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.
- ($5\%$) For speeds between $40$ and $65$ kilometers per hour, a truck gets $480/x$ kilometers for per liter of diesel gasoline when driven at a constant speed of $x$ kilometers per hour. Diesel gasoline costs $\$2.23$ per liter, and the driver is paid $\$15.10$ per hour. What is the most economical constant speed between $40$ and $65$ kilometers per hour at which to drive the truck?
- $52.5$ kilometers per hour
- $57$ kilometers per hour
- $65$ kilometers per hour
- None of the above
- ($5\%$) What is the value of $\displaystyle\int_1^2x^3\ln\left(\sqrt{x}\right)dx$?
- $0.713$
- $0.865$
- $0.918$
- None of the above
- ($5\%$) Which can be the solution form of $A\left(t\right)$ if
$A'\left(t\right)=A^2\left(t\right)+A\left(t\right)-1$, and $A\left(0\right)=0$.
- $A\left(t\right)=a+be^{ct}+d$
- $A\left(t\right)=\exp\left(a+be^{ct}\right)+d$
- $A\left(t\right)=\left(a+be^{ct}\right)^{-1}+d$
- None of the above
- ($5\%$) Consider a cash-or-nothing call option which pays the option holder an amount $K$ at the maturity $T$ only if the asset price at $T$ is larger than $K$. The pricing formula for this option is as follows.
$c=Ke^{-rT}N\left(d\right)$,
where$\displaystyle d=\frac{\displaystyle\ln\left(\frac{S}K\right)+\left(r-\frac{\sigma^2}2\right)T}{\sigma\sqrt{T}}$,
and $S$ is the current asset price, $r$ is a constant risk-free interest rate, $\sigma$ is the volatility of the asset price, and $N\left(\cdot\right)$ is the cumulative distribution function of the standard normal distribution defined as$\displaystyle N\left(d\right)=\int_{-\infty}^dn\left(x\right)dx=\int_{-\infty}^d\frac1{\sqrt{2\pi}}e^{-\frac{x^2}2}dx$.
Derive the Delta (i.e., $\displaystyle\frac{\partial c}{\partial S}$) of this option.- $\displaystyle\frac1{\sqrt{2\pi}\sigma\sqrt{T}}e^{-\frac{\left(d+\sigma\sqrt{T}\right)^2}2}$
- $\displaystyle\frac{K}{\sqrt{2\pi}S\sigma\sqrt{T}}e^{-\frac{d^2}2}$
- $0$
- None of the above
- ($5\%$) For the pricing formula of the cash-or-nothing call option in Question 4, derive the vega (i.e., $\displaystyle\frac{\partial c}{\partial\sigma}$) of this option.
- $\displaystyle\frac{K}{\sqrt{2\pi}}e^{-rT+\frac{d^2}2}\left(\sqrt{T}+\frac{d}{\sigma}\right)$
- $\displaystyle\frac{K}{\sqrt{2\pi}}e^{-rT+\frac{d^2}2}\left(\sqrt{T}-\frac{d}{\sigma}\right)$
- $\displaystyle\frac{-K}{\sqrt{2\pi}}e^{-rT-\frac{d^2}2}\left(\sqrt{T}+\frac{d}\sigma\right)$
- None of the above
- ($5\%$) What is the value of $\displaystyle\lim_{x\to0}\left(\frac1{1-\cos x}-\frac2{x^2}\right)$?
- $-\infty$
- $1/3$
- $1/6$
- $\infty$
- ($5\%$) Determine whether the following statement is true or false.
“Suppose $a_n$ is positive for all $n$. If $\displaystyle\sum_{n=1}^{\infty}a_n$ converges then $\displaystyle\sum_{n=1}^{\infty}\sqrt{a_na_{n+1}}$ also converges.”
- False
- True
- Uncertain (need more information)
- None of the above
- ($5\%$) A community is laid out as a rectangular grid in relation to two main streets that intersect at the city center. Each point in the community has coordinates $\left(x,y\right)$ in this grid, for $-10\leq x\leq10$, $-8\leq y\leq8$ with $x$ and $y$ measured in kilometers. Suppose the value of the land located at the point $\left(x,y\right)$ is $V$ thousand dollars, where
$V\left(x,y\right)=\left(250+17x\right)e^{-0.01x-0.05y}$.
Estimate the value of the block of land occupying the rectangular region $1\leq x\leq3$, $0\leq y\leq2$.- $759$ thousand dollars
- $859$ thousand dollars
- $959$ thousand dollars
- None of the above
- If a positive variable $\lambda$ follows a gamma distribution, i.e., $\lambda\sim\text{Gamma}\left(\alpha,\beta\right)$, where $\alpha$ and $\beta$ are positive real numbers. Consequently, the probability density function for $\lambda$ is as follows.
$\displaystyle f\left(\lambda\right)=\beta^{\alpha}\frac1{\Gamma\left(\alpha\right)}\lambda^{\alpha-1}e^{-\beta\lambda}$,
where $\Gamma\left(Z\right)$ is the gamma function and defined as$\displaystyle\Gamma\left(Z\right)=\int_0^{\infty}e^{-t}t^{Z-1}dt$,
if $Z$ is a complex number with a positive real part. What is the $\text{var}\left(\lambda\right)$ given $\alpha=1$ and $\beta=2$? ($5\%$)- $1/4$
- $1/2$
- $1$
- None of the above
- ($5\%$) The death rate and birth rate of many animal and plant species fluctuate periodically with the seasons. The population $P\left(t\right)$ of such a species at time $t$ changes at a rate that may be modeled by a differential equation of the form
$\displaystyle\frac{dP}{dt}=\left(1+\cos2\pi t\right)P$.
Given the initial population $P\left(0\right)$ be $1000$, what is the population $P\left(1.5\right)$?- $1482$
- $2482$
- $3482$
- None of the above
訣竅
依據題意列出成本函數，由此求導計算出最小值發生的速度。解法
設總距離為 $L$，而當時速為 $x$ 公里時，其成本為$\displaystyle C\left(x\right)=2.23\cdot\frac{Lx}{480}+15.10\cdot\frac{L}x=L\left(\frac{2.23x}{480}+\frac{15.1}x\right)$
為了找出 $C$ 的最小值，我們應解方程式 $C'\left(x\right)=0$，即$\displaystyle C'\left(x\right)=L\left(\frac{2.23}{480}-\frac{15.1}{x^2}\right)=0$
如此有 $\displaystyle x^2=\frac{1510\cdot480}{223}\approx3250$，因此 $x\approx57$。容易注意到 $\displaystyle C''\left(x\right)=\frac{30.2L}{x^3}>0$，故 $C$ 在 $x\approx57$ 時達到絕對最小值，故選(b)。訣竅
運用分部積分法求解即可。解法
運用分部積分法可知$\displaystyle\begin{aligned}\int_1^2x^3\ln\left(\sqrt{x}\right)dx&=\frac18\int_1^2\ln xdx^4=\left.\frac18x^4\ln x\right|_1^2-\frac18\int_1^2x^3dx=2\ln2-\left.\frac{x^4}{32}\right|_1^2\\&=2\ln2-\frac{15}{32}\approx2\cdot0.69314718056-0.46875=0.91754436112\end{aligned}$
故選(c)。訣竅
使用分離變量法積分求解即可。解法
移項整理有$\displaystyle\frac1{\sqrt5}\left(\frac1{\displaystyle A+\frac{1-\sqrt5}2}-\frac1{\displaystyle A+\frac{1+\sqrt5}2}\right)dA=\frac{dA}{\displaystyle\left(A+\frac12\right)^2-\left(\frac{\sqrt5}2\right)^2}=\frac{dA}{A^2+A-1}=dt$
同取積分有$\displaystyle\frac1{\sqrt5}\ln\left|\frac{2A+1-\sqrt5}{2A+1+\sqrt5}\right|=t+C$
由初值條件 $\displaystyle C=\frac1{\sqrt5}\ln\left(\frac{\sqrt5-1}{\sqrt5+1}\right)=\frac2{\sqrt5}\ln\frac{\sqrt5-1}2$。如此整理可解得$\displaystyle A\left(t\right)=-\frac{1+\sqrt5}2+4\sqrt5\left[4-\left(\sqrt5-1\right)^2e^{\sqrt5t}\right]^{-1}=\left[\frac1{\sqrt5}-\frac{\left(\sqrt5-1\right)^2}{4\sqrt5}e^{\sqrt5t}\right]^{-1}-\frac{1+\sqrt5}2$
此與(c)之形式相同，其中 $\displaystyle a=\frac1{\sqrt5}$、$\displaystyle b=-\frac{\left(\sqrt5-1\right)^2}{4\sqrt5}$、$c=\sqrt5$、$\displaystyle d=-\frac{1+\sqrt5}2$。訣竅
使用連鎖律求導即可。解法
由連鎖律可知$\displaystyle\frac{\partial c}{\partial S}=Ke^{-rT}\frac{\partial}{\partial S}N\left(d\right)=Ke^{-rT}\cdot n\left(d\right)\cdot\frac{\partial d}{\partial S}=Ke^{-rT}n\left(d\right)\cdot\frac1{\sigma S\sqrt{T}}=\frac{Ke^{-rT}}{\sqrt{2\pi}S\sigma\sqrt{T}}e^{-\frac{d^2}2}$
前三個選項皆與之不同，故選(d)。訣竅
運用連鎖律求導即可。解法
直接求導有$\displaystyle\begin{aligned}\frac{\partial c}{\partial\sigma}&=Ke^{-rT}\frac{\partial}{\partial\sigma}N\left(d\right)=Ke^{-rT}\cdot n\left(d\right)\cdot\frac{\partial d}{\partial\sigma}\\&=Ke^{-rT}n\left(d\right)\cdot\left(-\sqrt{T}-\frac{\displaystyle\ln\left(\frac{S}K\right)+\left(r-\frac{\sigma^2}2\right)T}{\sigma^2\sqrt{T}}\right)\\&=Ke^{-rT}\cdot\frac1{\sqrt{2\pi}}e^{-\frac{d^2}2}\left(-\sqrt{T}-\frac{d}{\sigma}\right)=\frac{-K}{\sqrt{2\pi}}e^{-rT-\frac{d^2}2}\left(\sqrt{T}+\frac{d}{\sigma}\right)\end{aligned}$
故選(c)。訣竅
通分後使用羅必達法則即可。解法
通分後使用羅必達法則可得$\displaystyle\begin{aligned}\lim_{x\to0}\left(\frac1{1-\cos x}-\frac2{x^2}\right)&=\lim_{x\to0}\frac{x^2-2+2\cos x}{x^2-x^2\cos x}=\lim_{x\to0}\frac{2x-2\sin x}{2x-2x\cos x+x^2\sin x}\\&=\lim_{x\to0}\frac{2-2\cos x}{2-2\cos x+4x\sin x+x^2\cos x}=\lim_{x\to0}\frac{2\sin x}{6\sin x+6x\cos x-x^2\sin x}\\&=\lim_{x\to0}\frac{2\cos x}{12\cos x-8x\sin x-x^2\cos x}=\frac16\end{aligned}$
故選(c)。訣竅
有技巧地使用柯西不等式即可獲得證明。解法
由柯西不等式可知$\displaystyle\left(\sum_{n=1}^k\sqrt{a_na_{n+1}}\right)^2\leq\sum_{n=1}^ka_n\cdot\sum_{n=1}^ka_{n+1}\leq\sum_{n=1}^{\infty}a_n\cdot\sum_{n=1}^{\infty}a_{n+1}<\infty$
故 $\displaystyle\sum_{n=1}^k\sqrt{a_na_{n+1}}$ 有均勻上界，因此該級數隨 $k$ 遞增有上界，故其極限存在，因此 $\displaystyle\sum_{n=1}^{\infty}\sqrt{a_na_{n+1}}<\infty$。因此命題成立，選(b)。【註】 原題有些許錯字，原敘述如下：($5\%$) Determine whether the following statement is true or false.
“Suppose $a_n$ is positive for all $n$. If $\displaystyle\sum_{i=1}^{\infty}a_n$ converges then $\displaystyle\sum_{i=1}^{\infty}\sqrt{a_na_{n+1}}$ also converges.”
其標號於本題中無意義。訣竅
按其意義作重積分計算即可。解法
直接作重積分$\displaystyle\begin{aligned}\text{Value}&=\int_1^3\int_0^2V\left(x,y\right)dxdy=\left(\int_0^2e^{-0.05y}dy\right)\left[\int_1^3\left(250+17x\right)e^{-0.01x}dx\right]\\&=-20e^{-0.05y}\Big|_0^2\cdot-100\int_1^3\left(250+17x\right)de^{-0.01x}\\&=-2000\left(1-e^{-0.1}\right)\left[\left(250+17x\right)e^{-0.01x}\Big|_1^3-17\int_1^3e^{-0.01x}dx\right]\\&=-2000\left(1-e^{-0.1}\right)\left(301e^{-0.03}-267e^{-0.01}+1700e^{-0.01x}\Big|_1^3\right)\\&=-2000\left(1-e^{-0.1}\right)\left(2001e^{-0.03}-1967e^{-0.01}\right)\approx1059\end{aligned}$
故選(d)。訣竅
回憶變異數與期望值等定義，並使用分部積分法計算其值。解法
當 $\alpha=1$ 且 $\beta=2$ 時，其機率密度函數為$\displaystyle f\left(\lambda\right)=2e^{-2\lambda}$
那麼分別使用分部積分法計算有$\displaystyle E\left[\lambda\right]=\int_0^{\infty}\lambda f\left(\lambda\right)d\lambda=2\int_0^{\infty}\lambda e^{-2\lambda}d\lambda=-\lambda e^{-2\lambda}\Big|_0^{\infty}+\int_0^{\infty}e^{-2\lambda}d\lambda=\left.-\frac{e^{-2\lambda}}2\right|_0^{\infty}=\frac12$
以及$\displaystyle E\left[\lambda^2\right]=\int_0^{\infty}\lambda^2f\left(\lambda\right)d\lambda=2\int_0^{\infty}\lambda^2e^{-2\lambda}d\lambda=-\lambda^2e^{-2\lambda}\Big|_0^{\infty}+2\int_0^{\infty}\lambda e^{-2\lambda}d\lambda=\frac12$
因此變異數為$\displaystyle\text{var}\left(\lambda\right)=E\left[\lambda^2\right]-\left(E\left[\lambda\right]\right)^2=\frac12-\left(\frac12\right)^2=\frac14$
故選(a)。訣竅
運用分離變量法解微分方程即可。解法
移項有$\displaystyle\frac{dP}P=\left(1+\cos2\pi t\right)dt$
同取不定積分可得$\displaystyle\ln\left(P\left(t\right)\right)=t+\frac{\sin2\pi t}{2\pi}+C$
又因 $P\left(0\right)=1000$，故有 $C=\ln1000=3\ln10$。因此有$\displaystyle P\left(t\right)=1000\exp\left(t+\frac{\sin\left(2\pi t\right)}{2\pi}\right)$
取 $t=1.5$，可得$\displaystyle P\left(1.5\right)=1000\exp\left(1.5\right)\approx1000\cdot2.718^{1.5}\approx4481$
- ($10\%$) Find $dy/dx$ given that $y^3+y^2-5y-x^2=-4$.
- ($10\%$) Evaluate
$\displaystyle\int_0^1\text{arcsin}\left(x\right)dx$.
- ($10\%$) Find the interval of convergence for $\displaystyle\sum_{k=1}^{\infty}\frac{2^kx^k}k$.
- 當 $\displaystyle x=-\frac12$ 時級數寫為 $\displaystyle\sum_{k=1}^{\infty}\frac{\left(-1\right)^k}k$，由交錯級數審歛法可知收斂；
- 當 $\displaystyle x=\frac12$ 時級數寫為 $\displaystyle\sum_{k=1}^{\infty}\frac1k$，此時調和級數可知其發散。
- ($10\%$) Find the surface area of the portion of the plane $x+y+z=1$ that lies in the first octant (where $x\geq0$, $y\geq0$, $z\geq0$).
- ($10\%$) Suppose $X$ is a metric space and $K$ is a subset of $X$. Describe and explain the definition of $K$ being a compact set.
訣竅
運用隱函數微分求解即可。解法
使用隱函數微分，直接對 $x$ 求導有$\displaystyle3y^2\frac{dy}{dx}+2y\frac{dy}{dx}-5\frac{dy}{dx}-2x=0$
整理可解得$\displaystyle\frac{dy}{dx}=\frac{2x}{3y^2+2y-5}$
訣竅
運用分部積分法直接計算即可。解法
使用分部積分法可知$\displaystyle\int_0^1\text{arcsin}\left(x\right)dx=x\text{arcsin}\left(x\right)\Big|_0^1-\int_0^1x\cdot\frac1{\sqrt{1-x^2}}dx=\frac\pi2+\sqrt{1-x^2}\Big|_0^1=\frac\pi2-1$
訣竅
運用比值審歛法的概念計算收斂半徑，隨後決定端點的收斂性。解法
使用比值審歛法可知$\displaystyle R=\lim_{k\to\infty}\left|\frac{a_k}{a_{k+1}}\right|=\lim_{k\to\infty}\left|\frac{2^k}{k}\cdot\frac{k+1}{2^{k+1}}\right|=\frac12$
故確定在 $\displaystyle x\in\left(-\frac12,\frac12\right)$ 上收斂。現檢查端點如下訣竅
由三角形面積公式計算；亦可運用曲面積分計算之。解法一
容易注意到平面 $x+y+z=1$ 與三軸的交點分別為 $\left(1,0,0\right)$、$\left(0,1,0\right)$ 與 $\left(0,0,1\right)$，容易注意到該三角形為邊長為 $\sqrt2$ 的正三角形，故其面積為 $\displaystyle\frac{\sqrt3}4\cdot\sqrt2^2=\frac{\sqrt3}2$。解法二
容易注意到平面 $x+y+z=1$ 與三軸的交點分別為 $A\left(1,0,0\right)$、$B\left(0,1,0\right)$ 與 $C\left(0,0,1\right)$，那麼三角形面積為$\displaystyle\text{Area}=\frac12\left|\vec{AB}\times\vec{AC}\right|=\frac12\left|\left(-1,1,0\right)\times\left(-1,0,1\right)\right|=\frac12\left|\left(1,1,1\right)\right|=\frac{\sqrt3}2$
解法三
設 $D=\left\{\left(x,y\right)\in\mathbb{R}^2:\,x+y\leq1,x,y\geq0\right\}$，而平面可寫為 $z=f\left(x,y\right):=1-x-y$，故曲面面積可表達為$\displaystyle\text{Area}=\iint_D\sqrt{f_x^2\left(x,y\right)+f_y^2\left(x,y\right)+1}dA=\iint_D\sqrt3dA=\frac{\sqrt3}2$
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