115年大同高中教甄-數學詳解

 臺北市立大同高級中學 115 學年度第1次教師甄選

一、 填充題(每題 6 分)

解答:$$f(0)=c=0 \Rightarrow f(x)=x^3+ax^2+bx \Rightarrow f(2)=8+4a+2b=2 \Rightarrow b=-3-2a \\ \Rightarrow f(x)=x^3+ax^2-(2a+3)x \Rightarrow f(6)=198+24a \Rightarrow 420\le 198+24a \le 460\\ \Rightarrow {37\over 4} \le a\le {131\over 12} \Rightarrow a=10 \Rightarrow b=-23 \Rightarrow f(x)=x^3+10x^2-23x \Rightarrow f(1)= \bbox[red, 2pt]{-12}$$

解答:$$A=({12\over 5},{9\over 5}) 為新橢圓長軸頂點 \Rightarrow a=\overline{OA} = \sqrt{({12\over 5})^2+ ({9\over 5})^2} =3 \\ \Rightarrow 旋轉前頂點座標(3,0) \xrightarrow{旋轉\theta後} ({12\over 5},{9\over 5}) \Rightarrow \begin{bmatrix}\cos \theta& -\sin \theta\\ \sin \theta& \cos \theta \end{bmatrix} \begin{bmatrix}3\\0 \end{bmatrix} = \begin{bmatrix}12/5\\ 9/5 \end{bmatrix} \Rightarrow \cases{\cos \theta=4/5\\ \sin \theta=3/5}\\ 假設P'(0,y_0) \in \Gamma' \Rightarrow 29\cdot 0^2-24\cdot 0\cdot y_0 +36y_0^2=180 \Rightarrow y_0^2=5 \Rightarrow y_0= \sqrt 5 \Rightarrow P'(0,\sqrt 5) \\ \Rightarrow P=(x,y) \Rightarrow  \begin{bmatrix} \cos(-\theta)& -\sin(-\theta) \\ \sin(-\theta)& \cos (-\theta) \end{bmatrix} \begin{bmatrix}0\\ \sqrt 5 \end{bmatrix} = \begin{bmatrix}3\sqrt 5/5\\ 4\sqrt 5/5\end{bmatrix} \Rightarrow P= \bbox[red, 2pt]{\left( {3\sqrt 5\over 5}, {4\sqrt 5\over 5} \right)}$$

解答:$$正面機率p={2\over 3} \Rightarrow 反面機率q=1-p={1\over 3} \\\Rightarrow 擲100次正面次數為k的機率P(X=k)= {100\choose k} p^k q^{100-k} \\ 取f(p,q)=(p+q)^{100}=1^{100} = P(X=0) +P(X=1)+ P(X= 2)+ \cdots+P(X= 100)\\ \Rightarrow f(-p,q) = \left( -{1\over 3} \right)^{100}=P(X=0)-P(X=1)+P(X=2)-P(X=3)+ \cdots+ P(X=100) \\ \Rightarrow f(p,q)+ f(-p+q) =1+ \left( {1\over 3} \right)^{100} =2 \left[P(X=0) +P(X=2)+ \cdots +P(X=100) \right]\\ \Rightarrow 正面次數為偶數的機率 =P(X=0) +P(X=2)+ P(X=4)+ \cdots+ P(X=100) \\={1\over 2} \left(  1+ \left( {1\over 3} \right)^{100}\right)  \Rightarrow {1+(1/3)^{100} \over 2} ={1+b^{100} \over a} \Rightarrow (a,b) = \bbox[red, 2pt]{\left( 2,{1\over 3} \right)}$$

解答:$$\bbox[cyan,2pt]{試題有誤}, z=3+4i \Rightarrow \cases{|-7/5+24i/5-z^3|=10\sqrt{149} \\|-7/5+24i/5-z|=2\sqrt 5} \Rightarrow 兩者不相等\\ 題目若改成\left|{-7\over 5}+{24\over 5}i-{z^3\over 25} \right| =\left|{-7\over 5}+{24\over 5}i-z \right|, 答案就是z=3+4i$$

解答:$$\lim_{n\to \infty}\sum_{k=1}^n  {k^p\over n^{p+1}} =\lim_{n\to \infty}\sum_{k=1}^n  \left( {1\over n} \cdot ({k\over n})^p \right) = \int_0^1 x^p\,dx = \left. \left[ {1\over p+1}x^{p+1} \right] \right|_0^1=\bbox[red, 2pt]{1\over p+1}$$

解答:$$圓內接多邊形的某個角是90度，代表對應的弦是直徑。\\ 在10個等分點可以配對成5條直徑，因此選出的四個點，必須包含一組對應圓心的「對應點」\\10個點任選4點，有C^{10}_4=210種選法。要讓四邊形沒有直角，代表不能選出「對應點」\\ 在5組「對應點」中挑出4組，再從4組中各挑1點，有C^5_4\times 2^4= 80種選法\\ 因此至少有一個直角的機率 ={210-80\over 210}= \bbox[red, 2pt]{13\over 21}, 但公布的答案是\bbox[cyan,2pt]{1\over 3}$$

解答:$$k= [\sqrt n] \Rightarrow k\le \sqrt n\lt k+1 \Rightarrow k^2 \le n\lt(k+1)^2 \Rightarrow 共有((k+1)^2-1)-k^2+1=2k+1個\\ \Rightarrow S=\sum_{n=1}^\infty {1\over 2^{[\sqrt n]}} = \sum_{k=1}^\infty  \left( \sum_{n=k^2}^{(k+1)^2-1} {1\over 2^k}\right) = \sum_{k=1}^\infty {2k+1\over 2^k}  \Rightarrow {1\over 2}S =\sum_{k=1}^\infty {k+1/2\over 2^{k}} \\ \Rightarrow S-{1\over 2}S= \sum_{k=1}^\infty {k+1/2\over 2^{k}}= \sum_{k=1}^\infty {k\over 2^{k}} +\sum_{k=1}^\infty {1\over 2^{k+1}}$$

解答:

$$以A為旋轉中心\Rightarrow \cases{P \xrightarrow{逆時針旋60^\circ} P'\\  B \xrightarrow{逆時針旋60^\circ} C} \Rightarrow \cases{\overline{P'C} =\overline{PB}=5\\ \overline{PA}=\overline{P'A} =3 \\ \angle PAP'=60^\circ } \Rightarrow \triangle APP'為邊長3的正\triangle\\ \Rightarrow \cos \angle PP'C ={\overline{PP'}^2+ \overline{P'C}^2- \overline{PC}^2 \over 2\cdot \overline{PP'} \cdot \overline{P'C}} ={3^2+5^2-7^2\over 2\cdot 3\cdot 5} =-{1\over 2} \Rightarrow \angle PP'C=120^\circ \\ \Rightarrow \angle AP'C=120^\circ-60^\circ=60^\circ \Rightarrow \cos \angle AP'C={1\over 2} ={3^2+5^2-\overline{AC}^2 \over 2\cdot 3\cdot 5} \Rightarrow \overline{AC}^2=19 \\ \Rightarrow 正\triangle ABC面積 ={\sqrt 3\over 4}\cdot 19 =\bbox[red, 2pt]{19\sqrt 3\over 4}$$

$$\bbox[cyan,2pt]{另一種情況}: A,P,B在一直線上，如上圖，則\overline{AB}=8 \Rightarrow \triangle ABC= {\sqrt 3\over 4} \cdot 8^2= \bbox[cyan,2pt]{16\sqrt 3}$$

二、計算題(52 分)

解答:$$\textbf{(1) }\bbox[red,2pt]{正確}\\ \textbf{(2) }\bbox[red, 2pt]{錯誤}:二項分布的前提是每次試驗必須獨立且機率固定,本題在「不放回」的情況下，\\\qquad 第二次抽到紅球的機率會受到第一結果的影響，不適用二項分布。$$

解答:$$\cases{球心O(0,0,0)\\ A(5,0,0)} \Rightarrow 球半徑R=\overline{OA}=5 \Rightarrow 球方程式:x^2+y^2+z^2= 25 \\ \Rightarrow \overline{AB} =\sqrt{2^2+ 4^2+0^2}= 2\sqrt 5 \\C=(x,y,z) 在球面上\Rightarrow  \stackrel{\Large \frown}{AB} =\stackrel{\Large \frown}{AC}=\stackrel{\Large \frown}{BC}  \Rightarrow \overline{AB}^2= \overline{AC}^2= \overline{BC}^2  \Rightarrow \cases{x^2+ y^2+z^2 =25 \\ (x-5)^2+ y^2+z^2=20\\ (x-3)^2 +(y-4)^2+z^2=20}\\ \Rightarrow \cases{x=3\\ y=3/2\\ z= \pm \sqrt{55}/2} \Rightarrow C= \bbox[red, 2pt]{\left( 3,{3\over 2}, \pm {\sqrt{55}\over 2} \right)}$$

解答:$$假設[x^3]=k=4x+3 為一整數 \Rightarrow x={k-3\over 4} \Rightarrow 4x+3\le x^3 \lt (4x+3)+1 \\ \Rightarrow k\le \left( {k-3\over 4} \right)^3 \lt k+1 \Rightarrow 64k\le k^3-9k^2+27k-27\lt 64k+64 \\ \Rightarrow 左半部: 64k\le k^3-9k^2+27k-27  \Rightarrow (k+1)(k^2-10k-27)\ge 0\\  \Rightarrow \cases{k\ge5+\sqrt{52} \Rightarrow k=13,14,\dots\\ 5-\sqrt{52}\le k\le -1 \Rightarrow k=-1,-2}\\ 右半部:k^3-9k^2+27k-27\lt 64k+64 \Rightarrow f(k)=k^3-9k^2-37k-91\lt 0 \\ \Rightarrow \cases{f(-1)=-64\lt 0\\ f(-2)=-61\lt 0\\ f(13) =104\not\lt 0, f(k)遞增, k\ge 13} \Rightarrow \cases{k=-1 \Rightarrow x=-1\\ k=-2 \Rightarrow -5/4} \Rightarrow x=\bbox[red, 2pt]{-{5\over 4}或-1}$$

解答:$$\overline{AC} =\sqrt{\overline{AB}^2+ \overline{BC}^2} =\sqrt{1^2+(\sqrt 2)^2} = \sqrt 3\\ 假設\cases{A(0,0,0) \\ \overline{AC}在x軸上\\ D在\overline{AC} 的垂足為K\\B 在\overline{AC} 的垂足為H} \Rightarrow \cases{\overline{AK}=2\sqrt 3/3\\ \overline{DK} =\sqrt 6/3\\ \overline{AH}=\sqrt 3/3\\ \overline{BH}= \sqrt 6/3} \Rightarrow \cases{D({2\sqrt 3\over 3},{\sqrt 6\over 3},0) \\ B({\sqrt 3\over 3},-{\sqrt 6\over 3},0)} \Rightarrow B'({\sqrt 3\over 3}, {\sqrt 6\over 3}\cos \theta, {\sqrt 6\over 3}\sin \theta) \\ \Rightarrow \overline{B'D}^2 = \left( {\sqrt 3\over 3}-{2\sqrt 3\over 3} \right)^2 + \left( {\sqrt 6\over 3}\cos \theta-{\sqrt 6\over 3} \right)^2 + \left( {\sqrt 6\over 3}\sin \theta-0 \right)^2 ={5\over 3}-{4\over 3} \cos\theta \\ \Rightarrow \overline{B'B}^2 = \left( {\sqrt 3\over 3}-{\sqrt 3\over 3} \right)^2 + \left( {\sqrt 6\over 3}\cos \theta+{\sqrt 6\over 3} \right)^2 + \left( {\sqrt 6\over 3}\sin \theta-0 \right)^2 = {4\over 3}+{4\over 3} \cos \theta \\ \theta=60^\circ \Rightarrow \cases{\overline{B'D}^2 =1 \\ \overline{B'B}^2 =\sqrt 2} \Rightarrow \overline{B'B} \gt \overline{B'D}\;違背題意\\ \theta =120^\circ \Rightarrow \cases{ \overline{B'D}^2 =7/3\\ \overline{B'B}^2 =2/3} \Rightarrow \overline{B'D} \gt \overline{B'B} 符合題意\Rightarrow \overline{B'D}=\bbox[red, 2pt]{\sqrt{21} \over 3}, 公布的答案是\bbox[cyan,2pt]1$$

解答:

$$假設\cases{圓半徑R\\ \angle OBA=\angle OBC=\theta}, 由於\overline{OC}=R = \overline{OB} \Rightarrow \angle OCB= \angle OBC=\theta \Rightarrow \angle OCP=\theta \\ \triangle OCB及\triangle PBC皆為等腰三角形 \Rightarrow \overline{PQ}為\overline{BC}的中垂線 \Rightarrow \cases{\overline{OQ} =R\sin \theta \\ \overline{BQ}= R\cos \theta} \Rightarrow \overline{PQ}=R\sin \theta+{R\over 2}\\ \Rightarrow \tan \angle PBQ= \tan 2\theta ={\overline{PQ} \over \overline{BQ}} \Rightarrow {\sin 2\theta\over \cos 2\theta} ={R\sin \theta+R/2\over R\cos \theta} \Rightarrow {2\sin \theta\cos \theta\over 1-2\sin^2\theta}={\sin \theta+1/2\over \cos \theta} \\ \Rightarrow 2\sin^2\theta+ 2 \sin \theta-1=0 \Rightarrow \sin \theta={\sqrt 3-1\over 2} \Rightarrow \cos\theta= {\sqrt{2\sqrt 3}\over 2}\\ 又\overline{OA}= \overline{OB}=R \Rightarrow \angle OAB=\theta \Rightarrow \angle AOP=180^\circ-\angle OAP-\angle OPA=180^\circ-\theta-(90^\circ+2\theta) \\ \Rightarrow \angle AOP=90^\circ-3\theta \Rightarrow \sin \angle AOP= \cos 3\theta= 4\cos^3\theta-3\cos\theta= {\sqrt{2\sqrt 3}\over 2}(2\sqrt 3-3) \\= \bbox[red, 2pt]{\sqrt{2\sqrt 3} \left( \sqrt 3-{3\over 2} \right)}$$

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