朱式幸福: 113年桃園陽明高中教甄-數學詳解

桃園市立陽明高中 113 學年度教師甄選數學科筆試測驗題目卷

一、填充題： (12 題，每題 5 分，共 60 分)

解答：$$\cases{L_1:3x-4y=6\\ L_2:11x+2y=22} \Rightarrow A=L_1\cap L_2=(2,0) \Rightarrow \overleftrightarrow{AD}: y=2x-4\\ L_3\bot \overleftrightarrow{AD} \Rightarrow L_3斜率=-{1\over 2} \Rightarrow L_3:y=-{1\over 2}(x+2)-8 \Rightarrow \cases{B=L_2\cap L_3 =(4,-11)\\ C=L_2\cap L_3=(-6,-6)} \\ \Rightarrow \cases{\overline{BC}= 5\sqrt 5\\ \overline{AD}=4\sqrt 5} \Rightarrow \triangle ABC={1\over 2}\cdot 5\sqrt 5\cdot 4\sqrt 5= \bbox[red, 2pt]{50}$$

解答：$$\sum_{k=1}^n k^4={1\over 30}n(n+1)(2n+1)(3n^2+3n-1) \Rightarrow n^5係數= {1\over 5}\\ \lim_{n\to \infty} {80 \over n}\sum_{k=1}^n \left( {5k-2\over n}\right)^4 = \lim_{n\to \infty} {50000 \over n^5}\sum_{k=1}^n \left( k-{2\over 5}\right)^4 =50000\times {1\over 5}= \bbox[red, 2pt]{10000}$$

解答：$$\text{數列中沒有23，因此 }23=\text{兩個數的平均，此兩數有 }(6,40),(10,36) , (21,25) \\\text{符合條件的情形：}\cases{A,6,40,B \Rightarrow \cases{A=3\\ B=42} \Rightarrow 只有1種\\ A,10,36,B \Rightarrow \cases{A=3,6,7\\ B=40,42} \Rightarrow 有3\times 2=6種\\ A,21,25,B \Rightarrow \cases{A=3,6,7,10,11,14\\ B=33,36,40,42} \Rightarrow 有6\times 4=24種} \\ 因此共有1+6+24-31種，機率為{31\over C^{12}_4} = \bbox[red, 2pt]{31 \over 495}$$

解答：$$3B3C的排列數={6!\over 3!3!}=20,詳如下表\\ \begin{array}{c}編號&字串 \\\hline 1& BCBCBC \\ 2& CBCBCB \\ \hdashline 3 &BCCBCB \\ 4&CBCBBC\\ 5& BCBCCB\\6& CBBCBC\\ \hdashline 7&BCCBBC\\8& CCBBCB\\9& BCBBCC\\10& CCBCBB\\11& BBCCBC\\12 &CBBCCB\\13& BBCBCC\\ 14 &CBCCBB\\ \hdashline 15&CBBBCC\\16& CCBBBC\\17& BBCCCB\\ 18& BCCCBB\\ \hdashline 19& CCCBBB\\20& BBBCCC\\ \hline \end{array} \Rightarrow \cases{1-2:沒有同字相鄰，有7空隔插入4個A，2C^7_4=70\\ 3-6:有1個同字相鄰，剩下6空隔插入3個A，4C^6_3= 80\\ 7-14:有2個同字相鄰，剩下5空隔插入2個A，8C^5_2= 80 \\15-18: 有3個同字相鄰，剩下4空隔插入1個A，4C^4_1= 16 \\19-20：有4個同字相鄰，4個A剛好用完，2} \\ 因此同字不相鄰共有70+80+80+16+2= \bbox[red, 2pt]{248}$$

解答：$$529＝23^2 \Rightarrow 22^{17}+24^{17} =(23-1)^{17}+ (23+1)^{17} =\sum_{k=0}^{17} \left( C^{17}_k\cdot 23^k(-1)^{17-k} +C^{17}_k\cdot 23^k \right) \\=2 \sum_{k=1}^9 C^{17}_{2k-1} 23^{2k-1} \equiv 2C^{17}_1 \cdot 23 \text{ mod }529 \equiv \bbox[red, 2pt]{253} \text{ mod } 529$$

解答：$$\cases{A(1,-4,4)\\ B(3,-2,2)\\ C(4,2,-2)} \Rightarrow \cases{\overrightarrow{AB}= (2,2,-2) \\\overrightarrow{AC}= (3,6,-6) }\Rightarrow (1,b,c) \parallel (3,6,-6) \Rightarrow \cases{b=2\\ c=-2} \Rightarrow E:x+2y-2z+d=0\\ 又\cases{ \overleftrightarrow{AB}:{x-1\over 1}={y+4\over 1}={z-4\over -1} \Rightarrow P(t+1,t-4,-t+4) \\ \overleftrightarrow{AC}: {x-1\over 1} ={y+4\over 2} ={z-4\over -2} \Rightarrow Q(s+1,2s-4,-2s+4)}, s,t\in \mathbb R \\ \cases{P在E上\\ Q在E上} \Rightarrow \cases{d=15-5t\\ d=15-9s} \Rightarrow 5t=9s \Rightarrow s={5\over 9}t\\ \cases{\triangle ABC面積=|\overrightarrow{AB} \times \overrightarrow{AC}| =|(0,6,6)| =6\sqrt 2 \\ \triangle APQ面積=|\overrightarrow{AP} \times \overrightarrow{AQ}|=(0,st,st)=st\sqrt 2 } =6\sqrt 2=30\cdot st\sqrt 2 \Rightarrow st={1\over 5} \\ \Rightarrow {5\over 9}t^2={1\over 5} \Rightarrow t={3\over 5}\;(若t=-{3\over 5}\Rightarrow P \not \in \overline{AB} \Rightarrow 不合)，\\ \Rightarrow P({8\over 5},-{17\over 5},{17\over 5}) \in E \Rightarrow {8\over 5}-{34\over 5}-{34\over 5}+d=0 \Rightarrow d=\bbox[red, 2pt]{12}$$

解答：$$x^2+(m+1)x-3(m+1)=0 \Rightarrow 判別式D=(m+1)^2+12(m+1) =(m+1)(m+13)\\ 由於係數為整數且有整數根，因此D\ge 0且D為一完全平方數 \Rightarrow \cases{D=0\\ D=4\times 16 =(-4)\times (-16)} \\ \Rightarrow \bbox[red, 2pt]{\cases{m=-1,-3\\ m=3,-17}}$$

解答：$$\cases{P在單位圓上\\ \overline{OP}與x軸正向夾角\theta} \Rightarrow P(\cos \theta, \sin \theta) \Rightarrow \cases{P旋轉3\theta為Q_1( \cos 4\theta,\sin 4\theta) \\ P對y=x鏡射得Q_2=(\sin \theta, \cos \theta)} \\ \Rightarrow Q=Q_1 =Q_2 \Rightarrow \cases{\sin \theta=\cos 4\theta\\ \cos \theta= \sin 4\theta} \Rightarrow \cos 4\theta \cos \theta -\sin \theta \sin 4\theta =0 \Rightarrow \cos 5\theta=0 \\ \Rightarrow \theta ={\pi\over 10},{ 3\pi\over 10},{5 \pi\over 10},{7 \pi\over 10},{9\pi\over 10},共\bbox[red, 2pt]{5}個$$

解答：$$$$

解答：$$T＝\begin{bmatrix}P(A\to A)& P(B\to A)& P(C\to A) & P(D\to A) & P(E\to A) \\ P(A\to B)& P(B\to B)& P(C\to B) & P(D\to B) & P(E\to B) \\ P(A\to C)& P(B\to C)& P(C\to C) & P(D\to C) & P(E\to C) \\ P(A\to D)& P(B\to D)& P(C\to D) & P(D\to D) & P(E\to D) \\ P(A\to E)& P(B\to E)& P(C\to E) & P(D\to E) & P(E\to E) \end{bmatrix} \\\qquad =\begin{bmatrix} 0 & 1/4 & 1/4& 1/4 & 0 \\1/3 &0 & 1/4 & 1/4& 1/3\\ 1/3& 1/4& 0& 1/4& 1/3 \\ 1/3& 1/4& 1/4 & 0 & 1/3 \\ 0 & 1/4& 1/4& 1/4 & 0\end{bmatrix} \\ T\mathbf x=\mathbf x \Rightarrow \begin{bmatrix} 0 & 1/4 & 1/4& 1/4 & 0 \\1/3 &0 & 1/4 & 1/4& 1/3\\ 1/3& 1/4& 0& 1/4& 1/3 \\ 1/3& 1/4& 1/4 & 0 & 1/3 \\ 0 & 1/4& 1/4& 1/4 & 0\end{bmatrix} \begin{bmatrix}A \\B \\C\\D\\E \end{bmatrix} = \begin{bmatrix}A \\B \\C\\D\\E \end{bmatrix} \\ \Rightarrow \cases{{1\over 4}(B+C+D)=A\\ {1 \over 4}(B+C+D)=E}且A+B+C +D+E=1 \Rightarrow {3\over 2}(B+C+D)=1 \\ \Rightarrow B+C+D={2\over 3} \Rightarrow A={1\over 4}\cdot {2\over 3} =\bbox[red, 2pt]{1\over 6}$$

解答：

$$拋物線\Gamma:y=2x-x^2=x(2-x) \Rightarrow \Gamma與x軸交於O(0,0),A(2,0)\\ \Rightarrow \Gamma與x軸所圍面積=\int_0^2 (2x-x^2)\,dx ={4\over 3}\\ \Gamma 與直線L:y=mx交於P(2-m,2m-m^2)\\ \Rightarrow \Gamma 與L所圍面積= \int_0^{2-m} (2x-x^2-mx)\,dx = \left. \left[ x^2-{1\over 3}x^3-{m\over 2}x^2\right] \right|_0^{2-m} \\=(2-m)^2 ({1\over 3}-{1\over 6}m) ={1\over 6}(2-m)^3 ={2\over 3} \Rightarrow m= \bbox[red, 2pt]{2-\sqrt[3] 4}$$

解答：

$$假設圓A、圓B、圓C、圓D的圓心分別為C_A,C_B,C_C,C_D，並假設C_C與C_D的中點為P\\ 圓C的半徑為r \Rightarrow \cases{\overline{C_BC_C} =r+2\\ \overline{C_CP} = r} \Rightarrow \overline{C_BP} = \sqrt{(r+2)^2-r^2 } =\sqrt{4r+4} \Rightarrow \overline{C_AP}= \sqrt{4r+4}-1 \\ 又\overline{C_AC_C}= 3-r \Rightarrow (3-r)^2= (\sqrt{4r+4}-1)^2 +r^2 \Rightarrow \sqrt{4r+4} =5r-2 \Rightarrow r(25r-24)=0 \\ \Rightarrow r = \bbox[red, 2pt]{24\over 25}$$