107年臺北市中正高中教甄-數學詳解

 臺北市立中正高級中學107學年度第1次專任教師甄選

數學科初選試卷

解： $$\cases{(2^{1/2})^6= 2^3=8 \\ (3^{1/3})^6=3^2 =9} \Rightarrow 3^{1/3} > 2^{1/2}，又\cases{(2^{1/2})^{14}= 2^7=128 \\ (7^{1/7})^{14}=7^2 =49} \Rightarrow  2^{1/2} \gt 7^{1/7}，\\ 因此3^{1/3} > 2^{1/2}\gt 7^{1/7}；\\ (x-2^{1/2})(x-3^{1/3})(x-7^{1/7})= |(x-2^{1/2})(x-3^{1/3})(x-7^{1/7})| \\\Rightarrow (x-2^{1/2})(x-3^{1/3})(x-7^{1/7})=\begin{cases} (x-2^{1/2})(x-3^{1/3})(x-7^{1/7}) & \text{if }x\ge 3^{1/3} 或7^{1/7} \le x\le 2^{1/2}\\ -(x-2^{1/2})(x-3^{1/3})(x-7^{1/7})& \text{if }2^{1/2} \le x\le 3^{1/3}或 x\le 7^{1/7}\end{cases} \\ \Rightarrow \cases{x\ge 3^{1/3} 或7^{1/7} \le x\le 2^{1/2}\\ x=2^{1/2},3^{1/3},7^{1/7}} \Rightarrow \bbox[red,2pt]{x\ge 3^{1/3} 或7^{1/7} \le x\le 2^{1/2}}$$

解： $$\begin{array}{} u& v& x+y+z & 數量\\\hline 1 & 1 & 8 & H^3_5=21\\ & & 7 & H^3_4=15\\ & & 6 & H^3_3=10 \\ & & 5 &  H^3_2= 6\\ & & 4 & H^3_1=3 \\ & & 3 & 1 \\\hdashline 2 & 1 & 5 & H^3_2=6\\ & & 4 & H^3_1=3 \\ & & 3 & 1\\\hline \end{array} \\ \Rightarrow 共有21+15+10+6+3+1+6+3+1 =\bbox[red,2pt]{66}組正整數解$$

解： $$f(x)=x^2-2x+2 =(x-1)^2+1 \Rightarrow f(1)為最小值\Rightarrow \begin{cases} f(x)遞減, & x\le 1 \\ f(x)遞增, & x \ge 1\end{cases}\\ \Rightarrow U_{3n}= \sum_{i=1}^n{1\over n}f({i-1\over n}) +\sum_{i=n+1}^{3n}{1\over n}f({i\over n}) \\= {1\over n}\sum_{i=1}^n\left[({i-1\over n})^2 -2({i-1\over n})+2\right] +{1\over n}\sum_{i=n+1}^{3n}\left[ ({i\over n})^2-2({i\over n})+2\right] \\= {1\over n}\sum_{i=1}^n\left[{i^2-2i+1\over n^2} -2({i-1\over n})+2\right] +{1\over n}\sum_{i=n+1}^{3n}\left[ {i^2\over n^2}-2({i\over n})+2\right] \\= {1\over n}\left( {2n^3-3n^2+n \over 6n^2} -{n^2-n\over n} +2n\right) +{1\over n}\left({52n^3+24n^2+2n \over 6n^2} -{8n^2+2n\over n} +4n\right) \\ ={54n^3+21n^2+3n\over 6n^3} -{9n^2+n\over n^2}+6 = \left( 9+{7\over 2n} +{1\over 2n^2}\right)-\left(9+{1\over n} \right)+6 \\ = \bbox[red,2pt]{{5n+1\over 2n^2}+6}$$

解：

(A)四男三女排列，恰有二女相鄰的的排列數：

     三女排列數:3!，再乘上左2右1或左1右2兩種分組，因此共有$2\times 3!$排法;

      在二組女生中間一定要塞一個男生，剩下三男往三個空間(兩組女生的左外側、右外側或兩組中間)組合，組合數為$H^3_3=10$，再乘上四男排列數4!；

因此有男生排列數$\times$女生排列數=$(10\times 4!)\times (2\times 3!)=2880$

(B)阿正以外的三男三女排列，恰有二女相鄰

   方法同A，因此男生排列數$\times$女生排列數=$(H^3_2\times 3!)\times (2\times 3!)=432$

(C)條件如(B)，再加上阿正在最左邊或最右邊

   排列數為(B)的2倍，即$432\times 2=864$

因此所求之數為(A)-(C) =$2880-864= \bbox[red,2pt]{2016}$

解： $$令\cases{\angle BAD = \angle DAE= \angle DAC= \theta\\ \overline{AB}=a\\ \overline{AD}=b \\ \overline{AE}=c \\ \overline{AC}=d} \Rightarrow \triangle ABD: \triangle ADE: \triangle AEC= \overline{BD}: \overline{DE}: \overline{EC}= 3:4:8; \\ 又\cases{\triangle ABD ={1\over 2}ab \sin \theta \\ \triangle ADE={1\over 2}bc \sin \theta \\ \triangle AEC= {1\over 2}cd \sin \theta \\} \Rightarrow \cases{ab: bc= 3:4 \\ bc:cd =4:8 } \Rightarrow \cases{a:c=3:4 \\ b:d=1:2 } \Rightarrow \cases{a=3t\\ b=s\\ c=4t\\ d=2s},\text{where }s,t\in R\\ \cos \theta ={a^2+b^2-9\over 2ab} ={b^2+c^2-16\over 2bc}={c^2+d^2-64\over 2cd} \\\Rightarrow {9t^2+s^2-9\over 6st} ={s^2+16t^2-16\over 8st} ={16t^2+4s^2-64\over 16st} \\ \Rightarrow 72t^2+8s^2-72 =6s^2+96t^2-96 =48t^2+12s^2-192 \\ \Rightarrow \cases{s^2-12t^2+12=0 \\ s^2-8t^2-16=0 \\ s^2-6t^2-30=0} \Rightarrow \cases{s=6\sqrt 2\\ t=\sqrt 7} \Rightarrow \cos \theta ={9t^2+s^2-9\over 6st}={\sqrt{14}\over 4} \Rightarrow \sin \theta ={\sqrt 2\over 4} \\ \Rightarrow \triangle ABC ={1\over 2}ad\sin 3\theta = {1\over 2}(3\sqrt 7)(12\sqrt 2)(3\sin \theta -4\sin^3\theta) =18\sqrt{14}({3\sqrt 2\over 4}-{\sqrt 2\over 8})= \bbox[red,2pt]{{45\over 2}\sqrt 7}$$

6.台灣彩券最早發行的樂透彩(俗稱小樂透)的玩法是「 42 選 6」： 購買者從 01~42 中任選六個號碼，當這六個號碼與開出的六個號碼完全相同(不計次序)時即得頭獎；近期，台灣彩券發行新的樂透彩──雙贏彩，玩法是「 24 選 12」： 購買者從 01~24 中任選十二個號碼，當這十二個號碼與開出的十二個號碼完全相同或完全不同(皆不計次序)時均得頭獎。 假設原來的小樂透中頭獎的機率是 p，而新發行的雙贏彩中頭獎的機率是 q。試求比值 q/p為何？（答案請化為小數，四捨五入至小數點下第一位）

解： $$\cfrac{2\over C^{24}_{12}}{1\over C^{42}_6}= \cfrac{2C^{42}_6}{C^{24}_{12}} =\cfrac{1517}{391} =3.879 \approx \bbox[red,2pt]{3.9}$$

解： $$0 \lt x \lt 1 \Rightarrow f(x)= x-\lfloor x\rfloor =x-0 = x \Rightarrow f({2\times 1\over 2018})+\cdots +f({2 \times 1008\over 2018}) \\={2\times 1\over 2018}+\cdots +{2\times 1008\over 2018}\\ 1\le x \lt 2\Rightarrow f(x)=x-\lfloor x\rfloor =x-1 \Rightarrow f({2\times 1009\over 2018})+\cdots +f({2 \times 2017\over 2018}) \\={2\times 1009 \over 2018}+\cdots +{2\times 2017\over 2018}-1009\\ f(2)=2-2=0\Rightarrow f({2\times 2018\over 2018}) =0\\ 因此所求之值={2\over 2018} (1+2+\cdots +2017)-1009 =2017-1009 = \bbox[red,2pt]{1008}$$

解： $$\cases{a= \sqrt{10+ \sqrt{51}} \\b= \sqrt{10-\sqrt{51}}} \Rightarrow \cases{a^2+ b^2 = 20 \\ ab=7} \Rightarrow (a+b)^2 = a^2+b^2 +2ab = 20+14 =34\\ \Rightarrow a+b=\sqrt{34} \Rightarrow a^3+b^3 = (a+b)(a^2+ b^2-ab) =\sqrt{34}\times (20-7) =\bbox[red,2pt]{13\sqrt{34}}$$

解： $$L:{x-3\over 2} ={y+4\over -3} ={z-9\over 6} \Rightarrow \cases{L上的點可表示成(2t+3,-3t-4,6t+9),t\in R \\ L的方向向量\vec u=(2,-3,6)} \\ 假設A(11,5,-30)在L上的投影點A'(2t+3,-3t-4,6t+9) \Rightarrow \overrightarrow{A'A} \bot \vec u\\ \Rightarrow (2t-8,-3t-9, 6t+39) \cdot (2,-3,6)=0 \Rightarrow 4t-16+9t+27+36t+234=0 \\ \Rightarrow 49t+245=0 \Rightarrow t=-5 \Rightarrow A'(-7,11,-21);\\ 同理,B(-3,-16,5)在L上的投影點B'(2s+3,-3s-4,6s+9) \Rightarrow \overrightarrow{B'B} \bot \vec u \\ \Rightarrow (2s+6,-3s+12,6s+4) \cdot (2,-3,6)=0 \Rightarrow s=0 \Rightarrow B'(3,-4,9)\\ 因此\cases{\overline{AA'}= \sqrt{18^2 + (-6)^2 + (-9)^2} =21\\ \overline{BB'}= \sqrt{6^2+12^2 +4^2} =14}，由於\triangle PAA' \sim \triangle PBB' \Rightarrow {\overline{PA'} \over \overline{PB'}} ={\overline{AA'} \over \overline{BB'}}\\ ={21\over 14} ={3\over 2} \Rightarrow P=(2A'+3B')/5 = \left({ -14+9\over 5},{ 22-12\over 5},{ -42+27\over 5} \right) =\left(-1,2,-3 \right) \\ \Rightarrow \overline{PA}+\overline{PB} =\sqrt{12^2+3^2+27^2} +\sqrt{2^2+18^2+8^2} =21\sqrt 2+ 14\sqrt 2=35\sqrt 2\\ 答：(1)P=\bbox[red,2pt]{(-1,2,-3)},(2)\overline{PA}+\overline{PB} =\bbox[red,2pt]{35\sqrt 2}$$

解：

(1) $$\alpha = \sqrt[3]{-1-\sqrt 5\over 2} +\sqrt[3]{-1+\sqrt 5\over 2}-1 \Rightarrow (\alpha+1)^3= \left(\sqrt[3]{-1-\sqrt 5\over 2} +\sqrt[3]{-1+\sqrt 5\over 2} \right)^3 \\ \Rightarrow \alpha^3+3\alpha^2 +3\alpha + 1 ={-1-\sqrt 5\over 2}+{-1+ \sqrt 5\over 2} +3\cdot \sqrt[3]{-4\over 4}\left(\sqrt[3]{-1-\sqrt 5\over 2} +\sqrt[3]{-1+\sqrt 5\over 2} \right)\\=-1-3(\alpha+1)\\ \Rightarrow \alpha^3+ 3\alpha^2+6\alpha +5=0 \Rightarrow \bbox[red,2pt]{f(x)=x^3+3x^2+6x+5}$$ (2) $$\alpha,\beta,\gamma 為f(x)=x^3+3x^2+6x+5=0之三根\Rightarrow \cases{\alpha+\beta+\gamma =-3\\ \alpha\beta + \beta\gamma +\gamma\alpha =6 \\ \alpha\beta\gamma=-5 }\\ 又f(x)=(x-\alpha)(x-\beta)(x-\gamma) \Rightarrow f(3)=27+27+18+5=77 =(3-\alpha) (3-\beta)(3-\gamma)\\g(x)=x^4+4x^3+9x^2 +10x+8  =(x^3+3x^2+6x+5)(x+1)-x+3 = f(x)(x+1)-x+3\\ \Rightarrow \cases{g(\alpha)=f(\alpha)(\alpha+1)-\alpha+3 =-\alpha+3\\ g(\alpha)=f(\beta)(\beta +1)-\beta+3 =-\beta+3\\ g(\beta)=f(\gamma)(\gamma+1)-\gamma+3 =-\gamma+3 } \\\Rightarrow {1\over g(\alpha)} +{1\over g(\beta)} +{1\over g(\gamma)} ={1\over -\alpha+3} +{1\over -\beta+3}+ {1\over -\gamma+3} \\ = {\alpha\beta +\beta\gamma +\gamma\alpha -6(\alpha+\beta+\gamma)+27\over (3-\alpha)(3-\beta)(3-\gamma)} ={6-6\times(-3)+27\over 77} = \bbox[red,2pt]{51\over 77}$$

解：

(1) $$\lim_{n\to \infty}{f(n)\over n^5} =8 \Rightarrow f(x)=8x^5+ ax^4+bx^3 +cx^2+dx+e\\ 又\lim_{x\to 0}{f(x) \over x} =4 \Rightarrow \cases{d=4\\ e=0} \Rightarrow f(x)=8x^5+ ax^4+bx^3 +cx^2+4x\\ f''(0)= 2\Rightarrow 2c= 2\Rightarrow c=1 \Rightarrow f(x)=8x^5+ ax^4+bx^3 +x^2+4x\\ 最後\cases{f(1)=2\\ f(-1)=6} \Rightarrow \cases{8+a+b+1+4=2 \\ -8+a-b+1-4=6} \Rightarrow \cases{a+b=-11 \\a-b=17} \Rightarrow \cases{a=3\\ b=-14}\\ \Rightarrow \bbox[red,2pt]{f(x)= 8x^5+ 3x^4-14x^3 +x^2+4x}$$ (2) $$\int_{-1}^1 f(x)\;dx=  \left. \left[ {4\over 3}x^6 +{3\over 5}x^5- {7\over 2}x^4 +{1\over 3}x^3+2x^2\right] \right|_{-1}^1 = {23\over 30}-(-{11\over 10})= \bbox[red,2pt]{-{1\over 3}}$$

解： $$\cases{log_6 (x+2) \\ \log_6 (5-x)} \Rightarrow \cases{x+2 >0 \\ 5-x> 0} \Rightarrow -2 \lt x \lt 5 \equiv S=\{x \mid -2< x < 5,x\in R\}\\\log_6 (x+2)+\log_6 (5-x)= \log_6(a-x) \Rightarrow  (x+2)(5-x)=a-x \Rightarrow f(x)=x^2-4x+a-10=0\\ (A)f(x)=0有1解 \Rightarrow 判別式=0 \Rightarrow 16-4(a-10)=0 \Rightarrow a=14 \Rightarrow x=2 \in S\\ (B)f(x)=0有2實數解，恰有1解滿足-2\lt x\lt 5 \Rightarrow 16-4(a-10) > 0 \Rightarrow \cases{a\lt 14 \\ x_1 =2+ \sqrt{14-a}\\ x_2=2-\sqrt{14-a}}\\ \Rightarrow \cases{\cases{x_1\in S\\ x_2 \not \in S} \Rightarrow \cases{-2 < 2+\sqrt{14-a} < 5\\ 2-\sqrt{14-a} \le -2} \Rightarrow \cases{5 < a\le 14\\ a\le -2}不可能同時滿足\\ \cases{x_1\not \in S\\ x_2 \in S} \Rightarrow \cases{2+\sqrt{14-a} \ge 5\\ -2 < 2-\sqrt{14-a} < 5} \Rightarrow \cases{a\le 5\\ -2 < a \le 14}}\\ \Rightarrow -2 < a \le 5 \Rightarrow a=-1,0,1,2,3,4,5\\ 綜合(A)與(B)的結果a=-1,0,1,2,3,4,5,14 \Rightarrow \cases{m=-1+0+1+2+3+4+5+14 =28\\ n=8} \\ \Rightarrow (m,n)= \bbox[red,2pt]{(28,8)}$$

解：

先求拋物線的對稱軸:

    在拋物線上半部任找二點A、C，下半部任找一點B。作$\overleftrightarrow{CD} \parallel \overline{AB}$，且$\overleftrightarrow{CD}$交拋物線下半部於D。分別作圖找$\overline{AB}與\overline{CD}$的中點E與F，作直線$\overleftrightarrow{EF}$。在拋物線上半部任找一點G，作圖$\overleftrightarrow{GH}\bot \overleftrightarrow{EF}$且 $\overleftrightarrow{GH}$交拋物線下半部於H，則對稱軸$L=\overline{GH}$的中垂線。

再找焦點:

    對稱軸L與拋物線的交點為P(頂點)，於L上任找一點Q(在拋物線外)，過Q作直線L'，使得$L\bot L'$；在L'上找R,及S，使得$\overline{QP} =\overline{QR} =\overline{RS}$；作直線$\overleftrightarrow{PS}$交拋物於T，過T作直線垂直L並交L於F，F即為焦點。

國中方法: $$假設\cases{\overline{AB}=\overline{BC}=1\\ \overline{BP}=a} \Rightarrow \cases{\overline{FH}=\overline{CP}=1-a\\ \overline{AE}= \overline{EB}=\overline{HP}=1/2} \\ \overline{GH}\parallel \overline{AE} \Rightarrow {\overline{GH} \over \overline{AE}} ={\overline{FH} \over \overline{FE}} \Rightarrow {\overline{GH}\over 1/2} ={1-a\over 1} \Rightarrow \overline{GH}={1-a\over 2} \Rightarrow \overline{GP}= {1\over 2}+{1-a\over 2}=1-{a\over 2}\\ 直角\triangle AGQ:\cases{\angle PAB=\angle GPA=\theta \Rightarrow \overline{GA}=\overline{GP}=1-{a\over 2}\\ \overline{QG}=1-\overline{GP}= a/2\\ \overline{AQ}=1-a}\\ \Rightarrow (1-a/2)^2 =a^2+(a/2)^2 \Rightarrow a={\sqrt 5-1\over 2} \Rightarrow {\overline{BC} \over \overline{BP}}=1/a= {\sqrt 5+1\over 2}為黃金比例$$ 高中方法: $$\cases {\angle DFA = \angle FAB=2\theta \Rightarrow \tan \angle DFA = {\overline{AD}\over \overline{DF}} \Rightarrow \tan 2\theta = 2 \\ \tan \angle PAB= {\overline{BP} \over \overline{AB}} \Rightarrow \tan \theta = a} \\\Rightarrow \tan 2\theta = \cfrac{2\tan \theta}{ 1-\tan^2 \theta} = \cfrac{2a}{1-a^2}=2 \Rightarrow a^2+a-1=0 \Rightarrow a=\cfrac{-1+\sqrt 5}{2} \\ \Rightarrow \cfrac{1}{a} =\cfrac{\sqrt 5+1}{2}為黃金比例$$