112年專科學力鑑定-微積分詳解

教育部 112 年自學進修專科學校學力鑑定考試

專業科目(一)：微積分

解答: $$(A)\times: f(2)=0但f'(x)\ne 0,該點切線非水平線\\ (B)\times: \lim_{x\to 3^+}f(x)=3\ne -3\\ (C)\bigcirc: \lim_{x\to 1^+} f(x) =\lim_{x\to 1^-}f(x) = 2 \Rightarrow \lim_{x\to 1}f(x)=2\\ (D)\times: f(5)未定義\\故選\bbox[red, 2pt]{(C)}$$

解答: $$\int_{-2}^4 f(x)\,dx =\int_{-2}^0 f(x)\,dx +\int_{0}^4 f(x)\,dx  \Rightarrow 5= \int_{-2}^0 f(x)\,dx +8 \Rightarrow \int_{-2}^0 f(x)\,dx=-3\\ 因此\int_{-2}^0(3f(x)+x)\,dx = 3\int_{-2}^0 f(x) \,dx+\int_{-2}^0 x\,dx =3\cdot (-3)+ \left.\left[{1\over 2}x^2 \right] \right|_{-2}^0 \\=-9-2=-11，故選\bbox[red, 2pt]{(B)}$$

解答: $$f(x,y)=x^3+xy^2+6x^2+y^2 \Rightarrow \cases{f_x=3x^2+y^2+12x\\ f_y=2xy+2y} \Rightarrow \cases{f_{xx}=6x+12\\ f_{xy}=2y\\ f_{yy} =2x+2} \\ \Rightarrow d(x,y)=f_{xx}f_{yy}-f_{xy}^2 = (6x+12)(2x+2)-4y^2 =12(x+2)(x+1)-4y^2\\\Rightarrow \cases{f_{xx}(0,0)=0 \not \lt 0\\ f_{xx}(-1,-3)=6\gt 0\\ f_{xx}(-1,3)=6\gt 0\\ f_{xx}(-4,0)=-12\lt 0}且d(-4,0)=72\gt 0 \Rightarrow f(-4,0)為相對極大值，故選\bbox[red, 2pt]{(B)}$$

解答: $$令u={x\over 4},則\int_0^4{1\over 16+x^2}\,dx = \int_0^1{4\over 16+16u^2}\,du = {1\over 4}\int_0^1 {1\over 1+u^2}\,du \\ ={1\over 4}\left. \left[ \tan^{-1}u \right]\right|_0^1 ={1\over 4}\cdot {\pi \over 4}={\pi \over 16}，故選\bbox[red, 2pt]{(D)}$$

解答: $$\int_0^4 (5x-x^2)-x\,dx = \int_0^44x-x^2\,dx = \left. \left[ 2x^2-{1\over 3}x^3 \right] \right|_0^4 =32-{64\over 3}={32\over 3}，故選\bbox[red, 2pt]{(B)}$$

解答: $$f(x)=(x^2-1)^{2/3} \Rightarrow f'(x)={2\over 3}(x^2-1)^{-1/3}\cdot 2x ={4\over 3}\cdot {x\over \sqrt[3]{x^2-1}} \\ 若f'(x)\gt 0 \Rightarrow x\gt 1 或-1\lt x\lt 0，故選\bbox[red, 2pt]{(D)}$$

解答: $$f(x)=\sin(1-2x) \Rightarrow \cases{f'(x)=-2\cos(1-2x)\\ f''(x)=-2^2\sin(1-2x)\\ f'''(x)=2^3\cos(1-2x)\\ f^{(4)}(x)=2^4\sin(1-2x)} \Rightarrow f^{(2023)}(x)=2^{2023}\cos(1-2x) \\ \Rightarrow f^{(2023)}({1\over 2}) =2^{2023} \cos 0 =2^{2023}，故選\bbox[red, 2pt]{(C)}$$

解答: $$y=x^{\ln x} =e^{\ln x^{\ln x}} =e^{(\ln x)^2} \Rightarrow y'=2\ln x\cdot {1\over x}e^{(\ln x)^2} ={2\ln x\over x}\cdot x^{\ln x}，故選\bbox[red, 2pt]{(A)}$$

解答: $$f(3x)=x^2g(2x^3) \Rightarrow 3f'(3x)=2xg(2x^3)+x^2g'(2x^3)\cdot 6x = 2xg(2x^3)+6x^3g'(2x^3)\\ \Rightarrow 3f'(3)=2g(2)+6g'(2) =2\cdot 2+ 6\cdot (-1)=-2 \Rightarrow f'(3)=-{2\over 3}，故選\bbox[red, 2pt]{(B)}$$

解答: $$f(x)= xe^{\sin x}+1 \Rightarrow f'(x)=e^{\sin x}+ x\cos xe^{\sin x} \Rightarrow f'(\pi) =e^0+\pi \cdot(-1)\cdot e^0=1-\pi\\，故選\bbox[red, 2pt]{(B)}$$

解答: $$\cases{e^x= 1+x+x^2/2!+ x^3/3!+\cdots +x^n/n!+\cdots\\ e^{-x} =1-x+x^2/2!-x^3/3!+ \cdots+ (-1)^nx^n/n!+ \cdots} \\\Rightarrow e^x-e^{-x} =2(x+x^3/3!+x^5/5!+\cdots) \Rightarrow {e^x-e^{-x} \over 2}=x+x^3/3!+x^5/5!+\cdots \\ =\sum_{n=0}^\infty{x^{2n+1}\over (2n+1)!}，故選\bbox[red, 2pt]{(A)}$$

解答: $$\lim_{x\to 0^+}{\ln(e^x-1)\over \ln x} =\lim_{x\to 0^+}{\frac{d }{dx}\ln(e^x-1)\over \frac{d }{dx}\ln x} =\lim_{x\to 0^+}{e^x/(e^x-1)\over 1/x} =\lim_{x\to 0^+} {xe^x\over e^x-1} \\=\lim_{x\to 0^+} {\frac{d }{dx}(xe^x)\over \frac{d }{dx}(e^x-1)}  =\lim_{x\to 0^+}{e^x+xe^x\over e^x} ={1+0\over 1}=1，故選\bbox[red, 2pt]{(A)}$$

解答: $$\lim_{n\to \infty}{e^{1/n}+e^{2/n}+ \cdots+ e^{n/n} \over n} =\lim_{n\to \infty} \sum_{k=1}^n\left({1\over n}e^{k/n} \right) =\int_0^1e^x\,dx =e-1，故選\bbox[red, 2pt]{(B)}$$

解答: $${1\over 2}\int_0^{\pi/2} r^2\,d\theta ={1\over 2}\int_0^{\pi/2} 1+2\cos \theta +\cos^2\theta\,d\theta ={1\over 2}\int_0^{\pi/2} 1+2\cos \theta +{1\over 2}(\cos 2\theta+1)\,d\theta \\= {1\over 2}\int_0^{\pi/2}{3\over 2}+2\cos \theta +{1\over 2}\cos 2\theta \,d\theta ={1\over 2}\left. \left[ {3\over 2}\theta+2\sin \theta +{1\over 4}\sin 2\theta \right] \right|_0^{\pi/2} \\={1\over 2}({3\pi\over 4}+2) =1+{3\pi\over 8}，故選\bbox[red, 2pt]{(D)}$$

解答: $$u=5x \Rightarrow du=5dx \Rightarrow \int \ln(5x)\,dx =\int {1\over 5}\ln u\,du ={1\over 5}(u\ln u-u)+C \\={1\over 5}(5x\ln(5x)-5x)+C =x\ln(5x)-x+C，故選\bbox[red, 2pt]{(C)}$$

解答: $$x^3+2x-1=x(x^2-4)+6x-1 \Rightarrow {x^3+2x-1\over x^2-4} =x+{6x-1\over x^2-4} =x+{11\over 4(x-2)}+{13\over 4(x+2)}\\ \Rightarrow \int {x^3+2x-1\over x^2-4}\,dx = \int x\,dx +{11\over 4}\int {1\over x-2}\,dx +{13\over 4}\int {1\over x+2}\,dx \\={1\over 2}x^2+{11\over 4}\ln |x-2| +{13\over 4} \ln|x+2|+C，故選\bbox[red, 2pt]{(D)}$$

解答: $$\pi\int_1^2 ({2\over x})^2-1\,dx =\pi \left.\left[ {-4\over x}-x\right] \right|_1^2 =\pi，故選\bbox[red, 2pt]{(C)}$$

解答: $$y={2\over 3}x^{3/2} \Rightarrow y'=\sqrt x \Rightarrow 曲線長=\int_1^9 \sqrt{1+y'^2}\,dx =\int_1^9 \sqrt{1+x}\,dx =\left. \left[{2\over 3}(1+x)^{3/2} \right]\right|_1^9 \\={2\over 3}(10^{3/2}-2^{3/2}) ={2\over 3}(10\sqrt {10}-2\sqrt 2)，故選\bbox[red, 2pt]{(D)}$$

解答: $$F(x,y)={e^x\over \ln y+x^3} \Rightarrow \cases{F_x={e^x\over \ln y+x^3}-{3x^2e^x\over (\ln y+x^3)^2} ={e^x(\ln y+x^3-3x^2e^x)\over (\ln y+x^3)^2}\\ F_y=-{e^x\over y(\ln y+x^3)^2}}\\選項(C)與(Ｄ)皆差一個負號，故\bbox[red, 2pt]{(無解)},但公佈的答案是\bbox[cyan,2pt]{(C)}$$

解答: $$\int_0^9 \int_{\sqrt x}^3 {1\over y^3+1}\,dydx =\int_0^3\int_0^{y^2}{1\over y^3+1}dxdy =\int_0^3{y^2\over y^3+1}\,dy =\left. \left[{1\over 3}\ln(y^3+1) \right] \right|_0^3 \\={1\over 3}\ln 28，故選\bbox[red, 2pt]{(A)}$$

================= END =======================

解題僅供參考,其他歷年試題及詳解