108年專科學力鑑定考試--微積分詳解

108年專科學校畢業程度自學進修學力鑑定考試

專業科目(一)：微積分 詳解

解：$$(A) \bigcirc:f(x)=\sqrt{(-x+108)(x-2019)}  \Rightarrow (-x+108)(x-2019)\ge 0  \\ \qquad\Rightarrow (x-108)(x-2019)\le 0  \Rightarrow 108 \le x \le 2019\\ (B) \times:f(x)=\sqrt{(x-108)(x-2019)}  \Rightarrow (x-108)(x-2019)\ge 0   \Rightarrow x\ge 2019\;或\;x\le 108 \\(C) \times: f(x)=\sqrt{(x+108)(x+2019)}  \Rightarrow (x+108)(x+2019)\ge 0   \Rightarrow x\ge -108\;或\;x\le -2019\\(D) \times: f(x)=\sqrt{(x+108)(-x+2019)}  \Rightarrow (x+108)(-x+2019)\ge 0  \\\qquad  \Rightarrow (x+108)(x-2019) \le 0\Rightarrow -108\le x \le 2019\\， 故選：\bbox[red,2pt]{(A)}$$

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解：$$f(x)與g(x)在x=2019連續，並不代表在其它點也連續，因此(A)與(C)均錯誤\\若g(2019)=0，則\frac{f(x)}{g(x)}在x=0就不連續， 故選\bbox[red,2pt]{(B)}$$

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解：
$$令g(x)=\frac{(x+1)(x+2)(x+3)(x+4)}{x+5} \Rightarrow f(x)= \frac{x(x+1)(x+2)(x+3)(x+4)}{x+5} =xg(x)\\ \Rightarrow f'(x)=g(x)+xg'(x) \Rightarrow f'(0)=g(0)= \frac{1\cdot 2\cdot 3\cdot 4}{5} = \frac{24}{5} =4.8， 故選\bbox[red,2pt]{(C)}$$

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解：$$f(x)=\sin (\cos (x)) \Rightarrow f'(x)=\cos(\cos (x))(-\sin(x)) \Rightarrow f'({\pi\over 2})= \cos(\cos ({\pi\over 2}))(-\sin({\pi\over 2})) \\= \cos(0)(-1)= 1\times (-1)=-1，故選\bbox[red,2pt]{(D)}$$

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解：$$f(x)= \frac{1}{2019-x} =(2019-x)^{-1} \Rightarrow f'(x)=(2019-x)^{-2}\Rightarrow f''(x)= 2(2019-x)^{-3} \\\Rightarrow f'''(x)= 6(2019-x)^{-4} \Rightarrow f^{[4]}(x)= 24(2019-x)^{-5} \Rightarrow f^{[5]}(x)= 120(2019-x)^{-6} \\  \Rightarrow  f^{[5]}(2018)=120， 故選\bbox[red,2pt]{(A)}$$

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解：$$\begin{cases}f在[0,2019]嚴格遞增且凹向上\\f(0)>0 \end{cases}  \Rightarrow f^2在[0,2019]嚴格遞增且凹向上， 故選\bbox[red,2pt]{(B)}$$

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解：$$\lim_{x\to 2019}(2019-x)\tan( \frac{\pi x}{4038} ) =\lim_{x\to 2019} \frac{(2019-x)\sin\frac{\pi x}{4038}}{\cos \frac{\pi x}{4038}}  =\lim_{x\to 2019} \frac{f(x)}{g(x)} = \lim_{x\to 2019} \frac{f'(x)}{g'(x)}\\ =\lim_{x\to 2019} \frac{-\sin \frac{\pi x}{4038}+ (2019-x)\frac{\pi }{4038}\cos\frac{\pi x}{4038}}{-\frac{\pi }{4038}\sin \frac{\pi x}{4038}} = \frac{-\sin {\pi\over 2}}{-\frac{\pi }{4038}\sin {\pi\over 2}} = \frac{4038}{\pi} ， 故選\bbox[red,2pt]{(C)}$$

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解：$$y=f(x)在[1,3]凹向上 \Rightarrow y=f(x)在[1,3]有極值，即\exists c\in [1,3]使得f'(c)=0 ，故選\bbox[red,2pt]{(D)} $$

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解：$$令g(x)=(2x^{100}-1)^{2019} \Rightarrow F(x)=\int_{3^x}^{9^x}g(x)\;dx \\\Rightarrow F'(x)=g(9^x){d\over dx}9^x- g(3^x){d\over dx}3^x +\int_{3^x}^{9^x}g'(x)\;dx\\ =g(9^x)\ln 9-g(3^x)\ln 3 +(g(9^x)-g(3^x))\\ \Rightarrow F'(0)=g(1)\ln 9-g(1)\ln 3 +(g(1)-g(1)) =\ln 9-\ln 3=\ln 3，故選\bbox[red,2pt]{(A)} $$

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解：

$$區域R如上圖，因此x積分範圍由0.5y-1至-0.5y+1，y積分範圍由0至2，故選\bbox[red,2pt]{(B)}$$

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解：$$f(x,y)=x^y \Rightarrow  \begin{cases} \frac{\partial f}{\partial x} =yx^{y-1}\\\frac{\partial f}{\partial y} =x^y\ln x\end{cases}  \Rightarrow \frac{\partial f}{\partial x}+\frac{\partial f}{\partial y} =yx^{y-1}+ x^y\ln x ，故選\bbox[red,2pt]{(D)}$$

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解：$$ \frac{d}{d t} f(\cos t,\sin t)= f_x(\cos t,\sin t)\frac{d}{d t}\cos t+ f_y(\cos t,\sin t)\frac{d}{d t}\sin t \\=f_x(\cos t,\sin t)(-\sin t)+ f_y(\cos t,\sin t)\cos t\\ \Rightarrow \left. \frac{d}{d t} f(\cos t,\sin t)\right|_{t=\pi/2}=f_x(\cos \pi/2,\sin\pi/2)(-\sin \pi/2)+ f_y(\cos \pi/2,\sin \pi/2)\cos \pi/2 \\=f_x(0,1)(-1)+ f_y(0,1)\cdot 0 =-3，故選\bbox[red,2pt]{(B)}$$

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解：$$  \lim_{n\to\infty} \frac{2n-1}{3n+2+n^{-n}} =\frac{2}{3}\ne 0 \Rightarrow \sum_{n=1}^\infty \frac{2n-1}{3n+2+n^{-n}}發散，故選\bbox[red,2pt]{(D)}$$

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解：
$$  \lim_{n\to\infty} \frac{1+3+\cdots+(2n-1)}{2+4+\cdots+2n} =\lim_{n\to\infty} \frac{2n\times n \div 2}{(2n+2)\times n\div 2}=\lim_{n\to\infty}  \frac{n^2}{n^2+n}=1 ，故選\bbox[red,2pt]{(A)}$$

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解：
$$\int_0^\pi  \frac{1}{2} r^2\;d\theta =\int_0^\pi  \frac{1}{2} \theta^2\;d\theta = \left. \frac{1}{6}\theta^3 \right|_0^\pi =\frac{1}{6}\pi^3，故選\bbox[red,2pt]{(A)}$$

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解：

$$所圍區域如上圖(0\le x,y\le 1)，因此y=(x-1)^2 \Rightarrow x=1-\sqrt{y}\\\Rightarrow 繞y軸旋轉體積為\int_0^1 (1-\sqrt{y})^2\pi\;dy = \pi\int_0^1 1-2\sqrt{y}+y\;dy\\ = \pi\left. \left[ y-{4\over 3}y^{3/2}+ {1\over 2} y^2 \right] \right|_0^1=\pi(1-{4\over 3}+{1\over 2}) ={1\over 6}\pi，故選\bbox[red,2pt]{(A)}$$

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解：

$$f(x)=g(x) \Rightarrow 交點\begin{cases} A=(-1,1)\\ B=(3,9)\end{cases}，見上圖\\ \Rightarrow 所圍面積=\int_{-1}^3 f(x)-g(x)\;dx = \int_{-1}^3 2x+3-x^2\;dx = \left. \left[ x^2+3x-{1\over 3}x^3 \right] \right|_{-1}^3 =9+{5\over 3}\\= {32\over 3}，故選\bbox[red,2pt]{(C)}$$

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解：$$y=x^x=e^{x\ln x} \Rightarrow {dy\over dx}=(\ln x+1)e^{x\ln x}=(\ln x+1)x^x，故選\bbox[red,2pt]{(D)} $$

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解：$$\text{依萊布尼茲積分原則(Leibniz integral rule) }{d\over dx}\int_0^x f(t)dt=f(x)\\ \Rightarrow {d\over dx}\int_0^x  \frac{50t^{49}}{1+t^{100}} dt= \frac{50x^{49}}{1+x^{100}}，故選\bbox[red,2pt]{(C)}$$

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解：$$令f(x)=\sin x^3 \Rightarrow f(-x)=-f(x) \Rightarrow f(x)是奇函數 \Rightarrow \int_{-1}^1 f(x)dx=0，故選\bbox[red,2pt]{(B)}$$

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解題僅供參考