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2022年11月7日 星期一

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(1)False:f(x)=|x|f(0)=0,f(0)(2)True:fx=cf(c)limxcf(x)f(c)xc=f(c)limxc(f(x)f(c))=limxc((xc)f(x)f(c)xc)=limxc(xc)limxcf(x)f(c)xc=0f(c)limxc(xc)=0limxcx=c(3)False:f(x)={x2sin1x,if x00,if x=0f(x),f(c)(4)True:f(x)={x,if x>0x,if x0{limx0+f(x)=limx0+x=0limx0f(x)=limx0(x)=0limx0f(x)=0=f(0)(5)False:f(0)=limh0f(0+h)f(0)h=limh0|0+h||0|)h=limh0|h|h=±1(6)False::f(x)=x3f(0)=0,f(0)
解答f(x)=exdkdxnf(x)|x=0=e0=1k=0f[k](0)k!(x0)k=k=01k!xk
解答f(x,y)=x2+y2+2{fx=2xfy=2y{fxx=2fxy=0fyy=2d=fxxfyyf2xy=4>0{fx=0fy=0x,y)=(0,0)Sf(0,0)=2
解答(a)1x1/n1x=1x1/n(1x1/n)(1+x1/n+x2/n++x(n1)/n)=11+x1/n+x2/n++x(n1)/nlimx1(1x1/2)(1x1/3)(1x1/n)(1x)n1=limx1(1x1/21x1x1/31x1x1/n1x)=limx1(11+x1/211+x1/3+x2/311+x1/n+x2/n++x(n1)/n)=123n=1n!(b)limx2cos(π/x)x2=limx2(cos(π/x))(x2)=limx2πx2sin(π/x)1=π4
解答(a){P(x,y)=y/(x2+y2)Q(x,y)=x/(x2+y2){Py=1/(x2+y2)+2y2/(x2+y2)2Qx=1/(x2+y2)2x2/(x2+y2)2Green's theorem R(C)CPdx+Qdy=

 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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