台聯大轉學考微積分 108 A2 卷解答

知道平均的意義還有積分遇到絕對值如何處理即可。

$\begin{array}{cl} \text{AVG.} & =\frac{{\int_{ – 1}^3 {(\left| x \right| – 1)} dx}}{{3 – ( – 1)}} = \frac{{\int_{ – 1}^0 { – x – 1} dx + \int_0^3 {(x – 1)} dx}}{4} \\ & =\frac{{(\left. { – \frac{{x^2 }}{2} – x} \right|_{ – 1}^0 ) +(\left. {\frac{{x^2 }}{2} – x} \right|_0^3 )}}{4} \\ & =\frac{{0 – ( – \frac{1}{2} + 1) + (\frac{9}{2} – 3) – 0}}{4} = \frac{1}{4}　∎ \end{array}$

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此題考點：定積分直觀觀念

四大積分法：變數變換、三角置換、分部積分和部分分式為必考題，一定要熟悉。考前將一些三角函數的微分、積分結果有記熟可以節省很多計算的時間。

$\begin{array}{cl} \int {x\sec ^2 x} dx & =x\tan x – \int {\tan x} dx \\ & =x\tan x + \ln \left| {\cos x} \right| + C \\ & =x\tan x – \ln \left| {\sec x} \right| + C　∎ \end{array}$

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此題考點：分部積分

遇不易算的二重積分，可嘗試極座標變換或交換積分次序 (Fubini 定理)，本題使用交換積分次序。

$\begin{array}{cl} \int_0^2 {\int_{\frac{y}{2}}^1 {ye^{x^3 } } } dxdy & =\int_0^1 {e^{x^3 } } \int_0^{2x} y dydx \\ & =\int_0^1 {e^{x^3 } } \cdot \left. {\frac{{y^2 }}{2}} \right|_{y = 0}^{y = 2x} dx \\ & =\int_0^1 {e^{x^3 } } \cdot \frac{{4x^2 }}{2}dx \\&=2\int_0^1 {e^{x^3 } } x^2 dx\\&(\text{Let }u = x^3 \Rightarrow du = 3x^2 dx)\\&=\frac{2}{3}\int_0^1 {e^u } du\\&=\frac{2}{3} \cdot \left. {e^u } \right|_{u = 0}^{u = 1} = \frac{2}{3}(e – 1)　∎ \end{array}$

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此題考點：二變數函數的積分

本題希望能找到合適的 $a,b$ 使得分段函數 $f(x)$ 在 $x=0$ 可微，需要接點兩邊的極限要一樣之外，微分也要相同。

$\because\ f(x)$ is differential at $x=0$
$\therefore\ f(x)$ is continued at $x=0$
$\Rightarrow \mathop {\lim }\limits_{x \to 0^ – } f(x) = \mathop {\lim }\limits_{x \to 0^ + } f(x)$
$\Rightarrow \left\{ \begin{array}{l} \mathop {\lim }\limits_{x \to 0^ – } f(x) = \mathop {\lim }\limits_{x \to 0^ – } a\underbrace x_0 + b\underbrace {\cos x}_1 = b \\ {\mathop {\lim }\limits_{x \to 0^ + } f(x) = \mathop {\lim }\limits_{x \to 0^ + } \frac{{2\sin ^2 x}}{x} = \mathop {\lim }\limits_{x \to 0^ + } 2\underbrace {\sin x}_0 \cdot \underbrace {\frac{{\sin x}}{x}}_1 = 0} \end{array} \right.\Rightarrow b = 0$
$\Rightarrow f(x) = \left\{ {\begin{array}{c}\frac{{2\sin ^2 x}}{x}, & \text{if }x > 0 \\ {ax,} & \text{if }x \le 0 \\ \end{array}} \right.$
$\Rightarrow f'(x) = \left\{ {\begin{array}{c}\frac{{(4\sin x\cos x) \cdot x – 2\sin ^2 x}}{{x^2 }}, & \text{if }x > 0 \\ {a,} & \text{if }x \le 0 \\ \end{array}} \right.$
$\because\ f(x)$ is differentiable at $x=0$
$\therefore\ \mathop {\lim }\limits_{x \to 0^ – } f'(x) = \mathop {\lim }\limits_{x \to 0^ + } f'(x)$
$\Rightarrow \mathop {\lim }\limits_{x \to 0^ – } a = \mathop {\lim }\limits_{x \to 0^ + } (4 \cdot \underbrace {\frac{{\sin x}}{x}}_1 \cdot \underbrace {\cos x}_1 – 2 \cdot \underbrace {\frac{{\sin ^2 x}}{{x^2 }}}_1) = 4 – 2 = 2 \Rightarrow a = 2$
Thus $a=2,\ b=0　∎$

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此題考點：導數與微分的概念

本題要找隱函數在點 $(0,\pi)$ 的切線，我們只要對其微分找到斜率即可求出切線方程式。

將原式對 $x$ 微分
$\Rightarrow 2x \cdot \cos ^2 y + x^2 \cdot 2\cos y \cdot ( – \sin y) \cdot y’ – (\cos y) \cdot y’ = 0$
At $(x,y) = (0,\pi ) \Rightarrow \left. {y’} \right|_{(0,\pi )} = 0$

the tangent line: $y=\pi　∎$

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此題考點：切線專論

遇到旋轉體積分，要立刻想到圓盤法或剝殼法，本題使用圓盤法。

$\begin{array}{cl} V & =\int_0^2 {\underbrace {\pi ( – x^2 + 4x)^2 – \pi (x^2 )^2 }_{\text{area of disk}}}\underbrace {dx}_{\text{thickness}} \\ & =\pi \int_0^2 {x^4 – 8x^3 + 16x^2 – x^4 } dx \\ & =\pi \left[ { – 2x^4 + \frac{{16}}{3}x^3 } \right]_{x = 0}^{x = 2} \\&=\pi ( – 2 \times 16 + \frac{{16}}{3} \times 8)\\&=\pi (128 – 96) = \frac{{32}}{3}\pi　∎ \end{array}$

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此題考點：旋轉體積分

指數形式的函數求極限、微分時，碰到底數或指數有未知數時，第一個想法即是換成 $f(x)=e^{\ln f(x)}$。

$\begin{array}{cl} \mathop {\lim }\limits_{x \to 0^ + } (\sin x)^x & =\mathop {\lim }\limits_{x \to 0^ + } e^{\ln (\sin x)^x } \\ & =e^{\mathop {\lim }\limits_{x \to 0^ + } x\ln (\sin x)} \\ & =e^{\mathop {\lim }\limits_{x \to 0^ + } \frac{{\ln (\sin x)}}{{x^{ – 1} }}} \\&\mathop = \limits^{\rm{L}} e^{\mathop {\lim }\limits_{x \to 0^ + } \frac{{\frac{{\cos x}}{{\sin x}}}}{{( – x^{ – 2} )}}}\\&=e^{\mathop {\lim }\limits_{x \to 0^ + } – \frac{{x^2 \cos x}}{{\sin x}}}\\&=e^{\mathop {\lim }\limits_{x \to 0^ + } – \underbrace {(\frac{x}{{\sin x}})}_1 \cdot \underbrace x_0 \cdot \underbrace {\cos x}_1}\\&=e^0=1　∎ \end{array}$

附註：$\mathop = \limits^{\rm{L}}$ 表示此等號使用羅必達法則 (L’Hôpital’s rule)

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此題考點：極限運算題型 II、羅必達法則

處理級數和可以先觀察一般項能不能拆解，或者有沒有已知的關係式讓計算變容易。

$\because\ \ln (1 + \frac{1}{n}) = \ln (\frac{{n + 1}}{n}) = \ln (n + 1) – \ln n$
$\begin{array}{cl}\therefore \sum\limits_{n = 3}^\infty {\frac{{\ln (1 + \frac{1}{n})}}{{(\ln n)\ln (n + 1)}}} & =\sum\limits_{n = 3}^\infty {\frac{{\ln (n + 1) – \ln n}}{{(\ln n)\ln (n + 1)}}} \\ & =\mathop {\lim }\limits_{k \to \infty } \sum\limits_{n = 3}^k {(\frac{1}{{\ln n}} – \frac{1}{{\ln (n + 1)}})} \\ & =\mathop {\lim }\limits_{k \to \infty } \left[ {(\frac{1}{{\ln 3}} – \frac{1}{{\ln 4}}) + (\frac{1}{{\ln 4}} – \frac{1}{{\ln 5}}) + \cdots + (\frac{1}{{\ln k}} – \frac{1}{{\ln (k + 1)}})} \right] \\&=\mathop {\lim }\limits_{k \to \infty } \left[ {\frac{1}{{\ln 3}} – \frac{1}{{\ln (k + 1)}}} \right] = \frac{1}{{\ln 3}}　∎ \end{array}$

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此題考點：級數

依題意將條件列完後就和一般極值問題一樣，在微積分的考試中，有限制的極值問題先想到拉格朗日乘數法。

令盒子的長$=x$, 寬$=y$, 高$=z$
cost: $g(x,y,z) = 3xy + 2xz + 2yz = 36$
volume: $xyz=f(x,y,z)$

Let $\nabla f+\lambda \nabla g=\overset{\rightharpoonup}{0}$
$\Rightarrow (yz,xz,xy) + \lambda (3y + 2z,3x + 2z,2x + 2y) =\overset{\rightharpoonup}{0}$
$\Rightarrow \left\{ {\begin{array}{c} yz + 3\lambda y + 2\lambda z = 0 \cdots (1) \\ xz + 3\lambda x + 2\lambda z = 0 \cdots (2) \\ xy + 2\lambda x + 2\lambda y = 0 \cdots (3) \end{array}} \right.$
　　$3xy+2xz+2yz=36 \cdots (4)$
$\begin{array}{cl} (1)-(2) & \Rightarrow z(y – x) + 3\lambda (y – x) = 0 \\ & \Rightarrow (y – x) \cdot (z + 3\lambda ) = 0 \end{array}$
若 $z=-3\lambda$
$\mathop \Rightarrow \limits^{(1)} – 3\lambda y + 3\lambda y – 6\lambda ^2 = 0 \Rightarrow \lambda = 0 \Rightarrow z = 0$
$\mathop \Rightarrow \limits^{(3)(4)} \left\{ {\begin{array}{c}{xy = 0} & \\ {3xy = 36} & { \Rightarrow xy = 12} \end{array}} \right.\ (\to \leftarrow)$
$\Rightarrow z \ne – 3\lambda$
$y=x$
$\mathop \Rightarrow \limits^{(1)(3)(4)} \left\{ {\begin{array}{l} xz + 3\lambda x + 2\lambda z = 0 \cdots (6) \\ x^2 + 4\lambda x = 0 \Rightarrow x(x + 4\lambda ) = 0 \\ 3x^2 + 4xz = 36 \cdots (5) \end{array}} \right.\Rightarrow x=0\text{ or }x=-4\lambda$
$\mathop \Rightarrow \limits^{(5)} 0 = 36\ (\to \leftarrow)\ \Rightarrow x=-4\lambda$
$\mathop \Rightarrow \limits^{(5)(6)} \left\{ {\begin{array}{l} – 4\lambda z – 12\lambda ^2 + 2\lambda z = 0 \cdots (7) \\ 48\lambda ^2 – 16\lambda z = 36 \cdots (8) \end{array}} \right.$
$\mathop \Rightarrow \limits^{(7)} 12\lambda ^2 + 2\lambda z = 0$
$\Rightarrow 6\lambda + \lambda z = 0\Rightarrow \lambda=0\text{ or }=-6\lambda$
若 $x=0=y\mathop \Rightarrow \limits^{(5)} 0 = 36\ (\to\leftarrow)$
若 $z = – 6\lambda \Rightarrow (x,y,z) = ( – 4\lambda , – 4\lambda , – 6\lambda )$
$\mathop \Rightarrow \limits^{(8)} 48\lambda ^2 + 96\lambda ^2 = 36 \Rightarrow \lambda = \pm \frac{1}{2}$
$\Rightarrow (x,y,z) = ( – 2, – 2, – 3)\text{ or }(2,2,3)$
$\because\ x,y,z>0$
$\therefore\ (x,y,z) = (2,2,3) \Rightarrow \text{Max.}=2 \cdot 2 \cdot 3 = 12　∎$

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此題考點：Lagrange 乘數法

此題要求用定義計算 $x=0$ 時的微分，和用微分性質的結果作比較。

$\begin{array}{cl} g'(0) & =\mathop {\lim }\limits_{h \to 0} \frac{{g(0 + h) – g(0)}}{h} \\ & =\mathop {\lim }\limits_{h \to 0} \frac{{h^2 \sin \frac{1}{h} – 0}}{h} \\ & =\mathop {\lim }\limits_{h \to 0} h \cdot \sin \frac{1}{h} = 0 \end{array}$

For $x\ne 0$
$\begin{array}{cl} g'(x) & =2x \cdot \sin \frac{1}{x} + x^2 (\cos \frac{1}{x}) \cdot \frac{{ – 1}}{x} \\ & =2x\sin \frac{1}{x} – \cos \frac{1}{x} \end{array}$
$\Rightarrow \mathop {\lim }\limits_{x \to 0} g'(x) = \mathop {\lim }\limits_{x \to 0} (2x \cdot \sin \frac{1}{x} – \cos \frac{1}{x})$ D.N.E.
$\Rightarrow g'(0) \ne \mathop {\lim }\limits_{x \to 0} g'(x)　∎$

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此題考點：導數與微分的概念

級數和的歛散性是必考的題目，因此八大審斂法需要很熟悉。第一題和高斯積分有關，可以配合積分審斂法；第二題則可以用 $\mathop {\lim }\limits_{x\to 0} \frac{\sin x}{x}$ 的結論搭配極限比較審斂法。

〔1〕
◎ 高斯積分
Let ${\rm{I}} = \int_0^\infty {e^{ – x^2 } } dx$
$\begin{array}{cl} \rm{I}^2 & =\int_0^\infty {e^{ – y^2 } } dy\int_0^\infty {e^{ – x^2 } } dx \\ & =\int_0^\infty {\int_0^\infty {e^{ – x^2 } e^{ – y^2 } } dx} dy \\ & =\int_0^{\frac{\pi }{2}} {\int_0^\infty {e^{ – r^2 } } dr} d\theta\\&=\frac{\pi }{2} \cdot \frac{{\rm{1}}}{{\rm{2}}} = \frac{\pi }{4}  \end{array}$
$\Rightarrow \int_0^\infty {e^{ – x^2 } } dx = \frac{{\sqrt \pi }}{2}$

$\because\ \int_0^\infty {e^{ – x^2 } } dx = \frac{{\sqrt \pi }}{2}$
$\therefore\ \sum\limits_{n = 1}^\infty {e^{ – n^2 } }\text{ converges. (by integral test)}　∎$
(更嚴謹的做法請參考 A3, A4, A7 卷)

〔2〕
$\begin{array}{cl} \because & \mathop {\lim }\limits_{n \to \infty } \frac{{\sin \frac{1}{n}}}{{\frac{1}{n}}} = \mathop {\lim }\limits_{\theta \to 0^ + } \frac{{\sin \theta }}{\theta } = 1\ \text{Let }\theta=\frac{1}{n} \\ & \text{and }\sum\limits_{n = 1}^\infty {\frac{1}{n}}\ \text{diverges. }(\text{by }p\text{-series test}) \\ \therefore & \sum\limits_{n = 1}^\infty {\sin \frac{1}{n}}\ \text{diverges. (by limit comparison test)}　∎ \end{array}$

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此題考點：p-級數、極限比較審斂法