國立臺灣大學 114 學年度碩士班招生考試試題
科目: 微積分(D)
Part I: 15 points for each problem. Make sure your work is organized and logical.
解答:$$\textbf{(a) }x=3\sec \theta \Rightarrow dx= 3\sec \theta \tan \theta d\theta \Rightarrow I=\int_6^\infty {81\over x^3(x^2-9)}\,dx= \int_{\pi/3}^{\pi/2} {81(3\sec \theta \tan\theta) d\theta\over (27\sec^3\theta)(9\tan^2\theta)} \\\quad =\int_{\pi/3}^{\pi/2} {1\over \sec^2\theta \tan \theta}\,d\theta =\int_{\pi/3}^{\pi/2} {\cos^3\theta\over \sin \theta}\,d\theta =\int_{\sqrt 3/2}^1 {1-u^2\over u} \,du \;(u=\sin \theta \Rightarrow du=\cos \theta \,d\theta) \\\quad =\int_{\sqrt 3/2}^1 \left( {1\over u}-u \right)\,du = \left. \left[ \ln|u|-{1\over 2}u^2 \right] \right|_{\sqrt 3/2}^1 = \bbox[red, 2pt]{\ln 2-{1\over 2}\ln 3-{1\over 8}} \\\textbf{(b) } I=\int_6^\infty \left( -{1\over x}-{9\over x^3}+{1/2\over x-3}+{1/2\over x+3} \right) \,dx = \left. \left[ {1\over 2}\ln \left| {x^2-9\over x^2}\right|+{9\over 2x^2} \right] \right|_6^\infty = \bbox[red, 2pt]{\ln 2-{1\over 2}\ln 3-{1\over 8}}$$
解答:$$\cases{f=x^2+4y+6z+8w\\ g=x^2+y^2+z^2-5\\ h=z+w-2} \Rightarrow \cases{f_x= \lambda g_x+ \mu h_x \\f_y= \lambda g_y + \mu g_y\\f_z= \lambda g_z+ \mu h_z \\ g=0\\ h=0} \Rightarrow \cases{2x= \lambda(2x) \\ 4=\lambda(2y) \\6=\lambda(2z) +\mu(1) \\ 8=\mu(1) \\ x^2+y^2+z^2=5\\ z+w=2} \\ \Rightarrow (x,y,z,w)=(0,2,-1,3) ,(0,-2,1,1) \Rightarrow \cases{f(0,2,-1,3)=26\\ f(0,-2,1,1)=6} \Rightarrow \bbox[red, 2pt]{\cases{\max=26\\ \min=6}}$$
解答:$$\text{Let } \cases{P(x,y)=2xy/(x^2+y^2)^2\\ Q(x,y)=(y^2-x^2)/(x^2+y^2)^2} \Rightarrow \mathbf F(x,y)= P(x,y) \mathbf i+Q(x,y )\mathbf j \\ \Rightarrow P_y= {2x^3-6xy^2 \over (x^2+y^2)^3} = Q_x, \text{ everywhere except at (0,0)}\\\text{ Since the curve }C \text{ encloses the origin, we cannot apply Green's Theorem directly to the }\\\text{region strictly bounded by }C. \text{ Instead, we use the extended version of Green's Theorem} \\ \text{ for a multiply connected region. Let }C_a \text{ be a small circle of radius }a \text{ centered at the} \\ \text{origin, oriented counterclockwise, entirely contained within C. Then, we have} \\ \iint_D \left( Q_x-P_y \right)dA =\iint_D 0\,dA= \oint_C \mathbf F\cdot d\mathbf r- \oint_{C_a} \mathbf F\cdot d\mathbf r \\ \Rightarrow \oint_C \mathbf F\cdot d\mathbf r= \oint_{C_a} \mathbf F\cdot d\mathbf r =\oint_{C_a} P\,dx+Q\,dy \\= \int_0^{2\pi} \left( {2(a\cos t)(a\sin t)\over a^4}(-a\sin t)+{a^2 \sin^2 t-a^2 \cos^2 t\over a^4} (a\cos t) \right) \,dt \\= -{1\over a} \int_0^{2\pi} \cos t(\sin^2 t+ \cos^2 t)\,dt = \bbox[red, 2pt]0$$
Part II: 4 points for each blank. Make sure you use the labels for
the blanks.
解答:$$\textbf{(1) } L= \left[ \sqrt{1+{3\over x}} \right]^{\sqrt{x^2+1}} \Rightarrow \ln L=\sqrt{x^2+1} \ln \sqrt{1+{3\over x}} \Rightarrow \lim_{x\to \infty } \ln L=\lim_{x\to \infty } {\ln \sqrt{1+3/x} \over 1/\sqrt{x^2+1}} \\\qquad =\lim_{x\to \infty } {(\ln \sqrt{1+3/x})' \over (1/\sqrt{x^2+ 1})'} =\lim_{x\to \infty } {3(x^2+1)^{3/2} \over 2x^2(x+1)} ={3\over 2} \Rightarrow \lim_{x\to \infty} L= \bbox[red, 2pt]{e^{3/2}}\\ \textbf{(2) }m=2n \Rightarrow \lim_{n\rightarrow\infty} \sum_{k=1}^{2n} \frac{n}{4n^{2} +k^{2}}= \lim_{m\to \infty} \sum_{k=1}^m {m\over 2(m^2 +k^2)} = \lim_{m\to \infty} \sum_{k=1}^m {1/m\over 2(1^2+(k/m)^2)} \\\qquad = \int_0^1 {1\over 2(1+x^2)}\, dx= \left. \left[ {1\over 2}\tan^{-1 }x \right] \right|_0^1 = \bbox[red, 2pt]{\pi\over 8}$$
解答:$$\textbf{(3) }f(x,y,z) = (2x+3z)^y \Rightarrow \cases{f_x= 2y(2x+3z)^{y-1} \\ f_y = (2x+3z)^y \ln(2x+3z) \\f_z=3y(2x+3z)^{y-1}} \\ \Rightarrow \nabla f=(2y(2x+3z)^{y-1} , (2x+3z)^y \ln(2x+3z), 3y(2x+3z)^{y-1}) \Rightarrow \nabla f(1,1,1) =\bbox[red, 2pt]{(2,5\ln 5,3)} \\\textbf{(4) }\nabla f(1,1,1) \cdot ({dx\over dt}, {dy\over dt}, {dz\over dt}) =0 \Rightarrow (2, 5\ln 5, 3) \cdot( 1,-1,{dz\over dt}) =2-5\ln 5-3{dz\over dt}=0 \\\qquad \Rightarrow {dz\over dt}= \bbox[red, 2pt]{2-5\ln 5\over 3}$$
解答:$$\textbf{(9) }\int_{0}^{2}\int_{2y}^{4}e^{-x^{2}}dx\,dy= \int_0^4 \int_0^{x/2} e^{-x^2}\,dy \,dx =\int_0^4 {x\over 2}e^{-x^2}\,dx = \left. \left[ -{1\over 4}e^{-x^2} \right] \right|_0^4= \bbox[red, 2pt]{{1\over 4}(1-e^{-16})} \\ \textbf{(10) }z={1\over 2}\sqrt{4-x^2-y^2} \Rightarrow x^2+y^2+4z^2=4 \Rightarrow \cases{u=x \\v=y\\ w=2z} \Rightarrow J= {\partial(x,y,z) \over \partial (u,v,w)} = \begin{vmatrix} 1& 0& 0\\ 0& 1& 0\\ 0&0 &1/2\end{vmatrix} ={1\over 2} \\ \Rightarrow \int_{-2}^2 \int_0^{\sqrt{4-x^2}} \int_0^{{1\over 2}\sqrt{4-x^2-y^2}} (x^2+y^2+4z^2)\,dx\,dy\,dz =\int_{-2}^2 \int_0^{\sqrt{4-u^2}} \int_0^{{1\over 2}\sqrt{4-u^2-v^2}} (u^2+v^2+ w^2)\,du\,dv\,dw\\ = \int_0^\pi \int_0^{\pi/2} \int_0^2 \rho^2 \cdot {1\over 2}\rho^2\sin \phi \,d\rho \,d\phi\,d\theta ={1\over 2} \left( \int_0^{\pi}d\theta \right) \left( \int_0^{\pi/2} \sin\phi \,d\phi \right) \left( \int_0^2 \rho^4\,d\rho\right) \\ ={1\over 2} \cdot \pi\cdot 1 \cdot {32\pi\over 5} = \bbox[red, 2pt] {16\pi\over 5}$$
解答:$$\textbf{(7) }\sin(u)= \sum_{n=0}^\infty (-1)^n\cdot {u^{2n+1} \over (2n+1)!} \Rightarrow \sin(t^3) = \sum_{n=0}^\infty (-1)^n\cdot {t^{6n+3} \over (2n+1)!} \\ \qquad \Rightarrow {\sin(t^3) \over t} = \sum_{n=0}^\infty (-1)^n\cdot {t^{6n+2} \over (2n+1)!} \Rightarrow f(1)= \int_1^2 \left( \sum_{n=0}^\infty (-1)^n\cdot {t^{6n+2} \over (2n+1)!} \right)\,dt \\ \qquad =\sum_{n=0}^\infty {(-1)^n \over (2n+1)!} \int_1^2 t^{6n+2}\,dt = \sum_{n=0}^\infty {(-1)^n \over (2n+1)!} \cdot {2^{6n+3}-1 \over 6n+3} \Rightarrow f(1) = \bbox[red, 2pt]{\sum_{n=0}^\infty {(-1)^n (2^{6n+3}-1)\over (2n+1)! (6n+3)}} \\\textbf{(8) } f(x) = \int_x^{2x} \left( \sum_{n=0}^\infty {(-1)^n t^{6n+2} \over (2n+1)!} \right) \,dt =\sum_{n=0}^\infty {(-1)^n \over (2n+1)!} \int_x^{2x} t^{6n+2}\,dt = \sum_{n=0}^\infty {(-1)^n (2^{6n+3}-1) \over (2n+1)! (6n+3)!}x^{6n+3} \\\qquad 6n+3=2025 \Rightarrow n=337 \Rightarrow {f^{(2025)} (0)\over 2025!} ={(-1)^{337}(2^{6\cdot 337+3}-1) \over (2\cdot 337+1)! (6\cdot 337+3)} ={1-2^{2025}\over 675!\cdot 2025} \\\qquad \Rightarrow f^{(2025)}(0) ={1-2^{2025} \over 675!\cdot 2025}\cdot 2025! = \bbox[red, 2pt]{(1-2^{2025}) \cdot 2024!\over 675!}$$
解答: $$\textbf{(9) } \int_0^\infty {x^p\over e^x-1}\,dx =\int_0^1 {x^p\over e^x-1}\,dx +\int_1^\infty {x^p\over e^x-1}\,dx \\ \text{Let }g(x)= x^{p-1}, \text{ then } \int_0^1 g(x)\,dx \text{ converges iff }p\gt 0. \text{ Now, } \lim_{x\to 0^+}{x^p/(e^x-1) \over g(x)} = \lim_{x\to 0^+} {x\over e^x-1}=1. \\ \text{By Limit Comparison Test, } \int_0^1 {x^p\over e^x-1}\,dx \text{ converges.} \\ \text{Additionally, }\lim_{x\to \infty} {x^p\over e^x-1}=0 \Rightarrow \int_1^\infty {x^p\over e^x-1}\,dx \text{ converges for all real number }p \\\Rightarrow \int_0^\infty {x^p\over e^x-1}\,dx \text{ converges when }p \in \bbox[red, 2pt]{(0,\infty)} \\\textbf{(10) } \lim_{n \to \infty } \left| {a_{n+1} \over a_n}\right| =\lim_{n\to \infty} \left| { (2p+1)^{n+1} \over \sqrt{n+1} \ln(n+2)} \cdot {\sqrt n \ln(n+1) \over (2p+1)^n}\right| \\\qquad =|2p+1| \lim_{n\to \infty} \left| \sqrt{n\over n+1} \cdot {\ln(n+1) \over \ln(n+2)}\right| =|2p+1| \lt 1 \Rightarrow -1\lt p\lt 0 \\ \textbf{Case I }p=0 \Rightarrow \sum_{n=2}^\infty {(2p+1)^n \over \sqrt n \ln(n+1)} = \sum_{n=2}^\infty {1 \over \sqrt n \ln(n+1)} \gt\sum_{n=2}^\infty {1 \over \sqrt n \sqrt n} =\sum_{n=2}^\infty {1 \over n} \text{ diverges} \\ \qquad \Rightarrow \sum_{n=2}^\infty {(2p+1)^n \over \sqrt n \ln(n+1)} \text{ diverges. } \\\textbf{Case II }p=-1 \Rightarrow \sum_{n=2}^\infty {(2p+1)^n \over \sqrt n \ln(n+1)} =\sum_{n=2}^\infty {(-1)^n \over \sqrt n \ln(n+1)} \\\qquad b_n={(-1)^n \over \sqrt n \ln(n+1)} \Rightarrow \cases{b_n \gt 0, n\ge 2\\ b_n \text{ is monotonically decreasing} \\ \lim_{n\to \infty} b_n =0} \Rightarrow \sum_{n=2}^\infty {(2p+1)^n \over \sqrt n \ln(n+1)} \text{ converges. } \\ \Rightarrow \sum_{n=2}^\infty {(2p+1)^n \over \sqrt n \ln(n+1)} \text{ converges when }p \in \bbox[red, 2pt]{[-1,0)}$$
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解題僅供參考，碩士班歷年試題及詳解
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