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2021年9月29日 星期三

110年鐵路特考-工程數學詳解

110年特種考試交通事業鐵路人員考試試題

考 試 別: 鐵路人員考試
等 別: 高員三級考試
類 科 別: 電力工程、 電子工程
科 目: 工程數學
甲、申論題部分:(50分)

解答y+2y+2y=f(t)L{y}+2L{y}+2L{y}=L{f(t)}s2Y(s)sy(0)y(0)+2(sY(s)y(0))+2Y(s)=L{f(t)}Y(s)(s2+2s+2)1=L{f(t)}={2/s,if 0<tπ0, if t>π{t>πY(s)=1(s+1)2+1y(t)=etsin(t)0<tπY(s)=s+2s(s2+2s+2)=1ss+1(s+1)2+1y(t)=1etcos(t)y(t)={etsin(t),t>π1etcos(t),0<tπ
解答
(一)A=[544121112445]det(AλI)=0λ3+λ25λ+3=0(λ1)2(λ+3)=0A31,1,3(二)λ3+λ25λ+3=0A3+A25A+3I=0{A3+A2=5A3I(1)A5+A45A3+3A2=0A5+A4=5A33A2(2)A5+A44A3+4A24A+4I=5A33A24A3+4A24A+4I((2))=A3+A24A+4I=5A3I4A+4I((1))=A+I=[644121012446]
解答Γ(t)=3eitΓ(t)=3ieitdtΓIm(z)dz=π/203eiteit2i3ieitdt=92π/20(e2it1)dt=92[12ie2itt]|π/20=92(iπ2)=92i94π
解答
(一)F{f1(t)}=e5|t|ejωtdt=0e5tejωtdt+0e5tejωtdt=0e(5jω)tdt+0e(5+jω)tdt=[15jωe(5jω)t]|0+[1(5+jω)e(5+jω)t]|0=15jω+15+jω=5+jω25+ω2+5jω25+ω2=10ω2+25(二)cos(100t)=12(ej100t+ej100t)F{f2(t)}=(cos100t)e5|t|ejωtdt0(cos100t)e5tejωtdt+0(cos100t)e5tejωtdt=120e(5jω+j100)t+e(5jωj100)tdt+120e(5jω+j100)t+e(5jωj100)tdt=12[15jω+j100e(5jω+j100)t+15jωj100e(5jωj100)t]|0+12[15jω+j100e(5jω+j100)t+15jωj100e(5jωj100)t]|0=12[15jω+j100+15jωj100]+12[15jω+j100+15jωj100]=12[15j(ω100)+15j(ω+100)]+12[15+j(ω100)+15+j(ω+100)]=12[5+j(ω100)25+(ω100)2+5+j(ω+100)25+(ω+100)2+5j(ω100)25+(ω100)2+5j(ω+100)25+(ω+100)2]=525+(ω100)2+525+(ω+100)2
解答z(x,y)=35002x23y2{zx|(3,2)=4x|(3,2)=12zy|(3,2)=6y|(3,2)=12(12,12)(C)
解答203041560=|6012|=72(B)
解答u1=uv|v|2v=510[013]=[00.51.5]u2=uu1=[121][00.51.5]=[11.50.5](B)
解答A=[101120220103]ATA=[101120220103][120001120123]=[50550103505553514]r1+r3,r13r2+r4[5055010300000000]rank =2(B)
解答X[1000211100993055]det(XλI)=0λ4+4λ3λ24λ=λ(λ3+4λ2λ4)=λ(λ1)(λ2+5λ+4)=λ(λ1)(λ+1)(λ+4)=0λ=0,1,1,4(B)
解答(A)×:sin(xy)(B)×:{T(1,1,1,1,1)=(1,2,1,0,1)T(2,2,2,2,2)=(2,4,2,0,1)T(2,2,2,2,2)2T(1,1,1,1,1)(C)×:xy(D):(D)
解答A=[100010000]eA=[e000e0001]rank(eA)=3(C)
解答y=n=0an(x1)n=a0+a1(x1)+a2(x1)2+a3(x1)3{x2y=a0x2+a1x2(x1)+a2x2(x1)2+a3x2(x1)3+y=a1+2a2(x1)+3a3(x1)2+y(1)=1a0=1(y+x2y)|x=1=cos(π)=1a0+a1=1a1=2(y+x2y)|x=0=y(0)=cos(0)=1a12a2+3a34a4+=1(y+x2y)|x=2=y(2)+4y=cos(2π)=1n=0nan+n=04an+(C)
解答y1=C,Cdy1dx=13xy21+1xy16x0=C23x+Cx6xC2+3C18=0(C+6)(C3)=0C=3y=y1+1u=3+1u(B)
解答y(x)=(a+bx)e2x+ce2x+d{y(x)=(2a+b+2bx)e2x2ce2xy(x)=(4a+4b+4bx)e2x+4ce2xy2y=2be2x+8ce2x=6e2x4e2x{b=3c=1/2{y(0)=1y(0)=6{a+c+d=1d=3/22a+b2c=6a=1{a=1b=3c=1/2d=3/2(A)
解答f(t)δ(tT)dt=f(T)L{f(t)}=0cos(πt)δ(t1)estdt=cos(π)es=es(A)
解答L{f(t)}=L{Aetcos(2t)+Betsin(2t)}=AL{etcos(2t)}+BL{etsin(2t)}=As+1(s+1)2+4+B2(s+1)2+4=As+A+2Bs2+2s+5=2s+5s2+as+b{a=2b=5a/b=2/5=0.4(D)
解答f(t)={0for 0t<43for t4f(t)=3H(t4)y+4y=3H(t4)L{y}=s2Y(s)sy(0)y(0)=s2Y(s)sL{y+4y}=L{f(t)}(s2+4)Y(s)s=31se4sY(s)=ss2+4+3s(s2+4)e4s=ss2+4+34(1sss2+4)e4sy(t)=cos(2t)+34(1cos(2(t4)))H(t4)=cos(at)+b(ccos(d(t4)))H(t4){a=2b=3/43/2c=1d=2(B)
解答f(t)=5[H(t3)H(t11)]=x(t7),x(t)=5rect(t/8)F{x(t)}=5ω2sin(ω82)=10ωsin(4ω)F{x(t7)}=10ωej7ωsin(4ω){a=10b=1c=77d=4(C)
解答Res(f,0)=lim
解答\cases{f(z)=\sin(z)/(z^2(z^2+1))\\ C:|z-0.5i|=1} \Rightarrow z=0,z=i 皆在C內部\\ \Rightarrow \cases{Res(f,0)=\lim_{z\to 0}{d\over dz}z^2f(z)  =\lim_{z\to 0}{d\over dz} \sin(z)/(z^2+1)=1\\ Res(f,i)= \lim_{z\to i}(z-i)f(z) =\lim_{z\to i} \sin(z)/(z^2(z+i)) =i\sin(i)/2} \\ \Rightarrow \oint_C f(z)\;dz=2\pi i(Res(f,0)+Res(f,i)) =2\pi i(1+i\sin(i)/2)= 2\pi i-\pi \sin(i),故選\bbox[red,2pt]{(A)}
解答C_n=\left( {2i\over 5+i}\right)^n \Rightarrow {1\over R}=\left| {C_{n+1}\over C_n}\right| =\left| {2i\over 5+i}\right| =\left| {2+10i\over 26}\right| = \sqrt{\left({1\over 13} \right)^2 +\left({5\over 13} \right)^2} ={\sqrt{26}\over 13}\\ \Rightarrow 收斂半徑R={13\over \sqrt{26}} ={\sqrt{26}\over 2},故選\bbox[red,2pt]{(B)}
解答\cases{E(X)= \int_{-a}^a {x\over 2a} \;dx = 0\\ E(X^2) =\int_{-a}^a {x^2\over 2a} \;dx={a^2\over 3}} \Rightarrow Var(X)= E(X^2)-(E(X))^2 = {a^2\over 3}=4 \Rightarrow a=2\sqrt 3,故選\bbox[red,2pt]{(A)}
解答Z={X-\mu\over \sigma} ={X-1\over 2},故選\bbox[red,2pt]{(C)}
解答X\sim B(n=4,p=1/2)\\(A)\bigcirc: P(X=3)=C^4_3\cdot {1\over 16}=0.25\\ (B)\bigcirc: P(X=2)=C^4_2\cdot {1\over 16}={3\over 8}=0.375\\ (C)\bigcirc: E(X)=np=4\cdot {1\over 2}=2\\ (D)\times: Var(X)= np(1-p)= 4\cdot {1\over 2} \cdot {1\over 2}=1 \ne 1.5\\,故選\bbox[red,2pt]{(D)}

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