108年高考三級工程數學詳解

108 年公務人員高等考試三級考試試題

類 科 ：電力工程、電子工程、電信工程、醫學工程
科 目：工程數學
甲、申論題部分：(50分)

解： $$\begin{cases}a=(1,1,1)\\b=(3,4,5) \\ c=(2,3,4)\\ d=(1,2,2)\end{cases} \Rightarrow \begin{cases}\overrightarrow{ab}=(2,3,4)\\\overrightarrow{ac}=(1,2,3) \\ \overrightarrow{ad}=(0,1,1)\end{cases} \Rightarrow 體積=\frac{1}{6} \left\|\begin{matrix}2&3&4\\ 1&2&3\\0& 1&1\end{matrix}\right\|=\frac{1}{6}|4+4+0-0-3-6|=\bbox[red,2pt]{\frac{1}{6}}$$

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解：
(一) $$C包住z=0\Rightarrow \int_C{\frac{\cos{z}}{z(z^2+8)}\,dz}\equiv \int_C{\frac{f(z)}{z}\,dz}=2\pi i\times f(0)=2\pi i\times \frac{\cos{0}}{0^2+8}=2\pi i\times \frac{1}{8}=\bbox[red,2pt]{\frac{\pi i}{4}}$$
(二) $$C包住z=0\Rightarrow \int_C{\frac{\cosh{z}}{z^4}\,dz}\equiv \int_C{\frac{f(z)}{z^4}\,dz}=\frac{2\pi i}{3!}\times f^{(3)}(0)=\frac{2\pi i}{3!}\times \sinh{0}=\bbox[red,2pt]{0}$$

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解： $$2\sin{\left(y^2\right)}dx+xy\cos{\left(y^2\right)}dy=0\Rightarrow \begin{cases}M(x,y)=2\sin{\left(y^2\right)}\\ N(x,y)=xy\cos{\left(y^2\right)}\end{cases} \Rightarrow  \begin{cases}{\partial M\over \partial y}=4y\cos{\left(y^2\right)}\\ {\partial N\over \partial x}=y\cos{\left(y^2\right)}\end{cases} \\\Rightarrow {\partial M\over \partial y}\ne {\partial N\over \partial x} \Rightarrow 非正合\\ 假設F=F(x)為積分因子\Rightarrow  {\partial (MF)\over \partial y}= {\partial (NF)\over \partial x} \Rightarrow 4y\cos{\left(y^2\right)}F= y\cos{\left(y^2\right)}F+xy\cos{\left(y^2\right)}{dF\over dx} \\ \Rightarrow  3y\cos{(y^2)} F = xy\cos{\left(y^2\right)}{dF\over dx} \Rightarrow {1\over F}{dF\over dx}={3\over x} \Rightarrow \int{{1\over F}{dF\over dx}\,dx}=\int{{3\over x}\,dx} \Rightarrow \ln{F}=3\ln{x} \\\Rightarrow F=x^3 \Rightarrow \begin{cases}{\partial u\over\partial x}=MF=2x^3\sin{(y^2)}\\ {\partial u\over\partial y}NF={x^{4}y\cos{(y^2)}}\end{cases} \Rightarrow \begin{cases}u(x,y)=\int{2x^3\sin{(y^2)}\,dx}=\frac{1}{2}x^4\sin{(y^2)}+p(y)\\ u(x,y)=\int{{x^{4}y\cos{(y^2)}}\,dy}=\frac{1}{2}x^4\sin{(y^2)}+q(x)\end{cases}\\ \Rightarrow u(x,y)=0為其通解，即\frac{1}{2}x^4\sin{(y^2)}+C=0為其通解，再由y(2)=\sqrt{\pi\over 2} \Rightarrow 8\sin{(\pi/2)}+C=0 \\ \Rightarrow C=-8 \Rightarrow 其解為\bbox[red,2pt]{x^4\sin{(y^2)}=16}。$$

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解：(一) $$A=\left[ \begin{matrix} 0 & 0 & -2 \\ 1 & 2 & 1 \\ 1 & 0 & 3 \end{matrix} \right] \Rightarrow det(A-\lambda I)=0\Rightarrow \left| \begin{matrix} -\lambda  & 0 & -2 \\ 1 & 2-\lambda  & 1 \\ 1 & 0 & 3-\lambda  \end{matrix} \right| =0\Rightarrow -\lambda (\lambda -2)(\lambda -3)-2(\lambda -2)=0\\ \Rightarrow (\lambda -2)(-\lambda ^{ 2 }+3\lambda -2)=0\Rightarrow (\lambda -2)^2(\lambda-1)=0\Rightarrow \lambda =1,2，即A的特徵值為\bbox[red,2pt]{1及2}$$ (二) $$\lambda =1\Rightarrow A-\lambda I=\left[ \begin{matrix} -1 & 0 & -2 \\ 1 & 1 & 1 \\ 1 & 0 & 2 \end{matrix} \right] \Rightarrow \left[ \begin{matrix} -1 & 0 & -2 \\ 1 & 1 & 1 \\ 1 & 0 & 2 \end{matrix} \right] \left[ \begin{matrix} x_{ 1 } \\ x_{ 2 } \\ x_{ 3 } \end{matrix} \right] =0\Rightarrow \begin{cases} -x_{ 1 }-2x_{ 3 }=0 \\ x_{ 1 }+x_{ 2 }+x_{ 3 }=0 \\ x_{ 1 }+2x_{ 3 }=0 \end{cases}\Rightarrow \left[ \begin{matrix} x_{ 1 } \\ x_{ 2 } \\ x_{ 3 } \end{matrix} \right] =a\left[ \begin{matrix} -2 \\ 1 \\ 1 \end{matrix} \right] \\ \lambda =2\Rightarrow A-\lambda I=\left[ \begin{matrix} -2 & 0 & -2 \\ 1 & 0 & 1 \\ 1 & 0 & 1 \end{matrix} \right] \Rightarrow \left[ \begin{matrix} -2 & 0 & -2 \\ 1 & 0 & 1 \\ 1 & 0 & 1 \end{matrix} \right] \left[ \begin{matrix} x_{ 1 } \\ x_{ 2 } \\ x_{ 3 } \end{matrix} \right] =0\Rightarrow x_{ 1 }+x_{ 3 }=0\Rightarrow \left[ \begin{matrix} x_{ 1 } \\ x_{ 2 } \\ x_{ 3 } \end{matrix} \right] =b\left[ \begin{matrix} 1 \\ 0 \\ -1 \end{matrix} \right] +c\left[ \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \right] \\ \Rightarrow 特徵向量為\bbox[red,2pt]{\left[ \begin{matrix} -2 \\ 1 \\ 1 \end{matrix} \right] ,\left[ \begin{matrix} 1 \\ 0 \\ -1 \end{matrix} \right] ,\left[ \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \right] }$$

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乙、測驗題部分：(50分)

解： $$(A)少一個條件 T(ax)=aT(x)\\(B) T不只是函數，更重要的是線性\\ (D) 不一定找得到線性的函數T\\，故選\bbox[red,2pt]{(C)}$$

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解： $$(D)\begin{cases} T(2,2)=\left( 0,0,0 \right)  \\ T(1,1)=\left( 0,0,0 \right)  \end{cases}\Rightarrow T(1,1)=T(2,2)\Rightarrow 不是一對一，故選\bbox[red,2pt]{(D)}$$

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解： $$A^{ T }B=\left[ \begin{matrix} 1 & 2 \\ 1 & 2 \\ 1 & 2 \\ 1 & 2 \end{matrix} \right] \left[ \begin{matrix} 1 & 3 & 5 & 7 \\ 2 & 4 & 6 & 8 \end{matrix} \right] =\left[ \begin{matrix} 5 & 11 & 17 & 23 \\ 5 & 11 & 17 & 23 \\ 5 & 11 & 17 & 23 \\ 5 & 11 & 17 & 23 \end{matrix} \right] \xrightarrow { -r_{ 1 }+r_{ 2 },-r_{ 1 }+r_{ 3 },-r_{ 1 }+r_{ 4 }, } \left[ \begin{matrix} 5 & 11 & 17 & 23 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} \right] \\ \Rightarrow \text{rank}(A^TB)=1 ，故選\bbox[red,2pt]{(A)}$$

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解： $$\left[ \begin{matrix} \frac { \sqrt { 3 }  }{ 2 }  & -\frac { 1 }{ 2 }  \\ \frac { 1 }{ 2 }  & \frac { \sqrt { 3 }  }{ 2 }  \end{matrix} \right] =\left[ \begin{matrix} \cos { \frac { \pi  }{ 6 }  }  & -\sin { \frac { \pi  }{ 6 }  }  \\ \sin { \frac { \pi  }{ 6 }  }  & \cos { \frac { \pi  }{ 6 }  }  \end{matrix} \right] 為旋轉\frac { \pi  }{ 6 } 矩陣\\ \Rightarrow \left[ \begin{matrix} \frac { \sqrt { 3 }  }{ 2 }  & -\frac { 1 }{ 2 }  \\ \frac { 1 }{ 2 }  & \frac { \sqrt { 3 }  }{ 2 }  \end{matrix} \right] ^{ 100 }=\left[ \begin{matrix} \cos { \frac { 100\pi  }{ 6 }  }  & -\sin { \frac { 100\pi  }{ 6 }  }  \\ \sin { \frac { 100\pi  }{ 6 }  }  & \cos { \frac { 100\pi  }{ 6 }  }  \end{matrix} \right] =\left[ \begin{matrix} \cos { \frac { 2\pi  }{ 3 }  }  & -\sin { \frac { 2\pi  }{ 3 }  }  \\ \sin { \frac { 2\pi  }{ 3 }  }  & \cos { \frac { 2\pi  }{ 3 }  }  \end{matrix} \right] \\ =\left[ \begin{matrix} -\frac { 1 }{ 2 }  & -\frac { \sqrt { 3 }  }{ 2 }  \\ \frac { \sqrt { 3 }  }{ 2 }  & -\frac { 1 }{ 2 }  \end{matrix} \right] ，故選\bbox[red,2pt]{(D)}$$

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解： $$L(x,y,z)=\left( 2x+3y+z,3x+3y+z,2x+4y+z \right) =\left[ \begin{matrix} 2 & 3 & 1 \\ 3 & 3 & 1 \\ 2 & 4 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =AX\\ \Rightarrow A^{ -1 }=\left[ \begin{matrix} -1 & 1 & 0 \\ -1 & 0 & 1 \\ 6 & -2 & -3 \end{matrix} \right] \Rightarrow L^{ -1 }(x,y,z)=\left( -x+y,-x+z,6x-2y-3z \right) ，故選\bbox[red,2pt]{(A)}$$

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解： $$A=\left[ \begin{matrix} 5 & 32 & 17 \\ 0 & 12 & 2 \\ 0 & -2 & 7 \end{matrix} \right] \\ (A)det\left( A-5I \right) =\left| \begin{matrix} 0 & 32 & 17 \\ 0 & 7 & 2 \\ 0 & -2 & 2 \end{matrix} \right| =0\\ (B)det\left( A-8I \right) =\left| \begin{matrix} -3 & 32 & 17 \\ 0 & 4 & 2 \\ 0 & -2 & -1 \end{matrix} \right| =\left| \begin{matrix} -3 & 32 & 17 \\ 0 & 0 & 0 \\ 0 & -2 & -1 \end{matrix} \right| =0\\ (C)det\left( A-11I \right) =\left| \begin{matrix} -6 & 32 & 17 \\ 0 & 1 & 2 \\ 0 & -2 & -4 \end{matrix} \right| =\left| \begin{matrix} -6 & 32 & 17 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{matrix} \right| =0\\ (D)det\left( A-14I \right) =\left| \begin{matrix} -9 & 32 & 17 \\ 0 & -2 & 2 \\ 0 & -2 & -7 \end{matrix} \right| \neq 0 ，故選\bbox[red,2pt]{(D)}$$ 當然也可以按部就班的計算 $$det(A-\lambda I)=0 \Rightarrow \lambda^3-24\lambda^2+183\lambda-440=0 \Rightarrow (\lambda-5)(\lambda-8)(\lambda-11)=0\\ \Rightarrow 特徵值\lambda= 5,8,11$$

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解： $$z=1+i=\sqrt { 2 } \left( \frac { 1 }{ \sqrt { 2 }  } +i\frac { 1 }{ \sqrt { 2 }  }  \right) =\sqrt { 2 } \left( \cos { \frac { 9\pi  }{ 4 }  } +i\sin { \frac { 9\pi  }{ 4 }  }  \right) =\sqrt { 2 } e^{ i\frac { 9\pi  }{ 4 }  }\\ \Rightarrow f\left( z \right) =r^{ \frac { 1 }{ 3 }  }\cos { \frac { \theta +2\pi  }{ 3 }  } +ir^{ \frac { 1 }{ 3 }  }\sin { \frac { \theta +2\pi  }{ 3 }  } \\ =\sqrt { 2 } ^{ \frac { 1 }{ 3 }  }\cos { \frac { \frac { 9\pi  }{ 4 } +2\pi  }{ 3 }  } +i\sqrt { 2 } ^{ \frac { 1 }{ 3 }  }\sin { \frac { \frac { 9\pi  }{ 4 } +2\pi  }{ 3 }  } \\ =\sqrt [ 6 ]{ 2 } \left( \cos { \frac { 17\pi  }{ 12 }  } +i\sin { \frac { 17\pi  }{ 12 }  }  \right) =\sqrt [ 6 ]{ 2 } e^{ i\frac { 17\pi  }{ 12 }  } ，故選\bbox[red,2pt]{(C)}$$

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解： $$C=\left\{ t+t^{ 2 }i\mid t:0\to 2 \right\} \Rightarrow \begin{cases} x=t \\ y=t^{ 2 } \end{cases}\Rightarrow \begin{cases} dx=dt \\ dy=2tdt \end{cases}\Rightarrow \int _{ C }{ z^{ 2 }\, dz } =\int _{ C }{ (x+yi)^{ 2 }\, d(x+yi) } \\ =\int _{ C }{ (x^{ 2 }-y^{ 2 })dx-2xydy } +i\int _{ C }{ (x^{ 2 }-y^{ 2 })dy+2xydx } \\ =\int _{ 0 }^{ 2 }{ (t^{ 2 }-t^{ 4 })dt-2t^{ 3 }\cdot 2tdt } +i\int _{ 0 }^{ 2 }{ 2(t^{ 2 }-t^{ 4 })tdt+2t^{ 3 }dt } \\ =\int _{ 0 }^{ 2 }{ (t^{ 2 }-5t^{ 4 })dt } +i\int _{ 0 }^{ 2 }{ (4t^{ 3 }-2t^{ 5 })dt } =\left. \left[ \frac { 1 }{ 3 } t^{ 3 }-t^{ 5 } \right]  \right| _{ 0 }^{ 2 }+i\left. \left[ t^{ 4 }-\frac { 1 }{ 3 } t^{ 6 } \right]  \right| _{ 0 }^{ 2 }\\ =\left( \frac { 8 }{ 3 } -16-32 \right) +i\left( 16-\frac { 64 }{ 3 }  \right) =-\frac { 88 }{ 3 } -\frac { 16 }{ 3 } i，故選\bbox[red,2pt]{(A)}$$

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解： $$\frac { n^{ n } }{ n! } (z+i)^{ { 2n } }=\frac { n^{ n } }{ n! } (z-(-i))^{ { 2n } }\Rightarrow 令c_{ n }=\sqrt { \frac { n^{ n } }{ n! }  } \Rightarrow R=\lim _{ n\to \infty  }{ \frac { c_{ n } }{ c_{ { n+1 } } }  } \\ =\lim _{ n\to \infty  }{ \sqrt { \frac { n^{ n } }{ n! } \times \frac { (n+1)! }{ (n+1)^{ n+1 } }  }  } =\lim _{ n\to \infty  }{ \sqrt { { \left( \frac { n }{ n+1 }  \right)  }^{ n }\times \frac { n+1 }{ n+1 }  }  } \\ =\lim _{ n\to \infty  }{ \sqrt { { \left( \frac { n }{ n+1 }  \right)  }^{ n } }  } =\sqrt { \lim _{ n\to \infty  }{ { \left( \frac { n }{ n+1 }  \right)  }^{ n } }  } =\sqrt { \frac { 1 }{ e }  } =\frac { 1 }{ \sqrt { e }  } \\ \Rightarrow \begin{cases} 中心點-i \\ 收斂半徑R=\frac { 1 }{ \sqrt { e }  }  \end{cases}，故選\bbox[red,2pt]{(C)}$$

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解： $$y=a\cos { \left( 3x \right)  } +b\sin { \left( 3x \right)  } +c\cos { \left( 4x \right)  } \\ \Rightarrow y'=-3a\sin { \left( 3x \right)  } +3b\cos { \left( 3x \right)  } -4c\sin { \left( 4x \right)  } \\ \Rightarrow \begin{cases} y(0)=0 \\ y'(0)=3 \end{cases}\Rightarrow \begin{cases} a+c=0 \\ 3b=3 \end{cases}\Rightarrow \begin{cases} a+c=0 \\ b=1 \end{cases}\Rightarrow a+b+c=1，故選\bbox[red,2pt]{(A)}$$

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解： $$L\{t\sin{(at)} \}={2as\over (s^2+a^2)^2} \Rightarrow L\{{t\over 2\omega}\sin{(\omega t)} \}={1\over 2\omega}{2\omega s\over (s^2+\omega^2)^2} = { s\over (s^2+\omega^2)^2}，故選\bbox[red,2pt]{(C)}$$

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解：
$$y''-{4\over x}y'+{4\over x^2}y=x^2+1,先求齊次解,即y''-{4\over x}y'+{4\over x^2}y=0\\ 令y=x^m\Rightarrow y'=mx^{m-1},y''=m(m-1)x^{m-2}代回齊次式\\ \Rightarrow m(m-1)x^{m-2}-4mx^{m-2}+4x^{m-2}=0\Rightarrow m(m-1)-4m+4=0 \\ \Rightarrow m^2-5m+4=0\Rightarrow (m-4)(m-1)=0\Rightarrow y_h=c_1x+c_2x^4\\ \Rightarrow 只剩下選項(C),(D)正確\\ (C) y_p=a_1+a_2x^2 \Rightarrow y_p'=2a_2x,y_p''=2a_2代入原式不合，故選\bbox[red,2pt]{(D)}。$$ 若要按部就班，可用參數變換法求解: $$y_h=c_1x+c_2x^4 \Rightarrow y_p=\phi_1 x+\phi_2 x^4 \Rightarrow \left[\begin{matrix}x&x^4\\{d\over dx}x &{d\over dx}x^4 \end{matrix}\right]\left[\begin{matrix}\phi_1'\\\phi_2' \end{matrix}\right]= \left[\begin{matrix}0\\x^2+1 \end{matrix}\right] \\ \Rightarrow \left[\begin{matrix}x&x^4\\1 &4x^3 \end{matrix}\right] \left[ \begin{matrix} \phi_1'\\\phi_2' \end{matrix}\right]= \left[\begin{matrix}0\\x^2+1 \end{matrix}\right] \Rightarrow \begin{cases} x\phi_1'+x^4\phi_2'=0\\\phi_1'+4x^3\phi_2'=x^2+1 \end{cases} \Rightarrow \begin{cases} \phi_1'=-x^2/3-1/3\\\phi_2'=1/3x+ 1/3x^3 \end{cases}\\ \Rightarrow \begin{cases} \phi_1=\int{(-x^2/3-1/3)\,dx}=-{1\over 9}x^3-{1\over 3}x\\\phi_2=\int{(1/3x+ 1/3x^3)\,dx} ={1\over 3}\ln{x}-{1\over 6x^2} \end{cases} \Rightarrow y_p=\phi_1 x+\phi_2 x^4\\ = -{1\over 9}x^4-{1\over 3}x^2+{1\over 3}x^4\ln{x}-{1\over 6}x^2= -{1\over 9}x^4-{1\over 2}x^2+{1\over 3}x^4\ln{x}\\ \Rightarrow y=y_h+y_p=c_1 x+(c_2-1/9)x^4-{1\over 2}x^2+{1\over 3}x^4\ln{x}\equiv C_1x+C_2x^4+A_1x^2+A_2x^4\ln{x}$$

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解： $$y=e^{-2x}\cos{x} \Rightarrow \begin{cases}y'=-2e^{-2x}\cos{x}-e^{-2x}\sin{x}\\ y''=3e^{-2x}\cos{x}+4e^{-2x}\sin{x}\\ y'''=-2e^{-2x}\cos{x}-11e^{-2x}\sin{x} \end{cases}\Rightarrow Ay'''+By''+Cy'+Dy=0\\ \Rightarrow (-2A+3B-2C+D)e^{-2x}\cos{x}+ (-11A+4B-C)e^{-2x}\sin{x}=0\\ \Rightarrow \begin{cases}-2A+3B-2c+D=0\\ -11A+4B-C=0\end{cases} \\(A)\times:A=1,B=7,C=16,D=-10\Rightarrow -11A+4B-C=-11+28-16=1\ne 0\\ (B)\bigcirc:A=B=1,C=-7,D=15 \Rightarrow \begin{cases}-2A+3B-2c+D=0\\ -11A+4B-C=0\end{cases} \\(C)\times:A=0,B=1,C=8,D=17 \Rightarrow -11A+4B-C=0+4-8=-4\ne 0 \\ (D) \times: A=3,B=2,C =-8,D=-16\Rightarrow -11A+4B-C=-33+8+8=-17\ne 0 \\，故選\bbox[red,2pt]{(B)}$$

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解： $$(A)y=x\Rightarrow y''+Ay'+By=0+A+BX=0 \Rightarrow A=0且B=0\\(B)y=x^2 \Rightarrow y''+Ay'+By=2+2Ax+Bx^2=0，無法找出A,B滿足該式\\ (C)y=e^{x+1}\Rightarrow y''+Ay'+By= e^{x+1}+Ae^{x+1}+Be^{x+1}=(A+B+1)e^{x+1}=0\\ \Rightarrow A+B=-1\\ (D)y=e^x\cos{(2x+3)} \Rightarrow y''+Ay'+By= (A+B-3)e^x\cos{(2x+3)}-(2A+4)e^x\sin{(2x+3)} =0\\ \Rightarrow \begin{cases}A+B-3=0\\2A+4=0\end{cases}\\，故選\bbox[red,2pt]{(B)}$$

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解： $$(A)\bigcirc :u(t,x)=\sin{2t}\sin{x}\Rightarrow \begin{cases}\frac{\partial u}{\partial x}=\sin{2t}\cos{x}\\ \frac{\partial u}{\partial t}=2\cos{2t}\sin{x}\end{cases} \Rightarrow \begin{cases}\frac{\partial^2 u}{\partial x^2}=-\sin{2t}\sin{x}\\ \frac{\partial^2 u}{\partial t^2}=-4\sin{2t}\sin{x}\end{cases} \Rightarrow \frac{\partial^2 u}{\partial t^2}= 2^2\frac{\partial^2 u}{\partial x^2}\\(B)\bigcirc :u(t,x)=e^{-4t}\cos{3x} \Rightarrow \begin{cases}\frac{\partial u}{\partial x}=-3e^{-4t}\sin{3x} \Rightarrow \frac{\partial^2 u}{\partial x^2}=-9e^{-t}\cos{3x}\\ \frac{\partial u}{\partial t}= -4e^{-4t}\cos{3x} \end{cases} \Rightarrow \frac{\partial u}{\partial t}= (3/2)^2\frac{\partial^2 u}{\partial x^2}\\ (C)\times :u(t,x)=e^t\sin{3x} \Rightarrow \begin{cases}\frac{\partial u}{\partial x}=3e^t\cos{3t}\\ \frac{\partial u}{\partial t}= e^t\sin{3x} \end{cases} \Rightarrow \begin{cases}\frac{\partial^2 u}{\partial x^2}=-9e^t\sin{3x}\\ \frac{\partial^2 u}{\partial t^2}=e^t\sin{3x}\end{cases} \Rightarrow \frac{\partial^2 u}{\partial t^2}=- (1/3)^2\frac{\partial^2 u}{\partial x^2}\\ (D)\bigcirc :u(t,x)=e^{-t}\sin{3x} \Rightarrow \begin{cases}\frac{\partial u}{\partial x}=3e^{-t}\cos{3x} \Rightarrow \frac{\partial^2 u}{\partial x^2}=-9e^{-t}\sin{3x}\\ \frac{\partial u}{\partial t}= -e^{-t}\sin{3x} \end{cases} \Rightarrow \frac{\partial u}{\partial t}= 3^2\frac{\partial^2 u}{\partial x^2}\\，故選\bbox[red,2pt]{(C)}$$

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解： $$L\left\{t*e^{2t} \right\}=L\{t\}\cdot L\{e^{2t}\}=\frac{1}{s^2}\cdot\frac{1}{s-2}=\frac{1}{s^2(s-2)}，故選\bbox[red,2pt]{(C)}$$

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解： $$b_n=\frac{1}{5} \int_{-5}^5{F(x)\sin{\left(\frac{n\pi}{5}x\right)}\,dx} = \frac{1}{5} \int_{0}^5 {3 \sin{\left(\frac{n\pi}{5}x\right)}\,dx} = \frac{3}{5}\left.\left[-\frac{5}{n\pi}\cos{\left(\frac{n\pi}{5}x\right)}\right]\right|_0^5\\ =-\frac{3}{n\pi}(\cos{(n\pi)}-1)=\frac{3(1-\cos{(n\pi))}}{n\pi} ，故選\bbox[red,2pt]{(C)}$$

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解： $$P(0< X\le 1)=\int_0^1{\frac{x^2}{3}\,dx} = \left.\left[\frac{1}{9}x^3\right]\right|_0^1=\frac{1}{9}，故選\bbox[red,2pt]{(A)}$$

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解： $$P[4< X \le 12,0 < Y < \infty]=\int_0^\infty{\int_4^{12}{\frac{1}{12}e^{-(x/4)-(y/3)}\,dx}\,dy}= \int_0^\infty{\left. \left[-\frac{1}{3}e^{-(x/4)-(y/3)}\right]\right|_4^{12}\,dy}\\ = -\frac{1}{3}\left(\int_0^\infty{e^{-3-y/3}-e^{-1-y/3}\,dy}\right)=-\frac{1}{3}\left.\left[-3e^{-3-y/3}+3e^{-1-y/3}\right]\right|_0^\infty= -\frac{1}{3}\left(0-(-3e^{-3}+3e^{-1}) \right)\\ =-\frac{1}{3}\left(3e^{-3}-3e^{-1}\right)=e^{-1}-e^{-3}，故選\bbox[red,2pt]{(B)}$$

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解： $$E[X]=x\int{f(x)\,dx}=\int_0^1{2x^2\,dx}=\left.\left[\frac{2}{3}x^3\right]\right|_0^1={2\over 3}，故選\bbox[red,2pt]{(C)}$$

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考選部未公布非選答案，解題僅供參考