115年台大碩士班-基礎數學詳解

國立臺灣大學 115 學年度碩士班招生考試試題

科目: 基礎數學

解答:$$P^T=(P^TP)^T=P^TP=P \Rightarrow P^T=P \Rightarrow P為對稱矩陣\\P=P^T \Rightarrow P^2=P^TP=P \Rightarrow P^2=P \Rightarrow P為投影矩陣  \;\bbox[red, 2pt]{QED.}$$

解答:$$A = \begin{bmatrix}1& 1\\1& 1\\0& 1 \end{bmatrix} \Rightarrow P=A(A^TA)^{-1}A^T = \begin{bmatrix}1& 1\\1& 1\\0& 1 \end{bmatrix} \begin{bmatrix}2& 2\\2& 3 \end{bmatrix}^{-1} \begin{bmatrix}1& 1& 0\\ 1& 1& 1 \end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix}1/2& 1/2 & 0\\1/2& 1/2 & 0\\ 0& 0& 1 \end{bmatrix}} \\ b= a_1\times a_2= \bbox[red, 2pt]{(1,-1,0)}$$

解答:$$\det(A)=\begin{vmatrix}1& 1& 0& 0\\ 0& 1& 1&0\\ 0& 0& 1& 1\\ -1& 0& 0& -{1\over 2} \end{vmatrix} =\begin{vmatrix}    1& 1&0\\   0& 1& 1\\   0& 0& -{1\over 2} \end{vmatrix} +\begin{vmatrix}  1& 0& 0\\   1& 1&0\\   0& 1& 1  \end{vmatrix} =-{1\over 2}+1 ={1\over 2} \\ \Rightarrow \cases{\det(2A)=2^4\det(A)=8\\ \det(-A)=(-1)^4\det(A)={1\over 2} \\ \det(A^2) =\det(A)\cdot \det(A)={1\over 4} \\ 1=\det(I) =\det(A\cdot A^{-1}) =\det(A)\cdot \det(A^{-1}) \Rightarrow {\det(A^{-1})=2}} \Rightarrow \bbox[red, 2pt]{\cases{\det(2A)=8\\ \det(-A)={1\over 2} \\ \det(A^2) ={1\over 4} \\ \det(A^{-1})=2}}$$

解答:$$u=\sin(x) \Rightarrow du= \cos (x)\,dx \Rightarrow   \int{\cos(x)dx\over \sqrt[7]{\sin^{11}(x)}} = \int {du\over \sqrt[7]{u^{11}}} =\int u^{-11/7} \,du=-{7\over 4}u^{-4/7}+C \\=\bbox[red, 2pt]{  -{7\over 4} \sin^{-4/7}(x)+C } $$

解答:$$u=e^x \Rightarrow du=e^x \,dx \Rightarrow \int e^{2x} e^{e^x}\,dx =\int ue^u\,du= ue^u-e^u+C= \bbox[red, 2pt]{e^x e^{e^x}-e^{e^x}+C}$$

解答:$$u=\ln(x) \Rightarrow du={dx\over x} \Rightarrow \int u\,du={1\over 2}u^2+C= \bbox[red, 2pt]{{1\over 2}(\ln(x))^2+C}$$

解答:$$\lim_{x\to 0} {e^x-x-1\over x^2} =\lim_{x\to 0} {e^x- 1\over 2x} =\lim_{x\to 0} {e^x\over 2} =\bbox[red, 2pt]{1\over 2}$$

解答:$$e^x=1+x+{x^2\over 2!} +\cdots+{x^n\over n!}+ \cdots \Rightarrow \lim_{x\to \infty}{x^n\over e^x} =\lim_{x\to \infty} {x^n \over 1+x+{x^2\over 2!} +\cdots+{x^n\over n!}+ \cdots} =\bbox[red, 2pt]0$$

解答:$$f(x)={x(x^2-5)+5(x-1) \over x^3-1} ={x^3-5\over x^3-1} \Rightarrow f'(x)={3x^2\over x^3-1}-{3x^2(x^3-5) \over (x^3-1)^2} = \bbox[red, 2pt]{12x^2\over (x^3-1)^2}$$

解答:$$f(x)=(x^3+x)^4 \Rightarrow f'(x)=4(x^3+x)^3(3x^2+1) \Rightarrow f''(x)=12(x^3+x)^2(3x^2+1)^2+ 4(x^3+x)^3\cdot 6x \\\qquad = \bbox[red, 2pt]{12(x^3+x)^2(11x^4+8x^2+1)}$$

解答:$$\lim_{x\to -2}{x+2\over (3x^2+10x+8)^2} =\lim_{x\to -2}{x+2\over (x+2)^2 (3x+4) ^2} =\lim_{x\to -2}{1\over (x+2) (3x+4) ^2} \\由於\cases{ \lim_{x\to -2^+}{1\over (x+2) (3x+4) ^2} =\infty\\ \lim_{x\to -2^-}{1\over (x+2) (3x+4) ^2} =-\infty } \Rightarrow \lim_{x\to -2}{1\over (x+2) (3x+4) ^2}\bbox[red, 2pt]{不存在}$$

解答:

$$面積= \int_1^2 \left( -3x+9 -(x^2-1) \right) \,dx +\int_2^3 \left( x^2-1-(-3x+9) \right)\,dx \\=\int_1^2 (-x^2-3x+10)\,dx + \int_2^3 (x^2+3x-10)\,dx = \left. \left[ -{1\over 3}x^3-{3\over 2}x^2+10x \right] \right|_1^2 + \left. \left[ {1\over 3}x^3+{3\over 2}x^2-10x \right] \right|_2^3 \\={19\over 6} +{23\over 6} =\bbox[red, 2pt]7$$

解答:$$\det(A-\lambda I)=-\lambda^3+4\lambda =-\lambda(\lambda-2)(\lambda+2)=0 \Rightarrow \lambda=0,2,-1 \\ \lambda_1 =0 \Rightarrow (A-\lambda_1)v=0 \Rightarrow \begin{bmatrix} 2 & 0 & 0 \\1 & -4 & -4 \\1 & 2 & 2\end{bmatrix} \begin{bmatrix}x_1\\x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1=0\\ x_2+ x_3=0} \Rightarrow v= x_3 \begin{bmatrix}0\\-1\\1 \end{bmatrix} \\\qquad \text{Choosing }x_3=1, \text{ we get }v_1= \begin{bmatrix}0\\-1\\1 \end{bmatrix} \\ \lambda_2=2 \Rightarrow  (A-\lambda_2)v=0 \Rightarrow \begin{bmatrix} 0 & 0 & 0 \\1 & -6 & -4 \\1 & 2 & 0\end{bmatrix} \begin{bmatrix}x_1\\x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1=x_3\\ 2x_2+x_3=0} \Rightarrow v= x_3 \begin{bmatrix}1 \\-1/2\\ 1 \end{bmatrix} \\\qquad \text{Choosing }x_3=1, \text{ we get }v_2= \begin{bmatrix} 1\\-1/2\\ 1 \end{bmatrix} \\ \lambda_3=-1  \Rightarrow  (A-\lambda_3) v=0 \Rightarrow \begin{bmatrix} 4 & 0 & 0 \\1 & -2 & -4 \\1 & 2 & 4\end{bmatrix} \begin{bmatrix}x_1\\x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1=0\\ x_2+2x_3=0} \Rightarrow v=x_3 \begin{bmatrix}0\\-2\\1 \end{bmatrix} \\\qquad \text{Choosing }x_3=1, \text{ we get }v_3= \begin{bmatrix}0\\-2\\1 \end{bmatrix} \\ \Rightarrow 特徵值: \bbox[red, 2pt]{0,2,-1} 特徵向量 \bbox[red, 2pt]{\begin{bmatrix}0\\-1\\1 \end{bmatrix},\begin{bmatrix}1\\-1/2\\ 1 \end{bmatrix},  \begin{bmatrix}0\\-2\\1 \end{bmatrix}} \\\det(A-\lambda I)=-\lambda^3+4\lambda=0 \Rightarrow A^3=4A \Rightarrow A^9=(4A)^3=64A^3=64\cdot 4A=256A \\=256 \begin{bmatrix}2&0&0\\1&-4 &-4\\1& 2& 2 \end{bmatrix}\Rightarrow A^9 = \bbox[red, 2pt]{\begin{bmatrix}512& 0& 0\\ 256& -1024&-1024 \\ 256& 512& 512 \end{bmatrix}}$$

解答:$$\cases{P(A\to A)= P(A\to B)=P(A\to C) =1/3 \\ P(B\to A)=P(B\to B)=1/2\\ P(C\to A)=P(C\to C) =P(C\to D)=1/3\\ P(D\to C) =P(D\to D)=1/2} \Rightarrow 矩移矩陣 T= \begin{bmatrix}1/3& 1/2 & 1/3 &0\\ 1/3& 1/2& 0& 0\\ 1/3& 0& 1/3& 1/2\\ 0& 0& 1/3& 1/2 \end{bmatrix} \\ 穩定狀態\Rightarrow T(v)=v \Rightarrow (T-I)v=0 \Rightarrow \begin{bmatrix}-2/3& 1/2 & 1/3 &0\\ 1/3& -1/2& 0& 0\\ 1/3& 0& -2/3& 1/2\\ 0& 0& 1/3& -1/2 \end{bmatrix} \begin{bmatrix}x\\ y\\z\\w \end{bmatrix}=0 \\ 且x+y+z+w=1 \Rightarrow \cases{x=3/10\\ y=1/5 \\z=3/10 \\w=1/5} \Rightarrow \cases{A鄉鎮病原數=10000\times(3/10) =3000 \\B鄉鎮病原數=10000\times(1/5) =2000 \\ C鄉鎮病原數=10000\times(3/10) =3000 \\D鄉鎮病原數=10000\times(3/10) =2000  } \\ \Rightarrow \bbox[red, 2pt]{\cases{A鄉鎮病原數: 3000 隻\\B鄉鎮病原數:2000隻 \\ C鄉鎮病原數: 3000 隻\\D鄉鎮病原數:2000隻}}$$

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