114年台綜大轉學考-微積分B詳解

 臺灣綜合大學系統114學年度學士班轉學生聯合招生考試

科目名稱:微積分B

解答:$$g(x)=e^{x^{-2}} \Rightarrow g'(x)=-2x^{-3}e^{x^{-2}} \Rightarrow g''(x)=6x^{-4}e^{x^{-2}}+ 4x^{-6}e^{x^{-2}} \\ \Rightarrow \bbox[red, 2pt]{\cases{g'(1)=-2e\\ g''(1)=10e}}$$

解答:$$f(x)={x^2-3\over x^3} ={1\over x}-{3\over x^3}\Rightarrow f'(x)=-{1\over x^2}+{9\over x^4} ={9-x^2\over x^4} \\f'(x)=0 \Rightarrow x=\pm 3 \Rightarrow \cases{f(3)= 2/9\\ f(1) =-2\\ f(5)=22/125} \Rightarrow \bbox[red, 2pt]{\cases{\text{maximum}: 2/9\\ \text{minimum}:-2}}$$

解答:$$\cases{u=x\\ dv=e^{2x}\,dx} \Rightarrow \cases{du=dx\\ v={1\over 2}e^{2x}} \Rightarrow \int xe^{2x}\,dx ={1\over 2}xe^{2x}-{1\over 2}\int e^{2x}\,dx ={1\over 2}xe^{2x}-{1\over 4}  e^{2x} +C \\ \Rightarrow \int_0^1 xe^{2x}\,dx = \left. \left[ {1\over 2}xe^{2x}-{1\over 4}  e^{2x}  \right] \right|_0^1 = \bbox[red, 2pt]{{1\over 4}+{1\over 4}e^2}$$

解答:$$x=\sqrt 3 \tan \theta \Rightarrow dx =\sqrt 3\sec^2 \theta\,d\theta \Rightarrow \int {1\over \sqrt{x^2+3}} \,dx = \int{\sqrt 3\sec^2 \theta \over \sqrt{3\tan^2\theta+3}}\,d\theta =\int {\sqrt 3\sec^2 \theta \over \sqrt 3 \sec \theta}\,d\theta \\=\int \sec \theta \,d\theta =\ln|\tan \theta+\sec \theta|+C = \ln\left|{x\over \sqrt 3}+{{\sqrt{x^2+3}\over \sqrt 3}} \right|+C \\ \Rightarrow \int_0^1 {1\over \sqrt{x^2+3}}\,dx = \left. \left[ \ln\left|{x\over \sqrt 3}+{{\sqrt{x^2+3}\over \sqrt 3}} \right| \right] \right|_0^1 =\ln{3\over \sqrt 3} -\ln {\sqrt 3\over \sqrt 3} = \bbox[red, 2pt]{\ln \sqrt 3}$$

解答:$$f(x)=2x^{-1} \Rightarrow f'(x)=-2x^{-2} \Rightarrow f''(x)=4x^{-3} \Rightarrow \cdots \Rightarrow f^{[n]}(x) =2\cdot n!\cdot (-1)^n x^{-(n+1)} \\ \Rightarrow f^{[100]}(x) =2\cdot 100! x^{-101} \Rightarrow a_{100} ={f^{[100]}(2) \over 100!} =2\cdot 2^{-101} = \bbox[red, 2pt]{2^{-100}}$$

解答:$$y^3-x^2=4 \Rightarrow 3y^2y'-2x=0 \Rightarrow 6y(y')^2+3y^2y''-2=0\\y'={2x\over 3y^2} \Rightarrow y'(2,2) ={4\over 3\cdot 4} ={1\over 3} \Rightarrow y''={2-6y(y')^2\over 3y^2} \Rightarrow y''(2,2)={2-6\cdot 2\cdot (1/9) \over 3\cdot 4}={1\over 18}\\ \Rightarrow \bbox[red, 2pt]{\cases{dy/dx=1/3\\ d^2y/dx^2=1/18}}\text{ at }(x,y)=(2,2)$$

解答:$$\cases{x(t)=3\cos \sqrt{t+1} \\y(t) =3\sin\sqrt{t+1} \\z(t) =4\sqrt{t+1}} \Rightarrow \cases{x'(t)=- {3\over 2\sqrt{t+1}}\sin \sqrt{t+1} \\ y'(t) ={3\over 2\sqrt{t+1}} \cos\sqrt{t+1} \\z'(t)={2\over \sqrt{t+1}}} \\ \Rightarrow L= \int_0^1 \sqrt{x'(t)^2 +y'(t)^2 +z'(t)^2}\,dt =\int_0^1 \sqrt{{9\over 4(t+1)}+{4\over t+1}}\,dt =\int_0^1 \sqrt{{25\over 4(t+1)} }\,dt \\={5\over 2} \int_0^1 \sqrt{1\over t+1}\,dt ={5\over 2} \left. \left[2\sqrt{t+1} \right] \right|_0^1 = \bbox[red, 2pt]{5(\sqrt 2-1)}$$

解答:$$f(x,y) =x+\sin(x+2y) \Rightarrow \nabla f(x,y)=(f_x,f_y) =(1+\cos(x+2y), 2\cos(x+2y)) \\ \Rightarrow \nabla f(0,0)=(2,2) \Rightarrow \text{unit vector }\vec e= \bbox[red, 2pt]{\left({\sqrt 2\over 2},{\sqrt 2\over 2} \right)} \\ \Rightarrow \text{the rate of change of }f \text{ in }\vec e =|\nabla f(0,0)|=|(2,2)|= \bbox[red, 2pt]{2\sqrt 2}$$

解答:$$\cases{f(x,y)=x^2\\ g(x,y)=x+y-1} \Rightarrow \cases{f_x= \lambda g_x\\ f_y= \lambda g_y \\ g=0} \Rightarrow \cases{2x=\lambda\\ 0=\lambda\\ x+y=1} \Rightarrow x=0 \Rightarrow y=1 \Rightarrow f(0,1)=0  \\ \Rightarrow  \bbox[red, 2pt]{\text{the smallest value of }f \text{ is }f(0,1)=0}$$

解答:$$\iint_R (x+y)d A = \int_0^1 \int_{x^2}^x (x+y)\,dydx =\int_0^1 \left({3\over 2}x^2 -x^3-{1\over 2}x^4 \right)\,dx =\bbox[red, 2pt]{3\over 20}$$

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解題僅供參考，轉學考歷年試題及詳解