110 年特種考試地方政府公務人員考試
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分)
解答:f(z)=−3z+5z(z2−3z+2)=−3z+5z(z−1)(z−2)⇒{Res(f,z=0)=−3z+5(z−1)(z−2)|z=0=52Res(f,z=1)=−3z+5z(z−2)|z=1=−2⇒∮cf(z)dz=2πi(Res(f,z=0)+Res(f,z=1))=2πi(52−2)=πi
解答:a0=12π∫π−πx+πdx=12π⋅2π2=πan=1π∫π−π(x+π)cos(nx)dx=0bn=1π∫π−π(x+π)sin(nx)dx=1π×−2πn(−1)n=−2n(−1)n⇒f(x)=x+π=π+∞∑n=1−2n(−1)nsin(nx)=π+2(sin(x)−12sin(2x)+13sin(3x)−14sin(4x)+⋯)⇒x2=sin(x)−12sin(2x)+13sin(3x)−14sin(4x)+⋯取x=π2⇒π4=1+0−13+0+15+0−17+⋯⇒π4=1−13+15−17+⋯,故得證
解答:假設生產總量為N⇒{A機器產量為0.25N,其中瑕疵品數量為0.25N×0.05B機器產量為0.35N,其中瑕疵品數量為0.35N×0.03C機器產量為0.3N,其中瑕疵品數量為0.3N×0.04⇒瑕疵品數量為0.25N×0.05+0.35N×0.03+0.3N×0.04=0.035N⇒取出一瑕疵品,{此瑕疵品為A機器出產的機率為0.25N×0.05/0.035N=5/14此瑕疵品為B機器出產的機率為0.35N×0.03/0.035N=3/10此瑕疵品為C機器出產的機率為0.3N×0.04/0.035N=12/35⇒取出一瑕疵品,此瑕疵品為A、B、C機器出產的機率各是514,310,1235
乙、 測驗題部分: (50分)
解答:k1p1+k2p2+k3p3=0⇒(k1+2k2+k3)x2+(−2k1+ak2+k3)x+(k1−k2+bk3)=0⇒{k1+2k2+k3=0−2k1+ak2+k3=0k1−k2+bk3=0⇒A=[121−2a11−1b]⇒f(a,b)=det(A)=ab−a+4b+5⇒{f(2,1/2)=6≠0f(−1,−2)=f(1,−4/5)=f(0,−5/4)=0,故選(A)解答:T=[1−√3√31]=2[cos(π/3)−sin(π/3)sin(π/3)cos(π/3)]⇒{r=2θ=60∘,故選(C)
解答:det(A)=2−4=−2⇒det(A5)=(−2)5=−32,故選(D)
解答:v=[111]⇒Av=1⋅v⇒v是特徵向量,故選(A)
解答:z=x+iy⇒ˉz=x−iy⇒(ˉz)2=(x2−y2)−2xyi⇒e(ˉz)2=ex2−y2⋅e−2xyi=ex2−y2(cos(−2xy)+isin(−2xy))=ex2−y2(cos(2xy)−isin(2xy))⇒e(ˉz)2的虛部為−ex2−y2sin(2xy),故選(B)
解答:det(A−λI)=|1−λ21a−λ213b−λ|=−λ3+(b+1)λ2+(2a−b+7)λ+3a−2ab−2⇒{b+1=−c2a−b+7=−33a−2ab−2=2⇒{{a=−1/4b=19/2c=−21/2⇒4a+b+c=−2{a=−4b=2c=−3⇒4a+b+c=−17,故選(D)
解答:由泰勒級數f(x)=∞∑n=0f[n](π)n!(x−π)n⇒cos(x)=−1+12(x−π)2+⋯y=a0+a1(x−π)+a2(x−π)2+a3(x−π)3+⋯⇒y′=a1+2a2(x−π)+3a3(x−π)2+4a4(x−π)3+⋯⇒y′+cos(x)y=(a1−a0)+(−a1+2a2)(x−π)+(12a0−a2+3a3)(x−π)2+⋯=1由於y(π)=A=0=a0⇒{a1−a0=1−a1+2a2=012a0−a2+3a3=0⇒a1=1⇒a2=1/2⇒a3=1/6,故選(B)
解答:f(x)=e−2x⇒{f(−1)=e2=Af(0)=1=Bf(1)=e−2=C⇒A+B+C=e2+e−2+1,故選(D)
解答:
梯形AFDE面積△ABC面積=3/26/2=12,故選(C)
解答:∫∫fXY(x,y)dxdy=1⇒∫20∫30A(x+y)dydx=∫20A(3x+92)dx=15A=1⇒A=1/15⇒fY(y)=∫fXY(x,y)dx=∫20115(x+y)dx=115(2+2y),故選(B)
解答:A=[12232548−1−3−2−50204]−2r1+r2→[12230102−1−3−2−50204]r1+r3→[122301020−10−20204]r2+r3,−r2+r4→[1223010200000000]⇒rank(A)=2,故選(B)
解答:R3的基底只需考慮(C)與(D),其它數量不是3;(D)A=[122−121086]r1+r2→[122043086]−2r2+r3→[122043000]⇒Rank(A)=2,非基底,故選(C)
解答:(A)特徵值為三相異值⇒Rank(B)=3(C)B與BT有相同的特徵值且det(B)=det(BT)=0⋅1⋅2=0⇒det(BTB)=0(D)B的特徵值0,1,2⇒B+I的特徵值=−1,0,1⇒(B+I)−1與B+I有相同特徵值只有(B)尚難確定,故選(B)
解答:P為投影矩陣⇒PT=P=P2=P3=⋯=P123,故選(B)
解答:|z+i|<4√116=12,故選(A)
解答:y″
解答:y(t)+\int_0^t (t-\tau)y(\tau)\;d\tau=1 \Rightarrow \mathcal{L}\{ y(t) \} +\mathcal{L}\{ \int_0^t (t-\tau)y(\tau)\;d\tau \} =\mathcal{L}\{ 1 \} \\ \Rightarrow Y(s)+ \mathcal{L}\{ t \} \mathcal{L}\{ y(t) \}={1\over s} \Rightarrow Y(s) +{1\over s^2}Y(s) ={1\over s} \Rightarrow Y(s)={s\over s^2+1} \\ \Rightarrow y(t)= \mathcal{L}^{-1}\{ {s\over s^2+1} \} =\cos(t),故選\bbox[red,2pt]{(A)}
解答:令\cases{M(x,y)=-y\\ N(x,y)= x \\ 積分因子I(x,y)} \\ (A)I=1/x^2 \Rightarrow \cases{IM=-y/x^2 \\ IN=1/x} \Rightarrow \cases{{\partial \over \partial y}IM =-1/x^2\\ {\partial \over \partial x}IN =-1/x^2} \Rightarrow {\partial \over \partial y}IM ={\partial \over \partial x}IN \\ (B)I=1/xy \Rightarrow \cases{IM=-1/x \\ IN=1/y} \Rightarrow \cases{{\partial \over \partial y}IM =0\\ {\partial \over \partial x}IN =0} \Rightarrow {\partial \over \partial y}IM ={\partial \over \partial x}IN \\(C)I=1/(x^2+y^2) \Rightarrow \cases{IM=-y/(x^2+y^2) \\ IN=x/(x^2+y^2)} \Rightarrow \cases{{\partial \over \partial y}IM = (y^2-x^2)/(x^2+y^2)^2 \\ {\partial \over \partial x}IN =(y^2-x^2)/(x^2+y^2)^2} \\\qquad \Rightarrow {\partial \over \partial y}IM ={\partial \over \partial x}IN \\(D) I=xy \Rightarrow \cases{IM=-xy^2 \\ IN=x^2y} \Rightarrow \cases{{\partial \over \partial y}IM =-2xy \\ {\partial \over \partial x}IN =2xy} \Rightarrow {\partial \over \partial y}IM \color{blue}{\ne} {\partial \over \partial x}IN \\,故選\bbox[red,2pt]{(D)}
解答:X:黑桃牌的次數 \Rightarrow \cases{P(X=0)=({3\over 4})^3 \\P(X=1)=C^3_1({3\over 4})^2({1\over 4}) } \Rightarrow P(X\ge 2) =1-P(X=0)-P(X=1)\\ = 1-({3\over 4})^3 -C^3_1({3\over 4})^2({1\over 4}) =1-{27\over 64}-{ 27 \over 64} ={10\over 64} ={5\over 32},故選\bbox[red,2pt]{(D)}
解答:X\sim U[a,b] \Rightarrow \cases{E(X)={a+b\over 2}=0 \\\sigma(X)= {b-a\over \sqrt {12}} =\sqrt {12}} \Rightarrow \cases{a+b= 0\\ b-a=12} \Rightarrow \cases{a=-6\\ b=6}\\ \Rightarrow 區間為[14-a,14+b]=[14:00-0:06,14:00+0:06] =[13:54,14:06],故選\bbox[red,2pt]{(A)}
解答:A=[12232548−1−3−2−50204]−2r1+r2→[12230102−1−3−2−50204]r1+r3→[122301020−10−20204]r2+r3,−r2+r4→[1223010200000000]⇒rank(A)=2,故選(B)
解答:R3的基底只需考慮(C)與(D),其它數量不是3;(D)A=[122−121086]r1+r2→[122043086]−2r2+r3→[122043000]⇒Rank(A)=2,非基底,故選(C)
解答:(A)特徵值為三相異值⇒Rank(B)=3(C)B與BT有相同的特徵值且det(B)=det(BT)=0⋅1⋅2=0⇒det(BTB)=0(D)B的特徵值0,1,2⇒B+I的特徵值=−1,0,1⇒(B+I)−1與B+I有相同特徵值只有(B)尚難確定,故選(B)
解答:P為投影矩陣⇒PT=P=P2=P3=⋯=P123,故選(B)
解答:|z+i|<4√116=12,故選(A)
解答:y″
解答:y(t)+\int_0^t (t-\tau)y(\tau)\;d\tau=1 \Rightarrow \mathcal{L}\{ y(t) \} +\mathcal{L}\{ \int_0^t (t-\tau)y(\tau)\;d\tau \} =\mathcal{L}\{ 1 \} \\ \Rightarrow Y(s)+ \mathcal{L}\{ t \} \mathcal{L}\{ y(t) \}={1\over s} \Rightarrow Y(s) +{1\over s^2}Y(s) ={1\over s} \Rightarrow Y(s)={s\over s^2+1} \\ \Rightarrow y(t)= \mathcal{L}^{-1}\{ {s\over s^2+1} \} =\cos(t),故選\bbox[red,2pt]{(A)}
解答:令\cases{M(x,y)=-y\\ N(x,y)= x \\ 積分因子I(x,y)} \\ (A)I=1/x^2 \Rightarrow \cases{IM=-y/x^2 \\ IN=1/x} \Rightarrow \cases{{\partial \over \partial y}IM =-1/x^2\\ {\partial \over \partial x}IN =-1/x^2} \Rightarrow {\partial \over \partial y}IM ={\partial \over \partial x}IN \\ (B)I=1/xy \Rightarrow \cases{IM=-1/x \\ IN=1/y} \Rightarrow \cases{{\partial \over \partial y}IM =0\\ {\partial \over \partial x}IN =0} \Rightarrow {\partial \over \partial y}IM ={\partial \over \partial x}IN \\(C)I=1/(x^2+y^2) \Rightarrow \cases{IM=-y/(x^2+y^2) \\ IN=x/(x^2+y^2)} \Rightarrow \cases{{\partial \over \partial y}IM = (y^2-x^2)/(x^2+y^2)^2 \\ {\partial \over \partial x}IN =(y^2-x^2)/(x^2+y^2)^2} \\\qquad \Rightarrow {\partial \over \partial y}IM ={\partial \over \partial x}IN \\(D) I=xy \Rightarrow \cases{IM=-xy^2 \\ IN=x^2y} \Rightarrow \cases{{\partial \over \partial y}IM =-2xy \\ {\partial \over \partial x}IN =2xy} \Rightarrow {\partial \over \partial y}IM \color{blue}{\ne} {\partial \over \partial x}IN \\,故選\bbox[red,2pt]{(D)}
解答:X:黑桃牌的次數 \Rightarrow \cases{P(X=0)=({3\over 4})^3 \\P(X=1)=C^3_1({3\over 4})^2({1\over 4}) } \Rightarrow P(X\ge 2) =1-P(X=0)-P(X=1)\\ = 1-({3\over 4})^3 -C^3_1({3\over 4})^2({1\over 4}) =1-{27\over 64}-{ 27 \over 64} ={10\over 64} ={5\over 32},故選\bbox[red,2pt]{(D)}
解答:X\sim U[a,b] \Rightarrow \cases{E(X)={a+b\over 2}=0 \\\sigma(X)= {b-a\over \sqrt {12}} =\sqrt {12}} \Rightarrow \cases{a+b= 0\\ b-a=12} \Rightarrow \cases{a=-6\\ b=6}\\ \Rightarrow 區間為[14-a,14+b]=[14:00-0:06,14:00+0:06] =[13:54,14:06],故選\bbox[red,2pt]{(A)}
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