109年台聯大轉學考-微積分A3A4A6詳解

台灣聯合大學系統109學年度學士班轉學生考試

科目：微積分
類組別：A3/A4/A6
甲、填充題：共8題，每題8分，共64分

解答：$$\lim_{x\to \infty}{x+\cos x\over x-\cos x} =\lim_{x\to \infty}\left(1+{2\cos x\over x-\cos x} \right) =1+0= \bbox[red, 2pt]1$$

解答：$$F(x)= \int_0^x \cos(t^{3/2})\,dt \Rightarrow F'(x)= \cos(x^{3/2}) \Rightarrow F''(x)=-{3\over 2}\sqrt x\sin(x^{3/2})\\ 因此F''(x)=0 \Rightarrow x=0,x^{3/2}=n\pi, n\in \mathbb{Z};\\ 依題意找最小正值的x，因此x= \bbox[red, 2pt]{\pi^{2/3}}$$

解答：$$f(x,y)=  10xye^{-(x^2+y^2)} \Rightarrow \cases{f_x= (10y-20x^2y)e^{-(x^2+y^2)}  \\f_y = (10x-20xy^2) e^{-(x^2+y^2)} } \\\Rightarrow \cases{f_{xx}= (-60xy +40x^3y)e^{-(x^2+y^2)} \\f_{xy} =(10-20x^2-20y^2+ 40x^2y^2) e^{-(x^2+y^2)} \\f_{yy} = (-60xy +40xy^3)e^{-(x^2+y^2)}} \Rightarrow g(x,y)=f_{xx}f_{yy}-f_{xy}^2\\ 因此\cases{f_x=0 \Rightarrow 10y(1-2x^2)=0 \Rightarrow x=\pm 1/\sqrt 2或y=0\\ f_y=0 \Rightarrow 10x(1-2y^2)=0 \Rightarrow x=0 或y= \pm 1/\sqrt 2}\\\Rightarrow 局部極值的候選點\cases{A(0,0)\\ B(1/\sqrt 2,1/\sqrt 2)\\ C(1/\sqrt 2,-1/\sqrt 2)\\ D(-1/\sqrt 2,1/\sqrt 2)\\ E(-1/\sqrt 2,-1/\sqrt 2)} \Rightarrow \cases{g(A)= 10\gt 0\\ g(B) =g(C) =g(D)= g(E)=400/e^2 \gt 0 }\\ \Rightarrow \cases{f(A)=0\\ f(B)=f(E) =5/e \\ f(C)= f(D)=-5/e} \Rightarrow 共有\bbox[red, 2pt]3種不同的局部極值$$

解答：$$\cases{f(x,y,z)= x^2 +y^2+z^2 \\ g(x,y,z)= (x-2)^2 +(y-2)^2 +z^2} \Rightarrow \cases{\nabla f=(2x,2y,2z)\\ \nabla g=(2(x-2),2(y-2), 2z)} \\ \Rightarrow \cases{\nabla f|_{(1,1,1)} =(2,2,2)\\ \nabla g|_{(1,1,1)} =(-2,-2,2)} \Rightarrow (2,2,2)\times (-2,-2,2) = (8,-8,0)\\ \Rightarrow 切線參數式:\bbox[red, 2pt]{\cases{x(t)=1+8t\\ y(t)=1-8t\\ z(t)=1},t\in \mathbb{R}}$$

解答：$$令D=\{(x,y)\mid \delta^2\le x^2+y^2\le 1\}及\cases{x=r\cos\theta\\ y=r\sin \theta} \Rightarrow \iint_D \ln\sqrt{x^2+y^2}\,dA = \int_0^{2\pi} \int_\delta^1 \ln \sqrt{r^2} r\,drd\theta\\ = \int_0^{2\pi} \int_\delta^1 r\ln r \,drd\theta =2\pi \int_\delta^1 r\ln r \,drd\theta = 2\pi \left.\left[{1\over 2}r^2\ln r-{r^2\over 4} \right] \right|_\delta^1 =2\pi (-{1\over 4}-{1\over 2}\delta^2\ln \delta+{\delta^2\over 2})\\ \Rightarrow \iint_R \ln\sqrt{x^2+y^2} \,dA = \lim_{\delta\to 0^+} \left(2\pi (-{1\over 4}-{1\over 2}\delta^2\ln \delta+{\delta^2\over 2}) \right)= \bbox[red, 2pt]{-{\pi\over 2}}$$

解答：$$若只考慮z\ge 0，角錐z^2=x^2+y^2與z軸的角度為45^\circ，\\因此欲求體積的上半部=\int_0^{\pi/4} \int_0^{2\pi} \int_1^{\sqrt 2} r^2\sin \theta \;drd\phi d\theta ={2\pi \over 3} (2\sqrt 2-1)(1-{ \sqrt 2 \over 2}) \\ ={2\pi \over 3} ({5\over 2}\sqrt 2-3) \Rightarrow 欲求之體積=2\times {2\pi \over 3} ({5\over 2}\sqrt 2-3) =\bbox[red, 2pt]{{2\pi\over 3}(5\sqrt 2-6)}$$

解答：$$令\cases{x=2u\\ y=3v} \Rightarrow \begin{Vmatrix} x_u & x_v\\ y_u& y_v\end{Vmatrix} =\begin{Vmatrix} 2 & 0\\ 0& 3 \end{Vmatrix} = 6\\ \Rightarrow \iint_R e^{9x^2+4y^2}\,dA = \iint_S e^{36u^2 +36v^2 }\cdot 6\,dudv, 其中S=\{(u,v)\mid  u^2+v^2\le 1\}\\ 取\cases{u= r\cos\theta\\ v=r\sin \theta} \Rightarrow \iint_S e^{36u^2 +36v^2 }\cdot 6\,dudv =\int_0^{2\pi} \int_0^1 6re^{36r^2}\,drd\theta =2\pi \left.\left[{1\over 12}e^{36r^2} \right]\right|_0^1 \, \\ = \bbox[red, 2pt]{{\pi \over 6}(e^{36}-1)}$$

解答：$$\cases{x(t)= 2\cos t-\cos 2t\\ y(t)=2\sin t-\sin2t} \Rightarrow r^2= x^2+y^2 =5-4\cos t \Rightarrow 面積={1\over 2}\int_0^{2\pi} 5-4\cos t\,dt\\ ={1\over 2}\left.\left[5t- 4\sin t \right]\right|_0^{2\pi} = \bbox[red, 2pt]{5\pi}$$

乙、計算、證明題：共3題，每題12分，共36分

解答：$$令\cases{x=r\cos\theta \\ y= r\sin \theta},又(\cos^2\theta -\sin^2\theta)^2 \ge 0 \Rightarrow \cos^4\theta +\sin^4\theta \ge 2\cos^2\theta \sin^2\theta\\ \Rightarrow {(xy)^p\over x^4+y^4} ={ r^{2p}(\sin\theta \cos\theta)^p\over r^4(\cos^4\theta + \sin^4\theta)} \le { r^{2p}(\sin\theta \cos\theta)^p\over r^4(2\cos^2\theta   \sin^2\theta)} ={1\over 2}r^{2p-4} (\sin\theta \cos\theta)^{p-2} \\\Rightarrow \lim_{(r,\theta)\to (0,0)} =\begin{cases} 0,& p\gt 2\\ {1\over 2r^{4-2p}(\sin\theta \cos\theta)^{2-p}} \ne 0, &p\le 2\end{cases} \Rightarrow \cases{f\text{ is continuous for all(x,y), if } p\gt 2 \\ f\text{ is discontinuous at (0,0), if } p\le 2}\\ \bbox[red, 2pt]{故得證}$$

解答：$$\mathbf{a.}\;級數:\sum_{n=1}^\infty (-1)^n a_n,其中a_n=\ln(1+{1\over n})\\(1)\lim_{n\to \infty}a_n = \ln 1=0\\ (2)a_n-a_{n+1}= \ln({n+1\over n})- \ln({n+2\over n+1}) = \ln {(n+1)^2\over n(n+2)} =\ln{n^2+2n+1\over n^2+2n} =\ln(1+{1\over n^2+2n})\gt 0\\ \Rightarrow a_n\gt a_{n+1}\\ 由(1)及(2)可得\sum_{n=1}^\infty (-1)^n \ln(1+{1\over n})收斂;\\ 又 \lim_{n\to \infty} \sum_{k=1}^n \ln(1+{1\over k}) = \lim_{n\to \infty} \ln {2\cdot 3\cdot 4\cdots (n+1)\over 1\cdot 2\cdot 3\cdots n} = \infty  \\\Rightarrow \sum_{n=1}^\infty \left| (-1)^n \ln(1+{1\over n}) \right| =\sum_{n=1}^\infty \ln(1+{1\over n}) 發散，因此原級數\bbox[red, 2pt]{條件收斂}\\\mathbf{b.}\;\sum_{n=1}^\infty   a_n 收斂 \Rightarrow \lim_{n\to \infty}a_n=0 \\\Rightarrow \lim_{n\to \infty} \sqrt[n]{\left|\left( {3+\sin (a_n)\over 5} \right)^n\right|} =\lim_{n\to \infty} \left| {3+\sin (a_n)\over 5}   \right| \le \lim_{n\to \infty}{3+|\sin (a_n)|\over 5}={3\over 5}\lt 1\\ \Rightarrow \sum_{n=1}^\infty \left( {3+\sin (a_n)\over 5} \right)^n 絕對收斂，即收斂，\bbox[red,2pt]{故得證}$$

解答：$$依題意:在x_1p_1+x_2p_2=L的條件下最大化U(x_1,x_2)\\ 因此x_2={L-x_1p_1 \over p_2} \Rightarrow {\partial \over \partial x_1}x_2= -{p_1\over p_2} \Rightarrow U(x_1,x_2)= U(x_1,{L-x_1p_1 \over p_2}) \\\Rightarrow U'(x_1) =0 \Rightarrow {\partial \over \partial x_1}U+ {\partial \over \partial x_2}U {\partial \over \partial x_1}x_2 = 0  \Rightarrow {\partial \over \partial x_1}U- {\partial \over \partial x_2}U {p_1 \over p_2} = 0 \\ \Rightarrow {\partial U/\partial x_1 \over \partial U/\partial x_2}={p_1\over p_2}，\bbox[red, 2pt]{故得證}$$

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