106年公務人員高等考試三級考試
類 科 :電力工程、電子工程、電信工程、醫學工程
科 目:工程數學
科 目:工程數學
(一)[11121−1][x1x2]=[012]≡Ax=b⇒ATAˆx=ATb,其中ˆx為最小平方的解A=[11121−1]⇒AT=[11112−1]⇒ATA=[11112−1][11121−1]=[3226]⇒(ATA)−1=[3226]−1=114[6−2−23]⇒ˆx=(ATA)−1ATb=114[6−2−23][11112−1][012]=114[42814−5][012]=114[18−6]=[9/7−3/7]≈[1.29−0.43]
解:
(一)|1−λ−1−113−λ1−31−1−λ|=0⇒λ3−3λ2−4λ−12=0⇒(λ−3)(λ−2)(λ+2)=0⇒特徵值λ=3,±2(二)|1−λ−1−113−λ1−31−1−λ|=0⇒λ3−3λ2−4λ−12=0⇒(λ−3)(λ−2)(λ+2)=0⇒λ=3,±2λ1=3⇒[−2−1−1101−31−4][x1x2x3]=0⇒取特徵向量u1=[1−1−1]λ2=2⇒[−1−1−1111−31−3][x1x2x3]=0⇒取特徵向量u2=[10−1]λ3=−2⇒[3−1−1151−311][x1x2x3]=0⇒取特徵向量u2=[1−14]令P=[u1,u2,u3]=[111−10−1−1−14]⇒MP=P[λ1000λ2000λ3]⇒P−1MP=[λ1000λ2000λ3](三)P−1MP=[λ1000λ2000λ3]⇒(P−1MP)4=[λ1000λ2000λ3]4⇒P−1M4P=[λ1000λ2000λ3]4=[λ41000λ42000λ43]=[810001600016]⇒M4=P[810001600016]P−1=[111−10−1−1−14][810001600016][−1/5−1−1/51101/501/5]=[811616−810−16−81−1664][−1/51−1/51101/501/5]=[3−65−13138113136529]
解:13s3+45s2+52s+50s4+4s3+7s2+16s+12=13s3+45s2+52s+50(s+1)(s+3)(s2+4)=As+1+Bs+3+Cs+Ds2+4⇒13s3+45s2+52s+50=A(s+3)(s2+4)+B(s+1)(s2+4)+(Cs+D)(s+1)(s+3)⇒{A+B+C=133A+B+4C+D=454A+4B+3C+4D=5212A+4B+3D=50⇒{A=3B=2C=8D=2⇒13s3+45s2+52s+50s4+4s3+7s2+16s+12=3s+1+2s+3+8s+2s2+4⇒y(t)=L−1{Y(s)}=3L−1{1s+1}+2L−1{1s+3}+8L−1{8ss2+22}+L−1{2s2+22}⇒y(t)=3e−t+2e−3t+8cos2t+sin2t=yh+yp⇒{yh=e−t+2e−3typ=e−t+2e−3t⇒特徵方程式(λ+1)(λ+3)=0⇒λ2+4λ+3=0⇒y″+4y′+3y=0⇒{a=4b=3又{y(0)=y0=3+2+8=13y′(0)=y′0=−3−6+2=−7⇒{y0=13y′0=−7因此(一)的解為{a=4b=3y0=13y′0=−7,(二)的解為y(t)=3e−t+2e−3t+8cos2t+sin2t
解:F(x)在x0有極大值⇒{F(x0)F(x0+1)≥1F(x0)F(x0−1)≥1(一)F(x0)F(x0+1)=n!x0!(n−x0)!px0(1−p)n−x0n!(x0+1)!(n−x0−1)!px0+1(1−p)n−x0−1=(x0+1)(1−p)(n−x0)p≥1⇒x0≥np+p−1(二)F(x0)F(x0−1)=n!x0!(n−x0)!px0(1−p)n−x0n!(x0−1)!(n−x0+1)!px0−1(1−p)n−x0+1=(n−x0+1)p(1−p)x0≥1⇒x0≤np+p由(一)及(二)⇒np+p−1≤x0≤np+p⇒{x0=np+p,np+p−1若np+p為整數x0=⌊np+p⌋若np+p不是整數
解:{P1=(2,2,0)P2=(−1,0,2)P3=(0,4,3)⇒{→P1P2=(−3,−2,2)→P1P3=(−2,2,3)⇒△P1P2P3=12√|→P1P2|2|→P1P3|2−(→P1P2⋅→P1P3)2=12√(9+4+4)(4+4+9)−(6−4+6)2=12√225=152=7.5,故選(C)
解:(A)1×(3,1,−4)−3×(2,0,−1)+1×(3,−1,1)=(0,0,0)⇒×(B)三維空間只需要三個線性獨立的向量⇒×(C)a×(1,1,−2)+b×(0,1,0)+c×(2,1,−1)=(0,0,0)⇒a=b=c=0⇒◯(D)三維空間只需要三個線性獨立的向量⇒×,故選(C
解:(x+y+z)2+y2+z2≥0⇒(x2+y2+z2+2xy+2yz+2zx)+y2+z2≥0⇒x2+2y2+2z2+2xy+2yz+2zx≥0,故選(D)
解:A=[1−3−20]⇒|1−λ−3−2−λ|=0⇒λ2−λ−6=0⇒(λ−3)(λ+2)=0⇒λ=3,−2特徵值λ1=3⇒[−2−3−2−3][x1x2]=0⇒取特徵向量x1=[3−2]特徵值λ2=−2⇒[3−3−22][x1x2]=0⇒取特徵向量x2=[11]⇒A=[31−21][300−2][31−21]−1⇒cosA=[31−21][cos300cos(−2)][31−21]−1=[31−21][cos300cos(−2)][1/5−1/52/53/5]=15[3cos3cos(−2)−2cos3cos(−2)][1−123]=15[3cos3+2cos(−2)−3cos3+3cos(−2)−2cos3+2cos(−2)2cos3+3cos(−2)]=15[3cos3+2cos2−3cos3+3cos2−2cos3+2cos22cos3+3cos2],故選(C)
解:A=[110021003]⇒A2=[110021003][110021003]=[131045009]⇒f(A)=2[131045009]−6[110021003]+3[100010001]=[−1020−14003],故選(A)
解:cosh(a+bi)=coshacosb+isinhasinb⇒cosh(5−2i)=cosh5cos(−2)+isinh5sin(−2)=cosh5cos2−isinh5sin2,故選(B)
解:i=eπ2i⇒i1+i=(eπ2i)1+i=eπ2i−π2=ie−π2=ie−(2nπ+π2),n∈,故選(C)
解:ex=1+x1!+x22!+x33!+⋯⇒e3Z=1+3Z1!+(3Z)22!+(3Z)33!+⋯⇒e3ZZ4=1Z4+3Z3+322Z2+336Z+⋯⇒ResZ=0(e3ZZ4)=336=92⇒∮Ce3ZZ4dZ=2πi×ResZ=0(e3ZZ4)=2πi×92=9πi,故選(C)
解:y′+4y=cost⇒{yh=Ce−4typ=Asint+Bcost⇒y=yh+yp=Ce−4t+Asint+Bcost,故選(D)
解:y″−3y′−4y=0⇒λ2−3λ−4=0⇒(λ−4)(λ+1)=0⇒λ=4,−1⇒yh=C1e4t+C2e−ty″−3y′−4y=8x2⇒yp=Ax2+Bx+C⇒y″p−3y′p−4yp=2A−3(2Ax+B)−4(Ax2+Bx+c)=−4Ax2+(−6A−4B)x+(2A−3B−4C)=8x2⇒{−4A=8−6A−4B=02A−3B−4C=0⇒{A=−2B=3C=−13/4⇒y=yh+yp=C1e4t+C2e−t−2x2+3x−13/4⇒{y(0)=1y′(0)=2⇒{C1+C2−13/4=14C1−C2+3=2{C1=13/20C2=18/5⇒y″(0)=16C1+C2−4=16×1320+185−4=705−4==14−4=10,故選(A)
解:y=a0+a1x+a2x2+a3x3+⋯+anxn+⋯⇒y′=a1+2a2x+3a3x2+⋯+nanxn−1+⋯⇒y″=2a2+6a3x+12a4x2+⋯+n(n−1)xn−2+⋯xy″+2y′=6x⇒2a1+(6a2−6)x+12a3x2+⋯+n(n+1)anxn−1+⋯=0⇒{2a1=06a2−6=0an=0,n≥3⇒{a1=0a2=1an=0,n≥3⇒y=a0+x2{y(1)=1y′(1)=2⇒a0=0⇒y=x2⇒y′=2x⇒y′(−1)=−2,故選(D)
解:dy(t)dt−3y(t)=6⇒{yh=Ae3typ=B⇒y=yh+yp=Ae3t+B⇒3Ae3t−3(Ae3t+B)=6⇒B=−2⇒lim
解:\int _{ 0 }^{ 1+i }{ z^{ 2 }dz } =\left. \left[ \frac { 1 }{ 3 } z^{ 3 } \right] \right| _{ 0 }^{ 1+i }=\frac { 1 }{ 3 } \left( 1+i \right) ^{ 3 }=\frac { 2i-2 }{ 3 } ,故選\bbox[red,2pt]{(C)}
解:\lim _{ n\to \infty }{ \frac { { a }_{ n } }{ { a }_{ n+1 } } } =\lim _{ n\to \infty }{ \frac { \frac { n }{ 2^{ n }\left( 3n-1 \right) } }{ \frac { \left( n+1 \right) }{ 2^{ n+1 }\left( 3\left( n+1 \right) -1 \right) } } } =\lim _{ n\to \infty }{ \frac { 2n\left( 3n+2 \right) }{ \left( 3n-1 \right) \left( n+1 \right) } } =\frac { 6 }{ 3 } =2,故選\bbox[red,2pt]{(B)}
解:L\left\{ f\left( t \right) \right\} =F\left( s \right) \Rightarrow L\left\{ tf\left( t \right) \right\} =-\frac { d }{ ds } F\left( s \right) \Rightarrow tf\left( t \right) =-L^{ -1 }\left\{ \frac { d }{ ds } F\left( s \right) \right\} \\ \Rightarrow f\left( t \right) =-\frac { 1 }{ t } L^{ -1 }\left\{ \frac { d }{ ds } F\left( s \right) \right\} \\ L^{ -1 }\left\{ \ln { \frac { s+1 }{ s-1 } } \right\} =L^{ -1 }\left\{ \ln { \left( s+1 \right) } -\ln { \left( s-1 \right) } \right\} =L^{ -1 }\left\{ \ln { \left( s+1 \right) } \right\} -L^{ -1 }\left\{ \ln { \left( s-1 \right) } \right\} \\ =-\frac { 1 }{ t } L^{ -1 }\left\{ \frac { d }{ ds } \ln { \left( s+1 \right) } \right\} +\frac { 1 }{ t } L^{ -1 }\left\{ \frac { d }{ ds } \ln { \left( s-1 \right) } \right\} =-\frac { 1 }{ t } L^{ -1 }\left\{ \frac { 1 }{ s+1 } \right\} +\frac { 1 }{ t } L^{ -1 }\left\{ \frac { 1 }{ s-1 } \right\} \\ =-\frac { 1 }{ t } e^{ -t }+\frac { 1 }{ t } e^{ t }=\frac { 1 }{ t } \left( e^{ t }-e^{ -t } \right) =\frac { 2 }{ t } \cdot \frac { e^{ t }-e^{ -t } }{ 2 } =\frac { 2 }{ t } \sinh { t } ,故選\bbox[red,2pt]{(B)}
解:f\left( x \right) 為奇函數\Rightarrow a_{ n }=0,n=0\~ \infty \\ b_{ n }=\frac { 1 }{ 2 } \int _{ -2 }^{ 2 }{ f\left( x \right) \sin { \frac { n\pi x }{ 2 } } dx } =\frac { 1 }{ 2 } \int _{ -1 }^{ 1 }{ x\sin { \frac { n\pi x }{ 2 } } dx } =\frac { 1 }{ 2 } \left. \left[ -\frac { 2x }{ n\pi } \cos { \frac { n\pi x }{ 2 } } +{ \left( \frac { 2 }{ n\pi } \right) }^{ 2 }\sin { \frac { n\pi x }{ 2 } } \right] \right| _{ -1 }^{ 1 }\\ =\frac { 1 }{ 2 } \left[ \left( -\frac { 2 }{ n\pi } \cos { \frac { n\pi }{ 2 } } +{ \left( \frac { 2 }{ n\pi } \right) }^{ 2 }\sin { \frac { n\pi }{ 2 } } \right) -\left( \frac { 2 }{ n\pi } \cos { \frac { n\pi }{ 2 } } -{ \left( \frac { 2 }{ n\pi } \right) }^{ 2 }\sin { \frac { n\pi }{ 2 } } \right) \right] \\ =-\frac { 2 }{ n\pi } \cos { \frac { n\pi }{ 2 } } +{ \left( \frac { 2 }{ n\pi } \right) }^{ 2 }\sin { \frac { n\pi }{ 2 } } \\ \Rightarrow f\left( x \right) =\sum _{ n=1 }^{ \infty }{ b_{ n }\sin { \frac { n\pi x }{ 2 } } } =\sum _{ n=1 }^{ \infty }{ \left( -\frac { 2 }{ n\pi } \cos { \frac { n\pi }{ 2 } } +{ \left( \frac { 2 }{ n\pi } \right) }^{ 2 }\sin { \frac { n\pi }{ 2 } } \right) \sin { \frac { n\pi x }{ 2 } } } ,故選\bbox[red,2pt]{(B)}
解:P\left( x\le 5|x\ge 2 \right) =\frac { P\left( 2\le x\le 5 \right) }{ P\left( x\ge 2 \right) } =\frac { F\left( 5 \right) -F\left( 2 \right) }{ 1-F\left( 2 \right) } =\frac { \frac { 3 }{ 4 } -\frac { 1 }{ 4 } }{ 1-\frac { 1 }{ 4 } } =\frac { 2 }{ 3 } ,故選\bbox[red,2pt]{(D)}
解:
不退貨的條件:10台手機全是好的或只有1台是壞的,其機率為(0.99)^{10}+10\times 0.01 \times (0.99)^9 =1.09\times 0.99^9,因此退貨的機率為1-1.09\times 0.99^9=1-1.09\times 0.9135 = 0.004285 \approx 0.4\%,故選\bbox[red,2pt]{(B)}
解:E\left[ X^{ 2 }Y^{ 2 } \right] =\frac { 1 }{ 24 } \int _{ 0 }^{ 4 }{ \int _{ 0 }^{ 6 }{ x^{ 2 }y^{ 2 }dxdy } } =\frac { 1 }{ 24 } \int _{ 0 }^{ 4 }{ \left. \left[ \frac { 1 }{ 3 } x^{ 3 }y^{ 2 } \right] \right| _{ 0 }^{ 6 }dy } =3\int _{ 0 }^{ 4 }{ { y }^{ 2 }dy } \\ =3\left. \left[ \frac { 1 }{ 3 } y^{ 3 } \right] \right| _{ 0 }^{ 4 }=4^{ 3 }=64,故選\bbox[red,2pt]{(D)}
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