111海洋大學碩士班-微積分詳解

國立臺灣海洋大學111學年度碩士班考試入學招生考試試題

考試科目：微積分
學系組名稱：運輸科學系碩士班不分組

解答： $$\mathbf{(1)}\;\lim_{x\to 1}{x^2-\sqrt x\over \sqrt x-1} =\lim_{x\to 1}{(x^2-\sqrt x)' \over( \sqrt x-1)'} =\lim_{x\to 1}{2x- 1/2\sqrt x\over 1/2\sqrt x} =\lim_{x\to 1} (4x\sqrt x-1) =\bbox[red, 2pt]3 \\\mathbf{(2)}\;L=\lim_{x\to 0} (e^x+x)^{1/x} =\lim_{x\to 0} e^{{1\over x}\ln(e^x+x)} \Rightarrow \ln L =\lim_{x\to 0}{\ln(e^x+x) \over x} =\lim_{x\to 0}{ (e^x+1)/(e^x+x) \over 1} \\ \qquad =\lim_{x\to 0} {e^x+1\over e^x+x} =2 \Rightarrow L=\bbox[red,2pt]{e^2}$$

解答：

$$假設籬笆長x，寬y，即xy=216 \Rightarrow y=216/x \Rightarrow 籬笆總長L(x)＝3x+2y =3x+{432 \over x}\\ \Rightarrow L'(x)=3-{432\over x^2} \Rightarrow L''(x)={864\over x^3};\\ L'(x)=0 \Rightarrow x^2=144 \Rightarrow x=12 \Rightarrow L''(x)\gt 0 \Rightarrow L(12)=72為極小值，此時y=216/12=18\\ \Rightarrow \bbox[red,2pt]{\cases{外圈籬笆的長寬為12\times 18\\ 所需圍籬總長度=72}}$$

解答： $$x^3+z^3+ye^{xz}+ z\cos y=0 \Rightarrow \cases{3x^2 + 3z^2z_x + y(z+xz_x)e^{xz} + z_x\cos y=0\\ 3z^2z_y+ e^{xz} + xyz_ye^{xz} + z_y\cos y-z\sin y =0} \\ 將(0,0,0)代入上式\Rightarrow \cases{0+0+0 +z_x=0\\ 0+1+ 0+z_y-0=0} \Rightarrow \bbox[red,2pt]{ \cases{{\partial z\over \partial x}=0 \\{\partial z\over \partial y}=-1}}$$

解答： $$\int_0^1 \int_0^1 {y\over 1+xy}\,dx dy = \int_0^1 \left.\left[ \ln(1+xy)\right] \right|_0^1\,dy =\int_0^1 \ln(1+y)\,dy =\left. \left[ (1+y)\ln(1+y)-y\right]\right|_0^1 \\= \bbox[red,2pt]{2\ln 2-1}$$

解答： $$f(x)={1\over \sqrt{|x|}} 為偶函數\Rightarrow \int_{-1}^4 f(x)\,dx = 2\int_0^1 {1\over \sqrt{x}}\,dx + \int_1^4  {1\over \sqrt{x}}\, dx = 2\left. \left[ 2\sqrt x\right]\right|_0^1 +\left. \left[ 2\sqrt x\right]\right|_1^4 \\ = 4+(4-2)= \bbox[red,2pt]6$$

解答： $$xy= \cot(xy) \Rightarrow y+xy'=-(y+xy')\csc^2(xy) \Rightarrow (x+x\csc^2(xy))y'= -y\csc^2(xy)-y\\ \Rightarrow \\ \Rightarrow y'=-{y(1+\csc^2(xy))\over x(1+\csc^2(xy))} \Rightarrow \bbox[red,2pt]{{dy\over dx} =-{y\over x}}$$

解答： $$圖形投影到平面y=k上，皆為雙曲線；離原點最近的雙曲線即是投影在y=0上\\，此時離原點最近的點為(\pm 1,0,0)，距離為\bbox[red,2pt] 1$$

解答： $$\mathbf{(a)}\;y=f(x)={x^3+8\over x^2-4} ={x^2-2x+4\over x-2} =x+{4\over x-2} \Rightarrow \bbox[red,2pt]{\cases{\text{domain of }f=\{x\mid x\ne 2,x\in \mathbb R\} \\ 漸近線：\cases{y=x\\ x=2}}} \\\mathbf{(b)}\; f(x)= x+{4\over x-2} \Rightarrow f'(x)=1-{4\over (x-2)^2} ,因此\cases{ x=4,0\Rightarrow f'(x)=0\\ f'(2) 未定義} \\ \qquad \Rightarrow \bbox[red,2pt]{\text{critical points at }x=0,2,4}\\ \qquad又f''(x)={8\over (x-2)^3} 且\cases{\lim_{x\to 2^+} f''(x)\gt 0\\ \lim_{x\to 2^-} f''(x)\lt 0} \Rightarrow \bbox[red,2pt]{\text{inflection point at }x=2} \\\mathbf{(c)}\; \cases{f'(x)\ge 0 \Rightarrow (x-2)^2 \ge 4 \Rightarrow x\ge 4或x\le 0 \\ f'(x)\le 0 \Rightarrow (x-2)^2\le 4 \Rightarrow 0\le x\le 4} \Rightarrow \bbox[red, 2pt]{\cases{當x\ge 4或x\le 0時，f為遞增 \\ 當x\in [0,2)或 x\in (2,4]，f為遞減}}\\\mathbf{(d)}\; \cases{f''(0)= -1 \lt 0\\ f''(4)= 1 \gt 0} \Rightarrow f(0)=\bbox[red,2pt]{-2為極大值} \\\mathbf{(e)}\;f''(x)={8\over (x-2)^3} \Rightarrow \cases{x\gt 2 \Rightarrow f''(x)\gt 0\\ x\lt 2\Rightarrow \bbox[red,2pt]{f''(x)\lt 0} \Rightarrow \cases{f\text{ is concave up, if }x\gt 2\\ f\text{ is concave down, if }x\lt 2}} \\\mathbf{(f)}\;圖形如下：$$