115年台綜大轉學考-微積分A詳解

臺灣綜合大學系統 115 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:$$ \lim_{x \rightarrow4}\frac{4 -x}{5- \sqrt{x^{2}+9}} = \lim_{x \rightarrow4} \frac{(4-x)(5+\sqrt{x^{2}+9})}{(5-\sqrt{x^{2}+9})(5+\sqrt{x^{2}+9})} =  \lim_{x \rightarrow4} \frac{(4-x)(5 +\sqrt{x^{2} +9})}{(4-x) (4+x)}  \\  = \lim_{x \rightarrow4} \frac{5+ \sqrt{x^{2} +9}}{4 +x} = \frac{5+ \sqrt{16+ 9}}{4+4} = \frac{5+ 5}{8}  = \bbox[red, 2pt]{\frac{5}{4}}$$

解答:$$ \lim_{x\rightarrow0} \frac{f(2 +3x)+f(2+5x)}{x} = \lim_{x \rightarrow0}\frac{\frac{d}{dx}[f(2+3x) +f(2 +5x)]}{\frac{d}{dx}[x]} = \lim_{x\to 0}{3f'(2+3x)+ 5f'(2+5x) \over 1} \\=3f'(2)+5f'(2)= 8f'(2)=8 \cdot 7=\bbox[red, 2pt]{56}$$

解答:$$u=-1-x \Rightarrow dx=-du \Rightarrow I= \int_{-1}^0 {e^x\over e^x+e^{-1-x}}\,dx = \int_0^{-1} {e^{-1-u} \over e^{-1-u}+e^{-1-(-1-u)}}(-du) \\=\int_{-1}^0 {e^{-1-u} \over e^{-1-u}+ e^u}\,du = \int_{-1}^0 {e^{-1-x} \over e^{-1-x}+ e^x}\,dx  \Rightarrow I+I=\int_{-1}^0 {e^x\over e^x+e^{-1-x}}\,dx+ \int_{-1}^0 {e^{-1-x} \over e^{-1-x}+ e^x}\,dx \\ \Rightarrow 2I= \int_{-1}^0 1\,dx=1 \Rightarrow I=\bbox[red, 2pt]{1\over 2}$$

解答:$$a_k= {k!\over k^k} \Rightarrow  L = \lim_{k \rightarrow \infty}\left \vert{} \frac{a_{k+1}}{a_{k}} \right\vert{} = \lim_{k\rightarrow\infty} \frac{\frac{(k+1)!}{(k+1)^{k+1}}}{\frac{k!}{k^{k}}} = \lim_{k \to \infty} {1\over \left( 1+{1\over k} \right)^k} ={1\over e} \lt 1 \Rightarrow \bbox[red, 2pt]{\text{convergent}}$$

解答:$$\lim_{y\to 0} f(0,y,0)= \lim_{y\to 0} {0+0+0 \over 0+ y^2+0}= \lim_{y\to 0} {0\over y^2} =0 \\ \lim_{x\to 0} f(x,x^2,0)=\lim_{x\to 0} {x^4\over x^4+x^4}={1\over 2} \\ \Rightarrow \lim_{y\to 0} f(0,y,0) \ne \lim_{x\to 0} f(x,x^2,0) \Rightarrow \bbox[red, 2pt]{\text{ The limit does NOT exist.}}$$

解答:$$\cases{x= r\cos \theta\\ y= r\sin \theta} \Rightarrow 3x+4y^2=3r\cos \theta+4r^2\sin^2 \theta \\\Rightarrow \iint_D (3x+4y^2)\,dA = \int_0^\pi \int_1^2 (3r\cos \theta+4r^2\sin^2 \theta) r\,dr\,d\theta =\int_0^{\pi} (7\cos \theta+15\sin^2\theta)\,d\theta \\= \left. \left[ 7\sin\theta +{15\over 2} \left( \theta-{1\over 2}\sin 2\theta \right) \right] \right|_0^{\pi} =0+{15\pi\over 2} = \bbox[red, 2pt] {15\pi\over 2}$$

解答:$$f(x,y) = \ln(x^2+y^2) \Rightarrow \nabla f(x,y)=(f_x,f_y) = \left( {2x\over x^2+y^2} ,{2y\over x^2+y^2}\right) \\ \Rightarrow \nabla f(1,1) = \left( {2\over 1+1}, {2\over 1+1} \right) =(1, 1) \Rightarrow \vec u= \bbox[red, 2pt]{\left( {\sqrt 2\over 2}, {\sqrt 2\over 2} \right)}$$

解答:$$\text{horizontal tangent line } \Rightarrow {dy\over d\theta}=0 \Rightarrow {dy\over d\theta} ={d\over d\theta}(r\sin \theta) ={dr\over d\theta}\sin \theta+ r\cos \theta \\ r^2=2\cos(2\theta) \Rightarrow 2r{dr\over d\theta}=-4\sin(2\theta) \Rightarrow {dr\over d\theta}={-2\sin(2\theta) \over r} \Rightarrow {-2 \sin(2 \theta) \over r}\sin \theta+ r\cos \theta=0 \\ \Rightarrow -2\sin(2\theta) \sin \theta+ r^2\cos \theta=0 \Rightarrow -2(2\sin \theta \cos \theta) \sin \theta+  2 \cos(2 \theta)\cos \theta =0 \\ \Rightarrow 2\cos \theta(\cos^2\theta-3\sin^2 \theta)=0 \Rightarrow \cases{\cos \theta=0 \Rightarrow \theta=\pi/2 \Rightarrow r^2=2\cos(\pi)=-2 \text{ no solution} \\ \cos^2\theta-3\sin^2\theta=0 \Rightarrow \tan^2\theta =1/3 \Rightarrow \tan \theta=1/\sqrt 3 \Rightarrow \theta=\pi/6} \\ \Rightarrow r^2=2\cos(2\times {\pi\over 6})=1 \Rightarrow r=1 \Rightarrow (r,\theta) = \bbox[red, 2pt]{\left( 1,{\pi\over 6} \right)}$$

解答:$$f(x)=e^x+ \ln(x+1) \Rightarrow \cases{f(0) =1\\ f(1)=e+\ln 2} \\ 利用反函數定積分公式:\int_a^b f(x)\,dx +\int_{f(a)}^{f(b)} f^{-1}(y)\,dy =bf(b)-af(a) \\ 將\cases{a=0\\ b=1} 代入公式\Rightarrow \int_0^1 (e^x+\ln(x+1))\,dx + \int_{1}^{e+\ln 2}f^{-1}(y)\,dy = \bbox[red, 2pt]{e+\ln 2}$$

 解答:$$利用\text{Stokes' Theorem}, 因為球面是一個封閉曲面，它沒有邊界曲線,即 \partial S=0\\因此  \iint_{S} \text{curl} \mathbf{F} \cdot \mathbf{n} dS = \oint_{\partial S} \mathbf{F} \cdot d\mathbf{r} = \bbox[red, 2pt]0 $$

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