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

解： $$\lim_{x\to 109}\frac{\sqrt x-\sqrt {109}}{x-109} =\lim_{x\to 109}\frac{(\sqrt x-\sqrt {109})'}{(x-109)'} = \lim_{x\to 109}\frac{{1\over 2\sqrt x} }{1} =\lim_{x\to 109}{1\over 2\sqrt x} ={1\over 2\sqrt{109}}\\，故選\bbox[red,2pt]{(A)}$$

解： $$g(x)=\frac{\sqrt{x-109}}{\sqrt[3]{x-2020}} \Rightarrow \cases{x-109 \ge 0\\ x-2020\ne 0} \Rightarrow x\in [109,2020) \cup (2020,\infty)，故選\bbox[red,2pt]{(D)}$$

解： $$x^2-xy-y^2=1 \Rightarrow {dy\over dx}(x^2-xy-y^2)= {dy\over dx}(1) \Rightarrow 2x-y-x{dy\over dx}-2y{dy\over dx}=0 \\ \Rightarrow 2x-y=(x+2y){dy\over dx} \Rightarrow {dy\over dx}=\frac{2x-y}{x+2y} \Rightarrow \left. {dy\over dx} \right|_{(2,1)} =\frac{4-1}{2+2}={3\over 4}，故選\bbox[red,2pt]{(B)}$$

解： $$g(x)=x\sqrt{x^2+8} \Rightarrow g'(x)=\sqrt{x^2+8}+\frac{x}{2\sqrt{x^2+8}}\cdot 2x =\sqrt{x^2+8}+\frac{x^2}{\sqrt{x^2+8}} \\ \Rightarrow g'(1)=\sqrt 9+\frac{1}{\sqrt 9} =3+{1\over 3}={10\over 3}，故選\bbox[red,2pt]{(D)}$$

解： $$g(x)={\sin x\over x^2+1} \Rightarrow g'(x)= {\cos x\over x^2+1} -{2x\sin x\over (x^2+1)^2} \Rightarrow g'(0)={1\over 1}-0=1，故選\bbox[red,2pt]{(A)}$$

解： $$\lim_{x\to \infty} {x^{109} \over e^x} =\lim_{x\to \infty} {(x^{109})' \over (e^x)'} =\lim_{x\to \infty} {(109x^{108})' \over (e^x)'} =\lim_{x\to \infty} {(109\cdot 108x^{107})' \over (e^x)'}=\cdots =\lim_{x\to \infty} {109! \over e^x }=0\\，故選\bbox[red,2pt]{(A)}$$

解： $$h(x)={x^3\over 3}+{x^2\over 2}-2x \Rightarrow h'(x)=x^2+x-2=(x+2)(x-1) \\ 因此h'(x)=0 \Rightarrow (x+2)(x-1)=0 \Rightarrow x=1,-2，故選\bbox[red,2pt]{(B)}$$

解： $$f(x)={2\over x}+x \Rightarrow f'(x)=-{2\over x^2}+1 \Rightarrow f''(x)={4\over x^3}\\ 若f'(x)=0 \Rightarrow {2\over x^2}=1 \Rightarrow x=\pm \sqrt 2 \Rightarrow \cases{f''(\sqrt 2) > 0\\ f''(-\sqrt 2) < 0} \Rightarrow f(\sqrt 2)為最小值,x\in [1,3]\\，故選\bbox[red,2pt]{(C)}$$

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

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

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

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

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

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

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

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

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

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

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

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

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