114年成德高中教甄-數學詳解

 新竹市立成德高級中學 114 學年度 第一次教師甄試

第一部份 填充題（ 每題８分）， 共８０分

解答:$$f(\log_2 384)= f \left( \log_2 {384\over 4}+2 \right) = f \left( \log_2 {384\over 4}-2 \right) =f(\log_2 24) =f(\log_2 6+2) = f(\log_2 6-2) \\=f(\log_2 {3\over 2}) =2^{\log_2(3/2)}+{1\over 4}={3\over 2}+{1\over 4}=\bbox[red, 2pt]{7\over 4}$$

解答:

$$x,y\ge 1 \Rightarrow 取\cases{u=\log_a x\gt 0\\ v= \log_a y\gt 0} \Rightarrow u^2+v^2= \log_a ax^2+ \log_a ay^2=1+2u+1+2v \\ \Rightarrow u^2-2u+v^2-2v=2 \Rightarrow (u-1)^2+ (v-1)^2=4 為一圓,圓心在P(1,1), 半徑r=2 \\ 由於u,v \gt 0, 因此為第一象限的圓弧. 欲求之\log_a xy= \log_a x+\log_a y=u+v =k \\ u+v=k為一直線，k的極值發生在圓切點(1+\sqrt 2,1+\sqrt 2)及端點(1+\sqrt 3,0),(0, 1+\sqrt 3) \\ \Rightarrow \cases{M=1+\sqrt 2+1+\sqrt 2=2+2\sqrt 2\\ m=0+1+\sqrt 3 =1+\sqrt 3} \Rightarrow (M,m)= \bbox[red, 2pt]{(2+2\sqrt 2,1+\sqrt 3)}$$

解答:$${\overline{AB} \over \sin C}=2\cdot R \Rightarrow \sin C={5\over 16} \Rightarrow \begin{vmatrix} -1 & \cos C& \cos B\\ \cos C &-1 &\cos A\\ \cos B& \cos A& 1\end{vmatrix} ={1\over abc} \begin{vmatrix} -a & b\cos C& c\cos B\\ a\cos C &-b &c\cos A\\ a\cos B& b\cos A& c\end{vmatrix} \\ ={1\over abc} \begin{vmatrix} -a & b\cos C& -a+b\cos C+c\cos B\\ a\cos C &-b &-b+ a\cos C+c\cos A\\ a\cos B& b\cos A& c+ a\cos B+ b\cos A\end{vmatrix}  ={1\over abc} \begin{vmatrix} -a & b\cos C& -a+ a\\ a\cos C &-b &-b+ b\\ a\cos B& b\cos A& c+ c\end{vmatrix} \\= {1\over abc} \begin{vmatrix} -a & b\cos C& 0\\ a\cos C &-b &0\\ a\cos B& b\cos A& 2c\end{vmatrix} ={2c\over abc} \begin{vmatrix} -a & b\cos C \\ a\cos C &-b   \end{vmatrix} ={2c\over abc} \cdot (ab-ab\cos^2C) \\=2(1-\cos^2 C) =2\sin^2 C=2 \cdot \left( {5\over 16} \right)^2 =\bbox[red, 2pt]{25\over 128}$$

解答:$$假設\cases{A(0,3,2) \\B(-3,2,4) \\C(3,1,-1) \\ P(\cos \theta, \sin \theta, 0)} \Rightarrow \cases{\overrightarrow{BA} =(3,1,-2) \\ \overrightarrow{BC} =(6,-1,-5) \\ \overrightarrow{BP} =(\cos\theta +3, \sin \theta-2,-4)} \\ \Rightarrow 體積=  \left| \overrightarrow{BP} \cdot ( \overrightarrow{BA} \times \overrightarrow{BC}) \right| =\left|( \cos\theta +3, \sin \theta-2, -4) \cdot (-7,3,-9) \right| = |-7\cos\theta+3\sin \theta +9| \\=\sqrt{58} |\sin(\theta+\alpha)|+9 \Rightarrow 最大值= \bbox[red, 2pt]{9+\sqrt{58}}$$

解答:$$甲+乙+丙+丁=16 \Rightarrow 每人至少一球有H^4_{16-4} =C^{15}_{12} =455 \\ 考慮甲=乙的情形:\cases{甲=乙=1 \Rightarrow 丙+丁=14\Rightarrow 每人至少一球有H^2_{14-2} =13\\  甲=乙=2 \Rightarrow 丙+丁=12\Rightarrow 每人至少一球有H^2_{12-2} =11\\甲=乙=3 \Rightarrow 9\\甲=乙=4 \Rightarrow 7\\ \cdots \\ 甲=乙=7 \Rightarrow 1} \\ \Rightarrow 甲=乙有1+3+\cdots+13=49種 \\ 所有的分法可以分成三種\cases{甲\gt 乙\\ 乙\gt 甲\\ 甲=乙}, 其中「甲\gt 乙」與「乙\gt 甲」的方法數是相等的\\ \Rightarrow 2\times(甲\gt 乙的方法數) +(甲=乙的方法數)=455 \Rightarrow 2\times (甲\gt 乙的方法數)+49=455 \\ \Rightarrow 甲\gt 乙的方法數=(455-49)/2= \bbox[red, 2pt]{203}$$

解答:$$A=\begin{bmatrix}-1& \sqrt 3\\ -\sqrt 3&-1 \end{bmatrix} = 2\begin{bmatrix}-1 /2& \sqrt 3/2\\ -\sqrt 3/2 &-1/2 \end{bmatrix} =2 \begin{bmatrix}\cos {4\pi\over 3}& -\sin{4\pi\over 3} \\ \sin {4\pi\over 3} & \cos{4\pi\over 3} \end{bmatrix} \\ \Rightarrow A^{50} =2^{50} \begin{bmatrix}\cos {200\pi\over 3}& -\sin{200\pi \over 3} \\ \sin {200\pi\over 3} & \cos{200\pi\over 3} \end{bmatrix}=2^{50} \begin{bmatrix}-1/2& \sqrt 3/2\\ -\sqrt 3/2 &-1/2 \end{bmatrix} = \begin{bmatrix}-2^{49} &  2^{49} \sqrt 3\\ -2^{49} \sqrt 3& -2^{49} \end{bmatrix} \\ \Rightarrow {bc-ad \over a+b+c+ d} ={-3\cdot 2^{98}-2^{98} \over -2^{50}}=2^{50} \Rightarrow \log_2 {bc-ad\over a+b+c+d}= \bbox[red, 2pt]{50}$$

解答:

$$\cases{O(0,0) \\A(2\sqrt 5/3),-5/3)} \Rightarrow \cases{\overline{OA}=\sqrt 5 =a(半長軸) \\ L=\overleftrightarrow{OA}:y=-\sqrt 5x/2 \Rightarrow \Gamma'的長軸位於L上} \\\Rightarrow \tan(\pi/2-\theta) =-{\sqrt 5\over 2}  \Rightarrow \cases{\sin \theta=2/3\\ \cos \theta=\sqrt 5/3} , 其中\Gamma 旋轉\theta 後變為\Gamma' \\ x=0代入\Gamma' \Rightarrow y^2={9\over 2} \Rightarrow y=\pm {3\over \sqrt 2} \Rightarrow P'(0,{3\over \sqrt 2}) \Rightarrow P=P'順時針旋轉\theta \\ \Rightarrow P= \begin{bmatrix}\sqrt 5/3& 2/3\\ -2/3& \sqrt 5/3 \end{bmatrix} \begin{bmatrix}0\\ 3/\sqrt 2 \end{bmatrix} = \begin{bmatrix}\sqrt 2\\ \sqrt{10}/2 \end{bmatrix} \Rightarrow P= \bbox[red, 2pt]{(\sqrt 2, {\sqrt{10}\over 2})}$$

解答:$$\lim_{n\to \infty}{S_n\over n} = \lim_{n\to \infty} \sum_{k=1}^n \left( {1\over n} \sin^2{\pi x\over 3} \right) =\int_0^1 \sin^2 {\pi x\over 3}\,dx = {1\over 2}\int_0^1 \left(1-\cos{2\pi x\over 3}  \right)\,dx \\ ={1\over 2 } \left. \left[ x-{3\over 2\pi}\sin {2\pi x\over 3}\right] \right|_0^1 ={1\over 2} \left( 1-{3\over 2\pi}\cdot {\sqrt 3\over 2} \right) = \bbox[red, 2pt]{{1\over 2}-{3\sqrt 3\over 8\pi}}$$

解答:

$$\angle DAB= 2\angle CBA =36^\circ \Rightarrow \angle CBA=18^\circ \Rightarrow  \sin \angle CBA={\overline{AC}\over \overline{AB}} \Rightarrow \overline{AB}={1\over \sin 18^\circ} \\ \cos \angle DAB={\overline{AB} \over \overline{AD}} \Rightarrow \overline{AD}={\overline{AB} \over \cos 36^\circ} ={1/\sin 18^\circ \over \cos 36^\circ} ={1\over \sin 18^\circ \cos 36^\circ} ={1\over k} \\ \Rightarrow k=\sin 18^\circ \cos 36^\circ \Rightarrow k\cos 18^\circ={1\over 2}\sin 36^\circ \cos 36^\circ ={1\over 4} \sin 72^\circ ={1\over 4} \cos 18^\circ \Rightarrow k={1\over 4} \\ \Rightarrow \overline{AD}= {1\over 1/4} =\bbox[red, 2pt]4$$

解答:$$\sum_{i=1}^{2025} (a_i\times \log a_i) =a_1\log a_1 +a_2 \log a_2 +\cdots+a_{2025} \log a_{2025} =2 \\ \Rightarrow a_1^{a_1}a_2^{a_2}a_3^{a_3}\cdots a_{2025}^{a_{2025}} =10^2=100 \cdots(1)\\ \sum_{i=1}^{2025} (a_{2026-i} \times \log a_i)= a_{2025}\log a_1 +a_{2024} \log a_2 +\cdots +a_1 \log a_{2025}=-1 \\ \Rightarrow a_1^{a_{2025}}\cdot a_2^{a_{2024}} \cdot a_3^{a_{2023}} \cdots a_{2025}^{a_1} =10^{-1}={1\over 10} \cdots(2) \\ (1)\times (2) \Rightarrow a_1^{a_1 +a_{2025}} \cdot a_2^{a_2+a_{2024}} \cdot a_3^{a_3+a_{2023}} \cdots a_{2025}^{a_{2025}+a_1} =10 \\ \Rightarrow a_1^{a_1+a_1r^{2024}} \cdot (a_1r)^{a_1r +a_1r^{2023}} \cdot (a_1r^2)^{a_1r^2+a_1r^{2022}} \cdots (a_1r^{2024})^{a_1r^{2024} +a_1} \\=a_1^{2(a_1+a_1r+ a_1r^2+ \cdots+a_1r^{2024})} \cdot r^{2024(a_1 +a_1r+ a_1r^2 +\cdots+a_1r^{2024})} = \left( a_1^2r^{2024} \right)^{a_1 +a_1r +\cdots+ a_1r^{2024}} =10\\ 欲求(a_1\times a_2\times \cdots \times a_{2025})^{a_1 +a_2+ \cdots+a_{2025}} =(a_1\cdot a_1r\cdot a_1r^2\cdots a_1r^{2024})^{a_1+a_1r+ \cdots+ a_1r^{2024}} \\= \left( a_1^{2025} r^{1+2+\cdots +2024} \right)^{a_1+a_1r+ \cdots+ a_1r^{2024}} =\left( (a_1^2)^{2025/2} (r^{2024})^{2025/2} \right)^{a_1+a_1r+ \cdots+ a_1r^{2024}} \\= \left(  \left( a_1^2r^{2024} \right)^{a_1 +a_1r +\cdots+ a_1r^{2024}} \right)^{2025/2} =\bbox[red, 2pt]{10^{2025/2}}$$

第二部份 填充題（每題１０ 分）， 共２ ０分

解答:

$$假設\cases{P(a,b) \\A(-3,2) \\B(4,2) \\C(0,5)} \Rightarrow c=\overline{PA}+ \overline{PB}+ \overline{PC} \Rightarrow P為\triangle ABC的費馬點(\text{Fermat Point})\\ 以\overline{BC}為底向外作一個正\triangle PBC \Rightarrow \cases{\angle CBP=60^\circ\\ \overline{BP} =\overline{BC} =5} \\  \cases{\cos\angle CBA= 4/5\\ \sin \angle CBA=3/5} \Rightarrow \cos \angle ABP =\cos(\angle B+60^\circ)= {7^2+5^2-\overline{AP}^2 \over 2\cdot 7\cdot 5} \\ \Rightarrow \cos \angle B\cos 60^\circ- \sin \angle B\sin 60^\circ ={4\over 5}\cdot {1\over 2}-{3\over 5}\cdot {\sqrt 3\over 2} ={4-3\sqrt 3\over 10}={74-\overline{AP}^2 \over 70} \\ \Rightarrow \overline{AP}^2 =46+21\sqrt 3 \Rightarrow c的最小值=\overline{AP} = \bbox[red, 2pt]{\sqrt{46+21\sqrt 3}}$$

解答:$$正弦定理:{a\over \sin A} ={b\over \sin B} ={c\over \sin C}=2R \Rightarrow \cases{\sin A=a/2R\\ \sin C=c/2R} \\\Rightarrow \sin A+ \sin C={a+c\over 2R} ={a+c\over b/\sin B} ={a+c\over b}\sin B ={\sqrt 3\over 2} \left( {a+c\over b} \right) \cdots(1)\\ 餘弦定理: \cos B={a^2+c^2-b^2\over 2ac} \Rightarrow {1\over 2}= {a^2+c^2-9ac/2\over 2ac} \Rightarrow a^2+c^2={11\over 2}ac \\ \Rightarrow (a+c)^2={15\over 2}ac 代入(1) \Rightarrow \sin A+\sin C={\sqrt 3\over 2}\cdot {\sqrt{15ac/2} \over \sqrt{9ac/2}} ={\sqrt 3\over 2}\cdot {\sqrt{15} \over 3} = \bbox[red, 2pt]{\sqrt 5\over 2}$$

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