Processing math: 69%

2019年8月17日 星期六

108年國安三等考試_電子組(選試英文)--工程數學詳解


108年國家安全局國家安全情報人員考試試題
考試別:國家安全情報人員
等別:三等考試
類 科組 :電子組
科 目:工程數學


(e2xy)+(1+λ)e2xy=0e2xy2e2xy+(1+λ)e2xy=0y2y+(1+λ)y=0k22k+(1+λ)=0k=2±44(1+λ)2k=1±iλy(x)=C1excosλx+C2exsinλx{y(0)=0y(1)=0{C1=0C1ecosλ+C2esinλ=0C2esinλ=0sinλ=0λ=nπλ=n2π2,n=1,2,y=Cexsin(nπx),n=1,2,


DAlembert,:y(x,t)=12(f(xt)+f(x+t))+12x+txtg(z)dz{f(x)=y(x,0)=0g(x)=yt(x,0)y(x,t)=12x+txtg(z)dzg(x)y(x,t)=X(x)T(t){2yt2=XT2yx2=XT,2yt2=2yx2XT=XTXX=TT,XX=TT=ω2{X+ω2X=0T+ω2T=0{X(x)=Acosωx+BsinωxT(t)=Ccosωt+Dsinωtg(x)=yt(x,0)=X(x)T(0)=(Acosωx+Bsinωx)(ωCsin0+ωDcos0)=ωD(Acosωx+Bsinωx)=ω(Acosωx+Bsinωx)y(x,t)=12x+txtω[Acosωz+Bsinωz]dz=12[AsinωzBcosωz]|x+txt=12[(Asinω(x+t)Bcosω(x+t))(Asinω(xt)Bcosω(xt))]=12[A(sinω(x+t)sinω(xt))B(cosω(x+t)Bcosω(xt))]=12[A(2cosωxsinωt)B(2sinωxsinωt)]=(Acosωx+Bsinωx)sinωty(x,t)=(Acosωx+Bsinωx)sinωt=n=1(Ancosωnx+Bnsinωnx)sinωnt -- 待續 - -


A=[242633285]det(A)=30+962412+12048=162A1=1det(A)[|3385||4285||4233||6325||2225||2263||6328||2428||2463|]=1162[936182466422430]=[1/182/91/94/271/271/277/274/275/27]


乙、測驗題部分:(50分)

u(v×u)u(v×u)=0(A)


{(m,n,m2n)W(p,q,p2q)Wa(m,n,m2n)+b(p,q,p2q)=(am,an,am2an)+(bp,bq,bp2bq)=(am+bp,an+bq,am2an+bp2bq)=(am+bp,an+bq,am+bp2(an+bq))Wu,vWau+bvWWR3(D)




[a1a2a3b1b2b3c1c2c3]r2+r1[a1b1a2b2a3b3b1b2b3c1c2c3]3r2[a1b1a2b2a3b33b13b23b3c1c2c3]r2+r3[a1b1a2b2a3b33b13b23b3c13b1c23b2c33b3]det[a1a2a3b1b2b3c1c2c3]=3×det(C)=15(A)


[1011111220103215]r1+r2,2r1+r3,3r1+r4[1011012100120242]2r2+r4[1011012100120000]2r3+r2,r3+r1[1001010500120000]Rank(A)=3A43=1(B)


[85]=a[11]+b[21]{a+2b=8ab=5{a=2b=3L([85])=L(a[11])+L(b[21])=aL([11])+bL([21])=a[121]+b[012]=2[121]+3[012]=[242]+[036]=[274](B)


det(AλI)=0|5λ1302λ120313λ|=0λ=5,1,10A[xyz]=λ[xyz]{λ=5λ=1λ=10{[5x+y+3z2y12z3y13z]=[5x5y5z][5x+y+3z2y12z3y13z]=[xyz][5x+y+3z2y12z3y13z]=[10x10y10z]{[xyz]=[x00][xyz]=[7z/64zz][xyz]=[4z/15zz]{[x00]=[100][7z/64zz]=[7246][4z/15zz]=[41515](D)


limz0z|z|=±1limz0|z|2z=0(C)


z3=i=0+i={cosπ2+isinπ2cos5π2+isin5π2cos9π2+isin9π2z={cosπ6+isinπ6=32+12icos5π6+isin5π6=32+12icos9π6+isin9π6=0i=i(C)


|z|=2f(z)z1dz=2πi×f(1)=2πi×(1+25)=4πi(D)


y+3yyy3y=0λ3+3λ2λ3=0(λ1)(λ+1)(λ+3)=0λ=3,1,1y=k1e3t+k2et+k3et(A)


|z3+7i|=4|z(37i)|=4z(37i)4(C)


yh,y+0.5y0.5y=0λ2+0.5λ0.5=0(2λ1)(λ+1)=0λ=12,1yh=C1ex/2+C2exyp=aex+bsinx+ccosxyp=aex+bcosxcsinxyp=aexbsinxccosxaex+(32b12c)sinx+(32c+12b)cosx=ex+sinx+3cosx{a=132b12c=132c+12b=3{a=1b=0c=2y=yh+yp=C1ex/2+C2ex+ex2cosxy=12C1ex/2C2ex+ex+2sinx{y(0)=0y(0)=1.5{C1+C2+12=012C1C2+1=1.5{C1+C2=1C12C2=1{C1=1C2=0y=ex/2+ex2cosx(A)


L{f(t)}=0estf(t)dt=2π0estsintdt+2πest(2+cost)dt=2π0estsintdt+22πestdt+2πestcostdt=[est(costs2+1ssints2+1)]|2π0+2[1sest]|2π+[est(sints2+1scosts2+1)]|2π=1s2+1(1e2πs)+2se2πs+e2πs(ss2+1)=1s2+1+e2πs(2s+s1s2+1)(C)



\int_1^2{f(x)\,dx}=\int_1^2{(x-ix^2)\,dx}=\left. \left[\frac{1}{2}x^2-\frac{1}{3}ix^3 \right] \right|_1^2=(2-\frac{8}{3}i)-(\frac{1}{2}-\frac{1}{3}i)=\frac{3}{2}-\frac{7}{3}i\\,故選\bbox[red,2pt]{(A)}


只看常數項\\y=a-x^2-{x^4\over 3}-\cdots \Rightarrow y'=-2x-{4x^3\over 3}+\cdots \Rightarrow y''=-2-4x^2+\cdots\\ \Rightarrow x^2y''=-2x^2-4x^4+ \cdots \Rightarrow (1-x^2)y''=-2 -2x^2-2x^4 + \cdots \\ \Rightarrow -2xy'=4x^2+ \dots \Rightarrow (1-x^2)y''-2xy'+2y=(-2+2a)+0\cdot x^2+\cdots =0\\ \Rightarrow -2+2a=0 \Rightarrow a=1,故選\bbox[red,2pt]{(B)}


\int_0^t{f(\tau)d\tau} =\int_0^t{\sin(2\tau)d\tau} = {1\over 2}\left( 1-\cos{2t}\right)\\ \Rightarrow L\{\int_0^t{f(\tau)d\tau} \}= L\{{1\over 2}\left( 1-\cos{2t}\right)\}={1\over 2}\left( {1\over s}-{s\over s^2+4}  \right) ={1\over 2}\left( {s^2+4-s^2\over s(s^2+4)}  \right)\\ ={2\over s(s^2+4)},故選\bbox[red,2pt]{(D)}


f(x)=x^2={\pi^2\over 3}+4\sum_{n=1}^\infty{(-1)^n\over n^2}\cos{nx}\\ \Rightarrow \begin{cases} f(0)=0= {\pi^2\over 3}+4\left(-1+{1\over 2^2} -{1\over 3^2} +{1\over 4^2}-\cdots\right)\cdots(1)\\ f(\pi)=\pi^2={\pi^2\over 3}+4\left(1+{1\over 2^2} +{1\over 3^2} +{1\over 4^2}+\cdots\right)\cdots(2) \end{cases} \\ (2)-(1)\Rightarrow \pi^2=4\left(2+{2\over 3^2} +{2\over 5^2}+{2\over 7^2}\cdots\right)=8 \left(1+{1\over 3^2} +{1\over 5^2}+{1\over 7^2}\cdots\right)\\ \Rightarrow 1+{1\over 3^2} +{1\over 5^2}+{1\over 7^2}\cdots=\pi^2/8,故選\bbox[red,2pt]{(A)}


\begin{cases} E(X)=\int _{ 0 }^{ 1 }{ 2x^{ 2 }\, dx } =\frac { 2 }{ 3 }  \\ E(X^{ 2 })=\int _{ 0 }^{ 1 }{ 2x^{ 3 }\, dx } =\frac { 1 }{ 2 }  \end{cases}\Rightarrow Var(X)=E(X^{ 2 })-(E(X))^{ 2 }=\frac { 1 }{ 2 } -\frac { 4 }{ 9 } =\frac { 1 }{ 18 } ,故選\bbox[red,2pt]{(A)}


f(x)=2(1-x) \Rightarrow F(x)=\int{f(x)dx}=2x-x^2 \Rightarrow F(y)=2\sqrt{y}-(\sqrt{y})^2=2\sqrt{y}-y\\ \Rightarrow f(y)=F'(y)={1\over \sqrt{y}}-1,故選\bbox[red,2pt]{(A)}


E[XY]=\int\int{xyf(x,y)dxdy} =\int_0^1\int_0^1{x^2y(y+1.5)dxdy} =\int_0^1{\left. \left[ {1\over 3}x^3y(y+1.5) \right] \right|_0^1\,dy} \\ =\int_0^1{{1\over 3}y(y+1.5)\,dy} =\int_0^1{({1\over 3}y^2+{1\over 2}y)\,dy} = \left. \left[ {1\over 9}y^3+ {1\over 4}y^2 \right]\right|_0^1= {1\over 9}+{1\over 4}={13\over 36},故選\bbox[red,2pt]{(C)}


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