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2018年9月2日 星期日

103年專科學力鑑定考試--工程數學詳解


103年專科學校畢業程度自學進修學力鑑定考試
專業科目(一):工程數學 詳解

AB=[101020][123456]=[15262×32×4]=[4468](B)



L{f(t)}=F(s)L{eatf(t)}=F(sa)L{f(t)}=1essL{e2tf(t)}=F(s2)=1e(s2)s2(D)



ddy(x2+6xy+2y2+1)=6x+4y=ddx(3x2+4xy+4y2+1)(A)


{u=ij2kv=2i+jw=2uv=3j4kw|w|=35j45k(D)



2xy=3yy32xy=0a(x)=e32xdx=x32a(x)(y32xy)=0x32y32x52y=0(x32y)=0y=Cx32y(1)=4C=4y=4x32y(4)=423=32(D)



b3=1222f(x)sin3πx2dx=1220sin3πx2dx=12[23πcos3πx2]|20=12×(23π)×(2)=23π(C)


g(x)=ex+ex2g(x)=g(x)ex+ex21+ex+ex2g(x)=exex2g(x)=exex2=g(x)exex2ex,(B)


(a+3b)(a+3b)=|a|2+6ab+9|b|2=52+6ab+942=132ab=0=|a||b|cosθcosθ=0θ=90°(D)


C=A+B=[100020009]+[011104130]=[111124139]det(C)=18+3+42912=2(C)


ddy(xmenx(3y+4xy))=ddx(xmenx(2x))xmenx(3+4x)=xmenx(2m22nx){3=2m24=2n{m=52n=2m+n=92(A)


16y8y+y=016λ28λ+1=0(4λ1)2=0λ=14y=(A+Bx)e14xy=Be14x+14(A+Bx)e14x=(14A+B+14Bx)e14x{y(1)=44e=4e14y(1)=24e=2e14{(A+B)e14=4e14(14A+54B)e14=2e14{A+B=4A+5B=8{A=3B=1y=(3+x)e14xy(0)=3(C)


y=xm2x2y3xy3y=2x2m(m1)xm23xmxm13xm=0xm(2m(m1)3m3)=02m(m1)3m3=02m25m3=0(C)


y+4y7y=100sin3typ=Acos3t+Bsin3t(9Acos3t9Bsin3t)+4(3Asin3t+3Bcos3t)7(Acos3t+Bsin3t)=100sin3t(16A+12B)cos3t+(12A16B)sin3t=100sin3t{16A+12B=012A16B=100{4A=3B3A+4B+25=0{A=3B=4yp=3cos3t4sin3t(A)


a2=1πππf(x)cos(2x)dx=1π(0π|sinx|cos(2x)dx+π0|sinx|cos(2x)dx)=1π(0πsinxcos(2x)dx+π0sinxcos(2x)dx)=1π0πsinxcos(2x)dx+1ππ0sinxcos(2x)dx=1π[23sinxsin2x+13cosxcos2x]|0π+1π[23sinxsin2x+13cosxcos2x]|π0=1π×23+1π×(23)=43π(B)


{b=ka|b|=5{b1=79kb2=83kb21+b22=54981k2+649k2=2562581k2=25k2=2581625=5292252k=5925=95b1=7995=75(B)


a×b=(1,3,4)×(2,7,5)=(|3475|,|4152|,|1327|)=(43,13,1)=43i13jk(C)


A=[101102103104]det(A)=101×104102×103=(102.51.5)(102.5+1.5)(102.50.5)(102.5+0.5)=(102.521.52)(102.520.52)=0.521.52=2a=1042=52(A)


AλI=0[2+52321+56125][x1x2x3]=0[723246125][x1x2x3]=0{x1=x3x2=2x3[x1x2x3]=t[121],t0(A)


L1{2s+12s2+6s+13}=L1{2s+12(s+3)2+22}=L1{2(s+3)+3×2(s+3)2+22}=2L1{s+3(s+3)2+22}+3L1{2(s+3)2+22}=2e3tcos(2t)+3e3tsin(2t)=3e3t(2cos(2t)+3sin(2t))(B)


L1{1s2(s+1)}=L1{1s2+1s+11s}=L1{1s2}+L1{1s+1}L1{1s}=t+et1(D)

解題僅供參考

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