108年高師大轉學考-微積分詳解

國立高雄師範大學 108 學年度學士班轉學生招生考試試題

系所別:數學系及光通系二年級
科 目:微積分(全一頁)

解答： $$\mathbf{(1)}\;\bbox[red,2pt]{False}: f(x)=|x| \Rightarrow  f(0)=0,但f'(0)不存在\\ \mathbf{(2)}\;\bbox[red, 2pt]{True}: 假設f在x=c可微 \Rightarrow f'(c)存在\Rightarrow \lim_{x\to c}{f(x)-f(c)\over x-c}=f'(c) \\\Rightarrow \lim_{x\to c}  (f(x)-f(c))    =\lim_{x\to c} \left((x-c)\cdot {f(x)-f(c)\over x-c} \right) =\lim_{x\to c} (x-c)\cdot \lim_{x\to c} {f(x)-f(c)\over x-c} =0 \cdot f'(c)\\ \Rightarrow \lim_{x\to c} (x-c)  =0 \Rightarrow  \lim_{x\to c} x=c \\\mathbf{(3)}\;\bbox[red, 2pt]{False}:f(x)=\begin{cases} x^2\sin{1\over x},& \text{if }x\ne 0\\ 0,& \text{if } x=0\end{cases} \Rightarrow f(x)可微,但f'(c)不存在\\ \mathbf{(4)}\;\bbox[red, 2pt]{True}: f(x)=\begin{cases}x, &\text{if }x\gt 0\\ -x,& \text{if }x\le 0 \end{cases} \Rightarrow \cases{\lim_{x\to 0^+} f(x)= \lim_{x\to 0^+} x=0 \\ \lim_{x\to 0^-} f(x)= \lim_{x\to 0^-} (-x)=0 } \Rightarrow \lim_{x\to 0}f(x)=0 =f(0)\\ \mathbf{(5)}\; \bbox[red,2pt]{False}: f'(0)= \lim_{h\to 0}{f(0+h)-f(0)\over h} =\lim_{h\to 0}{|0+h|-|0|)\over h} =\lim_{h\to 0}{|h|\over h}= \pm 1不存在\\\mathbf{(6)}\;\bbox[red, 2pt]{False}: 也可能是反曲點，例:f(x)=x^3 \Rightarrow f'(0)=0,但f(0)不是極值$$

解答： $$f(x) =e^x \Rightarrow \left.{d^k \over d x^n} f(x) \right|_{x=0} =e^0=1 \Rightarrow \sum_{k=0}^\infty {f^{[k]}(0) \over k!}(x-0)^k = \bbox[red, 2pt]{\sum_{k=0}^\infty {1\over k!}x^k}$$

解答： $$f(x,y)=x^2+y^2+2 \Rightarrow \cases{f_x= 2x\\ f_y=2y} \Rightarrow \cases{f_{xx}=2\\ f_{xy}=0\\ f_{yy}=2} \Rightarrow d= f_{xx}f_{yy}-f_{xy}^2 =4\gt 0\\ 因此\cases{f_x=0\\ f_y=0} \Rightarrow （x,y)=(0,0)\in S \Rightarrow f(0,0)=2 為相對極小值也是最小值\\$$

解答： $$\mathbf{(a)}\;{1-x^{1/n} \over 1-x} ={1-x^{1/n} \over (1-x^{1/n} )(1+x^{1/n} +x^{2/n} +\cdots +x^{(n-1)/n})}  ={1\over  1+x^{1/n} +x^{2/n} +\cdots +x^{(n-1)/n}}\\ \Rightarrow \lim_{x\to 1} {(1-x^{1/2}) (1-x^{1/3})\cdots (1-x^{1/n}) \over (1-x)^{n-1}} = \lim_{x\to 1} \left({1-x^{1/2} \over 1-x} \cdot {1-x^{1/3} \over 1-x} \cdots {1-x^{1/n} \over 1-x} \right) \\ = \lim_{x\to 1} \left({1\over 1+x^{1/2}} \cdot {1\over 1+x^{1/3}+ x^{2/3}} \cdots {1\over  1+x^{1/n} +x^{2/n} +\cdots +x^{(n-1)/n}}\right) ={1\over 2\cdot 3\cdots n} = \bbox[red, 2pt]{1\over n!} \\\mathbf{(b)}\; \lim_{x\to 2} {\cos(\pi/x) \over x-2} = \lim_{x\to 2} {(\cos(\pi/x))' \over (x-2)'} =\lim_{x\to 2} {{\pi\over x^2}\sin(\pi/x) \over 1} = \bbox[red, 2pt]{ \pi\over 4}$$

解答： $$\mathbf{(a)}\;令\cases{P(x,y)= -y/(x^2+y^2)\\ Q(x,y)= x/(x^2+y^2)} \Rightarrow \cases{P_y= -1/(x^2+y^2)+ 2y^2/(x^2+y^2)^2\\ Q_x= 1/(x^2+y^2)-2x^2/(x^2+y^2)^2} \\ 依\text{Green's theorem }定理，若R(封閉曲線C所圍區域)不含原點，則\oint_C P\,dx+Q\,dy =\\ \iint_R Q_x-P_y\,dA = \iint_R \left({2\over x^2+y^2} -{2(x^2+y^2)\over (x^2+y^2)^2}\right)dA =\iint_R 0\,dA=\bbox[red, 2pt]0\\\mathbf{ (b)}\;取C'為一圓路徑，其圓心為原點且半徑為a，路徑方向為順時鐘，則\oint_{C-C'} Pdy+Qdx =0\\ \Rightarrow \oint_C Pdy+Qdx = \oint_{C'} Pdy+Qdx = \int_0^{2\pi} {-a\cos\theta\over a^2}(-a\sin \theta)+ {a\cos\theta \over a^2}\cdot a\cos \theta \,d\theta \\ \int_0^{2\pi}1\,d\theta= \bbox[red,2pt]{2\pi} \left(取\cases{x=a\cos\theta\\ y=a\sin \theta}  \Rightarrow \cases{dx= -a\sin\theta\, d\theta\\ dy= a\sin\theta \,d\theta \\ x^2+y^2=a^2} \right)$$

 解答：

$$\cases{u= xy\\ v=x^2-y^2} \Rightarrow \begin{vmatrix} u_x &u_y\\ v_x & v_y\end{vmatrix}= \begin{vmatrix} y & x\\ 2x & -2y\end{vmatrix} =-2(x^2+y^2) \Rightarrow \begin{vmatrix} x_u &x_v\\ y_u & y_v \end{vmatrix} =1/\begin{vmatrix} u_x &u_y\\ v_x & v_y\end{vmatrix} =-{1\over 2(x^2+y^2)} \\ 又(x^2+y^2)^2 = (x^2-y^2)^2 +4x^2y^2 =v^2+ 4u^2 \Rightarrow x^2+y^2 = \sqrt{v^2+4u^2} \\ \Rightarrow \begin{vmatrix} x_u &x_v\\ y_u & y_v \end{vmatrix} =-{1\over 2 \sqrt{v^2+4u^2} } \Rightarrow \iint_R (x^4-y^4)e^{xy}dA = \iint_R (x^2+y^2) (x^2-y^2)e^{xy}dA \\ =\int_1^4 \int_{\sqrt{81-v^2}/2}^{\sqrt{ 256-v^2}/2} \sqrt{v^2+4u^2}\cdot ve^{u}\cdot {-1\over 2\sqrt{v^2+4u^2}} \;dudv=\int_1^4 \int_{\sqrt{81-v^2}/2}^{\sqrt{ 256-v^2}/2} -{1\over 2} ve^u\;dudv \\ =\int_1^4 -{1\over 2}v\left( e^{\sqrt{256-v^2}/2} -e^{\sqrt{81-v^2}/2}\right)dv =\int_1^4 {1\over 2}ve^{\sqrt{81-v^2}/2} \,dv -\int_1^4{1\over 2}ve^{\sqrt{256-v^2}/2}\,dv \\ =\int_{80}^{65} -{1\over 4}e^{\sqrt s}\,ds +\int_{255}^{240} {1\over 4}e^{\sqrt t}\,dt,其中\cases{s=81-v^2\\ t=256-v^2} \Rightarrow \cases{ds =-2vdv\\ dt =-2vdv} \\ =-{1\over 4}\left. \left[ 2e^{\sqrt s}(\sqrt s-1)\right]\right|_{80}^{65} +{1\over 4} \left.\left[2e^{\sqrt t}(\sqrt t-1) \right]\right|_{255}^{240}\\ =\bbox[red, 2pt]{{1\over 2}\left(e^{\sqrt{80}} (\sqrt{80}-1) +e^{\sqrt{240}}( \sqrt{240}-1) \right) -{1\over 2}\left(e^{\sqrt{65}}(\sqrt{65}-1) +e^{\sqrt{255}} (\sqrt{255}-1) \right) }$$

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