109年公費留考-微積分詳解

教育部109年公費留學考試試題

科目：微積分

解答: 

$$\mathbf{(a)}\; 圖形如上，f(x)={\ln(x)\over x} \Rightarrow f'(x)={1-\ln(x)\over x^2} \Rightarrow f''(x)={2\ln(x)-3\over x^3}\\ 若f'(x)=0 \Rightarrow x=e \Rightarrow f''(e)=-{1\over e^3} \lt 0 \Rightarrow f(e)= {1\over e}為相對極大值\\ 若f''(x)=0 \Rightarrow x=e^{3/2} \Rightarrow \cases{\lim_{x\to (e^{3/2})^+}f''(x)\gt 0 \\\lim_{x\to (e^{3/2})^-}f''(x)\lt 0 } \Rightarrow x=e^{3/2}為反曲點\\ \\ 因此\bbox[red, 2pt]{\cases{\text{local extreme value: }f(e)=1/e\\ \text{point of inflection: }x=e^{3/2}\\ \text{asymptote: }x=0}}\\ \mathbf{(b)}\;泰勒展開式：e^x =1+x+ {x^2\over 2!}+{x^3\over 3!}+\cdots \gt 0 \Rightarrow e^x \gt 1+x \\ \Rightarrow e^{\pi/e-1} \gt 1+\left({\pi \over e}-1\right) ={\pi\over e} \Rightarrow {e^{\pi/e}\over e} \gt {\pi\over e} \Rightarrow e^{\pi/e}\gt \pi \Rightarrow \left(e^{\pi/e} \right)^e \gt \pi^e \Rightarrow \bbox[red, 2pt]{e^\pi\gt \pi^e} \\\mathbf{(c)} \; f(x)=x^{1/x} \Rightarrow f'(x)= {x^{1/x}(1-\ln x)\over x^2} \Rightarrow f''(x)={x^{1/x}\over x^4}(1-3x+2x\ln(x)+(\ln(x))^2-2\ln(x))\\ 若f'(x)=0 \Rightarrow x=e \Rightarrow f''(e)\lt 0 \Rightarrow f(e)為極大值\\ L=x^{1/x} \Rightarrow \ln L={\ln x\over x} \Rightarrow \lim_{x\to \infty} \ln L= \lim_{x\to \infty}{\ln x\over x}=0 \Rightarrow  \lim_{x\to \infty} x^{1/x}=e^0=1\\f(x)為連續且可微，又\cases{f為嚴格遞增，x\in (1,e)\\ f為嚴格遞減，x \in(e,\infty)} 且\cases{f(x)\in(1,e^{1/e}),x\in(1,e) \\ f(x)\in (1,e^{1/e}),x\in (e,\infty)}\\ 因此對任意a\in(1,e)存在唯一b\in (e,\infty)使得a^{1/a}=b^{1/b}，\bbox[red, 2pt]{故得證}$$

解答: 

$$假設\cases{直圓錐底面圓心為D\\ 直圓錐底面圓半徑為r\\直圓錐頂點P\\ Q為直圓錐與圓球切點\\球心O\\ \overline{PO}=a \\ \angle OPQ=\theta}\;如上圖\\ 則\overline{PQ} =\sqrt{a^2-R^2}且\tan \theta ={\overline{OQ} \over \overline{PQ}} ={\overline{DB} \over \overline{DP}} \Rightarrow {R\over \sqrt{a^2-R^2}} ={r\over a+R} \Rightarrow r={R(a+R)\over \sqrt{a^2-R^2}} \\ \Rightarrow 直圓錐體積={1\over 3}\pi r^2\cdot (a+R) ={1\over 3}\pi\cdot {R^2(a+R)^3\over a^2-R^2}\\ 令f(a)={(a+R)^3\over a^2-R^2} \Rightarrow f'(a)={3(a^2-R^2)(a+R)^2-2a(a+R)^3 \over (a^2-R^2)^2}\\ 若f'(a)=0 \Rightarrow (a+R)^2(3a^2-3R^2-2a^2-2aR)=0 \Rightarrow (a-3R)(a+R)^3=0\\ \Rightarrow a=3R \Rightarrow 直圓錐體積={1\over 3}\pi\cdot {R^2(4R)^3\over 9R^2-R^2} =\bbox[red, 2pt]{{8\over 3}\pi R^3}$$

解答: $$題目有\bbox[cyan,2pt]{疑義}，若0\lt x\le 1，則表面積為無窮大$$

解答: $$令f(a,b)=\int_0^{\pi/2} (ax+b-\sin x)^2\,dx =\int_0^{\pi/2} (a^2x^2 +2abx-2ax\sin x+b^2-2b\sin x+\sin^2 x)\,dx \\= {1 \over 24}a^2 \pi^3 +{1\over 2}b^2\pi+{1\over 4}\pi +{1\over 4}ab\pi^2 -2a-2b \Rightarrow \cases{f_a=(a\pi^3+3b\pi^2 -24)/12 \\ f_b=a\pi^2/4+b\pi-2} \\ 若\cases{f_a=0 \\ f_b=0} \Rightarrow \bbox[red, 2pt]{\cases{a=-24(\pi-4)/\pi^3\\ b= 8(\pi-3)/\pi^2} }$$

解答: 

$$兩圖形\cases{2y=x\\ x^2+4y^2=4}在第一象限的交點A(\sqrt 2,\sqrt 2/2)\\ 因此\iint_D{y\over x}dxdy = \int_0^\sqrt 2 \int_0^{x/2} {y\over x}\,dydx +\int_\sqrt 2^2 \int_0^\sqrt{(4-x^2)/4)} {y\over x}\,dydx \\= \int_0^\sqrt 2 \left. \left[ {1\over 2x}y^2 \right]\right|_0^{x/2}\,dx +\int_\sqrt 2^2 \left. \left[ {1\over 2x}y^2 \right] \right|_0^{\sqrt{(4-x^2)/4}}\,dx = \int_0^\sqrt 2 {1\over 8}x\,dx +\int_\sqrt 2^2 {1\over 2x}-{1\over 8}x\,dx \\= \left.\left[ {1\over 16} x^2\right] \right|_0^\sqrt 2 + \left.\left[{1\over 2}\ln x -{1\over 16}x^2\right] \right|_{\sqrt 2}^{2} ={1\over 8} +{1\over 4}\ln 2-{1\over 8} =\bbox[red, 2pt]{{1\over 4}\ln 2}$$

解答: $$請參考\href{https://www.gregschool.org/newtons-laws/2017/10/27/gravitational-force-exerted-by-a-sphere-geml5}{Greg School}$$

解答: $$令f(n)={1\over \sqrt 1}+ {1\over \sqrt 2}+ \cdots +{1\over \sqrt n} \\先用歸納法證明：f(n) \ge \sqrt n\\ n=1 \Rightarrow {1\over \sqrt{1}} =\sqrt 1  \Rightarrow 原式成立\\ 假設n=k時，原式亦成立，即f(k) \ge \sqrt k\\ 當n=k+1時，f(k+1)\ge \sqrt k +{1\over \sqrt{k+1}}  ={\sqrt k \cdot \sqrt{k+1}+1 \over \sqrt{k+1}} \ge{\sqrt k \cdot \sqrt{k}+1 \over \sqrt{k+1}} =\sqrt{k+1} \\ \Rightarrow f(k+1) \ge \sqrt{k+1}，原式亦成立\\再用歸納法證明: f(n)\le 2\sqrt n\\n=1時顯然成立，並假設n=k時原式成立，即f(k)\le 2\sqrt k\\ n=k+1時，f(k+1) \le 2\sqrt k+{1\over \sqrt{k+1}} \\= {2\sqrt k \cdot \sqrt{k+1}+1 \over \sqrt{k+1}} \le {2\sqrt{k^2+k+1/4}+1\over \sqrt{k+1}} ={2(k+1/2)+1\over \sqrt{k+1}} =2\sqrt{k+1} \\ \Rightarrow f(k+1) \le 2\sqrt{k+1}，原式亦成立\\ 因此{ 2\sqrt n\over n^p}\ge{f(n) \over n^p}  \ge {\sqrt n\over n^p}  \Rightarrow 2n^{1/2-p}\ge {f(n) \over n^p}\ge n^{1/2-p}\\ \Rightarrow 取\bbox[red, 2pt]{p={1\over 2}}\Rightarrow 2\ge \lim_{n\to \infty}{f(n) \over n^p} \ge 1\Rightarrow   \lim_{n\to \infty}{f(n) \over n^p} \text{ is finite and nonzero}$$

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