114年屏科實中高中部教甄-數學詳解

 國立屏科實驗高級中等學校114學年度第1次專任教師甄選

高中部 數學科 教師 初試試題

解答:$$\textbf{(1) }空間中\cases{\vec a=(x_1,y_1,0) \\ \vec b=(x_2,y_2, 0)} \Rightarrow \vec a\times \vec b= (0,0,x_1y_2-x_2y_1) \\ \Rightarrow 平行四邊形面積= \|\vec a\times \vec b \| =\sqrt{(x_1y_2-x_2y_1)^2} =|x_1y_2-x_2y_1| = \begin{vmatrix} x_1&y_1\\ x_2& y_2\end{vmatrix}的絕對值\;\bbox[red, 2pt]{QED} \\\textbf{(2) }平行六面體體積=|\vec a\cdot( \vec b\times \vec c)|= |(x_1,y_1,z_1)\cdot (y_2z_3-y_3z_2, x_3z_2-x_2z_3, x_2y_3-x_3y_2) | \\\quad= |x_1y_2z_3-x_1y_3z_2+x_3y_1z_2-x_2y_1z_3+x_2y_3z_1-x_3y_2z_1| \\ \quad  = \begin{vmatrix}x_1&y_1& z_1\\ x_2&y_2& z_2 \\x_3& y_3& z_3 \end{vmatrix} 的絕對值\quad \bbox[red, 2pt]{QED}$$

解答:$$\textbf{法一: }x^2-2x+y^2-4y=-4 \Rightarrow (x-1)^2+(y-2)^2=1 \Rightarrow \cases{x=\cos \theta+1\\ y=\sin \theta+2} \\ \Rightarrow 2x+y+3=2\cos\theta+2+\sin\theta+2+3= 2\cos \theta+\sin \theta+7 =\sqrt 5\sin(\theta+\alpha)+7 \\ \Rightarrow |\sqrt 5\sin(\theta+\alpha)+7|最小值為7-\sqrt 5 \\\textbf{法二: }任一點P(x,y)到直線L:2x+y+3=0的距離=d(P,L) ={|2x+y+3| \over \sqrt 5} \\ \quad \Rightarrow |2x+y+3|= \sqrt 5\times d(P,L) \\\quad 若P在圓上，則d(P,L)的最小值=圓心(1,2)至L距離,減去圓半徑1\\ \quad ={7\over \sqrt 5}-1 \Rightarrow |2x+y+3|的最小值= \sqrt 5 \left( {7\over \sqrt 5}-1 \right) =7-\sqrt 5$$

解答:$$\textbf{(1) }\cases{11x+9y=13\\ 13x+10y=16} \Rightarrow \cases{110x+90y=130\\ 117x+90y=144} \Rightarrow 7x=14 \Rightarrow x=2 \Rightarrow 22+9y=13\\\qquad \Rightarrow 9y=-9 \Rightarrow y=-1 \Rightarrow \bbox[red, 2pt]{\cases{x=2\\ y=-1}} \\\textbf{(2)} 克拉瑪公式幾何意義:\href{https://ghresource.k12ea.gov.tw/uploads/1678432181269jfCqnmWg.pdf}{參考資料}$$

解答:$${p+q +r\over 3}\ge \sqrt[3]{pqr} \Rightarrow {p^3+q^3+r^3\over 3} \ge \sqrt[3]{p^3q^3r^3} =pqr \Rightarrow pqr \le {1\over 3}(p^3+q^3+ r^3) \\ \Rightarrow  \cases{  \displaystyle {a\over \sqrt[3]{a^3+b^3}} \cdot {x\over \sqrt[3]{x^3+y^3}} \cdot {m\over \sqrt[3]{m^3+n^3}} \le {1\over 3} \left( {a^3\over a^3+b^3} +{x^3\over x^3+y^3} +{m^3\over m^3+n^3} \right) \\\displaystyle {b\over \sqrt[3]{a^3+b^3}} \cdot {y\over \sqrt[3]{x^3+y^3}} \cdot {n\over \sqrt[3]{m^3+n^3}} \le {1\over 3} \left( {b^3\over a^3+b^3} +{y^3\over x^3+y^3} +{n^3\over m^3+n^3} \right) }\\  兩式相加 \Rightarrow {axm+ byn\over \left( \sqrt[3]{a^3+b^3} \right) \cdot \left( \sqrt[3]{x^3+y^3} \right) \cdot \left( \sqrt[3]{m^3+n^3} \right)} \le {1\over 3}(1+1+1)=1 \\ \Rightarrow axm+byn \le  \left( \sqrt[3]{a^3+b^3} \right) \cdot \left( \sqrt[3]{x^3+y^3} \right) \cdot \left( \sqrt[3]{m^3+n^3} \right) \\ \Rightarrow (axm+byn)^3 \le (a^3+b^3)(x^3 +y^3) (m^3+n^3)\quad \bbox[red, 2pt]{QED.}$$

解答:$$\textbf{(1) }公正骰子擲出6的機率為p={1\over 6}，擲出不是6的機率q=1-p={5\over 6} \\事件A:\cases{甲在第一輪出局:甲第一輪就擲出6:機率=p\\甲在第二輪出局: 第一輪三人都沒擲出6; 接著第二輪甲擲出6,因此機率為q^3p,\\甲在第三輪出局:前兩輪都沒擲出6; 接著第三輪甲擲出6,因此機率為q^6p \\\cdots} \\ \Rightarrow P(A)=p+q^3p+ q^6p+ \cdots =p(1+q^3+q^6+ \cdots) =p\cdot {1\over 1-q^3} ={36\over 91}\\ 同理, P(B)=qp+q^4p+q^7p+\cdots =qP(A)={30\over 91} \Rightarrow P(C)=q^2P(A)= {25\over 91} \Rightarrow \bbox[red, 2pt]{\cases{P(A)=36/91\\ P(B)=30/91 \\P(C) =25/91} } \\\textbf{(2) } 考慮兩人擲骰的情況，先擲骰者出局的機率P_1=p+q^2p+ q^4p+ \cdots ={p\over 1-q^2}={6\over 11};\\ \qquad \Rightarrow 後擲骰者出局的機率P_2=1-P_1 ={5\over 11}\\ A是第二位出局的情況:\\ \textbf{Case I 乙先出局,然後甲出局: } 乙是第一位出局的機率P(B) , 之後擲股順序為丙甲丙甲...\\ \qquad \Rightarrow 甲為後擲骰者 \Rightarrow 甲比丙先出區的機率為P_2 \Rightarrow P(B)\times P_2= {30\over 91}\times {5\over 11} ={150\over 1001} \\ \textbf{Case II 丙先出局,然後甲出局: }丙是第一位出局的機率P(C) , 之後擲股順序為甲乙甲乙...\\ \qquad \Rightarrow 甲為先擲骰者 \Rightarrow 甲比乙先出局的機率為P_1 \Rightarrow P(C) \times P_1={25\over 91} \times {6\over 11} = {150\over 1001} \\ \Rightarrow P(D) ={150\over 1001} +{150\over 1001} = \bbox[red, 2pt]{300\over 1001}$$

解答:$$\textbf{(1) }k=1+2\omega+3\omega^2+ \cdots+ n\omega^{n-1} \Rightarrow \omega k=\omega+ 2\omega^2+ 3\omega^3+ \cdots+n\omega^n \\ \Rightarrow k-\omega k =1+\omega+\omega^2+\cdots+\omega^{n-1}-n\omega^n ={1-\omega^n\over 1-\omega}-n\omega^n \Rightarrow k= {1-\omega^n\over (1-\omega)^2}-{n\omega^n \over 1-\omega} \\ \Rightarrow k={1-\omega^n-n\omega^n+n\omega^{n+1} \over (1-\omega)^2} ={1-1-n+n\omega \over (1-\omega)^2} =\bbox[red, 2pt]{-n\over 1-\omega} \\ \textbf{(2) } |k| ={n\over |1-\omega|} \ge{n\over 1+|\omega|} ={n\over 2} \Rightarrow |k|\ge {n\over 2} \cdots(1) \\ 又|k| = |1+2\omega+3\omega^2+ \cdots+ n\omega^{n-1}|  \le 1+|2\omega|+ |3\omega^2|+ \cdots+ |n\omega^{n-1|} \\\qquad\le 1+2+3+ \cdots +n={n(n+1)\over 2} \Rightarrow |k| \le{n(n+1)\over 2}\cdots(2) \\由(1)與(2)可得 {n\over 2} \le |k| \le{n(n+1) \over 2}\quad \bbox[red, 2pt]{QED.}$$

解答:$$假設{x+y-3\over x-y+1}=k \Rightarrow 直線L:(k-1)x-(k+1)y+k+3=0 \Rightarrow y={k-1\over k+1}(x-1)+2   \\極值發生在L為圓\Gamma:x^2+(y+4)^2=1的切線\Rightarrow d(P,圓心C(0,-4)) =半徑=1 \\ \Rightarrow {|5k+7| \over \sqrt{2k^2+2}} =1 \Rightarrow (5k+7)^2=2k^2+2 \Rightarrow 23k^2+70k+47=0 \Rightarrow (23k+47)(k+1)=0 \\ \Rightarrow k=-{47\over 23},-1 \Rightarrow\bbox[red, 2pt]{ \cases{最大值:-1\\ 最小值:-{47\over 23}}}$$

解答:$$\bbox[cyan,2pt]{略}$$

解答:$$\bbox[cyan,2pt]{略}$$

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