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2019年12月19日 星期四

108年地方特考-三等-工程數學詳解


108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 :電子工程、電力工程
科目:工程數學


y=c1ex+c2xex+x2ex{yh=c1ex+c2xexyp=x2exyh=c1ex+c2xexyh{yh=c1ex+c2ex+c2xexyh=c1ex+2c2ex+c2xexpyh+qyh+ryh=0(p+q+r)c2xex+((p+q+r)c1+(2p+q)c2)ex=0{p+q+r=02p+q=0{q=2pr=pp(yh2yh+yh)=0yh2yh+yh=0y2y+y=p(x),yp=x2exyp2yp+yp=2ex+4xex+x2ex(4xex+2x2ex)+x2ex=2ex=p(x)y2y+y=2ex


f(t)2TL{f(t)}=11e2Ts2T0f(t)estdt=11e2TsT0cos(πtT)estdt=11e2Ts[ests2+(π/T)2(ssinπtT+πTcosπtT)]|T0=11e2Ts[esTs2+(π/T)2(πT)+1s2+(π/T)2(πT)]=11e2Ts[1+esTs2+(π/T)2(πT)]=π/T(1esT)(s2+(π/T)2)



cos(z)=eiz+eiz2cos(3+2i)=ei(3+2i)+ei(3+2i)2=e2+3i+e23i2=12(e2e3i+e2e3i)=12(e2(cos3+isin3))+e2(cos(3)+isin(3)))=12(e2(cos3+isin3))+e2(cos3isin3))=cos3(e2+e2)2+isin3(e2e2)2=cos3cosh2isin3sinh2=a+ib{a=cos3cosh2b=sin3sinh2


γf(z)dz=γz(z+2)(z4i)dz=2πi×(Res(f,2)+Res(f,4i))=2πi(zz4i|z=2+zz+2|z=4i)=2πi(22+4i+4i2+4i)=2πi



A=[121422]AT=[112242]{ATA=[112242][121422]=[6101024]ATB=[112242][327]=[90]ATAX=ATB[6101024][x1x2]=[90]{6x1+10x2=910x1+24x2=0{x1=54/11x2=45/22x=[54/1145/22]使||AxB||


乙、測驗題部分:(50分)

{u=(1,1,1)v=(3,4,6)w=(2,4,2)|(u×v)w|=|((1,1,1)×(3,4,6))(2,4,2)|=|(2,3,1)(2,4,2)|=18(C)


A=[111021223]det(AλI)=0|1λ1102λ1223λ|=0(λ3)(λ2)(1λ)2+2(2λ)2(1λ)=0(λ3)(λ2)(λ1)=0λ=1,2,33依 Rayleigh Principle xTAxxTxλ=3(C)




A=[51010105205510]det(AλI)=0λ(λ+5)(λ5)λ=0,15,15λ=0(AλI)X=0[51010105205510][xyz]=0{x=2zy=0u1=[201]λ=15(AλI)X=0[1010101010205525][xyz]=0{x=yz=0u2=[110]λ=15(AλI)X=0[201010102020555][xyz]=0{x=0y=zu3=[011]P=[u1u2u3]=[210011101]=[ab001110c]{a=2b=1c=1abc=0(C)


m=n=r(B)


A=[130310002]det(AλI)=0(λ4)(λ+2)2=0λ=4,2λ=4(AλI)X=0[330330006][xyz]=0{x=yz=0u1=[110]λ=2(AλI)X=0[330330000][xyz]=0{x+y=0u2=[110],u3=[001]u1,u2,u3(C)


A=[0213]A=PDP1=[2111][1002][1112]eA=[2111][e00e2][1112]=[2ee22e2e2e+e2e+2e2]=[a11a12a21a22]a11+a22=2ee2e+2e2=e2+e(B)


(A)(1i)2=2i(1i)4=(2i)2=464(B)(2i)2=3+4i(2i)4=(3+4i)264(C)(12i)2=3+4i(12i)4=(3+4i)264(D)(22i)2=8i(3+4i)4=(8i)2=64(D)


ln(1i3)=ln(2(12i32))=ln(2(cos(π3)+isin(π3)))=ln(2ei(π/3))=ln2+i(π3+2nπ)(A)


z=t+itdt=dt+idt=(1+i)dtφz2dz=20(t+it)2(1+i)dt=(1+i)320t2dt=83(1+i)3=83(2+i2)=163(i1)(B)


z=0CCf(z)dz=0(A)


y=k1eax+k2ebx+ecx{yh=k1eax+k2ebxyp=ecx{yh6yh+8yh=0yp6yp+8yp=3exyh6yh+8yh=0(a26a+8)k1eax+(b26b+8)k2ebx=0{a26a+8=0b26b+8=0{(a4)(a2)=0(b4)(b2)=0ab(a,b)={(2,4)(4,2)a+b=6yp6yp+8yp=3ex(c26c+8)ecx=3exc=1a+b+c=6+1=7(C)


cf(z)dz=cezz(πi/2)dz=2πi×Res(f,πi/2)=2πi×eπi/2=2πi(cos(π/2)+isin(π/2))=2πi×(i)=2π(A)


{x1=x2x2=1.01x10.2x2x2=1.01x10.2x2=1.01x20.2x2x2+0.2x2+1.01x2=0x2=eαt(Acosβt+Bsinβt),α=0.22=0.1x2=e0.1t(Acosβt+Bsinβt)lim



F(s)={s+1\over s^2(s^2+1)} ={a\over s} +{b\over s^2}+ {cs+d \over s^2+1} ={(a+c)s^3 +(b+d)s^2+as+b \over s^2(s^2+1)}\\ \Rightarrow  \begin{cases}a  = 1\\b =1 \\a+c=0 \\ b+d=0 \end{cases} \Rightarrow  \begin{cases}a  = 1\\b =1 \\c=-1 \\ d=-1 \end{cases}  \Rightarrow F(s)= {1\over s}+ {1\over s^2}+{-s-1\over s^2+1} \\\Rightarrow L^{-1}\{F(s)\} =L^{-1}\{{1\over s}\} +L^{-1}\{{1\over s^2}\} -L^{-1}\{{s\over s^2+1}\}  -L^{-1}\{{1\over s^2+1}\} =1+t-\cos t-\sin t\\,故選\bbox[red,2pt]{(C)}


題目的\begin{cases}K=a^2y\\ T=a^2x\end{cases}  應該是\begin{cases}K=\alpha^2y\\ T=\alpha^2x\end{cases},\text{直接用代的比較快:} \\ (B)u=c_1e^{\alpha^2y}\cosh 2\alpha x+c_2 e^{\alpha^2y}\sinh 2\alpha x \Rightarrow u_x=2\alpha c_1e^{\alpha^2y}\sinh 2\alpha x+2\alpha c_2e^{\alpha^2y}\cosh 2\alpha x\\ \Rightarrow u_{xx}=4\alpha^2c_1\cosh 2\alpha x+4\alpha^2 c_2e^{\alpha^2y}\sinh 2\alpha x =4u_y,故選\bbox[red,2pt]{(B)}




{1\over s^2}\left({s-1\over s+1} \right)= {a\over s}+ {b\over s+1} + {c\over s^2} = {(a+b)s^2+(a+c)s+c \over s^2(s+1)} \Rightarrow \begin{cases}a+b=0\\ a+c=1\\ c=-1 \end{cases}\Rightarrow \begin{cases}a=2\\ b=-2\\ c=-1 \end{cases} \\ \Rightarrow \pmb{L^{-1}}\left\{{1\over s^2}\left({s-1\over s+1} \right)\right\}=\pmb{L^{-1}}\left\{2\over s \right\} +\pmb{L^{-1}}\left\{-2\over s+1 \right\} +\pmb{L^{-1}}\left\{-1\over s^2 \right\} =2-2e^{-t}-t,故選\bbox[red,2pt]{(B)}


\left| X(jw)\right|=2(u(w+3)-u(w-3))= \begin{cases}0 & x \ge 3\\ 2 & -3\le x < 3 \\0 & x<-3 \end{cases}   \Rightarrow \left| X(jw)\right|= \begin{cases} 2 & -3\le x < 3 \\0 & \text{otherwise} \end{cases}\\  \Rightarrow x(t)={1\over 2\pi}\int_{-\infty}^\infty X(jw)e^{jwt}dw ={1\over 2\pi}\int_{-\infty}^\infty \left| X(jw)\right|e^{j\angle X(jw)}e^{jwt}dw  \\ ={1\over 2\pi}\int_{-3}^3 2e^{j(-3w/2+\pi)}e^{jwt}dw ={1\over \pi}e^{j\pi}\int_{-3}^3 e^{jw(-3/2+t)}dw = -{1\over \pi}\int_{-3}^3 e^{jw(-3/2+t)}dw\\ = -{1\over \pi} \left. \left[{1\over j(-3/2+t)} e^{jw(-3/2+t)} \right] \right|_{-3}^3 =-{1\over j(-3/2+t)\pi}\left( e^{j(-9/2+3t)} - e^{j(9/2-3t)}\right)\\ \Rightarrow x(t)=0 \Rightarrow  e^{j(-9/2+3t)} = e^{j(9/2-3t)} \Rightarrow -{9\over 2}+3t={9\over 2}-3t +2\pi \\ \Rightarrow 6t=9+2\pi \Rightarrow t={9\over 6}+{2\pi \over 6}= {3\over 2}+ {\pi \over 3},故選\bbox[red,2pt]{(C)}


30顆IC,劣品率為1/6 \Rightarrow 良品為30\times (1-1/6)=25 \Rightarrow 10顆IC都是良品機率: \frac{C^{25}_{10}}{C^{30}_{10}},故選\bbox[red,2pt]{(B)}


\begin{cases} E(X)=\int_0^\infty {1\over 4}xe^{-x/4}\;dx = \left. \left[-xe^{-x/4}-4e^{-x/4}  \right] \right|_0^\infty =0-(-4)= 4 \\ E(X^2)= \int_0^\infty {1\over 4}x^2e^{-x/4}\;dx = \left. \left[-x^2e^{-x/4}-8xe^{-x/4} -32e^{-x/4}  \right] \right|_0^\infty =0-(-32)= 32\end{cases} \\\\ \Rightarrow Var(X)=\sigma_X^2= E(X^2)-(E(X))^2 =32-4^2=16 \\\Rightarrow \sigma_Y^2 = Var(Y) = Var(3X-2)=3^2Var(X)= 9\times 16=144,故選\bbox[red,2pt]{(D)}


E(X)=\int_{-\infty}^\infty xf(x)\;dx =\int_0^1 2x^2\;dx= {2\over 3},故選\bbox[red,2pt]{(B)}


考選部未公布申論題答案,解題僅供參考

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