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2023年9月15日 星期五

112年台聯大轉學考-微積分A3詳解

 台灣聯合大學系統112學年度學士班轉學生考試

科目:微積分
類組別:A3/A4/A6
甲、填充題:共8題,每題8分

解答:L=(cos(2x))2/xlnL=2lncos(2x)xlimx0+lnL=limx0+2lncos(2x)x=limx0+ddx(2lncos(2x))ddxx=limx0+2sin(2x)/xcos(2x)=limx0+2tan(2x)x=limx0+ddx(2tan(2x))ddxx=limx0+2sec2(2x)/x1/2x=limx0+4sec2(2x)=4limx0+lnL=4limx0+L=e4=1e4
解答:g(x)=tan1(x2)+x21sec(t1)dtg(x)=2x1+x4+2xsec(x21){g(1)=tan1(1)+0=π/4g(1)=121=3f(x)=g(x)eg(x)f(1)=g(1)eg(1)=3eπ/4
解答:{x(t)=ln(sect+tant)sinty(t)=cost{x(t)=sectcosty(t)=sint(x(t))2+(y(t))2=sec2t2+cos2+sin2t=sec2t1 =tan2t=π/30(x(t))2+(y(t))2dt=π/30tantdt=[ln(cost)]|π/30=ln12=ln2
解答:e1(ln(x))2πdx=[π(x(lnx)22xln(x)+2x)]|e1=π(e2)
解答:r(t)=(cos3t,sin3t,3t)r(t)=(3sin3t,3cos3t,3){r(π/3)=(1,0,π)r(π/3)=(0,3,3)f(x,y,z)=xyzf=(fx,fy,fz)=(yz,xz,xy)=(3tsin3t,3tcos3t,(sin6t)/2))f(π/3)=(0,π,0)f(π/3)r(π/3)r(π/3)=(0,π,0)(0,12,12)=2π2
解答:8023x1y4+1dydx=20y301y4+1dxdy=20y3y4+1dy=[14ln(y4+1)]|20=14ln17
解答:

,{x=rcosθy=rsinθx2+(y1)2=1r2=2rsinθr=2sinθRx2+y2dA=2π/2022sinθr2drdθ=2π/20[13r3]|22sinθdθ=163π/20(1sin3θ)dθ=163[θ112(cos3θ9cosθ)]|π/20=163(π2+112(8))=83π329
解答:41dx|x|=10dxx+40dxx=2+4=6

乙、計算、證明題: 共3題,每題12分,共36分

解答:(a)bn=lnnnlnnlimnbn=limnddnlnnddn(nlnn)=limn1n1=0 ddnbn=1lnn(nlnn)2<0,n3;: lnnnlnn>1nlnn>1n調,,(b)limn|an+1an|=limn|(x+4)n+1(n+1)3n+1n3n(x+4)n|=limn|x+43nn+1|=|x+43|<17<x<1x=1n=1(x+4)nn3n=n=11n調x=7n=1(x+4)nn3n=n=1(1)n1n{limn1n=01n>1n+1, 7x<1
解答:(a)y=mxlim(x,y)(0,0)f(x,y)=limx0x2mxx3+m3x3=limx0m1+m3=m1+m3f(0,0)f(0,0), (b)fx=2xyx3+y3x2y3x2(x3+y3)2=2xy4x4y(x3+y3)2(c)fx=limh0f(0+h,0)f(0,0)h=limh00h=0
解答:f(x,y)=x2+3y2+2y{fx=2xfy=6y+2{fxx=2fxy=0fyy=6d(x,y)=fxxfyyf2xy=12>0{fx=0fy=0(x,y)=(0,1/3)f(0,1/3)=13  Lagrange :{f(x,y)=x2+3y2+2yg(x,y)=x2+y21{fx=λgxfy=λgyg=0{2x=λ(2x)6y+2=λ(2y)x2+y2=1{x=0y=±1λ=1y=1/2x=±3/2{f(0,1)=5f(0,1)=1f(±3/2,1/2)=1/255,13






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解題僅供參考,其他歷年試題及詳解


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