112年臺綜大轉學考-微積分B詳解

臺灣綜合大學系統112學年度學士班轉學生聯合招生

科目名稱：微積分B

解答：$$\mathbf{(a)}\;\lim_{x \to 2}{x^2+5 \over x^3-2x} ={4+5\over 8-4} =\bbox[red, 2pt]{9\over 4} \\\mathbf{(b)} \;\int_{-1}^1 6x^5+3x^2\,dx = \left. \left[ x^6+x^3 \right] \right|_{-1}^1 =2-0=\bbox[red, 2pt] 2$$

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

解答：$$\lim_{x\to -\infty}{\ln(8^x+9^x)\over 3x} =\lim_{x\to -\infty}{\frac{\partial  }{\partial x}\ln(8^x+9^x)\over \frac{\partial  }{\partial x}3x} ={1\over 3}\lim_{x\to -\infty} {8^x\ln 8+9^x \ln 9\over 8^x+9^x} \\={1\over 3}\lim_{x\to -\infty} { \ln 8+ (9/8)^x \ln 9\over 1+(9/8)^x }={1\over 3}\ln 8= \bbox[red, 2pt]{ \ln 2}$$

解答：$$f(x,y)=x^2+4xy+ y^3+5 \Rightarrow \cases{f_x=2x+4y\\ f_y=4x+3y^2} \Rightarrow (f_x(2,-1),f_y(2,-1) )=(0,11)\\ \Rightarrow 切線方程式: 11(y+1)=0 \Rightarrow \bbox[red, 2pt]{y=-1} $$

解答：$$f(x)={x\over (x-1)^2} \Rightarrow f'(x)=-{x+1\over (x-1)^3} \Rightarrow \begin{cases}f(x)\ge 0 & -1\le x\lt 1\\ f(x)\le 0 & -\infty\lt x\le -1,x\gt 1 \end{cases} \\ \Rightarrow \bbox[red, 2pt]{\cases{遞增區間: [-1,1)\\ 遞減區間: (-\infty,-1] \cup (1,\infty)}}$$

解答：$$\int_0^1 {1\over 2x+10\sqrt x+12}\,dx = \int_0^1 {1\over 2(\sqrt x+3)(\sqrt x+2)}\,dx ={1\over 2}\int_0^1 {1\over \sqrt x +2} -{1\over \sqrt x +3} \,dx \\=\int_0^1 {u\over u+2}-{u\over u+3}\,du (其中u=\sqrt x ) = \int_0^1 1-{2\over u+2}-1+{3\over u+3}\,du \\ =\int_0^1 {3\over u+3}- {2\over u+2}\,du = \left. \left[ 3\ln(u+3)-2\ln(u+2)\right] \right|_0^1 =\bbox[red,2pt]{8\ln 2-5\ln 3}$$

解答：

$$兩積分區域可合併，並將積分順序對調 \Rightarrow \int_0^{\pi/2} \int_y^{2y} {\sin x\over x}\, dx +\int_{\pi/2}^\pi \int_y^\pi {\sin x\over x}\,dx \\= \int_0^\pi \int_{x/2}^{x} {\sin x\over x}\,dy dx =\int_0^\pi {\sin x\over x}\cdot {x\over 2}\, dx ={1\over 2}\int_0^\pi \sin x\,dx ={1\over 2} \left. \left[ -\cos x\right] \right|_0^\pi =\bbox[red, 2pt] 1$$

解答：$$令\cases{f(x,y,z)=x-y+z\\ g(x,y,z)=2x^2+4y^2+4z^2-9} \Rightarrow \cases{f_x= \lambda g_x\\ f_y=\lambda g_y\\ f_z=\lambda g_z\\ g=0} \Rightarrow \cases{1=\lambda (4x)\\ -1=\lambda (8y)\\ 1=\lambda(8z)} \\ \Rightarrow \cases{x=-2y\\ z=-y} 代入g=0 \Rightarrow 8y^2+4y^2+4y^2-9=0 \Rightarrow y^2={9\over 16} \\\Rightarrow \cases{y=3/4 \Rightarrow \cases{x=-3/2 \\ z=-3/4} \\ y=-3/4 \Rightarrow \cases{x=3/2\\ z=3/4}} \Rightarrow \cases{f(-3/2,3/4,-3/4)=-3\\ f(3/2,-3/4,3/4)=3} \Rightarrow 最小值= \bbox[red,2pt]{-3}$$

解答：$$f(x,y,z)=x^2ze^y+xz^2 \Rightarrow \nabla f(x,y,z)=(2xze^y+z^2,x^2ze^y, x^2e^y+2xz) \\ \Rightarrow \nabla f(1,\ln 2,2)=(12, 4,6) \Rightarrow 令\vec u={\nabla f(1,\ln 2,2)\over \Vert \nabla(1,\ln 2,2)\Vert}=({6\over 7},{2\over 7}, {3\over 7})\\ \Rightarrow 在方向\bbox[red, 2pt]{({6\over 7},{2\over 7}, {3\over 7})}最化最大 \Rightarrow 最大變化率=(12,4,6)\cdot ({6\over 7},{2\over 7}, {3\over 7}) =\bbox[red, 2pt]{14}$$

解答：$$|x|\lt {3\over 2} \Rightarrow \left|{2x\over 3} \right|\lt 1\\ f(x)={2x^5\over 16x^4+81} ={2\over 81}x^5 \cdot {1\over 1-(-({2\over 3}x)^4)} \\= {2\over 81}x^5\left(1- ({2\over 3}x)^4 +({2\over 3}x)^8-({2\over 3}x)^{12} +({2\over 3}x)^{16}-({2\over 3}x)^{20}+ \cdots \right) \\ \Rightarrow x^{25}的係數=-{2\over 81}\cdot ({2\over 3})^{20} =\bbox[red, 2pt]{-{2^{21}\over 3^{24}}}$$

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解答僅供參考，其他歷年試題及詳解