109年台北大學轉學考-微積分詳解

國立臺北大學109學年度日間學士班暨進修學士班轉學生招生考試

學制系級：通訊工程資、訊工程日間學士班2年級
科目：微積分

解答： $$令\cases{f(x,y)=x^2+2y^2 \\ g(x,y)=x^2+y^2-1} \Rightarrow \cases{f_x= \lambda g_x\\ f_y = \lambda g_y\\ g=0} \Rightarrow \cases{ 2x=\lambda (2x) \cdots(1)\\ 4y=\lambda(2y) \cdots(2) \\ x^2+y^2=1 \cdots(3)}\\ {(1)\over (2)} \Rightarrow {x\over 2y} ={x\over y}\Rightarrow xy=0 \Rightarrow x=0 或y=0 代入(3) \\\Rightarrow (x,y)=\cases{ A(0,1) \\B(0,-1)\\ C(1,0)\\ D(-1,0)} \Rightarrow \cases{f(A)=f(B)= 2\\ f(C)=f(D)=1}  \Rightarrow 極值為:\bbox[red, 2pt]{1及2}$$

解答： $$x^y= y^x \Rightarrow e^{y\ln x} =e^{x\ln y} \Rightarrow  (e^{y\ln x})' = (e^{x\ln y})' \Rightarrow {y\over x}e^{y\ln x} =(\ln y+{xy'\over y})e^{x\ln y}\\ \Rightarrow  {y\over x}\cdot x^y = (\ln y+{xy'\over y})y^x \Rightarrow {xy' \over y} ={{y\over x}\cdot x^y  \over y^x} -\ln y ={x^{y-1}\over y^{x-1}} -\ln y\\ \Rightarrow y'=\left( {x^{y-1}\over y^{x-1}} -\ln y \right) \cdot {y\over x} = \bbox[red, 2pt]{{x^{y-2}\over y^{x-2}} -\ln y\cdot {y\over x}}$$

解答：

$$y=e^{-1/(x+1)}\\ \mathbf{(a)}\; \text{domain }D=\{x\mid x\ne -1, x\in \mathbb{R}\} \\\mathbf{(b)}\;y(0)= e^{-1} \Rightarrow y截距=1/e; e^{-1/(x+1)} \ne 0, \forall x \Rightarrow 無x截距\\ \mathbf{(c)}\; \cases{\lim_{x\to \infty} y=1 \\ \lim_{y\to \infty} x=-1} \Rightarrow 漸近線為y=1與x=-1\\\mathbf{(d)} \; 在兩區間 (-\infty, -1)及(1,\infty)，圖形為遞增;\\\mathbf{(e)}\;y'={1\over (x+1)^2}e^{-1/(x+1)} \ne 0 ,\forall x \Rightarrow 無局部極大值或局部極小值\\ \mathbf{(f)}\;y''=-{2\over (x+1)^3}e^{-1/(x+1)} +{1\over (x+1)^4} e^{-1/(x+1)} \Rightarrow \cases{y''\gt 0, x\lt -1\\ y'' \lt 0, x\gt -1} \Rightarrow \cases{凹向上,x\lt -1\\ 凹向下,x\gt -1} \\\mathbf{(g)}\; f''(x)\ne 0 \Rightarrow 無反曲點\\ \mathbf{(h)}\;圖形如上，不含(-1,0)$$

解答： $$\mathbf{(a)}\;令u=\sqrt x \Rightarrow du ={1\over 2\sqrt x}dx ={1\over 2u}dx \Rightarrow dx=2udu \Rightarrow \int_0^1{\sqrt x\over x+1}\,dx = \int_0^1 {2u^2\over u^2+1}du\\ =\int_0^1 \left(2-{2\over u^2+1} \right)\,du =\left. \left[ 2u-2\tan^{-1} u  \right] \right|_0^1 = \bbox[red, 2pt]{2-{\pi \over 2}} \\\mathbf{(b)}\;令\cases{u=\ln x \Rightarrow du ={1\over x}dx\\ dv={1\over x^2}dx \Rightarrow v=-{1\over x}} \Rightarrow \int_0^\infty {\ln x\over x^2}\,dx = \left.\left[-{1\over x}\ln x \right]\right|_0^\infty +\int_0^\infty {1\over x^2}\,dx \\ = \left.\left[-{1\over x}\ln x -{1\over x}\right]\right|_0^\infty  \to \bbox[red, 2pt]{-\infty}$$

解答：

$$\mathbf{(a)}\;繞y=2旋轉體積= \pi\int_0^4 (2-\sqrt x)^2\,dx =\bbox[red,2pt]{8\pi \over 3} \\\mathbf{(b)}\;繞x=4旋轉體積= \pi\int_0^2 (4-y^2)^2\,dy = \bbox[red, 2pt]{256\pi \over 15}$$

解答： $$\mathbf{(a)}\;f(t)={1\over \sqrt{1-t^2}} \Rightarrow f'(t)={t \over (1-t^2)^{3/2}} \Rightarrow f''(t)= {2t^2+1 \over (1-t^2)^{5/2}} \Rightarrow f'''(t)= {6t^3+9t \over (1-t^2)^{7/2}}\\ \Rightarrow f^{[4]}(t) = {24t^4 +72t^2+ 9 \over (1-t^2)^{9/2}} \Rightarrow f^{[5]}(t) = { 15t(8t^4+ 40t^2+ 15)\over (1-t^2)^{11/2}} \Rightarrow f^{[6]} ={45( 16t^6 +120t^4 + 90t^2+5)\over (1-t^2)^{13/2}}\\ \Rightarrow \cases{f(0)=1\\ f^{[奇數]}(0)=0 \\ f''(0)=1\\ f^{[4]}(0)= 9\\ f^{[6]}(0)= 225 }\Rightarrow \text{Maclaurin series =} \sum_{n=0}^\infty f^{[n]}(0){t^n \over n!}= \bbox[red, 2pt]{1+ {1\over 2}t^2 +{3\over 8}t ^4 +{5\over 16}t^6}+\cdots \\\mathbf{(b)}\; g(x)=\sin^{-1} x=\int_0^x {1\over \sqrt{1-t^2}}\,dt \Rightarrow \cases{g(0)=0\\ g'(x)=f(x) }\\\ \Rightarrow \text{Maclaurin series =} \sum_{n=0}^\infty g^{[n]}(0){t^n \over n!}=  f(0)x +{1\over 3!}f''(0)x^3 +{1\over 5!}f^{[4]}x^5  +{1\over 7!}f^{[6]}x^7+\cdots \\= \bbox[red, 2pt]{x+ {1\over 6}x^3 +{3\over 40}x^5 +{5\over 112}x^7}+\cdots$$
 =============== END ===================

解題僅供參考，其它轉學考試題及詳解