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2020年12月23日 星期三

109年地方特考-工程數學詳解

109年特種考試地方政府公務人員考試試題

等 別:三等考試
類 科:電力工程、電子工程
科 目:工程數學

甲、申論題部分:(50%)
:R=x2+y2{Rx=xx2+y2Ry=yx2+y2;Θ=tan1yx{Θx=yx2+y2Θy=xx2+y2J=|RxRyΘxΘy|=|xx2+y2yx2+y2yx2+y2xx2+y2|=x2+y2(x2+y2)3/2=1x2+y2=1Rf(R,Θ)=12πσ2exp[R22σ2]×RfR(r)=2π0r2πσ2exp[r22σ2]dθ=rσ2exp[r22σ2]fR(r)=rσ2exp[r22σ2],0r<
:
(一)eiz=cosz+isinzcosz=12(eiz+eiz)ω=eiz,cosz=212(ω+1ω)=2ω222ω+1=0ω=2±1eiz=2±1iz=ln(2±1)z=iln(2±1)z=iln(2±1)+2nπ,nZ(二)f(z)=z3+z2+4z4+4z2=z3+z2+4z2(z+2i)(z2i){z=0,±2iz=0z=±2i{Res(f,0)=ddz(z2f(z))|z=0=ddz(z3+z2+4z2+4)|z=0=3z2+2zz2+42z(z3+z2+4)(z2+4)2|z=0=14Res(f,2i)=(z2i)f(z)|z=2i=z3+z2+4z2(z+2i)|z=2i=12Res(f,2i)=(z+2i)f(z)|z=2i=z3+z2+4z2(z2i)|z=2i=12Cf(z)dz=2πi(Res(f,0)+Res(f,2i)+Res(f,2i))=2πi×54=52πi
:(一)y(t)+5y(t)=5x(t)=3e2tu(t)L{y}+5L{y}=3L{e2tu(t)}sY(s)y(0)+5Y(s)=3s+2Y(s)=1s+23s+3L1{Y(s)}=L1{1s+2}3L1{1s+3}y(t)=e2t3e3t(二)initial value x(0)=limssX(s)=limse2s2s2+1(s+2)2=0final value x()=lims0sX(s)=lims0e2s2s2+1(s+2)2=14{x(t)0x(t)1/4

:
(一)(a)Ax=bA=(A|b)=(12313321462610361111k)r4+r1,3r4+r2,2r4+r3(01203k012163k048162k1111k)r1+r2,4r1+r3(01203k000194k00016+2k1111k)94k=6+2kk=52(一)(b)
    由(a)可知 rank(A)4det(A)=0
(二){dx1dt=2x12x2+3x3dx2dt=x1+x2+x3dx3dt=x1+3x2x3[x1x2x3]=[223111131][x1x2x3]A=[223111131],det(AλI)=0|2λ2311λ1131λ|=0(λ1)(λ+2)(λ3)=0λ=1,2,3λ=1u1=[111],λ=2u2=[11114],λ=3u3=[111][x1x2x3]=C1[111]ex+C2[11114]e2x+C3[111]e3x{x1(0)=1x2(0)=0x3(0)=1/2{1=C1+11C2+C30=C1+C2+C31/2=C114C2+C3{C1=2/3C2=1/30C3=7/10{x1=23ex1130e2x+710e3xx2=23ex130e2x+710e3xx3=23ex+1430e2x+710e3x
乙、測驗題部分:(50%)
:det(AλI)=0|1λ020λ0204λ|=0λ(λ4)(λ1)+4λ=0λ(λ25λ+44)=0λ2(λ5)=0λ=0,0,5D=(λ1000λ2000λ3)=(000000005)(A)
:{det(A)=6det(B)=2det(AB1)=det(A)det(B1)=det(A)det(B)=62=3(C)
:(A):r=(xi+yj+zk)(xi+yj+zk)=xx+yy+zz=1+1+1=3(B):{r=(xi+yj+zk)a=(c1i+c2j+c3k)a×r=((c2zc3y)i+(c3xc1z)j+(c1yc2x)k)(a×r)=x(c2zc3y)+y(c3xc1z)+z(c1yc2x)=0+0+0=0(C):×r=((yzzy)i,(zxxz)j,(xyyx)k)=0(D)×:×(a×r)=×((c2zc3y)i+(c3xc1z)j+(c1yc2x)k)=(c1c1)i+(c2c2)j+(c3c3)k=0a(D)
:(2,4)(1,1)2xydx+x2dy=[x2y]|(2,4)(1,1)=161=15(B)
:A=(1001)At=(t00t)eAt=(et00et)(C)
:
只有(B)的rank是2,其它都是1,故選(B)

:{M(x,y)=x+yN(x,y)=xln(x)(A)×:{yxM=y(x2+xy)=xxxN=x(x2ln(x))=2xln(x)+xyxMxxN(B)×:{y3M=y(3x+3y)=3x3N=x(3xln(x))=3ln(x)+3y3Mx3N(C):{y1xM=y(1+yx)=1xx1xN=xln(x)=1xy1xM=x1xN(D)×:{y1x2M=y(1x+yx2)=1x2x1x2N=xln(x)x=ln(x)x2+1x2y1x2Mx1x2N(C)
:y6y+9y=t2e3tL{y}6L{y}+9L{y}=L{t2e3t}s2Y(s)sy(0)y(0)6(sY(s)y(0))+9Y(s)=2!(s3)3s2Y(s)2s66sY(s)+12+9Y(s)=2(s3)3Y(s)=2(s3)5+2s3=2(s3)4+2(s3)5=2(sc)d+2(sa)b{a=3b=5c=3d=4{a+b+c+d=15a+b+cd=7abc+d=1a+bc+d=33(D)
:f(x)=x3f(x)a0=0,an=0(A)
:y=c1ex+c2e2x+c3xe2x1,2,2λ2+aλ2+bλ+c=0{a=(122)=3b=1(2)+(2)(2)+(2)1=0c=1(2)(2)=4a+b+c=1(D)
:f(t)=cost+t0eτf(tτ)dτL{f(t)}=L{cost}+L{t0eτf(tτ)dτ}F(s)=ss2+1+F(s)s+1F(s)=s+1s2+1=ss2+1+1s2+1L1{F(s)}=L1{ss2+1}+L1{1s2+1}f(t)=cost+sint(D)
:ff(x)dx=110kx2dx=1[13kx3]|10=113k=1k=3{EX=10xf(x)dx=103x3dx=34EX2=10x2f(x)dx=103x4dx=35σ2=EX2(EX)2=35916=380(D)
:
:z_0=\pi不在圓周|z|=3內,而z_0=\pi/2則在圓周|z|=3內,\\因此\oint_C\left({\cosh z\over (z-\pi)^3} -{\sin^2 z\over (2z-\pi)^3} \right)dz =\oint_C\left( -{\sin^2 z\over (2z-\pi)^3} \right)dz =\oint_C\left( -{\sin^2 z\over 8(z-\pi/2)^3} \right)dz  \\=\oint_C  {f(z)\over (z-\pi/2)^3}  dz , 其中f(z)=-{1\over 8}\sin^2 z \Rightarrow f'(z)=-{1\over 4}\sin z\cos z =-{1\over 8} \sin (2z) \\\Rightarrow f''(z)= -{1\over 4}\cos (2z) \Rightarrow f''(\pi/2)=-{1\over 4}\cos \pi={1\over 4} \\\Rightarrow \oint_C  {f(z)\over (z-\pi/2)^3}  dz =(2\pi i)\times {1\over 2!}\times f''(\pi/2) = {\pi i\over 4},故選\bbox[red, 2pt]{(A)}
:{z\over (z+1)(z^2+1)}={z\over (z+1)(z+i)(z-i)} ={f(z)\over z-i} ={g(z)\over z+i},其中\cases{f(z)={z\over (z+1)(z+i) }\\ g(z)={z\over (z+1) (z-i)}}\\ -1不再橢圓內,而\pm i在橢圓內,因此\oint_C{z\over (z+1)(z^2+1)}\;dz = 2\pi i(f(i)+g(-i)) \\=2\pi i({i\over (1+i)(2i)} +{-i\over (1-i)(-2i)})  =2\pi i({1-i\over 4} +{1+i\over 4}) =\pi i,故選\bbox[red, 2pt]{(C)}
:z_0={1+i\over \sqrt 2} =\cos {\pi \over 4}+i\sin {\pi \over 4} =e^{i{\pi\over 4}} \Rightarrow \cases{z_0^2 =i\\z_0^4=-1\\ z_0^8=1} \\ \Rightarrow f(z_0) =z_0^{24}- 3z_0^{20}+4z_0^{12}-5z_0^6 =(z_0^{8})^3- 3(z_0^{4})^5+4(z_0^{4})^3-5z_0^4\cdot z_0^2\\ = 1+3-4+5i=5i,故選\bbox[red, 2pt]{(A)}
:\oint_\Gamma {\cos (z)\over z}\;dz = \oint_\Gamma {f(z)\over z}\;dz = 2\pi i\times f(0)=2\pi i\times 1 =2\pi i,故選\bbox[red, 2pt]{(D)}
:封閉路徑之線積分為0,故選\bbox[red, 2pt]{(C)}

:\cases{x=t^2\\ y=-t \\ z=t^2} \Rightarrow \cases{dx=2tdt\\ dy=-dt \\ dz=2tdt} \Rightarrow \int_C x^2dx-yzdy +e^zdz =\int_0^1 t^4(2tdt)-(-t^3)(-dt)+e^{t^2}(2tdt)\\ =\int_0^1 2t^5-t^3+2te^{t^2}\;dt = \left. \left[ {1\over 3}t^6-{1\over 4}t^4+e^{t^2}\right] \right|_0^1 =({1\over 12}+e)-1=e-{11\over 12},故選\bbox[red, 2pt]{(C)}

:\cases{p_X(1)=0+1/4=1/4\\ p_X(0)=1/4+1/4=1/2\\ p_X(-1)=1/4+0 =1/4} \Rightarrow E(X)=1\cdot{1\over 4} +0\cdot {1\over 2} +(-1)\cdot {1\over 4}=0\\ \cases{p_Y(1)=0+1/4 +1/4=1/2\\ P_Y(-1)=1/4+1/4+0 =1/2} \Rightarrow E(Y)=1\cdot {1\over 2}-1\cdot {1\over 2}=0\\ E(XY) = \sum xyp(x,y) \\= 1\cdot 1\cdot p(1,1)+ 1\cdot (-1)\cdot p(1,-1)+  (-1)\cdot 1\cdot p(-1,1)+  (-1)\cdot (-1)\cdot p(-1,-1) \\=0 -1/4-1/4+0 = -1/2\\ 因此 COV(X,Y)= E(XY)-EXEY =-1/2-0=-1/2,故選\bbox[red, 2pt]{(B)}

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  2. 不好意思,請問第二題的第二小題是否寫錯了?

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