112年高雄大學轉學考-微積分詳解

國立高雄大學 112 學年度轉學招生考試試題(轉二年級)

科目：微積分
考試時間：80 分鐘
系所：資訊工程學系(無組別)

一﹑選擇題 Choose the best answer for each question (每題 3 分，共 30 分)

解答： $$\lim_{x\to 0}{\tan x\over 1-e^{-x}} =\lim_{x\to 0}{{d\over dx}\tan x\over {d\over dx}(1-e^{-x})} =\lim_{x\to 0}{\sec ^2 x\over e^{-x}} =1，故選 \bbox[red, 2pt]{(b)}$$

解答： $$\lim_{x\to 0}{23x \over \sqrt{23x+9}-3} =\lim_{x\to 0}{{d\over dx}23x \over {d\over dx}(\sqrt{23x+9}-3)} = \lim_{x\to 0}{23\over 23/2\sqrt{23x+9}} =\lim_{x\to 0} 2\sqrt{23x+9} =6\\，故選 \bbox[red, 2pt]{(b)}$$

解答： $$f(x)=e^x+23x \Rightarrow f'(x)=e^x+23 \Rightarrow f'(0)=1+23=24\\ \Rightarrow 圖形在(0, 23)的切線斜率=24，故選 \bbox[red, 2pt]{(c)}$$

解答： $$f(x)=x\sin(3x) \Rightarrow f'(x)=\sin (3x)+3x\cos(3x) \Rightarrow f''(x)=3\cos(3x)+ 3\cos (3x)- 9x\sin(3x) \\=6\cos(3x)-9x\sin(3x) \Rightarrow f''(0)=6，故選 \bbox[red, 2pt]{(d)}$$

解答： $${d\over dx}23\sqrt{2x} ={23\over \sqrt{2x}} \Rightarrow {d\over dx}e^{23\sqrt{2x}} ={23\over \sqrt{2x}} e^{23\sqrt{2x}}，故選 \bbox[red, 2pt]{(c)}$$

解答： $${d\over dx}\arctan x={1\over 1+x^2} \Rightarrow {d\over dx} \arctan(x^3)={3x^2 \over 1+x^6}，故選 \bbox[red, 2pt]{(d)}$$

解答： $$\lim_{x\to -\infty}{23x+1\over \sqrt{x^2-x}}= \lim_{x\to -\infty}{23x+1\over |x|\sqrt{1-1/x}}=-23，故選 \bbox[red, 2pt]{(d)}$$

解答： $${d\over dx}F(x)= \sin\sqrt{x^4}\cdot {d\over dx}x^4 =4x^3 \sin x^2，故選 \bbox[red, 2pt]{(c)}$$

解答： $$\int {1\over x^2+8x+41}\,dx = \int {1\over (x+4)^2+25}\,dx = {1\over 25} \int {1\over ({x+4\over 5})^2+1}\,dx ={1\over 25}\cdot 5\arctan({x+4\over 5})+C\\ ={1\over 5}\arctan({x+4\over 5})+C，故選 \bbox[red, 2pt]{(c)}$$

解答： $$4\lt \sqrt{23}\lt 5 \Rightarrow \int_4^5 {2x\over (x^2-23)^2}\,dx =\infty，故選 \bbox[red, 2pt]{(a)}$$

二﹑填充題 Please fill in the following blanks (每題 5 分，共 25 分)

解答： $$f(x)=26-{25\over x} \Rightarrow f'(x)={25\over x^2}\Rightarrow \cases{f'(c)={25\over c^2}\\ {f(25)-f(1)\over 25-1}=1} \Rightarrow {25\over c^2}=1 \Rightarrow c=\bbox[red, 2pt]5\;(-5\not \in (1,25))$$

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

解答： $$\cos^2 x={1\over 2}(\cos 2x+1) \Rightarrow \cos^6 x={1\over 8}(\cos 2x+1)^3 ={1\over 8}(\cos^3(2x)+ 3\cos^2(2x)+ 3\cos(2x)+1) \\={1\over 8}(\cos(2x)(1-\sin^2(2x))+{3\over 2}(\cos(4x)+1) +3\cos(2x)+1)  \\={1\over 8}(4\cos(2x)-\cos(2x)\sin^2(2x) +{3\over 2}\cos(4x)+{5\over 2}) \\ \Rightarrow \int_0^\pi \cos^6 x\,dx = \int_0^\pi {1\over 2}\cos(2x) -{1\over 8}\cos(2x) \sin^2(2x) +{3\over 16}\cos(4x)+ {5\over 16} \,dx\\ =\left. \left[ {1\over 4}\sin(2x)-{1\over 48}\sin^3(2x)+{3\over 64}\sin(4x)+{5\over 16}x \right] \right|_0^\pi =\bbox[red, 2pt]{{5\over 16}\pi}$$

解答： $$u=\tan x \Rightarrow du=\sec^2x \,dx \Rightarrow \int_0^{\pi/6} 12(\sec x)^3\,dx =12\int_0^{1/\sqrt 3} \sqrt{1+u^2}\,du \\=12 \left. \left[{1\over 2}u\sqrt{1+u^2}+{1 \over 2}\ln(u+ \sqrt{1+u^2}) \right] \right|_0^{1/\sqrt 3} =12({1\over 3}+{1\over 4}\ln 3) =\bbox[red, 2pt]{4+3\ln 3}$$

解答： $$\cases{x=f(t)=\int_0^{2t} \sqrt{\sin z}\,dz \\ y=g(t)= \int_0^{2t} \sqrt{4+3\sin z}\,dz} \Rightarrow \cases{x'(t)=2\sqrt{\sin (2t)}\\ y'(t)= 2\sqrt{4+3\sin(2t)}} \\ \Rightarrow 曲線長 = \int_0^\pi \sqrt{x'(t)^2+y'(t)^2}\,dt =\int_0^\pi \sqrt{4 \sin(2t)+ 16+12\sin(2t)} \,dt =\int_0^\pi 4\sqrt{\sin(2t)+1}\,dt \\=\int_0^\pi 4\sqrt{\cos(\pi/2-2t)+1}\,dt=\int_0^\pi 4\sqrt{2\cos^2(\pi/4-t)}\,dt =4\sqrt 2 \int_0^{\pi} \cos({\pi\over 4}-t)\,dt \\ =4\sqrt 2\left. \left[ -\sin({\pi\over 4}-t) \right] \right|_0^\pi =4\sqrt 2\cdot \sqrt 2=\bbox[red, 2pt] 8$$

三﹑計算題 Problems (每題 15 分，共 45 分)

解答： $$f(x)=g(x) \Rightarrow x=0,4且f(x)\ge g(x), \forall x\in [0,4] \Rightarrow 繞x軸旋轉體積= \int_0^4 \pi(f^2(x)-g^2(x))\,dx \\= \pi\int_0^4 12\sqrt x-5x-{1\over 4}x^2 \,dx= \pi \left. \left[ 8x^{3/2}-{5\over 2}x^2-{1\over 12}x^3 \right] \right|_0^4 =\bbox[red, 2pt]{56\pi \over 3}$$

解答： $$2-x^2-y^2=2-y \Rightarrow x^2+(y-{1\over 2})^2=({1\over 2})^2 \Rightarrow 取\cases{x=r\cos \theta\\ y=r\sin \theta+1/2} \\ \Rightarrow 所圍體積=\int_0^{2\pi} \int_0^{1/2} ({1\over 4}-r^2)r \,drd\theta= \bbox[red, 2pt]{\pi \over 32}$$

解答： $$\cases{F_1=6xy\\ F_2=3x+3y} \Rightarrow {d\over dx}F_2-{d\over dy}F_1=3-6x \Rightarrow \int_C 6xy\,dx+(3x+3y)dy = \int_{-1}^1 \int_0^{1-x^2} (3-6x)\,dydx \\=\int_{-1}^1 (1-x^2)(3-6x)\,dx =\int_{-1}^1 3-6x-3x^2+6x^3\,dx = \left. \left[ 3x-3x^2-x^3 +{3\over 2}x^4 \right] \right|_{-1}^1 =\bbox[red, 2pt]4$$

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解題僅供參考，其他歷年試題及詳解