台聯大轉學考微積分 109 A2 卷解答

可觀察到 $x$ 控制這個極限，所以分子分母同除之。

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

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

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此題考點：夾擠定理

求水平漸近線的基本題，需注意絕對值。

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

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

horizontal asymptotes: $y=\pm 1　∎$

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此題考點：漸近線

積分上下界有未知數，想到微積分基本定理。

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

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

$\Rightarrow$ 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 II、微分求極值法

有限制條件求極值時，使用拉格朗日乘數法。

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：拉格朗日乘數法

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。

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

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

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

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)。

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

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此題考點：二變數函數的積分

本題使用二次微分檢驗法。

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

機率函數需要滿足其值介於 0 到 1 之間，及全部加總為 1。

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

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

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此題考點：變數變換法

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。

$U(x_1 ,x_2 ):$ 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：拉格朗日乘數法

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。

〔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

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。

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