x-1}\right)^{(n)}=\frac{1}{5}\frac{(-1)^n\cdot n!}{(x+2)^{n+1}}-\frac{2}{5}\frac{(-1)^n\cdot2^n\cdot n!}{(2x-1)^{n+1}} \\
                  =           & \frac{n!}{5}\left(\frac{(-1)^n}{(x+2)^{n+1}}-\frac{(-1)^n2^{n+1}}{(2x-1)^{n+1}}\right),
              \end{aligned}
          \]

          所以$y^{(n)}(0)=\frac{n!}{5}\left(\frac{(-1)^{n}}{2^{n+1}}+2^{n+1}\right)$.

    \item \textbf{Solution}.$y=\sin^{4} x-\cos^{4} x=(\sin^{2} x+\cos^{2} x)(\sin
              ^{2} x-\cos^{2} x)=-\cos 2x$,

          所以
          \[
              \begin{aligned}
                  y'=  & 2\sin 2x, \\
                  y''= & 4\cos 2x.
              \end{aligned}
          \]
\end{enumerate}

\section{综合题(每小题 6 分，共 30 分)}
\begin{enumerate}
    \setcounter{enumi}{10}
    \item \textbf{Solution}. 设切点$(x_{0},y_{0})$对应的参数为$t=t_{0}$，则$\left
              .\frac{\mathrm{d}y}{\mathrm{d}x}\right|_{t=t_0}=\left.\frac{4-2t}{2t}\right
              |_{t=t_0}=\frac{2-t_{0}}{t_{0}}$，

          曲线在$(x_{0},y_{0})$的切线方程为$y-y_{0}=\frac{2-t_{0}}{t_{0}}(x-x_{0})$，

          即$y-(4t_{0}-t_{0}^{2})=\frac{2-t_{0}}{t_{0}}(x-t_{0}^{2}-1)$化简得$y=\frac{2-t_{0}}{t_{0}}
              x+2t_{0}-\frac{2-t_{0}}{t_{0}}$，

          切线过$(-1,0)$，故代入方程得$t_{0}^{2}-t_{0}-2=0$，解得：$t_{0}=-2$（不合题意），$t
                  _{0}=1$，

          由$t_{0}=1$得$(x_{0},y_{0})=(2,3)$，故所求切线方程为$y=x+1$.

    \item \textbf{Solution}. 由题意得$f'(x)=1+(1+x)^{x}\left(\ln(1+x)+\frac{x}{1+x}
              \right)$，

          所以$f'(1)=2(\ln2+1)$.

          $f(1)=3$，$g'(3)=\frac{1}{f'(1)}=\frac{1}{2(\ln2+1)}$，

          $y'=g'(2+x^{2})\cdot 2x=2xg'(2+x^{2})$，$\left.y'\right|_{x=1}=g'(3)\cdot2=
              \frac{1}{\ln2+1}$.

    \item \textbf{Solution}.

          \[
              \begin{aligned}
                  x_{n+1}= & \frac{b_1}{1+b_1}+\frac{b_2}{(1+b_1)(1+b_2)}+\cdots+\frac{b_n}{(1+b_1)(1+b_2)\cdots(1+b_n)}+\frac{b_{n+1}}{(1+b_1)(1+b_2)\cdots(1+b_{n+1})} \\
                  =        & x_{n}+\frac{b_{n+1}}{(1+b_1)(1+b_2)\cdots(1+b_{n+1})}\geq x_{n},
              \end{aligned}
          \]

          故数列$\{x_{n}\}$单调递增.

          因$\frac{b_{k}}{(1+b_{1})(1+b_{2})\cdots(1+b_{k})}=\frac{1}{(1+b_{1})(1+b_{2})\cdots(1+b_{k-1})}
              -\frac{1}{(1+b_{1})(1+b_{2})\cdots(1+b_{k})}$，

          所以$x_{n}=1-\frac{1}{1+b_{1}}+\left(\frac{1}{1+b_{1}}-\frac{1}{(1+b_{1})(1+b_{2})}
              \right)+\cdots+\left(\frac{1}{(1+b_{1})(1+b_{2})\cdots(1+b_{n-1})}-\frac{1}{(1+b_{1})(1+b_{2})\cdots(1+b_{n})}
              \right)$，

          故$x_{n}=1-\frac{1}{(1+b_{1})(1+b_{2})\cdots(1+b_{n})}<1$.

          从而数列$\{x_{n}\}$有上界，由单调有界准则知数列得极限存在.

    \item \textbf{Solution}.

          当$x\neq0$时，$f'(x)=\frac{x(g'(x)+\mathrm{e}^{-x})-g(x)+\mathrm{e}^{-x}}{x^{2}}
              =\frac{xg'(x)+(x+1)\mathrm{e}^{-x}-g(x)}{x^{2}}$.

          当$x=0$时，
          \[
              \begin{aligned}
                  f'(0)= & \lim_{x\to0}\frac{\frac{g(x)-\mathrm{e}^{-x}}{x}-0}{x}=\lim_{x\to0}\frac{g(x)-\mathrm{e}^{-x}}{x^2}=\lim_{x\to0}\frac{g'(x)+\mathrm{e}^{-x}}{2x} \\
                  =      & \lim_{x\to0}\frac{g'(x)+1+\mathrm{e}^{-x}-1}{2x}=\lim_{x\to0}\frac{g'(x)-g'(0)+\mathrm{e}^{-x}-1}{2x}=\frac{1}{2}(g''(0)-1),
              \end{aligned}
          \]

          因
          \[
              \begin{aligned}
                  \lim_{x\to 0}f'(x)= & \lim_{x\to0}\frac{x(g'(x)+\mathrm{e}^{-x})}{x^2}=\lim_{x\to0}\left(\frac{g'(x)+\mathrm{e}^{-x}}{x}-\frac{g(x)-\mathrm{e}^{-x}}{x^2}\right)            \\
                  =                   & \lim_{x\to0}\frac{g'(x)+\mathrm{e}^{-x}}{x}=-\lim_{x\to0}\frac{g(x)-\mathrm{e}^{-x}}{x^2}=g''(0)-1-\frac{1}{2}(g''(0)-1)=\frac{1}{2}(g''(0)-1)=f'(0),
              \end{aligned}
          \]

          故$f'(x)$在$x=0$处连续.

    \item \textbf{Solution}. 设时刻$t$飞机离观察者的水平距离为$x\mathrm{km}$，望远镜视角为$\theta$弧度，则有
          \[
              \frac{\mathrm{d}x}{\mathrm{d}t}=-16\mathrm{km/min},\quad x=20\tan\theta,
          \]

          故$\theta=\arctan\frac{x}{20}$，于是$\frac{\mathrm{d}\theta}{\mathrm{d}t}=\frac{1}{20+\frac{x^2}{20}}
              \frac{\mathrm{d}x}{\mathrm{d}t}$.

          当$x=40\mathrm{km}$，$\frac{\mathrm{d}x}{\mathrm{d}t}=-16(\mathrm{km/min})$时，$\frac{\mathrm{d}\theta}{\mathrm{d}t}
              =\frac{1}{20+\frac{40^2}{20}}(-16)=-\frac{4}{25}\mathrm{rad/min}$.

          即望远镜视角改变的速率是$-\frac{4}{25}\mathrm{rad/min}$.
\end{enumerate}

\section{证明题(每小题 5 分，共 10 分)}
\begin{enumerate}
    \setcounter{enumi}{15}
    \item \textbf{Proof}.不妨设$a<x_{1}<x_{2}<\cdots<x_{n}<b$，显然$f(x)$在$[
                  x_{1},x_{n}]$上连续.

          故$f(x)$在$[x_{1},x_{n}]$上必取得最大值$M$和最小值$m$，即$m\leq f(x)\leq M,
              x\in[x_{1},x_{n}]$，从而有$m\leq f(x_{i})\leq M,i=1,2,\cdots,n$，故有
          \[
              m\leq\frac{1}{n}\left(f(x_{1})+f(x_{2})+\cdots+f(x_{n})\right)\leq M.
          \]

          由介值定理知至少存在一点$\xi\in[x_{1},x_{n}]\subset(a,b)$使得
          \[
              f(\xi)=\frac{1}{n}\left(f(x_{1})+f(x_{2})+\cdots+f(x_{n})\right).
          \]

          使得$f(\xi)=0$即$\alpha\xi=\tan\xi$，故方程$\alpha x=\tan x$在$\left(0,\frac{\pi}{2}
              \right)$内至少有一个实根.

    \item \textbf{Proof}.

          \textbf{法一.} 令$F(x)=f(x)-\frac{1}{2}x^{2}$，

          则$F(x)$在$[0,1]$上连续，在$(0,1)$内可导，且$F(0)=f(0)=0$，$F(1)=f(1)-\frac{1}{2}
              =0$，

          函数$F(x)$在$\left[0,\frac{1}{2}\right],\left[\frac{1}{2},1\right]$上分别应用Lagrange中值定理得，$\exists
              \,\eta\in\left(0,\frac{1}{2}\right),\xi\in\left(\frac{1}{2},1\right)$，使得
          \[
              \begin{gathered}
                  F\left(\frac{1}{2}\right)-F(0)=F'(\eta)\frac{1}{2},\qquad\text{即}\,2F\left(\frac{1}{2}\right)=f'(\eta)-\eta,\\
                  F(1)-F\left(\frac{1}{2}\right)=F'(\xi)\frac{1}{2},\qquad\text{即}\,-2F\left(\frac{1}{2}\right)=f'(\xi)-\xi,
              \end{gathered}
          \]

          两式相加得$f'(\eta)-\eta+f'(\xi)-\xi=0$，即$f'(\xi)+f'(\eta)=\xi+\eta$.

          \textbf{法二.} 令$G(x)=f(x)-f(1-x)$，则
          \[
              G'(x)=f'(x)+f'(1-x),\quad G(0)=f(0)-f(1)=-\frac{1}{2},\quad G\left(\frac{1}{2}
              \right)=0.
          \]

          应用Lagrange中值定理得，$\exists\,\xi\in\left(0,\frac{1}{2}\right)$，使
          \[
              G\left(\frac{1}{2}\right)-G(0)=G'(\xi)\left(\frac{1}{2}-0\right)\quad\text{即}
              \quad f'(\xi)+f'(1-\xi)=1.
          \]

          令$\eta=1-\xi\in\left(\frac{1}{2},1\right)$，即得$f'(\xi)+f'(\eta)=\xi+\eta$.
\end{enumerate}
\end{document}
