For Questions 1 to 10, select a correct answer for each question and mark the letter (A), (B), (C), or (D) on your answer card. ※注意：請用 2B 鉛筆作答於答案卡，並先詳閱答案卡上之「畫記說明」。
- ($5\%$) Suppose that the relationship between the tax rate $t$ on imported shoes and the total sales $S$ (in millions of dollars) is given by $S\left(t\right)=8-15\sqrt[3]t$. Find the tax rate $t$ that maximizes revenue for the government.
- $3.6\%$
- $6.4\%$
- $9.2\%$
- None of the above
- ($5\%$) The capital value of an asset (such as an oil well) that produces a continuous stream of income is the sum of the present values of all future earnings from the asset. Therefore, the capital value of an asset that produces income of $r\left(t\right)$ dollars at time $t$ (discounted at a continuous interest rate $i$) is
$\displaystyle\text{Capital Value}=\int_0^Tr\left(t\right)e^{-it}dt$,
where $T$ is the expected life (in years) of the asset. For an oil well that produces income of $r\left(t\right)=10000t^2$ for the following 10 years and the interest rate $i=5\%$, what is the capital value of this oil well?- $2,302,028$ dollars
- $99,346,934$ dollars
- $26,563,254$ dollars
- None of the above
- ($5\%$) What is the solution form of $A\left(t\right)$ if
$A'\left(t\right)=A^2\left(t\right)-1$, and $A\left(0\right)=0$.
- $A\left(t\right)=a+be^{ct}$
- $A\left(t\right)=\left(a+be^{ct}\right)^{-1}$
- $A\left(t\right)=\left(a+be^{ct}\right)^{-1}+d$
- None of the above
- ($5\%$) Solve the function $A\left(t\right)$ in Question 3. What is the value of $A\left(1\right)$?
- $1.2976$
- $-0.2976$
- $-0.7616$
- None of the above
- ($5\%$) A company's production is given by the Cobb-Douglas function $P=60L^{2/3}K^{1/3}$, where $L$ and $K$ are the number of units of labor and capital. Each unit of labor costs $\$25$ and each unit of capital costs $\$100$. The company wants to produce exactly $1920$ units. Find $L^*$ and $K^*$ that meet the production requirements at the lowest cost. What is the value of $L^*+K^*$?
- $64$
- $72$
- $80$
- None of the above
- What is the value of $\displaystyle\lim_{n\to\infty}\left(1+\frac2{n^2}\right)^n$? ($5\%$)
- $2$
- $1$
- $0$
- None of the above
- What is the value of $\displaystyle\sum_{n=2}^{\infty}\ln\left(1-\frac1{n^2}\right)$? ($5\%$)
- $\infty$
- $-\ln1$
- $-\ln2$
- $-\infty$
- What is the value of $\displaystyle\int_0^1\int_0^2xe^{xy}dxdy$? ($5\%$)
- $e^2-3$
- $\displaystyle\frac34e^4+\frac14$
- $e^2+2$
- None of the above
- If a variable $x$ is exponentially distributed with the following probability density function,
$f\left(x\right)=\begin{cases}2e^{-2x}&\text{if}~x\geq0\\0&\text{if}~x<0\end{cases},$
what is the expected value of $x$? ($5\%$)- $e^2$
- $2$
- $0.5$
- None of the above
- ($5\%$) Find the maximum value of $f\left(x,y\right)=\cos x+\cos y$ subject to the constraint condition $\displaystyle y-x=\frac\pi4$.
- $\displaystyle\cos\frac\pi4+\cos\frac\pi2$
- $\displaystyle\cos\frac{-\pi}4+\cos0$
- $\displaystyle\cos\frac{-3\pi}{16}+\cos\frac\pi{16}$
- None of the above
訣竅
按照題意先設定出稅收函數，並隨後求導計算極值可能發生的位置，最終確定最大值。解法
設稅收函數為 $R\left(t\right)$，那麼按照題意可知 $R\left(t\right)=tS\left(t\right)=8t-15t^{4/3}$，求導有$\displaystyle R'\left(t\right)=8-20t^{1/3}$
那麼解方程式 $R'\left(t\right)=0$，可得 $\displaystyle t=\frac8{125}$。再者可以看出當 $t\in\left(0,8/125\right)$ 有 $R'\left(t\right)>0$，而當 $t\in\left(8/125,\infty\right)$ 有 $R'\left(t\right)<0$，故在 $\displaystyle t=\frac8{125}=6.4\%$ 時能使 $R\left(t\right)$ 達到最大值，應選(b)。訣竅
使用分部積分法計算並透過自然指數的近似值估算之。解法
直接計算有$\displaystyle\begin{aligned}\text{Capital Value}&=\int_0^{10}10000t^2e^{-0.05t}dt=-200000t^2e^{-0.05t}\Big|_0^{10}+400000\int_0^{10}te^{-0.05t}dt\\&=-20000000e^{-0.5}-8000000te^{-0.05t}\Big|_0^{10}+8000000\int_0^{10}e^{-0.05t}dt\\&=−100000000e^{-0.5}-160000000e^{-0.05}\Big|_0^{10}=160000000−260000000e^{-0.5}\end{aligned}$
因 $e\approx2.718281828459$，故$\text{Capital Value}\approx160000000-260000000\cdot2.718281828459^{-1/2}\approx2302028.47471$
故選(a)。訣竅
運用分離變量法求解。解法
移項整理有$\displaystyle\frac12\left(\frac1{A-1}-\frac1{A+1}\right)dA=\frac{dA}{A^2-1}=dt$
同取積分有$\displaystyle\frac{\ln\left|A-1\right|-\ln\left|A+1\right|}2=t+C$
依初始條件有 $C=0$，整理有$\displaystyle A\left(t\right)=\frac{1-e^{2t}}{1+e^{2t}}=-1+\frac2{1+e^{2t}}=-1+2\left(1+e^{2t}\right)^{-1}=-1+\left(\frac12+\frac12e^{2t}\right)^{-1}$
此符合(c)的形式，其中 $\displaystyle a=b=\frac12$、$c=2$ 及 $d=-1$。訣竅
利用前一題的結論計算函數值，其中利用自然指數的近似值作計算。解法
取 $t=1$ 代入有$\displaystyle A\left(1\right)=-1+2\left(1+e^2\right)^{-1}\approx=-1+2\left(1+2.718^2\right)^{-1}\approx-0.7616$
故選(c)。訣竅
按照題意設定成本函數，隨後可利用算術幾何不等式求極值；亦可使用拉格朗日乘子法求解。解法一
設定成本函數為 $C=25L+100K$，而限制條件為 $60L^{2/3}K^{1/3}=1920$，即 $L^{2/3}K^{1/3}=32$，那麼由算術幾何不等式可知$\displaystyle\frac{C}3=\frac{12.5L+12.5L+100K}3\geq\sqrt[3]{\frac{625}4\cdot100L^2K}=800$
因此 $C\geq2400$，而此時等號成立條件為 $12.5L=100K$，從而有 $L^*=64$、$K^*=8$，故 $L^*+K^*=72$，應選(b)。解法二
設定成本函數為 $C=25L+100K$，而限制條件為 $60L^{2/3}K^{1/3}=1920$，即 $L^{2/3}K^{1/3}=32$，據此設定拉格朗日乘子函數為$F\left(L,K,\lambda\right)=25L+100K+\lambda\left(L^2K-32^3\right)=0$
依此解下列的聯立方程組$\left\{\begin{aligned}&F_L\left(L,K,\lambda\right)=25+2\lambda LK=0\\&F_K\left(L,K,\lambda\right)=100+\lambda L^2=0\\&F_{\lambda}\left(L,K,\lambda\right)=L^2K-32^3=0\end{aligned}\right.$
明顯可知 $\lambda,L,K$ 三數皆不為零，故有 $\displaystyle-\frac{2LK}{25}=\frac1{\lambda}=-\frac{L^2}{100}$。因此消去 $L$ 便得 $L=8K$，從而 $L^2K=32^3$ 便化為 $L^3=2^{18}$，因此 $L=64$，而 $K=8$，從而 $L^*+K^*=80$，應選(b)。訣竅
運用換底公式與羅必達法則即可。解法
使用換底公式與羅必達法則有$\displaystyle\begin{aligned}\lim_{n\to\infty}\left(1+\frac2{n^2}\right)^n&=\exp\left(\lim_{n\to\infty}n\ln\left(1+\frac2{n^2}\right)\right)=\exp\left[\lim_{n\to\infty}\frac{\displaystyle\ln\left(1+\frac2{n^2}\right)}{1/n}\right]\\&=\exp\left[\lim_{n\to\infty}\frac{\displaystyle\left(1+\frac2{n^2}\right)^{-1}\cdot-\frac4{n^3}}{\displaystyle-\frac1{n^2}}\right]=\exp\left[\lim_{n\to\infty}\frac{4n}{n^2+2}\right]=\exp\left(0\right)=1\end{aligned}$
故選(b)。訣竅
展開後使用分項對消來求部分和的極限。解法
直接展開計算如下$\displaystyle\sum_{n=2}^{\infty}\ln\left(1-\frac1{n^2}\right)=\lim_{k\to\infty}\sum_{n=2}^k\left[\ln\left(1-\frac1n\right)+\ln\left(1+\frac1n\right)\right]=\lim_{k\to\infty}\left[\ln\frac12+\ln\frac{k+1}k\right]=\ln\frac12=-\ln2$
應選(c)。訣竅
交換積分次序後直接計算之。解法
交換積分次序後直接計算可知$\displaystyle\int_0^1\int_0^2xe^{xy}dxdy=\int_0^2\int_0^1xe^{xy}dydx=\int_0^2e^{xy}\Big|_{y=0}^{y=1}dx=\int_0^2\left(e^x-1\right)dx=\left(e^x-x\right)\Big|_0^2=e^2-3$
故選(a)。訣竅
按照連續隨機變數的期望值的定義計算求解，其中需使用分部積分法。解法
隨機變數 $x$ 的期望值為$\displaystyle E\left[x\right]=\int_{\mathbb{R}}xf\left(x\right)=\int_0^{\infty}2xe^{-2x}dx=-xe^{-2x}\Big|_0^{\infty}+\int_0^{\infty}e^{-2x}dx=\left.-\frac{e^{-2x}}2\right|_0^{\infty}=\frac12$
選(c)。訣竅
代入消去後視為單變數函數求其極值；亦可使用拉格朗日乘子法求解。解法一
代入消去變數 $y$ 後有$\displaystyle g\left(x\right)=f\left(x,x+\frac\pi4\right)=\cos x+\cos\left(x+\frac\pi4\right)=\frac{\left(2+\sqrt2\right)\cos x-\sqrt2\sin x}2$
為了求極值，我們解方程式 $g'\left(x\right)=0$，即$\displaystyle\frac{-\left(2+\sqrt2\right)\sin x-\sqrt2\cos x}2=0$
即 $\left(1+\sqrt2\right)\sin x+\cos x=0$，或說 $\tan x=1-\sqrt2$，至終有 $x=\tan^{-1}\left(1-\sqrt2\right)+k\pi$，其中 $k$ 為整數。又留意到 $\displaystyle\tan^{-1}\left(1-\sqrt2\right)=-\frac\pi8$，因此 $\displaystyle x=\frac{7\pi}8$ 或 $\displaystyle\frac{15\pi}8$ 與它們的同界角，故最大最小值分別為$\displaystyle\cos\frac\pi8+\cos\frac{15\pi}8$, $\displaystyle\cos\frac{7\pi}8+\cos\frac{9\pi}8$
故選(d)。解法二
根據題意設定拉格朗日乘子函數如下$F\left(x,y,\lambda\right)=\cos x+\cos y+\lambda\left(4y-4x-\pi\right)$
據此解下列的聯立方程組$\left\{\begin{aligned}&F_x\left(x,y,\lambda\right)=-\sin x-4\lambda=0\\&F_y\left(x,y,\lambda\right)=-\sin y+4\lambda=0\\&F_{\lambda}\left(x,y,\lambda\right)=4y-4x-\pi=0\end{aligned}\right.$
前兩式相加有 $\sin x=\sin\left(-y\right)$，此表明 $x=-y+2k\pi$ 或 $x-y=\left(2k+1\right)\pi$，其中 $k$ 為整數。對於第二種情形會與第三式衝突，故僅有 $x=-y+2k\pi$，此時與第三式解聯立得 $\displaystyle\left(x,y\right)=\left(\frac{\left(8k-1\right)\pi}8,\frac{\left(8k+1\right)\pi}8\right)$。代回原式可知最大最小值分別為$\displaystyle\cos\frac\pi8+\cos\frac{15\pi}8$, $\displaystyle\cos\frac{7\pi}8+\cos\frac{9\pi}8$
故選(d)。- ($10\%$) Evaluate
$\displaystyle\lim_{h\to0^+}h^h$.
- ($10\%$) Compute
$\displaystyle\int_1^2\frac{dx}{x^3\left(9-x^3\right)^{1/3}}$
- ($10\%$) Calculate $\displaystyle\int_{-\pi}^{\pi}\cos mx\cos nxdx$ and $\displaystyle\int_{-\pi}^{\pi}\sin mx\sin nxdx$, where $m$ and $n$ are different positive integers.
- ($10\%$) Maximize the objective function $f\left(x_1,x_2\right)=x_1x_2$ subject to the constraint $x_1+4x_2=16$.
- ($10\%$) What is an open set? Show that the interval
$\left(0,1\right)\equiv\left\{x\in\mathbb{R}:\,0<x<1\right\}$
is an open set. - 若 $x\in\left(0,0.5\right)$，那麼 $\delta=x$，則 $\left(x-\delta,x+\delta\right)=\left(0,2x\right)$，此時 $2x<1$，故有 $\left(x-\delta,x+\delta\right)\subset\left(0,1\right)$。
- 若 $x=0.5$，那麼 $\delta=0.5$，則 $\left(x-\delta,x+\delta\right)=\left(0,1\right)$。
- 若 $x\in\left(0.5,1\right)$，那麼 $\delta=1-x$，則 $\left(x-\delta,x+\delta\right)=\left(2x-1,1\right)$，此時 $0<1-2x<1$，故有 $\left(x-\delta,x+\delta\right)\subset\left(0,1\right)$。
訣竅
運用換底公式，隨後使用羅必達法則。解法
運用換底公式並使用羅必達法則可知$\displaystyle\lim_{h\to0^+}h^h=\lim_{h\to0^+}\exp\left(h\ln h\right)=\exp\left(\lim_{h\to0^+}\frac{\ln h}{1/h}\right)=\exp\left(\lim_{h\to0^+}\frac{1/h}{-1/h^2}\right)=\exp\left(0\right)=1$
訣竅
使出相當技巧性的觀察，隨後由變數代換的概念直接計算。解法
將其中一個 $x$ 移動出三次方根外後便有$\displaystyle\begin{aligned}\int_1^2\frac{dx}{x^3\left(9-x^3\right)^{1/3}}&=\int_1^2x^{-4}\left(9x^{-3}-1\right)^{-1/3}dx=-\frac13\int_1^2\left(9x^{-3}-1\right)^{-1/3}dx^{-3}\\&=-\frac1{27}\int_1^2\left(9x^{-3}-1\right)^{-1/3}d\left(9x^{-3}-1\right)=\left.-\frac1{27}\cdot\frac32\left(9x^{-3}-1\right)^{2/3}\right|_1^2=-\frac1{18}\left(\frac14-4\right)=\frac5{24}\end{aligned}$
訣竅
運用積化和差公式改寫後直接積分。解法
由和角公式可知$\cos\left(mx-nx\right)=\cos mx\cos nx+\sin mx\sin nx$, $\cos\left(mx+nx\right)=\cos mx\cos nx-\sin mx\sin nx$
兩式相加減可得$\displaystyle\cos mx\cos nx=\frac{\cos\left(mx-nx\right)+\cos\left(mx+nx\right)}2$, $\displaystyle\sin mx\sin nx=\frac{\cos\left(mx-nx\right)-\cos\left(mx+nx\right)}2$
據此同取積分便有$\displaystyle\begin{aligned}&\int_{-\pi}^{\pi}\cos mx\cos nxdx=\frac12\int_{-\pi}^{\pi}\left[\cos\left(mx-nx\right)+\cos\left(mx+nx\right)\right]dx=\left.\frac12\left[\frac{\sin\left(mx-nx\right)}{m-n}+\frac{\sin\left(mx+nx\right)}{m+n}\right]\right|_{-\pi}^{\pi}=0\\&\int_{-\pi}^{\pi}\sin mx\sin nxdx=\frac12\int_{-\pi}^{\pi}\left[\cos\left(mx-nx\right)-\cos\left(mx+nx\right)\right]dx=\left.\frac12\left[\frac{\sin\left(mx-nx\right)}{m-n}-\frac{\sin\left(mx+nx\right)}{m+n}\right]\right|_{-\pi}^{\pi}=0\end{aligned}$
訣竅
運用代入消去與配方法即可；亦可初等不等式估算之；亦可使用拉格朗日乘子法計算求解。解法一
由於 $x_1=16-4x_2$，故給定的函數為$f\left(x_1,x_2\right)=f\left(16-4x_2,x_2\right)=x_2\left(16-4x_2\right)=-4x_2^2+16x_2=-4\left(x_2-2\right)^2+16\leq16$
而等號成立時 $x_2=2$，而 $x_1=8$。解法二
由於當 $x_1$ 與 $x_2$ 異號時有 $x_1x_2<0$，故為使其有最大值，我們僅需考慮 $x_1$ 與 $x_2$ 同號的情形。又搭配限制條件可知 $x_1$ 與 $x_2$ 皆為正號。據此考慮算術幾何不等式便有$\displaystyle8=\frac{x_1+4x_2}2\geq\sqrt{4x_1x_2}=2\sqrt{x_1x_2}$
因此有 $x_1x_2\leq16$，而等號成立條件為 $x_1=8$、$x_2=2$。因此最大值為 $8$。解法三
設定拉格朗日乘子函數如下$F\left(x_1,x_2,\lambda\right)=x_1x_2+\lambda\left(x_2+4x_2-16\right)$
據此解下列的聯立方程組$\left\{\begin{aligned}&F_{x_1}\left(x_1,x_2,\lambda\right)=x_2+\lambda=0\\&F_{x_2}\left(x_1,x_2,\lambda\right)=x_1+4\lambda=0\\&F_{\lambda}\left(x_1,x_2,\lambda\right)=x_1+4x_2-16=0\end{aligned}\right.$
第一式乘以 $4$ 與第二式相加並搭配第三式可知 $8\lambda+16=0$，故 $\lambda=-2$，從而 $x_2=2$、$x_1=8$，此時 $x_1x_2=16$，此即最大值。訣竅
回憶開集的定義並循此建構符合該定義的數。解法
我們稱集合 $E$ 為 $\mathbb{R}$ 中的開集，其定義為「若 $x\in E$，則存在一正數 $\varepsilon$ 使得 $\left(x-\delta,x+\delta\right)\subseteq E$ 成立。」若 $x\in\left(0,1\right)$，那麼我們取 $\delta=\min\left\{x,1-x\right\}$，則有
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