115年台聯大轉學考A3A4A5-微積分詳解

 台灣聯合大學系統 115 學年度學士班轉學生考試

類組別： A3/A4/A6

甲) 填充題：共 10 題，每題 8 分

解答:$$\text{Let the equation of the tangent line be }y=mx+c \\ f(x)=x^2=mx+c \Rightarrow x^2-mx-c=0 \Rightarrow \Delta=0 \Rightarrow m^2+4c=0 \Rightarrow c=-{m^2 \over 4} \\ g(x)={1\over 2}x^2-2x+7=mx+c \Rightarrow {1\over 2}x^2-(m+2)x+7-c=0 \Rightarrow \Delta= (m+2)^2-2(7-c)=0 \\ \Rightarrow m^2+4m-10+2c=0 \Rightarrow m^2+4m-10+2\cdot \left( -{m^2\over 4} \right)=0 \Rightarrow m^2+8m-20=0\\ \Rightarrow (m+10)(m-2)=0 \Rightarrow \cases{m=2 \Rightarrow c=-1 \Rightarrow \bbox[red, 2pt]{y=2x-1}\\ m=-10 \Rightarrow c=-25 \Rightarrow \bbox[red, 2pt]{y=-10x-25}}$$

解答:$$x^2y+\sin(\pi xy)=y^3+3 \Rightarrow 2xy+x^2y'+ (\pi y+ \pi xy')\cos(\pi xy)=3y^2y' \\ \Rightarrow (x^2+\pi x\cos(\pi xy)-3y^2)y'=-2xy-\pi y\cos(\pi xy) \Rightarrow y'=-{2xy+\pi y\cos(\pi xy) \over x^2+\pi x \cos(\pi xy)-3y^2} \\ \Rightarrow y'(2,1)=-{4+\pi \cos2\pi\over 4+ 2\pi\cos (2\pi)-3} = \bbox[red, 2pt]{-{4+\pi\over 1+ 2\pi}}$$

解答:$$P(x)= x^{27}+3x^{26}-x^2+23=x^{27} \left( 1+{3\over x}-{1\over x^{25}}+{23\over x^{27}} \right) \\\Rightarrow  P(x)^{1/27} =x  \left( 1+{3\over x}-{1\over x^{25}}+{23\over x^{27}} \right) ^{1/27}\\ \text{Let }u={1\over x}. \text{ As }x\to \infty , u\to 0^+. \text{ Then } \\L=\lim_{x\to \infty} P(x)^{1/27} -x =\lim_{u\to 0^+} {(1+3u-u^{25}+23u^{27})^{1/27}-1\over u} \\=\lim_{u\to 0^+} {((1+3u-u^{25}+23u^{27})^{1/27}-1)'\over (u)'} \\=\lim_{u\to 0^+} {1\over 27}(1+3u-u^{25}+23u^{27})^{-26/27} (3-25u^{24}+ 23\cdot 27u^{26}) = \bbox[red, 2pt]{1\over 9}$$

解答:$$\int_0^1 {1\over 1+e^x}\,dx =\int_0^1 \left( 1-{e^x\over 1+e^x} \right)\,dx = \left. \left[ x-\ln(1+e^x) \right] \right|_0^1=1-\ln(1+e)+\ln 2= \bbox[red, 2pt]{1+\ln{2\over 1+e}}$$

解答:$${d\over dx} \left( \tan^{-1}x \right)= {d\over dx} \left( \int_0^{x^2} {f(t) \over t^3}\,dt  \right) \Rightarrow {1\over 1+x^2}=2x\cdot {f(x^2) \over x^6} \Rightarrow f(x^2)={x^5\over 2(1+x^2)} \\ \Rightarrow f(4)=f(2^2)={2^5\over 2(1+2^2)}= \bbox[red, 2pt]{16\over 5}$$

解答:$$f(x)=x+ \sin x=0 \Rightarrow x=0 \Rightarrow f'(x)=1+\cos x \Rightarrow f'(0)=2 \\ \Rightarrow (f^{-1})'(0)={1\over f'(0)}= \bbox[red, 2pt]{1\over 2}$$

解答:$$a_n={(\ln x)^n\over n\cdot 2^n} \Rightarrow \lim_{n\to \infty} \left|{a_{n+1}\over a_n} \right| = \lim_{n\to \infty} \left|{(\ln x)^{n+1} \over (n+1)\cdot 2^{n+1}} \cdot{n\cdot 2^n\over (\ln x)^n} \right|= \left| {\ln x\over 2}\right| \lim_{n\to \infty}{n\over n+1} \\=\left| {\ln x\over 2}\right| \lt 1 \Rightarrow -2\lt \ln x\lt 2 \Rightarrow e^{-2}\lt x\lt e^2\\ x=e^2 \Rightarrow \sum_{n= 1}^\infty{(\ln x)^n \over n\cdot 2^n} = \sum_{n=1}^\infty{1\over n}.  \text{ This is the harmonic series, which diverges.}\\ x=e^{-2} \Rightarrow \sum_{n= 1}^\infty{(\ln x)^n \over n\cdot 2^n} = \sum_{n= 1}^\infty {(-1)^n\over n}. \text{ It converges by the Alternating Series Test } \\ \Rightarrow \text{The series converges for all real values of }\bbox[red, 2pt]{x \in [e^{-2},e^2)}$$

解答:$$\text{Change the order of inntegration: }I= \int_0^4 \int_{\sqrt x}^2 {x\over y^5+1}\,dy dx = \int_0^2  \int_0^{y^2} {x\over y^5+1}\, dx\,dy \\= \int_0^2  {1\over y^5+1}\left. \left[ {x^2\over 2} \right] \right|_0^{y^2} \,dy = \int_0^2 {y^4\over 2(y^5+1)}\,dy = \left. \left[{1\over 10} \ln(y^5+1) \right] \right|_0^2 =\bbox[red, 2pt]{\ln 33 \over 10}   $$

解答:$$f(x,y)=x^3-3x+xy^2 \Rightarrow \cases{f_x=3x^2-3+y^2\\ f_y=2xy} \Rightarrow \cases{f_{xx}= 6x\\ f_{xy} = 2y\\ f_{yy} =2x} \\ \Rightarrow \text{The discriminant  is given by }D=f_{xx}f_{yy}-(f_{xy})^2=12x^2-4y^2 \\ \cases{f_x=0\\ f_y=0} \Rightarrow \cases{2x^2-3+y^2=0\\ 2xy=0} \Rightarrow \cases{x=0 \Rightarrow y^2=3 \Rightarrow y=\pm \sqrt 3\\ y=0 \Rightarrow 3x^2-3=0 \Rightarrow x= \pm 1} \\ \Rightarrow \cases{D(0,\pm\sqrt 3) =-12\lt 0 \Rightarrow \text{ saddle points} \\D(\pm 1,0)=12 \gt 0 \Rightarrow \cases{f_{xx}(1,0)=6\gt 0 \Rightarrow (1,0): \text{local minimum} \\f_{xx}(-1,0)=-6\lt 0 \Rightarrow (-1,0):\text{ local maxmimum}}} \\ \Rightarrow \bbox[red, 2pt]{\cases{(0, \pm 3): \text{saddle points}\\ (1,0): \text{ local minnimum} \\ (-1,0): \text{local maximum}}}$$

解答:$$\cases{M(x,y,z)=xyz^2 \\ g(x,y,z)= x^2+y^2+z^2-4} \Rightarrow \cases{M_x= \lambda g_x\\ M_y= \lambda g_y\\ M_z=\lambda g_z\\ g=0} \Rightarrow \cases{yz^2 =2\lambda x\\ xz^2=2\lambda y\\ 2xyz =2\lambda z\\ x^2+y^2+z^2=4} \Rightarrow \cases{yz^2/xz^2=2\lambda x/2\lambda y\\ 2z(\lambda -xy)=0} \\ \Rightarrow \cases{z=0 \Rightarrow M=0\\ x^2=y^2\\ \lambda=xy \Rightarrow z^2=2x^2=2y^2} \Rightarrow x^2=1 \Rightarrow (x,y,z)=(\pm 1,\pm 1,\pm \sqrt 2) \Rightarrow \cases{M(1,1,\pm \sqrt 2)=2\\ M(-1,-1,\pm \sqrt 2)=2 \\ M(1,-1,\pm\sqrt 2)=-2\\M(-1,1,\pm \sqrt 2)=-2} \\ \Rightarrow \text{The points where the magnetic field is stronget are: }\bbox[red, 2pt]{(1,1,\pm \sqrt 2), (-1,-1,\pm \sqrt 2)}$$

乙) 計算題：共 2 題，每題 10 分

解答:$$取\cases{u= x+y\\ v=x-y} \Rightarrow \cases{x=(u+v)/2\\ y=(u-v)/2} \Rightarrow |J|= {\partial(x,y) \over \partial(u,v)} = \begin{Vmatrix} 1/2& 1/2\\1/2&-1/2 \end{Vmatrix} =\left|-{1\over 2}\right|={1\over 2} \\ \Rightarrow I= \iint_R (x+y)e^{x^2-y^2}\,dA=  \int_1^3 \int_0^1 ue^{uv} {1\over 2}\,dvdu = \int_1^3 \left. \left[{1\over 2} e^{uv}  \right] \right|_{0}^1\,du ={1\over 2}\int_1^3 \left( e^u-1 \right)\,du \\={1\over 2} \left. \left[ e^u-u \right] \right|_1^3 = \bbox[red, 2pt]{{1\over 2}(e^3-e -2)}$$

解答:$$f(x,y)=x^2+2y^2-y \Rightarrow \cases{f_x=2x =0 \Rightarrow x=0\\ f_y=4y-1=0 \Rightarrow y=1/4} \Rightarrow (0,{1\over 4})\in D \Rightarrow f(0,{1\over 4})=-{1\over 8} \\\text{Boundary }B_1:y=0\text{ and }-1\le x\le 1 \Rightarrow f(x,0)=x^2 \Rightarrow \cases{f(0,0)=0\\ f(\pm 1,0)=1} \\ \text{Boundary }B_2:x^2+y^2=1 \text{ and } y\ge 0 \Rightarrow x^2=1-y^2 \Rightarrow g(y)= (1-y^2)+2y^2-y=y^2-y+1 \\\qquad =(y-1/2)^2+3/4 \Rightarrow \cases{g(1/2)=3/4\\ g(0)=1\\ g(1)=1} \\ \Rightarrow \bbox[red, 2pt]{\text{absolute max: 1, absolute min: }-{1\over 8}}$$

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