111年台綜大轉學考-微積分A詳解

臺灣綜合大學系統111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： $$L=\lim_{x\to 0}\left( {e^{ax}-1\over x^2}-ax +{b\over x}\right) =\lim_{x\to 0}\left( {e^{ax}-1-ax^3+bx\over x^2} \right) =\lim_{x\to 0}\left( ({e^{ax}-1-ax^3+bx)'\over (x^2)'} \right) \\ =\lim_{x\to 0}\left( {ae^{ax} -3ax^2+b \over 2x } \right) \Rightarrow a+b=0 \cdots(1)\\ L= \lim_{x\to 0}\left( ({ae^{ax} -3ax^2+b)' \over (2x)' } \right) =\lim_{x\to 0}\left( {a^2e^{ax} -6ax  \over 2 } \right) ={1\over 2}\Rightarrow a^2=1 \Rightarrow a= \pm 1 \Rightarrow b=\mp 1\\ \Rightarrow (a,b)=\bbox[red, 2pt]{(1,-1)或(-1,1)}$$

解答： $$(a)\; 令x=\sec^2 y \Rightarrow \sec y=\pm \sqrt x \Rightarrow y= \sec^{-1}( \sqrt x) \Rightarrow f^{-1}(x)= \sec^{-1}(\sqrt x) \\ \Rightarrow (f^{-1})'= {1\over \sqrt x \sqrt{x-1}} \cdot {1\over 2\sqrt x} \Rightarrow (f^{-1})'(4)= \bbox[red, 2pt]{1\over 8\sqrt 3}\\(b)\;\cases{x=rs\\ y=r+s} \Rightarrow \cases{{\partial x\over \partial s}= r\\ {\partial y\over \partial s}=1} \Rightarrow {\partial h\over \partial s} (r=1,s=1)= {\partial h\over \partial x}{\partial x\over \partial s}(r=1,s=1) +{\partial h\over \partial y}{\partial y\over \partial s}(r=1,s=1)\\ ={\partial h\over \partial x}(x=1,y=1+1=2) +{\partial h\over \partial y}(1,2)={\partial f\over \partial x}( 1, 2) +{\partial f\over \partial y}(1,2)=2+1= \bbox[red, 2pt]3$$

解答： $$f(x)=\int_0^{x^2} (1-t^2)e^{t^2}\,dt \Rightarrow f'(x)= 2x(1-x^4)e^{x^4} \ge 0 \Rightarrow x(1-x^4)\ge 0\\ \Rightarrow x(1-x)(1+x) (1+x^2) \ge 0\Rightarrow x(1-x)(1+x) \ge 0 \Rightarrow x(x+1)(x-1)\le 0\\ \Rightarrow x\in \bbox[red,2pt]{[0,1]\cup (-\infty, -1]}$$

解答： $$令x=a\tan t \Rightarrow dx=a\sec^2 t\,dt \Rightarrow I=\int_0^a {x^2\over (x^2+a^2)^{3/2}}\,dx =\int_0^{\pi/4} {a^2 \tan^2 t\over (a^2 \tan^2 t+a^2)^{3/2}}\cdot a \sec^2 t\,dt \\ =\int_0^{\pi/4} {a^2 \tan^2 t\over a^3 \sec^3 t  }\cdot a \sec^2 t\,dt =\int_0^{\pi/4} { \tan^2 t\over \sec t}\,dt = \int_0^{\pi/4} {\sec^2 t-1\over \sec t}\,dt = \int_0^{\pi/4} (\sec t-\cos t)\,dt\\ =\left.\left[ \ln|\sec t+\tan t|-\sin t\right] \right|_0^{\pi/4} = \bbox[red, 2pt]{(\ln (\sqrt 2+1)-{\sqrt 2\over 2})}$$

解答：

$$\cases{x=r\cos \theta\\ y=r\sin \theta} \Rightarrow \cases{r=4\sin \theta = 4y/r \Rightarrow r^2=4y \Rightarrow x^2+y^2-4y=0 \Rightarrow x^2+(y-2)^2=4\\ r=4 \Rightarrow x^2+y^2=4}\\ \Rightarrow \cases{\Gamma_1: \cases{圓心B(0,2) \\圓半徑r=2} \\ \Gamma_2:\cases{圓心D(0,0) \\圓半徑r=2} } \Rightarrow \Gamma_1與\Gamma_2 交集為上圖黃色區域;\\ 又\cases{\triangle ABD與\triangle BCD 皆為邊長為2的正三角形，面積均為{\sqrt 3\over 4}\cdot 2^2= \sqrt 3\\ 扇形DABC面樍為三分之一圓面積={4\over 3}\pi} \\ \Rightarrow 黃色面積=2({4\over 3}\pi-2\sqrt 3)+ 2\sqrt 3 = {{8\over 3}\pi-2\sqrt 3} \Rightarrow 欲求藍色面積=4\pi-({8\over 3}\pi-2\sqrt 3)\\ =\bbox[red, 2pt]{{4\over 3}\pi+2\sqrt 3}$$

解答： $$\int_3^\infty {1\over x\ln x} \,dx = \left.\left[ \ln \ln x \right]\right|_3^\infty = \infty \Rightarrow \sum_{n=3}^\infty {1\over n\ln n} 發散\\ \text{Case I: } 0\le p\le 1 \Rightarrow \sum_{n=3}^\infty {1\over n^p \ln n} \ge \sum_{n=3}^\infty {1\over n  \ln n} \Rightarrow \sum_{n=3}^\infty {1\over n^p \ln n} 發散 \Rightarrow \sum_{n=2}^\infty {1\over n^p \ln n} 發散 \\ \text{Case II: } p\gt 1 \Rightarrow \sum_{n=3}^\infty {1\over n^p \ln n}\le \sum_{n=3}^\infty {1\over n^p }\lt \infty \Rightarrow \sum_{n=3}^\infty {1\over n^p \ln n}收斂 \Rightarrow \sum_{n=2}^\infty {1\over n^p \ln n}收斂\\ \Rightarrow \bbox[red, 2pt]{p\gt 1}該級數收斂$$

解答： $$觀察\sum_{n=0}^\infty {x^{2n}\over (n+3)!} ={1\over 3!} +{x^2\over 4!}  +{x^4\over 5!}  +{x^6\over 6!}  +{x^8\over 7!}  +{x^{10}\over 8!} +{x^{12}\over 9!} +\cdots\\ 由e^x = 1+x +{x^2\over 2!}+ {x^3\over 3!} +{x^4\over 4!}+\cdots \Rightarrow e^{x^2} =1+x^2 +{x^4\over 2!}+ {x^6\over 3!} +{x^8\over 4!}+\cdots \\ \Rightarrow {1\over x^6}e^{x^2}= {1\over x^6}+ {1\over x^4} +{1\over 2!x^2}+ {1\over 3!} +{x^2\over 4!} +{x^4\over 5!}+{x^6\over 6!}  +{x^8\over 7!}\cdots \\ \Rightarrow {1\over (x+2)^6}e^{(x+2)^2} ={1\over (x+2)^6}+ {1\over (x+2)^4} +{1\over 2(x+2)^2}+ {1\over 3!} +{(x+2)^2 \over 4!} +{(x+2)^4 \over 5!}+{(x+2)^6\over 6!}  +{(x+2)^8\over 7!}\cdots \\ \Rightarrow \sum_{n=0}^\infty {(x+2)^{2n} \over (n+3)!} =\bbox[red, 2pt]{{1\over (x+2)^6}e^{(x+2)^2}-{1\over (x+2)^6}- {1\over (x+2)^4} -{1\over 2(x+2)^2}}$$

解答： $$令\cases{f(x,y,z)= x^2+y^2+z^2 \\ g(x,y,z)= x^2+ 2y^2+3z^2-1} \Rightarrow \cases{f_x=\lambda g_x \\f_y=\lambda g_y \\f_z=\lambda g_z \\ g=0} \Rightarrow \cases{2x= \lambda (2x) \Rightarrow 2x(1-\lambda)=0 \cdots(1)\\ 2y= \lambda (4y)  \Rightarrow 2y(1-2\lambda)=0 \cdots(2)\\ 2z= \lambda (6z)  \Rightarrow 2z(1-3\lambda)=0 \cdots(3)\\ x^2+ 2y^2+3z^2=1 \cdots(4)} \\ \Rightarrow (\lambda,x,y,z)=(1,\pm 1,0,0),(1/2,0,\pm {1\over \sqrt 2},0), (1/3,0,0,\pm {1\over \sqrt 3}) \Rightarrow \cases{f(\pm 1,0,0)=1\\ f(0,\pm{1\over \sqrt 2},0) ={1\over 2} \\ f(0,0,\pm{1\over \sqrt 3})={1\over 3}} \\ \Rightarrow \bbox[red, 2pt]{\cases{\text{global maximum= }1 \\ \text{global minimum= }1/3} }$$

解答： $$令\cases{x=u/v\\ y=v} \Rightarrow \begin{vmatrix} x_u & x_v\\ y_u & y_v\end{vmatrix} =\begin{vmatrix} 1/v & -u/v^2\\ 0 & 1\end{vmatrix} ={1\over v}\\ 因此\cases{y=x \Rightarrow u/v=v \Rightarrow u=v^2 \\ y=3x \Rightarrow 3u/v=v \Rightarrow 3u=v^2 \\ xy=1 \Rightarrow u=1\\ xy=3 \Rightarrow u=3},又(x,y)在第一象限\Rightarrow \cases{u/v\gt 0\\ v\gt 0} \Rightarrow \cases{u\gt 0\\ v\gt 0}\\ \Rightarrow \iint_R {y\over x}e^{xy}\,dA = \int_1^3 \int_{\sqrt u}^{\sqrt{3u}} {v^2\over u} e^u{1\over v}\,dvdu =\int_1^3 \int_{\sqrt u}^{\sqrt{3u}} {v\over u} e^u \,dvdu = \int_1^3 e^u\,du =\bbox[red, 2pt]{e^3-e}$$

解答： $$C:\cases{x=2\cos \theta \\ y=2\sin \theta},\theta=0-\pi \Rightarrow \cases{dx= -2\sin\theta d\theta\\ dy = 2\cos\theta d\theta}\\ \Rightarrow \int_C (1-y^3)dx +(x^3+e^{-y^2})dy =\int_0^\pi (1-8\sin^3\theta)(-2\sin \theta d\theta)+ (8\cos^3\theta +e^{-4\sin^2\theta})(2\cos\theta d\theta) \\ \int_0^\pi 2\cos\theta e^{-4\sin^2\theta} +16(\sin^4\theta +\cos^4\theta )-2\sin\theta \,d\theta = \int_0^\pi 0+16(\sin^4\theta +\cos^4\theta )-2\sin\theta \,d\theta\\=16\cdot {3\pi\over 4} -2\cdot 2= \bbox[red, 2pt]{12\pi-4}$$

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