114年台聯大轉學考-微積分A2詳解

台灣聯合大學系統114學年度學士班轉學生考試

科目:微積分 類組別:A2

甲、填充題：共8題，每題8分，共64分

解答:$$\lim_{n\to \infty}{\sqrt{n+1}+\sqrt n\over n(\sqrt{n+1}-\sqrt n)} =\lim_{n\to \infty}{(\sqrt{n+1}+\sqrt n)^2 \over n(\sqrt{n+1}-\sqrt n)(\sqrt{n+1}+\sqrt n)} =\lim_{n\to \infty}{2n+1+2\sqrt{n^2+n} \over n } \\=\lim_{n\to \infty}{2n+1+2n\sqrt{1+1/n} \over n } =\bbox[red, 2pt]4$$

解答:$$F(x)=f(\ln(2\sin x)) \Rightarrow F'(x)={\cos x\over \sin x}f'(\ln(2\sin x)) \\ \Rightarrow F'(\pi/6) ={\sqrt 3/2\over 1/2} f'(\ln 1) =\sqrt 3 f'(0)= \sqrt 3\cdot 2= \bbox[red, 2pt]{2\sqrt 3}$$

解答:$$u=x^3 \Rightarrow du=3x^2\,dx \Rightarrow \int e^{x^3}x^2\,dx = \int {1\over 3}e^u\,du ={1\over 3}e^u+C= \bbox[red, 2pt]{{1\over 3}e^{x^3}+C, C\text{ is constant}}$$

解答:$$f(x,y)=x^3-6xy-y^2 \Rightarrow \cases{f_x=3x^2-6y\\ f_y=-6x-2y} \Rightarrow \cases{f_{xx}= 6x\\ f_{xy} =-6\\ f_{yy} =-2} \Rightarrow d(x,y)=f_{xx}f_{yy}-(f_{xy})^2 =-12x-36 \\ \cases{f_x=0\\ f_y=0} \Rightarrow \cases{x^2=2y\\ y=-3x} \Rightarrow x^2+6x=0 \Rightarrow \cases{x=0 \Rightarrow y=0\\ x=-6 \Rightarrow y=18} \Rightarrow \cases{d(0,0)=-36\lt 0\\ d(-6,18)=36\gt 0} \\ \Rightarrow f_{xx}(-6,18)=-36 \lt 0 \Rightarrow \bbox[red, 2pt]{\cases{(0,0) \text{ is a saddle point} \\ f(-6,18)=108 \text{ is a local maxima}}}$$

解答:$$x=2.\overline{21} =2.212121 \cdots \Rightarrow 100x=221.212121\cdots \Rightarrow 100x-x=99x=218 \\ \Rightarrow x= \bbox[red, 2pt]{218\over 99}$$

解答:

$$x^2=4-x^2 \Rightarrow 2x^2=4 \Rightarrow x=\pm \sqrt 2 \Rightarrow \text{enclosed area }=4\int_0^\sqrt 2 (4-x^2-2)\,dx \\=4\left. \left[ 2x-{1\over 3}x^3\right] \right|_0^\sqrt 2 = \bbox[red, 2pt]{{16\over 3}\sqrt{2}}$$

解答:$$\int_0^3 \int_0^2 20x^2 e^{-y}\,dy\,dx =(1-e^{-2}) \int_0^320x^2\,dx = \bbox[red, 2pt]{180(1-e^{-2})}$$

解答:$$f(x)={\sqrt{4-x^2} \over \ln(x-1)} \Rightarrow \cases{4-x^2\ge 0\\x-1\gt 0\\ \ln(x-1)\ne 0} \Rightarrow \cases{-2\le x\le 2\\ x\gt 1\\ x\ne 2} \Rightarrow \bbox[red, 2pt]{1\lt x\lt 2}$$

乙、計算、證明題：共2題，每題10分，共20分

解答:$$f(x)={x\over 3+2x} ={x\over 3}\cdot {1\over 1-(-2x/3)} ={x\over 3}(1-{2x\over 3}+{4x^2\over 9}-{8x^3\over 27}+ \cdots) ={x\over 3}-{2x^2\over 9}+{4x^3\over 27} -{8x^4\over 81}+ \cdots\\ \Rightarrow \bbox[red, 2pt]{f(x) =\sum_{n=1}^\infty(-1)^{n+1}{2^{n-1} \over 3^n}x^{n-1}}\Rightarrow \lim_{n\to \infty}|{a_{n+1}\over a_n}| = \lim_{n\to \infty} \left| (-1)^{n+2}{2^{n}x^n \over 3^{n+1}}/(-1)^{n+1}{2^{n-1} x^{n-1}\over 3^n}\right|\\ ={2\over 3}|x|\lt 1 \Rightarrow |x|\lt {3\over 2} \Rightarrow \bbox[red, 2pt]{\text{ radius of convergence: }{3\over 2}}$$

解答:$${dQ\over dt}=2(100-Q) \Rightarrow {dQ\over 100-Q} =2dt \Rightarrow \int {1\over 100-Q}dQ =\int 2\,dt \Rightarrow -\ln(100-Q)=2t+ c_1 \\ \Rightarrow \ln(100-Q)=-2t+c_2 \Rightarrow 100-Q=e^{-2t+c_2} =c_3e^{-2t} \Rightarrow Q=100-c_3e^{-2t} \\ \text{initial condition }Q(0)=3 \Rightarrow 100-c_3=3 \Rightarrow c_3=97 \Rightarrow \bbox[red, 2pt]{Q=100-97e^{-2t}}$$

解答:$$V =x(10-2x)^2 \Rightarrow V'=(10-2x)^2-4x(10-2x) =0 \Rightarrow (10-2x)(10-6x) =0\\ \Rightarrow x={5\over 3} \Rightarrow \text{the dimensions of the box: } \bbox[red, 2pt]{\cases{\text{side}=20/3 \\\text{height}=5/3}}$$

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