114年永春高中教甄-數學詳解

 臺北市立永春高中 114 學年度第 1 次正式教師甄選

一、填充題：每題 7 分，共 70 分

解答: $$假設S_1:\cases{甲有1白1黑球 \\ 乙有3白3黑球}, S_2:\cases{甲有2黑球 \\ 乙有4白2黑球},S_3:\cases{甲有2白球 \\ 乙有2白4黑球} \\ \Rightarrow \cases{P(S_1\to S_1)={1\over 2}\cdot{3\over 6} +{1\over 2} \cdot{3\over 6} ={1\over 2} \\ P(S_1\to S_2) ={1\over 2}\cdot {1\over 2}={1\over 4} \\P(S_1\to S_3)={1\over 2}\cdot {1\over 2}={1\over 4}}, \cases{P(S_2\to S_1)= {4\over 6}={2\over 3}\\  P(S_2\to S_2)= {2\over 6}={1\over 3}} , \cases{P(S_3\to S_1)={4\over 6}={2\over 3} \\ P(S_3\to S_3)={2\over 6}={1\over 3}} \\ \Rightarrow 轉換矩陣A=\begin{bmatrix} 1/2& 2/3& 2/3\\ 1/4& 1/3& 0\\ 1/4& 0& 1/3\end{bmatrix} ,穩定後Ax=x \Rightarrow \begin{bmatrix} 1/2& 2/3& 2/3\\ 1/4& 1/3& 0\\ 1/4& 0& 1/3\end{bmatrix} \begin{bmatrix} x_1\\ x_2\\  x_3\end{bmatrix}=\begin{bmatrix} x_1\\ x_2\\  x_3\end{bmatrix} \\ \Rightarrow \cases{x_1=4/7\\ x_2=3/14\\ x_3=3/14} \Rightarrow P(S_1)=x_1=\bbox[red, 2pt]{4\over 7}$$

解答: $$\text{INSTITUTIONALIZED }17個字母  \Rightarrow \cases{\text{I}\times 4 \\ \text{T}\times 3\\ \text{N}\times 2 \\其它8個字母各1} \\ \Rightarrow \cases{4同:只1種\\ 3同1異:C^2_1C^{10}_1 \times 排列數4=80\\ 2同2同:C^3_2\times 排列數{4!\over 2!2!}=18\\ 2同2異:C^3_1C^{10}_2 \times 排列數{4!\over 2!} =1620 \\ 4異:C^{11}_4\times 排列數4!=7920} \Rightarrow 合計1+80+18+ 1620+7920= \bbox[red, 2pt]{9639}$$

解答: $$(\sqrt{x^2+x+1}+ \sqrt{2x^2+x+5} )^2= (\sqrt{x^2-3x+13})^2 \\ \Rightarrow 3x^2+2x+6+ 2\sqrt{(x^2+x+1)(2x^2+x+5)} =x^2-3x+13 \\ \Rightarrow 2\sqrt{(x^2+x+1)(2x^2+x+5)} =-2x^2-5x+7 \Rightarrow 4(x^2+x+1)(2x^2+x+5) =(-2x^2-5x+7)^2 \\ \Rightarrow 4(2x^4+ 3x^3+ 8x^2+  6x +5) =(4x^4+25x^2+49+2(10x^3-35x-14x^2)) \\ \Rightarrow 8x^4+12x^3+32x^2+24x+20=4x^4+20x^3-3x^2-70x+49 \\ \Rightarrow 4x^4-8x^3+35x^2+94x-29=0 \Rightarrow (2x^2+3x-1)(2x^2-7x+29)=0\\ \Rightarrow x= \bbox[red, 2pt]{-3\pm \sqrt{17} \over 4},{7\pm \sqrt{183}i\over 4}(非實根,不合)\\ \bbox[red, 2pt]{另解: }假設x^2-3x+13=a(x^2+x+1)+b(2x^2+x+5 ) \Rightarrow \cases{a+2b=1\\ a+b=-3\\a+5b= 13} \Rightarrow \cases{a=-7\\ b=4}\\ 原式\sqrt{x^2+x+1} +\sqrt{2x^2+x+5}=\sqrt{-7(x^2+x+1)+4(2x^2+x+5)} \\ \equiv \sqrt{A}+\sqrt{B}=\sqrt{-7A+4B} \Rightarrow A+B+2\sqrt{AB}=-7A+4B \Rightarrow 2\sqrt{AB} =-8A+3B \\ \Rightarrow 4AB= 64A^2-48AB+9B^2 \Rightarrow 64A^2 -52AB+9B^2=0 \Rightarrow (16A-9B)(4A-B)=0 \\ \Rightarrow \cases{16A=9B\\ 4A=B} \Rightarrow \cases{2x^2-7x+29=0 無實數解\\ 2x^2+3x-1=0} \Rightarrow x= \bbox[red, 2pt]{-3\pm \sqrt{17} \over 4}$$

解答: $$x=r\cos \theta = (1+\cos \theta)\cos \theta = \cos^2+ \cos \theta = (\cos \theta+{1\over 2})^2-{1\over 4} \Rightarrow 最小值: \bbox[red, 2pt]{-{1\over 4}}$$

解答: $$\sum P(X)=1 \Rightarrow \sum_{n=1}^\infty\left({a\over 2^n} +{b\over 3^n} \right) =a\cdot {1/2\over 1-1/2} +b\cdot {1/3\over 1-1/3} =a+{1\over 2}b=1 \cdots(1) \\  令 f(x)={1\over 1-x} =\sum_{n=0}^\infty x^n \Rightarrow f'(x)={1\over (1-x)^2} =\sum_{n=1}^\infty nx^{n-1} \\\Rightarrow 取g(x) =xf'(x) ={x\over (1-x)^2} =\sum_{n=1}^\infty nx^{n} \\ \Rightarrow EX= \sum_{n=1}^\infty\left({na\over 2^n} +{nb\over 3^n} \right) =a g({1\over 2})+bg({1\over 3}) =a\cdot 2+b\cdot {3\over 4} ={15\over 8} \cdots(2) \\ 由式(1)及(2)可得 (a,b)= \bbox[red, 2pt]{\left({3\over 4},{1\over 2} \right)}$$

解答:

$$\overline{AP} =d(y=3x,y=3x+2) ={2\over \sqrt{10}} \Rightarrow \overline{OA}= \overline{AB} ={2\over \sqrt{10}} \cdot {2\over \sqrt 3} ={4\over \sqrt{30}} \\ A在y=3x上\Rightarrow A(a,3a) \Rightarrow \overline{OA}^2=10a^2 ={16\over 30} \Rightarrow a={2\over 5\sqrt 3} \Rightarrow A({2\over 5\sqrt 3},{6\over 5\sqrt 3}) \\ \Rightarrow 圓心Q=A逆時針旋轉60^\circ =\begin{bmatrix}1/2& -\sqrt 3/2 \\ \sqrt 3/2 & 1/2 \end{bmatrix} \begin{bmatrix} {2\over 5\sqrt 3}\\{6\over 5\sqrt 3}\end{bmatrix} =\begin{bmatrix} {\sqrt 3-9\over 15}\\{\sqrt 3+1\over 5 }\end{bmatrix} \\\Rightarrow Q= \bbox[red, 2pt] {\left(  {\sqrt 3-9\over 15},{\sqrt 3+1\over 5 } \right)}$$

解答:

$$在\overline{BC}外取一點Q，使得\overline{BQ}=6且\angle ABC=PBQ，則{\overline{AB} \over \overline{BC}} ={\overline{BP} \over \overline{BQ}} ={4\over 6} \Rightarrow \triangle ABC \sim \triangle PBQ \\ \Rightarrow \overline{PQ}=5又\cases{\angle ABP=\angle CBQ \\ {\overline{PB}\over \overline{BA}}= {\overline{BQ} \over \overline{BC}} ={1\over k}} \Rightarrow \triangle PBA \sim \triangle QBC \Rightarrow \overline{QC}=12 \\ \Rightarrow \cos \angle BQP= {5^2+6^2-4^2\over 2\cdot 5\cdot 6} = {3\over 4} \Rightarrow \sin \angle BQP = {\sqrt 7\over 4}\\又 \triangle PQC三邊長為5,12,13 \Rightarrow \angle PQC=90^\circ \Rightarrow \cos \angle APB = \cos \angle BQC = \cos (\angle BQP+90^\circ) \\=-\sin \angle BQP = \bbox[red, 2pt]{-{\sqrt 7\over 4}}$$

解答: $$A= \begin{bmatrix}7&-8\\ -7&8 \end{bmatrix} \Rightarrow \det(A-\lambda I) =\lambda^2-15\lambda  =0 \Rightarrow \lambda=0,15\\ 取f(\lambda) =(1+\lambda)^n = (\lambda^2-15\lambda) p(\lambda)+ a\lambda+b \Rightarrow \cases{f(0)=1=b\\ f(15) =16^n=15a+b} \Rightarrow \cases{a=(16^n-1)/15 \\b=1} \\ \Rightarrow f(A)=(1+A)^n =aA+bI={1\over 15}(16^n-1)A+I \Rightarrow a_n= \bbox[red, 2pt]{{1\over 15}(16^n-1)}$$

解答: $$a_n = {1\over (\sqrt n+\sqrt{n+1}) (\sqrt {n+1}+\sqrt{n+2}) (\sqrt n+\sqrt{n+2}) } \\= {1\over \sqrt{n+2}-\sqrt n}\left({1\over (\sqrt n+\sqrt{n+1}) (\sqrt {n}+\sqrt{n+2})  } -{1\over (\sqrt{n}+ \sqrt{n+ 2})(\sqrt{n +1}+\sqrt{n+2})} \right) \\={1\over (\sqrt{n+2}-\sqrt n)(\sqrt n+\sqrt{n+2})} \left({1\over \sqrt n+\sqrt{n+1}   } -{1\over \sqrt{n+1}+\sqrt{n+2}} \right) \\={1\over 2} \left({1\over \sqrt n+\sqrt{n+1}   } -{1\over \sqrt{n+1}+\sqrt{n+2}} \right) \\ \Rightarrow \sum_{k=1}^\infty a_k ={1\over 2} \left( {1\over \sqrt 1+\sqrt 2}-{1\over \sqrt 2+\sqrt 3} +{1\over \sqrt 2+\sqrt 3}-{1\over \sqrt 3+\sqrt 4} + \cdots\right) \\ ={1\over 2}\cdot {1\over 1+\sqrt 2} = \bbox[red, 2pt]{\sqrt 2-1\over 2}$$

解答:

$$f(x)=\ln(x+1) \Rightarrow  f'(x)={1\over x+1} \Rightarrow f'(1)={1\over 2} \Rightarrow 法線斜率:-2 \Rightarrow 法線L:y=-2(x-1)+\ln(2) \\ \Rightarrow x=1+{1\over 2}(\ln 2-y)\Rightarrow L與y軸交於B(0,2+\ln 2) \Rightarrow R=R_1\cup R_2,如上圖\\ \Rightarrow \cases{R_1繞y軸旋轉體積= \int_{\ln 2}^{2+\ln 2} \left(1+{1\over 2}(\ln 2-y) \right)^2 \pi\,dy ={2\over 3}\pi\\ R_2繞y軸旋轉體積= \int_0^{\ln 2} \left( e^y-1 \right)^2\pi \,dy =\pi\left(\ln 2-{1\over 2} \right)} \Rightarrow {2\over 3}\pi+ \pi\left(\ln 2-{1\over 2} \right) \\= \bbox[red, 2pt]{\pi\left({1\over 6}+\ln 2 \right)}$$

二、計算證明題：每題 10 分，共 30 分

解答: $$假設\cases{\overline{BC}=a\\ \overline{AC}=b\\ \overline{AB}=c}, 又O為外心 \Rightarrow \cases{\overrightarrow{AO} \cdot \overrightarrow{AB} =\overrightarrow{BO} \cdot \overrightarrow{BA}=c^2/2\\ \overrightarrow{AO} \cdot \overrightarrow{AC} = \overrightarrow{CO} \cdot \overrightarrow{CA}= b^2/2\\ \overrightarrow{BO} \cdot \overrightarrow{BC} =\overrightarrow{CO} \cdot \overrightarrow{CB}= a^2/2 }\\ \overrightarrow{AO} \cdot \overrightarrow{BC} =3\overrightarrow{BO} \cdot \overrightarrow{AC} +4\overrightarrow{CO} \cdot \overrightarrow{BA} \\ \Rightarrow \overrightarrow{AO} \cdot (\overrightarrow{BA} +\overrightarrow{AC}) =3\overrightarrow{BO} \cdot (\overrightarrow{AB} +\overrightarrow{BC}) + 4\overrightarrow{CO} \cdot (\overrightarrow{BC}+ \overrightarrow{CA}) \\ \Rightarrow -c^2+b^2 =3(-c^2+a^2)+4(-a^2+b^2) \Rightarrow a^2+2c^2= 3b^2 \\ \Rightarrow \cos B={a^2+c^2-b^2 \over 2ac} ={a^2+c^2-(a^2+2c^2)/3\over 2ac} ={(2/3)a^2+(1/3)c^2 \over 2ac} \ge {2\sqrt{(2/9)a^2c^2}\over 2ac} ={\sqrt 2\over 3} \\ \Rightarrow 最小值為\bbox[red, 2pt]{\sqrt 2\over 3}$$

解答: $$柯西不等式: \left(\sum_{i=1}^n|x_i-\bar x|^2 \right)  (1^2+1^2+ \cdots+1^2) \ge \left(\sum_{i=1}^n |x_i-\bar x| \right)^2 \\ \Rightarrow {1\over n}\sum_{k=1}^n |x_i-\bar x|^2 \ge {1\over n^2}\left(\sum_{i=1}^n |x_i-\bar x| \right)^2 \\ \Rightarrow \sqrt{{1\over n}\sum_{i=1}^n (x_i^2-2\bar x x_i+(\bar x)^2)} =\sqrt{{1\over n} \sum_{i=1}^n x_i^2-{2\over n^2}(\sum_{i=1}^n x_i)^2+{1\over n^2}(\sum_{i=1}^n x_i)^2  } \ge {1\over n}\sum_{i=1}^n |x_i-\bar x| \\ \Rightarrow \sqrt{{1\over n} \sum_{i=1}^nx_i^2-{1\over n^2}(\sum_{i=1}^n x_i)^2} =\sqrt{{1\over n} \sum_{i=1}^nx_i^2-(\bar x)^2}\ge {1\over n}\sum_{i=1}^n |x_i-\bar x| \quad \bbox[red, 2pt]{QED.}$$

解答: $$不失一般性，假設\cases{A(0,0,0) \\B(a,a,0) \\C(a,0,a)\\ D(0, a,a)} \Rightarrow \cases{P=(A+C)/2= (a/2,0,a/2)\\ Q=(B+C)/2 =(a,a/2,a/2) \\R=(B+D)/2 =(a/2,a, a/2)} \Rightarrow \cases{\overrightarrow{PQ} =(a/2,a/2,0) \\ \overrightarrow{PR} =(0,a,0)} \\ \Rightarrow \vec n= \overrightarrow{PQ} \times \overrightarrow{PR} =(0,0,a^2/2) \parallel (0,0,1) \Rightarrow \cases{E=\triangle PQR:z={a\over 2} \\ \overleftrightarrow{AD}:(0,t,t),t\in\mathbb R} \Rightarrow S=(0,a/2,a/2) \\ \Rightarrow \cases{\overline{PQ} =\overline{QR} =\overline{RS} =\overline{PS} =a/\sqrt 2 \\ \overrightarrow{PQ}\cdot \overrightarrow{PS}=(a/2,a/2,0)\cdot (-a/2,a/2,0)=0 } \Rightarrow PQRS為正方形\quad \bbox[red, 2pt]{QED.}$$

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