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須列計算過程，否則不予計分
總計十題，每題十分
- Given $m=p/q$ with both $p$ and $q$ are integers, find $\displaystyle\lim_{x\to1}\left(\frac{x^m-1}{x-1}\right)=$ .
- Given $\displaystyle Y=\frac{x_1}c\left(b-e^{cz_1}\right)+ae^{-rt_1}\frac{x_2}c\left(b-e^{-cz_2}\right)$, where $a,b,c$ are constants, $Y,x_1,x_2,z_1,z_2$ are variables, and $z_i=f_i\left(x_i\right)$, $i=1,2$, it can be derived that
$\displaystyle\frac{\partial Y}{\partial x_1}=G\left(z_1\right)+H\left(x_1,Y\right)\frac{\partial z_1}{\partial x_1}$
Please find the expressions of $G\left(z_1\right)$ and $H\left(x_1,Y\right)$. - Find the values of $x$ for which the series
$1-2\left(x-1\right)+3\left(x-1\right)^2+\cdots+n\left(-1\right)^{n-1}\left(x-1\right)^{n-1}+\cdots$
converges. - 當 $x=0$ 時級數寫為 $\displaystyle\sum_{n=1}^{\infty}n$，明顯發散；
- 當 $x=2$ 時級數寫為 $\displaystyle\sum_{n=1}^{\infty}n\left(-1\right)^{n-1}$，明顯發散。
- If the probability density function (p.d.f.) of a random variable $X$ is as follows, please find the value of $c$, the mean and the variance of $X$.
$f\left(x\right)=\begin{cases}-cx,&-1\leq x\leq0,\\6cx^3,&0<x\leq1.\end{cases}$
- Let $y=\left(3x-2\right)\left(x-1\right)^{-1}$, $z=\left(y^2-1\right)^{\frac13}$, find $\displaystyle\frac{dz}{dx}=$ .
- If $\displaystyle\lim_{x\to\infty}\left(\left(x^n+a_1x^{n-1}+a_2x^{n-2}+\cdots+a_{n-1}x+a_n\right)^{\frac1n}-\left(x+1\right)\right)=b$, then $a_1=$ .
- $\displaystyle\int\frac1{\sqrt{4x^2+4x+7}}dx=$ .
- $\displaystyle\int_0^{\infty}e^{-x}\left(x^2+x\right)dx=$ .
- Let $f\left(x,y\right)=4x^2+4x-2y^2+6$. Find the maximum and minimum values of the function $f\left(x,y\right)$ on the region $x^2+y^2\leq3$.
- 若 $y=0$，則由第三式可知 $x^2=3$，即 $x=\pm\sqrt3$；
- 若 $\lambda=2$，則第一式可給出 $12x+4=0$，即 $\displaystyle x=-\frac13$，則第三式得 $\displaystyle y^2=\frac{26}9$，即 $\displaystyle y=\pm\frac{\sqrt{26}}3$。
- Let $f\left(x\right)=\sqrt{3x^2+\sqrt{3x^2+\sqrt{3x^2+\sqrt{\cdots}}}}$. Find the relation between $f\left(x\right)$ and $f'\left(x\right)$.
訣竅
消去因子來求解；亦可辨識出此為特定函數的導數值來求解。解法一
將分子分解可注意到下列的關係式$\displaystyle\begin{aligned}\lim_{x\to1}\frac{x^m-1}{x-1}&=\lim_{x\to1}\frac{\left(x^{1/q}-1\right)\left(x^{\left(p-1\right)/q}+x^{\left(p-2\right)/q}+\cdots+x^{1/q}+1\right)}{x-1}\\&=\lim_{x\to1}\frac{\left(x^{1/q}-1\right)\left(x^{\left(q-1\right)/q}+\cdots+1\right)\left(x^{\left(p-1\right)/q}+\cdots+1\right)}{\left(x-1\right)\left(x^{\left(q-1\right)/q}+\cdots+1\right)}\\&=\lim_{x\to1}\frac{x^{\left(p-1\right)/q}+\cdots+1}{x^{\left(q-1\right)/q}+\cdots+1}=\frac{p}q=m\end{aligned}$
解法二
設 $f\left(x\right)=x^m$，那麼可知 $f'\left(x\right)=mx^{m-1}$，那麼可以觀察到所求為$\displaystyle\lim_{x\to1}\frac{x^m-1}{x-1}=\lim_{x\to1}\frac{f\left(x\right)-1}{x-1}=f'\left(1\right)=m$
訣竅
運用多變函數的連鎖律計算即可求解。解法
對 $x_1$ 求偏導可知$\displaystyle\frac{\partial Y}{\partial x_1}=\frac1c\left(b-e^{cz_1}\right)+\frac{x_1}c\cdot-e^{cz_1}\cdot c\frac{dz_1}{dx_1}$
因此 $\displaystyle G\left(z_1\right)=\frac{b-e^{cz_1}}c$，而 $\displaystyle H\left(x_1,Y\right)=-x_1e^{cf_1\left(x_1\right)}$。訣竅
運用比值審歛法確認出收斂半徑，隨後檢查端點即可。解法
設給定的級數為 $\displaystyle F\left(x\right)=\sum_{n=1}^{\infty}n\left(-1\right)^{n-1}\left(x-1\right)^{n-1}$，那麼由比值審歛法可知收斂半徑為$\displaystyle R=\lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|=\lim_{n\to\infty}\frac{n}{n+1}=1$
故可確定級數至少在 $\left(0,2\right)$ 上收斂。現確認端點如下訣竅
根據機率密度以及連續機率的平均與變異數的定義計算之。解法
按照機率密度的定義可知
$\displaystyle1=\int_{-1}^1f\left(x\right)dx=\int_{-1}^0\left(-cx\right)dx+\int_0^16cx^3dx=\left.-\frac{cx^2}2\right|_{-1}^0+\left.\frac{3cx^4}2\right|_0^1=\frac{c}2+\frac{3c}2=2c$
因此 $\displaystyle c=\frac12$。進一步，我們使用期望值（平均）的定義計算如下
$\displaystyle\mbox{E}\left(X\right)=\int_{-1}^1xf\left(x\right)dx=\int_{-1}^0x\cdot-\frac{x}2dx+\int_0^1x\cdot3x^3dx=\left.-\frac{x^3}6\right|_{-1}^0+\left.\frac{3x^5}5\right|_0^1=\frac{13}{30}$
為了計算變異數，我們先計算下述的值
$\displaystyle\mbox{E}\left(X^2\right)=\int_{-1}^1x^2f\left(x\right)dx=\int_{-1}^0x^2\cdot-\frac{x}2dx+\int_0^1x^2\cdot3x^3dx=\left.-\frac{x^4}8\right|_{-1}^0+\left.\frac{x^6}2\right|_0^1=\frac58$
按照變異數的一個定義可知所求為$\displaystyle\mbox{Var}\left(X\right)=\mbox{E}\left(X^2\right)-\left(\mbox{E}\left(X\right)\right)^2=\frac58-\left(\frac{13}{30}\right)^2=\frac{787}{1800}$
訣竅
運用連鎖律計算即可。解法
直接使用連鎖律求導可知$\displaystyle\frac{dz}{dx}=\frac13\left(y^2-1\right)^{-\frac23}\cdot2\frac{dy}{dx}=\frac23\left(y^2-1\right)^{-\frac23}\cdot\left[3\left(x-1\right)^{-1}-\left(3x-2\right)\left(x-1\right)^{-2}\right]=-\frac2{3\left(x-1\right)^2\left(y^2-1\right)^{\frac23}}$
訣竅
利用極限的四則運算定理求解。解法
注意到極限式$\displaystyle\lim_{x\to\infty}\left[x\left(1+\frac{a_1}x+\cdots+\frac{a_n}{x^n}\right)^{\frac1n}-x\left(1+\frac1x\right)\right]=b$
整理後觀察羅必達法則可知$\displaystyle\begin{aligned}b&=\lim_{x\to\infty}\frac{\displaystyle\left(1+\frac{a_1}x+\cdots+\frac{a_n}{x^n}\right)^{\frac1n}-\left(1+\frac1x\right)}{\displaystyle\frac1x}\\&=\lim_{x\to\infty}\frac{\displaystyle\frac1n\left(1+\frac{a_1}x+\cdots+\frac{a_n}{x^n}\right)^{-\frac{n-1}n}\left(-\frac{a_1}{x^2}-\cdots-\frac{na_n}{x^{n+1}}\right)+\frac1{x^2}}{\displaystyle-\frac1{x^2}}\\&=-\frac1n\cdot1\cdot\left(-a_1\right)-1=\frac{a_1}n-1\end{aligned}$
這表明 $a_1=n\left(1+b\right)$。訣竅
配方後使用三角代換來處理即可。解法
將分母配方有 $4x^2+4x+7=\left(2x+1\right)^2+6$，據此令 $2x+1=\sqrt6\tan\theta$，如此求導有 $2dx=\sqrt6\sec^2\theta d\theta$，從而所求的不定積分可改寫如下$\displaystyle\int\frac1{\sqrt{4x^2+4x+7}}dx=\int\frac1{\sqrt{6\tan^2\theta+6}}\cdot\frac{\sqrt6}2\sec^2\theta d\theta=\frac12\int\sec\theta d\theta=\frac12\ln\left|\sec\theta+\tan\theta\right|+C$
現可注意到 $\displaystyle\sec\theta=\sqrt{1+\tan^2\theta}=\sqrt{1+\frac{\left(2x+1\right)^2}6}$，因此所求為$\displaystyle\int\frac1{\sqrt{4x^2+4x+7}}dx=\ln\left|\sqrt{1+\frac{\left(2x+1\right)^2}6}+\frac{2x+1}{\sqrt6}\right|+C$
訣竅
使用分部積分法求解即可。解法
使用分部積分法可知$\displaystyle\begin{aligned}\int_0^{\infty}e^{-x}\left(x^2+x\right)dx&=-\int_0^{\infty}\left(x^2+x\right)de^{-x}=-e^{-x}\left(x^2+x\right)\Big|_0^{\infty}+\int_0^{\infty}e^{-x}\left(2x+1\right)dx=-\int_0^{\infty}\left(2x+1\right)de^{-x}\\&=-e^{-x}\left(2x+1\right)\Big|_0^{\infty}+2\int_0^{\infty}e^{-x}dx=1-\left(2e^{-x}\right)\Big|_0^{\infty}=3\end{aligned}$
訣竅
在圓盤內部上使用一階偏導為零來尋求極值，而在圓盤邊界上可使用拉格朗日乘子法來解極值。解法
在圓盤內部時，我們解下列的聯立方程組
$\left\{\begin{aligned}&f_x\left(x,y\right)=8x+4=0\\&f_y\left(x,y\right)=-4y=0\end{aligned}\right.$
如此得座標 $\displaystyle\left(x,y\right)=\left(-\frac12,0\right)$。在邊界上時，我們運用拉格朗日乘子法，設定拉格朗日乘子函數如下
$F\left(x,y,\lambda\right)=4x^2+4x-2y^2+6+\lambda\left(x^2+y^2-3\right)$
據此解下列的聯立方程組$\left\{\begin{aligned}&F_x\left(x,y,\lambda\right)=8x+4+2\lambda x=0\\&F_y\left(x,y,\lambda\right)=-4y+2\lambda y=0\\&F_{\lambda}\left(x,y,\lambda\right)=x^2+y^2-3=0\end{aligned}\right.$
由第二式可知 $y=0$ 或 $\lambda=2$。$\displaystyle f\left(-\frac12,0\right)=5,\quad f\left(\pm\sqrt3,0\right)=18\pm4\sqrt3,\quad f\left(-\frac13,\pm\frac{\sqrt{26}}3\right)=-\frac23$
故最大值發生在 $\left(\sqrt3,0\right)$，其值為 $18+4\sqrt3$，而最小值發生在 $\displaystyle\left(-\frac13,\pm\frac{\sqrt{26}}3\right)$，其值為 $\displaystyle-\frac23$。訣竅
遞迴的表達出 $f$ 的關係，如此求導便可獲得函數與導函數的關係。解法
注意到 $f\left(x\right)=\sqrt{3x^2+f\left(x\right)}$，因此求導可知$\displaystyle f'\left(x\right)=\frac1{2\sqrt{3x^2+f\left(x\right)}}\cdot\left(6x+f'\left(x\right)\right)$
因此移項可知$\displaystyle\left[1-\frac1{2\sqrt{3x^2+f\left(x\right)}}\right]f'\left(x\right)=\frac{3x}{\sqrt{3x^2+f\left(x\right)}}$
因此 $f\left(x\right)$ 與 $f'\left(x\right)$ 的關係式如下$\displaystyle f'\left(x\right)=\frac{6x}{2\sqrt{3x^2+f\left(x\right)}-1}$
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