國立嘉義大學 108 學年度轉學生招生考試試題
科目：微積分 ＜請將答案寫在答案卷上＞
一、填充題：（每題 6 分，共 60 分）
解答：$$\cases{-1/x \le \cos x/x \le 1/x\\ -1/x \le \sin x/x \le 1/x} \Rightarrow \cases{\lim_{x\to \infty} \cos x/x=0 \\ \lim_{x\to \infty} \sin x/x=0} \\ 因此\lim_{x\to \infty} {x+\cos x\over x-\sin x}= \lim_{x\to \infty} {1+\cos x/x\over 1-\sin x/x} =\bbox[red, 2pt]1$$
解答：$$取x=my^2,m為常數 \Rightarrow \lim_{(x,y)\to (0,0)}{xy^2\over x^2+y^4} =\lim_{x\to 0}{my^4\over (m^2+1) y^4} ={m\over m^2+1}非定值\Rightarrow 極限\bbox[red,2pt]{不存在}$$
解答：$$\int {1\over x^2-5x+6}\,dx =\int {1\over (x-3)(x-2)}\,dx =\int {1\over x-3}-{1\over x-2}\,dx = \bbox[red, 2pt]{\ln(x-3)-\ln(x-2)+C}$$
解答：$$令\cases{u=\ln(x+1)\\ dv =xdx} \Rightarrow \cases{du =dx/(x+1) \\ v=x^2/2} \Rightarrow \int x\ln(x+1)\,dx = {1\over 2}x^2 \ln(x+1)-{1\over 2}\int {x^2\over x+1}\,dx \\ ={1\over 2}x^2 \ln(x+1)-{1\over 2}\int x-1+{1\over x+1}\,dx={1\over 2}x^2 \ln(x+1)-{1\over 2}\left( {1\over 2}x^2-x+ \ln(x+1) \right)+C\\ =\bbox[red, 2pt]{{1\over 2}(x^2-1) \ln(x+1)-{1\over 4}x^2+{1\over 2}x+C}$$
解答：$$x^2-xy^2+y^3=1 \Rightarrow 2x-y^2-2xyy'+3y^2y'=0 \Rightarrow {dy\over dx }=y'= \bbox[red, 2pt]{2x-y^2 \over 2xy-3y^2}$$
解答：$$y=\sin^{-1}(x^2+1) \Rightarrow {dy\over dx} ={1\over \sqrt{1-(x^2+1)^2}} \cdot (2x) =\bbox[red, 2pt]{2x\over \sqrt{-x^2-2x}}$$
解答：$${d\over dx} \int_1^{x^2} {t\over 1+t^3}\,dt = {x^2 \over 1+x^6}\cdot (2x) = \bbox[red, 2pt]{2x^3\over 1+x^6}$$
解答：$$\lim_{x\to 0}{e^x-1-x\over x^2} =\lim_{x\to 0}{e^x-1 \over 2x } =\lim_{x\to 0}{e^x \over 2 } = \bbox[red,2pt]{1\over 2}$$
解答：$$x^2+y^2=25 \Rightarrow 2x+2yy'=0 \Rightarrow y'=-{x \over y} \Rightarrow 切線斜率=\left. y'\right|_{(3,4)} =-{3\over 4} \\ \Rightarrow 過(3,4)斜率為－{3\over 4} 的切線：y=-{3\over 4}(x-3)+4 \Rightarrow \bbox[red,2pt]{3x+4y= 25}$$
解答：$$\int {x-9\over x^2+3x-10}\,dx =\int {x-9\over (x+5)(x-2)}\,dx = \int {2\over x+5}-{1\over x-2}\,dx \\= \bbox[red, 2pt]{2\ln (x+5)-\ln(x-2)+C}$$
二、計算題：（每題 10 分，共 40 分；請寫計算過程）
解答：$$e^x的泰勒級數＝\sum_{n=0}^\infty {1\over n!}x^n \Rightarrow x^2e^x的泰勒級數＝\bbox[red, 2pt]{\sum_{n=0}^\infty {1\over n!}x^{n+2} }$$
解答：$$\int_0^8 \int_{\sqrt[3] y}^2 e^{x^4}\,dxdy =\int_0^2 \int_0^{x^3} e^{x^4}\,dydx =\int_0^2 x^3e^{x^4}\,dx =\left. \left[ {1\over 4}e^{x^4} \right]\right|_0^2 = \bbox[red,2pt]{{1\over 4}(e^{16}-1)}$$
解答：$$L=\lim_{n\to \infty}\left| {a_{n+1}\over a_n}\right| =\lim_{n\to \infty}\left| {(-1)^{n+1}x^{n+1}\over n+2} \cdot {n+1\over (-1)^n x^n}\right| =\lim_{n\to \infty} {n+1\over n+2}|x| =|x| \lt 1\\ 依比值審斂法,當\bbox[red, 2pt]{|x|\lt 1}時，冪級數收斂$$
解答：$$\int_0^1 \int_x^1 \sin(y^2)\,dy dx=\int_0^1 \int_0^y \sin(y^2)\,dx dy= \int_0^1 y\sin(y^2)\,dy =\left.\left[ -{1\over 2}\cos(y^2) \right] \right|_0^1 =-{1\over 2}(\cos 1-1)\\ =\bbox[red,2pt]{{1\over 2}(1-\cos 1)}$$
==================== END =======================
沒有留言:
張貼留言