115年政大金融轉學考-微積分詳解

 國立政治大學 115 學年度學士班轉學生招生考試

考試科目 微積分 
系所別 金融學系二年級

1 Multiple Choice (5 points each, 50 points total)

解答:$$f(x)={x\over 1+\sqrt x} \Rightarrow f'(x)= {1\over 1+\sqrt x}-{x /2\sqrt x\over (1+\sqrt x)^2}= {1+\sqrt x/2\over (1+\sqrt x)^2} \\g(x)={2-\sin x\over 2-\cos x}  \Rightarrow g'(x)={-\cos x\over 2-\cos x}-{(2-\sin x)(\sin x) \over (2-\cos x)^2} ={-2\cos x-2\sin x+1\over (2-\cos x)^2} \\ \Rightarrow  \text{The correct statement is } \bbox[red, 2pt]{(b)}.$$

解答:$$\cases{\lim_{x\to c^-} f(x)=a+bc^2\\ \lim_{x\to c^+}f(x)={1\over c}} \Rightarrow a+bc^2= {1\over c} \cdots(1) \\ \cases{{d\over dx}(a+bx^2) =2bx \Rightarrow \lim_{x\to c^-}f'(x)=2bc\\ {d\over dx}({1\over x}=-{1\over x^2}) \Rightarrow \lim_{x\to c^+}f'(x)=-{1\over c^2}} \Rightarrow 2bc=-{1\over c^2} \cdots(2)\\ (2) \Rightarrow b=-{1\over 2c^3} \Rightarrow a+ \left( -{1\over 2c^3} \right) c^2 ={1\over c} \Rightarrow a={3\over 2c} \Rightarrow \cases{a=3/2c\\ b=-1/2c^3}\\ \text{This corresponds to statement }\bbox[red, 2pt]{(d)}.$$

解答:$$(a) \times:x\gt 0 \Rightarrow \int_0^x |t|\,dt = \int_0^x t\,dt ={1\over 2}x^2 \ne {1\over 2}|x| \\(b)\bigcirc:x\gt 0 \Rightarrow \int_0^x (t+|t|)^2 \,dt = \int_0^x (2t)^2\,dt =\int_0^x 4t^2\,dt =  {4\over 3}x^3 ={2\over 3}x^2(x+|x|) \\\qquad x\lt 0 \Rightarrow  \int_0^x (t+|t|)^2\,dt =\int_0^x 0^2\,dt =0 \\(c)\times: x\lt 0 \Rightarrow \int_0^x |t|\,dt = \int_0^x -t\,dt  =-{1\over 2}x^2 \ne {1\over 2}x^2 \\(d)\times:  x\gt 0 \Rightarrow  {3\over 4}x^2(x+|x|)={3\over 2}x^3 \ne{4\over 3} x^3\\  \Rightarrow \text{The correct statement is }\bbox[red, 2pt]{(b)}.$$

解答:$$u=\cos x \Rightarrow du=-\sin x\,dx \Rightarrow I_1= \int_0^{\pi/2} \cos^2 x\sin x\,dx =\int_1^0 -u^2\,du ={1\over 3} \Rightarrow \cases{\text{(a) is correct}\\\text{(c) is incorrect}} \\u=x^2+2x+3 \Rightarrow du= 2(x+1)\,dx \Rightarrow I_2= \int_2^3 {x+1\over \sqrt{x^2+2x+3}}\,dx = \int_{11}^{18}{1\over 2\sqrt u}\,du\\= \left. \left[ \sqrt u \right] \right|_{11}^{18} =\sqrt{18}-\sqrt{11} \Rightarrow \text{(b) is correct} \\\cases{u=x^2\\ dv= \cos x\,dx} \Rightarrow \cases{du =2x\,dx\\ v=\sin x} \Rightarrow I_3=\int_0^{\pi/2} x^2\cos x\,dx = \left. \left[ x^2\sin x \right] \right|_0^{\pi/2}- \int_0^{\pi/2} 2x\sin x\,dx \\={\pi^2\over 4}-2 \int_0^{\pi/2} x\sin x\,dx ={\pi^2\over 4}-2 \left. \left[ \sin x-x\cos x \right] \right|_0^{\pi/2}={\pi^2\over 4}-2 \Rightarrow \text{(d) is correct} \\ \text{Conclusion: The incorrect statement is }\bbox[red, 2pt]{(c)}.$$

解答:$$\cases{u= \sin(\log x) \\ dv =dx} \Rightarrow \cases{du =\cos(\log x)/x \,dx\\ v=x} \Rightarrow I_1=\int f(x)\,dx =x\sin (\log x)-\int\cos(\log x)\,dx \\ \cases{u=\cos(\log x)\\ dv=dx} \Rightarrow \cases{du=-\sin(\log x)/x \,dx\\ v=x} \Rightarrow \int\cos(\log x)\,dx=x\cos(\log x)+I_1 \\ \Rightarrow I_1=x\sin(\log x)-x\cos(\log x)-I_1 \Rightarrow I_1={1\over 2} \left( x\sin(\log x)-x\cos(\log x) \right) +C\\ \text{Using the same way, we have }I_2=\int g(x)\,dx = {1\over 2}(x\sin (\log x)+ x\cos(\log x))+C \\ \text{The correct statement: }\bbox[red, 2pt]{(d)}.$$

解答:$$\int f(x)\,dx = \int \left( {3/2\over x-1}-{1/2\over x+1}-{1\over (x+1)^2} \right)\,dx ={3\over 2}\log |x-1|-{1\over 2}\log|x+1|+{1\over x+1}+C \\ \int g(x)\,dx = \int \left( {1/2\over x}+{2\over x-1}-{1/2\over x+2} \right) \,dx ={1\over 2}\log|x|+2\log|x-1|-{1\over 2}\log|x+2|+C \\ \Rightarrow \text{The correct statement: }\bbox[red, 2pt]{(b)}$$

解答:$$\text{Assume $\sum a_n$ converges and $\sum b_n$ diverges. }\\ \text{Suppose }\sum(a_n+b_n ) \text{ converges} \Rightarrow \left( \sum(a_n+b_n ) -\sum a_n  \right) =\sum b_n \text{  converges, which contradicts} \\ \text{the initail premise that $\sum b_n$ diverges.} \Rightarrow \text{The false statement: }\bbox[red, 2pt]{(b)}$$

解答:$$\text{Let }S_k= \sum_{n=1}^k a_n = \sum_{n=1}^k (b_n-b_{n+1}) =b_1-b_{k+1} \Rightarrow \sum_{n=1}^\infty a_n = \lim_{k\to \infty} S_k =b_1-\lim_{k\to \infty} b_{k+1} \\ \Rightarrow \bbox[red,2pt]{(c)} \text{ is incorrect}$$

解答:$$\cases{u=x\\ dv=ke^{-kx}\,dx } \Rightarrow \cases{du=dx \\ v=-e^{-kx}} \Rightarrow E(X) =\int_0^\infty x(ke^{-kx})\,dx = \left. \left[ -xe^{-kx} \right] \right|_0^\infty - \int_0^\infty -e^{-kx}\,dx \\=0+\int_0^\infty e^{-kx}\,dx = \left. \left[ -{1\over k}e^{-kx} \right] \right|_0^\infty ={1\over k}=10 \Rightarrow k=0.1 \Rightarrow \bbox[red, 2pt]{(c)}$$

解答:$$f(x)={1\over 1+{1\over x}} ={x\over x+1} \Rightarrow f'(x)={1\over x+1}-{x\over (x+1)^2} ={1\over (x+1)^2} \\ g(x)={1\over 1+{1\over f(x)}} ={1\over 1+{x+1\over x}} ={x\over 2x+1} \Rightarrow g'(x)={1\over 2x+1}-{ 2x\over(2x+1)^2} ={1\over (2x+1)^2} \\ \Rightarrow \text{The correct statement: }\bbox[red, 2pt]{(b)}$$

2 Calculation (50 points total)

解答:$$\textbf{(a) }{d\over dx} \left[ \int_0^{x^2} f(t)\,dt \right] ={d\over dx}(x^2(1+x)) \Rightarrow 2xf(x^2)=2x+3x^2 \Rightarrow f(x^2)=1+{3\over 2}x \\ \quad\Rightarrow f(2)= \bbox[red, 2pt]{1+{3\over 2}\sqrt 2} \\ \textbf{(b) } {d\over dx} \left[ \int_0^{x^2(1+x)} f(t)\,dt \right]= {d\over dx}x \Rightarrow (2x+3x^2)f(x^2+x^3)=1 \Rightarrow f(x^2+x^3)={1\over 2x+3x^2} \\ \quad \Rightarrow f(2)=f(1+1) ={1\over 2+3}= \bbox[red, 2pt]{1\over 5}$$

解答:$$\cases{f(x,y) = 2x^2+y^2\\ g(x,y)= x+y-1} \Rightarrow \cases{f_x= \lambda g_x\\ f_y= \lambda g_y\\ g=0} \Rightarrow \cases{4x=\lambda\\ 2y=\lambda\\x+y=1} \Rightarrow y=2x \Rightarrow x+2x=3x=1\Rightarrow x={1\over 3} \\ \Rightarrow y={2\over 3} \Rightarrow f \left( {1\over 3},{2\over 3} \right)= \bbox[red, 2pt]{2\over 3}$$

解答:$$f(x)=xe^{-x^2} \Rightarrow f(-x)=-f(x)\Rightarrow f \text{ is odd} \Rightarrow \int_{-\infty}^\infty f(x)\,dx = \bbox[red, 2pt]0\\ \text{The geometric interpretation of the result $0$ is that the positive area bounded by the curve in the }\\ \text{first quadrant perfectly cancels out the negative area bounded by the curve in the third quadrant.}\\ \text{The net area is zero.}$$

解答:$$I= \int_0^2 \int_x^{2x} (x^2+y^2) \,dy\,dx = \int_0^2 {10x^3\over 3}\,dx =\bbox[red, 2pt]{40\over 3}$$

解答:$$\alpha={1+\sqrt 5\over 2} \text{ is the positive root of the equation }x^2-x-1=0 \Rightarrow \alpha^2=\alpha+1 \\ n=1 \Rightarrow a_1= 1\lt {1+\sqrt 5\over 2} =\alpha\Rightarrow \text{ the statement holds for }n=1\\ n=2 \Rightarrow \cases{a_2=2\\ \alpha^2=\alpha+1=(3+\sqrt 5)/2} \Rightarrow a_2\lt \alpha^2 \Rightarrow \text{ the statement holds for }n=2\\ \text{Assume that the inequality holds for all integers $k$ such that $1 \le k \le n$ for some integer $n \ge 2$.} \\ \Rightarrow a_{n+1}= a_n+a_{n-1} \lt \alpha^n +\alpha^{n-1} =\alpha^{n-1}(\alpha+1) =\alpha^{n-1}\cdot \alpha^2=\alpha^{n+1} \Rightarrow a_{n+1} \lt \alpha^{n+1}\\ \text{By mathematical induction, the inequality: }a_n \lt \left( {1+\sqrt 5\over 2} \right)^n \text{ is true for }n\ge 1. \\ \bbox[red, 2pt]{QED.}$$

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