115年嘉義大學應數碩士班-微積分詳解

 國立嘉義大學 115 學年度應用數學系碩士班(甲組)招生考試

科目: 微積分 (每題 10 分，共 100 分)

解答:$$ \lim_{x\to 1} \left( \sqrt{x^2 + 5x - 1} - \sqrt{x^2 - 4x + 8} \right) = \sqrt 5-\sqrt 5=\bbox[red, 2pt]0$$

解答:$$y= \sin(x^2+\sqrt{x^2+1}) \Rightarrow y'=\cos (x^2+\sqrt{x^2+1})\cdot {d\over dx}(x^2+\sqrt{x^2+1}) \\=\cos (x^2+\sqrt{x^2+1})\cdot (2x+{2x\over 2\sqrt{x^2+1}}) = \bbox[red, 2pt]{\left( 2x+{x\over \sqrt{x^2+1}} \right) \cos (x^2+\sqrt{x^2+1})}$$

解答:$$y=(x^2+1)\tan^{-1}e^{x+1} \Rightarrow y'={d\over dx}(x^2+1)\cdot \tan^{-1}e^{x+1}(2x) + (x^2+1) \cdot {d\over dx} \tan^{-1}e^{x+1}\\= \bbox[red, 2pt]{2x\tan^{-1}e^{x+1}+(x^2+1)\cdot {e^{x+1}\over 1+e^{2x+2}}  }$$

解答:$$ x^3 + x^2y + xy^2 + y^3 = 1000 + xy \Rightarrow 3x^2+2xy+ x^2y'+y^2+2xyy'+3y^2y'=y+xy' \\ \Rightarrow  y'(x^2 + 2xy + 3y^2 - x) = y - 3x^2 - 2xy - y^2 \Rightarrow  y' =\bbox[red, 2pt]{ \frac{y - 3x^2 - 2xy - y^2}{x^2 + 2xy + 3y^2 - x}}$$

解答:$$\cases{f(x,y)=4x+3y\\ g(x,y)= x^2+y^2-2} \Rightarrow \cases{f_x= \lambda g_x\\ f_y= \lambda g_y\\ g=0} \Rightarrow \cases{4=\lambda \cdot 2x \\3=\lambda\cdot 2y\\ x^2+y^2=2} \Rightarrow \cases{x=2/\lambda\\ y=3/2\lambda} \Rightarrow  \left(\frac{2}{\lambda}\right)^2 + \left(\frac{3}{2\lambda}\right)^2 = 2 \\\Rightarrow  \lambda^2 = \frac{25}{8} \implies \lambda =   \pm \frac{5 \sqrt{2}}{4} \Rightarrow \cases{\lambda=5\sqrt 2/4 \Rightarrow (x,y)=({4\sqrt 2\over 5}, {3\sqrt 2\over 5}) \\ \lambda=-5\sqrt 2/4 \Rightarrow (x,y) =(-{4\sqrt 2\over 5}, -{3\sqrt 2\over 5})} \\ \Rightarrow \cases{f( {4\sqrt 2\over 5}, {3\sqrt 2\over 5} ) = 5\sqrt 2\\ f(-{4\sqrt 2\over 5}, -{3\sqrt 2\over 5}) =-5\sqrt 2} \Rightarrow \bbox[red, 2pt]{\cases{極大值: 5\sqrt 2\\ 極小值:-5\sqrt 2}}$$

解答:$$f(x) =2xe^{-3x^2} \Rightarrow f(-x)=-2xe^{-3x^2}=-f(x) \Rightarrow f(x)為奇函數 \Rightarrow \int_{-\infty}^\infty f(x,y) = \bbox[red, 2pt]0$$

解答:$$\left\lfloor x \right\rfloor = \begin{cases}0,& 0\le x\lt 1\\ 1,& 1\le x\lt 2\\ 2, & 2\le x\lt 3 \end{cases} \Rightarrow \int_0^3 {\left\lfloor x \right\rfloor\over x}\,dx =  \int_0^1 \frac{0}{x} dx + \int_1^2 \frac{1}{x} dx + \int_2^3 \frac{2}{x} dx  \\=0+\ln 2+ 2\ln 3-2\ln 2=2\ln 3-\ln 2= \bbox[red, 2pt] {\ln {9\over 2}}$$

解答:$$u=e^x+e^{-x} \Rightarrow du=(e^x-e^{-x})dx \Rightarrow I= \int \frac{2(e^x - e^{-x})}{(e^x + e^{-x})^2} dx = \int {2\over u^2}du =-{2\over u}+C \\ =\bbox[red, 2pt]{-\frac{2}{e^x + e^{-x}} + C}$$

解答:$$u=x^2+1 \Rightarrow du=2x\,dx \Rightarrow  \int x^5 (x^2 + 1)^{30} dx =  \int (u - 1)^2 u^{30} \cdot \frac{1}{2} du = \frac{1}{2} \int (u^2 - 2u + 1) u^{30} du \\= \frac{1}{2} \int (u^{32} - 2u^{31} + u^{30}) du  = \frac{1}{2} \left( \frac{u^{33}}{33} - \frac{2u^{32}}{32} + \frac{u^{31}}{31} \right) + C \\= \bbox[red, 2pt]{\frac{(x^2 + 1)^{33}}{66} - \frac{(x^2 + 1)^{32}}{32} + \frac{(x^2 + 1)^{31}}{62} + C }$$

解答:$$u=\log_3 x={\ln x\over \ln 3} \Rightarrow du={1\over x\ln3}dx \Rightarrow \int {\log_3 x\over x}\,dx = \ln 3\int u \,du ={1\over 2}\ln 3\cdot u^2+C \\= \bbox[red, 2pt]{\frac{\ln 3}{2} (\log_3 x)^2 + C}$$

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解題僅供參考，碩士班歷年試題及詳解