2020年11月10日 星期二

109年專科學力鑑定-工程數學詳解

教育部109年自學進修專科學校學力鑑定考試
專業科目(一):工程數學
(A)×:y4y=0(B)×:y3(C)×:y2(D):(D)

f(t)=12cos(2t)3e2tL{f(t)}=12L{cos(2t)}3L{e2t}=12ss2+2231s+2=s2s2+83s+2(A)

|231a01513|=3a159a+2=38a=16a=2(A)

{3x+2y+z=62x+3y+z=7x+2y+3z=2{9x+6y+3z=18(1)6x+9y+3z=21(2)x+2y+3z=2(3){(1)(2)(2)(3){3x3y=35x+7y=19{x=y1(4)5x+7y=19(5),(4)(5)5y5+7y=1912y=24y=2(4)x=21=1x+y=1+2=3(C)

y=exy=ex+C;y(0)=1e0+C=1C=0y=exy(1)=e(B)

{u=(a,2,1)v=(1,b,2)w=(2,1,c)u+2vw=(a,2,1)+(2,2b,4)+(2,1,c)=(a,2b+1,3c)=(1,5,2){a=12b+1=53c=2{a=1b=2c=1a+2b=1+4=5(D)

{s=(2,3,1)t=(a,2,5)st=2a+65=5a=2(C)

{u=(2,a,5)v=(1,0,2)u×v=(|a502|,|5221|,|2a10|)=(2a,9,a)=(6,9,3)a=3(C)

(mx3y+3xy2)dx+(x4+nx2y)dy=0{M(x,y)=mx3y+3xy2N(x,y)=x4+nx2yMy=Nxmx3+6xy=4x3+2nxy{m=46=2nn=33m+2n=12+6=18(??)
公告的答案為(B)=>最後送分

A=[2354]det(AλI)=0|2λ354λ|=0λ26λ+815=0(λ7)(λ+1)=0λ=7,17(1)=8(D)

(B)0(B)

(A)×:f1(1)=0f(1)=2(B)×:f2(π/2)=sin(π/2)=1sin(π/2)=0(C)×:f3(π/2)=π2/4f3(π/2)=π2/4(D):f4(x)=f4(x)=x2+1(D)


{A=[a124]B=[2103]C=[1352](A+2B)C=([a124]+[4206])[1352]=[a+43210][1352]=[11a3a+184826]=[8274826]a=3(A)

f(x)=xf(x)an=0bn=1πππxsin(nx)dx=1π[1nxcos(nx)+1n2sin(nx)]|ππ=1π(2πncos(nπ))=2n(1)n+1f(x)=n=1bnsin(nx)=n=12n(1)n+1sin(nx)(A)

2s5(s+2)(s1)=as+2+bs1=a(s1)+b(s+2)(s+2)(s1)=(a+b)sa+2b(s+2)(s1){a+b=2a+2b=5{a=3b=1L1{2s5(s+2)(s1)}=L1{3s+21s1}=3L1{1s+2}L1{1s1}=3e2tet(B)

L1{2s+6s2+9}=L1{(2)ss2+32+23s2+32}=2L1{ss2+32}+2L1{3s2+32}=2cos3t+2sin3t(A)

:y2y+5y=0λ22λ+5=0λ=1±2iyh=ex(Acos(2x)+Bsin(2x))yp=Cx+Dyy=yh+yp=Aexcos(2x)+Bexsin(2x)+Cx+D(B)

:y+4y=0yh=Acos(2x)+Bsin(2x)y=yh+yp=Acos(2x)+Bsin(2x)+14x(C)

H(t)=t0e20τ(tτ)19dτL1{H(t)}=L1{e20t}L1{t19}=1s20×19!s20(C)

y=xry=rxr1y=r(r1)xr2x2y2xy+2y=0r(r1)xr2rxr+2xr=0xr(r2r2r+2)=0r23r+2=0(r2)(r1)=0y=C1x+C2x2(D)


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