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:
(一)A=[5−4412−11124−45]⇒det(A−λI)=0⇒λ3+λ2−5λ+3=0⇒(λ−1)2(λ+3)=0⇒A的3個特徵值為1,1,−3(二)λ3+λ2−5λ+3=0⇒A3+A2−5A+3I=0⇒{A3+A2=5A−3I⋯(1)A5+A4−5A3+3A2=0⇒A5+A4=5A3−3A2⋯(2)因此欲求之A5+A4−4A3+4A2−4A+4I=5A3−3A2−4A3+4A2−4A+4I(式(2)代入)=A3+A2−4A+4I=5A−3I−4A+4I(式(1)代入)=A+I=[6−4412−10124−46]
解答:Γ(t)=3eit⇒Γ′(t)=3ieitdt⇒∫ΓIm(z)dz=∫π/203⋅eit−e−it2i⋅3ieitdt=92∫π/20(e2it−1)dt=92[12ie2it−t]|π/20=92(i−π2)=92i−94π
解答:
解答:Γ(t)=3eit⇒Γ′(t)=3ieitdt⇒∫ΓIm(z)dz=∫π/203⋅eit−e−it2i⋅3ieitdt=92∫π/20(e2it−1)dt=92[12ie2it−t]|π/20=92(i−π2)=92i−94π
解答:
(一)F{f1(t)}=∫∞−∞e−5|t|⋅e−jωtdt=∫0−∞e5t⋅e−jωtdt+∫∞0e−5t⋅e−jωtdt=∫0−∞e(5−jω)tdt+∫∞0e−(5+jω)tdt=[15−jωe(5−jω)t]|0−∞+[1−(5+jω)e−(5+jω)t]|∞0=15−jω+15+jω=5+jω25+ω2+5−jω25+ω2=10ω2+25(二)cos(100t)=12(ej100t+e−j100t)⇒F{f2(t)}=∫∞−∞(cos100t)e−5|t|⋅e−jωtdt∫0−∞(cos100t)e5t⋅e−jωtdt+∫∞0(cos100t)e−5t⋅e−jωtdt=12∫0−∞e(5−jω+j100)t+e(5−jω−j100)tdt+12∫∞0e(−5−jω+j100)t+e(−5−jω−j100)tdt=12[15−jω+j100e(5−jω+j100)t+15−jω−j100e(5−jω−j100)t]|0−∞+12[1−5−jω+j100e(−5−jω+j100)t+1−5−jω−j100e(−5−jω−j100)t]|∞0=12[15−jω+j100+15−jω−j100]+12[−1−5−jω+j100+−1−5−jω−j100]=12[15−j(ω−100)+15−j(ω+100)]+12[15+j(ω−100)+15+j(ω+100)]=12[5+j(ω−100)25+(ω−100)2+5+j(ω+100)25+(ω+100)2+5−j(ω−100)25+(ω−100)2+5−j(ω+100)25+(ω+100)2]=525+(ω−100)2+525+(ω+100)2
解答:z(x,y)=3500−2x2−3y2⇒{∂z∂x|(−3,2)=−4x|(−3,2)=12∂z∂y|(−3,2)=−6y|(−3,2)=−12⇒(12,−12),故選(C)
解答:‖203041560‖=|−60−12|=72,故選(B)
解答:u1=u⋅v|v|2v=510[013]=[00.51.5]⇒u2=u−u1=[121]−[00.51.5]=[11.5−0.5],故選(B)
解答:A=[101120220103]⇒ATA=[101120220103][120001120123]=[50550103505553514]−r1+r3,−r1−3r2+r4→[5055010300000000]⇒rank =2,故選(B)
解答:X[10002−1−1100−9930−55]⇒det(X−λI)=0⇒λ4+4λ3−λ2−4λ=λ(λ3+4λ2−λ−4)=λ(λ−1)(λ2+5λ+4)=λ(λ−1)(λ+1)(λ+4)=0⇒λ=0,1,−1,−4,故選(B)
解答:(A)×:sin(x−y)非線性轉換(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非線性轉換(D)◯:均為線性運算,故選(D)
解答:A=[100010000]⇒eA=[e000e0001]⇒rank(eA)=3,故選(C)
解答:y=∞∑n=0an(x−1)n=a0+a1(x−1)+a2(x−1)2+a3(x−1)3⋯⇒{x2y=a0x2+a1x2(x−1)+a2x2(x−1)2+a3x2(x−1)3+⋯y′=a1+2a2(x−1)+3a3(x−1)2+⋯初始值y(1)=1⇒a0=1;又(y′+x2y)|x=1=cos(π)=−1⇒a0+a1=−1⇒a1=−2又(y′+x2y)|x=0=y′(0)=cos(0)=1⇒a1−2a2+3a3−4a4+⋯=1(y′+x2y)|x=2=y′(2)+4y=cos(2π)=1⇒∞∑n=0nan+∞∑n=04an+,故選(C)
解答:令非齊次解y1=C,C為常數⇒dy1dx=13xy21+1xy1−6x⇒0=C23x+Cx−6x⇒C2+3C−18=0⇒(C+6)(C−3)=0取C=3⇒y=y1+1u=3+1u,故選(B)
解答:y(x)=(a+bx)e2x+ce−2x+d⇒{y′(x)=(2a+b+2bx)e2x−2ce−2xy″(x)=(4a+4b+4bx)e2x+4ce−2x⇒y″−2y′=2be2x+8ce−2x=6e2x−4e−2x⇒{b=3c=−1/2又初始值{y(0)=−1y′(0)=6⇒{a+c+d=−1⇒d=−3/22a+b−2c=6⇒a=1⇒{a=1b=3c=−1/2d=−3/2,故選(A)
解答:∫∞−∞f(t)δ(t−T)dt=f(T)⇒L{f(t)}=∫∞0cos(πt)δ(t−1)e−stdt=cos(π)e−s=−e−s,故選(A)
解答:L{f(t)}=L{Ae−tcos(2t)+Be−tsin(2t)}=AL{e−tcos(2t)}+BL{e−tsin(2t)}=A⋅s+1(s+1)2+4+B⋅2(s+1)2+4=As+A+2Bs2+2s+5=2s+5s2+as+b⇒{a=2b=5⇒a/b=2/5=0.4,故選(D)
解答:f(t)={0for 0≤t<43for t≥4⇒f(t)=3H(t−4)⇒y″+4y=3H(t−4)L{y″}=s2Y(s)−sy(0)−y′(0)=s2Y(s)−s⇒L{y″+4y}=L{f(t)}⇒(s2+4)Y(s)−s=3⋅1se−4s⇒Y(s)=ss2+4+3s(s2+4)e−4s=ss2+4+34(1s−ss2+4)e−4s⇒y(t)=cos(2t)+34(1−cos(2(t−4)))H(t−4)=cos(at)+b(c−cos(d(t−4)))H(t−4)⇒{a=2b=3/4≠3/2c=1d=2,故選(B)
解答:f(t)=5[H(t−3)−H(t−11)]=x(t−7),x(t)=5⋅rect(t/8)F{x(t)}=5ω⋅2⋅sin(ω⋅82)=10ωsin(4ω)⇒F{x(t−7)}=10ωe−j7ωsin(4ω)⇒{a=10b=1c=7≠−7d=4,故選(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)}
解答:z(x,y)=3500−2x2−3y2⇒{∂z∂x|(−3,2)=−4x|(−3,2)=12∂z∂y|(−3,2)=−6y|(−3,2)=−12⇒(12,−12),故選(C)
解答:‖203041560‖=|−60−12|=72,故選(B)
解答:u1=u⋅v|v|2v=510[013]=[00.51.5]⇒u2=u−u1=[121]−[00.51.5]=[11.5−0.5],故選(B)
解答:A=[101120220103]⇒ATA=[101120220103][120001120123]=[50550103505553514]−r1+r3,−r1−3r2+r4→[5055010300000000]⇒rank =2,故選(B)
解答:X[10002−1−1100−9930−55]⇒det(X−λI)=0⇒λ4+4λ3−λ2−4λ=λ(λ3+4λ2−λ−4)=λ(λ−1)(λ2+5λ+4)=λ(λ−1)(λ+1)(λ+4)=0⇒λ=0,1,−1,−4,故選(B)
解答:(A)×:sin(x−y)非線性轉換(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非線性轉換(D)◯:均為線性運算,故選(D)
解答:A=[100010000]⇒eA=[e000e0001]⇒rank(eA)=3,故選(C)
解答:y=∞∑n=0an(x−1)n=a0+a1(x−1)+a2(x−1)2+a3(x−1)3⋯⇒{x2y=a0x2+a1x2(x−1)+a2x2(x−1)2+a3x2(x−1)3+⋯y′=a1+2a2(x−1)+3a3(x−1)2+⋯初始值y(1)=1⇒a0=1;又(y′+x2y)|x=1=cos(π)=−1⇒a0+a1=−1⇒a1=−2又(y′+x2y)|x=0=y′(0)=cos(0)=1⇒a1−2a2+3a3−4a4+⋯=1(y′+x2y)|x=2=y′(2)+4y=cos(2π)=1⇒∞∑n=0nan+∞∑n=04an+,故選(C)
解答:令非齊次解y1=C,C為常數⇒dy1dx=13xy21+1xy1−6x⇒0=C23x+Cx−6x⇒C2+3C−18=0⇒(C+6)(C−3)=0取C=3⇒y=y1+1u=3+1u,故選(B)
解答:y(x)=(a+bx)e2x+ce−2x+d⇒{y′(x)=(2a+b+2bx)e2x−2ce−2xy″(x)=(4a+4b+4bx)e2x+4ce−2x⇒y″−2y′=2be2x+8ce−2x=6e2x−4e−2x⇒{b=3c=−1/2又初始值{y(0)=−1y′(0)=6⇒{a+c+d=−1⇒d=−3/22a+b−2c=6⇒a=1⇒{a=1b=3c=−1/2d=−3/2,故選(A)
解答:∫∞−∞f(t)δ(t−T)dt=f(T)⇒L{f(t)}=∫∞0cos(πt)δ(t−1)e−stdt=cos(π)e−s=−e−s,故選(A)
解答:L{f(t)}=L{Ae−tcos(2t)+Be−tsin(2t)}=AL{e−tcos(2t)}+BL{e−tsin(2t)}=A⋅s+1(s+1)2+4+B⋅2(s+1)2+4=As+A+2Bs2+2s+5=2s+5s2+as+b⇒{a=2b=5⇒a/b=2/5=0.4,故選(D)
解答:f(t)={0for 0≤t<43for t≥4⇒f(t)=3H(t−4)⇒y″+4y=3H(t−4)L{y″}=s2Y(s)−sy(0)−y′(0)=s2Y(s)−s⇒L{y″+4y}=L{f(t)}⇒(s2+4)Y(s)−s=3⋅1se−4s⇒Y(s)=ss2+4+3s(s2+4)e−4s=ss2+4+34(1s−ss2+4)e−4s⇒y(t)=cos(2t)+34(1−cos(2(t−4)))H(t−4)=cos(at)+b(c−cos(d(t−4)))H(t−4)⇒{a=2b=3/4≠3/2c=1d=2,故選(B)
解答:f(t)=5[H(t−3)−H(t−11)]=x(t−7),x(t)=5⋅rect(t/8)F{x(t)}=5ω⋅2⋅sin(ω⋅82)=10ωsin(4ω)⇒F{x(t−7)}=10ωe−j7ωsin(4ω)⇒{a=10b=1c=7≠−7d=4,故選(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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