113年鳳新高中教甄-數學詳解

 國立鳳新高級中學 113 學年度第 1 次教師甄選

一、 計算證明題 (第 1 題至第 10 題每題 8 分， 第 11 題至第 12 題每題 10 分)

解答:

$$\overrightarrow{OI} ={7\over 3+6+7} \overrightarrow{OA} +{6\over 3+6+7} \overrightarrow{OB} +{3\over 3+6+7} \overrightarrow{OC}  ={7\over 16} \overrightarrow{OA} +{6\over 16} \overrightarrow{OB} +{3\over 16} \overrightarrow{OC} \\ \Rightarrow \overrightarrow{OI} \cdot \overrightarrow{BC} =  {7\over 16} \overrightarrow{OA} \cdot \left( \overrightarrow{BA} +  \overrightarrow{AC} \right)+{6\over 16} \overrightarrow{OB} \cdot \overrightarrow{BC}+{3\over 16} \overrightarrow{OC} \cdot \overrightarrow{BC}  \\= {7\over 16} ({1\over 2} \cdot 3^2-{1\over 2} \cdot 6^2) +{6\over 16}(-{1\over 2}\cdot 7^2) +{3\over 16}({1\over 2}\cdot 7^2) = \bbox[red, 2pt]{-{21\over 2}}$$

解答: $$f(\theta) =\sin 2\theta +2\sin \theta \Rightarrow f'(\theta)=0 \Rightarrow 2\cos 2\theta+2\cos \theta=0 \Rightarrow 4\cos^2\theta+ 2\cos \theta-2=0\\ \Rightarrow 2(2\cos \theta-1)(\cos \theta+1)=0 \Rightarrow \cases{\cos \theta=1/2 \Rightarrow \theta=\pi/3\\ \cos \theta=-1 \Rightarrow \theta=\pi,不合} \Rightarrow f(\pi/3) ={\sqrt 3\over 2}+\sqrt 3={3\sqrt 3\over 2} \\ \Rightarrow \bbox[red, 2pt]{ \cases{最大值:3\sqrt 3/2 \\ \theta=\pi/3}}$$

解答: $$五個盒子挑一個(假設編號1)放2個球，其他盒子(編號2-5)各放1個球\\ \text{Case I }其他盒子中有兩個編號與小球相同:6\times 14=84\\  \text{Case II }其他盒子中有三個編號與小球相同:4\times 11=44 \\ 共有5(84+44)= \bbox[red, 2pt]{640}$$

解答:

$$假設\overline{BC}的中點D，及\overline{AD}=高=h \Rightarrow \overline{BD} =\overline{CD} =\sqrt{4-h^2} \Rightarrow \overline{CD}=2\sqrt{4-h^2} \\ \cos \angle BP_iC= \cos \pi= -1={\overrightarrow{P_iB} \cdot \overrightarrow{P_iC} \over \overline{BP_i} \cdot \overline{CP_i}} \Rightarrow \overline{BP_i} \cdot \overline{CP_i} =-\overrightarrow{P_iB} \cdot \overrightarrow{P_iC} = -(\overrightarrow{BA} +\overrightarrow{AP_i}) \cdot (\overrightarrow{CA} +\overrightarrow{AP_i}) \\ \Rightarrow m_i= \overline{AP_i}^2+ \overline{BP_i}\cdot \overline{CP_i} =(\overrightarrow{AP_i} \cdot \overrightarrow{AP_i})-(\overrightarrow{BA} +\overrightarrow{AP_i}) \cdot (\overrightarrow{CA} +\overrightarrow{AP_i}) \\=-\overrightarrow{BA}\cdot \overrightarrow{CA} -\overrightarrow{BA}\cdot \overrightarrow{AP_i}-\overrightarrow{AP_i}\cdot \overrightarrow{CA} =-\overrightarrow{BA}\cdot \overrightarrow{CA} -(\overrightarrow{BD} +\overrightarrow{DA}) \cdot   \overrightarrow{AP_i} - \overrightarrow{AP_i}\cdot (\overrightarrow{CD} +\overrightarrow{DA}) \\= -\overrightarrow{BA}\cdot \overrightarrow{CA} -2 \overrightarrow{AP_i} \cdot \overrightarrow{DA} \quad ( \because\overrightarrow{BD} + \overrightarrow{CD}=0) =-\overrightarrow{AB} \cdot \overrightarrow{AC} +2\overrightarrow{AP_i} \cdot \overrightarrow{AD} \\=-\overline{AB}\cdot \overline{AC} \cos \angle A+ 2h^2 =-4\cdot {4+4-(2\sqrt{4-h^2})^2\over 2\cdot 2\cdot 2} +2h^2 =4 \\ \Rightarrow m_1+\cdots+m_{100} =100\times 4=\bbox[red, 2pt]{400}$$

$$\bbox[cyan,2pt]{另解}\quad 假設\angle A=90^\circ \Rightarrow \overline{AD}= \overline{CD}=\overline{DB} =\sqrt 2 \\ \Rightarrow m_i=\overline{AP_i}^2+ \overline{BP_i}\cdot \overline{CP_i} = (2+\overline{DP_1}^2) +(\sqrt 2-\overline{DP_i})(\sqrt 2+\overline{DP_i}) =4 \Rightarrow \sum_{i=1}^{100} m_i =\bbox[red, 2pt]{400}$$

解答:

$$假設E為\overline{AD}與\overline{BC}的交點，由於\overline{AD}為\angle A的角平分線，因此{\overline{BE} \over \overline{EC}} ={\overline{AB} \over \overline{AC}} ={3\over 5} \Rightarrow \cases{\overline{BE} =21/8\\ \overline{CE}= 35/8} \\ 又\cos \angle A={3^2+5^2-7^2\over 2\cdot 3\cdot 5}=-{1\over 2} \Rightarrow \angle A=120^\circ \Rightarrow \angle BAE =60^\circ\\ \Rightarrow \cos \angle BAE={1\over 2} ={3^2+\overline{AE}^2-(21/8)^2 \over 2\cdot 3\cdot \overline{AE}}  \Rightarrow \overline{AE} ={15 \over 8} \\ \angle ABE=\angle ADC \Rightarrow \triangle ABE \sim \triangle CDE (AAA) \Rightarrow {\overline{AE} \over \overline{BE}} ={\overline{CE}\over  \overline{DE}}  \Rightarrow \overline{DE} ={49\over 8}   \Rightarrow \overline{AD}={15 \over 8} +{49 \over 8} =8\\ 又{\overline{BE} \over \overline{AB}}  ={\overline{EI} \over  \overline{AI}} \Rightarrow \overline{AI} ={8\over 15}\cdot \overline{AE} =1 \Rightarrow{\overline{ID} \over \overline{AD}} ={8-1\over 8} =\bbox[red, 2pt]{7\over 8}$$

解答: $$動差母函數M_X(t) =E(e^{tX}) =\sum_{k=0}^n e^{tk}p^k q^{n-k} = (pe^t +q)^n, p+q=1\\ \Rightarrow M_{X}'(t)  =n(pe^t+q)^{n-1} pe^t \Rightarrow M'_X(0)=np \\ \Rightarrow M''_X(t) =M'_X(t) +n(n-1)(pe^t+q)^{n-2} p^2e^{2t} \Rightarrow M_X''(0) =np+n(n-1)p^2 \\ \Rightarrow M_X'''(t) =M_X''(t) + 2n(n-1)(pe^t+q)^{n-2} p^2e^{2t}+ n(n-1)(n-2)(pe^t+q)^{n-3}p^3e^{3t} \\ \Rightarrow E(X^3) =M_X'''(0) =np+n(n-1)p^2 +2n(n-1)p^2 +n(n-1)(n-2)p^3 \\=1+0.9 +1.8 +0.72 = \bbox[red, 2pt]{4.42}$$

解答: $$\cases{k=(x+\sqrt{x^2-2024}) (y+\sqrt{y^2-2024}) =xy +x\sqrt{y^2-2024} +y\sqrt{x^2-204} +\sqrt{(x^2-2024) (y^2-2024)} \\ 2024=(x-\sqrt{x^2-2024}) (y-\sqrt{y^2-2024}) =xy -x\sqrt{y^2-2024} -y\sqrt{x^2-204} +\sqrt{(x^2-2024) (y^2-2024)}} \\ \Rightarrow \cases{兩式相乘:2024k=2024\cdot 2024 \Rightarrow k=2024\\兩式相減:k-2024=2(x\sqrt{y^2-2024 }+y \sqrt{x^2-2024}  )\\} \Rightarrow x\sqrt{y^2-2024}=-y\sqrt{x^2-2024} \\ \Rightarrow \cases{x=y \Rightarrow x=\sqrt{2024} \\ x=-y \Rightarrow (x-\sqrt{x^2-2024}) (-x-\sqrt{x^2-2024}) =-2024\ne 2024,不合} \\ \Rightarrow x=y= \sqrt{2024} \Rightarrow 3x^2-2y^2+3x-3y-2023=x^2-2023=\bbox[red, 2pt]1$$

解答: $$f(x)= 5x^3-5(k+1)x^2 +(71k-1)x+1-66k =(x-1)(5x^2-5kx+66k-1) \\\qquad =(x-1)5(x-\alpha)(x-\beta) \Rightarrow 三根為1,\alpha,\beta \\ \Rightarrow f({66\over 5}) =({66\over 5}-1)(5 \cdot  ({66\over 5})^2-66k+66k-1)=({66\over 5}-1)5({66\over 5}-\alpha)({66\over 5}-\beta) \\ \Rightarrow {4356 \over 5}-1 =5({66\over 5}-\alpha)({66\over 5}-\beta) \Rightarrow 4351 =19\times 229=(66-5\alpha) (66-5\beta) \\ \Rightarrow \cases{5\alpha-66=19\\ 5\beta-66=229} \Rightarrow \cases{\alpha= 17\\ \beta=59} 或\cases{\alpha=59\\ \beta=17} \Rightarrow k=\alpha+\beta= \bbox[red, 2pt]{76}$$

解答:

$$\cases{\Gamma_1:y=x^2+a\\ \Gamma_2: y=-x^2-a\\ \Gamma_3: y^2=x-a\\ \Gamma_4: y^2=-x-a} \Rightarrow  \cases{\Gamma_1與\Gamma_3對稱直線L_1:x=y \\\Gamma_2與\Gamma_4對稱直線L_1 \\ \Gamma_1與\Gamma_4對稱直線L_2:x=-y \\ \Gamma_2與\Gamma_3對稱直線L_2 }\Rightarrow \cases{\Gamma_1與\Gamma_3的切點在L_1 \\\Gamma_2與\Gamma_4的切點在L_1 \\ \Gamma_1與\Gamma_4的切點在L_2 \\ \Gamma_2與\Gamma_3的切點在L_2} \\ \Rightarrow x^2-x+a=0恰有一根 \Rightarrow 1-4a=0 \Rightarrow a={1\over 4} \Rightarrow 切點\cases{(1/2,1/2)\\ (1/2, -1/2)\\ (-1/2,-1/2)\\ (-1/2,1/2)} \\ \Rightarrow 所圍面積=8\int_0^{1/2} (x^2+a-x)\,dx =8\cdot {1\over 24} =\bbox[red, 2pt]{1\over 3}$$

解答: $$\cases{L_1:{x-3\over 1} ={y\over 2} ={z+2\over -2} \Rightarrow 方向向量\vec u=(1,2,-2) \\L_2:{x\over 3} ={y-2\over 1} ={z+1\over -2} \Rightarrow 方向向量\vec v=(3,1,-2)} \Rightarrow \vec n= \vec u\times \vec v=(-2,-4,-5) \\ \Rightarrow 包含L_2的平面E: -2x-4(y-2)-5(z+1)=0 \Rightarrow 2x+4y+5z=3 \\ A(3,0,-2) \in L_1 \Rightarrow 兩歪斜線距離=d(A,E) ={7\over 3\sqrt 5} =\triangle PQR 的高 \Rightarrow S_{\triangle PQR} ={1\over \sqrt 3} \left( {7\over 3\sqrt 5}\right)^2 \\= \bbox[red, 2pt]{49\sqrt 3\over 135}$$

解答:

$$\cases{|z_1|=|z_1+z_2|=3 \\ |z_2-z_1|=3\sqrt 3} \Rightarrow \cases{z_1=3e^{i\theta} \\z_2=3e^{i(\theta+2\pi/3)}} \Rightarrow \cases{z_1\overline{z_2} =9e^{i(-2\pi/3)} \\ \overline{z_1} z_2 = 9e^{i(2\pi/3)}} \\ \Rightarrow \left( z_1\overline{z_2} \right)^{2000} +\left( \overline{z_1} z_2 \right)^{2000} =9^{2000} \left( e^{i(-4000\pi/3)} +e^{i(4000\pi/3)}\right) =3^{4000} \cdot 2\cos {4000\over 3}\pi =3^{4000} \cdot 2\cos {4\over 3}\pi \\=-3^{4000} \Rightarrow \log \left| \left( z_1\overline{z_2}\right)^{2000}+\left( \overline{z_1} z_2 \right)^{2000} \right| = \bbox[red, 2pt] {\log 3^{4000} =4000\log 3}$$

解答: $$柯西不等式: \\\left(({1\over n+1})^2+ ({1\over n+2})^2+ \cdots ({1\over n+n})^2  \right)(1^2+\cdots +1^2) \ge \left( {1\over n+1} +{1\over n+2}+ \cdots +{1\over 2n}\right)^2 \\ 而左式: \left(({1\over n+1})^2+ ({1 \over n+2})^2+ \cdots ({1\over n+n})^2 \right) n \\\qquad \qquad \le \left( {1\over n(n+1)}+ {1\over (n+1)(n+2)} + \cdots +{1\over (2n-1)(2n)}\right)n \\= \left( {1\over n}-{1\over n+1}+{1\over n+1}-{1\over n+2 }+ \cdots +{1\over 2n-1}-{1\over 2n} \right)n =\left( {1\over n}-{1\over 2n}\right)n ={1\over 2}\\ 因此\left( {1\over n+1} +{1\over n+2}+\cdots +{1\over 2n} \right)^2 \le {1\over 2} \Rightarrow {1\over n+1} +{1\over n+2}+\cdots +{1\over 2n}\le {\sqrt 2\over 2}, \bbox[red, 2pt]{故得證}$$

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