台聯大轉學考微積分 109 A3/A4/A6 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

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

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

這題要求用參數式表達空間中圖形的切線。作圖後會明確很多，再利用題目資訊找出所求切線的方向向量。

令 $E_1$ 為過 $P$ 點與 $C$ 相切之平面，其法向量為 $\overset{\rightharpoonup}{n_1}$
令 $E_2$ 為 $C$ 所在平面，其法向量為 $\overset{\rightharpoonup}{n_2}$
$\Rightarrow \overset{\rightharpoonup}{n_1} \mathbin{/ \mkern-5mu/} \left. {(2x,2y,2z)} \right|_{(1,1,1)} = (2,2,2)$
取 $\overset{\rightharpoonup}{n_1}=(1,1,1)$
取 $\overset{\rightharpoonup}{n_2} \mathbin{/ \mkern-5mu/} \left. {(2x-2,2(y-2),2z)} \right|_{(1,1,1)} = (-2,-2,2)$
取 $\overset{\rightharpoonup}{n_2}=(1,1,-1)$
$\Rightarrow \overset{\rightharpoonup}{L}\mathbin{/ \mkern-5mu/} (\overset{\rightharpoonup}{n_1}\times \overset{\rightharpoonup}{n_2})=(-2,2,0)$ 取 $\overset{\rightharpoonup}{L}=(1,-1,0)$
$\Rightarrow L:\left\{ \begin{array}{l}
{x = 1 + t} \\
{y = 1 – t} \\
{z = 1} \\
\end{array} \right.　∎$

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此題考點：方向導數

不易算的二重積分可以嘗試極座標變換或交換積分次序(Fubini 定理)。本題被積函數與積分範圍皆和圓有關，因此使用極座標變換。

$\begin{array}{cl} \iint_{R}\ln \sqrt {x^2 + y^2 } &=\int_0^{2\pi } {\int_0^1 {\ln \sqrt {r^2 } } } rdrd\theta \\& =\int_0^{2\pi } {(\left. {\frac{{r^2 }}{2}\ln r} \right|_0^1\int_0^1 {\frac{r}{2}} dr)} d\theta \\ & =-\frac{1}{2}\int_0^{2\pi } {\int_0^1 r dr} d\theta \\ & =-\frac{1}{2}\int_0^{2\pi } {\frac{1}{2}} d\theta = – \frac{\pi }{2}　∎ \end{array}$

◎ $\left. {\frac{{r^2 }}{2}\ln r} \right|_0^1 =0 – \mathop {\lim }\limits_{r \to 0} \frac{{r^2 }}{2}\ln r=\mathop {\lim }\limits_{r \to 0} \frac{{\ln r}}{{2r^{ – 2} }}\mathop = \limits^{\rm{L}} \mathop {\lim }\limits_{r \to 0} \frac{{\frac{1}{r}}}{{2( – 2)r^{ – 3} }}=\mathop {\lim }\limits_{r \to 0} \frac{{r^2 }}{{ – 4}} = 0$

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此題考點：二重積分座標變換

本題要算同心球夾層裡面，限制範圍的體積。首先把示意圖畫出來才好標示積分範圍。此題積分出現很多平方，用球座標來標示會比較容易。

$\begin{array}{cl} V=2\iiint_{\Omega}dV &=2\int_0^{2\pi } {\int_0^{\frac{\pi }{4}} {\int_1^{\sqrt 2 } {r^2 \sin \rho } } } drd\rho d\theta \\& =2\int_0^{2\pi } {\int_0^{\frac{\pi }{4}} {\sin \rho \left[ {\frac{{r^3 }}{3}} \right]_{r = 1}^{r = \sqrt 2 } } } d\rho d\theta \\ & =\frac{{2(2\sqrt 2 – 1)}}{3}\int_0^{2\pi } {\left[ { – \cos \rho } \right]_{\rho = 0}^{\rho = \frac{\pi }{4}} } d\theta \\ & =\frac{{2(2\sqrt 2 – 1)}}{3} \times (1 – \frac{{\sqrt 2 }}{2}) \times \underbrace {\int_0^{2\pi } {d\theta } }_{2\pi }\\&=\frac{{2\pi }}{3}(5\sqrt 2 – 6) 　∎ \end{array}$

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此題考點：柱座標與球座標

本題二重積分的積分區域與被積函數皆與橢圓有關，因此使用極座標變換。

Let $\left\{ {\begin{array}{c}
{u = \frac{x}{2}} \\
{v = \frac{y}{3}} \\
\end{array}}\right.\Rightarrow \left\{ {\begin{array}{c}
{x = 2u} \\
{y = 3v} \\
\end{array}} \right.\Rightarrow \left\{ {\begin{array}{c}
{dx = 2du} \\
{dy = 3dv} \\
\end{array}} \right.$

$\begin{array}{cl} \iint_{R}e^{9x^2+4y^2}dA &=\iint_{u^2v^2\leq 1}e^{36u^2+36v^2}(2du)(3dv) \\& =6\iint_{u^2+v^2\leq 1}e^{36u^2+36v^2}dudv\ \\&\text{let }\left\{ {\begin{array}{c}
{u = r\cos \theta } \\
{v = r\sin \theta } \\
\end{array}} \right. \\ & = 6\int_0^{2\pi } {\int_0^1 {e^{36r^2 } } } rdrd\theta \\& \text{let }\omega =36r^2\Rightarrow d\omega =72rdr \\ & =6\int_0^{2\pi } {\int_0^{36} {e^\omega } } \frac{{d\omega }}{{72}}d\theta \\&=\frac{1}{{12}}\int_0^{2\pi } {\left. {e^\omega } \right|_{\omega = 0}^{\omega = 36} } d\theta = \frac{\pi }{6}(e^{36} – 1)　∎ \end{array}$

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此題考點：二重積分座標變換

遇到包含 sine 、 cosine 的參數式並非都是極座標，計算面積時需考慮函數的走向，有些時候會差負號。

◎f不自交的曲線下面積：令 $x=g(t)$, $y=f(t)$
若圖形隨 $t$ 增加為逆時針方向，則面積$= \int_a^b {f(t)} g'(t)dt$ ；
若為順時針方向則面積$= -\int_a^b {f(t)} g'(t)dt$

$\begin{array}{cl} \text{Area} &= – 2\int_0^\pi {y(t)} \cdot \underbrace {x'(t)dt}_{dx} \\& = – 2\int_0^\pi {(2\sin t – \sin 2t)( – 2\sin t + 2\sin 2t)} dt\\&= – 2\int_0^\pi {( – 4\sin ^2 t – 2\sin ^2 2t + 6\sin 2t\sin t)} dt\\&= – 2\int_0^\pi { – 2(1 – \cos 2t) – (1 – \cos 4t) – 3(\cos 3t – \cos t)} dt\\&= – 2\int_0^\pi {( – 3 + 2\underbrace {\cos 2t}_0 + \underbrace {\cos 4t}_0 – 3\underbrace {\cos 3t}_0 + 3\underbrace {\cos t}_0)} dt\\&= – 2( – 3\pi ) = 6\pi　∎ \end{array}$

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此題考點：曲線分析(上)

題目已要求使用極座標判斷此極限的連續狀況。

Let $\left\{ {\begin{array}{l}
{x = r\cos \theta } \\
{y = r\sin \theta } \\
\end{array}} \right.$
$\Rightarrow f(x,y)=\left\{ {\begin{array}{l}
{\frac{{r^{2p} \cos ^p \theta \sin ^p \theta }}{{r^4 \cos ^4 \theta + r^4 \sin ^4 \theta }},} & {(x,y) \ne (0,0)} \\{0,} & {(x,y) = (0,0)} \\ \end{array}} \right.$

$\mathop {\lim }\limits_{(x,y) \to (0,0)} f(x,y)$
$=\mathop {\lim }\limits_{(x,y) \to (0,0)} \frac{{r^{2p} \cos ^p \theta \sin ^p \theta }}{{r^4 (\cos ^4 \theta + \sin ^4 \theta )}}$
$=\mathop {\lim }\limits_{(x,y) \to (0,0)} r^{2p – 4} \cdot \frac{{\cos ^p \theta \sin ^p \theta }}{{(\cos ^4 \theta + \sin ^4 \theta )}}$
$=\mathop {\lim }\limits_{(x,y) \to (0,0)} r^{2p – 4} \cdot \underbrace {\cos ^{p – 2} \theta \sin ^{p – 2} \theta }_{\text{between }-1\text{ to }1} \cdot \underbrace {\frac{{\cos ^p \theta \sin ^p \theta }}{{(\cos ^4 \theta + \sin ^4 \theta )}}}_{\text{between }\rm{ 0 }\text{ to }\frac{1}{2}}$
$=\left\{ {\begin{array}{l}
{{\rm{0,}}} & {p > 2}\\{{\rm{D}}{\rm{.N}}{\rm{.E}}{\rm{.}},} & {p \le 2} \end{array}} \right.$

◎ $(\cos ^2 \theta – \sin ^2 \theta )^2 \ge 0$
◎ $\Rightarrow \cos ^4 \theta + \sin ^4 \theta \ge 2\cos ^2 \theta \sin ^2 \theta$
◎ $\Rightarrow \frac{{\cos ^2 \sin ^2 \theta }}{{\cos ^4 \theta + \sin ^4 \theta }} \le \frac{1}{2}$

If $p>1$, $\mathop {\lim }\limits_{(x,y) \to (0,0)} f(x,y) = 0 = f(0,0)$
$\Rightarrow f(x,y)$ is continuous at $(0,0)$.

If $p\leq 2$, $\mathop {\lim }\limits_{(x,y) \to (0,0)} f(x,y) \ne f(0,0)$
$\Rightarrow f(x,y)$ is discontinuous at $(0,0).　∎$

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此題考點：二變數函數極限特殊求法

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