臺灣綜合大學系統 113 年微積分B 轉學考考古題｜附 AI 解答

Question 1

Given function $f(x) = (x^2 - x + 1)^{100}$.

(a) Find the first derivative $f'(0)$.

(b) Find the second derivative $f''(0)$.

Accepted Answer

**(a)** $f'(x) = 100(x^2-x+1)^{99}\cdot(2x-1)$

$$f'(0) = 100\cdot(0-0+1)^{99}\cdot(0-1) = 100\cdot 1\cdot(-1) = \boxed{-100}$$

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**(b)** 對 $f'(x) = 100(x^2-x+1)^{99}(2x-1)$ 再用乘積法則：

$$f''(x) = 100\left[99(x^2-x+1)^{98}(2x-1)^2 + (x^2-x+1)^{99}\cdot 2\right]$$

代入 $x=0$（此時 $x^2-x+1=1$，$2x-1=-1$）：

$$f''(0) = 100\left[99\cdot 1^{98}\cdot(-1)^2 + 1^{99}\cdot 2\right] = 100(99+2) = \boxed{10100}$$

Question 2

Evaluate the limit. $\lim\limits_{x 	o 0} \frac{\sqrt{3 + \cos 3x} - 2}{x^2}$

Accepted Answer

分子分母趨近 $0$，用 Taylor 展開處理。

$$\cos(3x) = 1 - \frac{(3x)^2}{2} + O(x^4) = 1 - \frac{9x^2}{2} + O(x^4)$$

$$3 + \cos(3x) = 4 - \frac{9x^2}{2} + O(x^4)$$

$$\sqrt{4 - \frac{9x^2}{2}} = 2\sqrt{1 - \frac{9x^2}{8}} \approx 2\left(1 - \frac{9x^2}{16}ight) = 2 - \frac{9x^2}{8}$$

故分子：
$$\sqrt{3+\cos 3x} - 2 \approx -\frac{9x^2}{8}$$

$$\lim_{x	o 0}\frac{-\frac{9x^2}{8}}{x^2} = \boxed{-\dfrac{9}{8}}$$

Question 3

(a) Given function $f(x) = 2^{x^2}$. Find the derivative $f'(2)$.

(b) Given function $g(x) = x^{2x}$. Find the derivative $g'(2)$.

Accepted Answer

**(a)** $f(x) = 2^{x^2} = e^{x^2\ln 2}$，鏈鎖律：

$$f'(x) = 2^{x^2}\cdot\ln 2\cdot 2x$$

$$f'(2) = 2^4\cdot\ln 2\cdot 4 = 16\cdot 4\ln 2 = \boxed{64\ln 2}$$

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**(b)** $g(x) = x^{2x}$，取對數：$\ln g = 2x\ln x$

兩邊對 $x$ 微分：
$$\frac{g'}{g} = 2\ln x + 2x\cdot\frac{1}{x} = 2\ln x + 2$$

$$g'(x) = x^{2x}(2\ln x+2)$$

$$g'(2) = 2^4\cdot(2\ln 2+2) = 16\cdot 2(\ln 2+1) = \boxed{32(\ln 2+1)}$$

Question 4

Evaluate the definite integral. $\displaystyle\int_2^3 \frac{x}{x^2 - 5x + 4} dx$

Accepted Answer

**Step 1：因式分解分母**

$$x^2-5x+4 = (x-1)(x-4)$$

**Step 2：部分分式分解**

$$\frac{x}{(x-1)(x-4)} = \frac{A}{x-1}+\frac{B}{x-4}$$

$x = A(x-4)+B(x-1)$

- $x=1$：$1 = -3A \Rightarrow A = -\dfrac{1}{3}$
- $x=4$：$4 = 3B \Rightarrow B = \dfrac{4}{3}$

**Step 3：積分**

$$\int_2^3\left(\frac{-1/3}{x-1}+\frac{4/3}{x-4}ight)dx = \left[-\frac{1}{3}\ln|x-1|+\frac{4}{3}\ln|x-4|ight]_2^3$$

代入上限 $x=3$：$-\dfrac{1}{3}\ln 2 + \dfrac{4}{3}\ln 1 = -\dfrac{\ln 2}{3}$

代入下限 $x=2$：$-\dfrac{1}{3}\ln 1 + \dfrac{4}{3}\ln 2 = \dfrac{4\ln 2}{3}$

$$	ext{結果} = -\frac{\ln 2}{3} - \frac{4\ln 2}{3} = \boxed{-\dfrac{5\ln 2}{3}}$$

Question 5

Evaluate the improper integral. $\displaystyle\int_3^{\infty} e^{-\sqrt{3x}} dx$

Accepted Answer

**Step 1：換元**

令 $u = \sqrt{3x}$，則 $u^2 = 3x$，$x = \dfrac{u^2}{3}$，$dx = \dfrac{2u}{3}\,du$。

積分限：$x=3 \Rightarrow u=3$；$x	o\infty \Rightarrow u	o\infty$。

$$\int_3^\infty e^{-\sqrt{3x}}\,dx = \int_3^\infty e^{-u}\cdot\frac{2u}{3}\,du = \frac{2}{3}\int_3^\infty u e^{-u}\,du$$

**Step 2：分部積分**

$$\int u e^{-u}\,du = -ue^{-u} + \int e^{-u}\,du = -ue^{-u}-e^{-u} = -(u+1)e^{-u}$$

**Step 3：代入上下限**

$$\frac{2}{3}\Big[-(u+1)e^{-u}\Big]_3^\infty = \frac{2}{3}\left[0-\big(-(3+1)e^{-3}\big)ight] = \frac{2}{3}\cdot 4e^{-3}$$

$$= \boxed{\dfrac{8}{3e^3}}$$

Question 6

Given the Taylor series of the function as below. Find the values of $c_1, c_2, c_3$. $$\sqrt[5]{32 + x} = 2 + c_1 x + c_2 x^2 + c_3 x^3 + \cdots$$

Accepted Answer

令 $f(x) = (32+x)^{1/5}$，則 $c_k = \dfrac{f^{(k)}(0)}{k!}$。

**計算各階導數在 $x=0$ 的值：**

利用 $32^{1/5} = 2$，且 $32^{n/5} = 2^n$：

$$f'(x) = \frac{1}{5}(32+x)^{-4/5} \Rightarrow f'(0) = \frac{1}{5}\cdot 32^{-4/5} = \frac{1}{5}\cdot\frac{1}{2^4} = \frac{1}{80}$$

$$f''(x) = \frac{1}{5}\cdot\left(-\frac{4}{5}\right)(32+x)^{-9/5} \Rightarrow f''(0) = -\frac{4}{25}\cdot\frac{1}{2^9} = -\frac{4}{25\cdot 512} = -\frac{1}{3200}$$

$$f'''(x) = \frac{1}{5}\cdot\left(-\frac{4}{5}\right)\cdot\left(-\frac{9}{5}\right)(32+x)^{-14/5} \Rightarrow f'''(0) = \frac{36}{125}\cdot\frac{1}{2^{14}} = \frac{36}{125\cdot 16384} = \frac{9}{512000}$$

**求係數：**

$$c_1 = \frac{f'(0)}{1!} = \boxed{\dfrac{1}{80}}$$

$$c_2 = \frac{f''(0)}{2!} = \frac{-1/3200}{2} = \boxed{-\dfrac{1}{6400}}$$

$$c_3 = \frac{f'''(0)}{3!} = \frac{9/512000}{6} = \boxed{\dfrac{3}{1024000}}$$

Question 7

(a) Given function $f(x) = \sin(3x)$. Find the higher derivative $f^{(111)}(0)$.

(b) Given function $g(x) = \sin(x^3)$. Find the higher derivative $g^{(111)}(0)$.

Accepted Answer

利用 Taylor 級數中係數與高階導數的關係：$f^{(n)}(0) = n!\cdot[x^n 	ext{ 係數}]$

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**(a)** $f(x) = \sin(3x) = \displaystyle\sum_{n=0}^\infty \frac{(-1)^n(3x)^{2n+1}}{(2n+1)!} = \sum_{n=0}^\infty \frac{(-1)^n 3^{2n+1}}{(2n+1)!}x^{2n+1}$

$x^{111}$ 對應 $2n+1=111$，即 $n=55$：

$$[x^{111}	ext{ 係數}] = \frac{(-1)^{55}\cdot 3^{111}}{111!} = -\frac{3^{111}}{111!}$$

$$f^{(111)}(0) = 111!\cdot\left(-\frac{3^{111}}{111!}ight) = \boxed{-3^{111}}$$

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**(b)** $g(x) = \sin(x^3) = \displaystyle\sum_{n=0}^\infty \frac{(-1)^n x^{3(2n+1)}}{(2n+1)!} = \sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}x^{6n+3}$

$x^{111}$ 對應 $6n+3=111$，即 $n=18$：

$$[x^{111}	ext{ 係數}] = \frac{(-1)^{18}}{(2\cdot 18+1)!} = \frac{1}{37!}$$

$$g^{(111)}(0) = 111!\cdot\frac{1}{37!} = \boxed{\dfrac{111!}{37!}}$$

Question 8

Given function $f(x,y) = (4x^2 + y^2)e^{-2x}$. Find all critical points and determine their types (local maximum/local minimum/saddle point).

Accepted Answer

**Step 1：求偏導數並令其為零**

$$f_x = e^{-2x}\big[8x - 2(4x^2+y^2)\big] = e^{-2x}(8x-8x^2-2y^2) = 0$$

由於 $e^{-2x}>0$，得：$4x - 4x^2 - y^2 = 0 \quad\cdots(i)$

$$f_y = 2ye^{-2x} = 0 \Rightarrow y = 0 \quad\cdots(ii)$$

代 $y=0$ 入 $(i)$：$4x(1-x)=0 \Rightarrow x=0$ 或 $x=1$

**臨界點：$(0,0)$ 與 $(1,0)$**

**Step 2：計算二階偏導數**

$$f_{xx} = e^{-2x}(16x^2-32x+4y^2+8)$$
$$f_{yy} = 2e^{-2x}$$
$$f_{xy} = -4ye^{-2x}$$

**Step 3：判別式 $D = f_{xx}f_{yy} - (f_{xy})^2$**

**在 $(0,0)$：**
$$f_{xx}=8,\quad f_{yy}=2,\quad f_{xy}=0$$
$$D = 8\cdot 2 - 0 = 16 > 0,\quad f_{xx}=8>0$$
$\Rightarrow$ **(0, 0) 為局部極小值**，$f(0,0)=0$

**在 $(1,0)$：**
$$f_{xx} = e^{-2}(16-32+8) = -8e^{-2}$$
$$f_{yy} = 2e^{-2},\quad f_{xy}=0$$
$$D = (-8e^{-2})(2e^{-2}) = -16e^{-4} < 0$$
$\Rightarrow$ **(1, 0) 為鞍點（saddle point）**

Question 9

The Cobb-Douglas production function is $P(x,y) = x^{1/3}y^{2/3}$ ($x$: capital, $y$: labour) subject to budget constraint $3x^{1/2} + 5y^{1/2} = 45$. Use the method of Lagrange multiplier to find the values of $x, y$ such that $P$ is maximized.

Accepted Answer

**Step 1：建立 Lagrange 條件 $
abla P = \lambda
abla g$**，其中 $g = 3\sqrt{x}+5\sqrt{y}-45$

$$P_x = \frac{1}{3}x^{-2/3}y^{2/3} = \lambda\cdot\frac{3}{2\sqrt{x}} \quad\cdots(1)$$
$$P_y = \frac{2}{3}x^{1/3}y^{-1/3} = \lambda\cdot\frac{5}{2\sqrt{y}} \quad\cdots(2)$$

**Step 2：消去 $\lambda$（(1)÷(2)）**

$$\frac{P_x}{P_y} = \frac{y}{2x} = \frac{3\sqrt{y}}{5\sqrt{x}\cdot 2}\cdot\frac{2\sqrt{y}}{5\sqrt{x}}$$

更直接地：$(1)\div(2)$：

$$\frac{(1/3)x^{-2/3}y^{2/3}}{(2/3)x^{1/3}y^{-1/3}} = \frac{3/(2\sqrt{x})}{5/(2\sqrt{y})} \implies \frac{y}{2x} = \frac{3\sqrt{y}}{5\sqrt{x}}$$

$$\frac{\sqrt{y}}{2\sqrt{x}} = \frac{3}{5} \implies \sqrt{\frac{y}{x}} = \frac{6}{5} \implies \frac{y}{x} = \frac{36}{25}$$

故 $y = \dfrac{36x}{25}$。

**Step 3：代入約束條件**

$$3\sqrt{x}+5\sqrt{\frac{36x}{25}} = 45$$
$$3\sqrt{x}+5\cdot\frac{6\sqrt{x}}{5} = 45$$
$$3\sqrt{x}+6\sqrt{x} = 45$$
$$9\sqrt{x} = 45 \implies \sqrt{x}=5 \implies \boxed{x=25}$$

$$y = \frac{36\cdot 25}{25} = \boxed{y=36}$$

此時 $P(25,36) = 25^{1/3}\cdot 36^{2/3}$。

Question 10

Evaluate the double integral. $\displaystyle\int_0^4 \int_{\sqrt{x}}^2 \sqrt{x}\cdot\frac{\sin(y^2)}{y^2} dy dx$

Accepted Answer

**Step 1：確認積分域**

原積分：$0\leq x\leq 4$，$\sqrt{x}\leq y\leq 2$

等價描述：$0\leq y\leq 2$，$0\leq x\leq y^2$

**Step 2：交換積分順序**

$$\int_0^4\int_{\sqrt{x}}^2\sqrt{x}\cdot\frac{\sin(y^2)}{y^2}\,dy\,dx = \int_0^2\int_0^{y^2}\sqrt{x}\cdot\frac{\sin(y^2)}{y^2}\,dx\,dy$$

**Step 3：先對 $x$ 積分**

$$\int_0^{y^2}\sqrt{x}\,dx = \left[\frac{2}{3}x^{3/2}\right]_0^{y^2} = \frac{2}{3}(y^2)^{3/2} = \frac{2y^3}{3}$$

**Step 4：再對 $y$ 積分**

$$\int_0^2\frac{\sin(y^2)}{y^2}\cdot\frac{2y^3}{3}\,dy = \int_0^2\frac{2y}{3}\sin(y^2)\,dy$$

令 $u=y^2$，$du=2y\,dy$：

$$= \int_0^4\frac{1}{3}\sin u\,du = \frac{1}{3}\Big[-\cos u\Big]_0^4 = \frac{1}{3}(-\cos 4+\cos 0)$$

$$= \boxed{\dfrac{1-\cos 4}{3}}$$

臺灣綜合大學系統 113 年度微積分B

這份試題整理

微積分B 其他年度考古題

臺灣綜合大學系統 113 年度 微積分B

這份試題整理

微積分B 其他年度考古題

臺灣綜合大學系統 113 年度微積分B