109年台聯大轉學考-微積分A2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：A2

甲、簡答題：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： $$f(x)={|x|x\over x^2+1} \Rightarrow \begin{cases}f(x)={x^2\over x^2+1} =1-{1\over x^2+1},& \text{if }x\ge 0\\f(x)={-x^2\over x^2+1} =-1+{1\over x^2+1},& \text{if }x\le 0  \end{cases} \Rightarrow \begin{cases} \lim_{x\to \infty} f(x)=1, & \text{if }x\ge 0 \\\lim_{x\to \infty} f(x)=-1, & \text{if }x\le 0\end{cases}\\ \Rightarrow 水平漸近線為\bbox[red, 2pt]{y=\pm 1}$$

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： $$假設長寬高分為x,y,z \Rightarrow \cases{xyz= 12000 \Rightarrow z=12000/xy\\ \text{cost}=2xy+ 8xz +6yz} \\ \Rightarrow \text{cost}=f(x,y)=2xy+8x\cdot{12000\over xy} +6y\cdot {12000\over xy}= 2xy+{1\over y}96000+{1\over x}72000\\ 欲求f(x,y)的極小值，即 \cases{f_x=0\\ f_y=0} \Rightarrow \cases{2y-72000\cdot {1\over x^2}=0\\ 2x-96000\cdot {1\over y^2}=0} \Rightarrow \cases{2x^2y = 72000 \cdots(1)\\ 2xy^2= 96000 \cdots(2)}\\ {(1)\over (2)} \Rightarrow {x\over y}={3\over 4} \Rightarrow x={3\over 4}y 代入(2) \Rightarrow y^3=64000 \Rightarrow y=40 \Rightarrow x=30\\ \Rightarrow f(30, 40)=2400 +{96000\over 40} +{72000\over 30} =\bbox[red, 2pt]{7200}$$

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

解答： $$\int_0^{\ln 5} \int_{e^x}^5 {1\over \ln y}\,dydx = \int_1^5\int_0^{\ln y} {1\over \ln y}\,dx dy = \int_1^5 1\,dy =\bbox[red, 2pt]4$$

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： $$\int_{0}^\infty \int_{0}^\infty kxye^{-(x^2+y^2)} \,dxdy= \int_{0}^\infty  \left.\left[ -{1\over 2}kye^{-(x^2+y^2)} \right] \right|_{0}^\infty\;dy = \int_{0}^\infty{1\over 2}kye^{-y^2}\,dy= \left.\left[ -{1\over 4}ke^{-y^2}\right ]\right|_0^\infty\\ ={1\over 4}k=1 \Rightarrow k= \bbox[red, 2pt]4$$

乙、計算、證明題：共3題，每題12分，共36分

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

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： $$水槽長度固定，因此容量大小取決於截面積；\\截面積=一個長形及兩個全等三角形= \cos\theta +2\times {1\over 2}\sin\theta \cos\theta\\ \Rightarrow 截面積f(\theta)=\cos\theta +\sin\theta \cos \theta \Rightarrow f'(\theta)=-\sin\theta +\cos^2\theta-\sin^2\theta= -2\sin^2\theta-\sin\theta+1 \\ 因此f'(\theta)=0 \Rightarrow (2\sin\theta-1)(\sin\theta+1)=0 \Rightarrow \sin\theta = 1/2(-1不合, \theta\ne 3\pi/2) \\ \Rightarrow \theta= \bbox[red,2pt]{30^\circ}有最大的容量$$

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