106年台中文華高中教甄-數學詳解

臺中市立文華高級中等學校106學年度第2次教師甄選

一、填充題

解答：$$f(x)=(x^2+2x+3)(x^2+2x-2)+5x^2+10x+4 \\ =(x^2+2x)^2+(x^2+2x)-6 +5(x^2+2x)+4\\ =(x^2+2x)^2+6(x^2+2x)-2 \\ \Rightarrow f'(x)= 2(x^2+2x)(2x+2)+ 6(2x+2)\\ 因此f'(x)=0 \Rightarrow x=-1 \Rightarrow f(-1)=1-6-2=-7\\ \Rightarrow (\alpha,m)= \bbox[red, 2pt]{(-1,-7)}$$

解答：$$\log_2 x+\log_x 8-6=0 \Rightarrow {\log x\over \log 2} +{3\log 2\over \log x}-6=0 \Rightarrow (\log x)^2-6\log 2(\log x)+ 3(\log 2)^2=0\\ \Rightarrow \cases{\log \alpha +\log \beta = 6\log 2\\ \log \alpha \log \beta = 3 (\log 2)^2} \Rightarrow (\log \alpha)^2 +(\log \beta)^2 =(\log \alpha+\log \beta)^2 -2\log \alpha \log \beta =30(\log 2)^2 \\ \Rightarrow \log_\alpha \beta +\log_\beta \alpha = {\log \beta \over \log \alpha} +{\log \alpha \over \log \beta} ={(\log \alpha)^2 +(\log \beta)^2 \over \log \alpha \log \beta} ={30(\log 2)^2 \over 3(\log 2)^2} =\bbox[red,2pt]{10}$$

解答：$$\log B= I +f ，其中I是首數，f是尾數 \Rightarrow \log A= (I-1)+ (f+\log 5) \\ \Rightarrow A= 10^{I+f+\log 5-1} =B10^{\log 5-1} =B\times {5\over 10}={B\over 2} \Rightarrow A-B+{15\over 2} =-{B\over 2}+{15\over 2}=0\\ \Rightarrow B=15 \Rightarrow A= {15\over 2} \Rightarrow A+B ={15\over 2}+15 = \bbox[red,2pt]{45\over 2}$$

解答：$$(m^2+1)x^2 -4mx + 2=0 \Rightarrow x={ 2m\pm \sqrt{2m^2-2}\over m^2+1} \Rightarrow \cases{\alpha={ 2m+ \sqrt{2m^2-2}\over m^2+1} \\\beta ={ 2m- \sqrt{2m^2-2}\over m^2+1}} \\ \Rightarrow \cases{\alpha-3\beta = { -4m+4 \sqrt{2m^2-2}\over m^2+1} \\ 2\alpha\beta= {4\over m^2+1}} ；由題意知: 2\alpha\beta = \alpha-3\beta \Rightarrow -m+ \sqrt{2m^2-2} =1 \\ \Rightarrow 2m^2-2 =(m+1)^2 =m^2+2m+1 \Rightarrow m^2-2m-3=0 \\\Rightarrow m=\bbox[red,2pt]{3} ( m=-1 \Rightarrow 2x^2+4x+2 =0 \Rightarrow x=-1，違反正根)$$

解答：$$(1+2x)^n = \sum_{k=0}^n C^n_k2^kx^k \Rightarrow x^3的係數為C^n_32^3 ={4n(n-1)(n-2)\over 3} \\\Rightarrow {1\over a_n} ={3\over 4}\cdot {1\over n(n-1)(n-2)} ={3\over 8}\left({1\over (n-1)(n-2)}-{1\over n(n-1)}  \right) \\ \Rightarrow \sum_{k=3}^{100} {1\over a_n} ={3\over 8}\left({1\over 2\cdot 1}- {1\over 3\cdot 2} + {1\over 3\cdot 2}- {1\over 4\cdot 3} + \cdots +{1\over 99\cdot 98}- {1\over 100\cdot 99} \right)\\ ={3\over 8}\left({1\over 2}-{1\over 9900} \right) ={3\over 8} \cdot {4949\over 9900} =\bbox[red,2pt]{4949\over 26400}$$

解答：$$\cases{全塗:3^4+3=84\\ 一區不塗:4\times 3^2\times 4= 144\\ 相鄰二區不塗: 4\times 3\times 4=48\\ 對面二區不塗: 4^2\times 2=32} \Rightarrow 共有84+ 144+48+32= \bbox[red,2pt]{308}種塗法$$

解答：$$x+y+z+w=20, 1\le x,y,z,w \le 6 \Rightarrow x+y+z+w = 16, 0\le x,y,z,w \le 5\\ \begin{array}{c|l} (x,y,z,w) & 排列數\\ \hline(5,5,5,1) & 4!/3!=4\\ (5,5,4,2) & 4!/2=12 \\ (5,5,3,3) & 4!/(2!2!) =6 \\ (5,4,4,3) & 4!/2!=12\\ (4,4,4,4) & 1\\\hline \end{array} \Rightarrow 共有4+12+6+12+1= \bbox[red,2pt]{35}種買法$$

解答：$$任取5數，a_1-a_5，滿足\cases{ a_3=6 \\ (a_1+\cdots +a_5)\div 5=6}\\ \begin{array}{} a_1 & a_2 &a_3 & a_4 & a_5\\ \hline 1 & 2 & 6 & 10 & 11  \\\hdashline 1 & 3& 6 & 9 & 11\\\hdashline 1 & 4& 6 &8 & 11\\ & & & 9 & 10\\\hdashline
2 & 3 & 6 & 8& 11\\ & & & 9 & 10 \\\hdashline 1 & 5 & 6 & 7 & 11\\ & & & 8 & 10\\\hdashline 2 & 4 & 6 & 7 & 11\\ & & & 8 & 10 \\\hdashline 2 & 5 & 6 & 7 & 10 \\ & & & 8 & 9\\\hdashline 3 & 4 & 6 & 7 & 10 \\& & & 8 & 9 \\\hdashline 3 & 5 & 6 & 7 & 9\\\hdashline 4 & 5 & 6 & 7 & 8\\\hline\end{array}，共有16組數字\\，則{中位數為6且平均數也為6\over 中位數為6} ={16\over C^5_2 \times C^5_2} = \bbox[red, 2pt]{4\over 25}$$

解答：$$f(x)=\begin{vmatrix}x+2 & 2 & 2\\2017 & 2x+2018 & 2019\\ 2017^2 & 2018^2 &3x+2019^\color{blue}{2}\end{vmatrix} \Rightarrow f(x)的x^3係數為6，則abc=-{f(0)\over 6}；\\ 而f(0)= \begin{vmatrix}2 & 2 & 2\\2017 & 2018 & 2019\\ 2017^2 & 2018^2 &2019^2\end{vmatrix} = 2\begin{vmatrix}1 & 1 & 1\\2017 & 2018 & 2019\\ 2017^2 & 2018^2 &2019^2\end{vmatrix} = 2\begin{vmatrix}1 & 1 & 1\\0 & 1 & 2\\ 0 & 4035 & 8072 \end{vmatrix} \\ =2\begin{vmatrix}1 & 1 & -1\\0 & 1 & 0\\ 0 & 4035 & 2 \end{vmatrix} =2(2-0)=4 \Rightarrow abc=-{4\over 6} =\bbox[red,2pt]{-{2\over 3}}\\註:原題行列式最後一個元素3x+2019 應修訂為3x+2019^2，才能與公布之答案對應$$

解答：$$y=3 \sin^2x + 4\sqrt 3 \sin x\cos x-\cos^2 x = 3-4\cos^2x+ 2\sqrt 3\sin (2x)\\=3-2(\cos (2x)+1)+ 2\sqrt 3\sin (2x)= 2\sqrt 3\sin (2x)-2 \cos (2x)+1 \\=4({\sqrt 3\over 2}\sin(2x)-{1\over 2}\cos (2x))+1 =4\sin(2x-{\pi \over 6})+1\\ \Rightarrow 當2x-{\pi\over 6}={3\pi \over 2}，即x={5\over 6}\pi 時y有最小值-4+1=-3；因此(a,m)=\bbox[red, 2pt]{({5\over 6},-3)}$$

解答：

$$\sin\theta={4\over 5} \Rightarrow \cos \theta = {3\over 5} \Rightarrow \cos 2\theta=\cos^2\theta-\sin^2\theta= -{7\over 25}\\假設\angle POX =\alpha， 則 \cases{\angle XOP_1= \alpha,其中P_1為P以\overline{OX}的對稱點\\ \angle P_2OY= \theta-\alpha,其中P_2為P以\overline{OY}的對稱點}；\\因此\angle P_1OP_2 = (\theta-\alpha)+\theta + \alpha=2\theta \Rightarrow \cos \angle P_1OP_2 = {10^2+10^2-\overline{P_1P_2}^2 \over 2\times 10\times 10} \\ \Rightarrow \overline{P_1P_2}= 10\sqrt{2-2\cos 2\theta} =10\sqrt{2+{14\over 25}}=  10\sqrt{64\over 25}=16 \\ \triangle PQR周長= \overline{PR} +\overline{PQ} +\overline{QR} = \overline{P_2R} +\overline{P_1Q} +\overline{QR} =\overline{P_1P_2} =\bbox[red,2pt]{16}$$

解答：

$$圓C:(x-7)^2 +(y+4)^2= 5 \Rightarrow \cases{圓心O(7,-4)\\ 半徑r=\sqrt 5}；\\A(5,2)與y軸的對稱點A'，則A'=(-5,2)，\overline{A'O}與y軸的交點即為P、與圓C的交點即為Q；\\ 因此\overline{PA}+ \overline{PQ} = \overline{PA'} +\overline{PQ} = \overline{A'O}- r =\sqrt{12^2+6^2}-\sqrt 5 =\bbox[red,2pt]{5\sqrt 5}$$

解答：

$$\cases{\overset{\large\rightharpoonup}{PA}+ 2\overset{\large\rightharpoonup}{PB} +3\overset{\large\rightharpoonup}{PC} =\overset{\large\rightharpoonup}{CA} =\overset{\large\rightharpoonup}{CP} +\overset{\large\rightharpoonup}{PA} \\ \overset{\large\rightharpoonup}{QA}+ 2\overset{\large\rightharpoonup}{QB} +3\overset{\large\rightharpoonup}{QC} =2\overset{\large\rightharpoonup}{AB} =2\overset{\large\rightharpoonup}{AQ} +2\overset{\large\rightharpoonup}{QB} \\ \overset{\large\rightharpoonup}{RA}+ 2\overset{\large\rightharpoonup}{RB} +3\overset{\large\rightharpoonup}{RC} =3\overset{\large\rightharpoonup}{BC} =3\overset{\large\rightharpoonup}{BR} +3\overset{\large\rightharpoonup}{RC} } \Rightarrow \cases{ \overset{\large\rightharpoonup}{PB} +2\overset{\large\rightharpoonup}{PC} =0 \\ \overset{\large\rightharpoonup}{QA} +\overset{\large\rightharpoonup}{QC} =0 \\ \overset{\large\rightharpoonup}{RA}+ 5\overset{\large\rightharpoonup}{RB} =0} \\ \Rightarrow \cases{|\overset{\large\rightharpoonup}{PB}|=2|\overset{\large\rightharpoonup}{PC}| \\ |\overset{\large\rightharpoonup}{QA}|=|\overset{\large\rightharpoonup}{QC}| \\|\overset{\large\rightharpoonup}{RA}|=5|\overset{\large\rightharpoonup}{RB}| } \Rightarrow \cases{ \triangle AQR : \triangle ABC=1\cdot 5 :2\cdot 6=5:12 \\ \triangle BPR: \triangle ABC=1\cdot 2:6\cdot 3=1:9\\ \triangle CPQ: \triangle ABC=1\cdot 1:2\cdot 3=1:6 } \\ \Rightarrow {\triangle ABC \over \triangle PQR}= {1\over 1-{5\over 12}-{1 \over 9}-{1 \over 6}} ={1\over 1-{25\over 36}} = \bbox[red,2pt]{36\over 11}$$

解答：

$$\cases{0\le \alpha \le \pi/3\\ \pi/2 \le \beta \le 3\pi/4} \Rightarrow \cases{0\le \sin \alpha \le \sqrt 3/2\\ -1/\sqrt 2\le \cos \beta \le 0}\\因此\cases{x=5\sin \alpha+\cos \beta\\ y=4\cos\beta +\sin \alpha} \Rightarrow \cases{\sin \alpha= (4x-y)/19\\ \cos \beta = (5y-x)/19} \Rightarrow \cases{0\le (4x-y)/19\le \sqrt 3/2\\ -1/\sqrt 2\le (5y-x)/19 \le 0}\\ 令\cases{L_1:4x=y\\ L_2 :4x-y=19\sqrt 3/2\\ L_3: x=5y\\ L_4: 5y-x=-19/\sqrt 2} \Rightarrow \cases{d=\text{dist}(L_1,L_2)={19\sqrt 3\over 2\sqrt {17}}\\ L_1,L_2與L_3,L_4分別交於P(0,0),Q(-1/\sqrt 2),R,S} \\ \Rightarrow \overline{PQ}= {\sqrt{17}\over \sqrt 2} \Rightarrow PQRS面積=d \times \overline{PQ}={19\sqrt 3\over 2\sqrt {17}}\times {\sqrt{17}\over \sqrt 2}= \bbox[red,2pt]{19\sqrt 6\over 4}$$

解答：

$$邊長為1的正六邊形，任取3頂點的三角形面積有三種尺寸，如上圖；\\\cases{面積為\sqrt 3/4的有6個\\ 面積為\sqrt 3/2的有12個\\ 面積為3\sqrt 3/4的有2個 } \Rightarrow 期望值{\sqrt 3\over 4} \times {6\over 20} +{\sqrt 3\over 2} \times {12\over 20} +{3\sqrt 3\over 4} \times {2\over 20} = \bbox[red,2pt]{{9\over 20}\sqrt 3}$$

解答：$$f(x)=x^4-2x^3+3x -\left( \int_2^x (3t^3-7t^2+5t-1)\;dt\right)-6 \\ \Rightarrow \cases{f(2)=16-16+6-0-6=0 \\ f'(x)=4x^3-6x^2+3 -(3x^3-7x^2+5x-1) \Rightarrow f'(2)=11-(5)=6}\\ \lim_{h\to 0} {f(2+3h)\over 4h} =\lim_{h\to 0} {f(2+3h)-f(2)\over 4h} = \lim_{h\to 0} {3\over 4}\cdot {f(2+3h)-f(2)\over 3h} ={3\over 4}f'(2) ={3\over 4}\cdot 6=\bbox[red,2pt]{9\over 2}$$

解答：$$f(x)=\sqrt{3-x} +\sqrt{5x-4} \Rightarrow f'(x)={-1\over 2\sqrt{3-x}} +{5\over 2\sqrt{5x-4}} ={5\sqrt{3-x}-\sqrt{5x-4}\over 2\sqrt{3-x}\cdot \sqrt{5x-4}}\\ f'(x)=0 \Rightarrow 5\sqrt{3-x}=\sqrt{5x-4} \Rightarrow 75-25x=5x-4 \Rightarrow x={79\over 30} \\ \Rightarrow f({79\over 30}) =\sqrt{3-{79\over 30}} +\sqrt{{79\over 6}-4} =\sqrt{11\over 30} +\sqrt {55\over 6}  =\sqrt{11\over 30} +\ {5\sqrt{11}\over \sqrt{30}} ={6\sqrt{11} \over \sqrt{30}} = \bbox[red,2pt]{\sqrt{330}\over 5}$$

解答：

$$\triangle ABC 三邊長\cases{\overline{BC}= x\\ \overline{AC}= y\\ \overline{AB}= z}，及邊長上的高\cases{\overline{AF}=5\\ \overline{BE}=6\\ \overline{CD}=10}，如上圖；\\則x,y,z滿足題意\cases{x= \sqrt{y^2-25}+ \sqrt{z^2-25} \\ y= \sqrt{x^2-36}+ \sqrt{z^2-36} \\z= \sqrt{x^2-100}+ \sqrt{y^2-100} }；\\ \triangle ABC面積= {1\over 2}\cdot 5\cdot x= {1\over 2}\cdot6\cdot y= {1\over 2}\cdot10 \cdot z =k \Rightarrow \cases{x=2k/5\\ y=k/3\\ z=k/5}\\ 令s= (x+y+z)/2 = 7k/15，則\triangle ABC面積 = \sqrt{s(s-x)(s-y)(x-z)} \\=\sqrt{{7k\over 15} \cdot {k\over 15} \cdot {2k\over 15} \cdot {4k\over 15} \cdot }  ={2\sqrt{14} \over 15^2}k^2 =k \Rightarrow k={15^2 \over 2\sqrt{14}} \Rightarrow x+2y+4z= {2k\over 5}+ {2k\over 3}+ {4k\over 5} \\={28 \over 15}k ={28 \over 15}\cdot {15^2 \over 2\sqrt{14}} = \bbox[red,2pt]{15\sqrt{14}}$$

解答：$$A=\begin{bmatrix} 1 & -5 \\ -5 & 1\end{bmatrix} =P\begin{bmatrix} 6 & 0\\ 0 & -4\end{bmatrix}P^{-1}，其中P=\begin{bmatrix} 1 & 1 \\ -1 & 1\end{bmatrix},P^{-1}=\begin{bmatrix} 1/2 & -1/2 \\ 1/2 & 1/2\end{bmatrix}\\ \Rightarrow A^n= \begin{bmatrix} 1 & 1 \\ -1 & 1\end{bmatrix}\begin{bmatrix} 6^n & 0\\ 0 & (-4)^n\end{bmatrix} \begin{bmatrix} 1/2 & -1/2 \\ 1/2 & 1/2\end{bmatrix} =\begin{bmatrix} 6^n & (-4)^n\\ -6^n & (-4)^n\end{bmatrix} \begin{bmatrix} 1/2 & -1/2 \\ 1/2 & 1/2\end{bmatrix} \\ =\begin{bmatrix} (6^n+(-4)^n)/2 & (-6^n+(-4)^n)/2\\ (-6^n +(-4)^n)/2& (6^n+(-4)^n)/2\end{bmatrix} =\begin{bmatrix} a & b \\ c& d\end{bmatrix} \Rightarrow b=\bbox[red,2pt]{-6^n+(-4)^n\over 2}$$

解答：$$\overline{AB}=\sqrt{9^2+12^2} =15，令\overline{BP}=a，則\overline{PA}=54-15-a=39-a \\ \Rightarrow \triangle ABP = \sqrt{s(s-a)(s-39+a)(s-15)},s=54/2=27 \\\Rightarrow \triangle ABP=\sqrt{27(27-a)(a-12)(12)}\\ 令f(a)=(27-a)(a-12) =-a^2 +39a -324 \Rightarrow f(39/2)為極大值\\ \Rightarrow \triangle ABP最大值=\sqrt{ 27(27-{39\over 2})({39\over 2}-12)(12)} =\sqrt{27\cdot {15\over 2} \cdot {15\over 2}\cdot 12} =18\times {15\over 2}=\bbox[red,2pt]{135}$$

解答：$$方程式a^{2|x|}-8x^2 +5|x|-1 對稱y軸，因此\alpha+\beta =0 \Rightarrow \alpha-2\beta= \alpha+2\alpha=3\alpha={3\over 4} \\ \Rightarrow \alpha={1\over 4} \Rightarrow a^{1/2}-{1\over 2}+{5\over 4}=1 \Rightarrow \sqrt a={1\over 4} \Rightarrow a=\bbox[red, 2pt]{1\over 16}$$