教育部 110 年自學進修專科學校學力鑑定考試
解答：$$f(x)={x^4-1\over x^2-1} ={(x^2+1)(x^2-1)\over (x^2-1)} =x^2+1 \Rightarrow \cases{f(1)=2 =a\\ f(-1)=2=b} \Rightarrow {a\over b}=1，故選\bbox[red,2pt]{(A)}$$
解答：$$f(x)=\ln(1+x) \Rightarrow f'(x)=(1+x)^{-1} \Rightarrow f''(x)=-(1+x)^{-2}\Rightarrow f'''(x)=2(1+x)^{-3}\\f^{[4]}(x)=-3!(1+x)^{-4} \Rightarrow f^{[n]}=(-1)^{n-1}(n-1)!(1+x)^{-n} \Rightarrow f^{[2021]}(0)=2020!，故選\bbox[red,2pt]{(C)}$$
解答：$$x^y= y^x \Rightarrow \ln(x^y)=\ln(y^x) \Rightarrow y\ln x=x\ln y \Rightarrow (y\ln x)'= (x\ln y)' \\ \Rightarrow y'\ln x+{y\over x}=\ln y+x{y'\over y} \Rightarrow y'(\ln x-{x\over y})=\ln y-{y\over x} \Rightarrow {dy\over dx}= y'={\ln y-{y\over x}\over \ln x-{x\over y}}\\ ={xy\ln y-y^2\over xy\ln x-x^2}，故選\bbox[red,2pt]{(B)}$$
解答：$$f(x)=4x^3-8x^2+7x-2 \Rightarrow \cases{f'(x) =12x^2-16x+7\\f(1)= 1\\ f(0)=-2} \Rightarrow f'(c)={f(1)-f(0)\over 1-0}\\ \Rightarrow 12c^2-16c+7={1-(-2)\over 1}=3 \Rightarrow 12c^2-16c+4=0 \Rightarrow 3c^2-4c+1=0\\ \Rightarrow (3c-1)(c-1)=0 \Rightarrow c={1\over 3}(c=1不合,因為0\lt c\lt 1)，故選\bbox[red,2pt]{(B)}$$
解答：$$f(x)=x+{a\over x-1} \Rightarrow f'(x)=0 \Rightarrow 1-{a\over (x-1)^2}=0 \Rightarrow (x-1)^2=a \Rightarrow x=\sqrt a+1 \\ \Rightarrow f(\sqrt a+1)=3 \Rightarrow \sqrt a+1+ {a\over \sqrt a}=3 \Rightarrow 2\sqrt a=2 \Rightarrow a=1，故選\bbox[red,2pt]{(A)}$$
解答：$$f(x)=3x-(x-1)^{3/2} \Rightarrow f'(x)=3-{3\over 2}\sqrt{x-1}，因此f'(x)=0 \Rightarrow \sqrt{x-1}=2 \Rightarrow x=5\\ \Rightarrow f(5)=15-4^{3/2}=15-8=7，故選\bbox[red,2pt]{(D)}$$
解答：$$\displaystyle\lim_{x\to 0^+} \cfrac{e^{-1/x}}{x} = \lim_{x\to 0^+} \cfrac{1/x }{ e^{1/x}} = \lim_{x\to 0^+} \cfrac{-1/x^2 }{(-1/x^2) e^{1/x}} = \lim_{x\to 0^+} \cfrac{1 }{e^{1/x}}=0，故選\bbox[red,2pt]{(C)}$$
解答：$$\displaystyle\lim_{x\to 0^+} ({1\over 2}x^2-x^2\ln x-{1\over 2})= \lim_{x\to 0^+} ({1\over 2}x^2-x^2\ln x)-{1\over 2}\\ 而\lim_{x\to 0^+} ({1\over 2}x^2-x^2\ln x) =\lim_{x\to 0^+} \cfrac{{1\over 2} - \ln x}{1/x^2} =\lim_{x\to 0^+} \cfrac{({1\over 2} - \ln x)'}{(1/x^2)'}= \lim_{x\to 0^+}\cfrac{x^2}{2}=0\\ 因此 \lim_{x\to 0^+} ({1\over 2}x^2-x^2\ln x-{1\over 2})=0-{1\over 2}=-{1\over 2}，故選\bbox[red,2pt]{(C)}$$
解答：$${d\over dx}\int_0^{x^2} \sqrt{1+t^{110}}\;dt =(x^2)'\sqrt{1+(x^2)^{110}} =2x\sqrt{1+x^{220}}，故選\bbox[red,2pt]{(D)}$$
解答：$$\int (2^x+\ln e^{x^2})\;dx=\int (2^x+x^2)\;dx ={2^x\over \ln 2}+{1\over 3}x^3+C，故選\bbox[red,2pt]{(B)}$$
解答：$$令u=\sqrt x \Rightarrow du ={dx\over 2\sqrt x} \Rightarrow \int \cfrac{e^{\sqrt x}}{\sqrt x}\;dx = \int 2{e^u}\;du = 2e^u+C = 2e^{\sqrt x}+C，故選\bbox[red,2pt]{(C)}$$
解答：$$\cases{u=x \\dv=\cos xdx} \Rightarrow \cases{du=dx\\ v=\sin x} \Rightarrow \int_0^{\pi/2}x\cos x\;dx =\left.\left[ x\sin x-\int \sin x\;dx \right] \right|_0^{\pi/2} \\ =\left.\left[ x\sin x+\cos x \right] \right|_0^{\pi/2} ={\pi\over 2}-1，故選\bbox[red,2pt]{(A)}$$
解答：$${2\over (x-1)(x-2)(x-3)} ={a\over x-1}+{b\over x-2}+{c\over x-3} \\\Rightarrow a(x-2)(x-3)+ b(x-1)(x-3)+ c(x-1)(x-2)=2\\ \Rightarrow \cases{a=1\\b=-2\\c =1}\Rightarrow \int {2\over (x-1)(x-2)(x-3)} \;dx =\int ({1\over x-1}-{2\over x-2}+{1\over x-3})\;dx\\ =\ln|x-1|-2\ln|x-2|+\ln |x-3|+C，故選\bbox[red,2pt]{(D)}$$ 其實不用算，只有(D)符合a+b+c=0；
解答：
$$x^2=x+2 \Rightarrow x^2-x-2=0 \Rightarrow (x-2)(x+1)=0 \Rightarrow x=-1,2\\ \Rightarrow 兩曲線\cases{y=x^2\\ y=x+2}交於\cases{P(-1,1)\\ Q(2,4)} \\\Rightarrow 所圍區域面積=\int_{-1}^2 x+2-x^2\;dx = \left.\left[ {1\over 2}x^2+2x-{1\over 3}x^3 \right]\right|_{-1}^2 ={9\over 2}，故選\bbox[red,2pt]{(B)}$$
解答：$$\int_1^2 y^2\pi\;dx =\int_1^2 (x+{1\over x})^2\pi\;dx =\int_1^2 (x^2+{1\over x^2}+2) \pi\;dx = \pi\left. \left[ {1\over 3}x^3-{1\over x}+2x \right] \right|_1^2 \\ =({8\over 3}-{1\over 2}+4-{1\over 3}+1-2)\pi= {29\over 6}\pi，故選\bbox[red,2pt]{(C)}$$
解答：$$r=e^\theta \Rightarrow {dr\over d\theta}=e^\theta \Rightarrow 弧長= \int_1^2 \sqrt{r^2+({dr\over d\theta})^2}\;d\theta = \int_1^2 \sqrt{e^{2\theta}+e^{2\theta}}\;d\theta= \int_1^2 \sqrt 2e^\theta\;d\theta\\ =\left.\left[ \sqrt 2e^\theta \right]\right|_1^2 =\sqrt 2(e^2-e)，故選\bbox[red,2pt]{(B)}$$
解答：$$f(x,y)={xy\over x+y^2+1} \Rightarrow f_x(x,y)={y\over x+y^2+1} -{xy\over (x+y^2+1)^2} \Rightarrow f_x(2,1)={1\over 4}-{2\over 16}={1\over 8}\\，故選\bbox[red,2pt]{(A)}$$
解答：$$f(x,y)=x^3+y^3-12x-3y \Rightarrow \cases{f_x=0\\ f_y=0} \Rightarrow \cases{3x^2-12=0\\ 3y^2-3=0} \Rightarrow \cases{x=\pm 2\\ y=\pm 1} \\ \Rightarrow 極值發生在(x,y)=(\pm 2,\pm 1)\\ 又\cases{f_{xx}=6x\\ f_{yy}=6y\\ f_{xy}=0}，並令g(x,y)=f_{xx}f_{yy}-(f_{xy})^2 \Rightarrow \cases{f_{xx}(-2,-1) \lt 0\\ g(-2,-1)\gt 0} \\\Rightarrow f(-2,-1)為相對極大值，故選\bbox[red,2pt]{(D)}$$
解答：$$\cases{x=r\cos \theta\\ y=r\sin \theta} \Rightarrow \iint_R e^{1+x^2+y^2}\;dxdy =\int_0^{2\pi}\int_0^1 e^{1+r^2}r\;drd\theta =\int_0^{2\pi} \left.\left[{1\over 2}e^{1+r^2} \right]\right|_0^1 \;d\theta \\ =\int_0^{2\pi } {1\over 2}(e^2-e)\;d\theta = \pi(e^2-e)，故選\bbox[red,2pt]{(A)}$$
解答：$$f(x)=3^x \Rightarrow f'(x)=\ln 3\cdot 3^x \Rightarrow f''(x)=(\ln 3)^2\cdot 3^x \Rightarrow f^{[n]}(x)=(\ln 3)^n\cdot 3^x\\ \Rightarrow f(x)的泰勒級數=\sum_{n=0}^\infty {f^{[n]}(0)\over n!}x^n = \sum_{n=0}^\infty{(\ln 3)^n\over n!} x^n，故選\bbox[red,2pt]{(D)}$$
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解答：$$\int_1^2 y^2\pi\;dx =\int_1^2 (x+{1\over x})^2\pi\;dx =\int_1^2 (x^2+{1\over x^2}+2) \pi\;dx = \pi\left. \left[ {1\over 3}x^3-{1\over x}+2x \right] \right|_1^2 \\ =({8\over 3}-{1\over 2}+4-{1\over 3}+1-2)\pi= {29\over 6}\pi，故選\bbox[red,2pt]{(C)}$$
解答：$$r=e^\theta \Rightarrow {dr\over d\theta}=e^\theta \Rightarrow 弧長= \int_1^2 \sqrt{r^2+({dr\over d\theta})^2}\;d\theta = \int_1^2 \sqrt{e^{2\theta}+e^{2\theta}}\;d\theta= \int_1^2 \sqrt 2e^\theta\;d\theta\\ =\left.\left[ \sqrt 2e^\theta \right]\right|_1^2 =\sqrt 2(e^2-e)，故選\bbox[red,2pt]{(B)}$$
解答：$$f(x,y)={xy\over x+y^2+1} \Rightarrow f_x(x,y)={y\over x+y^2+1} -{xy\over (x+y^2+1)^2} \Rightarrow f_x(2,1)={1\over 4}-{2\over 16}={1\over 8}\\，故選\bbox[red,2pt]{(A)}$$
解答：$$f(x,y)=x^3+y^3-12x-3y \Rightarrow \cases{f_x=0\\ f_y=0} \Rightarrow \cases{3x^2-12=0\\ 3y^2-3=0} \Rightarrow \cases{x=\pm 2\\ y=\pm 1} \\ \Rightarrow 極值發生在(x,y)=(\pm 2,\pm 1)\\ 又\cases{f_{xx}=6x\\ f_{yy}=6y\\ f_{xy}=0}，並令g(x,y)=f_{xx}f_{yy}-(f_{xy})^2 \Rightarrow \cases{f_{xx}(-2,-1) \lt 0\\ g(-2,-1)\gt 0} \\\Rightarrow f(-2,-1)為相對極大值，故選\bbox[red,2pt]{(D)}$$
解答：$$\cases{x=r\cos \theta\\ y=r\sin \theta} \Rightarrow \iint_R e^{1+x^2+y^2}\;dxdy =\int_0^{2\pi}\int_0^1 e^{1+r^2}r\;drd\theta =\int_0^{2\pi} \left.\left[{1\over 2}e^{1+r^2} \right]\right|_0^1 \;d\theta \\ =\int_0^{2\pi } {1\over 2}(e^2-e)\;d\theta = \pi(e^2-e)，故選\bbox[red,2pt]{(A)}$$
解答：$$f(x)=3^x \Rightarrow f'(x)=\ln 3\cdot 3^x \Rightarrow f''(x)=(\ln 3)^2\cdot 3^x \Rightarrow f^{[n]}(x)=(\ln 3)^n\cdot 3^x\\ \Rightarrow f(x)的泰勒級數=\sum_{n=0}^\infty {f^{[n]}(0)\over n!}x^n = \sum_{n=0}^\infty{(\ln 3)^n\over n!} x^n，故選\bbox[red,2pt]{(D)}$$
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