2022年8月25日 星期四

111年台聯大轉學考-微積分A3A4A6詳解

台灣聯合大學系統111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,共64分

解答limn(1nn+1+1nn+2++1nn+n)=limnnk=11nn+k=limnnk=11nnn+k=limnnk=11n11+kn=1011+xdx=211udu,u=x+1=[2u]|21=222
解答{x=ρsinϕcosθy=ρsinϕsinθz=ρcosϕ=2π0π/3021/cosϕρ2sinϕdρdϕdθ=2π0π/30sinϕ((8cos3ϕ1)sinϕ3cos3ϕ)dϕdθ=2π056dθ=53π
解答4x23x+24x24x+3dx=(1+x14x24x+3)dx=x+(188x44x24x+31214x24x+3)dx=x+18ln|4x24x+3|181x2x+3/4dx=x+18ln|4x24x+3|181(x1/2)2+1/2dx=x+18ln|4x24x+3|141(2(x1/2))2+1dx=x+18ln|4x24x+3|142tan1(2(x1/2))+C=x+18(ln(4x24x+3)2tan1(2(x1/2))+C):4x24x+3=(2x1)2+2>0,xR
解答101xsin(y2)dydx=10y0sin(y2)dxdy=10ysin(y2)dy=[12cos(y2)]|10=12cos1+12=12(12sin212)+12=sin212
解答{x=rcosθy=rsinθx2+y2=4{0θ2π0r2;y+2z=2z=2y2{xz=0yz=1/2=2π02002+(1/2)2+1rdrdθ=522π020rdrdθ=52π01,dθ=25π
解答u=3xdu=3dx31cos3xxdx=93cosuu/313du=93cosuudu=F(9)F(3)
解答y=3xx2y=32x(a,y(a))=(a,3aa2)=32ay=(32a)(xa)+3aa2{A(0,a2)B(a22a3,0)OAB=f(a)=12¯OA¯OB=12a2a22a3=a44a6f(a)=4a34a64a4(4a6)2f(a)=04a3(3a6)(4a6)2=0a=2(a=0=(0,0))f(2)=1686=8
解答Lagrange's 算子求極值:{f(x,y,z)=4x2+y2+4z216T(x,y,z)=8x2+4yz16z+600{Tx=λfxTy=λfyTz=λfzf=0{16x=λ8x4z=λ2y4y16=λ8z{(λ2)x=0(1)2z=λy(2)y4=2λz(3)(1):λ=2x=0Cases I: λ=2{z=yy4=4zy=z=43f=0x=±43Cases II: x=0f=0y2+4z2=16(4)(2)(3)2zy4=y2z4z2=y24y(4)y2+y24y=16(y4)(y+2)=0{y=4z=0y=2z=±3Cases I Cases II {A(4/3,4/3,4/3)B(4/3,4/3,4/3)C(0,4,0)D(0,2,3)E(0,2,3)T(x,y,z){T(A)=600T(B)=1928/3=642.7()T(C)=600T(D)=600243T(E)=600+243641.57:(43,43,43)

乙、計算、證明題:共3題,每大題12分,共36分


解答f(x)={e1/x2,if x00,if x=0f(0)=f(0)=f(0)=0f(x)f(0)+f(0)x+12f(0)x2=0
解答ba0bnan02bnan+bnbnn2bnna2+b2nbn21/nbna2+b2blimn21/nb=bblimnna2+b2b:limnna2+b2=bb

解答
a.{y=0lim(x,y)(1,0)xey1xey1+y=limx1x1x1=1y=lnxlim(x,y)(1,0)xey1xey1+y=limx1x21x21+lnx=limx12x2x+1/x=2/3b.{y=xlim(x,y)(0,0)sin(xy)|x|+|y|=limx0sin(0)2|x|=0y=xlim(x,y)(0,0)sin(xy)|x|+|y|=limx0+sin(2x)2x=1

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3 則留言:

  1. 第三題錯了,答案1/4根號2,不是根號2

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    1. 前面有個1/8, 所以變成根號2, 答案沒錯!!

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  2. 乙部分第2題,開n次方根號後為何會變成a平方+b平方?

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