台聯大轉學考微積分 109 A2 卷解答

平易近人的考卷,題目中規中矩不過要小心細節,大部分計算量不算多,和 A3A4A6 卷同樣是適合考前練習的考古題。

本解答由張旭老師及南享老師共同完成。

建議使用電腦瀏覽,手機或平板的算式可能會超出畫面。

甲. 簡答題

第一題 (8 分)
Find the value of $\mathop {\lim }\limits_{x \to \infty } \frac{{x + \cos x}}{{x – \cos x}}$.

提示

可觀察到 $x$ 控制這個極限,所以分子分母同除之。

解答

$\begin{array}{cl} ◎\ &\because\ 1 \le \cos x \le 1 \\ & \therefore \frac{{ – 1}}{x} \le \frac{{\cos x}}{x} \le \frac{1}{x}\ (\text{for }x>0) \\ &\because \mathop {\lim }\limits_{x \to \infty } ( – \frac{1}{x}) = \mathop {\lim }\limits_{x \to \infty } \frac{1}{x} = 0 \\&\therefore \mathop {\lim }\limits_{x \to \infty } \frac{{\cos x}}{x} = 0\ (\text{By sandwich lemma}) \end{array}$

$\mathop {\lim }\limits_{x \to \infty } \frac{{x + \cos x}}{{x – \cos x}}=\mathop {\lim }\limits_{x \to \infty } \frac{{1 + \frac{{\cos x}}{x}}}{{1 – \frac{{\cos x}}{x}}} = 1 ∎$

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此題考點夾擠定理

第二題 (8 分)
Find all horizontal asymptotes of graph of the function $f(x) = \frac{{\left|x \right|x}}{{x^2 + 1}}$.

提示

求水平漸近線的基本題,需注意絕對值。

解答

$\mathop {\lim }\limits_{x \to \infty }\frac{{\left| x \right|x}}{{x^2 + 1}} = \mathop {\lim }\limits_{x \to \infty }\frac{{x^2 }}{{x^2 + 1}} = 1$

$\mathop {\lim }\limits_{x \to – \infty } \frac{{ – x^2 }}{{x^2 + 1}} = – 1$

horizontal asymptotes: $y=\pm 1 ∎$

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此題考點漸近線

第三題 (8 分)
Find the smallest positive $(x>0)$ inflection point of $F(x) = \int_0^x {\cos (t^{\frac{3}{2}} )} dt$.

提示

積分上下界有未知數,想到微積分基本定理。

解答

$F'(x) = \cos (x^{\frac{3}{2}} ) \Rightarrow F^{\prime \prime} (x) = – \frac{3}{2}x^{\frac{1}{2}} \cdot \sin (x^{\frac{3}{2}} )\mathop = \limits^{{\rm{Let}}} 0$

$\Rightarrow x^{\frac{3}{2}} = k\pi$, $k=1,2,3,\dots$

$\Rightarrow$ the smallest positive inflection point appears at $x=\pi ^{\frac{2}{3}}$
$\Rightarrow$ the inflection point is $(\pi ^{\frac{2}{3}} ,\int_0^{\pi ^{\frac{2}{3}} } {\cos (t^{\frac{3}{2}} )} dt) ∎$

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此題考點微積分基本定理微分求極值法

第四題 (8 分)
A building in the shape of a rectangular box is to have a volume of $12,000\ ft^3$. It is estimated that the annual heating and cooling costs will be \$ 2 / square foot for the top, \$ 4 / square foot for the front and back, and $ 3 / square foot for the sides. What is the minimal annual heating and cooling cost?

提示

有限制條件求極值時,使用拉格朗日乘數法。

解答

法一 Lagrange 乘數法

Let $f(x,y,z)=2xy+6yz+8xz$
constraint: $xyz=12000$. Let $g(x,y,z)=xyz$
Let $\nabla f+\lambda \nabla g=0$
$\Rightarrow \left\{ {\begin{array}{c}
{2y + 8z + \lambda yz = 0} \\
{2x + 6z + \lambda xz = 0} \\
{6y + 8x + \lambda xy = 0} \\
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{c}
{2xy + 8xz + \lambda xyz = 0} \\
{2xy + 6yz + \lambda xyz = 0} \\
{6yz + 8xz + \lambda xyz = 0} \\
\end{array}} \right.$
$\Rightarrow \left\{ {\begin{array}{c}
{8xz = 6yz \Rightarrow x:y = 3:4} \\
{2xy = 8xz \Rightarrow y:z = 4:1} \\
{2xy = 6yz \Rightarrow x:z = 3:1} \\ \end{array}} \right.\ \Rightarrow x:y:z = 3:4:1$ $\Rightarrow\text{ Let} \left\{ {\begin{array}{c}{x = 3t} \\{y = 4t} \\{z = t} \\ \end{array}} \right.$, $t>0$
$\because xyz=12000$
$\therefore (3t)(4t)(t)=12000$
$\Rightarrow 12t^3 = 12000 \Rightarrow t = 10 \Rightarrow \left\{ {\begin{array}{c}
{x = 30} \\
{y = 40} \\
{z = 10} \\
\end{array}} \right.$
$\begin{array}{cl} \Rightarrow \text{minimum} & =f(30,40,10) \\ & =2400+2400+2400=7200 ∎ \end{array}$

法二 算幾不等式(簡答題或驗算時使用)

$\begin{array}{cl} \frac{{{\rm{2}}xy{\rm{ + 6}}yz + 8xz}}{3} & \ge \sqrt[3]{{(2xy)(6yz)(8xz)}} \\ 2xz + 6yz + 8xz & \ge 3 \cdot \sqrt[3]{{2^2 \cdot 3 \cdot 2^3 \cdot 10^6 \cdot 4^2 \cdot 3^2 }} \\ & =3 \cdot 2 \cdot 10^2 \cdot 3 \cdot 4 = 7200 ∎ \end{array}$

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此題考點Lagrange 乘數法

第五題 (8 分)
Find the values of $p$ for which the function $f(x,y) = \left\{ {\begin{array}{c}{\frac{{(xy)^p }}{{x^4 + y^4 }},} &{{\rm{if }}(x,y) \ne (0,0)} \\ {0,} & {{\rm{if }}(x,y) = (0,0)} \\ \end{array}} \right.$ is discontinuous at $(0,0)$.

提示

多變數函數尋找不連續點, 也就是計算極限,這題判斷是否收斂,需要妥善地處理分子跟分母的次方。

解答

$\begin{array}{cl} \mathop {\lim }\limits_{(x,y) \to (0,0)} f(x,y)&=\mathop {\lim }\limits_{(x,y) \to (0,0)} \frac{{(xy)^p }}{{x^4 + y^4 }} \\ & =\mathop {\lim }\limits_{(x,y) \to (0,0)} \frac{{(xy)^{p – 2} (xy)^2 }}{{x^4 + y^4 }} \\ & =\mathop {\lim }\limits_{(x,y) \to (0,0)} (xy)^{p – 2} \frac{{(xy)^2 }}{{x^4 + y^4 }} \end{array}$

$\begin{array}{cll} ◎\ & \because &(x^2 – y^2 )^2 = x^4 + y^4 – 2x^2 y^2 \\ &&\text{and }(x^2-y^2)\ge 0 \\ & \therefore & x^4+y^4\ge 2x^2y^2 \\&&\Rightarrow \frac{{x^2 y^2 }}{{x^4 + y^4 }} \le \frac{1}{2} \end{array}$

So, if $p>2$,
$\mathop {\lim }\limits_{(x,y) \to (0,0)} f(x,y) = \mathop {\lim }\limits_{(x,y) \to (0,0)} (xy)^{p – 2} \cdot \frac{{x^2 y^2 }}{{x^4 + y^4 }} = 0$
i.e. $f(x,y)$ is continuous at $(0,0)$.

if $p\leq 2$,
$\mathop {\lim }\limits_{(x,y) \to (0,0)} f(x,y) = \mathop {\lim }\limits_{(x,y) \to (0,0)} (xy)^{p – 2} \cdot \frac{{x^2 y^2 }}{{x^4 + y^4 }}$ D.N.E.
i.e. $f(x,y)$ is discontinuous at $(0,0)$.

Thus $f(x,y)$ is discontinuous at $(0,0)$ when $p\leq 2 ∎$

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此題考點二變數函數的極限

第六題 (8 分)
Evaluate $\int_0^{\ln 5} {\int_{e^x }^5 {\frac{1}{{\ln y}}} } dydx$.

提示

遇到不好算的二重積分,可以嘗試極座標變換或交換積分順序 (Fubini 定理)。

解答

$\begin{array}{cl} \int_0^{\ln 5} {\int_{e^x }^5 {\frac{1}{{\ln y}}} } dydx &=\int_1^5 {\int_0^{\ln y} {\frac{1}{{\ln y}}} } dxdy\\&=\int_1^5 {\frac{1}{{\ln y}}} \int_0^{\ln y} {dx} dy\\&= \int_1^5 1 dy=4 ∎ \end{array}$

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此題考點二變數函數的積分

第七題 (8 分)
How many local extreme values does the function $f(x,y) = 10xye^{ – (x^2 + y^2 )}$ have ?

提示

本題使用二次微分檢驗法。

解答

$\begin{array}{cl} \nabla f =& (10ye^{ – (x^2 + y^2 )} + 10xye^{ – (x^2 + y^2 )} \cdot ( – 2x), \\ & 10xe^{ – (x^2 + y^2 )} + 10xye^{ – (x^2 + y^2 )} \cdot ( – 2y)) \end{array}$
$\begin{array}{cl} \Rightarrow \nabla f =& (10ye^{ – (x^2 + y^2 )} \cdot (1 – 2x^2 ), \\ & 10xe^{ – (x^2 + y^2 )} \cdot (1 – 2y^2 )) \mathop = \limits^{{\rm{Let}}} (0,0) \end{array}$
$\Rightarrow \left\{ {\begin{array}{c}
{y(1 – 2x^2 ) = 0 \Rightarrow y = 0{\text{ or }}x = \pm \frac{1}{{\sqrt 2 }}} \\
{x(1 – 2y^2 ) = 0 \Rightarrow x = 0{\text{ or }}y = \pm \frac{1}{{\sqrt 2 }}} \\
\end{array}} \right.$
$\Rightarrow$ critical points: $(0,0)$, $\pm (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$, $\pm (\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$

$\begin{array}{cl} \text{H}f & =\left[ {\begin{array}{c}
{10ye^{ – (x^2 + y^2 )} \cdot ( – 2x)(1 – 2x^2 + 2)} & {(1 – 2y^2 )10e^{ – (x^2 + y^2 )} \cdot (1 – 2x^2 )} \\
{(1 – 2x^2 )10e^{ – (x^2 + y^2 )} \cdot (1 – 2y^2 )} & {10xe^{ – (x^2 + y^2 )} \cdot ( – 2y)(1 – 2y^2 + 2)} \\
\end{array}} \right] \\ & =10e^{ – (x^2 + y^2 )} \left[ {\begin{array}{c}
{ – 2xy(3 – 2x^2 )} & {(1 – 2y^2 )(1 – 2x^2 )} \\
{(1 – 2x^2 )(1 – 2y^2 )} & { – 2xy(3 – 2y^2 )} \\
\end{array}} \right]\\ \end{array}$

${\rm{D}}f(0,0) = {\rm{det}}(10\left[ {\begin{array}{c}
0 & 1 \\
1 & 0 \\
\end{array}} \right]) = – 100 < 0$
$\Rightarrow (0,0,f(0,0))$ is a saddle point.
${\rm{D}}f( \pm (\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }})) = {\rm{det}}(10d^{ – 1} \left[ {\begin{array}{c}
{ – 2} & 0 \\
0 & { – 2} \\
\end{array}} \right]) = 40e^{ – 1} > 0$
$\Rightarrow f( \pm (\frac{1}{{\sqrt 2 }}, – \frac{1}{{\sqrt 2 }}))$ is a local extreme.
${\rm{D}}f( \pm (\frac{1}{{\sqrt 2 }}, – \frac{1}{{\sqrt 2 }})) = {\rm{det}}(10e^{ – 1} \left[ {\begin{array}{c}
2 & 0 \\
0 & 2 \\
\end{array}} \right]) = 40e^{ – 1} > 0$
$\Rightarrow f( \pm (\frac{1}{{\sqrt 2 }}, – \frac{1}{{\sqrt 2 }}))$ is a local extreme.
$\text{There are }4\text{ local extreme values.} ∎$

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此題考點微分求極值

第八題 (8 分)
Let $f(x,y) = kxye^{ – (x^2 + y^2 )}$ be a joint probability density function on $D = \left\{ {0 < x < \infty ,0<y<\infty} \right\}$, then $k=?$

提示

機率函數需要滿足其值介於 0 到 1 之間,及全部加總為 1。

解答

$\iint_{D}f(x,y)dA=1$
$\Rightarrow \int_0^\infty {\int_0^\infty {kxye^{ – (x^2 + y^2 )} } } dxdy = 1$
$\Rightarrow \int_0^\infty {\int_0^\infty {xye^{ – x^2 } e^{ – y^2 } } } dxdy = \frac{1}{k}$
$\Rightarrow \int_0^\infty {ye^{ – y^2 } } (\underbrace {\int_0^\infty {xe^{ – x^2 } } dx}_{\frac{1}{2}})dy = \frac{1}{2}\int_0^\infty {ye^{ – y^2 } } dy = \frac{1}{4}$
$\Rightarrow k=4 ∎$

◎ $\int_{\rm{0}}^\infty {xe^{ – x^2 } } dx=\int_0^\infty {e^{ – u} } \frac{{du}}{2}= \left[ { – \frac{1}{2}e^{ – u} } \right]_{u = 0}^{u = \infty }= \frac{1}{2}\left[ {0 – ( – 1)} \right] = \frac{1}{2}$

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此題考點變數變換

乙.計算、證明題

第一題 (12 分)
Goods 1 and 2 are available at prices (in dollars) of $p_1$ per unit of good 1 and $p_2$ per unit of good 2. A unility function $U(x_1,x_2)$ is a function representing the unity or benefit for comsuming $x_j$ units of goods $j$. The marginal unity of the $j$th good is $\frac{{\partial U}}{{\partial x_j }}$, the rate of increase in utility per unit increase in the $j$th good. Prove the following law of economics: Given a budget of $L$ dollars, utility is maximized at the consumption level $(a,b)$ where the ratio of marginal utility is equal to the ratio of prices:

$\frac{{{\text{Marginal utility of good }}1}}{{{\text{Marginal utility of good }}2}} = \frac{{\partial U/\partial x_1 }}{{\partial U/\partial x_2 }} = \frac{{p_1 }}{{p_2 }}$

提示

這題是限制條件的問題,要求利益函數的最大值,這時候要使用的就是 Lagrange 乘數法。

解答

$U(x_1 ,x_2 ):$ 目標函數
constraint: Let $g(x_1 ,x_2 ) = x_1 p_1 + x_2 p_2 = L$
Let $\nabla U + \lambda \nabla g = 0$
$\Rightarrow (\frac{{\partial U}}{{\partial x_1 }},\frac{{\partial U}}{{\partial x_2 }}) + \lambda (p_1 ,p_2 ) = (0,0)$
$\Rightarrow \left\{ {\begin{array}{c}
{\frac{{\partial U}}{{\partial x_1 }} + \lambda p_1 = 0} \\
{\frac{{\partial U}}{{\partial x_2 }} + \lambda p_2 = 0} \\
\end{array} \Rightarrow \left\{ {\begin{array}{c}
{\frac{{\partial U}}{{\partial x_1 }} = – \lambda p_1 \cdots (1) } \\
{\frac{{\partial U}}{{\partial x_2 }} = – \lambda p_2 \cdots (2) } \\
\end{array}} \right.} \right.$
$\frac{(1)}{(2)}\Rightarrow \frac{{\partial U/\partial x_1 }}{{\partial U/\partial _2 }} = \frac{{ – \lambda p_1 }}{{ – \lambda p_2 }} = \frac{{p_1 }}{{p_2 }} ∎$

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此題考點Lagrange 乘數法

第二題 (12 分)
〔1〕Determine whether the series $\sum\limits_{n = 1}^\infty {( – 1)^n \ln (1 + \frac{1}{n})}$ diverges or converges conditionally or converges absolutely and give reasons for your answer. (6 分)

〔2〕Show that if $\sum\limits_{n = 1}^\infty {a_n }$ converges, then $\sum\limits_{n = 1}^\infty {(\frac{{3 + \sin (a_n )}}{5})^n }$ converges. (6 分)

提示

兩題為判別級數和收斂的基本題,八大審斂法與絕對、條件收斂需要很熟悉。

解答

〔1〕
$\because \mathop {\lim }\limits_{n \to \infty } \ln (1 + \frac{1}{n}) = 0$ and $\ln (1 + \frac{1}{n})$ decreases.
$\therefore \sum\limits_{n = 1}^\infty {( – 1)^n \ln (1 + \frac{1}{n})}$ converges. (by alternating series)

$\sum\limits_{n = 1}^\infty {\left| {( – 1)^n \ln (1 + \frac{1}{n})} \right|} = \sum\limits_{n = 1}^\infty {\ln (1 + \frac{1}{n})}$
$\begin{array}{rl} \because & \mathop {\lim }\limits_{n \to \infty } \frac{{\ln (1 + \frac{1}{n})}}{{\frac{1}{n}}}=\mathop {\lim }\limits_{t \to 0^ + } \frac{{\ln (1 + t)}}{t}=\mathop = \limits^{\rm{L}} \mathop {\lim }\limits_{t \to 0^ + } \frac{{\frac{{\rm{1}}}{{{\rm{1 + }}t}}}}{{\rm{1}}} = 1\ (\text{let }t=\frac{1}{n}) \\ & \text{and } \sum\limits_{n = 1}^\infty {\frac{1}{n}}\text{ diverges.(by }p\text{-series test}) \\ \therefore &\sum\limits_{n = 1}^\infty {\ln (1 + \frac{1}{n})} \text{ diverges. (by limit comparison test)}  ∎\end{array}$
附註:$\mathop = \limits^{\rm{L}}$ 表示此等號使用羅必達法則 (L’Hôpital’s rule)

〔2〕
$\because \sum\limits_{n = 1}^\infty {a_n }$ converges.
$\therefore \mathop {\lim }\limits_{n \to \infty } a_n = 0$

$\because \mathop {\lim }\limits_{n \to \infty } \left| {(\frac{{3 + \sin (a_n )}}{5})^n } \right|^{\frac{1}{n}} = \mathop {\lim }\limits_{n \to \infty } \frac{{3 + \overbrace {\sin (a_n )}^0}}{5} = \frac{3}{5} < 1$
$\therefore \sum\limits_{n = 1}^\infty {(\frac{{3 + \sin (a_n )}}{5})^n }$ converges. (by root test)$ ∎$

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此題考點絕對收斂與條件收斂

第三題 (12 分)
A trough with a trapezoidal cross section is to be constructed with a 1-foot base and sides that are 20 feet long and 1 foot wide, as shown in the figure. Only the angle $\theta$ can be varied. What value of $\theta$ will maximize the trough’s volume ?

提示

這題要求梯形容積最大的時機,因此只要把容積用 $\theta$ 表示好,微分找最大值對應的 $\theta$ 即可。

解答

area of trapezoidal $=\frac{{(2\sin \theta + 1) \cdot \cos \theta }}{2}= (1 + \sin \theta ) \cdot \cos \theta \mathop = \limits^{{\rm{Let}}} f(\theta )$
$\begin{array}{cl} \Rightarrow f'(\theta) & =\cos \theta \cdot \cos \theta + (1 + \sin \theta ) \cdot ( – \sin \theta ) \\ & =\cos ^2 \theta – \sin ^2 \theta – \sin \theta \\ & =1 – 2\sin ^2 \theta – \sin \theta \mathop = \limits^{{\rm{Let}}} 0 \end{array}$
$\Rightarrow 2\sin ^2 \theta + \sin \theta – 1 = (2\sin \theta – 1)(\sin \theta + 1) = 0$
$\Rightarrow \sin \theta = \frac{1}{2}$ or $\sin \theta=-1$ (負不合 $0^{\circ}\leq \theta \leq 90^{\circ}$)

For $\sin \theta =\frac{1}{2}$
$f(\theta ) = (1 + \frac{1}{2}) \times \frac{{\sqrt 3 }}{2} = \frac{{3\sqrt 3 }}{4}$
For $\theta =0^{\circ}$
$f(\theta)=1=\frac{4}{4}$

When $\sin\theta =\frac{1}{2}$ (i.e. $\theta =30^{\circ}$) the volume reaches its max.$ ∎$

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此題考點微分求極值法

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作者:

數學老師張旭

中央大學數學學士 臺灣師範大學數學碩士 --- 文心創媒股份有限公司負責人 數學老師張旭線上教學品牌創辦人 張旭無限教室線上課程平台創辦人 張旭教育革命軍線上學習社群創辦人 --- 乾果多股份有限公司負責人 乾果多線上學院線上課程平台創辦人

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