Processing math: 31%

網頁

2021年1月7日 星期四

106年南二中教甄-數學詳解

國立臺南第二高級中學106學年度第1次教師甄選

一、填充題
{A(1,1,2)B(1,0,3)C(3,8,1){u=AB=(2,1,1)v=AC=(2,7,1)ABC=12|u|2|v|2(uv)2=12654144=35;n=u×v=6(1,0,2)ABC:(x1)+2(z2)=0E:x+2z=5L:{2x+yz=3xy+7z=9{x=42zy=5z5LD(42t,5t5,t),tRdist(D,E)=|45|5=15ABCDE=13×ABC×dist(D,E)=13×35×15=1
{¯AD=5¯AB=3¯BD=4{ABDABD=90h=3×45=125BZΓ=h2π=14425πΓDA=13×144π25ׯAD=48π25×5=48π5
π2<x32π1cosx0y=5cosx3cosx=5+153cosx5+153(1)y5+15305+153(1)y5+153054y0
α=x+yi,x,yR{β=(1+i)α=(xy)+(xy)i|α3|=1(x3)2+y2=1(x,y)=(cosθ+3,sinθ){u=OA=(x,y)v=OB=(xy,xy)OAB=12|u|2|v|2(uv)2=12(x2+y2)(2(x2+y2))(x2+y2)2=12(x2+y2)=12((cosθ+3)2+sin2θ))=12(10+6cosθ){θ=08θ=1802+=10
f(x)=x+16+xx2=x+1(x3)(x+2){(x+1)(x3)(x+2)0x3x21x<3x<2
7242409=703×101×102+103(724240)10=10k=0C10k(703101102)k(103)10k(103)10mod(101102);使(103)10=(103103)5=((102+1)(101+2))5=(101102+307)53075mod(101102)3075=(307307)2307=(9101102+1531)230715312307mod(101102)15312307=(45101102+6427)153164271531mod(101102)64271531=(63102+1)(15101+16)1663102+15101+16mod(101102)1663102+15101+16=10101102+13271327mod(101102)
f(x)=2x^6-3x^5+4x^4-3x^3+4x^2-3x+2\\ f(\pm 1)\ne 0,f(\pm 2)\ne 0,試x^3=\pm 1,可得x^2+x+1及x^2-x+1皆為f(x)之因式\\ \Rightarrow f(x)=(x^2+x+1) (x^2-x+1)(2x^2-3x+2)\\ 因此f(x)=0 \Rightarrow x={-1\pm \sqrt 3i\over 2},{ 1\pm \sqrt 3i\over 2}, {3\pm \sqrt 7i\over 4} \\ \Rightarrow 在第一象限的解為\bbox[red,2pt]{{ 1+ \sqrt 3i\over 2}, {3+ \sqrt 7i\over 4}}
答案是\bbox[red,2pt]{381654729},詳細說明可參考以下超連結\href{https://mathcenter.ck.tp.edu.tw/Resources/Ctrl/ePaper/ePaperOpenFileX.ashx?autoKey=1024}{按這裡}
a_n=n^3+2n^2-200n = n((n+1)^2-201) \Rightarrow \cases{a_n < 0,\text{if }n\le 13\\ a_n > 0,\text{if }n\ge 14}; \\因此 \sum_{n=1}^{20}|a_n| =\sum_{n=1}^{13}(200n-2n^2-n^3)+ \sum_{n=14}^{20}(n^3+2n^2-200n)\\ =200\cdot {14\times 13\over 2}-2\cdot {13\times 14\times 27\over 6}-\left({14\times 13\over 2}\right)^2+ \left[\left({21\times 20\over 2}\right)^2 -\left({14\times 13\over 2}\right)^2\right]\\\qquad +2\left[{20\times 21\times 41\over 6}-{13\times 14\times 27\over 6} \right]-200\cdot {34\times 7\over 2} \\ =18200-1638-8281+35819 +2 \times 2051-23800 =\bbox[red,2pt]{24402}

 \cases{A:萬位數字\\B:千位數字\\C:百位數字\\D:十位數字\\ E:個位數字} \Rightarrow A+B+C+D+E \le 10且A,B,C,D,E \in \{0,..,9\} \\ \Rightarrow A+B+C+D+E +F= 10且A,B,C,D,E,F \in \{0,..,9\} \\ \Rightarrow 共有H^6_{10}= C^{15}_{10}=3003 組解,但需扣除A=10,B=10,...,F=10,共6組\\,再加上十萬數字100000這1組;因此k=3003-6+1 = \bbox[red,2pt]{2998}

令G為重心\Rightarrow \overrightarrow{GH}= 2\overrightarrow{OG} \Rightarrow \overrightarrow{GA} + \overrightarrow{AH}= 2(\overrightarrow{OA} +\overrightarrow{AG}) \Rightarrow  \overrightarrow{AH}= -2 \overrightarrow{AO}+3  \overrightarrow{AG} \\ =-2({2\over 5} \overrightarrow{AB}+{1\over 4} \overrightarrow{AC})+3\cdot {2\over 3}({1\over 2} \overrightarrow{AB}+{1\over 2} \overrightarrow{AC}) ={1\over 5} \overrightarrow{AB}+{1\over 2} \overrightarrow{AC} \Rightarrow (x,y)= \bbox[red,2pt]{({1\over 5},{1\over 2})}


x+\log_2(kx^2) = x+2^{|x|} \Rightarrow \log_2(kx^2) = 2^{|x|},相當於求兩圖形\cases{y=f(x)=\log_2(kx^2) \\y =g(x)=2^{|x|}}的交點;\\由於兩圖形皆對稱於y軸,即\cases{f(x)=f(-x)\\ g(x)=g(-x)},因此交點也對稱y軸;\\令原函數交點為\cases{A(a, a+ \log_2(ka^2))\\ B(-a,-a+\log_2(ka^2))} \Rightarrow \overline{AB}= \sqrt{4a^2+4a^2} =6\sqrt 2 \Rightarrow 8a^2=72 \\ \Rightarrow a^2=9 \Rightarrow A(3,3+\log_2(9k))代入另一函數\Rightarrow 3+\log_2(9k)=3+2^{|3|} \\ \Rightarrow \log_2(9k) = 8 \Rightarrow k={2^8\over 9} =\bbox[red,2pt]{256\over 9}


2人猜拳分出勝負次數的期望值E_2: \\\Rightarrow E_2=P(平手)(再猜一次)+P(分出勝負)\cdot (只猜1次) ={1\over 3}(E_2+1)+ {2\over 3}\cdot 1\\ \Rightarrow E_2 ={3\over 2}\\3人猜拳分出勝負次數的期望值E_3: \\\Rightarrow P(平手)(再猜一次)+ P(分出勝負)\cdot (只猜1次) +P(1人輸,2人勝)\cdot (再猜E_2次) \\ \Rightarrow E_3={1\over 3}(E_3+1)+{1\over 3}\cdot 1+{1\over 3}(1+E_2(={3\over 2})) \Rightarrow E_3=\bbox[red,2pt]{9\over 4}
a_n ={(2n)^2\over (2n-1)(2n+1)} ={4n^2 \over 4n^2-1} =1+{1\over 4n^2-1}=1+{1\over 2}({1\over 2n-1}-{1\over 2n+1})\\ \Rightarrow S= \sum_{n=1}^{1008}\left(1+{1\over 2}({1\over 2n-1}-{1\over 2n+1}) \right) \Rightarrow 2S= \sum_{n=1}^{1008} \left(2+ {1\over 2n-1}-{1\over 2n+1} \right) \\=2\times 1008+{1\over 1}-{1\over 3} +{1\over 3}-{1\over 5}+ \cdots + {1\over 2015} -{1\over 2017} =2016+1-{1\over 2017} \approx\bbox[red,2pt]{2017}

二、計算與證明題


(1)\overline{PS}= \overline{OP}\sin \theta = 3\sin \theta\\ \overline{RS}= \overline{OS}-\overline{OR}= 3\cos \theta-\overline{OQ}\cos {\pi\over 3} = 3\cos \theta-\overline{QR}\csc{\pi\over 3}\cos {\pi\over 3} \\= 3\cos\theta -3\sin \theta \times {2\over \sqrt 3}\times {1\over 2} =3\cos\theta -\sqrt 3\sin \theta\\ \Rightarrow \bbox[red,2pt]{\cases{\overline{RS}= 3\cos \theta -\sqrt 3\sin \theta\\ \overline{PS}=3\sin \theta}}(2)面積= \overline{RS}\times \overline{PS} =( 3\cos \theta -\sqrt 3\sin \theta)3\sin \theta =9\sin \theta\cos \theta-3\sqrt 3\sin^2\theta \\ ={9\over 2}\sin 2\theta- 3\sqrt 3({1-\cos 2\theta \over 2}) ={9\over 2}\sin 2\theta +{3\sqrt 3\over 2}\cos 2\theta -{3\sqrt 3\over 2} \\ \Rightarrow 最大值為\sqrt{({9\over 2})^2 +({3\sqrt 3\over 2})^2}-{3\sqrt 3\over 2} = \sqrt{108\over 4}-{3\sqrt 3\over 2} ={6\sqrt 3\over 2}-{3\sqrt 3\over 2} =\bbox[red,2pt]{{3\sqrt 3\over 2}}

(1)\cases{a_3=4\times 3\times 2=24\\ a_4= 4\cdot 3\cdot 2\cdot 2+ 4\cdot 3\cdot 3\cdot 1=84} \Rightarrow \bbox[red,2pt]{\cases{a_3=24\\ a_4=84}}(2)a_n+a_{n-1} =k(k-1)^{n-1} = 4\cdot 3^{n-1} \Rightarrow a_n=(k-1)^n+ (-1)^n(k-1)=3^n+3(-1)^n\\ \Rightarrow \bbox[red,2pt]{a_n= 3^n+3(-1)^n,n\ge 3}

f(x)=\sqrt{x^4-3x^2+4} +\sqrt{x^4-3x^2-8x+20} =\sqrt{x^2 +(x^2-2)^2} +\sqrt{(x-4)^2 +(x^2-2)^2}\\ \Rightarrow f(x)= \overline{PA} +\overline{PB},其中\cases{A(0,2)\\ B(4,2) \\ P(x,x^2)} \Rightarrow f(x)的最小值出現在直線L: \overleftrightarrow{AB} 與拋物線\Gamma:y=x^2的交點上;\\將L:y=2代入\Gamma \Rightarrow x^2=2 \Rightarrow x= \sqrt 2 (\sqrt 2 不合,\because 0 \lt x\lt 4) \Rightarrow f(\sqrt 2)= \sqrt{2}+ \sqrt{(\sqrt 2-4)^2} \\ =\sqrt 2+ 4-\sqrt 2= 4 \Rightarrow \bbox[red,2pt]{最小值為4,此時x=\sqrt 2}


3 則留言:

  1. 您好:請問第12題的交點對稱y軸,那是不是這兩個交點的y座標要一樣呢?可是跟假設不合,請問為什麼呢?謝謝

    回覆刪除
    回覆
    1. 原來的兩圖形y1,y2並沒對稱y軸,只有A、B的x坐標對稱y軸,但y坐標各異;只是求原兩圖形的交點,相當於求兩個對稱y軸圖形的交點,依此特性來斷定A(a,y1(a)), B(-a,y2(-a)).

      刪除