111台聯大轉學考-微積分A2詳解

台灣聯合大學系統111學年度學士班轉學生考試

科目:微積分
類組別:A2

甲、簡答題：共8題，每題8分，共64分

解答：$$f(x)=x^{\ln x} =e^{\ln(x^{\ln x })} =e^{\ln x\ln x} \Rightarrow f'(x)={2\over x}\ln x\cdot x^{\ln x} \Rightarrow 切線斜率f'(e)={ 2\over e}e = \bbox[red, 2pt]2$$

解答：$$\lim_{x\to 0} {\sin x-x\cos x\over \tan^3 x} =\lim_{x\to 0} {(\sin x-x\cos x)' \over (\tan^3 x)'} =\lim_{x\to 0} { x\sin x\over 3\tan^2 x \sec^2 x} = \lim_{x\to 0} { x\cos^4 x\over 3 \sin  x}\\ =\lim_{x\to 0} { (x\cos^4 x)'\over (3 \sin  x)'} =\lim_{x\to 0} { \cos^4 x-4x\cos^3x \sin x\over 3 \cos  x} = \bbox[red, 2pt]{1\over 3}$$

解答：$$\iint_R f(x,y)\,dA =\int_0^1 \int_0^{1-y} e^{x+2y}\,dxdy =\int_0^1  e^{1+y}-e^{2y }\,dy =\left. \left[e^{1+y}-{1\over 2}e^{2y} \right] \right|_0^1 =\bbox[red, 2pt]{{1\over 2}e^2-e+{1\over 2}}$$

解答：$$f(x)= x^{1/2}(1-x)^3 \Rightarrow f'(x)={1\over 2\sqrt x}(1-x)^3 -3\sqrt x(1-x)^2 =(1-x)^2\cdot {1-7x\over 2\sqrt x} \\ f'(x)=0 \Rightarrow x=0,{1 \over 7} \Rightarrow \cases{f(0)=0 \\ f(1/7)=6^3 \sqrt 7/7^4 \\ f(1)=0} \Rightarrow 最大值為\bbox[red, 2pt]{216 \sqrt 7\over 2401}$$

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

解答：$$\int_0^{\ln 2} x^{-2}e^{-1/x}\,dx =\left. \left[ e^{-1/x}\right] \right|_0^{\ln 2} = \bbox[red, 2pt]{e^{-1/\ln 2}}\\ 註: \lim_{x\to 0^+ }e^{-1/x}=0$$

解答：$$f(x,y)=30x^{2/3}y^{1/3} \Rightarrow df=20x^{-1/3}y^{1/3}\,dx+ 10x^{2/3}y^{-2/3}\,dy\\ \Rightarrow f(123,29)-f(125,27) \approx df|_{(x,y)=(125,27),(dx,dy)=(-2,2)}\\=20\cdot 125^{-1/3}\cdot 27^{1/3}\cdot (-2)+10\cdot 125^{2/3}\cdot 27^{-2/3}\cdot 2= \bbox[red, 2pt]{284\over 9}$$

解答：$$L=\lim_{n\to \infty}\left|{(-1)^{n+2}(x+2)^{n+1}\over (n+1)2^{n+1}} \cdot {n2^n\over (-1)^{n+1}(x+2)^n}\right| =\lim_{n\to \infty} \left|{ (-1)(x+2) n\over (n+1) 2} \right| =|x+2|\lim_{n\to \infty}{n \over 2(n+1)}\\ ={1\over 2}|x+2| \lt 1 \Rightarrow 收斂半徑R=\bbox[red, 2pt]2$$

乙、計算、證明題：共3題，每題12分，共36分。

解答：$$(a)\ln(4e^n-1)-\ln(2e^n+1) =\ln \left({4e^n-1\over 2e^n+1}\right)= \ln(2-{3\over 2e^n+1}) \Rightarrow \lim_{n\to \infty} \ln(2-{3 \over 2e^n+1})=\ln 2 \ne 0\\ \Rightarrow \sum_{n=0}^\infty \left(\ln(4e^n-1)-\ln(2e^n+1)\right) \bbox[red, 2pt]{發散}\\(b)\int_2^\infty {1\over x(\ln x)^{3/2}}\,dx =\left.\left[ -{2 \over \sqrt{\ln x}}\right]\right|_2^\infty ={2\over \sqrt{\ln 2}} \Rightarrow \sum_{n=2}^\infty {1\over n(\ln n)^{3/2}} \bbox[red, 2pt]{收斂}$$

解答：$$f(x,y)= e^{x^2-y^2} \Rightarrow \cases{f_x=2xe^{x^2-y^2} \\ f_y=-2y e^{x^2-y^2}} \Rightarrow \cases{f_{xx}=2e^{x^2-y^2}+ 4x^2e^{x^2-y^2} \\ f_{xy}=-4xy e^{x^2-y^2}\\ f_{yy}= -2e^{x^2-y^2} +4y^2e^{x^2-y^2}}  \Rightarrow D(x,y)= f_{xx}f_{yy}-f_{xy}^2 \\ f_x=f_y=0 \Rightarrow \text{critical point: }\bbox[red, 2pt]{(0,0)} \\ 又D(0,0)=f_{xx}(0,0)f_{yy}(0,0)-f_{xy}(0,0)^2 =2\cdot (-2)-0=-4\lt 0 \Rightarrow (0,0)為\bbox[red, 2pt]{\text{鞍點(saddle point)}}$$

解答：

$$積分區域介於圖形y=x^3與x軸之間且x\in [0,1]，如上圖;\\\int_0^1 \int_{\sqrt[3]y}^1 {2\pi \sin(\pi x^2) \over x^2}\,dxdy =\int_0^1\int_0^{x^3} {2\pi \sin(\pi x^2) \over x^2}\,dy dx =\int_0^1 2\pi x\sin(\pi x^2)\,dx =\left.\left[ -\cos(\pi x^2)\right] \right|_0^1 \\ =-\cos \pi +\cos 0= \bbox[red, 2pt]2$$