111年台聯大轉學考-微積分A3A4A6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：A3/A4/A6

甲、填充題：共8題，每題8分，共64分

解答： $$\lim_{n \to \infty} \left( {1\over \sqrt n\,\sqrt{n+1}} +{1\over \sqrt n\,\sqrt{n+2}} +\cdots +{1\over \sqrt n\,\sqrt{n+n}} \right) =\lim_{n \to \infty}\sum_{k=1}^n{1\over \sqrt n\,\sqrt{n+k}}\\ =\lim_{n \to \infty} \sum_{k=1}^n {1\over   n} \cdot {\sqrt n\over \sqrt{n+k}}  =\lim_{n \to \infty} \sum_{k=1}^n {1\over   n} \cdot {1\over \sqrt{1+{k\over n}}} =\int_0^1 {1\over \sqrt{1+x}} \,dx =\int_1^2 {1\over \sqrt u}\,du,其中u=x+1\\ =\left.\left [ 2\sqrt u\right] \right|_1^2 =\bbox[red, 2pt]{2\sqrt 2-2}$$

解答： $$利用球坐標\cases{x=\rho \sin \phi \cos \theta\\ y= \rho\sin \phi \sin \theta\\ z=\rho \cos \phi} \Rightarrow 較小的體積=\int_0^{2\pi} \int_0^{\pi/3} \int_{1/\cos\phi}^2 \rho^2\sin \phi\, d\rho d\phi d\theta \\ =\int_0^{2\pi} \int_0^{\pi/3} \sin \phi\left( {(8\cos^3\phi-1)\sin \phi\over 3\cos^3\phi}\right) d\phi d\theta =\int_0^{2\pi} {5\over 6}d\theta =\bbox[red, 2pt]{{5\over 3}\pi}$$

解答： $$\int {4x^2-3x+2 \over 4x^2-4x+3}\,dx =\int \left(1+{x-1 \over 4x^2-4x+3}\right)\,dx \\=x +\int \left( {1\over 8}\cdot {8x-4\over 4x^2-4x+3} -{1\over 2}\cdot {1\over 4x^2-4x+3}\right)\,dx \\ =x+ {1\over 8}\ln|4x^2-4x+3| -\int {1\over 8}\cdot {1\over x^2-x+3/4}\,dx\\ =x+ {1\over 8}\ln|4x^2-4x+3| -{1\over 8}\int {1\over (x-1/2)^2 +1/2}\,dx\\ =x+ {1\over 8}\ln|4x^2-4x+3| -{1\over 4}\int {1\over (\sqrt 2(x-1/2))^2 +1}\,dx \\=x+ {1\over 8}\ln|4x^2-4x+3| -{1\over 4\sqrt 2}\tan^{-1}(\sqrt 2(x-1/2))+C \\ = \bbox[red, 2pt]{x+{1\over 8}\left(\ln(4x^2-4x+3)-\sqrt 2 \tan^{-1}(\sqrt 2(x-1/2))+C \right)}\\ 註: 4x^2-4x+3 =(2x-1)^2+2 \gt 0,\forall x\in \mathbb{R}$$

解答： $$\int_0^1 \int_x^1 \sin(y^2)\,dydx =\int_0^1\int_0^y \sin (y^2)\,dxdy =\int_0^1 y\sin(y^2)\,dy =\left.\left[ -{1\over 2} \cos(y^2)\right] \right|_0^1 = -{1\over 2}\cos 1+{1\over 2}\\ =-{1\over 2}(1-2\sin^2 {1\over 2})+{1\over 2} =\bbox[red, 2pt]{\sin^2 {1\over 2}}$$

解答： $$令\cases{x=r\cos \theta\\ y=r\sin \theta}，因此x^2+y^2=4 \Rightarrow \cases{0\le \theta \le 2\pi\\ 0\le r\le 2};\\ 又y+2z=2 \Rightarrow z={2-y\over 2} \Rightarrow \cases{{\partial \over \partial x}z =0 \\ {\partial \over \partial y}z =-1/2} \Rightarrow 截面積=\int_0^{2\pi} \int_0^2 \sqrt{0^2+(-1/2)^2+1}\,r\,drd\theta \\= {\sqrt 5\over 2}\int_0^{2\pi} \int_0^2  r\,drd\theta =\sqrt 5 \int_0^{2\pi } 1,d\theta =\bbox[red, 2pt]{2\sqrt 5 \pi}$$

解答： $$令u=3x \Rightarrow du =3dx \Rightarrow \int_1^3 {\cos 3x\over x}dx = \int_3^9 {\cos u\over u/3}{1\over 3}du =\int_3^9 {\cos u\over u}du = \bbox[red, 2pt]{F(9)-F(3)}$$

解答： $$y=3x-x^2 \Rightarrow y'=3-2x \Rightarrow 切點(a,y(a))=(a,3a-a^2)的斜率=3-2a\\ \Rightarrow 切線方程式y=(3-2a)(x-a)+3a-a^2 \Rightarrow 切線與坐標軸的交點\cases{A(0, a^2) \\B({a^2 \over 2a-3},0)} \\ \Rightarrow \triangle OAB面積=f(a)={1\over 2}\cdot \overline{OA}\cdot \overline{OB} ={1\over 2}\cdot a^2 \cdot {a^2\over 2a-3} ={ a^4\over 4a-6} \Rightarrow  f'(a)= {4a^3\over 4a-6}-{4a^4\over (4a-6)^2}\\ 因此f'(a)=0 \Rightarrow {4a^3(3a-6)\over (4a-6)^2} =0 \Rightarrow a=2(a=0 \Rightarrow 切點=(0,0)不合)\\ \Rightarrow f(2)={16\over 8-6}= \bbox[red, 2pt]8$$

解答： $$利用\text{Lagrange's 算子求極值}: 令 \cases{f(x,y,z)=4x^2+y^2+4z^2-16 \\T(x,y,z)= 8x^2+ 4yz-16z+ 600} \Rightarrow \cases{T_x =\lambda f_x\\ T_y =\lambda f_y\\ T_z= \lambda f_z\\f=0} \\ \Rightarrow \cases{16x= \lambda \cdot 8x\\ 4z= \lambda\cdot 2y\\ 4y-16= \lambda\cdot 8z} \Rightarrow \cases{(\lambda-2)x=0 \cdots(1)\\ 2z=\lambda y\cdots(2)\\ y-4= 2\lambda z \cdots(3)}，由(1)知:\lambda=2 或x=0\\ \text{Cases I: } \lambda=2 \Rightarrow \cases{z=y\\ y-4=4z} \Rightarrow y=z=-{4\over 3}代入f=0 \Rightarrow x=\pm {4\over 3}\\ \text{Cases II: }x=0 代入f=0 \Rightarrow y^2+4z^2=16\cdots(4)，再由{(2)\over (3)} \Rightarrow {2z\over y-4}={y\over 2z}  \Rightarrow 4z^2=y^2-4y 代入(4)\\ \Rightarrow y^2+y^2-4y=16 \Rightarrow (y-4)(y+2)=0 \Rightarrow \cases{y=4 \Rightarrow z=0\\ y=-2 \Rightarrow z=\pm \sqrt 3}\\ 將\text{Cases I }及\text{Cases II }的極值點\cases{A(4/3,-4/3,-4/3)\\ B(-4/3,-4/3,-4/3)\\ C(0,4,0)\\ D(0,-2,\sqrt 3)\\ E(0,-2,-\sqrt 3)} 代入T(x,y,z)\\ \Rightarrow \cases{T(A)= 600 \\T(B)=1928/3=642.7 (最大)\\ T(C)=600\\ T(D)=600-24\sqrt 3\\T(E)= 600+24\sqrt 3 \approx 641.57} \Rightarrow 最熱的點:\bbox[red,2pt]{(-{4\over 3}, -{4\over 3}, -{4\over 3})}$$

乙、計算、證明題：共3題，每大題12分，共36分

解答： $$f(x)=\begin{cases} e^{-1/x^2}, & \text{if }x\ne 0\\ 0, & \text{if }x=0\end{cases} \Rightarrow f''(0)=f'(0)=f(0)=0\\ \Rightarrow f(x)\approx f(0)+f'(0)x+ {1\over 2}f''(0)x^2 = \bbox[red,2pt]0$$

解答： $$b\ge a\ge 0 \Rightarrow b^n \ge a^n \ge 0 \Rightarrow 2b^n \ge a^n+b^n \ge b^n \Rightarrow \sqrt[n]{2b^n} \ge \sqrt[n]{a^2 +b^2} \ge \sqrt[n]{b^n} \\ \Rightarrow 2^{1/n} b\ge  \sqrt[n]{a^2 +b^2} \ge b，由於\lim_{n\to \infty}2^{1/n} b=b，因此 b\ge \lim_{n\to \infty}\sqrt[n]{a^2 +b^2} \ge b\\ 由夾擠定理可知:\lim_{n\to \infty}\sqrt[n]{a^2 +b^2} =b，也就是該數列\bbox[red,2pt]{收斂}，且極限值為\bbox[red, 2pt]b$$

解答：

a. $$\cases{y=0 \Rightarrow  \lim_{(x,y)\to (1,0)} {xe^y-1\over xe^y-1+y} =\lim_{x\to 1}{x -1\over x-1} =1\\ y=\ln x \Rightarrow  \lim_{(x,y)\to (1,0)} {xe^y-1\over xe^y-1+y}  =\lim_{x\to 1}{x^2-1\over x^2-1+\ln x} =\lim_{x\to 1}{2x\over 2x+1/x}=2/3}  \Rightarrow 極限不存在$$ b. $$\cases{y=x \Rightarrow  \lim_{(x,y)\to (0,0)} {\sin(x-y)\over |x|+|y|} =\lim_{x\to 0}{\sin(0)\over 2|x|} =0\\ y=-x \Rightarrow \lim_{(x,y)\to (0,0)} {\sin(x-y)\over |x|+|y|} =\lim_{x\to 0^+}{\sin(2x)\over 2x}=1}  \Rightarrow 極限不存在$$

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