113年台師大全球經營碩士班-微積分詳解

 國立臺灣師範大學113學年度碩士班招生考試

科目:微積分
適用系所:全球經營與策略研究所

解答:$$f(x,y)=1-(x^2+y^2)^{1/3} \Rightarrow \cases{f_x=-{2x\over 3}(x^2+y^2)^{-2/3} \\ f_y=-{2y\over 3} (x^2 +y^2)^{-2/3}} \Rightarrow \cases{f_{xx}= {2(x^2-3y^2)\over 9(x^2+y^2)^{5/3}} \\ f_{xy} ={8xy\over 9(x^2+ y^2)^{5/3}} \\f_{yy} ={2(-3x^2+y^2) \over 9(x^2+y^2)^{5/3}}} \\ \Rightarrow d(x,y)= f_{xx}f_{yy}-f_{xy}^2 =-{4\over 27(x^2+y^2)^{4/3}} \\ \cases{f_x=0\\ f_y=0} \Rightarrow (x,y)=(0,0) \Rightarrow d(0,0) 不存在,無法分類\\ 但x^2+y^2 \ge0 \Rightarrow f(x,y)=1-(x^2+y^2)^{1/3}\le 1 \Rightarrow f(0,0)=\bbox[red, 2pt]{1為相對極大值}$$

解答:$$f(x,y)=e^{-(x^2+y^2-4y)} \Rightarrow \cases{f_x=-2xe^{-(x^2+y^2-4y)} \\ f_y=2(2-y) e^{-(x^2+y^2-4y)}} \Rightarrow \cases{f_{xx}= (4x^2-2)e^{-(x^2+y^2-4y)} \\ f_{xy} = 4x(y-2) e^{-(x^2+y^2-4y)} \\f_{yy} = 
 (4(y-2)^2- 2)e^{-(x^2+y^2-4y)}} \\ \Rightarrow d(x,y)= f_{xx}f_{yy}-f_{xy}^2 = (-8x^2-8y^2+32y-28)e^{-(x^2+y^2-4y)} \\ \cases{f_x=0\\ f_y=0} \Rightarrow (x,y)=(0,2) \Rightarrow \cases{d(0,2) =4e^8 \gt 0\\ f_{xx}(0,2) =-2e^4 \lt 0} \Rightarrow \bbox[red, 2pt]{f(0,2)=e^4 為相對極大極值}$$

解答:$$f(x)=-3x^5+ 5x^3 \Rightarrow f'(x)=-15x^4+15x^2 \Rightarrow f''(x)=-60x^3+30x\\ f'(x)=0 \Rightarrow 15x^2(x^2-1)=0 \Rightarrow x=0,\pm 1 \Rightarrow \cases{f''(0)=0\\ f''(1)=-30\lt 0\\ f''(-1) =30\gt 0} \\ f''(x)=0 \Rightarrow 30x(1-2x^2 )=0 \Rightarrow x=0, \pm {\sqrt 2\over 2} \Rightarrow \cases{f(0)=0\\ f(\sqrt 2/2) ={7\over 8}\sqrt 2 \\ f(-\sqrt 2/2) =-{7\over 8}\sqrt 2}\\\Rightarrow \bbox[red, 2pt]{\cases{反曲點(0,0),(\sqrt 2/2,7\sqrt 2/8),(-\sqrt 2/2, -7\sqrt 2/8) \\ 相對極大值f(1)=2\\ 相對極小值f(-1)=-2}}$$

解答:$$\textbf{(1)}\;\cases{u=\ln(2x-3) \\dv=xdx} \Rightarrow \cases{du={2\over 2x-3} dx\\ v={1\over 2}x^2} \\\Rightarrow I=\int_2^6 x\ln(2x-3)\,dx = \left. \left[{1\over 2}x^2\ln(2x-3) \right] \right|_2^6-\int_2^6 {x^2\over 2x-3}\,dx \\=18\ln 9- \int_2^6 \left({1\over 2}x+{3\over 4}+{9/4\over 2x-3} \right)\,dx =18\ln 9-\left. \left[ {1\over 4}x^2+{ 3\over 4}x+{9\over 8} \ln(2x-3)\right] \right|_2^6 \\=18\ln 9-\left(11+{9\over 8} \ln 9 \right)= \bbox[red, 2pt]{{135\over 4}\ln 3 -11} \\\textbf{(2)}u=3-x \Rightarrow du=-dx \Rightarrow \int_0^3 {1\over \sqrt{3-x}}\,dx = \int_0^3 {1\over \sqrt u}\,du = \left. \left[ 2\sqrt u  \right] \right|_0^3 = \bbox[red, 2pt]{2\sqrt 3} \\\textbf{(3)} \lim_{x\to 3}{1\over (x-3)^2} 不存在 \Rightarrow \int_0^4 {1\over (x-3)^2}\,dx \bbox[red, 2pt]{不存在}$$

解答:$$\textbf{(1)} \lim_{n\to \infty} {1\over n}[({1\over n})^4 +({2\over n})^4 + \cdots +({n\over n})^4 ] =\lim_{n\to \infty} \sum_{k=1}^n{1\over n} ({k\over n})^4 =\bbox[red, 2pt]{\int_0^1 x^4\,dx} \\\textbf{(2)}  \lim_{n\to \infty}  {1+\sqrt 2+\sqrt 3+\cdots +\sqrt n\over \sqrt{n^3}} =\lim_{n\to \infty} \sum_{k=1}^n{\sqrt k\over \sqrt{n^3}} =\lim_{n\to \infty} \sum_{k=1}^n {1\over n}\sqrt{k\over n} = \bbox[red, 2pt]{\int_0^1 \sqrt x\,dx}$$

解答:$$\text{beta function }B(m,n)=\int_0^1 x^{m-1}(1-x)^{n-1}\,dx ={(m-1)!(n-1)! \over (m+n-1)!} \Rightarrow B(m,n)= B(n,m) \\\cases{I= \int_0^1 x^m(1-x)^n\,dx = B(m+1,n+1) \\ J=\int_0^1 (1-x)^m x^n\,d x=B(n+1,m+1)} \Rightarrow B(m+1,n+1) =B(n+1,m+1) \Rightarrow I=J \Rightarrow \bbox[red, 2pt]真$$

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解題僅供參考，碩士班歷年試題及詳解