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2020年11月9日 星期一

109年專科學力鑑定-微積分詳解

教育部109年自學進修專科學校學力鑑定考試
專業科目(一): 微積分 詳解
limx109x109x109=limx109(x109)(x109)=limx10912x1=limx10912x=12109(A)

g(x)=x1093x2020{x1090x20200x[109,2020)(2020,)(D)

x2xyy2=1dydx(x2xyy2)=dydx(1)2xyxdydx2ydydx=02xy=(x+2y)dydxdydx=2xyx+2ydydx|(2,1)=412+2=34(B)

g(x)=xx2+8g(x)=x2+8+x2x2+82x=x2+8+x2x2+8g(1)=9+19=3+13=103(D)

g(x)=sinxx2+1g(x)=cosxx2+12xsinx(x2+1)2g(0)=110=1(A)

limxx109ex=limx(x109)(ex)=limx(109x108)(ex)=limx(109108x107)(ex)==limx109!ex=0(A)

h(x)=x33+x222xh(x)=x2+x2=(x+2)(x1)h(x)=0(x+2)(x1)=0x=1,2(B)

f(x)=2x+xf(x)=2x2+1f

g(x)=\sin x \Rightarrow g'(x)=\cos x \Rightarrow g''(x)=-\sin x \Rightarrow g^{[3]}(x)=-\cos x \Rightarrow g^{[4]}(x)= g(x)=\sin x\\ \Rightarrow g^{[n]}(0)=\cases{(-1)^{k} \cos x,n=2k+1\\ 0,n=2k},k=0,1,2,...\\ \Rightarrow g(x)=\sin x= \sum_{n=0}^\infty {g^{[n]}(0)\over n!} x^n ={1\over 1!} x-{1\over 3!}  x^3+{1\over 5!}x^5 -\cdots\\ \Rightarrow f(x)=\sin x^4 ={1\over 1!} x^4-{1\over 3!}  x^{12}+{1\over 5!}x^{20} -\cdots  \Rightarrow \cases{a_1=0\\ a_4=1 \ne 0\\ a_8=0\\ a_{12}=-1/6 \ne 0}\\,故選\bbox[red,2pt]{(C)}

\Omega=\{(x,y)\mid x^2+y^2\le 1\} \Rightarrow \cases{x=r\cos \theta\\ y=r\sin \theta},0\le r \le 1, 0\le \theta \le 2\pi \\\Rightarrow \iint_{\Omega}{1\over 1+x^2+y^2}dx dy =\int_0^{2\pi}\int_0^1{r\over 1+r^2}drd\theta =\int_0^{2\pi}\left .\left [ {1\over 2}\ln(1+r^2)\right] \right|_0^1\;d\theta =\int_0^{2\pi} {1\over 2}\ln 2\;d\theta \\ ={1\over 2}\ln 2\times 2\pi =\pi \ln 2,故選\bbox[red,2pt]{(B)}

\lim_{n\to \infty} \frac{1^5+2^5+\cdots +n^5}{n^6} =\lim_{n\to \infty} \frac{n^2(n+1)^2(2n^2 +2n-1)/12}{n^6} =\lim_{n\to \infty} \frac{2n^6 /12}{n^6} = {1\over 6},故選\bbox[red,2pt]{(B)}

令\cases{u=\cot^{-1}x \\ dv = dx} \Rightarrow \cases{du =-{1\over 1+x^2} dx\\ v=x} \Rightarrow \int \cot^{-1}x\;dx= x\cot^{-1}x +\int {x\over 1+x^2}\;dx \\=x\cot^{-1}x+{1\over 2}\ln (1+x^2) \Rightarrow \int_0^1 \cot^{-1}x\;dx= \left. \left[ x\cot^{-1}x+{1\over 2}\ln (1+x^2)\right] \right|_0^1\\= \cot^{-1}1+{1\over 2}\ln 2 ={\pi \over 4}+\ln \sqrt 2,故選\bbox[red,2pt]{(D)}

\int_{-1}^2 {1\over x^2+7x+12}\;dx =\int_{-1}^2 {1\over (x+3)(x+4)}\;dx = \int_{-1}^2 {1\over x+3}-{1\over x+4}\;dx = \left. \left[ \ln|x+3|-\ln|x+4|\right] \right|_{-1}^1 \\ =\ln 4-\ln 5-\ln 2+\ln 3 =\ln {4\over 5}-\ln {2\over 3} =\ln ({4\over 5}\times{3\over 2}) =\ln {6\over 5},故選\bbox[red,2pt]{(C)}

x=\sqrt 2\tan u \Rightarrow dx = \sqrt 2\sec^2u \;du \Rightarrow \int_{-\sqrt 2}^{\sqrt 2} {(2+x^2)^{-3/2}}\;dx= \int_{-\pi/4}^{\pi/4} (2+2\tan^2 u)^{-3/2}\cdot \sqrt 2\sec^2 u\;du \\ = \sqrt 2\int_{-\pi/4}^{\pi/4} (2\sec^2u  )^{-3/2}\cdot \sec^2u \;du = {1\over  2}\int_{-\pi/4}^{\pi/4} (\sec^2 u)^{-1/2}\;du ={1\over 2}\int_{-\pi/4}^{\pi/4} \cos u\;du = {1\over 2} \left. \left[ \sin u\right] \right|_{-\pi/4}^{\pi/4}\\={1\over 2}\times \sqrt 2 ={1\over \sqrt 2},故選\bbox[red,2pt]{(C)}

\int_{-\pi/2}^{\pi/2}{1\over 1+\cos x}\;dx =\int_{-\pi/2}^{\pi/2}{1\over 1+2\cos^2 (x/2)-1}\;dx = \int_{-\pi/2}^{\pi/2}{1\over 2}\sec^2 (x/2) \;dx = \left. \left[ \tan(x/2)\right] \right|_{-\pi/2}^{\pi/2} \\ =1-(-1) =2,故選\bbox[red,2pt]{(B)}

\int_0^{\pi/2} \sin^2(x)\cos^3(x)\;dx =\int_0^{\pi/2} (1-\cos^2(x))\cos^3(x)\;dx = \int_0^{\pi/2} \cos^3(x)-\cos^5(x)\;dx \\= \left. \left[ (\sin(x)-{1\over 3}\sin^3 (x))- (\sin(x)-{2\over 3}\sin^3 (x)+{1\over 5}\sin^5 (x))\right] \right|_0^{\pi/2} \\= \left. \left[  {1\over 3}\sin^3 (x) -{1\over 5}\sin^5 (x)\right] \right|_0^{\pi/2} ={1\over 3}-{1\over 5}= {2\over 15}\\ \Rightarrow f(x)在\left[0,{\pi\over 2}\right]的平均值= {\int_0^{\pi/2}f(x)\;dx \over \pi/2} ={2/15 \over \pi/2} ={4\over 15\pi},故選\bbox[red,2pt]{(A)}

\cases{x=r\cos \theta \\ y=r\sin \theta} \Rightarrow r=4\cos \theta \Rightarrow r^2=4r\cos \theta \Rightarrow x^2+y^2=4x \Rightarrow (x-2)^2+y^2 =2^2\\ \Rightarrow \cases{圓心(2,0)\\ 圓半徑=2} \Rightarrow 此圓一半在第一象限,面積=4\pi\times {1\over 2}=2\pi,故選\bbox[red,2pt]{(B)}

{1\over 2}V =\int_0^\pi 2\pi x\sin x\;dx =2\pi\left. \left[ \sin x-x\cos x\right] \right|_0^\pi =2\pi^2 \Rightarrow V=4\pi^2,故選\bbox[red,2pt]{(D)}

g(s,t)=f(x(s,t),y(s,t))= f(s+t^2,e^t-s^2) =(s+t^2)^{e^t-s^2} =e^{(e^t-s^2)\ln (s+t^2)} \\ \Rightarrow g_t(0,1)=(e^t\ln(s+t^2) +(e^t-s^2){2t\over s+t^2}) e^{(e^t-s^2)\ln (s+t^2)} \\ \Rightarrow g_t(0,1) = (e^1\ln(1) +(e^1-0){2\over 0+1}) e^{(e^1-0)\ln (1)} =(0+2e)e^0=2e,故選\bbox[red,2pt]{(??)}註:公布的答案為(C)

f(x,y)={1\over x}-xy+{8\over y} \Rightarrow \cases{f_x=-{1\over x^2}-y=0\\ f_y=-x-{8\over y^2}=0} \Rightarrow \cases{y=-{1\over x^2}\\ x=-{8\over y^2}} \Rightarrow \cases{x=-1/2\\ y=-4} \\ \Rightarrow f(-1/2,-4)=-2-2-2=-6,故選\bbox[red,2pt]{(A)}


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