113年台綜大轉學考-微積分A詳解

 臺灣綜合大學系統113學年度學士班轉學生聯合招生考試

科目名稱:微積分A

解答:$$\lim_{x\to 1^+} {|x^2-1|\over x-1} =\lim_{x\to 1^-} {x^2-1\over x-1} = \lim_{x\to 1^+} {(x+1)(x-1)\over x-1} =\lim_{x\to 1^+} (x+1)=\bbox[red, 2pt]{2}\\ \lim_{x\to 1^-} {|x^2-1|\over x-1} =\lim_{x\to 1^-} {1-x^2\over x-1} = \lim_{x\to 1^-} {(1-x)(1+x)\over x-1} =\lim_{x\to 1^-} -(1+x)=\bbox[red, 2pt]{-2}$$

解答:$$f(x)= \int_{\tan x}^{\sec x} {1\over \sqrt{1+t^4}} \,dt \Rightarrow f'(x)=  {1\over \sqrt{1+(\sec x)^4}}\sec x\tan x- {1\over \sqrt{1+(\tan x)^4}} \sec^2 x \\ \Rightarrow f'(0)=0-1=\bbox[red, 2pt] {-1}$$

解答:$$f(x)= {x\over x^2-x+1} \Rightarrow f'(x)= -{(x-1)(x+1) \over (x^2-x+1)^2} \\ f'(x)=0 \Rightarrow \cases{x=-1\\ x=1 \not \in [-2,0]} \Rightarrow \cases{f(-2) =-2/7\\ f(-1) =-1/3\\ f(0) =0} \Rightarrow \cases{\text{absolute max: }\bbox[red, 2pt]0\\ \text{absolute min: }\bbox[red, 2pt]{-1/3}}$$

解答:$$y=\sqrt{1-4x^2} \Rightarrow 4x^2+y^2=1 \Rightarrow {x^2\over 1/4}+ {y^2\over 1}=1為一橢圓,其中\cases{a=1\\ b=1/2} \\ \Rightarrow 橢圓面積=ab\pi ={1\over 2}\pi \Rightarrow \int_0^{1/2} \sqrt{1-4x^2}\,dx = {1\over 4}橢圓面積=\bbox[red, 2pt]{{1\over 8}\pi}$$

解答:$$\cases{u=\ln x\\ dv =xdx} \Rightarrow \cases{du= dx/x\\ v=x^2/2} \Rightarrow \int x\ln x\,dx = {1\over 2}x^2\ln x-\int {1\over 2}x \,dx ={1\over 2}x^2\ln x-{1\over 4}x^2+C \\ \Rightarrow \int_0^1 x\ln x\,dx =\bbox[red, 2pt]{-{1\over 4}}$$

解答:$$a_n= {x^n\over \sqrt n} \Rightarrow \lim_{n \to \infty}\left|{a_{n+1}\over a_n} \right| =\lim_{n \to \infty}\left|{x^{n+1}\over \sqrt{n+1}} \cdot {\sqrt n\over x^n}\right| = |x| \lt 1 \Rightarrow -1\lt x\lt 1\\ x=1 \Rightarrow \sum_{n=1}^\infty a_n= \sum_{n=1}^\infty {1\over \sqrt n},由於\int_1^\infty {1\over \sqrt n}\,dx \to \infty \Rightarrow \sum_{n=1}^\infty {1\over \sqrt n}發散\\ x=-1 \Rightarrow \sum_{n=1}^\infty a_n= \sum_{n=1}^\infty {(-1)^n\over \sqrt n},由於\lim_{n\to \infty}{1\over \sqrt n}=0且{1\over \sqrt n} \gt {1\over \sqrt{n+1}} \Rightarrow \sum_{n=1}^\infty {(-1)^n\over \sqrt n}收斂\\因此收斂區間:\bbox[red, 2pt]{[-1,1)}$$

解答:$$\text{The value of }f(x,y) \text{ decreases the fastest in the directio of }-\nabla f =-(f_x,f_y) = (-{1\over x},-{1\over y}) \\ \Rightarrow -\nabla f(1,2) =\bbox[red, 2pt]{\left(-1, -{1\over 2} \right)} \Rightarrow \text{the rate of decrease: }-|\nabla f|=-\sqrt{1+{1\over 4}} = \bbox[red, 2pt]{-{\sqrt 5\over 2}}$$

解答:$$\cases{x(t)=\cos(2t) \\y(t) =\sin(2t) \\z(t)=t} \Rightarrow \cases{x'(t)=-2\sin(2t) \\ y'(t)= 2\cos(2t) \\z'(t)= 1} \Rightarrow \text{ arc length }=\int_0^\pi \sqrt{x'(t)^2+ y'(t)^2 +z'(t)^2}\, dt \\= \int_0^\pi  \sqrt{4\sin^2(2t)+ 4\cos^2(2t)+1}\,dt = \int_0^\pi \sqrt 5\,dt= \bbox[red, 2pt]{\sqrt 5 \pi}$$

解答:$$y=\sqrt{x-x^2} \Rightarrow x^2-x+y^2=0 \Rightarrow (x-{1\over 2})^2+y^2 ={1\over 4} \Rightarrow 積分區域為一半圓,圓心({1\over 2},0), 半徑{1\over 2} \\ 取\cases{x= r\cos \theta  \\ y=r \sin \theta} \Rightarrow  x^2+y^2=x \Rightarrow r^2=r\cos \theta \Rightarrow r=\cos \theta \\ \Rightarrow \int_0^1 \int_0^{\sqrt{x-x^2}} \sqrt{x^2+ y^2}\, dydx = \int_0^{\pi/2} \int_0^{\cos \theta} \sqrt{r^2\cos^2 \theta+ r^2 \sin^2 \theta} r \, drd\theta = \int_0^{\pi/2} \int_0^{\cos \theta} r^2\,dr d\theta \\=  \int_0^{\pi/2}{1\over 3}\cos^3 \theta \,d \theta = \left. \left[ {1\over 4}\sin \theta+ {1\over 36} \sin 3\theta \right] \right|_0^{\pi/2} ={1\over 4}-{1\over 36} =\bbox[red, 2pt]{2\over 9}$$

解答:$$4x^2+9y^2=36 \Rightarrow {x^2\over 9}+{y^2 \over 4} =1 \Rightarrow \cases{x(t)=3\cos t\\ y(t) =2\sin t}, 0\le t\le 2\pi \Rightarrow \cases{x'(t)=-3\sin t\\ y'(t) =2\cos t} \\ \Rightarrow \int_C {y\over x^2+y^2}dx -{x\over x^2+y^2}\,dy = \int_0^{2\pi} \left( {2\sin t\over 9\cos^2 t+4\sin^2 t}\cdot (-3\sin t)-{3\cos t\over 9\cos^2t +4\sin^2 t} \cdot 2\cos t \right)dt \\= \int_0^{2\pi} {-6\over 4+5\cos^2 t}\,dt =-6\times { \pi \over 3} =\bbox[red, 2pt]{-2\pi}\\ \bbox[cyan,2pt]{註}:z=e^{i\theta} \Rightarrow dz=ie^{i\theta}d\theta \Rightarrow d\theta={dz\over iz},又\cos \theta={1\over 2}(e^{i\theta}+ e^{-i\theta}) ={1\over 2}(z+{1\over z}) \\\Rightarrow \cos^2 \theta={1\over 4}(z^2+2+ {1\over z^2})\Rightarrow \int_0^{2\pi} {1\over 4+5 \cos^2 \theta}\,d\theta =\oint_C {1\over 4+{5\over 4} (z^2+2+{1\over z^2})}\cdot {dz\over iz}\\ ={1\over i} \oint_C {4z\over 5z^4+26z^2+5}\,dz, 其中C為單位圓,即|z|=1 \\={1\over i} \oint_C{4z\over (5z^2+1)(z^2+5)}\,dz, 其中\cases{5z^2+1=0 \Rightarrow z=\pm {1\over \sqrt 5}i在C內\\ z^2+5=0 \Rightarrow z=\pm \sqrt 5i不在C內}\\ 令f(z)={4z/5\over (z^2+1/  5)(z^2+5)}\Rightarrow res(f,\pm {i\over \sqrt 5}) ={1\over 12} \Rightarrow \int_0^{2\pi} {1\over 4+5 \cos^2 \theta}\,d\theta \\={1\over i}(res(f,{i\over \sqrt 5}) + res(f, -{i\over \sqrt 5})) \times 2\pi i=({1\over 12}+ {1\over 12})2\pi ={\pi\over 3} \\ \Rightarrow \int_0^{2\pi} {-6\over 4+5 \cos^2 t}\,dt = -6\times{\pi\over 3}=-2\pi$$

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