臺灣綜合大學系統108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：$$\mathbf{(a)}\lim_{x \to 3} {x^2+x-12\over x-3} =\lim_{x \to 3} {(x-3)(x+4)\over x-3} =\lim_{x \to 3} (x+4)= \bbox[red, 2pt]7 \\\mathbf{(b)} \lim_{x \to 0} {x^2 \over \sec x-1} = \lim_{x \to 0} {(x^2)' \over (\sec x-1)' } = \lim_{x \to 0} {2x \over \tan x\sec x} =\lim_{x \to 0} {2x \cos^2 x\over \sin x}\\\quad =\lim_{x \to 0} {2\cos^2 x-4x\cos x\sin x\over \cos x} =\bbox[red, 2pt]2$$
解答：$$\cases{f_x(0,0) =\lim_{h\to 0}{f(h,0)-f(0,0)\over h} =\lim_{h\to 0}{3h/2 \over h} ={3\over 2} \\ f_y(0,0)=\lim_{h\to 0}{f(0,h)-f(0,0)\over h}=\lim_{h\to 0}{ 0\over h}=0} \Rightarrow \bbox[red,2pt]{\cases{\left.{\partial f\over \partial x} \right|_{(0,0)}={3\over 2} \\ \left.{\partial f\over \partial y} \right|_{(0,0)}= 0} }$$
解答：$$F(x)=\int_{-1}^{2ax+2} f(t)\,dt \Rightarrow F'(x)= 2af(2ax+2) =\cases{2a,2ax+2\le 0\\ 2a(1-2ax-2)=-2a(1+2ax),2ax\gt 0} \\ 又 F'(-2a)=0 \Rightarrow -2a(1-4a^2)=0 \Rightarrow a=\bbox[red,2pt]{1\over 2}$$
解答：$$令\cases{A(0,2)\\ B(1,0)}，及P在x^2+{y^2\over 4}=1上，即P(\cos\theta, 2\sin \theta) \Rightarrow \cases{\overrightarrow {PA}=(-\cos\theta,2-2\sin\theta) \\ \overrightarrow{PB}=(1-\cos\theta,-2\sin\theta)} \\ \Rightarrow \triangle PAB面積= {1\over 2} \begin{Vmatrix} -\cos\theta & 2-2\sin\theta\\ 1-\cos\theta & -2\sin\theta \end{Vmatrix} = |\cos\theta +\sin \theta-1| \Rightarrow \theta=-{\pi\over 4}有極大值\bbox[red ,2pt]{\sqrt 2+1}$$
解答：$$u=2e^x \Rightarrow du=2e^xdx \Rightarrow \int_{\ln(1/4)}^{\ln(1/2)}{e^x \over \sqrt{1-4e^{2x}}}\,dx =\int_{1/2}^1 {1/2\over \sqrt{1-u^2}}du =\left.\left[ {1\over 2}\sin^{-1} u \right] \right|_{1/2}^1 \\={1\over 2}({\pi\over 2}-{\pi\over 6})= \bbox[red,2pt]{\pi \over 6}$$
解答：$$\int_0^{\sqrt 2/2} \int_y^{\sqrt{1-y^2}} e^{x^2+y^2}\,dxdy =\int_0^{\pi/4} \int_0^1 re^{r^2}\,drd\theta =\int_0^{\pi/4} \left. \left[{1\over 2}e^{r^2} \right] \right|_0^1 \;d\theta ={1\over 2}(e-1)\times{\pi \over 4} \\=\bbox[red,2pt]{{\pi \over 8}(e-1)}$$
解答：$$f(x)=\ln\left({1+2x\over 1-2x}\right) \Rightarrow f'(x)={4\over 1-4x^2} =4(1+4x^2+ (4x^2)^2+ (4x^2)^3+\cdots ) \\= 4+4^2 x^2+ 4^3x^4 +4^4 x^6+\cdots +4^{n+1}x^{2n}+\cdots =\sum_{n=0}^\infty 4^{n+1}x^{2n}\\\Rightarrow f''(x)= \sum_{n=1}^\infty 2n\cdot 4^{n+1}x^{2n-1} \Rightarrow f(x)=f(0)+f'(0)x +{f''(0)\over 2!}x^2 +\cdots \\ =0+4x +{16\over 3} x^3 +{64\over 5}x^5 +\cdots +{4^k\over 2k-1}x^{2k-1}+\cdots \Rightarrow f(x)= \bbox[red,2pt]{\sum_{k=1}^\infty {4^k\over 2k-1}x^{2k-1}}$$
解答：$$T(x,y)=1+x^2-y^2 \Rightarrow \nabla T=(2x,-2y) \Rightarrow -\nabla T(\gamma (t))=(-2x(t),2y(t)) =\gamma'(t)=(x'(t),y'(t))\\ \Rightarrow \cases{-2x(t)= x'(t) \\ 2y(t)= y'(t)} \Rightarrow \cases{x(t)=C_1e^{-2t}\\ y(t)= C_2e^{2t}}，又\gamma(0)=(x(0),y(0))= (1,4)，因此\cases{x(0)=C_1= 1\\ y(0)= C_2=4} \\ \Rightarrow \bbox[red, 2pt]{\gamma(t)= (e^{-2t}, 4e^{2t})}$$
解答：$$令F(x,y,z)={x\over 2}-{y\over 4}+{\sin(2z)\over 4} \Rightarrow \cases{F_x=1/2\\ F_y= -1/4\\ F_z=\cos(2z)/2}\\ \Rightarrow 過P(1,a,b)之切平面: F_x(P)(x-1) +F_y(P)(y-a)+ F_z(P)(z-b)=0\\ \Rightarrow {1\over 2}(x-1)-{1\over 4}(y-a)+{\cos (2b)\over 2} (z-b)=0 \Rightarrow 2(x-1)-(y-a)+2\cos(2b)(z-b)=0 \cdots(1)\\ 又\cases{\ell_1方向向量\vec u=(1/2,1,0)\\ \ell_2方向向量\vec v=(0,2,1)} \Rightarrow \vec n=\vec u\times \vec v=(1,-1/2,1)\\ \Rightarrow 切平面方程式:(x-1)-{1\over 2}(y-a)+(z-b)=0 \Rightarrow 2(x-1)-(y-a)+2(z-b)=0 \cdots(2) \\ 因此(1)=(2) \Rightarrow \cos(2b)=1 \Rightarrow b=0\Rightarrow F(1,a,0)=0 \Rightarrow {1\over 2}-{a\over 4}+0=0 \Rightarrow a=2 \\ \Rightarrow (a,b)=\bbox[red, 2pt]{(2,0)}$$
解答：$$\cases{C_1=A\to B\to C\to A為一loop \\ C_2=\overline{AD} =\{(-t,0)\mid t=0\to 1\}}\\ 依\text{Green's Theorem}: \vec F=(P(x,y),Q(x,y)) =(4x+5y,e^{\cos y}+7x) \\\Rightarrow \int_{C_1}\vec F\cdot d\vec r = \int \left({\partial \over \partial x}Q -{\partial \over \partial y}P\right)\;dA =\int (7-5)\,dA = \int 2\,dA = 2\times {\pi\over 4}={\pi \over 2}\\ 又\int_{C_2} \vec F\cdot d\vec r =\int_0^1 (-4t,1-7t)\cdot (-1,0)dt =\int_0^1 4t\,dt =2\\ 因此 \int_L \vec F\cdot d\vec r= \bbox[red,2pt]{{\pi \over 2}+2}$$
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