- The symbol $0.\bar{9}=0.999999\dots$ means there are infinitely many $9$ after the decimal point, then $0.\bar{9}=1$.
- If $f'\left(x\right)$ is continuous on $\left[0,\infty\right)$ and $\displaystyle\lim_{x\to\infty}f\left(x\right)=0$, then $\displaystyle\int_0^{\infty}xf'\left(x\right)dx+\int_0^{\infty}f\left(x\right)dx=0$.
- There exists a power series $\displaystyle\sum_{n=0}^{\infty}a_nx^n$ which is convergent on $\left(-1,2\right)$ and divergent on $\left(-\infty,-1\right]\cup\left[2,\infty\right)$.
- Student A knew the formula: $2\sin x\cos y=\sin\left(x+y\right)+\sin\left(x-y\right)$. He has an idea that if we view $x$ as a fixed number and view $y$ as a variable, then we can differentiate this formula with respect to $y$ on both sides to get
$-2\sin x\sin y=\cos\left(x+y\right)-\cos\left(x-y\right).$
Is this process correct and is this new formula correct as well? - Consider the limit $\displaystyle\lim_{n\to\infty}\left(\frac1{n^2}+\frac2{n^2}+\cdots+\frac{n}{n^2}\right)$. Student A gives the following argument: Since $\displaystyle\lim_{n\to\infty}\frac{k}{n^2}=0$ for every $k=1,2,\dots,n$ by the Limit Sum Law, we have
$\displaystyle\lim_{n\to\infty}\left(\frac1{n^2}+\frac2{n^2}+\cdots+\frac{n}{n^2}\right)=\lim_{n\to\infty}\frac1{n^2}+\lim_{n\to\infty}\frac2{n^2}+\cdots+\lim_{n\to\infty}\frac{n}{n^2}=0+0+\cdots+0=0.$
Is this argument correct? - Let $f\left(x\right)=\begin{cases}x^2&\mbox{if}~x<1\\2x&\mbox{if}~x\geq1\end{cases}$. Find $f'\left(x\right)$. Student A gives the following argument:
- If $x<1$, then $f'\left(x\right)=2x$.
- If $x>1$, then $f'\left(x\right)=2$.
- If $x=1$, then $\displaystyle\lim_{x\to1^-}f'\left(x\right)=\lim_{x\to1^+}f'\left(x\right)=2$. So $f'\left(1\right)=2$.
- Student A has an idea to find the definite integral $\displaystyle\int_0^{\frac\pi2}\sin^2\theta d\theta$. First, from the fact $\sin^2\theta+\cos^2\theta=1$, it implies $\displaystyle\int_0^{\frac\pi2}\left(\sin^2\theta+\cos^2\theta\right)d\theta=\frac\pi2$. Next, from the fact $\displaystyle\cos\theta=\sin\left(\frac\pi2-\theta\right)$, by the Substitution Rule and the concept of dummy variable, we have $\displaystyle\int_0^{\frac\pi2}\cos^2\theta d\theta=\int_0^{\frac\pi2}\sin^2\left(\frac\pi2-\theta\right)d\theta=-\int_{\frac\pi2}^0\sin^2\varphi d\varphi=\int_0^{\frac\pi2}\sin^2\theta d\theta$. Hence $\displaystyle\int_0^{\frac\pi2}\sin^2\theta d\theta=\frac\pi4$. Is this idea correct?
- Consider one loop of the four-leaved rose curve $r=\cos2\theta$, $\displaystyle-\frac\pi4\leq\theta\leq\frac\pi4$, in polar equation. The enclosed area formula of this curve can be written as
Enclosed Area $\displaystyle=\int_{\frac\pi4}^{-\frac\pi4}y\left(\theta\right) dx\left(\theta\right)=\int_{\frac\pi4}^{-\frac\pi4}\cos2\theta\sin\theta d\left(\cos2\theta\cos\theta\right).$
- Consider an infinite sequence $\left\{a_n\right\}_{n=1}^{\infty}$, where $\displaystyle a_{n+1}=\left(-1\right)a_n+\frac1n$ and $a_1=-1$. Student A gives the following argument: Let $\displaystyle\lim_{n\to\infty}a_n=L$, then
$\displaystyle\lim_{n\to\infty}a_{n+1}=\left(-1\right)\lim_{n\to\infty}a_n+\lim_{n\to\infty}\frac1n\quad\Rightarrow\quad L=-L+0\quad\Rightarrow\quad L=0.$
Is this argument correct? - Student A computed the indefinite integral $\displaystyle\int2\sin\theta\cos\theta d\theta=2\int\sin\theta d\sin\theta=\sin^2\theta+C$.
Student B computed the indefinite integral $\displaystyle\int2\sin\theta\cos\theta d\theta=-2\int\cos\theta d\cos\theta=-\cos^2\theta+C$.
Student C computed the indefinite integral $\displaystyle\int2\sin\theta\cos\theta d\theta=\int\sin2\theta d\theta=-\frac12\cos2\theta+C$.
All students computed the indefinite integral correctly.
訣竅
明白無窮循環小數的定義即可。解法
此敘述正確。因為記號 $0.\bar{9}$ 表達了一個無窮級數，如此可計算如下$\displaystyle0.\bar{9}=\sum_{k=1}^{\infty}\frac9{10^k}=\lim_{n\to\infty}\sum_{k=1}^n\frac9{10^k}=\lim_{n\to\infty}\left(1-\frac1{10^n}\right)=1$
訣竅
注意這些條件無法保證瑕積分存在。解法
本敘述不正確。考慮函數 $\displaystyle f\left(x\right)=\frac1{x+1}$，明顯這個函數的導函數在 $\left[0,\infty\right)$ 上有定義且連續，並滿足 $\displaystyle\lim_{x\to\infty}f\left(x\right)=0$。但瑕積分 $\displaystyle\int_0^{\infty}f\left(x\right)dx=\infty$ 不存在，故該式無意義。訣竅
由冪級數具有的收斂半徑概念即可回答。解法
本敘述錯誤。若此冪級數之收斂半徑為 $R$，那麼以原點為中心可知只要 $x\in\left(-R,R\right)$ 便能使該冪級數收斂且 $x\in\left(-\infty,-R\right)\cup\left(R,\infty\right)$ 發散。但按照條件明顯知道不存在這樣的 $R$，故沒有這樣的冪級數。註：原題敘述敘述如下
There exists a power series $\displaystyle\sum_{n=0}^{\infty}a_nx^n$ which is convergent on $\left(-1,2\right)$ and divergent on $\left(\infty,-1\right]\cup\left[2,\infty\right)$.
這樣的敘述明顯有誤。訣竅
由多變數函數的偏導計算即可理解。解法
這是正確的，其中該過程中的右邊第二項也注意到使用連鎖律求導。訣竅
留意極限的四則運算定理的限制。解法
因為該求和為連加 $n$ 項，但該極限的四則運算定理是針對確定項數進行改寫的，故這個論證不正確。正確的計算方法如下：$\displaystyle\lim_{n\to\infty}\left(\frac1{n^2}+\frac2{n^2}+\cdots+\frac{n}{n^2}\right)=\lim_{n\to\infty}\frac{n\left(n+1\right)}{2n^2}=\lim_{n\to\infty}\frac{n+1}{2n}=\frac12$
訣竅
導函數未必是連續函數。解法
這個論證不正確，因為透過取左右極限不見得可確認該點的函數值存在。事實上函數 $f$ 在 $x=1$ 不連續，從而該點不可導。訣竅
這是一個很巧妙的的論證，主要的技巧是注意到兩函數具有累積平均相同的特性。解法
這是正確的論證。訣竅
留意變量之間的轉換。解法
較為嚴謹的處理這個面積，我們分為第一象限與第四象限的部分。首先注意到第一象限的部分為 $\displaystyle\theta\in\left[0,\frac\pi4\right]$，而第四象限的部分則為 $\displaystyle\left[-\frac\pi4,0\right]$。所求的面積為$\displaystyle\begin{aligned}\mbox{Enclosed Area}&=\int_{x=0}^{x=1}y_+dx+\int_{x=0}^{x=1}\left(-y_-\right)dx\\&=\int_{\theta=\frac\pi4}^{\theta=0}\left(\cos2\theta\sin\theta\right)d\left(\cos2\theta\cos\theta\right)-\int_{\theta=-\frac\pi4}^{\theta=0}\cos2\theta\sin\theta d\left(\cos2\theta\cos\theta\right)\\&=\int_{\frac\pi4}^{-\frac\pi4}\cos2\theta\sin\theta d\left(\cos2\theta\cos\theta\right)\end{aligned}$
訣竅
應先確認極限存在才能進行這樣的計算。解法
此論證不正確。因為該生未先確認此極限存在，事實上該數列之一般項能計算如下$\displaystyle a_n=\left(-1\right)^{n-1}\left(-1+\sum_{k=1}^{n-1}\frac{\left(-1\right)^k}k\right)$
然而此級數發散。訣竅
留意到三角恆等式即可。解法
這些計算皆正確。這是因為 $\sin^2\theta+\cos^2\theta=1$，故 $\sin^2\theta+C=\left(1-\cos^2\theta\right)+C=-\cos^2\theta+\left(C+1\right)$，故第一位學生與第二位學生的計算結果本質是相同的。又注意到 $\displaystyle\sin^2\theta=\frac{1-\cos2\theta}2$，因此第一位學生的計算結果可改寫如下$\displaystyle\sin^2\theta+C=\frac{1-\cos2\theta}2+C=-\frac12\cos2\theta+\left(C+\frac12\right)$
故也與第三位學生的計算結果相同。- $\displaystyle\frac{d}{dx}\left(x^2e^{\sin\left(x^3-x\right)}\right)=$ (11) .
- Suppose that $f\left(x\right)$ is continuous on $\left(0,\infty\right)$ and $\displaystyle F\left(x\right)=\int_{\frac1x}^{\ln x}\left(x-f\left(t\right)\right)dt$, then $F'\left(x\right)=$ (12) .
- Reverse the order of the following integration (integrating first with respect to $x$ and then $y$):
$\displaystyle\int_0^1\int_0^{\sqrt{x}}f\left(x,y\right)dydx+\int_1^2\int_0^{2-x}f\left(x,y\right)dydx=$ (13) .
- Find the limit $\displaystyle\lim_{x\to\infty}\frac{\ln\left(1+4e^{2x}\right)}{\sqrt{2+3x^2}}=$ (14) .
- Let $f\left(x,y\right)=x^2y+2xy-3$ and $P=\left(1,1\right)$. Find the direction (vector $v$ with unit length) in which $f\left(x,y\right)$ decreases most rapidly at $P$. $v=$ (15) .
訣竅
使用乘法的微分公式與連鎖律等計算即可。解法
直接使用微分公式計算如下$\displaystyle\begin{aligned}\frac{d}{dx}\left(x^2e^{\sin\left(x^3-x\right)}\right)&=e^{\sin\left(x^3-x\right)}\frac{d}{dx}x^2+x^2\frac{d}{dx}e^{\sin\left(x^3-x\right)}=2xe^{\sin\left(x^3-x\right)}+x^2e^{\sin\left(x^3-x\right)}\cdot\frac{d}{dx}\sin\left(x^3-x\right)\\&=2xe^{\sin\left(x^3-x\right)}+x^2e^{\sin\left(x^3-x\right)}\cos\left(x^3-x\right)\cdot\left(3x^2-1\right)\\&=\left[\left(3x^4-x^2\right)\cos\left(x^3-x\right)+2x\right]e^{\sin\left(x^3-x\right)}\end{aligned}$
訣竅
使用微積分基本定理與連鎖律計算求解即可。解法
首先可以將 $F$ 表達如下$\displaystyle F\left(x\right)=x\int_{\frac1x}^{\ln x}1dt-\int_{\frac1x}^{\ln x}f\left(t\right)dt=x\ln x-1-\int_{\frac1x}^{\ln x}f\left(t\right)dt$
兩邊求導可知$\displaystyle F'\left(x\right)=\ln x+1-f\left(\ln x\right)\cdot\frac1x+f\left(\frac1x\right)\cdot-\frac1{x^2}=1+\ln x-\frac{f\left(\ln x\right)}x-\frac{f\left(1/x\right)}{x^2}$
訣竅
利用 Fubini 定理的概念交換積分次序即可。解法
首先兩個雙重積分的積分區域分別可表達如下$\left\{\left(x,y\right)\in\mathbb{R}^2:\,0\leq x\leq1,\,0\leq y\leq\sqrt{x}\right\}\cup\left\{\left(x,y\right)\in\mathbb{R}^2:\,1\leq x\leq2,\,0\leq y\leq2-x\right\}$
容易注意到該集合的邊界為 $y=0$、$y=\sqrt{x}$ 與 $y=2-x$ 所圍成，故積分區域可改寫為$\left\{\left(x,y\right)\in\mathbb{R}^2:\,y^2\leq x\leq2-y,\,0\leq y\leq1\right\}$
因此交換積分次序後的重積分為 $\displaystyle\int_0^1\int_{y^2}^{2-y}f\left(x,y\right)dxdy$。訣竅
運用適當的改寫後使用羅必達法則求極限。解法
注意到所求之極限可計算如下$\displaystyle\begin{aligned}\lim_{x\to\infty}\frac{\ln\left(1+4e^{2x}\right)}{\sqrt{2+3x^2}}&=\lim_{x\to\infty}\frac{\ln\left(1+4e^{2x}\right)}{\displaystyle x\sqrt{3+\frac2{x^2}}}=\lim_{x\to\infty}\frac1{\displaystyle\sqrt{3+\frac2{x^2}}}\cdot\lim_{x\to\infty}\frac{\ln\left(1+4e^{2x}\right)}x\\&=\frac1{\sqrt3}\lim_{x\to\infty}\frac{8e^{2x}/\left(1+4e^{2x}\right)}1=\frac8{\sqrt3}\lim_{x\to\infty}\frac1{4+e^{-2x}}=\frac2{\sqrt3}=\frac{2\sqrt3}3\end{aligned}$
訣竅
使用方向導數的概念求解即可。解法
計算函數 $f$ 的梯度為$\nabla f\left(x,y\right)=\left(f_x\left(x,y\right),f_y\left(x,y\right)\right)=\left(2xy+2y,x^2+2x\right)$
那麼在 $P$ 減少最快的方向為 $-\nabla f\left(P\right)=-\left(4,3\right)$，因此所求為 $\displaystyle v=\frac{-\left(4,3\right)}5=\left(-0.8,0.6\right)$。- Consider the ellipse $E:13x^2-10xy+13y^2=72$.
- (4%) Find an equation of the tangent line to the ellipse $E$ at the point $\displaystyle P\left(0,\frac6{13}\sqrt{26}\right)$.
- (8%) Write a polar equation of the ellipse $E$. Find the area of the ellipse $E$ by integrating this polar equation. You may need to use the substitution $\varphi=2\theta$ and $\displaystyle t=\tan\left(\frac\varphi2\right)$, where $\theta$ is the polar angle.
- (8%) Find the area of the ellipse $E$ by changing of variables in multiple integrals. You need to compute the Jacobian of the transformation.
- (10%) Find the area of the ellipse $E$ by the Lagrange multiplier method. Here is the idea to realize the method: Notice that the center of the ellipse $E$ is $O\left(0,0\right)$. We can find the maximum distance and minimum distance from the point $P\left(x,y\right)$ on the ellipse $E$ to the center $O\left(0,0\right)$. They will correspond to the length of the semi-major $a$ and the semi-minor axis $b$, respectively. The area of the ellipse is $ab\pi$.
- 使用隱函數微分可得
$\displaystyle26x-10y-10x\frac{dy}{dx}+26y\frac{dy}{dx}=0$
取 $x=0$ 與 $\displaystyle y=\frac6{13}\sqrt{26}$ 可得 $\displaystyle\left.\frac{dy}{dx}\right|_P=\frac5{13}$。因此所求的切線方程式為$\displaystyle y-\frac6{13}\sqrt{26}=\frac5{13}\left(x-0\right)$
或寫為 $5x-13y+6\sqrt{26}=0$。 - 由極座標變換，令 $\left\{\begin{aligned}&x=r\cos\theta\\&y=r\sin\theta\end{aligned}\right.$，代入可得
$13r^2\cos^2\theta-10r^2\cos\theta\sin\theta+13r^2\sin^2\theta=72$
即 $\displaystyle r^2=\frac{72}{13-5\sin2\theta}$，或寫為 $\displaystyle r=\frac{6\sqrt2}{\sqrt{13-5\sin2\theta}}$。那麼所求的面積為$\displaystyle A=\frac12\int_0^{\pi}r^2\left(\theta\right)d\theta=36\int_0^{2\pi}\frac{d\theta}{13-5\sin2\theta}=72\int_{-\frac\pi2}^{\frac\pi2}\frac{d\theta}{13-5\sin2\theta}$
按照提示，我們令 $t=\tan\theta$，那麼有- 當 $\displaystyle x\to\left(\frac\pi2\right)^-$ 有 $u\to\infty$；
- 當 $\displaystyle x\to\left(-\frac\pi2\right)^+$ 有 $\displaystyle u\to-\infty$；
- 整理有 $\displaystyle\sin2\theta=2\sin\theta\cos\theta=\frac{2t}{1+t^2}$，此外求導可知 $\displaystyle d\theta=\frac{dt}{1+t^2}$。
$\displaystyle\begin{aligned}A&=72\int_{-\infty}^{\infty}\frac{\displaystyle\frac{dt}{1+t^2}}{\displaystyle13-5\frac{2t}{1+t^2}}=72\int_{-\infty}^{\infty}\frac{dt}{13-10t+13t^2}\\&=\frac{72}{13}\int_{-\infty}^{\infty}\frac{dt}{\displaystyle\left(t-\frac5{13}\right)^2+\left(\frac{12}{13}\right)^2}=\left.\frac{72}{13}\cdot\frac{13}{12}\tan^{-1}\frac{13t-5}{12}\right|_{-\infty}^{\infty}=6\pi\end{aligned}$
- 設 $D=\left\{\left(x,y\right)\in\mathbb{R}^2:\,13x^2-10xy+13y^2\leq72\right\}$。將給定的算式配方法可知
$\displaystyle\left(x-\frac{5y}{13}\right)^2+\left(\frac{12y}{13}\right)^2=\frac{72}{13}$
故令 $\displaystyle u=x-\frac{5y}{13}$ 而 $\displaystyle v=\frac{12y}{13}$，那麼 $\left(u,v\right)$ 為以 $\left(0,0\right)$ 為中心而半徑為 $\displaystyle\frac{6\sqrt2}{\sqrt{13}}$ 的圓，即$D'=\left\{\left(u,v\right)\in\mathbb{R}^2:\,u^2+v^2\leq72/13\right\}$
而此變換所對應的 Jacobian 行列式能計算如下$\displaystyle\left|J\right|=\Big|\frac{\partial\left(x,y\right)}{\partial\left(u,v\right)}\Big|=\Big|\frac{\partial\left(u,v\right)}{\partial\left(x,y\right)}\Big|^{-1}=\Big|\begin{vmatrix}\displaystyle\frac{\partial u}{\partial x}&\displaystyle\frac{\partial u}{\partial y}\\[3mm]\displaystyle\frac{\partial v}{\partial x}&\displaystyle\frac{\partial v}{\partial y}\end{vmatrix}\Big|^{-1}=\Big|\begin{vmatrix}1&\displaystyle-\frac5{13}\\[3mm]0&\displaystyle\frac{12}{13}\end{vmatrix}\Big|^{-1}=\frac{13}{12}$
故所求之面積為$\displaystyle A=\iint_D1dA=\iint_{D'}\left|J\right|dA'=\frac{13}{12}\left|D'\right|=\frac{13}{12}\cdot\frac{72}{13}\pi=6\pi$
- 設與原點的距離平方函數為 $d\left(x,y\right)=x^2+y^2$，那麼設定拉格朗日乘子函數為
$F\left(x,y,\lambda\right)=x^2+y^2+\lambda\left(13x^2-10xy+13y^2-72\right)$
據此解聯立方程組$\left\{\begin{aligned}&F_x\left(x,y,\lambda\right)=2x+\lambda\left(26x-10y\right)=0\\&F_y\left(x,y,\lambda\right)=2y+\lambda\left(-10x+26y\right)=0\\&F_{\lambda}\left(x,y,\lambda\right)=13x^2-10xy+13y^2-72=0\end{aligned}\right.$
由於 $\left(0,0\right)$ 明顯不為第三式之解，故前兩式 $\left\{\begin{aligned}&\left(2+26\lambda\right)x-10\lambda y=0\\&-10\lambda x+\left(2+26\lambda\right)y=0\end{aligned}\right.$ 的行列之值為 $0$，即 $\begin{vmatrix}2+26\lambda&-10\lambda\\-10\lambda&2+26\lambda\end{vmatrix}=0$，即 $\left(2+26\lambda\right)^2=100\lambda^2$，故 $2+26\lambda=\pm10\lambda$，可解得 $\displaystyle\lambda=-\frac18$ 或 $\displaystyle\lambda=-\frac1{18}$。- 當 $\displaystyle\lambda=-\frac18$ 時，第一式給出 $x=y$，故第三式可寫為 $16x^2=72$，因此 $\displaystyle x=\pm\frac{3\sqrt2}2$，故得 $\displaystyle\left(x,y\right)=\pm\left(\frac{3\sqrt2}2,\frac{3\sqrt2}2\right)$；
- 當 $\displaystyle\lambda=-\frac1{18}$ 時，第一式給出 $x+y=0$，故第三式可寫為 $36x^2=72$，故 $x=\pm\sqrt2$，從而 $\left(x,y\right)=\pm\left(\sqrt2,-\sqrt2\right)$。
- (6%) Find the Maclaurin series for $f\left(x\right)=\sin^2x$. Hint: Half-angle formula $\displaystyle\sin^2x=\frac{1-\cos2x}2$.
- (4%) Find $f^{\left(106\right)}\left(0\right)$ and $f^{\left(2017\right)}\left(0\right)$.
- 利用半角公式與餘弦函數的泰勒展開式可知
$\displaystyle\sin^2x=\frac{1-\cos2x}2=\frac{\displaystyle1-\sum_{n=0}^{\infty}\frac{\left(-1\right)^n}{\left(2n\right)!}\left(2x\right)^{2n}}2=-\frac12\sum_{n=1}^{\infty}\frac{\left(-1\right)^n}{\left(2n\right)!}4^nx^{2n}=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}2^{2n-1}}{\left(2n\right)!}x^{2n}$
- 由於 $f^{\left(n\right)}\left(0\right)$ 為 $f$ 所對應的泰勒級數中的 $n$ 次項係數乘以 $n!$，故取 $n=53$ 並注意到缺項可知
$\displaystyle f^{\left(106\right)}\left(0\right)=106!\cdot\frac{2^{106-1}}{106!}=2^{105}$, $f^{\left(2017\right)}\left(0\right)=0$。
- (10%) Solve the initial-value problem:
$\left(x^2+1\right)y'+\left(x+1\right)^2y=x^2+1$, $y\left(0\right)=1$.
訣竅
依序利用隱函數微分求導並由點斜式獲得切線方程；利用極座標變換獲得極座標方程式並由此計算面積；亦可使用多變量的變數變換獲得面積；最後利用長軸與短軸的概念求面積。解法
訣竅
按照常見函數的泰勒展開式求解；並由泰勒級數計算高階導函數之值。解法
訣竅
運用積分因子法求解。解法
首先同除以 $x^2+1$ 可得$\displaystyle y'+\frac{\left(x+1\right)^2}{x^2+1}y=1$
那麼積分因子為$\displaystyle\exp\left[\int\frac{\left(x+1\right)^2}{x^2+1}dx\right]=\exp\left[\int\left(1+\frac{2x}{x^2+1}\right)dx\right]=\exp\left(x+\ln\left(x^2+1\right)\right)=\left(x^2+1\right)e^x$
故兩邊同乘以積分因子可將微分方程改寫如下$\displaystyle\left[e^x\left(x^2+1\right)y\left(x\right)\right]'=e^x\left(x^2+1\right)y'+e^x\left(x+1\right)^2=e^x\left(x^2+1\right)$
故同取積分有$\displaystyle e^x\left(x^2+1\right)y\left(x\right)-1=\int_0^xe^t\left(t^2+1\right)dt=\left(t^2e^t-2te^t+3e^t\right)\Big|_0^x=x^2e^x-2xe^x+3e^x-3$
因此所求函數為$\displaystyle y\left(x\right)=\frac{x^2-2x+3-2e^{-x}}{x^2+1}$
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