This exam consists of $10$ multiple-choice questions. Each question worths $10$ points. There is no penalty for guessing. However, students who only make correct choice but do not provide detailed and correct derivations and elaborations will not earn any point. Please write down your answer and derivations on the answer sheets.
- Let
$\displaystyle a_n=n\sin\left(\frac1n\right)+\left(-1\right)^n\frac{\cos\left(n\right)}n$
for $n=1,2,\cdots$. Which statement is true of the sequence $\left\{a_n\right\}$?- It is bounded but does not converge.
- It converges to $0$.
- It converges to a positive number.
- It diverges to infinity.
- It is unbounded and contains both arbitrarily large positive and arbitrarily large negative terms.
- Let $f:\mathbb{R}\to\mathbb{R}$ be a function with Taylor series converging to $f\left(x\right)$ for all real numbers $x$. If $f\left(0\right)=2$, $f'\left(0\right)=2$, and $f^{\left(n\right)}\left(0\right)=3$ for $n\geq2$, then $\displaystyle f\left(x\right)=$
- $3e^x+2x-1$
- $e^{3x}+2x+1$
- $e^{3x}-x+1$
- $3e^x-x-1$
- $3e^x+5x+5$
- Let $f$ be a function for which
$\displaystyle I=\int_2^4\int_x^{2x}f\left(x,y\right)dydx$
exists. Which of the following expressions is equal to $I$ with the order of integration reversed?- $\displaystyle\int_2^4\int_2^yf\left(x,y\right)dxdy+\int_4^8\int_{y/2}^4f\left(x,y\right)dxdy$
- $\displaystyle\int_x^{2x}\int_2^4f\left(x,y\right)dxdy$
- $\displaystyle\int_2^8\int_{y/2}^yf\left(x,y\right)dxdy$
- $\displaystyle\int_2^4\int_{y/2}^4f\left(x,y\right)dxdy+\int_4^8\int_2^yf\left(x,y\right)dxdy$
- $\displaystyle\int_2^4\int_2^8f\left(x,y\right)dxdy$
- Let $F\left(x\right)$ be a strictly decreasing continuously differentiable function on $\left[a,b\right]$. Then $\displaystyle\int_a^b\left|F'\left(x\right)\right|dx$ must equal to
- $\left|F\left(b\right)\right|-\left|F\left(a\right)\right|$
- $F\left(a\right)-F\left(b\right)$
- $F\left(b\right)-F\left(a\right)$
- $\left|F\left(a\right)\right|-\left|F\left(b\right)\right|$
- $F\left(-b\right)-F\left(-a\right)$
- A population grows exponentially. At $10$ years, the population is $1,000$. At $20$ years, it is $2,000$. What was the approximate population at $5$ years?
- $140$
- $250$
- $500$
- $700$
- $750$
- A rectangle with one side lying along the $x$-axis is to be inscribed in the closed region of the $xy$-plane bounded by the lines $y=0$, $y=3x$, and $y=30-2x$. What is the largest possible area of such a rectangle?
- $135/8$
- $45$
- $135/2$
- $90$
- $270$
- What is the cosine of the angle between the vectors $\begin{pmatrix}0\\-6\\8\end{pmatrix}$ and $\begin{pmatrix}1\\1\\1\end{pmatrix}$?
- $-3/4$
- $1/150$
- $3^{1/2}/15$
- $1/2$
- $2$
- The coordinates of an object moving through $\mathbb{R}^3$ are $x=a\sin\left(t\right)$, $y=b\cos\left(t\right)$, and $\displaystyle z=\frac12ct^2$, for some $t>0$, where $a$, $b$, and $c$ are constants. What is the speed of the object at time $t$?
- $\displaystyle\sqrt{a^2\sin^2\left(t\right)+b^2\cos^2\left(t\right)+\frac14c^2t^4}$
- $\displaystyle\sqrt{a^2\cos^2\left(t\right)+b^2\sin^2\left(t\right)+4c^2t^2}$
- $\sqrt{a^2+b^2+c^2t^2}$
- $\displaystyle\sqrt{a^2\cos^2\left(t\right)-b^2\sin^2\left(t\right)+4c^2t^2}$
- $\displaystyle\sqrt{a^2\cos^2\left(t\right)+b^2\sin^2\left(t\right)+c^2t^2}$
- Let $f\left(x\right)=e^{\left(x^3+x^2+x\right)}$ for any real number $x$, and let $g$ be the inverse function for $f$. What is $g'\left(e^3\right)$?
- $\displaystyle\frac1{34e^{30}}$
- $\displaystyle\frac1{6e^3}$
- $\displaystyle\frac16$
- $6$
- $6e^3$
- Which expression below is equal to
$\displaystyle\lim_{n\to\infty}\sum_{i=1}^n\frac{n}{\left(n+i\right)^2}$
- $\displaystyle\int_1^2\frac1xdx$
- $\displaystyle\int_0^1\frac1{x^2}dx$
- $\displaystyle\int_1^2\frac1{x^2}dx$
- $\displaystyle\int_{-1}^0\frac1xdx$
- $\displaystyle\int_2^3\frac1{x^2}dx$
訣竅
給定的數列可分為兩部分分析，前者為經典極限的變形，後者可利用夾擠定理處理。解法
對於極限 $\displaystyle\lim_{n\to\infty}n\sin\left(\frac1n\right)$，我們可令 $\displaystyle t=\frac1n$，那麼有
$\displaystyle\lim_{n\to\infty}n\sin\left(\frac1n\right)=\lim_{t\to0^+}\frac{\sin t}t=1$
另一方面，注意到不等式$\displaystyle0\leq\left|\left(-1\right)^n\frac{\cos\left(n\right)}n\right|\leq\frac1n$
由 $\displaystyle\lim_{n\to\infty}\frac1n=0=\lim_{n\to\infty}0$ 以及夾擠定理可知 $\displaystyle\lim_{n\to\infty}\left(-1\right)^n\frac{\cos\left(n\right)}n=0$。由以上兩部分的分析可知數列 $a_n$ 當 $n$ 趨於無窮時趨近於 $1$，故選 C.
訣竅
利用條件組織出函數本身，並利用經典函數的冪級數來改寫。解法
因為函數 $f$ 對應的泰勒級數會收斂到 $f$ 本身，故有$\displaystyle f\left(x\right)=\sum_{n=0}^{\infty}\frac{f^{\left(n\right)}\left(0\right)}{n!}x^n=2+2x+\sum_{n=2}^{\infty}\frac{3}{n!}x^n=2+2x+3\left(\sum_{n=0}^{\infty}\frac{x^n}{n!}-1-x\right)$
留意到指數函數產生的冪級數為 $\displaystyle e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$，故所求的函數 $f$ 可寫為$f\left(x\right)=2+2x+3\left(e^x-1-x\right)=3e^x-x-1$
故選 D.訣竅
利用不等式表達出交換次序後的積分範圍。解法
原積分範圍為 $\left\{\begin{aligned}&2\leq x\leq4\\&x\leq y\leq2x\end{aligned}\right.$，由第二式可以注意到$\displaystyle\frac{y}2\leq x\leq y$
再者由第一式 $x\in\left[2,4\right]$，故搭配第二式有 $y\in\left[2,8\right]$。然而當 $y\in\left[2,4\right]$ 時有 $x\in\left[2,y\right]$；而當 $y\in\left[4,8\right]$ 時有 $x\in\left[y/2,4\right]$。至此我們有$\displaystyle\begin{aligned}&\,\left\{\left(x,y\right)\in\mathbb{R}^2:~2\leq x\leq4,~x\leq y\leq2x\right\}\\=&\,\left\{\left(x,y\right)\in\mathbb{R}^2:~2\leq x\leq y,~2\leq y\leq4\right\}\cup\left\{\left(x,y\right)\in\mathbb{R}^2:~\frac{y}2\leq x\leq4,~4\leq y\leq8\right\}\end{aligned}$
由此可知重積分應改寫為$\displaystyle\int_2^4\int_x^{2x}f\left(x,y\right)dydx=\int_2^4\int_2^yf\left(x,y\right)dxdy+\int_4^8\int_{y/2}^4f\left(x,y\right)dxdy$
因此選 A.訣竅
利用單調性判斷導函數的符號，隨後使用微積分基本定理計算該定積分。解法
因為 $F$ 嚴格遞減，故 $F'\left(x\right)\leq0$，因此 $\left|F'\left(x\right)\right|=-F'\left(x\right)$，從而所求為$\displaystyle\int_a^b\left|F'\left(x\right)\right|dx=-\int_a^bF'\left(x\right)dx=-F\left(x\right)\Big|_a^b=F\left(a\right)-F\left(b\right)$
故選 B.訣竅
按題目寫出人口成長函數，並透過條件確定其中的參數。解法
設 $P\left(t\right)$ 表示在時刻 $t$ 時（單位：年）的人口數。由於人口指數成長，即有 $P\left(t\right)=ke^{at}$，其中 $k$ 與 $a$ 為待定的常數。那麼由題意有$P\left(10\right)=ke^{10a}=1000,\qquad P\left(20\right)=ke^{20a}=2000$
將兩式相除可知 $e^{10a}=2$，即有 $e^{5a}=\sqrt2$。此外也可知 $k=500$。那麼所求 $P\left(5\right)=ke^{5a}=500\sqrt2\approx500\cdot1.414=707$，故選 D.訣竅
按照題意考慮長方形的頂點位置，從而表達出面積函數進而使用配方法求極值。解法
由於兩直線 $y=3x$ 與 $y=30-2x$ 交於 $\left(6,18\right)$，故考慮變量 $a\in\left(0,6\right)$。設長方形落在 $y=3x$ 的頂點為 $\left(a,3a\right)$，那麼落在 $y=30-2x$ 的頂點為 $\displaystyle\left(15-\frac{3a}2,3a\right)$，而落在 $y=0$ 上的頂點則分別為 $\left(a,0\right)$ 與 $\displaystyle\left(15-\frac{3a}2,0\right)$。此時長方形的面積函數可表示如下$\displaystyle A\left(a\right)=3a\cdot\left(15-\frac{5a}2\right)=\frac{15}2\left(-a^2+6a\right)=-\frac{15}2\left(a-3\right)^2+\frac{135}2$
故當 $a=3$ 時有最大的長方形面積，面積為 $\displaystyle\frac{135}2$，應選 C.訣竅
運用內積即可求夾角的餘弦值。解法
設 $\vec{u}=\left(0,-6,8\right)$、$\vec{v}=\left(1,1,1\right)$，那麼向量 $\vec{u}$ 與 $\vec{v}$ 的夾角 $\theta$ 的餘弦值為$\displaystyle\cos\theta=\frac{\vec{u}\cdot\vec{v}}{\left|\vec{u}\right|\left|\vec{v}\right|}=\frac{0\cdot1+\left(-6\right)\cdot1+8\cdot1}{\sqrt{0^2+\left(-6\right)^2+8^2}\cdot\sqrt{1^2+1^2+1^2}}=\frac{2}{10\sqrt3}=\frac{\sqrt3}{15}$
故選 C.訣竅
速度大小為位置向量函數微分後的長度。解法
將物體位置表達成位置向量函數如下$\displaystyle\vec{s}\left(t\right)=\left(x\left(t\right),y\left(t\right),z\left(t\right)\right)=\left(a\sin\left(t\right),b\cos\left(t\right),\frac12ct^2\right)$
那麼速度向量函數為$\displaystyle\vec{v}\left(t\right)=\frac{d\vec{s}}{dt}=\left(\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}\right)=\left(a\cos\left(t\right),-b\sin\left(t\right),ct\right)$
故在時刻 $t$ 的速率（即速度大小）為$\displaystyle\left|\vec{v}\left(t\right)\right|=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2}=\sqrt{\left(a\cos\left(t\right)\right)^2+\left(-b\sin\left(t\right)\right)^2+\left(ct\right)^2}=\sqrt{a^2\cos^2\left(t\right)+b^2\sin^2\left(t\right)+c^2t^2}$
故選 E.訣竅
運用反函數的定義與連鎖律求解即可。解法
按照反函數的定義有 $g\left(f\left(x\right)\right)=x$，那麼使用連鎖律可知$g'\left(f\left(x\right)\right)f'\left(x\right)=1$
由給定的函數可注意到 $f\left(1\right)=e^3$，故取 $x=1$ 可得$\displaystyle g'\left(e^3\right)=\frac1{f'\left(1\right)}$
而 $f$ 的導函數為 $f'\left(x\right)=\left(3x^2+2x+1\right)e^{x^3+x^2+x}$，因此 $f'\left(1\right)=6e^3$，如此所求即為 $\displaystyle g'\left(e^3\right)=\frac1{f'\left(1\right)}=\frac1{6e^3}$，選 B.訣竅
將之視為黎曼和，故其極限可化為定積分表示。解法
該極限式可改寫為$\displaystyle\lim_{n\to\infty}\sum_{i=1}^n\frac{n}{\left(n+i\right)^2}=\lim_{n\to\infty}\frac1n\sum_{i=1}^n\frac1{\left(1+\frac{i}n\right)^2}$
此可視為函數 $\displaystyle f\left(x\right)=\frac1{x^2}$ 在 $\left[1,2\right]$ 上作 $n$ 等分割產生的黎曼和，故其極限可表示為 $\displaystyle\int_1^2\frac1{x^2}dx$，應選 C.
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