107年專門職業及技術人員高等考試
建築師、技師、第二次食品技師考試暨普通考試不動產經紀人、記帳士考試試題
建築師、技師、第二次食品技師考試暨普通考試不動產經紀人、記帳士考試試題
等別:高等考試
類別:電子工程技師
程目:工程數學
解:y″−3y′+2y=2x+ex+cos(e−x)=r(x)先求齊次解,λ2−3λ+2=0⇒(λ−2)(λ−1)=0⇒λ=1,2⇒yh=C1ex+C2e2x令y1=ex,y2=e2x⇒W=|y1y2y′1y′2|=|exe2xex2e2x|=2e3x−e3x=e3x⇒y=−y1∫y2r(x)Wdx+y2∫y1r(x)Wdx=−ex∫e2x(2x+ex+cos(e−x))e3xdx+e2x∫ex(2x+ex+cos(e−x))e3xdx=−ex∫2x+ex+cos(e−x)exdx+e2x∫2x+ex+cos(e−x)e2xdx=−ex∫(2xe−x+1+e−xcos(e−x))dx+e2x∫(2xe−2x+e−x+e−2xcos(e−x))dx=−ex(−2xe−x−2e−x+x−sin(e−x))+e2x(−xe−2x−12e−2x−e−x−e−xsin(e−x)−cos(e−x))=ex(2xe−x+2e−x−x+sin(e−x))−e2x(xe−2x+12e−2x+e−x+e−xsin(e−x)+cos(e−x))=2x+2−xex+exsin(e−x)−x−12−ex−exsin(e−x)−e2xcos(e−x)=32+x−ex−xex−e2xcos(e−x)⇒y=yh+yp=C1ex+C2e2x+32+x−xex−e2xcos(e−x)
解:
f(x+25)=f(x)⇒L{f(x)}=∫250e−stf(t)dt1−e−25s=∫1055e−stdt1−e−25s=5[−1se−st]|1051−e−25s=51−e−25s(−1se−10s+1se−5s)=5e−5ss(1−e−25s)(1−e−5s)
解:{z=a+bi⇒e2z=e2a+2bi=e2ae2bi=e2a(cos(2b)+isin(2b))3+2i=√13(3√13+i2√13)=√13(cosθ+isinθ)⇒{e2a=√132b=θ=cos−13√13⇒{a=14ln13b=12cos−13√13⇒z=14ln13+i12cos−13√13
解:f(z)=2iz+sin(z)z3+z=2iz+sin(z)z(z+i)(z−i)=g(z)z+i⇒z=−i的殘值為lim
解:
撲克牌有四種花色,各有13張
(一)取到同花的次數為4\times C^{13}_5,52張抽5張有C^{52}_5種抽法,因此取到同花的機率為\bbox[red,2pt]{\frac{4\times C^{13}_5}{C^{52}_5}}
(二)「順」的情況有(1-5), (2-6),..., (10,11,12,13,A) 共有10種情況,因此同花順的機率為\frac{4\times 10}{C^{52}_5}=\bbox[red,2pt]{\frac{40}{C^{52}_{5}}}
解:令T=\left[ \begin{matrix} a & b \\ c & d \end{matrix} \right] \\ \begin{cases} T\left( \left[ \begin{matrix} 2 \\ 0 \end{matrix} \right] \right) =\left[ \begin{matrix} 2 \\ -4 \end{matrix} \right] \\ T\left( \left[ \begin{matrix} 0 \\ 1 \end{matrix} \right] \right) =\left[ \begin{matrix} -1 \\ 1 \end{matrix} \right] \end{cases}\Rightarrow \begin{cases} \left[ \begin{matrix} a & b \\ c & d \end{matrix} \right] \left[ \begin{matrix} 2 \\ 0 \end{matrix} \right] =\left[ \begin{matrix} 2 \\ -4 \end{matrix} \right] \\ \left[ \begin{matrix} a & b \\ c & d \end{matrix} \right] \left[ \begin{matrix} 0 \\ 1 \end{matrix} \right] =\left[ \begin{matrix} -1 \\ 1 \end{matrix} \right] \end{cases}\Rightarrow \begin{cases} 2a=2 \\ 2c=-4 \\ b=-1 \\ d=1 \end{cases}\\ \Rightarrow \begin{cases} a=1 \\ c=-2 \\ b=-1 \\ d=1 \end{cases}\Rightarrow \bbox[red,2pt]{T=\left[ \begin{matrix} 1 & -1 \\ -2 & 1 \end{matrix} \right] }\\ T\left( \left[ \begin{matrix} 3 \\ 6 \end{matrix} \right] \right) =\left[ \begin{matrix} 1 & -1 \\ -2 & 1 \end{matrix} \right] \left[ \begin{matrix} 3 \\ 6 \end{matrix} \right] =\left[ \begin{matrix} 3-6 \\ -6+6 \end{matrix} \right] =\bbox[red,2pt]{\left[ \begin{matrix} -3 \\ 0 \end{matrix} \right] }
解:A=\left[ \begin{matrix} 2 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 1 & 0 & -3 & 0 \\ -1 & -2 & 0 & -3 \end{matrix} \right] \Rightarrow det\left( A \right) =\left| \begin{matrix} 2 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 1 & 0 & -3 & 0 \\ -1 & -2 & 0 & -3 \end{matrix} \right| =2\times \left| \begin{matrix} 2 & 0 & 0 \\ 0 & -3 & 0 \\ -2 & 0 & -3 \end{matrix} \right| \\=2\times 2\times \left( -3 \right) \times \left( -3 \right) =\bbox[red,2pt]{36}\\ det\left( A-\lambda I \right) =0\Rightarrow \left| \begin{matrix} 2-\lambda & 0 & 0 & 0 \\ 0 & 2-\lambda & 0 & 0 \\ 1 & 0 & -3-\lambda & 0 \\ -1 & -2 & 0 & -3-\lambda \end{matrix} \right| =0\Rightarrow { \left( \lambda -2 \right) }^{ 2 }{ \left( \lambda +3 \right) }^{ 2 }=0\\\Rightarrow 特徵值\lambda =\bbox[red,2pt]{2,-3}\\ \lambda =2\Rightarrow \left[ \begin{matrix} 2-\lambda & 0 & 0 & 0 \\ 0 & 2-\lambda & 0 & 0 \\ 1 & 0 & -3-\lambda & 0 \\ -1 & -2 & 0 & -3-\lambda \end{matrix} \right] \left[ \begin{matrix} x_{ 1 } \\ x_{ 2 } \\ x_{ 3 } \\ x_{ 4 } \end{matrix} \right] =0\Rightarrow \left[ \begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & -5 & 0 \\ -1 & -2 & 0 & -5 \end{matrix} \right] \left[ \begin{matrix} x_{ 1 } \\ x_{ 2 } \\ x_{ 3 } \\ x_{ 4 } \end{matrix} \right] =0\\\Rightarrow \begin{cases} x_{ 1 }=5x_{ 3 } \\ x_{ 1 }+2x_{ 2 }+5x_{ 4 }=0 \end{cases} \Rightarrow 取特徵向量u_{ 1 }=\bbox[red,2pt]{\left[ \begin{matrix} 5 \\ 0 \\ 1 \\ -1 \end{matrix} \right] }及u_{ 2 }= \bbox[red,2pt]{ \left[ \begin{matrix} 0 \\ -5 \\ 0 \\ 2 \end{matrix} \right] }\\ \lambda =-3\Rightarrow \left[ \begin{matrix} 2-\lambda & 0 & 0 & 0 \\ 0 & 2-\lambda & 0 & 0 \\ 1 & 0 & -3-\lambda & 0 \\ -1 & -2 & 0 & -3-\lambda \end{matrix} \right] \left[ \begin{matrix} x_{ 1 } \\ x_{ 2 } \\ x_{ 3 } \\ x_{ 4 } \end{matrix} \right] =0\Rightarrow \left[ \begin{matrix} 5 & 0 & 0 & 0 \\ 0 & 5 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ -1 & -2 & 0 & 0 \end{matrix} \right] \left[ \begin{matrix} x_{ 1 } \\ x_{ 2 } \\ x_{ 3 } \\ x_{ 4 } \end{matrix} \right] =0\\\Rightarrow \begin{cases} x_{ 1 }=0 \\ x_{ 2 }=0 \\ x_{ 1 }+2x_{ 1 }=0 \end{cases} \Rightarrow 取特徵向量u_{ 3 }=\bbox[red,2pt]{\left[ \begin{matrix} 0 \\ 0 \\ 1 \\ 0 \end{matrix} \right]} 及u_{ 4 }=\bbox[red,2pt]{\left[ \begin{matrix} 0 \\ 0 \\ 0 \\ 1 \end{matrix} \right] }
解:
(一)A=\left[\begin{array}{rrr|c}1&0&1&0\\0&1&2&0\\2&-1&0&0\\1&-1&-1&0\end{array}\right]\xrightarrow{(-2)\times r_1+r_3,\,(-1)\times r_1+r_4}\left[\begin{array}{rrr|c}1&0&1&0\\0&1&2&0\\0&-1&-2&0\\0&-1&-2&0\end{array}\right] \\\xrightarrow{r_2+r_3,\, r_2+r_4}\left[\begin{array}{rrr|c}1&0&1&0\\0&1&2&0\\0&0&0&0\\0&0&0&0\end{array}\right]\Rightarrow Ax=0\Rightarrow x=\alpha\left[\begin{array}{r}1\\2\\-1\end{array}\right]\Rightarrow \bbox[red,2pt]{\left\{(1,2,-1)\right\}}可為Ax=0之x一組基底(二)A=\left[\begin{array}{rrr|c}1&0&1&b_1\\0&1&2&b_2\\2&-1&0&b_3\\1&-1&-1&b_4\end{array}\right]\xrightarrow{(-2)\times r_1+r_3,\,(-1)\times r_1+r_4}\left[\begin{array}{rrr|l}1&0&1&b_1\\0&1&2&b_2\\0&-1&-2&b_3-2b_1\\0&-1&-2&b_4-b_1\end{array}\right] \\\xrightarrow{r_2+r_3,\, r_2+r_4}\left[\begin{array}{rrr|l}1&0&1&b_1\\0&1&2&b_2\\0&0&0&b_2+b_3-2b_1\\0&0&0&b_2+b_4-b_1\end{array}\right]\Rightarrow Ax=b\Rightarrow \begin{cases}b_2+b_3-2b_1=0\\b_2+b_4-b_1=0\end{cases} \\\Rightarrow b=\alpha\left[\begin{array}{}1\\0\\2\\1\end{array}\right]+\beta\left[\begin{array}{}1\\1\\1\\0\end{array}\right] \Rightarrow \bbox[red,2pt]{\left\{(1,0,2,1),(1,1,1,0)\right\}}可作為b之一組基底
解題僅供參考
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