112年台聯大轉學考-微積分A3詳解

 台灣聯合大學系統112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

解答:$$L=(\cos(2\sqrt x))^{2/x} \Rightarrow \ln L={2\ln \cos(2\sqrt x)\over x} \Rightarrow \lim_{x\to 0^+} \ln L= \lim_{x\to 0^+}{2\ln \cos(2\sqrt x)\over x}\\ = \lim_{x\to 0^+}{{d\over dx}(2\ln \cos(2\sqrt x))\over {d\over dx}x} = \lim_{x\to 0^+}{-2\sin(2\sqrt x)/\sqrt x\over \cos(2\sqrt x)} =  \lim_{x\to 0^+}{-2\tan(2\sqrt x) \over \sqrt x} \\=  \lim_{x\to 0^+}{{d\over dx}(-2\tan(2\sqrt x)) \over {d\over dx} \sqrt x} =\lim_{x\to 0^+}{-2\sec^2 (2\sqrt x)/\sqrt x\over 1/2\sqrt x}=\lim_{x\to 0^+} -4\sec^2(2\sqrt x)=-4\\ \Rightarrow \lim_{x\to 0^+} \ln L= -4 \Rightarrow \lim_{x\to 0^+} L=e^{-4} =\bbox[red, 2pt]{1\over e^4}$$

解答:$$g(x)=\tan^{-1}(x^2)+ \int_1^{x^2} \sec(t-1)\,dt \Rightarrow g'(x)={2x\over 1+x^4} +2x\sec(x^2-1)\\ \Rightarrow \cases{g(-1)= \tan^{-1}(1)+0= \pi/4\\ g'(-1)=-1-2\cdot 1=-3}  \Rightarrow f'(x)=g'(x)e^{g(x)} \Rightarrow f'(-1)= g'(-1) e^{g(-1)} = \bbox[red, 2pt]{-3e^{\pi/4}}$$

解答:$$\cases{x(t)=\ln(\sec t+\tan t)-\sin t\\ y(t)=\cos t} \Rightarrow \cases{x'(t)= \sec t-\cos t\\ y'(t)=-\sin t} \\ \Rightarrow (x'(t))^2+ (y'(t))^2 =\sec^2 t-2+\cos^2+\sin^2 t= \sec^2 t-1　=\tan^2 t\\ \Rightarrow 曲線長=\int_0^{\pi/3} \sqrt{(x'(t))^2+ (y'(t))^2 }\,dt =\int_0^{\pi/3} \tan t\,dt= \left. \left[ -\ln(\cos t) \right]\right|_0^{\pi/3} =-\ln{1\over 2} =\bbox[red, 2pt]{\ln 2}$$

解答:$$\int_1^e (\ln(x))^2\pi\,dx =\left. \left[\pi(x(\ln x)^2-2x\ln(x)+2x) \right] \right|_1^e =\bbox[red, 2pt]{\pi(e -2)}$$

解答:$$位置r(t)=(\cos 3t,\sin  3t,3t) \Rightarrow 速度r'(t)=(-3\sin 3t,3\cos 3t,3) \Rightarrow \cases {r(\pi/3)=(1,0,\pi) \\ r'(\pi/3)=(0,-3,3)}\\ 又f(x,y,z)=xyz \Rightarrow \nabla f=(f_x,f_y,f_z) =(yz,xz,xy) =(3t\sin 3t,3t\cos 3t,(\sin 6t)/2))\\ \Rightarrow \nabla f(\pi/3)=(0,-\pi,0) \Rightarrow \nabla f(\pi/3)\cdot {r'(\pi/3)\over \Vert r'(\pi/3) \Vert} =(0,-\pi,0)\cdot (0,-{1\over \sqrt 2},{1\over \sqrt 2}) =\bbox[red, 2pt]{\sqrt 2\pi\over  2}$$

解答:$$\int_0^8 \int_{\sqrt[3]x}^2 {1\over y^4+1}\,dydx= \int_0^2 \int_0^{y^3} {1\over y^4+1}\,dx dy = \int_0^2 {y^3\over y^4+1}dy =\left. \left[{1\over 4}\ln(y^4+1) \right]\right|_0^2 = \bbox[red, 2pt]{{1\over 4}\ln 17}$$

解答:

$$積分區域如上圖著色區域,令\cases{x= r\cos \theta\\ y=r\sin \theta} \Rightarrow x^2+(y-1)^2=1 \Rightarrow  r^2=2r\sin \theta \Rightarrow r=2\sin \theta \\ 因此 \iint_R \sqrt{x^2+y^2}\,dA =2\int_0^{\pi/2} \int_{2\sin\theta}^2 r^2\,drd\theta =2\int_0^{\pi/2} \left. \left[ {1\over 3}r^3 \right] \right|_{2\sin \theta}^2 \,d\theta \\ ={16\over 3 }\int_0^{\pi/2} (1-\sin^3 \theta)\,d\theta= {16\over 3}\left. \left[ \theta-{1\over 12}(\cos 3\theta-9\cos \theta) \right] \right|_0^{\pi/2} ={16\over 3}({\pi\over 2}+{1\over 12}(-8)) \\=\bbox[red, 2pt]{{8\over 3}\pi -{32\over 9}}$$

解答:$$\int_{-1}^4 {dx\over \sqrt {|x|}} = \int_0^1{dx\over \sqrt x} +\int_0^4{dx\over \sqrt x}=2+4 = \bbox[red, 2pt]6$$

乙、計算、證明題: 共３題，每題１２分，共３６分

解答:$$\mathbf{(a)}\;令b_n={\ln n\over n-\ln n} \Rightarrow \lim_{n \to \infty}b_n = \lim_{n \to \infty}{\frac{\text{d}}{\text{d}n} \ln n\over \frac{\text{d}}{\text{d}n}(n-\ln n)} = \lim_{n \to \infty}{1\over n-1} =0\\　又\frac{\text{d}}{\text{d}n}b_n ={1-\ln n\over (n-\ln n)^2} \lt 0, \forall n \ge 3; 因此由交錯級數審定法知該級數收斂\\ 現在要判斷是絕對收斂或條件收斂:　由於{\ln n\over n-\ln n} \gt {1\over n-\ln n} \gt {1\over n}\\ 而調和級數發散,因此原級數絕對值發散,即\bbox[red,2pt]{條件收斂} \\ \mathbf{(b)}\;\lim_{n\to \infty}\left|{a_{n+1} \over a_n} \right| =\lim_{n\to \infty}\left|{(x+4)^{n+1} \over (n+1)3^{n+1}} \cdot { n3^n\over (x+4)^n} \right| =\lim_{n\to \infty}\left|{x+4 \over 3} \cdot {n\over n+1} \right| =\left|{x+4 \over 3}  \right| \lt 1 \\ \Rightarrow -7\lt x\lt -1 \\ x=-1 \Rightarrow \sum_{n=1}^\infty {(x+4)^n \over n3^n} =\sum_{n=1}^\infty {1 \over n} 為調和級數 \Rightarrow 發散\\ x=-7 \Rightarrow \sum_{n=1}^\infty {(x+4)^n \over n3^n} =\sum_{n=1}^\infty (-1)^n {1\over n} \Rightarrow \cases{\lim_{n\to \infty}{1\over n}=0 \\ {1\over n}\gt {1\over n+1}},交錯級數審定法 \Rightarrow 收斂\\　因此級數收斂範圍為\bbox[red, 2pt]{-7\le x\lt -1}$$

解答:$$\mathbf{(a)}\;取y=mx \Rightarrow \lim_{(x,y)\to(0,0)} f(x,y)= \lim_{x\to 0} {x^2\cdot mx\over x^3+m^3x^3} =\lim_{x\to 0} {m\over 1+m^3} ={m\over 1+m^3} \ne f(0,0) \\ \quad \Rightarrow f(0,0) 不連續,\bbox[red, 2pt]{故得證}　\\\mathbf{(b)}\; {\partial f\over \partial x} ={2xy \over x^3+y^3}-{x^2y \cdot 3x^2\over (x^3+y^3)^2} =\bbox[red, 2pt]{2xy^4-x^4y \over (x^3+y^3)^2}\\ \mathbf{(c)}\;\frac{\partial f}{\partial x} =\lim_{h\to 0}{f(0+h,0)-f(0,0)\over h} =\lim_{h\to 0}{0\over h}= \bbox[red, 2pt]0$$

解答:$$f(x,y)=x^2+3y^2+2y \Rightarrow \cases{f_x= 2x\\ f_y= 6y+2} \Rightarrow \cases{f_{xx}=2\\ f_{xy} =0\\ f_{yy}=6} \Rightarrow d(x,y)=f_{xx}f_{yy}-f_{xy}^2=12 \gt 0\\ 若\cases{f_x= 0\\ f_y= 0} \Rightarrow (x,y)=(0,-1/3) \Rightarrow f(0,-1/3)= -{1\over 3}為極小值\\　再利用\text{ Lagrange }算子求極大值: \cases{f(x,y)=x^2+3y^2+2y\\ g(x,y)=x^2+y^2-1} \Rightarrow \cases{f_x= \lambda g_x\\ f_y=\lambda g_y\\ g=0} \Rightarrow \cases{2x=\lambda(2x) \\ 6y+2=\lambda(2y) \\ x^2+y^2=1} \\ \Rightarrow \cases{x=0 \Rightarrow y=\pm 1\\ \lambda=1 \Rightarrow y=-1/2 \Rightarrow x=\pm \sqrt 3/2} \Rightarrow \cases{f(0,1)=5\\ f(0,-1)=1\\ f(\pm\sqrt 3/2,-1/2)=1/2} \Rightarrow 最大值5 \\ \Rightarrow \bbox[red, 2pt]{最大值5,最小值-{1\over 3}}$$

==================== END =========================

解題僅供參考,其他歷年試題及詳解