111年政大轉學考-微積分詳解

國立政治大學111學年度轉學生招生考試

科目:微積分

系所別:應用數學系三年級

解答: $$\textbf{(a) }\lim_{(x,y)\to (0,0)} {x^2y \sin y\over 3x^4+y^2} = \lim_{(x,y)\to (0,0)} \left({x^2 y^2 \over 3x^4+y^2}  \cdot {\sin y\over y}\right) =\lim_{(x,y)\to (0,0)}{x^2 y^2 \over 3x^4+y^2} \cdot \lim_{y\to 0} {\sin y\over y} \\\quad =\lim_{(x,y)\to (0,0)}{x^2 y^2 \over 3x^4+y^2} \cdot 1 =\lim_{(x,y)\to (0,0)}{x^2 y^2 \over 3x^4+y^2} \\ 0\le {x^2 y^2 \over 3x^4+y^2} \le {x^2 y^2 \over  y^2} = x^2=0 (\text{as }x \text{ to 0}) \Rightarrow \lim_{(x,y)\to (0,0)}{x^2 y^2 \over 3x^4+y^2}=0 \Rightarrow \lim_{(x,y)\to (0,0)} {x^2y \sin y\over 3x^4+y^2} = \bbox[red, 2pt] 0\\ \textbf{(b) }L=\lim_{x \to 0} {\sin^3 x\tan^2(3x) \over x(1-\cos(5x))^2} = \lim_{x \to 0}  {\left(x-{x^3\over 3!}+ \cdots \right)^3 \left(3x+{1\over 3}(3x)^3+ \cdots \right)^2 \over x\left( {(5x)^2\over 2!}-{(5x)^4\over 4!} + \cdots\right)^2} \\ \text{Considering the coefficients of }x^5, L=  {1\cdot 3^2 \over 1\cdot {625\over 4}} = \bbox[red, 2pt]{36\over 625}$$

解答: $$\textbf{(a) }\sec u={x-2\over 2} \Rightarrow \cases{ \tan u \sec u\,du =dx/2\\ 2\tan u=\sqrt{x^2-4x}} \Rightarrow \int {1\over \sqrt{x^2-4x}} \,dx = \int {2\tan u \sec u\over 2\tan u}\,du \\= \int \sec u\,du = \ln(|\tan u+\sec u|) +C = \bbox[red, 2pt]{\ln\left({1\over 2}\left|\sqrt{x^2-4x}+x-2 \right| \right) +C }$$

$$\textbf{(c) }\cases{u=y+x\\ v=y-x} \Rightarrow \cases{x=(u-v)/2\\ y=(u+v)/2} \Rightarrow |J|=\left|{\partial (x,y) \over \partial(u,v)} \right| =\begin{Vmatrix}1/2 & -1/2\\ 1/2& 1/2 \end{Vmatrix} ={1\over 2} \\ \Rightarrow \iint_R \cos\left({y-x\over y+x} \right)\,dA =\int_1^2 \int_{-u}^u {1\over 2}\cos{v\over u} \,dvdu =\int_1^2  \left. \left[ {u\over 2} \sin {v\over u} \right] \right|_{-u}^u\,d u =\int_1^2 u\sin 1\,du \\= \sin 1\left. \left[{1\over 2}u^2  \right] \right|_1^2 = \bbox[red, 2pt]{{3\over 2}\sin 1}$$

$$\textbf{(d) }\cases{P(x,y)=5x+\cos y^2\\ Q(x,y) =3y+e^{\sqrt x}} \Rightarrow \cases{P_x=5\\ Q_y=3} \\\quad \text{Using Green Theorem, }\int_C Q\,dx+P\,dy= \int_D (P_x-Q_y)dA =\int_D(5-3)\,dA =2\cdot {1\over 3}= \bbox[red, 2pt]{2\over 3} \\ \quad \text{Area of }D= \int_0^1 (\sqrt x-x^2)\,dx ={1\over 3}$$

解答: $$\textbf{(a) } 1-{1\over 2n} ={2n-1\over 2n}\lt p \lt 1  \Rightarrow \left(1-{1\over 2n} \right)^{3n\ln n} \lt p^{3n\ln n} \lt p^{3n} \lt p^n, \text{ for }p\ge 3\\ \sum p^n \text{ is convergent}\Rightarrow \sum \left(1-{1\over 2n} \right)^{3n\ln n} \text{ is } \bbox[red, 2pt]{\text{convergent}}\\ \textbf{(b) } \text{Let }a_n={1\over n^{1+1/n}}, b_n={1\over n}(\sum b_n  \text{ is divergent}), \text{ then }\lim_{n\to \infty}{a_n\over b_n} =\lim_{n\to \infty} {1\over n^{1/n}} =1\\\quad \text{By limit comparison test}, \sum b_n \text{ and }\sum a_n \text{ are both divergent}. \text{ That is, } \bbox[red, 2pt]{\sum_{n=3}^\infty {1\over n^{1+1/n}} \text{ is divergent}}$$

解答:

$$\text{area of }R = \overline{AB}\cdot \overline{OB}\cdot \pi = \bbox[red, 2pt]{9\sqrt{10}\pi}$$

解答: $$V=\int_{-1}^1 {4\over 3}\pi (1-u^2)^{3/2}\,du \\ u=\sin \theta \Rightarrow du= \cos \theta \,d\theta \Rightarrow V={4\over3}\pi \int_{-\pi/2}^{\pi/2} \cos^4 \theta \,d \theta ={4\over3}\pi \int_{-\pi/2}^{\pi/2} \cos^2 \theta(1-\sin^2 \theta)  \,d \theta \\={4\over3}\pi \int_{-\pi/2}^{\pi/2} \cos^2 \theta \,d\theta-{4\over3}\pi \int_{-\pi/2}^{\pi/2} \cos^2 \theta \sin^2\theta \,d\theta ={2\over 3} \pi \int_{-\pi/2}^{\pi/2} (\cos 2\theta+1) \,d\theta-{1\over 3}\pi\int_{-\pi/2}^{\pi/2} \sin^2 2\theta\,d\theta \\={2\over 3}\pi \left. \left[ {1\over 2}\sin 2\theta+ \theta \right] \right|_{-\pi/2}^{\pi/2} -{1\over 6}\pi \int_{-\pi/2}^{\pi/2} (1-\cos 4\theta)\,d\theta ={2\over 3}\pi^2-{1\over 6}\pi \left. \left[ \theta-{1\over 4}\sin 4\theta \right] \right|_{-\pi/2}^{\pi/2} \\={2\over 3}\pi^2 -{1\over 6}\pi^2 =\bbox[red, 2pt]{{1\over 2}\pi^2}$$

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解題僅供參考，轉學考歷年試題及詳解