115年中央大學數學碩士班-微積分詳解

國立中央大學115學年度碩士班考試入學

系所:數學系碩士班
科目:微積分

解答:$$ f(x) = 4\sin x - 4x + \frac{2}{3}x^3 \Rightarrow  f'(x) = 4\cos x - 4 + 2x^2 \Rightarrow f''(x) = -4\sin x + 4x \\ \Rightarrow f'''(x) = -4\cos x + 4 = 4(1 - \cos x) \gt 0, \text{ for }x\gt 0 \Rightarrow f''(x) \text{ is strictly increasing} \\ \Rightarrow f'(x) \text{ is strictly increasing and }f'(0)=0 \Rightarrow f(x)\text{ is strictly increasing, for }x \ge 0 \\ \Rightarrow f(x)\gt 0, x\in (0,\infty) \Rightarrow  4\sin x - 4x + \frac{2}{3}x^3\gt 0 \Rightarrow  4x-\frac{2}{3}x^3 \lt 4 \sin x \quad \bbox[red, 2pt]{QED.}$$

解答:$$ \frac{d}{dx} \left( x^5 \int_{\pi/4}^{x^2} \tan^{-1} t dt \right) = \frac{d}{dx} x^5 \cdot \left(  \int_{\pi/4}^{x^2} \tan^{-1} t dt \right)+x^5 \cdot {d\over dx}\left( x^5 \int_{\pi/4}^{x^2} \tan^{-1} t dt \right) \\=5x^4\int_{\pi/4}^{x^2} \tan^{-1} t dt +x^5 (\tan^{-1}x^2) \cdot 2x = \bbox[red, 2pt]{5x^4\int_{\pi/4}^{x^2} \tan^{-1} t dt +2x^6 (\tan^{-1}x^2)}$$

解答:$$\textbf{(A) }f(x)={\sin x\over 1+2x^2+3x^4} \Rightarrow f(-x)=-f(x) \Rightarrow f(x) \text{ is odd} \Rightarrow \int_{-1}^1 f(x)\,dx =\bbox[red, 2pt]0 \\ \textbf{(B) }u= \sqrt{1+ \sqrt{2+x}} \Rightarrow u^2 =1+\sqrt{2+x} \Rightarrow (u^2-1)^2 =2+x \Rightarrow x=(u^2-1)^2-2 \\ \qquad \Rightarrow dx = (4u^3-4u)\,du \Rightarrow \int \sqrt{1+ \sqrt{2+x}} \,dx = \int u(4u^3-4u)\,du =\int (4u^4-4u^2)\,du \\\qquad ={4\over 5}u^5-{4\over 3}u^3+C= \frac{4}{5}(\sqrt{1 + \sqrt{2 + x}})^5 - \frac{4}{3}(\sqrt{1 + \sqrt{2 + x}})^3 + C \\\qquad = \bbox[red, 2pt]{\frac{4}{5}(1 + \sqrt{2 + x})^{5/2} - \frac{4}{3}(1 + \sqrt{2 + x})^{3/2} + C}$$

解答:$$b_n= {1\over n^{5/4}} \Rightarrow \sum_{n=1}^\infty b_n \text{ converges, since }{5\over 4}\gt 1\\ a_n={\ln n\over n^{3/2}} \Rightarrow  \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{\ln n}{n^{3/2}}}{\frac{1}{n^{5/4}}} =   \lim_{n \to \infty} \frac{\ln n}{n^{1/4}} =   \lim_{n \to \infty} \frac{1/n}{{1\over 4}n^{-3/4}} =\lim_{n \to \infty} \frac{4}{n^{1/4}} =0 \\ \Rightarrow \text{By Limit Comparison Test, }\sum_{n=1}^\infty a_n \bbox[red, 2pt]{\text{ converges}}$$

解答:$$\textbf{(A)}  \frac{\partial g}{\partial x}(0,0) = \lim_{h \to 0} \frac{g(h,0) - g(0,0)}{h}  = \lim_{h \to 0} \frac{{2h^2\cdot 0\over h^4+0^4}  - g(0,0)}{h}  = \lim_{h \to 0} \frac{0 - 0}{h} =0\\  \frac{\partial g}{\partial y}(0,0) = \lim_{k \to 0} \frac{g(0,k) - g(0,0)}{k} = \lim_{k \to 0} \frac{ {2\cdot 0^2\cdot k^2\over 0^4+k^4} - g(0,0)}{k}=\lim_{k \to 0} \frac{0 - 0)}{k} =0 \\ \Rightarrow \bbox[red, 2pt]{\frac{\partial g}{\partial x}(0,0)=0, \frac{\partial g}{\partial y}(0,0) =0} \\ \textbf{(B) } \lim_{x \to 0} g(x,0) = \lim_{x \to 0} \frac{2x^2(0^2)}{x^4+0^4} = \lim_{x \to 0} 0 = 0 ,  \lim_{x \to 0} g(x,x) = \lim_{x \to 0} \frac{2x^2(x^2)}{x^4+x^4} = \lim_{x \to 0} \frac{2x^4}{2x^4} = \lim_{x \to 0} 1 = 1 \\ \quad \Rightarrow  \lim_{x \to 0} g(x,0) \ne \lim_{x \to 0} g(x,x) \Rightarrow \lim_{(x,y) \to (0,0)} g(x,y) \text{ does not exist} \\\Rightarrow \bbox[red, 2pt]{\text{No, $g(x,y)$ is NOT differentiable at $(0,0)$)} }$$

解答:$$h(x,y) = 2 \tan^{-1}\left(\frac{y}{x}\right) + 2\sqrt{3} \sin^{-1}\left(\frac{xy}{2}\right) \Rightarrow \nabla h(x,y)=(h_x,h_y) \\= \left(   \frac{-2y}{x^2 + y^2} + \frac{\sqrt{3}y}{\sqrt{1 - \frac{x^2y^2}{4}}},   \frac{2x}{x^2 + y^2} + \frac{\sqrt{3}x}{\sqrt{1 - \frac{x^2y^2}{4}}} \right) \Rightarrow \nabla h(1,1) =(1,3) \\ v= \langle 3,-2\rangle \Rightarrow u={v \over |\vec v|} = \left\langle \frac{3}{\sqrt{13}}, \frac{-2}{\sqrt{13}} \right\rangle \Rightarrow  D_u h(1,1) = \nabla h(1,1) \cdot u = \langle 1, 3 \rangle \cdot \left\langle \frac{3}{\sqrt{13}}, \frac{-2}{\sqrt{13}} \right\rangle \\= \bbox[red, 2pt]{-3\over \sqrt{13}}$$

解答:$$\cases{ f(x,y,z) = x^2 + y^2 + z^2 \\ g(x,y,z) = y + 2z - 12 = 0 \\h(x,y,z) = x + y - 6 = 0  } \Rightarrow \text{ Using Lagrange Multipliers, } \nabla f = \lambda \nabla g + \mu \nabla h  \\ \Rightarrow \langle 2x, 2y, 2z \rangle = \lambda \langle 0, 1, 2 \rangle + \mu \langle 1, 1, 0 \rangle  \Rightarrow \cases{ 2x = \mu \\2y = \lambda + \mu \\2z = 2\lambda  } \Rightarrow z=\lambda \Rightarrow  2y=z+2x \Rightarrow z=2y-2x\\ \Rightarrow \cases{g(x,y,2y-2z)=5y-4x-12=0 \\ h(x,y,z)=x+y-6=0} \Rightarrow \cases{x=2\\ y=4} \Rightarrow z=4 \Rightarrow (x,y,z)= \bbox[red, 2pt]{(2,4,4)}$$

解答:$$r=3\sin \theta=1+\sin \theta \Rightarrow \sin \theta={1\over 2} \Rightarrow \theta={\pi\over 6}, {5\pi\over 6} \\ \Rightarrow \text{area }A  = \frac{1}{2} \int_{\pi/6}^{5\pi/6} \left[ (3 \sin \theta)^2 - (1 + \sin \theta)^2 \right] d\theta   = \frac{1}{2} \int_{\pi/6}^{5\pi/6} (3 - 4 \cos(2\theta) - 2 \sin \theta) d\theta \\= {1\over 2} \left. \left[ 3\theta-2\sin (2\theta)+2 \cos \theta \right] \right|_{\pi/6}^{5\pi/6}= {1\over 2} \left( {5\pi\over 2}-{\pi\over 2} \right) =\bbox[red, 2pt] \pi$$

解答:$$D_1:y\gt 2x \Rightarrow \max\{4x^2,y^2\}=y^2 \Rightarrow  I_1=\int_{0}^{1} \int_{0}^{2} e^{\max\{4x^2, y^2\}} dy dx = \int_{0}^{1} \int_{2x}^{2} e^{  y^2 } dy dx \\ = \int_{0}^{2} \int_{0}^{y/2} e^{y^2} dx dy = \int_0^2 {y\over 2}e^{y^2}\,dy= \left. \left[ {1\over 4}e^{y^2} \right] \right|_0^2={1\over 4} \left( e^4-1 \right) \\ D_2:y\lt 2x \Rightarrow \max\{ 4x^2,y^2\}=4x^2 \Rightarrow I_2 =\int_{0}^{1} \int_{0}^{2} e^{\max\{4x^2, y^2\}} dy dx = \int_{0}^{1} \int_{0}^{2x} e^{ 4x^2 } dy dx \\ \qquad = \int_0^1 2xe^{4x^2}\,dx = \left. \left[ {1\over 4}e^{4x^2} \right] \right|_0^1={1\over 4} \left( e^4-1 \right) \\ \Rightarrow \int_{0}^{1} \int_{0}^{2} e^{\max\{4x^2, y^2\}} dy dx =I_1+I_2 =  \bbox[red, 2pt]{{1\over 2}\left( e^4-1 \right)}$$

解答:$$\cases{x =r\cos \theta\\ y=r\sin \theta} \Rightarrow  \int_0^{\sin\beta} \int_{y\cot\beta}^{\sqrt{1-y^2}} \ln(x^2 + y^2) dx dy = \int_0^\beta \int_0^1 (2\ln r) \cdot r dr d\theta = \int_0^\beta \int_0^1 2r\ln r dr d\theta \\=  \left( \int_0^\beta d\theta \right) \cdot \left( \int_0^1 2r\ln r dr \right)  = \beta \cdot \left. \left[ r^2\ln r-{1\over 2}r^2 \right] \right|_0^1 = \beta \left( -{1\over 2}-0 \right)= \bbox[red, 2pt]{-{\beta \over 2}}\\  \text{PS.}\lim_{r \to 0^+} r^2\ln r = \lim_{r \to 0^+} \frac{\ln r}{r^{-2}} = \lim_{r \to 0^+} \frac{1/r}{-2r^{-3}} = \lim_{r \to 0^+} \frac{-r^2}{2} = 0$$

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解題僅供參考，碩士班歷年試題及詳解