115年台中二中教甄2-數學詳解

 臺中市立臺中第二高級中等學校115學年度第2次教師甄選

一、 填充題：

解答:$$|\vec a\times \vec b|= |\vec a||\vec b|\sin \theta \Rightarrow {2\over 3}=1\cdot 1\cdot \sin \theta \Rightarrow \sin \theta={2\over 3} \Rightarrow \cos \theta=\pm {\sqrt 5\over 3} \\取f(x)= |\vec a+x \vec b|= \sqrt{(\vec a+x\vec b) \cdot (\vec a+x\vec b)} = \sqrt{|\vec a|^2 +2x(\vec a\cdot \vec b)+ x^2|\vec b|^2} =\sqrt{1+2x(a\cdot b)+x ^2}\\ \Rightarrow f'(x)= {\vec a\cdot \vec b+x\over \sqrt{1+2x(\vec a\cdot \vec b)+x^2}} \Rightarrow \lim_{x\to 0}{|\vec a +x\vec b|-|\vec a|\over x} =f'(0) ={\vec a\cdot \vec b\over 1} = |\vec a||\vec b|\cos \theta= \bbox[red, 2pt]{\pm {\sqrt 5\over 3}}$$

解答:$$Q\in L:{x-25\over -15}={y-1\over 2}={z-18\over 11} \Rightarrow \cases{Q=(-15t+25,2t+1,11t+18) \\ \vec u=(-15,2,11) \Rightarrow |\vec u|= \sqrt{350}}\\ \Rightarrow \overrightarrow{QP}=(15t-10,-2t-15,-11t+5) \Rightarrow \cos 60^\circ ={\overrightarrow{QP}\cdot (-15,2,11) \over |\overrightarrow{QP}||(-15,2,11)|} \Rightarrow 3t(t-1)=0\\  \textbf{Case I }t=0\Rightarrow Q=(25,1,18) \Rightarrow |\overrightarrow{QP}|= \sqrt{350} \Rightarrow \overline{QR} =2\sqrt{350} =2|\vec u| \Rightarrow \overrightarrow{QR} =\pm 2\vec u \\ \qquad \Rightarrow \cases{ \overrightarrow {QR}=2\vec u \Rightarrow \overrightarrow{QP} \cdot \overrightarrow{QR} =350 \gt 0 \\ {QR}= -2\vec u \Rightarrow \overrightarrow{QP} \cdot \overrightarrow{QR} =-350\lt 0 與\cos 60^\circ 矛盾} \Rightarrow R=Q+2\vec u=(-5,5,40) \\ \textbf{Case II }t=1 \Rightarrow Q=(10,3,29),同理可得R=Q-2\vec u =(40,-1,7) \\ \Rightarrow R=\bbox[red, 2pt]{(-5,5,40), (40,-1,7)}$$

解答:$$\lim_{x\to \infty} \left( \sqrt{x^2+x+1}-x \right) =\lim_{x\to \infty} {\left( \sqrt{x^2+x+1}-x \right)  \left( \sqrt{x^2+x+1}+ x \right) \over \sqrt{x^2+x+1}+x}= \lim_{x\to \infty} {x+1 \over \sqrt{x^2+x+1}+x} \\= \lim_{x\to \infty} {1+1/x \over \sqrt{1+1/x+1/x^2}+1} ={1\over 2} \\ \lim_{x\to \infty} \left( \sqrt[3]{x^3+ 2x^2+1}-x \right) = \lim_{x\to \infty} { (x^3+2x^2+1)-x^3 \over (\sqrt[3]{x^3+2x^2+1 })^2+ x \sqrt[3]{x^3+2x^2+1}+x^2} \\=  \lim_{x\to \infty} { 2x^2+1  \over (\sqrt[3]{x^3+2x^2+1 })^2+ x \sqrt[3]{x^3+2x^2+1}+x^2} ={2\over 1+1+1}={2\over 3} \\ 因此欲求\lim_{x\to \infty}  \left( \sqrt{x^2+x+1}- \sqrt[3]{x^3+2x^2+1} \right) =\lim_{x\to \infty}  \left[\left( \sqrt{x^2+x+1}- x\right) - \left( \sqrt[3]{x^3+2x^2+1} -x \right)\right]  \\={1\over 2}-{2\over 3} =\bbox[red, 2pt]{-{1\over 6}}$$

解答:$$狀況一: 取完球的順序為黃\to 紅\to 白，最後一球是白球，機率為{4\over 9}\\\qquad 剩下2紅3黃，紅球排最後的機率為{2\over 5}。因此狀況一的機率為{4\over 9}\times {2\over 5}= {8\over 45} \\ 狀況二:取完球的順序為白\to 紅\to 黃，最後一球是黃球，機率為{3\over 9}={1\over 3} \\\qquad 剩下2紅4白，紅球排最後機為{2\over 6}={1\over 3}。因此狀況二的機率為{1\over 3}\times {1\over 3}={1\over 9} \\ 機率合計={8\over 45}+{1\over 9}= \bbox[red, 2pt]{13\over 45}$$

解答:$$假設m的各個位數乘積為P(m) \Rightarrow 0\le P(m)\le m \Rightarrow 0\le m^2-20m-19\le m\\ m^2-20m-19\ge 0 \Rightarrow m\ge 10+\sqrt{119} (m\not \le 10-\sqrt{119}\lt 0) \Rightarrow m\ge 21\\ m^2-20m-19\le m \Rightarrow m^2-21m-19\le 0 \Rightarrow m\le {21+\sqrt{517}\over 2} \approx 21.8 \Rightarrow m\le 21\\ 因此m= \bbox[red, 2pt]{21}$$

解答:$$\cases{\vec u-\vec v =(-2,-3,0) \\ \vec v-\vec w= (3,1,2)} \Rightarrow (\vec u-\vec v)+(\vec v-\vec w)= \vec u-\vec w=(1,-2, 2) \\ \Rightarrow  (\vec u-\vec v) \cdot (\vec v- \vec w)=(-2,-3,0) \cdot(3,1,2) \Rightarrow  \vec u\cdot \vec v-\vec u\cdot \vec w-|\vec v|^2+\vec v\cdot \vec w  =-6-3\\\Rightarrow -|\vec v|^2=-9 \Rightarrow |\vec v|=3 \\ 同理可得\cases{(\vec u-\vec v)\cdot (\vec u-\vec w) =|\vec u|^2=4 \Rightarrow |\vec u|= 2\\ (\vec v-\vec w)\cdot (\vec u-\vec w)=|\vec w|^2=5 \Rightarrow |\vec w|=\sqrt 5} \Rightarrow 體積=3\times 2\times \sqrt 5= \bbox[red, 2pt]{6\sqrt 5}$$

解答:$$假設在投擲第 k 號硬幣後的錢數為 M_k，初始金額為 M_0 = 1,\\且隨機變數X_k代表錢數變化的倍率,即\cases{X_k=2, 出現正面\\ X_k=1, 出現反面} \Rightarrow M_k=M_{k-1} \times X_k \\ \Rightarrow M_{2026} =M_0\times X_1\times X_2\times \cdots\times X_{2026} = \prod_{k=1}^{2026}X_k \\ E(X_k)= 2\times 出現正面機率+1\times 出現反面機率=2\times {1\over k}+1\times \left( 1-{1\over k} \right) ={k+1\over k} \\ \Rightarrow E(M_{2026}) = E \left( \prod_{k=1}^{2026}X_k \right) =\prod_{k=1}^{2026}E(X_k) =\prod_{k=1}^{2026}{k+1\over k}={2\over 1} \times {3\over 2} \times {4\over 3}\times \cdots \times {2027\over 2026} =\bbox[red, 2pt]{2027}$$

解答:$$f(x)=x^3-6x^2-5x+4 \Rightarrow f'(x)=3x^2-12x-5 \Rightarrow f''(x)=6x-12\\ f''(x)=0 \Rightarrow x=2 \Rightarrow f(2)=-22 \Rightarrow 反曲點(2,-22) \\ 對反曲點(\alpha,f(\alpha)) 滿足f(\alpha-n)+f(\alpha+n)=2 f(\alpha), 因此f(2-n)+f(2+n)=2f(2)\\ \Rightarrow a=b=2 \Rightarrow f(a)=f(2)= \bbox[red, 2pt]{-22}$$

解答:$$5個X,3個Y,2個Z的排列數={10!\over 5!3! 2!}=2520 \Rightarrow  4個連串的情形:\\\cases{X有2串,Y有1串, Z有1串 \Rightarrow 排列數{4!\over 2! }=6 \Rightarrow \cases{5個X放入2區塊有4種\\ 3個Y放入1區塊有1種\\ 2個Z放入1區塊有1種} \Rightarrow 共6\cdot 4\cdot 1\cdot 1=24種\\X有1串,Y有2串, Z有1串 \Rightarrow 排列數6\Rightarrow \cases{5個X放入1區塊有1種\\ 3個Y放入1區塊有2種\\ 2個Z放入1區塊有1種} \Rightarrow 共6\cdot 1\cdot 2\cdot 1=12種\\X有1串,Y有1串, Z有2串 \Rightarrow 排列數6\Rightarrow \cases{5個X放入1區塊有1種\\ 3個Y放入1區塊有1種\\ 2個Z放入2區塊有1種} \Rightarrow 共6\cdot 1\cdot 1\cdot 1=6種} \\ \Rightarrow 機率={24+12+6\over 2520}= \bbox[red, 2pt]{1 \over 60}$$

解答:$$(x-m)^2=4(y-mh) \Rightarrow 2(x-m)=4y' \Rightarrow y'={x-m\over 2} \Rightarrow 切點P(5,3)的切線斜率m_1={5-m\over 2}\\ 圖形沿著y=mx平移 \Rightarrow 新圖形(x-t-m)^2=4(y-mt-mh) \Rightarrow 2(x-t-m)=4y' \\ \Rightarrow y'={x-m-t\over 2} \Rightarrow 切點P(5,3)的切線斜率m_2={5-m-t\over 2} \Rightarrow m_1+m_2={10-2m-t\over 2}=1 \\ \Rightarrow t=8-2m \cdots(1) \\ \cases{P在舊圖形上\Rightarrow (5-m)^2=4(3-mh) \cdots(2)\\ P在新圖形上\Rightarrow ((5-m)-t)^2=4(3-mh)-4mt \cdots(3)} \Rightarrow -2t(5-m)+t^2=-4mt \\ \Rightarrow t=10-6m 代入(1) \Rightarrow 10-6m=8-2m \Rightarrow m= \bbox[red, 2pt]{1\over 2}$$

二、 計算證明題

解答:

$$假設\angle ABC=\angle ACB= \angle ECD = \angle EDC=\theta，並取\cases{C為原點\\ \overline{CE} =\overline{DE}=5k} \\ \Rightarrow \cases{C(0,0)\\ A(-5\cos \theta, 5\sin \theta) \\ B(-10\cos \theta,0) \\E(5k\cos \theta, 5k\sin \theta)\\ D(10k\cos \theta,0)} \Rightarrow \cases{\overrightarrow{AD} =((10k+5)\cos \theta,-5\sin \theta) \\ \overrightarrow{BE}= ((5k+10)\cos \theta, 5k\sin \theta)}  \Rightarrow \overrightarrow{AD} \cdot \overrightarrow{BE}=0 \\ \Rightarrow \tan^2 \theta=5+2k+{2\over k} \ge 5+ 2\sqrt{2k\cdot {2\over k}}=9 \Rightarrow 極值發生在k=1 \Rightarrow \tan \theta=3 \\\Rightarrow \cases{\sin \theta =3/\sqrt{10} \\ \cos \theta=1/\sqrt{10}}  \Rightarrow \triangle ABC + \triangle ECD =2\triangle ABC= 10\cos \theta\times 5\sin \theta=\bbox[red, 2pt]{15}$$

解答:$$x^{10}+(115x-1)^{10}=0 \Rightarrow 1+ \left( {115x-1\over x} \right)^{10}=0 \Rightarrow \left( {1\over x}-115 \right)^{10}=-1 \Rightarrow z^{10}=-1\\ \Rightarrow z^{10}+1=0 \Rightarrow z_{m},m=1,2,\dots,10為其根\Rightarrow \cases{|z_m|=1 \Rightarrow |z_m|^2=z_m \bar z_m=1\\ \sum_{m=1}^{10} z_m=0 \; (z^9係數為0)} \\ \Rightarrow {1\over r_m}-115=z_m  \Rightarrow {1\over r_1},{1\over \bar r_1},{1\over r_2},{1\over \bar r_2}, \dots, {1\over r_5}, {1\over \bar r_5}相對應的就是z_1+115,z_2+115,\dots,z_{10}+115 \\ \Rightarrow {1\over r_k\bar r_k}={1\over |r_k|^2} = |z_k+115|^2 \Rightarrow {1\over r_1 \bar r_1}+\cdots+{1\over r_5 \bar r_5}= {1\over 2} \sum_{m=1}^{10} |z_m+115|^2 \\= {1\over 2} \sum_{m=1}^{10} (z_m+115)(\bar z_m+115) = {1\over 2} \sum_{m=1}^{10}  (13226+115(z_m+ \bar z_m))= {1\over 2} \sum_{m=1}^{10}  (13226+ 0) = \bbox[red, 2pt]{66130}$$

解答:

$$\triangle ADF: \cos A={81+\overline{AD}^2-81\over 18\overline{AD}} \Rightarrow \overline{AD}=18\cos A, 同理\triangle CEF: \overline{CE}=24\cos C \\ 假設\cases{外接圓半徑R \\ \overline{BC}=a\\ \overline{AC}=b\\ \overline{AB}=c},圓冪定理: \cases{(\overline{OA}+R) (\overline{OA}-R) =\overline{AD}\cdot \overline{AB}  \\ (\overline{OC}+R) (\overline{OC}-R) =\overline{CB}\cdot \overline{CE}  } \\ \Rightarrow \cases{\overline{OA}^2-R^2=c\cdot 18\cos A \\ \overline{OC}^2-R^2=a\cdot 24\cos C} \Rightarrow \cases{\overline{OA}^2=R^2+18c\cdot {21^2+c^2-a^2\over 42c} \\ \overline{OC}^2=R^2+24a\cdot {21^2+a^2-c^2\over 42a}} \\ 利用\text{Stewart's Theorem, }\triangle OAC: \overline{AF}\cdot \overline{OC}^2+ \overline{FC}\cdot \overline{OA}^2= \overline{AC}\cdot \overline{OF}^2 +\overline{AF} \cdot \overline{FC} \cdot \overline{AC} \\ \Rightarrow 4 \overline{OA}^2 +3\overline{OC}^2 =4 \left( R^2+{3\over 7}(441+c^2-a^2) \right) +3 \left( R^2+{4\over 7}(441+a^2-c^2) \right) =2331 \\ \Rightarrow 7R^2+ 1512=2331 \Rightarrow R^2=117 \Rightarrow R= \bbox[red, 2pt]{3\sqrt{13}}$$

解答:$$\lim_{n\to \infty} \sum_{k=n}^{4n}{n^3+k^3\over n^2k^2} =\lim_{n\to \infty} \sum_{k=n}^{4n} \left( {n\over k^2}+{k\over n^2} \right)  =\lim_{n\to \infty} \sum_{k=n}^{4n} {1\over n}\left( ({n\over k})^2+{k\over n} \right)  = \int_1^4  \left( {1\over x^2}+x \right)\,dx \\= \left. \left[ -{1\over x}+{1\over 2}x^2 \right] \right|_1^4 =\bbox[red, 2pt]{33\over 4}$$

解答:

$$依題意\cases{2c=2\sqrt 7\\ 2b^2/a=9/2} \Rightarrow \cases{c=\sqrt 7 \\b^2=9a/4} \Rightarrow a^2-b^2=a^2-{9\over 4}a=7 \Rightarrow a=4 \Rightarrow b^2=9 \\ \Rightarrow 橢圓方程式: {x^2\over 16}+{y^2\over 9} =0 \Rightarrow {2x\over 16}+{2yy'\over 9}=0 \Rightarrow y'=-{9x\over 16y}\\由於\cases{\overline{AP} \bot \overline{BP} \\ \overline{PA}=\overline{PB}} \Rightarrow 過P(x_0,y_0)之切線斜率為1 \Rightarrow 1=-{9x_0\over 16y_0} \Rightarrow x_0=-{16\over 9}y_0 \\ \Rightarrow { ({-{16\over 9}y_0})^2\over 16}+{y_0^2\over 9}=1 \Rightarrow y_0^2={81\over 25} \Rightarrow y_0={9\over 5}\; (P在第二象限, y_0\gt 0) \Rightarrow x_0=-{16\over 5} \\ \Rightarrow P= \bbox[red, 2pt]{\left( -{16\over 5},{9\over 5} \right)}$$

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