112年羅東高中教甄-數學詳解

國立羅東高級中學 112 學年度第 1 次教師甄選數學科試題

一、填充題（請在答案卷第一頁依題號順序作答並標明題號，每題 5 分，共 60 分）：

解答：$$取a=x+{1\over x} \Rightarrow a^2=x^2+{1\over x^2}+2 \Rightarrow x^2+{1\over x^2}=a^2-2\\ 又a^3=(x+{1\over x})^3 =x^3+{1\over x^3}+3(x+{1\over x}) \Rightarrow a^3=4\sqrt 7+3a\\ \Rightarrow a^3-3a+4\sqrt 7=0 \Rightarrow a=\sqrt 7 \Rightarrow x^2+{1\over x^2}=a^2-2=7-2=\bbox[red, 2pt]5$$

解答：$$11=2\cdot 5\cdot 11,因此令Ｓ=\{x\mid 1\le x\le 110, x\in \mathbb N\}，則取\cases{S_2=\{x\mid  x是2的倍數且x\in S\} \\S_5=\{x\mid  x是5的倍數且x\in S\} \\S_{11}=\{x\mid  x是11的倍數且x\in S\} }\\ \Rightarrow \cases{|S|=110\\ |S_2|=110/2=55\\ |S_5|=110/5=22\\ |S_{11}|= 110/11=10} \Rightarrow \cases{|S_2\cap S_5|=110/10=11\\ |S_2\cap S_{11}|=110/22=5\\ |S_5\cap S_{11}=110/55=2 \\ |S_2\cap S_5\cap S_{11}=110/110=1}\\ \Rightarrow 在Ｓ中與110互質的個數有110-(55+22+10)+(11+5+2)-1=40個\\ 與110互質的自然數由小至大為\langle a_k\rangle\\ \Rightarrow \cases{ a_1=1, a_2=3, a_3=7, \dots,a_{40}=109\\ a_{41}=110+1,a_{42}=110+3,\dots, a_{80}=110+ 109\\ \dots}\\ 2023=40\times 50+23 \Rightarrow a_{2023}=110\times 50+a_{23} =5500+61=\bbox[red,2pt]{5561}\\ \langle a_k\rangle =1 ,3 ,7 ,9 ,13 ,17 ,19 ,21 ,23 ,27 ,29 ,31 ,37 ,39 ,41 ,43 ,47 ,49 ,51 ,53 ,57 ,59 ,61 ,63,\dots$$

解答：$$f(\theta)={3 \cos \theta+2\over 2\sin \theta +3} \Rightarrow f'(\theta)=-{9 \sin \theta+ 4 \cos \theta+6\over (2\sin \theta+3)^2}\\若 f'(\theta)=0\Rightarrow 9 \sin \theta+ 4 \cos \theta+6=0 ，此時取x=\tan{\theta\over 2} \Rightarrow \cases{\sin \theta=2x/(1+x^2)\\ \cos \theta = (1-x^2)/(1+x^2)}\\ \Rightarrow g(x)={18x\over 1+x^2}+{4-4x^2\over 1+x^2}+6=0 \Rightarrow x^2+9x+5=0 \Rightarrow x={-9\pm \sqrt{61}\over 2}\\ 當x={-9+\sqrt{61}\over 2}\Rightarrow \cases{\sin \theta=(-18+2\sqrt{61})/(73-9\sqrt{61})\\ \cos \theta=(-69+9\sqrt{61}/(73-9\sqrt{61})} \\ \Rightarrow {3\cos \theta+2\over 2\sin \theta+3} ={-61+9\sqrt{61}\over 183-23\sqrt{61}} ={6+\sqrt{61}\over 5}\\ 同理，當x={-9-\sqrt{61}\over 2} \Rightarrow {3\cos \theta+2\over 2\sin \theta+3} ={6-\sqrt{61}\over 5}\\因此(M,m)=\bbox[red,2pt]{\left({6+\sqrt{61}\over 5},{6-\sqrt{61}\over 5} \right)}$$

解答：$$a為z^7-1=0的虛根 \Rightarrow 1+a+a^2+a^3+a^4+a^5+a^6=0 \\ 令z=a+a^2+a^4 \Rightarrow \bar z=a^6+a^5+a^3 \Rightarrow z\bar z=(a+a^2+a^4)(a^6+a^5+a^3) \\ =a^7+a^6+a^4+a^8 +a^7+a^5+a^{10}+a^9+a^7\\ =a^6+a^5+a^4+a^3+ a^2+a+3=-1+3=2  \Rightarrow z\bar z=2 \Rightarrow \bar z={2\over z}\\ 又z+\bar z=a^6+a^5+a^4+a^3+ a^2+a=-1 \\ \Rightarrow z+\bar z=-1\Rightarrow z+{2\over z}=-1 \Rightarrow z^2+z+2=0 \Rightarrow z=\bbox[red,2pt]{-1\pm \sqrt{7}i\over 2}$$

解答：$$\cases{k=1 \Rightarrow y=x^5-x^4+x^3 -3x+1 \cdots(1)\\ k=-2 \Rightarrow y=x^5-x^4+4x^3-6x^2+1 \cdots(2)}\\ (1)=(2) \Rightarrow x^3-3x=4x^3-6x^2 \Rightarrow 3x^3-3x^2=0 \Rightarrow x=0或1 \\ \Rightarrow \cases{x=0代入(1) \Rightarrow y=1\\ x=1代入(1) \Rightarrow y=-1} \Rightarrow 兩定點為\bbox[red, 2pt]{(0,1),(1,-1)}$$

解答：

$$\sqrt{x^4-3x^2+4} +\sqrt{x^4-3x^2-8x+20} =\sqrt{(x^2-2)^2 +(x-0)^2}+\sqrt{(x^2-2)^2+(x-4)^2} \\= \overline{PA}+\overline{PB}，其中\cases{P(x^2,x)\in \Gamma:x=y^2\\ A(2,0)\\ B(2,4)}\\ 因此\overline{PA}+\overline{PB}的最小值=\overline{AB} ＝\bbox[red,2pt]4,此時P=\overline{AB} \cap \Gamma$$

解答：$$\cases{L_1:{x\over 2} ={y-1\over 1}={2-z\over 1} \\ L_2:{x+1\over 1} ={y+2\over 2} ={1-z\over 3}} \Rightarrow \cases{L_1的方向向量\vec u=(2,1,-1)\\ L_2的方向向量\vec v=(1,2,-3} \Rightarrow \vec n=\vec u\times \vec v=(-1,5,3)\\ 若\cases{P_1\in L_1\\ P_2\in L_2} \Rightarrow \cases{P_1(2t,t+1,2-t)\\ P_2(s-1,2s-2,-3s+1)} \\\Rightarrow P(x,y,z)=(2P_1+P_2)/3 =({4t+s-1\over 3}, {2t+2s \over 3},{-2t-3s+5\over 3}) \\法向量為\vec n且過P的平面方程式：-x+5y+3z={16\over 3} \Rightarrow \bbox[red, 2pt]{3x-15y-9z+16=0}$$

解答：$$\bbox[blue,2pt]{送分}$$

解答：$$\color{blue}{12\cdot 6^6}-28\cdot 6^4-14\cdot 6^3+17\cdot 6^2-16\cdot 6-\color{blue}{67\cdot 6^5}\\=5\cdot 6^5-28\cdot 6^4-14\cdot 6^3+17\cdot 6^2-16\cdot 6 =2\cdot 6^4-14\cdot 6^3+17\cdot 6^2-16\cdot 6 \\=-2\cdot 6^3+17\cdot 6^2-16\cdot 6 =5\cdot 6^2-16\cdot 6=14\cdot 6=\bbox[red, 2pt]{84}$$

解答：$$\overline{AB}=1\Rightarrow \overline{AC}=\overline{BC}=1/\sqrt 2 \Rightarrow 取\cases{C(0,0,0)\\ A(1/\sqrt 2,0,0)\\ B(0,1/\sqrt 2,0)} \Rightarrow S(1/2\sqrt 2,1/2\sqrt 2,k)\\ 又\overline{SC}=1 \Rightarrow {1\over 8}+{1\over 8}+k^2=1 \Rightarrow k={\sqrt 3\over 2}\\ 令O(a,b,c) \Rightarrow \cases{\overline{OC}=\overline{OA}\\ \overline{OC}=\overline{OB}\\ \overline{OC}=\overline{OS}} \Rightarrow O({1\over 2\sqrt 2},{1\over 2\sqrt 2},{1\over 2\sqrt 3})\\ 平面E＝\triangle ABC:z=0 \Rightarrow d(O,E)={1\over 2\sqrt 3}=\bbox[red,2pt]{\sqrt 3\over 6}$$

解答：$$兩圖形\cases{y=x\\ y=f(x)=16/2^x}的交點為(\alpha,\alpha),兩圖形\cases{y=x\\ y=g(x)=4-2^x}的交點為(\beta,\beta)\\ 由於兩圖形\cases{y=f(x)\\ y=g(x)}對稱於P(2,2),因此\beta =4-\alpha \Rightarrow \alpha+\beta=\bbox[red, 2pt]4$$

解答：$${\overline{AD}+\sqrt 3\cdot \overline{BC}+\overline{AC}/2 \over 3} \ge \sqrt[3]{{\sqrt 3\over 2}\cdot \overline{AD}\cdot \overline{BC}\cdot \overline{AC}} \Rightarrow 1\ge \sqrt[3]{{\sqrt 3\over 2}\cdot \overline{AD}\cdot \overline{BC}\cdot \overline{AC}} \\ \Rightarrow{\sqrt 3\over 2}\cdot \overline{AD}\cdot \overline{BC}\cdot \overline{AC} \le 1 \Rightarrow \overline{AD}\cdot \overline{BC}\cdot \overline{AC}\le {2\over \sqrt 3} \cdots(1)\\ 四面體體積\le {1\over 3}\cdot \triangle ABC\cdot \overline{AD} ={1\over 6}\overline{AC}\cdot \overline{BC}\sin 60^\circ \cdot \overline{AD}={\sqrt 3\over 12}\cdot \overline{AD}\cdot \overline{BC}\cdot \overline{AC} \\ \Rightarrow \overline{AD}\cdot \overline{BC}\cdot \overline{AC} \ge {1\over 6}\cdot {12\over \sqrt 3}={2\over \sqrt 3} \cdots(2) \\ 由(1)及(2)得\overline{AD}\cdot \overline{BC}\cdot \overline{AC}={2\over \sqrt 3 },此時\overline{AD}= \sqrt 3\cdot \overline{BC}={  \overline{AC}\over 2}＝1且 \overline{AD}與\triangle ABC垂直 \\ \Rightarrow \cases{\overline{AC}=2\\ \overline{AD}=1} \Rightarrow \overline{DC}=\sqrt{\overline{AD}^2 +\overline{AC}^2} =\sqrt{1+4}=\bbox[red, 2pt]{\sqrt 5}$$

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