112年政大轉學考-微積分詳解

國立政治大學112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:$$\textbf{(a) } \lim_{n\to \infty} {(-1)^n n^2 \over n^2+n+1} = \lim_{n\to \infty} {(-1)^n   \over (1+1/n+1/n^2)} = \pm 1 \Rightarrow \bbox[red, 2pt]{\text{ the limit does not exist}}\\ \textbf{(b) } \lim_{x\to -\infty}{\sqrt{4x^6+7} \over x^3+1} = \lim_{x\to -\infty}{2|x^3| \sqrt{1 +7/(4x^6)} \over x^3+1} = \bbox[red, 2pt]{-2} \\\textbf{(c) }\cases{x=r\cos \theta\\ y=r\sin \theta} \Rightarrow \lim_{(x,y) \to (0,0)} {2xy+3x^3\over \sqrt{x^2+y^2}} =\lim_{r\to 0} {2r^2\cos \theta \sin \theta+3r^3\cos^3\theta\over r}\\\quad =\lim_{r\to 0} (r\sin 2\theta+3r^2\cos^3\theta) = \bbox[red, 2pt]0$$

解答:$$xy+e^{xz}+yz^2=0 \Rightarrow y+(z+xz_x)e^{xz}+2yzz_x=0 \Rightarrow \frac{\partial z}{\partial x}=z_x= -{y+ze^{xz}\over xe^{xz}+2yz} \\ \Rightarrow \frac{\partial z}{\partial x}(0,1,2) =-{1+2\over 0+4} = \bbox[red, 2pt]{-{3\over 4}}$$

解答:$$\textbf{(a) }\cases{u=\sec x\\ dv=\sec^2 x\,dx} \Rightarrow \cases{du = \tan x \sec x\,dx\\ v=\tan x} \Rightarrow I=\int {1\over \cos^3 x}\,dx = \int \sec^3\,dx \\=  \tan x\sec x-\int \tan^2 x\sec x\,dx =  \tan x\sec x-\int ( \sec^2 x-1) \sec x\,dx =\tan x\sec x-I+\int \sec x\,dx \\ \Rightarrow 2I=\tan x\sec x+  \ln |\tan x+\sec x| +C\Rightarrow \int {2\over \cos^3 x} =2I= \bbox[red, 2pt]{\tan x\sec x+  \ln |\tan x+\sec x| +C} \\\textbf{(b) }u=e^x \Rightarrow du=e^x\,dx \Rightarrow \int e^{x+e^x}\,dx = \int e^{e^x}\cdot e^x\,dx =\int e^u\,du =e^u+C= \bbox[red, 2pt]{e^{e^x}+C}$$

解答:$$f'(c)={f(b)-f(a)\over b-a} \Rightarrow -1\le {f(5)-f(1)\over 5-1} \le 3 \Rightarrow -1\le {4-f(1) \over 4} \le 3 \Rightarrow -8\le f(1)\le 8\\ \Rightarrow \text{the largest value }f(x)\text{ is }\bbox[red, 2pt]8$$

解答:$$\cases{f(x,y)=x^2y \\g(x,y)=x^2+y^2-1} \Rightarrow   \cases{f_x= \lambda g_x\\ f_y= \lambda g_y\\ g=0} \Rightarrow \cases{2xy=\lambda(2x) \\ x^2=\lambda(2y) \\ x^2+y^2=1} \Rightarrow {2y\over x}={x\over y} \Rightarrow x^2=2y^2\\ \Rightarrow 2y^2+y^2=1 \Rightarrow y=\pm {1\over \sqrt 3} \Rightarrow x=\pm \sqrt{2\over 3} \Rightarrow \cases{f(\sqrt{2\over 3},\sqrt{1\over 3}) ={2\sqrt 3\over 9} \\ f(\sqrt{2\over 3},-\sqrt{1\over 3}) =-{2\sqrt 3\over 9} \\ f(-\sqrt{2\over 3},\sqrt{1\over 3}) ={2\sqrt 3\over 9} \\ f(-\sqrt{2\over 3},-\sqrt{1\over 3}) =-{2\sqrt 3\over 9} } \\ \Rightarrow \bbox[red, 2pt]{\cases{\text{max:} {2\sqrt 3\over 9} \\\text{min:} -{2\sqrt 3\over 9}}}$$

解答:$$g(x)=1+x+x^2+ \cdots = 1+x(1+x+ x^2+ \cdots) =1+xg(x) \\ \Rightarrow g(x)={1\over 1-x}, |x|\lt 1 \Rightarrow f(x)=g(-x^2) ={1\over 1+x^2} =1-x^2+ x^4-x^6+ \cdots \\ \Rightarrow f(x)={1\over 1+x^2} = \sum_{n=0}^\infty (-1)^n x^{2n}, |-x^2|\lt 1 \Rightarrow |x|\lt 1\\f(x)={1\over 1+x^2} =\sum_{n=0}^\infty (-1)^n x^{2n} \Rightarrow \int f(x)\,dx =\int {1\over 1+x^2}\,dx =\int \sum_{n=0}^\infty (-1)^n x^{2n} \\ =\arctan(x)+C= C+ \sum_{n=0}^\infty {(-1)^n \over 2n+1} x^{2n+1} \Rightarrow \arctan(x) =\sum_{n=0}^\infty {(-1)^n \over 2n+1} x^{2n+1} \\ \Rightarrow \arctan({1\over \sqrt 3}) =\sum_{n=0}^\infty {(-1)^n \over (2n+1)(\sqrt 3)^{2n+1}}  \Rightarrow \sum_{n=0}^\infty {(-1)^n \over (2n+1)(\sqrt 3)^{2n+1}}= \bbox[red, 2pt]{\pi \over 6}$$

解答:

$$\cases{P(x,y)=e^x+y^2\\ Q(x,y)=e^y+x^2} \Rightarrow Q_x-P_y=2x-2y\\ \text{By Green's Theorem,} \int_C P\,dx +Q\,dy = \int_D(Q_x-P_y)\,dxdy= \int_0^2 \int_{x^2}^4 (2x-2y)\,dydx \\=\int_0^2 (x^4-2x^3+8x-16)\,dx = \bbox[red, 2pt]{-{88\over 5}}$$

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解題僅供參考，轉學考歷年試題及詳解