113年台聯大轉學考-微積分A2詳解

 台灣聯合大學系統113學年度學士班轉學生考試

科目:微積分
類組別:A2

一、填充題:共８題，每題８分，總計64分

解答:$$I=\int_0^\infty \int_y^\infty y^2 e^{-x^2}\,dxdy = \int_0^\infty \int_0^x y^2e^{-x^2} \,dydx= {1\over 3}\int_0^\infty  x^3e^{-x^2}\,dx \\ 取u=x^2,則du=2xdx \Rightarrow I={1\over 6}\int_0^\infty ue^{-u}\,du ={1\over 6} \left. \left[ -ue^{-u}-e^{-u} \right] \right|_0^\infty \\={1\over 6}(0-(-1))= \bbox[red, 2pt]{1\over 6}$$

解答:$$f(x)=e^x \sin x \Rightarrow f'(x)=e^x \sin x+e^x \cos x =e^x(\sin x+\cos x) \Rightarrow f''(x)=2e^x \cos x\\ f''(x)=0 \Rightarrow \cos x=0 \Rightarrow x={\pi\over 2},{3\pi\over 2} \Rightarrow \cases{f(\pi/2)=e^{\pi/2} \\f(3\pi/2)=-e^{3\pi/2}} \\ \Rightarrow \text{inflection points: }\bbox[red, 2pt]{\left({\pi\over 2}, e^{\pi/2} \right), \left( {3\pi\over 2},-e^{3\pi/2} \right)}$$

解答:$$\lim_{x\to 0} {x\sin x\over 1-\cos x} =\lim_{x\to 0} {(x\sin x)' \over (1-\cos x)'} =\lim_{x\to 0} {\sin x+ x\cos x\over \sin x} = \lim_{x\to 0} {(\sin x+ x\cos x)' \over (\sin x)'} \\=\lim_{x\to 0} {2\cos x -x\sin x\over \cos x} =\bbox[red, 2pt]2$$

解答:$$a_n={(-1)^n(n+2) \over 3^n}(x-3)^n \Rightarrow \lim_{n\to \infty}\left|{a_{n+1} \over a_n} \right|\\ = \lim_{n\to \infty}\left|{(-1)^{n+1}(n+3)(x-3)^{n+1} \over 3^{n+1}} \cdot {3^n\over (-1)^n(n+2)(x-3)^n} \right| =\lim_{n\to \infty}\left|{(n+3)(x-3)  \over 3(n+2)}   \right| \\={1\over 3}|x-3| \Rightarrow -1\lt {1\over 3}|x-3| \lt 1 \Rightarrow 0\lt x\lt 6 \Rightarrow \text{interval of convergence: }\bbox[red, 2pt]{(0,6)}$$

解答:$$\cases{\cases{\lim_{x \to 0^-} f(x)=\sin 0=0\\ \lim_{x\to 0^+} f(x)=b-3} \Rightarrow b-3=0 \Rightarrow  b=3\\[1ex] \cases{\lim_{x \to 0^-} f'(x) =\lim_{x \to 0^-} 2\cos(2x)=2\\ \lim_{x\to 0^+} f'(x)=\lim_{x \to 0^+} 4x+a =a} \Rightarrow a=2}  \Rightarrow (a,b)= \bbox[red, 2pt]{(2,3)}$$

解答:

$$圓錐體體積相當於直線\;y=-{r\over h}x+r繞x軸旋轉所得體積,即\\ V=\pi \int_0^h \left( -{r\over h}x+r \right)^2 \,dx=\pi \int_0^h \left( {r^2\over h^2}x^2-{2r^2\over h}x+r^2 \right) \,dx\\ =\pi \left.\left[{r^2\over 3h^2}x^3-{r^2\over h}x^2+r^2 x \right] \right|_0^h =\bbox[red, 2pt]{{1\over 3}{r^2h\pi}}$$

解答:$$f(x,y)=x^3-12xy +8y^3 \Rightarrow \cases{f_x=3x^2-12y\\ f_y=-12x+24y^2} \Rightarrow \cases{f_{xx}=6x\\ f_{xy}=-12 \\f_{yy} =48y} \\ \Rightarrow D(x,y)= f_{xx}f_{yy}-f_{xy}^2= 288 xy-144\\ 若\cases{f_x=0\\ f_y=0} \Rightarrow \cases{(x,y)=(0,0) \\ (x,y)=(2,1)} \Rightarrow \cases{D(0,0)=-144 \lt 0 \Rightarrow (0,0) \text{ is a saddle point}\\ D(2,1)=432 \gt 0 \Rightarrow f_{xx}(2,1)=12 \gt 0} \\ \Rightarrow f(2,1)=-8 \Rightarrow  \bbox[red, 2pt]{\text{relative minimum: -8}}$$

解答:$$x^2+xy=\sin x \Rightarrow 2x+y+ xy'=\cos x \Rightarrow y'={dy\over dx}= \bbox[red, 2pt]{\cos x-2x-y\over x}$$

二、計算、證明題: 共３題，每題１２分，總計３６分

解答:$$\text{By integral test, }I=\int_3^\infty{1\over x(\ln x)^p}\,dx = \int_{\ln 3}^\infty {1\over u^p}\,du =\left. \left[ {1\over 1-p} u^{1-p} \right] \right|_{\ln 3}^\infty \\ \textbf{Case I } p\gt 1 : I=-{1\over 1-p}(\ln 3)^{1-p} \Rightarrow \text{ convergent} \\ \textbf{Cases II }p\lt 1 : I =\infty \Rightarrow \text{ divergent} \\ \textbf{Case III }p=1: I= \int_{\ln 3}^\infty {1\over u}\,du =\ln \infty-\ln(\ln 3)=\infty \Rightarrow \text{ divergent} \\ \Rightarrow \bbox[red, 2pt]{ \cases{\text{convergent }, p\gt 1\\ \text{divergent }, 0\le p\le 1}}$$

解答:$$假設M(t)為鹽含量\Rightarrow {dM\over dt} =2\times 4-{M(t)\over 20-(4-4)t}\times 4 =8-{M(t)\over 5} \\ \Rightarrow M'=8-{1\over 5}M, M(0)=0 \Rightarrow M'+{1\over 5}M=8 \Rightarrow e^{t/5}M'+{1\over 5}M'e^{t/5}=8e^{t/5} \\ \Rightarrow \left( e^{t/5}M \right)'=8e^{t/5} \Rightarrow e^{t/5}M= \int 8e^{t/5}\,dt= 40e^{t/5}+c_1 \\ \Rightarrow M(t)=40+c_1e^{-t/5}  \Rightarrow M(0)=40+c_1=0 \Rightarrow c_1=-40 \\ \Rightarrow M(t)=40-40e^{-t/5} \Rightarrow M(10)= \bbox[red, 2pt]{40(1-{1\over e^2})}$$

解答:$$f(x)=g(x) \Rightarrow e^x+6e^{-x}=5 \Rightarrow e^{2x}-5e^x+6=0 \Rightarrow (e^x-3)(e^x-2)=0\\ \Rightarrow x=\ln 3,\ln 2 \Rightarrow 面積=\int_{\ln 2}^{\ln 3} (f(x)-g(x))\,dx  =\int_{\ln 2}^{\ln 3} (-e^x-6e^{-x}+5)\,dx \\ = \left. \left[ -e^x+6e^{-x}+5x \right] \right|_{\ln 2}^{\ln 3} = \bbox[red, 2pt]{5\ln{3\over 2}-2}$$

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