國立臺灣大學一百零五學年度研究所碩士班入學考試試題：微積分(C)

For Questions 1 to 5, show your calculations in detail on the answer sheet.

1. ($10\%$) Find $\displaystyle\frac{dF\left(x\right)}{dx}$ if $F\left(x\right)=\sqrt{x^2+1}$.
訣竅
運用連鎖律求解即可。
解法

直接求導可得

$\displaystyle F'\left(x\right)=\frac12\left(x^2+1\right)^{-1/2}\cdot2x=\frac{x}{\sqrt{x^2+1}}$



2. ($10\%$) If $f\left(x,y\right)=x^3+x^2y^3-2y^2$, find $f_x\left(2,1\right)$ and $f_y\left(2,1\right)$.
訣竅
直接計算偏導函數並取特定點演算之。
解法
首先計算一階偏導函數有
$f_x\left(x,y\right)=3x^2+2xy^3$,　　$f_y\left(x,y\right)=3x^2y^2-4y$
故取 $\left(x,y\right)=\left(2,1\right)$ 代入可得偏導函數值為 $f_x\left(2,1\right)=16$、$f_y\left(2,1\right)=8$。


3. ($10\%$) Find $\displaystyle\frac{d}{dx}\left[\ln\left|\cos\left(x\right)\right|\right]$.
訣竅
使用連鎖律求導即可。
解法

直接使用連鎖律求導有

$\displaystyle\frac{d}{dx}\left[\ln\left|\cos\left(x\right)\right|\right]=\frac1{\left|\cos x\right|}\cdot\frac{\left|\cos x\right|}{\cos x}\cdot-\sin x=-\tan x$



4. ($10\%$) If $w=x^3y^4$ and $x=t^2-2$, $y=5t-3$, find $\displaystyle\frac{dw}{dt}$ at $t=1$.
訣竅
運用多變數函數的連鎖律計算即可；亦可利用代入後使用連鎖律求導。
解法一

由多變數函數的連鎖律有

$\displaystyle\frac{dw}{dt}=\frac{\partial w}{\partial x}\frac{dx}{dt}+\frac{\partial w}{\partial y}\frac{dy}{dt}=3x^2y^4\cdot2t+4x^3y^3\cdot5$

當 $t=1$ 時有 $x=-1$、$y=2$，故

$\displaystyle\frac{dw}{dt}\left(1\right)=\left[3\cdot\left(-1\right)^2\cdot2^4\right]\cdot2+\left[4\cdot\left(-1\right)^3\cdot2^3\right]\cdot5=-64$

解法二
運用代入消去有
$w\left(t\right)=\left(t^2-2\right)^3\left(5t-3\right)^4$
故求導有
$w'\left(t\right)=3\left(t^2-2\right)^2\left(5t-3\right)^4\cdot2t+4\left(t^2-2\right)^3\left(5t-3\right)^3\cdot5$
取 $t=1$ 代入有 $w'\left(1\right)=-64$。


5. ($10\%$) Find $\displaystyle\int\ln xdx$.
訣竅
使用分部積分法計算即可。
解法

取不定積分可知

$\displaystyle\int\ln xdx=x\ln x-\int x\cdot\frac1xdx=x\ln x-x+C$

For Questions 6 to 15, select a correct answer for each question and mark the letter (A), (B), (C), or (D) on your answer card.

6. ($5\%$) What is the value of $x+2^2x^2+\cdots+n^2x^n$ if $x=3$ and $n=12$?
   1. $106022478$
   2. $108070187$
   3. $108080187$
   4. None of the above
訣竅
取特定之值並使用級數和的方式表達之，從而求解。
解法
當 $x\neq1$ 時考慮等比級數有
$\displaystyle\frac{1-x^{n+1}}{1-x}=\sum_{k=0}^nx^k$
求導並乘以 $x$ 有
$\displaystyle\frac{x-\left(n+1\right)x^{n+1}+nx^{n+2}}{\left(1-x\right)^2}=\frac{-\left(n+1\right)x^n}{1-x}+\frac{1-x^{n+1}}{\left(1-x\right)^2}=\sum_{k=1}^nkx^k$
接著再次求導並乘以 $x$ 可得
$\displaystyle\frac{x-\left(n+1\right)^2x^{n+1}+n\left(n+2\right)x^{n+2}}{\left(1-x\right)^2}+\frac{2\left[x^2-\left(n+1\right)x^{n+2}+nx^{n+3}\right]}{\left(1-x\right)^3}=\sum_{k=1}^nk^2x^k$
取 $x=3$ 與 $n=12$ 代入便有
$\displaystyle\begin{aligned}\sum_{k=1}^{12}k^2\cdot3^k&=\frac{3-13^2\cdot3^{13}+12\cdot14\cdot3^{14}}{\left(1-3\right)^2}+\frac{2\left[9-13\cdot3^{14}+12\cdot3^{15}\right]}{\left(1-3\right)^3}\\&=\frac{3+335\cdot3^{13}}4-\frac{9+69\cdot3^{13}}4=\frac{-3+133\cdot3^{13}}2=106022478\end{aligned}$
故選(A)。


7. ($5\%$) For the summation equation in Question 6, what is its value if $x=0.5$ and $n\to\infty$?
   1. $5$
   2. $6$
   3. $7$
   4. None of the above
訣竅
可延續前一小題的過程來進一步使用；亦可透過無窮等比級數的概念來求解。
解法一
沿用前一小題的過程取 $x=0.5$ 有
$\displaystyle\sum_{k=1}^n\frac{k^2}{2^k}=6-\frac{n-1}{2^{n-2}}-\frac{n^2+n+1}{2^{n-1}}+\frac{n^2+2n}{2^n}$
取 $n$ 趨於無窮得 $\displaystyle\sum_{k=1}^{\infty}\frac{k^2}{2^k}=6$，故選(B)。
解法二
由無窮等比級數可知
$\displaystyle\frac1{1-x}=\sum_{k=0}^{\infty}x^k$
求導後乘以 $x$ 有
$\displaystyle\frac{x}{\left(1-x\right)^2}=\sum_{k=1}^{\infty}kx^k$
再進行一次相同的操作有
$\displaystyle\frac{x}{\left(1-x\right)^2}+\frac{2x^2}{\left(1-x\right)^3}=\sum_{k=1}^{\infty}k^2x^k$
取 $x=0.5$ 代入便有 $\displaystyle\sum_{k=1}^{\infty}k^2x^k=6$，選(B)。


8. ($5\%$) Determine whether the following statement is true or false.

   “Suppose $a_n$ is a series. If $\displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=c$, then $\displaystyle\lim_{n\to\infty}\sqrt[n]{\left|a_n\right|}=c$.”

   1. False
   2. True
   3. Uncertain (need more information)
   4. None of the above
訣竅
由極限式的定義與夾擠定理證明，此外該結論表明比值審歛法可使用的場合則根式審歛法也必定可以使用。
解法
假若 $c\neq0$，則由極限式的定義可知，對任何給定的 $c>\varepsilon>0$，我們有正整數 $N$ 使得「不等式 $n>N$ 蘊含 $\displaystyle\left|\left|\frac{a_{n+1}}{a_n}\right|-c\right|<\varepsilon$。」據此當 $n>N$ 時便有
$\displaystyle\left|a_n\right|=\left|a_n\right|=\frac{\left|a_n\right|}{\left|a_{n-1}\right|}\cdot\frac{\left|a_{n-1}\right|}{\left|a_{n-2}\right|}\cdots\frac{\left|a_{N+1}\right|}{\left|a_N\right|}\cdot\left|a_N\right|<\left(c+\varepsilon\right)^{n-N}\left|a_N\right|$
類似地也有 $\left|a_n\right|>\left(c-\varepsilon\right)^{n-L}\left|a_N\right|$，因此 $\displaystyle\lim_{n\to\infty}\left|a_n\right|^{1/n}=c$，其中我們使用了極限式 $\displaystyle\lim_{n\to\infty}\left|a_N\right|^{1/n}=1$。而對於 $c=0$ 的情形則使用 $\displaystyle0<\left|a_n\right|^{1/n}<\varepsilon^{\left(n-N\right)/n}\left|a_N\right|^{1/n}$ 則可獲得結論。因此本敘述正確，選(B)。


9. ($5\%$) The Gaussian Quadrature method is a numerical way to generate the approximation of the definite integral of any function $f\left(x\right)$ as follows.

   $\displaystyle\int_{-1}^1f\left(x\right)dx\approx\sum_{i=1}^nw_if\left(x_i\right)$.

   Both the abscissas, $x_i$ and the weighting coefficients, $w_i$, are free parameters and they can be solved based on the following $2n$ equations

   $\displaystyle\int_{-1}^1x^ldx=\sum_{i=1}^nw_ix_i^l$ for $l=0,1,\dots,2n-1$.

   If $n=2$, what is the value of $w_1+w_2+x_1+x_2$?
   1. $0$
   2. $1$
   3. $2$
   4. None of the above
訣竅
按照題意解聯立方程組即可。
解法
由於 $n=2$，我們取 $l=0,1,2,3$ 則有下列四式
$\displaystyle\left\{\begin{aligned}&2=\int_{-1}^1x^0dx=w_1+w_2\\&0=\int_{-1}^1x^1dx=w_1x_1+w_2x_2\\&\frac23=\int_{-1}^1x^2dx=w_1x_1^2+w_2x_2^2\\&0=\int_{-1}^1x^3dx=w_1x_1^3+w_2x_2^3\end{aligned}\right.$
由第一式可知 $w_2=2-w_1$，如此第二式至第四式可寫為
$\left\{\begin{aligned}&w_1\left(x_1-x_2\right)+2x_2=w_1x_1+\left(2-w_1\right)x_2=0\\&w_1\left(x_1^2-x_2^2\right)+2x_2^2=w_1x_1^2+\left(2-w_1\right)x_2^2=\frac23\\&w_1\left(x_1^3-x_2^3\right)+2x_2^3=w_1x_1^3+\left(2-w_1\right)x_2^3=0\end{aligned}\right.$
那麼有 $w_1\left(x_1-x_2\right)=-2x_2$。分別同乘以 $x_1+x_2$ 與 $x_1^2+x_1x_2+x_2^2$ 可得
$\displaystyle -2\left(x_1+x_2\right)x_2+2x_2^2=\frac23$,　$-2\left(x_1^2+x_1x_2+x_2^2\right)x_2+2x_2^3=0$.
即 $\displaystyle x_1x_2=-\frac13$ 與 $\left(x_1+x_2\right)x_1x_2=0$。容易知道 $x_1+x_2=0$ 且由 $\displaystyle x_1x_2=-\frac13$ 得 $\displaystyle\left(x_1,x_2\right)=\left(\frac1{\sqrt3},-\frac1{\sqrt3}\right)$ 或 $\displaystyle\left(x_1,x_2\right)=\left(-\frac1{\sqrt3},\frac1{\sqrt3}\right)$。若 $\displaystyle\left(x_1,x_2\right)=\left(\frac1{\sqrt3},-\frac1{\sqrt3}\right)$，那麼可知
$\displaystyle w_1=-\frac{2x_2}{x_1-x_2}=1$
從而 $w_2=2-w_1=1$。類似地，對於 $\displaystyle\left(x_1,x_2\right)=\left(-\frac1{\sqrt3},\frac1{\sqrt3}\right)$ 也可得到 $w_1=w_2=1$。綜上可知
$\displaystyle w_1+w_2+x_1+x_2=2$
故選(C)。


10. ($5\%$) Given the solutions of $w_1$, $w_2$, $x_1$, and $x_2$ in Question $9$, use the Gaussian Quadrature method (with $n=2$) to approximate $\displaystyle\int_{-1}^1x\ln\left(x^2\right)dx$. What is the absolute value of the approximation error?
    1. $\ln\left(1/3\right)$
    2. $\sqrt{1/3}$
    3. $0$
    4. None of the above
訣竅
沿用前一題的結論並直接計算該積分便得精確的誤差量。
解法
由前一小題的結論可得近似值為
$\displaystyle\int_{-1}^1x\ln\left(x^2\right)dx\approx-\frac1{\sqrt3}\ln\frac13+\frac1{\sqrt3}\ln\frac13=0$
而給定的定積分由對稱性也知為 $0$，故誤差為兩者之差仍為零，故選(C)。


11. ($5\%$) Suppose that a random variable $S$ follows a lognormal distribution and its probability density function is

    $\displaystyle\frac1{S\sigma\sqrt{2\pi}}e^{-\frac{\left(\ln S-\mu\right)^2}{2\sigma^2}}$,

    where $\mu$ and $\sigma$ represent the mean and standard deviation of the lognormal distribution, respectively. Define $N\left(d\right)$ as the cumulative probability of the standard normal distribution from $-\infty$ to a constant $d$. What is the expected value of $S$ conditional on $S\geq K$, where $K$ is a nonnegative constant?
    1. $\displaystyle e^{\mu+0.5\sigma^2}N\left(\frac{\mu+\sigma^2-\ln K}{\sigma}\right)$
    2. $\displaystyle e^{\mu+\sigma^2}N\left(\frac{\mu+0.5\sigma^2-\ln K}{\sigma}\right)$
    3. $\displaystyle e^{\mu+0.5\sigma^2}N\left(\frac{\mu+0.5\sigma^2-\ln K}\sigma\right)$
    4. None of the above
訣竅
先按照常態對數分配的定義寫出對應的條件期望值，如此運用標準常態分配及其累積分配函數表達之。
解法
運用條件期望值的定義有
$\displaystyle E\left[S\mid S\geq K\right]=\frac{\displaystyle\int_K^{\infty}S\cdot\frac1{S\sigma\sqrt{2\pi}}e^{-\frac{\left(\ln S-\mu\right)^2}{2\sigma^2}}dS}{\displaystyle P\left[S\mid S\geq K\right]}$
令 $\displaystyle t=\frac{\ln S-\mu}\sigma$，那麼
• 當 $S=K$ 時有 $\displaystyle t=\frac{\ln K-\mu}\sigma$；
• 當 $S\to\infty$ 時有 $t\to\infty$；
• 求導有 $\displaystyle dt=\frac{dS}{S\sigma}$，或寫為 $dS=\sigma e^{\mu+\sigma t}dt$。
如此所求的分子可改寫並計算如下
$\displaystyle\begin{aligned}\int_K^{\infty}S\cdot\frac1{S\sigma\sqrt{2\pi}}e^{-\frac{\left(\ln S-\mu\right)^2}{2\sigma^2}}dS&=\frac1{\sqrt{2\pi}}\int_{\frac{\ln K-\mu}\sigma}^{\infty}e^{-t^2/2}\cdot e^{\mu+\sigma t}dt=\frac{e^{\mu+0.5\sigma^2}}{\sqrt{2\pi}}\int_{\frac{\ln K-\mu}\sigma}^{\infty}e^{-\left(t-\sigma\right)^2/2}dt\\&=\frac{e^{\mu+0.5\sigma^2}}{\sqrt{2\pi}}\int_{\frac{\ln K-\mu-\sigma^2}{\sigma}}^{\infty}e^{-s^2/2}ds=\frac{e^{\mu+0.5\sigma^2}}{\sqrt{2\pi}}\int_{-\infty}^{\frac{\mu+\sigma^2-\ln K}{\sigma}}e^{-s^2/2}ds\\&=e^{\mu+\sigma^2}N\left(\frac{\mu+\sigma^2-\ln K}{\sigma}\right)\end{aligned}$
又
$\displaystyle P\left[S\mid S\geq K\right]=\int_K^{\infty}\frac1{S\sigma\sqrt{2\pi}}e^{-\frac{\left(\ln K-\mu\right)^2}{2\sigma^2}}dS=\frac1{\sqrt{2\pi}}\int_{\frac{\ln K-\mu}{\sigma}}^{\infty}e^{-t^2/2}dt=1-N\left(\frac{\ln K-\mu}{\sigma}\right)$
故所求為
$\displaystyle E\left[S\mid S\geq K\right]=\frac{e^{\mu+\sigma^2}N\left(\left(\mu+\sigma^2-\ln K\right)/\sigma\right)}{1-N\left(\left(\ln K-\mu\right)/\sigma\right)}$
故選(D)。


12. ($5\%$) A company grows in value by $10\%$ each year, and also gains $20\%$ of a growing market estimated at $100e^{0.1t}$ million dollars, where $t$ is the number of years that the company has been in business. Therefore, the value $y\left(t\right)$ of the company (in millions of dollars) satisfies

    $y'=0.1y+20e^{0.1t}$ and $y\left(0\right)=1$ (million dollars).

    What is the value of this company after $5$ years?
    1. $173$ million dollars
    2. $167$ million dollars
    3. $180$ million dollars
    4. None of the above
訣竅
運用積分因子法求解微分方程。
解法
同乘以 $e^{-0.1t}$ 可得
$\left(y\left(t\right)e^{-0.1t}\right)'=e^{-0.1t}y'\left(t\right)-0.1e^{-0.1t}y\left(t\right)=20$
同取積分並透過初值條件有
$e^{-0.1t}y\left(t\right)-1=20t$
因此所求函數為 $y\left(t\right)=e^{0.1t}\left(1+20t\right)$，故 $y\left(5\right)=101e^{0.5}\approx166.5$，故選(B)。


13. ($5\%$) Let $f\left(x,y\right)=2x^2+y^2+2xy+4x+2y+7$. Find the minimum value of the function $f\left(x,y\right)$ subject to the constraint $4x^2+4xy=1$.
    1. $4$
    2. $5$
    3. $6$
    4. None of the above
訣竅
將原問題化為單變數極值問題，隨後求導計算極值候選點；亦可使用拉格朗日乘子法求解。
解法一
由於限制條件明顯沒有 $x=0$ 的情形，故 $\displaystyle y=\frac{1-4x^2}{4x}=-x+\frac1{4x}$。因此代入給定的雙變數函數中有
$\displaystyle\begin{aligned}g\left(x\right)&=f\left(x,-x+\frac1{4x}\right)=\frac{15}2+\left(-x+\frac1{4x}\right)^2+4x+2\left(-x+\frac1{4x}\right)\\&=x^2+2x+7+\frac1{2x}+\frac1{16x^2}\end{aligned}$
求導有 $\displaystyle g'\left(x\right)=2x+2-\frac1{2x^2}-\frac1{8x^3}$。解方程式 $g'\left(x\right)=0$ 即 $16x^4+16x^3-4x-1=0$，因式分解為 $\left(2x-1\right)\left(2x+1\right)^3=0$，故知 $\displaystyle x=\frac12$ 或 $\displaystyle x=-\frac12$。又因 $\displaystyle g'\left(x\right)=\frac{\left(2x-1\right)\left(2x+1\right)^3}{8x^3}$，因此在 $\displaystyle x=-\frac12$ 與 $\displaystyle x=\frac12$ 皆有局部極小，其值分別為 $\displaystyle f\left(-\frac12\right)=\frac{11}2$、$\displaystyle f\left(\frac12\right)=\frac{19}2$。因此局部最小值為 $\displaystyle\frac{13}2$，故選(D)。
解法二
設拉格朗日乘子函數為
$F\left(x,y,\lambda\right)=2x^2+y^2+2xy+4x+2y+7+\lambda\left(4x^2+4xy-1\right)$
據此解聯立方程組
$\left\{\begin{aligned}&F_x\left(x,y,\lambda\right)=4x+2y+4+\lambda\left(8x+4y\right)=0\\&F_y\left(x,y,\lambda\right)=2y+2x+2+\lambda\cdot4x=0\\&F_{\lambda}\left(x,y,\lambda\right)=4x^2+4xy-1=0\end{aligned}\right.$
由第一式有 $\left(4x+2y\right)\left(1+2\lambda\right)+4=0$，由此可知 $4x+2y\neq0$，因此 $\displaystyle\lambda=-\frac12-\frac1{2x+y}$。代入後有
$\displaystyle2y+2-\frac{4x}{2x+y}=0$
整理便有 $y^2+y+2xy=0$，即 $y\left(2x+y+1\right)=0$。
• 若 $y=0$，則由第三式可知 $4x^2=1$，故 $\displaystyle x=\pm\frac12$，因此有 $\displaystyle\left(x,y\right)=\left(\pm\frac12,0\right)$；
• 若 $2x+y+1=0$，則由第三式有 $4x^2+4x+1=0$，因此 $\displaystyle x=-\frac12$，故 $y=0$。
兩者分別代回計算有
$\displaystyle f\left(\frac12,0\right)=\frac{19}2$,　　$\displaystyle f\left(-\frac12,0\right)=\frac{11}2$.
因此局部最小值為 $\displaystyle\frac{11}2$，故選(D)。


14. ($5\%$) Evaluate the following double integral for a specified region $A$.

    $\displaystyle\iint_Ae^{y^3}dA$,

    where $A$ is the region bounded by $y=\sqrt{x}$, $y=1$, and $x=0$.
    1. $\displaystyle\frac12\left(e-1\right)$
    2. $\displaystyle\frac13\left(e-2\right)$
    3. $1$
    4. None of the above
訣竅
適當的表達積分區域後運用迭代積分計算之。
解法
給定的積分區域 $A$ 可表達為 $\left\{\begin{aligned}&0\leq x\leq y^2\\&0\leq y\leq1\end{aligned}\right.$，如此所求的重積分可表達並計算如下
$\displaystyle\iint_Ae^{y^3}dA=\int_0^1\int_0^{y^2}e^{y^3}dxdy=\int_0^1y^2e^{y^3}dy=\left.\frac{e^{y^3}}3\right|_0^1=\frac{e-1}3$
選(D)。


15. ($5\%$) What is the value of $\displaystyle\int_1^{\infty}\frac{\ln x}{x^2}dx$?
    1. $0$
    2. $\infty$
    3. $1$
    4. None of the above
訣竅
運用分部積分法求解。
解法
直接由分部積分法求解有
$\displaystyle\int_1^{\infty}\frac{\ln x}{x^2}dx=\left.-\frac{\ln x}x\right|_1^{\infty}+\int_1^{\infty}\frac{dx}{x^2}=\left.-\frac1x\right|_1^{\infty}=1$
選(C)。