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2020年12月18日 星期五

106年新北市高中教甄聯招-數學詳解

 新北市立高級中等學校106學年度教師聯合甄選

一、填充題: 72%,每題 6 分。

:$$54x+21y=906 \Rightarrow 18x+7y=302 經過(2,38)\Rightarrow 該直線可表示成(7t+2,-18t+38)\\ 由於\cases{7t+2 \in N\\ -18t+38 \in N} \Rightarrow \cases{7t+2 \ge 1 \\ -18t+38\ge 1} \Rightarrow \cases{t\ge -1/7\\ t\le 37/18} \Rightarrow t= 2,1,0\\ \Rightarrow 共\bbox[red,2pt]{3}組正整數解$$



:$$\begin{array}{cc|cc}方式& 組員& 方式 & 組員\\\hline a & 女(3,1,1) & b & 女(1,3,1)\\ & 男(1,2,2) & &男(3,0,2)\\\hdashline c & 女(1,1,3) & d & 女(2,2,1)\\ & 男(3,2,0) & & 男(2,1,2)\\\hdashline e & 女(2,1,2) & f & 女(1,2,2) \\ & 男(2,2,1) & & 男(3,1,1)\\ \hline \end{array}\\ 由於不分組別,所以方式b與方式c是相同的、方式d與方式e也是相同的;\\因此只要考慮其它4組情況;\\ 方式a: C^5_3 \times C^5_1C^4_2  =300  \\ 方式b:C^5_1C^4_3\times C^5_3=200\\ 方式d:C^5_2C^3_2\times C^5_2C^3_1=900\\ 方式f:C^5_1C^4_2\times C^5_3=300\\ 共有300+200+900+300 =\bbox[red,2pt]{1700}種$$



:$$x^6+ x^5+x^4-28x^3 +x^2+x+1=0 \Rightarrow x^3+x^2+x-28+{1\over x}+{1\over x^2} +{1\over x^3}=0 \cdots(1)\\ 令a=x+{1\over x} \Rightarrow \cases{x^2+{1\over x^2} =a^2-2 \\ x^3+{1\over x^3}=(x+{1\over x})(x^2-1+{1\over x^2})=a(a^2-2-1)=a^3-3a} \\ 式(1) \Rightarrow a^3-3a+a^2-2+a-28=0 \Rightarrow a^3+a^2-2a-30=0 \Rightarrow (a-3)(a^2+4a+10)=0 \\ \Rightarrow a=3 \Rightarrow x+{1\over x}=3 \Rightarrow x^2-3x+1=0 \Rightarrow x= \bbox[red,2pt]{3\pm \sqrt{5}\over 2}$$






$$假設\cases{\triangle ABC邊長為1\\ A(0,0) \\ \overline{BD}與\overline{FG}交點為P} \Rightarrow \cases{B(1,0)\\ C({1\over 2},{\sqrt 3\over 2})} \Rightarrow D=(A+2C)/2 = ({1\over 3},{\sqrt 3\over 3})\\ \Rightarrow P=(B+D)/2 =({2\over 3},{\sqrt 3\over 6}) \\ \Rightarrow \overleftrightarrow{BD} 斜率=-{\sqrt 3\over 2} \Rightarrow \overleftrightarrow{FG} 斜率={2\over \sqrt 3}  \Rightarrow \overleftrightarrow{FG}方程式: y={2\over \sqrt 3}(x-{2\over 3})+{\sqrt 3\over 6} \\\Rightarrow \overleftrightarrow{FG}與x軸交於F({5\over 12},0) \Rightarrow {\overline{AF}\over \overline{AB}} =\bbox[red,2pt]{5\over 12}$$

:$$令\omega =\cos {2\pi \over 7} +i\sin {2\pi \over 7}  \Rightarrow  \cases{\omega^2 =\cos {4\pi \over 7} +i\sin {4\pi \over 7}\\ \omega^3 =\cos {6\pi \over 7} +i\sin {6\pi \over 7}\\ \omega^4 =\cos {8\pi \over 7} +i\sin {8\pi \over 7} =\cos {6\pi \over 7} -i\sin {6\pi \over 7}\\\omega^5=\cos {10\pi \over 7} +i\sin {10\pi \over 7} =\cos {4\pi \over 7} -i\sin {4\pi \over 7} \\\omega^6= \cos {12\pi \over 7} +i\sin {12\pi \over 7} =\cos {2\pi \over 7} -i\sin {2\pi \over 7} \\ \omega^7=1} \\ \Rightarrow \cases{\alpha=\omega+\omega^6 = 2\cos{2\pi\over 7} \\\beta= \omega^2 +\omega^5= 2\cos{4\pi\over 7} \\\gamma= \omega^3 + \omega^4 = 2\cos{6\pi \over 7}  \\ \omega^6 +\omega^5 +\omega^4 +\omega^3 +\omega^2 +\omega +1=0} \Rightarrow \cases{a=\alpha/2\\ b=\beta/2 \\ c=\gamma/2}\\ \Rightarrow \cases{\alpha+\beta+\gamma= \omega^6 +\omega^5 +\omega^4 +\omega^3 +\omega^2 +\omega = -1\\ \alpha\beta +\beta\gamma +\gamma\alpha= 2(\omega^6 +\omega^5 +\omega^4 +\omega^3 +\omega^2 +\omega)=-2\\ \alpha\beta\gamma = \omega^6 +\omega^5 +\omega^4 +\omega^3 +\omega^2 +\omega+2=1} \\  \Rightarrow \alpha,\beta,\gamma 為x^3+x^2-2x-1=0之三根\\ 因此a+{1\over a} +b+{1\over b}+c+{1\over c}= a+b+c + {ab+bc+ ca \over abc} ={1\over 2}(\alpha+\beta +\gamma)+\cfrac{(\alpha\beta+ \beta\gamma + \gamma\alpha)/4}{\alpha\beta\gamma /8} \\ =-{1\over 2}+\cfrac{-1/2}{1/8} =\bbox[red,2pt]{-{9\over 2}}$$另解$$abc =\cos{2\pi \over 7} \cos{4\pi \over 7} \cos{6\pi \over 7} =-\cos{\pi \over 7}\cos{2\pi \over 7} \cos{4\pi \over 7}  \\ =-2\sin{\pi \over 7}\cos{\pi \over 7}\cos{2\pi \over 7} \cos{4\pi \over 7}/ (2\sin{\pi \over 7}) = -\sin{2\pi \over 7} \cos{2\pi \over 7} \cos{4\pi \over 7}/ (2\sin{\pi \over 7})\\ = -2\sin{2\pi \over 7} \cos{2\pi \over 7} \cos{4\pi \over 7}/ (4\sin{\pi \over 7}) = -\sin{4\pi \over 7}  \cos{4\pi \over 7}/ (4\sin{\pi \over 7}) \\= -2\sin{4\pi \over 7}  \cos{4\pi \over 7}/ (8\sin{\pi \over 7}) =-\sin{8\pi \over 7}/ (8\sin{\pi \over 7}) \\=\sin{ \pi \over 7}/ (8\sin{\pi \over 7})={1\over 8}\\ \quad\\a+b+c=\cos{2\pi \over 7} +\cos{4\pi \over 7} +\cos{6\pi \over 7} \\=(2\sin{\pi\over 7}\cos{2\pi \over 7} +2\sin{\pi\over 7}\cos{4\pi \over 7} +2\sin{\pi\over 7} \cos{6\pi \over 7})/(2\sin{\pi\over 7}) \\=(\sin{3\pi \over 7}-\sin{\pi \over 7} +\sin{5\pi \over 7}-\sin{3\pi \over 7}+ \sin{7\pi \over 7}-\sin{5\pi \over 7}) /(2\sin{\pi\over 7}) \\= -\sin{\pi \over 7} /(2\sin{\pi\over 7}) =-{1\over 2}\\ \quad \\ab+bc+ca =\cos{2\pi \over 7} \cos{4\pi \over 7}+ \cos{4\pi \over 7}\cos{6\pi \over 7} +\cos{6\pi \over 7} \cos{2\pi \over 7} \\ ={1\over 2}(2\cos{2\pi \over 7} \cos{4\pi \over 7}+ 2\cos{4\pi \over 7}\cos{6\pi \over 7} +2\cos{6\pi \over 7} \cos{2\pi \over 7}) \\ ={1\over 2}(\cos{2\pi \over 7} +\cos{6\pi \over 7} +\cos{2\pi \over 7} +\cos{10\pi \over 7} +\cos{4\pi \over 7} +\cos{8\pi \over 7}) \\ ={1\over 2}(\cos{2\pi \over 7} -\cos{\pi \over 7} +\cos{2\pi \over 7} +\cos{4\pi \over 7} +\cos{4\pi \over 7} -\cos{\pi \over 7}) \\ =-\cos{\pi \over 7} +\cos{2\pi \over 7} +\cos{4\pi \over 7} \\ =(-2\sin{\pi \over 7}\cos{\pi \over 7} +2\sin{\pi \over 7}\cos{2\pi \over 7} +2\sin{\pi \over 7}\cos{4\pi \over 7})/(2\sin{\pi \over 7}) \\ =(-\sin{2\pi \over 7}+\sin{3\pi \over 7}-\sin{\pi \over 7}+\sin{5\pi \over 7}-\sin{3\pi \over 7})/(2\sin{\pi \over 7}) \\ =(-\sin{2\pi \over 7}+\sin{3\pi \over 7}-\sin{\pi \over 7}+\sin{2\pi \over 7}-\sin{3\pi \over 7})/(2\sin{\pi \over 7}) \\=-\sin{\pi \over 7}/(2\sin{\pi \over 7})  =-{1\over 2}\\ 因此a+{1\over a}+b+{1\over b} +c+ {1\over c} =a+b+c+{ab+bc+ca\over abc} =-{1\over 2}+{-1/2\over 1/8}= \bbox[red,2pt]{-{9\over 2}}$$


$$令g(x)=xf(x)-1,則1,2,\dots,2018 為g(x)=0的根\Rightarrow g(x)=xf(x)-1 \\=k(x-1)(x-2)\cdots (x-2018) \Rightarrow g(0)= -1 = k\cdot 2018! \Rightarrow k=-{1\over 2018!}\\因此g(2020)= 2020f(2020) -1= k\cdot 2019\cdot 2018\cdots 2 =k\cdot 2019! = -2019 \\ \Rightarrow f(2020)={-2019+1\over 2020} =-{2018\over 2020} =\bbox[red,2pt]{-{1009\over 1010}}$$

$$f_n(x)=x^n+n (n=1,2,3,4) \Rightarrow \cases{f_1(x)=x+1\\ f_2(x)=x^2+2 \\ f_3(x)=x^3+3\\ f_4(x)=x^4+4} \Rightarrow \cases{f_1(1)=2\\ f_2(1)=3\\ f_3(1)=4\\ f_4(1)=5} \\ \Rightarrow \cases{兩兩相乘之和=2(3+4+5) +3(4+5)+4\cdot 5= 71\\ 三三相乘之和= 2\cdot 3(4+5) + 2\cdot 4\cdot 5+  3\cdot 4\cdot 5= 154 \\ 四數相乘= 2 \cdot 3 \cdot 4\cdot 5 =120} \\ \Rightarrow 總和=71+154+120 = \bbox[red,2pt]{345}$$


$$延長\overline{BP}交\overline{AC}於D,如上圖,並令\cases{\angle PAD=\theta\\ \overline{PD}=a},則\angle PDA= 180^\circ - \angle ABD-\angle A= 60^\circ \\\Rightarrow \angle DPC=30^\circ \Rightarrow \overline{DC} =\overline{DP}=a \Rightarrow \cos \angle PDC = \cos 120^\circ = -{1\over 2}={a^2 +a^2-\overline{CP}^2 \over 2a^2}\\ \Rightarrow \overline{CP}= \sqrt 3a\\ \cases{\triangle ADP: {a\over \sin \theta}={\overline{AP} \over \sin 60^\circ} \Rightarrow \overline{AP}={\sqrt 3a \over 2\sin \theta} \cdots(1)\\ \triangle ABP: {\overline{AP} \over \sin 80^\circ} ={\overline{BP} \over \sin(40^\circ-\theta)} \Rightarrow \overline{AP} ={\sin 80^\circ \over \sin(40^\circ-\theta)}\overline{BP} \cdots(2)\\ \triangle BCP: {\sqrt 3a \over \sin 20^\circ}={\overline{BP} \over \sin 10^\circ} \Rightarrow \overline{BP}= {\sin10^\circ \over \sin 20^\circ}\sqrt 3a \cdots(3) }\\ 將(3)代入(2) \Rightarrow \overline{AP}= {\sin 80^\circ \over \sin(40^\circ-\theta)} \times {\sin10^\circ \over \sin 20^\circ}\sqrt 3a ={\sqrt 3a \over 2\sin \theta} \Rightarrow \sin \theta = {\sin 20^\circ \over 2\sin 10^\circ \sin 80^\circ}\sin(40^\circ-\theta) \\ ={\sin 20^\circ \over \cos 70^\circ -\cos 90^\circ }\sin(40^\circ-\theta) ={\sin 20^\circ \over \cos 70^\circ  }\sin(40^\circ-\theta) ={\sin 20^\circ \over \sin 20^\circ  }\sin(40^\circ-\theta) =\sin(40^\circ-\theta) \\ \Rightarrow \sin \theta = \sin(40^\circ-\theta) \Rightarrow \theta=20^\circ \Rightarrow \angle APC = (180^\circ-60^\circ-\theta)+ 30^\circ = \bbox[red,2pt]{130}度$$




$${x^2\over 36} +{y^2\over 32}=1 \Rightarrow \cases{a=6\\ b=4\sqrt 2} \Rightarrow c=2 \Rightarrow \cases{F_1(-2,0)=A\\ F_2(2,0)} \Rightarrow \overline{PA}+\overline{PF_2}= 2a=12\\ 因此\overline{PA} +\overline{PB} = 12-\overline{PF_2}+ \overline{PB} 此值要最大,即P-F_2-B在一直線上;因此\overline{PB}-\overline{PF_2}= \overline{BF_2}\\ \Rightarrow 最大值為12+\overline{BF_2} = 12+\sqrt{6^2+8^2} =12+10 =\bbox[red,2pt]{22}$$



此題相當於用3種顏色著色問題,在一長條形分隔成左右相鄰的12格子,相鄰格子需不同顏色,著色數為\(3\times 2^{12-1}=3\cdot 2^{11}\);但三種顏色都必須用到,因此要扣除只用二種顏色的著色數,即\(C^3_2\times 2)\);
因此本題答案為\(3\cdot 2^{11}-C^3_2\times 2= 6144-6=\bbox[red,2pt]{6138}\)



:$$P在L:x-1=y={z \over \sqrt 2}\; 上 \Rightarrow P(t+1,t,\sqrt 2t),t\in R \Rightarrow \cases{\overline{PA} = \sqrt{t^2+(t-1)^2+(\sqrt 2t)^2} \\ \overline{PB}= \sqrt{t^2 +(t-2)^2+ (\sqrt 2t)^2}} \\ \Rightarrow \cases{\overline{PA} = \sqrt{4t^2 -2t+ 1} =2\sqrt{(t-{1\over 4})^2+{3\over 16} }\\ \overline{PB}= \sqrt{4t^2 -4t+4} =2\sqrt{(t-{1\over 2})^2 +{3\over 4}}}\\ \Rightarrow \overline{PA} +\overline{PB} =2\left(\sqrt{(t-{1\over 4})^2+{3\over 16} } + \sqrt{(t-{1\over 2})^2 +{3\over 4}} \right) \\ =2(\overline{P'A'} +\overline{P'B'}),其中\cases{P'(t,0)\\ A'({1\over 4},{\sqrt 3\over 4}) \\B'({1\over 2},{\sqrt 3\over 2})},P'在x軸上且A',B'皆在第1象限;\\ 因此取A'與x軸的對稱點A''({1\over 4},-{\sqrt 3\over 4}),\\則\overline{P'A'} +\overline{P'B'}的最小值等於\overline{A''B'} =\sqrt{({1\over 4})^2+ ({3\sqrt 3\over 4})^2}={\sqrt 7\over 2} \\ \Rightarrow \overline{PA} +\overline{PB} 的最小值= 2\times {\sqrt 7\over 2} =\bbox[red,2pt]{\sqrt 7}$$




依題意,上圖A、B、C、D為交點,其它是端點;紅色線才是「相鄰交點所連成的線段」,其它線段並不符合要求;
兩條直線相交於一點,會產生4條線段;
三條直線相交於一點,會產生6條線段;
因此k個兩線交點及m-k個三線交點,會產生\(4k+6(m-k)\)條線段,但並非均符合「相鄰交點所連成的線段」;
有相交的n條直線一定至少產生2n條線段,因此\([(4k+6(m-k)-2n]\div 2=\bbox[red,2pt]{3m-k-n}\)才是符合要求的線段數量。

二、計算證明題
略(未公布答案)

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解題僅供參考




2 則留言:

  1. 你好:請問第2題,方式a是不是少乘C(2,1),方式f也是不是少乘C(2,1)呢?謝謝

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