110年專科學力鑑定-工程數學詳解

教育部 110 年自學進修專科學校學力鑑定考試

解答：$$(A)\times: \cases{f(x)為偶函數\\ \sin(nx)為奇函數} \Rightarrow f(x)\sin(nx)為奇函數 \Rightarrow \int_{-1}^1 f(x)\sin(nx)\;dx=0 \\(B)\times: \cases{f(x)為偶函數\\ \cos(nx)為偶函數} \Rightarrow f(x)\sin(nx)為偶函數 \Rightarrow \int_{-1}^1 f(x)\sin(nx)\;dx\ne 0\\(C)\times: f(x)為偶函數\Rightarrow f(-x)=f(x)\ne -f(x)\\(D)\bigcirc: f(x)為偶函數\Rightarrow \int_{-1}^1 f(x)\;dx = \int_{-1}^0 f(x)\;dx +\int_0^{1} f(x)\;dx =2\int_0^{1} f(x)\;dx \\，故選\bbox[red,2pt]{(D)}$$

解答：$$\cases{\vec u=(1,2,a)\\ \vec v=(2,1,3)\\ \vec n=\vec u\times \vec v=(5,-1,b)} \Rightarrow \cases{\vec u\times \vec n=0\\ \vec v\times \vec n=0} \Rightarrow \cases{5-2+ab=0\\ 10-1+3b=0} \Rightarrow \cases{a=1\\ b=-3}\\ \Rightarrow a+b= 1-3=-2，故選\bbox[red,2pt]{(A)}$$

解答：$$\cases{\vec u=(2,1,2)\\ \vec v=(2,6,a)\\ \vec w=(1,3,5) } \Rightarrow \vec u\cdot (\vec v\times \vec w)=(2,1,2)\cdot ((2,6,a ) \times (1,3,5)) =(2,1,2)\cdot (30-3a,a-10,0)\\ =60-6a+a-10= 50-5a=0 \Rightarrow a=10，故選\bbox[red,2pt]{(D)}$$

解答：$$2A+3B-C = 2\begin{bmatrix}1 & 1\\ 2& 3 \end{bmatrix} +3\begin{bmatrix}2 & 1\\ 3& 4 \end{bmatrix}-\begin{bmatrix}3 & 2\\ 1& 6 \end{bmatrix} = \begin{bmatrix}2 & 2\\ 4& 6 \end{bmatrix} + \begin{bmatrix}6 & 3\\ 9& 12 \end{bmatrix}-\begin{bmatrix}3 & 2\\ 1& 6 \end{bmatrix} \\ = \begin{bmatrix}8 & 5\\ 13& 18 \end{bmatrix}  -\begin{bmatrix}3 & 2\\ 1& 6 \end{bmatrix} = \begin{bmatrix}5 & 3\\ 12& 12 \end{bmatrix} ，故選\bbox[red,2pt]{(B)}$$

解答：$$ \begin{vmatrix}10 & 6 & a\\ 2& 1 & 6 \\ 5 & 3 & 2 \end{vmatrix} + \begin{vmatrix}-10 & -6 & -1\\ 2& 1 & 6 \\ 5 & 3 & 2 \end{vmatrix}  =(a-4)+3=5 \Rightarrow a=6，故選\bbox[red,2pt]{(D)}$$

解答：$$ \begin{bmatrix}5 &  a\\ b& 1 \end{bmatrix}   \begin{bmatrix}1 &  2\\ 2& 5 \end{bmatrix}  =\begin{bmatrix}5+2a &  10+5a\\ b+2& 2b+5 \end{bmatrix} =\begin{bmatrix}1 &  0\\ 0& 1 \end{bmatrix} \Rightarrow \cases{a=-2\\ b=-2} \\ \Rightarrow a+2b=-2-4=-6，故選\bbox[red,2pt]{(A)}$$

解答：$$y'={x+1\over y^2} \Rightarrow y^2y'=x+1 \Rightarrow \int y^2dy=\int x+1\;dx \Rightarrow {1\over 3}y^3={1\over 2}x^2+x+C\\ 將y(0)=1代入上式\Rightarrow {1\over 3}=C \Rightarrow {1\over 3}y^3={1\over 2}x^2+x+{1\over 3} \Rightarrow y^3={3\over 2}x^2+3x+1 \\ \Rightarrow y=\sqrt[3]{1.5x^2+3x+1}，故選\bbox[red,2pt]{(B)}$$

解答：$$(2\vec u-\vec v)\cdot \vec w=((6,0,2)-(2,3,1))\cdot (a,2,5)=(4,-3,1)\cdot (a,2,5) =4a-6+5=23\\ \Rightarrow 4a=24 \Rightarrow a=6，故選\bbox[red,2pt]{(B)}$$

解答：$$可以簡化成y'=f({y\over x})，稱為齊次\\(A)y'={xy+y^2\over x^2} ={y\over x}+({y\over x})^2 \Rightarrow 齊次，故選\bbox[red,2pt]{(A)}$$

解答：$$W(x)=\begin{vmatrix} f_1& f_2\\ f_1' & f_2'\end{vmatrix}\\(A)\times: W(x)=\begin{vmatrix} \sin(2x) & \sin(x)\cos (x)\\ 2\cos(2x) & \cos(2x) \end{vmatrix} =\sin(2x)\cos(2x)- \cos(2x)\sin(2x)=0 \\(B)\times: W(x)=\begin{vmatrix} \ln(x^2) & \ln(x)\\ 2/x & 1/x \end{vmatrix}={\ln(x^2)\over x}-{2\ln(x)\over x} ={2\ln(x )\over x}-{2\ln(x)\over x}=0 \\(C)\bigcirc:  W(x)=\begin{vmatrix} \sin(x) &  \cos (x)\\ \cos(x) & -\sin(x) \end{vmatrix} = -\sin^2(x)-\cos^2(x)=-1 \ne 0\\ (D)\times: W(x)=\begin{vmatrix} x^2+x & 4x(x+1)\\ 2x+1 & 4(2x+1)  \end{vmatrix} =4x(x+1)(2x+1)-4x(x+1)(2x+1)=0\\，故選\bbox[red,2pt]{(C)}$$

解答：$$(e^x-\sin y)dx+ \cos y\;dy=0 \Rightarrow 令\cases{M(x,y)= e^x-\sin y\\N(x,y)=\cos y}\\(A)\bigcirc: \cases{e^{-x}M(x,y)=1-e^{-x} \sin y \\ e^{-x}N(x,y)= e^{-x}\cos y} \Rightarrow \cases{{d\over dy}e^{-x}M(x,y) =-e^{-x}\cos y \\ {d\over dx} e^{-x}N(x,y)=-e^{-x}\cos y}，兩者相同\\ (B)\times:  \cases{\sin (x)M(x,y)=\sin (x)e^{x}-\sin (x) \sin y \\ \sin (x)N(x,y)= \sin (x)\cos y} \Rightarrow \cases{{d\over dy}\sin (x)M(x,y)= -\sin (x) \cos y \\ {d\over dx}\sin (x)N(x,y)= \cos (x)\cos y}，兩者不同\\ (C)\times:  \cases{e^xM(x,y)=e^{2x}-e^x\sin y \\ e^x N(x,y)= e^x\cos y} \Rightarrow \cases{{d\over dy}e^xM(x,y)= -e^x\cos y \\ {d\over dx}e^xN(x,y)= e^x\cos y}，兩者不同 \\(D)\times:  \cases{\cos (x)M(x,y)=\cos (x)e^{x}-\cos (x) \sin y \\ \cos (x)N(x,y)= \cos (x)\cos y} \Rightarrow \cases{{d\over dy}\cos (x)M(x,y)= -\cos (x) \cos y \\ {d\over dx}\cos (x)N(x,y)= -\sin (x)\cos y}，兩者不同\\，故選\bbox[red,2pt]{(A)}$$

解答：$$\cases{x+y=4\\ x-y=-2} \Rightarrow \cases{x=1\\ y=3} 代入\cases{ax+2y=8\\ 2x+by=5} \Rightarrow \cases{a+6=8\\ 2+3b=5} \Rightarrow \cases{a=2\\b=1}\\ \Rightarrow a+b=3，故選\bbox[red,2pt]{(C)}$$

解答：$$y''+y'-6y=0 \Rightarrow \lambda^2+\lambda-6=0 \Rightarrow (\lambda+3)(\lambda-2)=0 \Rightarrow \lambda=-3,2 \Rightarrow y=C_1e^{-3x}+C_2e^{2x}\\，故選\bbox[red,2pt]{(B)}$$

解答：$$(A)\times:\mathcal{L}\{e^{at}\}={1\over s-a} \ne {s\over s-a}\\(B)\times: \mathcal{L}\{\sin(t)\}={1\over s^2+1} \ne{ s\over s^2+1} \\(C)\bigcirc: \mathcal{L}\{e^{at}f(t)\} =F(s-a)\\ (D)\times: \mathcal{L}\{f(at)\}={1\over a}F({s\over a}) \ne aF(as)\\，故選\bbox[red,2pt]{(C)}$$

解答：$$AB^t=\begin{bmatrix}3 & 1 & 2\\ 2 & 5 & 3\end{bmatrix}\begin{bmatrix}2 & 3\\ 1 & 5 \\ 2& 0\end{bmatrix} = \begin{bmatrix}11 & 14  \\ 15 & 31\end{bmatrix}，故選\bbox[red,2pt]{(A)}$$

解答：$$\mathcal{L}\{f*g\}=\mathcal{L}\{f \}\times\mathcal{L}\{ g\} \Rightarrow \mathcal{L}\{e^t*\sin(t)\} =\mathcal{L}\{e^t\} \times \mathcal{L}\{\sin(t)\} ={1\over s-1}\times {1\over s^2+1}，故選\bbox[red,2pt]{(D)}$$

解答：$$\int \sin(nx)\sin(nx)\;dx=\int \sin^2(nx)\;dx \gt 0，故選\bbox[red,2pt]{(C)}$$

解答：$$(A-\lambda I)X=0 \Rightarrow \begin{bmatrix} 2-7 & a\\ 5 & 4-7\end{bmatrix}\begin{bmatrix}3\\5 \end{bmatrix} = \begin{bmatrix} -5 & a\\ 5 & -3\end{bmatrix}\begin{bmatrix}3\\5 \end{bmatrix} =0 \\ \Rightarrow -15+5a=0 \Rightarrow a=3，故選\bbox[red,2pt]{(C)}$$

解答：$$\det(A-\lambda I)=0 \Rightarrow \begin{vmatrix} 5-\lambda & 1 & 0\\ 1 & 2-\lambda & 3\\ 3 & 1 & 2-\lambda \end{vmatrix}=0 \Rightarrow \lambda^3-9\lambda^2+20\lambda-12=0\\ \Rightarrow (\lambda-2)(\lambda^2-7\lambda+6)=0 \Rightarrow (\lambda-2)(\lambda-1)(\lambda-6)=0 \Rightarrow \lambda=1,2,6，故選\bbox[red,2pt]{(D)}$$

解答：$$g(t)=\cases{0,t\lt a\\ f(t-a),t\ge a} =u(t-a)f(t-a)  \Rightarrow \mathcal{L}\{g(t)\} =\mathcal{L}\{u(t-a)f(t-a)\} =e^{-as}F(s)\\，故選\bbox[red,2pt]{(A)}$$

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