台灣聯合大學系統 112 年微積分 A2 轉學考考古題｜附 AI 解答

Question 1

Find the limit $\lim\limits_{x 	o 0^+} \left(\frac{1}{x} - \frac{1}{e^x - 1}ight)$.

Accepted Answer

$$\frac{1}{x}-\frac{1}{e^x-1}=\frac{e^x-1-x}{x(e^x-1)}.$$
當 $x\to0^+$，可用泰勒展開 $e^x=1+x+\frac{x^2}{2}+o(x^2)$，故
$$e^x-1-x=\frac{x^2}{2}+o(x^2),\qquad e^x-1=x+\frac{x^2}{2}+o(x^2).$$
因此
$$\frac{e^x-1-x}{x(e^x-1)}=\frac{\frac{x^2}{2}+o(x^2)}{x\left(x+\frac{x^2}{2}+o(x^2)\right)}\to\frac12.$$
答案：$\boxed{\frac12}$.

Question 2

Find $\displaystyle \frac{d^{2023}}{dx^{2023}}(x \sin x)$.

Accepted Answer

用萊布尼茲公式：對任意可微 $f$，有
$$\frac{d^n}{dx^n}(x f(x))=x f^{(n)}(x)+n f^{(n-1)}(x).$$
取 $f(x)=\sin x$，且 $\sin^{(m)}x=\sin\left(x+\frac{m\pi}{2}\right)$。

因此
$$\frac{d^{2023}}{dx^{2023}}(x\sin x)=x\sin\left(x+\frac{2023\pi}{2}\right)+2023\sin\left(x+\frac{2022\pi}{2}\right).$$
因 $2023\equiv3\pmod4$、$2022\equiv2\pmod4$，故
$$\sin\left(x+\frac{2023\pi}{2}\right)=\sin\left(x+\frac{3\pi}{2}\right)=-\cos x,$$
$$\sin\left(x+\frac{2022\pi}{2}\right)=\sin(x+1011\pi)=-\sin x.$$
所以
$$\boxed{\frac{d^{2023}}{dx^{2023}}(x\sin x)=-x\cos x-2023\sin x.}$$

Question 3

Evaluate the integral $\displaystyle\int_{-1}^{4} \frac{dx}{\sqrt{|x|}}$.

Accepted Answer

分段積分（注意 $\sqrt{|x|}$）：
$$\int_{-1}^{4}\frac{dx}{\sqrt{|x|}}=\int_{-1}^{0}\frac{dx}{\sqrt{-x}}+\int_{0}^{4}\frac{dx}{\sqrt{x}}.$$
第一段令 $u=-x$，得
$$\int_{-1}^{0}\frac{dx}{\sqrt{-x}}=\int_{0}^{1}u^{-1/2}\,du=2.$$
第二段
$$\int_0^4 x^{-1/2}\,dx=\left[2\sqrt{x}\right]_0^4=4.$$
合併得答案 $\boxed{6}$.

Question 4

Evaluate the integral $\displaystyle\int_0^8 \int_{\sqrt[3]{x}}^2 \frac{1}{y^4 + 1} dy dx$.

Accepted Answer

積分區域：$0\le x\le 8$ 且 $\sqrt[3]{x}\le y\le 2$。
由 $\sqrt[3]{x}\le y \iff x\le y^3$，且在 $0\le y\le2$ 時有 $y^3\le8$，故換序後
$$0\le y\le2,\qquad 0\le x\le y^3.$$

因此
$$\int_0^8\int_{\sqrt[3]{x}}^2\frac{1}{y^4+1}\,dy\,dx
=\int_0^2\int_0^{y^3}\frac{1}{y^4+1}\,dx\,dy
=\int_0^2\frac{y^3}{y^4+1}\,dy.$$
令 $u=y^4+1$，$du=4y^3dy$，得
$$\int_0^2\frac{y^3}{y^4+1}\,dy=\frac14\int_{1}^{17}\frac{1}{u}\,du=\frac14\ln 17.$$
答案：$\boxed{\frac14\ln 17}$.

Question 5

Consider the region bounded by the graphs of $y = \ln x$, $y = 0$, and $x = e$. Find the volume of the solid formed by revolving the region about the $x$-axis.

Accepted Answer

旋轉軸為 $x$ 軸，用圓盤法。
區域由 $y=\ln x$、$y=0$、$x=e$ 圍成；且 $\ln x=0\iff x=1$，故 $x\in[1,e]$。

體積
$$V=\pi\int_{1}^{e}(\ln x)^2\,dx.$$
利用公式
$$\int (\ln x)^2 dx = x\bigl((\ln x)^2-2\ln x+2\bigr)+C.$$
代入得
$$\int_{1}^{e}(\ln x)^2\,dx= e\,(1-2+2)-1\,(0-0+2)=e-2.$$
故
$$\boxed{V=\pi(e-2)}.$$

Question 6

Let $y = (\sin x)^x$, $\sin x > 0$. Find $\frac{dy}{dx}$.

Accepted Answer

令 $y=(\sin x)^x$（且 $\sin x>0$ 使對數有意義）。取對數：
$$\ln y = x\ln(\sin x).$$
微分得
$$\frac{y'}{y}=\ln(\sin x)+x\cdot\frac{\cos x}{\sin x}=\ln(\sin x)+x\cot x.$$
因此
$$\boxed{\frac{dy}{dx}=(\sin x)^x\bigl(\ln(\sin x)+x\cot x\bigr).}$$

Question 7

Assume that constants $a$ and $b$ are positive. Find equations for all horizontal asymptotes for the graph of $y = \frac{\sqrt{ax^2 + 4}}{x - b}$.

Accepted Answer

求水平漸近線看 $x	o\pm\infty$：
$$y=\frac{\sqrt{ax^2+4}}{x-b}.$$
當 $|x|	o\infty$ 時，$\sqrt{ax^2+4}=|x|\sqrt{a+4/x^2}\sim |x|\sqrt{a}$。

1) $x	o\infty$ 時 $|x|=x$，
$$y\sim \frac{x\sqrt{a}}{x-b}=\frac{\sqrt{a}}{1-b/x}	o \sqrt{a}.$$
2) $x	o-\infty$ 時 $|x|=-x$，
$$y\sim \frac{-x\sqrt{a}}{x-b}=\frac{-\sqrt{a}}{1-b/x}	o -\sqrt{a}.$$

故所有水平漸近線為 $\boxed{y=\sqrt{a}}$ 與 $\boxed{y=-\sqrt{a}}$.

Question 8

We say that the two commodities are substitute commodities if a decrease in the demand for one results in an increase in the demand for the other. Conversely, two commodities are referred to as complementary commodities if a decrease in the demand for one results in a decrease in the demand for the other as well. Suppose that the demand equations that relate the quantities demanded $x$ and $y$ to the unit prices $p$ and $q$ of the commodities A and B respectively are given by $x = f(p, q) = \frac{q^2}{q + p^2}$ and $y = g(p, q) = e^{-2q+p}$. Are A and B substitute, complementary or neither?

Accepted Answer

判斷替代品/互補品通常看交叉偏導（價格上升是否使另一方需求上升）：
- 若 $\partial x/\partial q>0$ 且 $\partial y/\partial p>0$，則為替代品；
- 若交叉偏導皆 <0，則為互補品。

已知
$$x=f(p,q)=\frac{q^2}{q+p^2},\qquad y=g(p,q)=e^{-2q+p}.$$
計算
$$\frac{\partial x}{\partial q}=\frac{2q(q+p^2)-q^2}{(q+p^2)^2}=
\frac{q^2+2qp^2}{(q+p^2)^2}=\frac{q(q+2p^2)}{(q+p^2)^2}.$$
若價格 $p,q>0$，則 $\partial x/\partial q>0$。
又
$$\frac{\partial y}{\partial p}=e^{-2q+p}>0.$$

因此 A、B 為替代品（substitutes）。

Question 9

Consider the function $f(x) = \begin{cases} x \sin \left(\frac{1}{x}ight), & x 
eq 0 \ 0, & x = 0 \end{cases}$.

(a) Show that $f$ is continuous at $x = 0$.

(b) Find $f'(x)$ for $x 
eq 0$.

(c) Use the limit definition of the derivative to show that $f$ is not differentiable at $x = 0$.

Accepted Answer

(a) 當 $x
e0$，有
$$|f(x)|=\left|x\sin\frac{1}{x}ight|\le |x|\cdot 1=|x|	o0.$$
故 $\lim_{x	o0}f(x)=0=f(0)$，所以 $f$ 在 $0$ 連續。

(b) $x
e0$ 時用乘積與鏈鎖律：
$$f'(x)=\sin\frac{1}{x}+x\cos\frac{1}{x}\cdot\left(-\frac{1}{x^2}ight)
=\sin\frac{1}{x}-\frac{1}{x}\cos\frac{1}{x}.$$

(c) 由導數定義
$$f'(0)=\lim_{h	o0}\frac{f(h)-f(0)}{h}=\lim_{h	o0}\frac{h\sin(1/h)}{h}=\lim_{h	o0}\sin\frac{1}{h}.$$
但 $\sin(1/h)$ 當 $h	o0$ 不存在極限（在 $[-1,1]$ 間振盪），故 $f'(0)$ 不存在，
因此 $f$ 在 $0$ 不可微。

Question 10

(a) Use the integral test to determine if the series $\sum\limits_{n=1}^{\infty} \frac{n}{n^2 + 4}$ converges or diverges.

(b) Find all values of $x$ for which $\sum\limits_{n=1}^{\infty} \left(\sqrt{n+1} - \sqrt{n}\right)(x - 3)^n$ converges absolutely.

Accepted Answer

(a) 用積分判別法，令 $f(x)=\dfrac{x}{x^2+4}$（$x\ge1$）正且遞減。
$$\int_1^\infty \frac{x}{x^2+4}\,dx=\frac12\int_1^\infty \frac{2x}{x^2+4}\,dx
=\frac12\left[\ln(x^2+4)\right]_1^\infty=\infty.$$
故級數 $\sum_{n=1}^\infty \frac{n}{n^2+4}$ 發散。

(b) 設 $a_n=\sqrt{n+1}-\sqrt{n}$。注意
$$a_n=\frac{1}{\sqrt{n+1}+\sqrt{n}}\sim \frac{1}{2\sqrt{n}},$$
故 $\limsup_{n\to\infty}|a_n|^{1/n}=1$，收斂半徑 $R=1$。
因此當 $|x-3|<1$ 時絕對收斂；當 $|x-3|>1$ 時發散。

檢查端點：
- $x=4$（$x-3=1$）時，級數為 $\sum (\sqrt{n+1}-\sqrt{n})$，部分和望遠鏡為 $\sqrt{N+1}-1\to\infty$，發散。
- $x=2$（$x-3=-1$）時，級數為 $\sum (-1)^n(\sqrt{n+1}-\sqrt{n})$，交錯且項遞減至 0，故條件收斂；但不絕對收斂。

所以絕對收斂的 $x$ 為 $\boxed{2<x<4}$.

Question 11

Suppose $x$ units of labor and $y$ units of capital are required to produce $$f(x, y) = 100x^{3/4}y^{1/4}$$ units of a certain product. If each unit of labor costs \$200, each unit of capital costs \$300, and a total of \$60,000 is available for production, determine how many units of labor and how many units of capital should be used to maximize production.

Accepted Answer

最大化 $f(x,y)=100x^{3/4}y^{1/4}$，限制條件為成本
$$200x+300y=60000,\qquad x,y\ge0.$$
可等價最大化 $x^{3/4}y^{1/4}$。用拉格朗日乘數：
$$
abla(x^{3/4}y^{1/4})=\lambda
abla(200x+300y).$$
計算偏導：
$$\frac{\partial}{\partial x}(x^{3/4}y^{1/4})=\frac34 x^{-1/4}y^{1/4}=200\lambda,$$
$$\frac{\partial}{\partial y}(x^{3/4}y^{1/4})=\frac14 x^{3/4}y^{-3/4}=300\lambda.$$
兩式相除得
$$\frac{\frac34 x^{-1/4}y^{1/4}}{\frac14 x^{3/4}y^{-3/4}}=\frac{200}{300}
\Rightarrow 3\frac{y}{x}=\frac23\Rightarrow x=\frac{9}{2}y.$$
代入成本：
$$200\cdot\frac{9}{2}y+300y=900y+300y=1200y=60000\Rightarrow y=50.$$
故 $x=\frac{9}{2}\cdot 50=225$。

答案：$\boxed{x=225	ext{（勞動）},\ y=50	ext{（資本）}}$.

台灣聯合大學系統 112 年度微積分 A2

這份試題整理

微積分 A2 其他年度考古題

$甲、填充題$

$乙、計算、證明題$

台灣聯合大學系統 112 年度 微積分 A2

這份試題整理

微積分 A2 其他年度考古題

甲、填充題

乙、計算、證明題

台灣聯合大學系統 112 年度微積分 A2

$甲、填充題$

$乙、計算、證明題$