113年復興高中教甄-數學詳解

 臺北市立復興高級中學113學年度第一次專任教師甄選數學科 試題

說明： 本試 卷 測 驗時間 9 0 分 鐘， 共 7 題 。

解答:$$25^x-4\times 5^x+a=(5^x)^2-4(5^x)+a=0 二根皆為正數 \Rightarrow 5^x \gt 1\Rightarrow \cases{1-4+a \gt 0 \\ 判別式:16-4a\ge 0} \\ \Rightarrow \bbox[red, 2pt]{3\lt a\le 4}$$

解答:$$\alpha為x^4+2x^3+x^2+2x+1=0的根 \Rightarrow \alpha^4+ 2\alpha^3+\alpha^2+ 2\alpha+1=0 \\ \Rightarrow (\alpha^2+\alpha+1)^2-2\alpha^2=0 \Rightarrow (\alpha^2+\alpha+1)^2=2\alpha^2 \Rightarrow \alpha^2+\alpha+1=\sqrt 2 \alpha\\ 同理,\cases{\beta^2+\beta+1 =\sqrt 2\beta\\ \gamma^2+ \gamma+1=\sqrt 2\gamma\\ \delta^2+ \delta+1=\sqrt 2 \delta} \Rightarrow (\alpha^2+\alpha+1) (\beta^2+\beta+1) (\gamma^2+ \gamma+1)( \delta^2+ \delta+1) \\=4\alpha \beta\gamma \delta= \bbox[red, 2pt]4$$

解答:$$\cases{遺失的牌是紅心的機率=1/4 \\ 遺失的牌不是紅心的機率=3/4}\\ \Rightarrow 剩餘51張選2張皆是紅心機率={1\over 4} \times {{12\choose 2}\over {51\choose 2}} +{3\over 4}\times {{13\choose 2} \over {51\choose 2}} \\ \Rightarrow 遺失的牌是紅心的機率=\cfrac{{1\over 4} \times {{12\choose 2}\over {51\choose 2}}}{{1\over 4} \times {{12\choose 2}\over {51\choose 2}} +{3\over 4}\times {{13\choose 2} \over {51\choose 2}}} =\bbox[red, 2pt]{11\over 50}$$

解答:$$此題相當於求\text{ bubble sort }交換次數的期望值,直接代公式{1\over 2}C^n_2={1\over 2}C^5_2= \bbox[red, 2pt]5\\ \href{https://math.stackexchange.com/questions/245018/bubble-sorting-question?noredirect=1&lq=1}{公式來源}$$

解答:$$假設\cases{圓A半徑為a\\ 圓B半徑為b\\ 圓C半徑為c\\ 圓D半徑為d},則依題意\cases{\overline{AB} =7\\ \overline{BC}=8 \\ \overline{CD} =5\\ \overline{DA}=4} \Rightarrow \cases{a+b=7\\ b+c=8\\ c+d=5\\ d+a=4} \Rightarrow \cases{c=a+1\\ b=d+3} \\ \Rightarrow a^2 +b^2+c^2+d^2 =a^2+(d+3)^2+ (a+1)^2+ d^2= 2a^2+2a+2d^2+6d+10 \\=2(a+{1\over 2})^2 +2(d+{3\over 2})^2 +5\\柯西不等式:\left((a+{1\over 2})^2 +(d+{3\over 2})^2 \right) (1^2+1^2) \ge (a+{1\over 2}+d+{3\over 2})^2 =(4+2)^2=36 \\ \Rightarrow (a+{1\over 2})^2 +(d+{3\over 2})^2 \ge 18 \Rightarrow 四圓面積和\ge (2\times 18+5)\pi= \bbox[red, 2pt]{41\pi}$$

解答:$$\cases{A(3,-1,2) \\B(5,-3,3) \\C(0,0,4) \\ D(1,2,3)} \Rightarrow \cases{\overrightarrow{AB} =(2,-2,1) \\ \overrightarrow{CD} =(1,2,-1)} \Rightarrow \cases{L_1= \overleftrightarrow{AB}: {x-3\over 2} ={y+1\over -2} ={z-2\over 1} \\ L_2= \overleftrightarrow{CD}: {x\over 1}={y\over 2}={z-4\over -1} } \\ \Rightarrow \vec n= \overrightarrow{AB} \times \overrightarrow{CD} = (0,3,6) \Rightarrow E:y+1+2(z-2)=0 \Rightarrow y+2z=3 \Rightarrow d(C,E)=\bbox[red, 2pt] {\sqrt 5}$$

解答:$$請參閱下方學校提供的方法$$

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解題僅供參考,其他歷年試題及詳解