國立彰化高級中學 115 學年度第一次教師甄選
一、 填充題(每題 5 分，共 85 分)
解答:$$假設\cases{ 數值為 -1 的項共有 x 個\\ 數值為 0 的項共有 y 個 \\ 數值為 1 的項共有 z 個\\數值為 2 的項共有 w 個} \Rightarrow \cases{ (-1)x + 0\cdot y+1\cdot z+ 2\cdot w=16 \\(-1)^2x+ 0^2y+ 1^2z+ 2^2w=50 \\ (-1)^3x+ 0^3y+1^3z+2^3 w=70} \Rightarrow \cases{ -x + z + 2w &= 16 \\ x + z + 4w &= 50 \\ -x + z + 8w &= 70} \\ \Rightarrow \cases{x=8\\z=6\\w=9} \Rightarrow \sum_{k=1}^n |a_k| =|-1|x+ |0|y+ |1|z+ |2|w=x+z+2w= \bbox[red, 2pt]{32}$$
解答:$$f(x)=(x-1)+{1\over x-1}+3\\ \textbf{Case I }x-1\gt 0 \Rightarrow x-1+{1\over x-1} \ge 2\sqrt{(x-1)\cdot {1\over x-1}}=2 \Rightarrow f(x)\ge 2+3=5 \\ \textbf{Case II }x-1\lt 0 \Rightarrow t=-(x-1 \gt 0 ) \Rightarrow f(x)= - \left( t+{1\over t} \right)+3\le -2+3=1 \\ \Rightarrow \bbox[red, 2pt]{f(x)\ge 5或f(x) \le 1}$$
解答:$$\cases{f(115)= 115\\ f(n)+f(n+3)=n^2} \Rightarrow f(n)=n^2-f(n+3) \Rightarrow f(70) =70^2-f(73) =70^2-(73^2-f(76)) \\=70^2-73^2+f(76) = 70^2-73^2+ 76^2-f(79) = \cdots \\=(70^2-73^2)+(76^2-79^2)+ \cdots+(106^2-109^2) +112^2-f(115) \\=-3(143+ 155+ \cdots+ 215 )+112^2-115 =-3759+112^2-115 =\bbox[red, 2pt]{8670}$$
解答:$$18個頂點形成{18\choose 3}=816個三角形\\ \textbf{Case I }直角三角形 \Rightarrow 18 個等分點可連出 9 條直徑，每條直徑搭配剩下的 16 個點 \\ \qquad \Rightarrow 直角三角形個數=9\times 16=144\\ \textbf{Case II }鈍角三角形 \Rightarrow 三個頂點都在同一半圓內 \Rightarrow 對任一頂點在半圓內有8個頂點任選二點\\ \qquad 組合成一個鈍角三角形\Rightarrow 共有18\times {8\choose 2}=504 \\ \Rightarrow 銳角三角形個數=816-144-504= \bbox[red, 2pt]{168}$$
解答:$$x,y \in \mathbb N \Rightarrow \cases{A=\sqrt{x+3} \in \mathbb N\\ B= \sqrt{x+2007} \in \mathbb N} \Rightarrow \cases{A+B=y\\ A^2-B^2=2004} \Rightarrow (A+B)(A-B)=2004 \\ \Rightarrow \cases{A+B 為偶數\\ A-B為偶數} \Rightarrow 取\cases{A+B= 2m\\ A-B=2n}, m\gt n \Rightarrow 4mn=2004 \Rightarrow mn=501=3\times 167 \\ \Rightarrow \cases{m=501,n=1 \Rightarrow y=A+B=2m=1002 \Rightarrow B=500\Rightarrow x=500^2-3=249997\\ m=167,n=3 \Rightarrow y=334 \Rightarrow x=164^2-3=26893} \\ \Rightarrow (x,y)= \bbox[red, 2pt]{(26893,334), (249997,1002)}$$
解答:$$兩塊相同隔板與10個人排列，排列數={(10+2)!\over 2!}=\bbox[red, 2pt]{12!\over 2} \\ 兩塊隔板把人群分成三組，每一組代表不同的出口，因此排列數即為所求$$
解答:$$己知\cases{n=10\\\sum x_i=20\\ \sum y_i=300\\ \sum x_i^2=50} \Rightarrow \cases{\mu_x=20/10=2\\ \mu_y=300/10=30} \Rightarrow \sum(x_i+ \mu_x)^2= \sum(x_i+2)^2 =\sum x_i^2 +4\sum x_i+40\\=50+4\cdot 20+ 40 =170 \Rightarrow \sum (x_i+\mu_x)(y_i+\mu_y) =\sum (x_i+2)(y_i+ 30) \\ =\sum x_iy_i +30 \sum x_i +2 \sum y_i+\sum 60 = \sum x_iy_i+30\cdot 20+2\cdot 300+60\cdot 10= \sum x_iy_i+ 1800\\ \Rightarrow 錯誤的斜率={\sum x_iy_i+1800 \over 170} ={230\over 17} \Rightarrow \sum x_iy_i =2300-1800=500 \\ \Rightarrow 正確的斜率={\sum(x_i-\mu_x)(y_i- \mu_y) \over \sum (x_i-\mu_x)^2} = {\sum x_iy_i -n \mu_x \mu_y \over \sum x_i^2 -n\mu_x^2} ={500-10\cdot 2\cdot 30\over 50-10\cdot 2^2}= {-100\over 10} =\bbox[red, 2pt]{-10}$$
解答:$$假設\cases{x= \triangle PBC面積\\ y= \triangle PAC面積\\ z= \triangle PAB面積} \Rightarrow \triangle ABC面積 =x+y+z \\\Rightarrow {\overline{AP} +\overline{PD}\over \overline{PD}} ={x+y+z \over x} \Rightarrow 1+{\overline{AP} \over \overline{PD}} =1+{y+z\over x} \Rightarrow {\overline{AP} \over \overline{PD}} ={y+z\over x} \\ 同理可得: {\overline{BP} \over \overline{PE} } ={z+x\over y}及 {\overline{CP} \over \overline{PF}} ={x+y\over z} \\ 依題意: 已知{y+z\over x}+{z+x\over y}+ {x+y\over z} =115 \Rightarrow {yz(y+z) +zx(z+x) +xy(x+y) \over xyz} =115 \\ \Rightarrow {(x+y+z) (xy+yz+zx)-3xyz\over xyz} ={(x+y+z) (xy+yz+zx)\over xyz}-3=115\\欲求{y+z\over x}\cdot {z+x\over y}\cdot {x+y\over z} = {(x+y+z)(xy+yz+ zx)\over xyz}-1 =115+3-1= \bbox[red, 2pt]{117}$$
解答:$$假設t=2^x+2^{-x} \Rightarrow \cases{2^{x+1}+ 2^{1-x} =2(2^x+2^{-x}) =2t\\ 2^{2x+1}+2^{1-2x}=2(2^{2x}+2^{-2x}) =2(t^2-2) =2t^2-4} \\ 取\alpha=\log_3 y,則原式: \alpha^2+2t\alpha+2t^2-4 =0 有實根\Rightarrow 4t^2-4(2t^2-4)\ge 0 \Rightarrow t^2\le 4 \Rightarrow -2\le t\le 2 \\ 由於 t=2^x+2^{-x} \ge 2\sqrt{2^x \cdot 2^{-x}} =2 \Rightarrow t\ge 2又同時滿足-2\le t\le 2 \Rightarrow t=2 \\ \Rightarrow 2^x+2^{-x}=2 \Rightarrow x=0 \Rightarrow \alpha^2+ 4\alpha+4=0 \Rightarrow (\alpha+2)^2=0 \Rightarrow \alpha =-2 \Rightarrow \log_3 y=-2\\ \Rightarrow y={1\over 9} \Rightarrow (x,y) = \bbox[red, 2pt]{ \left( 0,{1\over 9} \right)}$$
解答:
$$y=f(x) =2|x-1|+3|mx-5| \Rightarrow \lim_{x\to \pm \infty} f(x)= \infty \Rightarrow 圖形為開口向上的折線圖\\ y=f(x)與y=6有兩個交點 \Rightarrow \min f(x)\lt 6\Rightarrow f(x)的最小值發生在x=1或x={5\over m} \\\Rightarrow \cases{f(1)=3|m-5| \lt 6 \Rightarrow 3\lt m\lt 7\\f(5/m)= 2|{5\over m}-1| ={2|m-5|\over |m|} \lt 6 \Rightarrow 8m^2+10m-25\gt 0 \Rightarrow m\gt 5/4或m\lt -5/2} \\取兩者的聯集 \Rightarrow \bbox[red, 2pt]{m\gt {5\over 5}或m\lt -{5\over 2}}$$
解答:$$n=5代入公式: \prod_{k=1}^n \sin{k\pi\over 2n+1} ={\sqrt{2n+1} \over 2^n} \Rightarrow \prod_{k=1}^5 \sin{k\pi\over 11} =\bbox[red, 2pt]{\sqrt{11} \over 32} \\ 若不記得公式就慢慢算: {z^{11}-1\over z-1} =1+z+z^2+ \cdots+z^{10} =0的10個複數根z_k \\ \Rightarrow z_k=e^{2k\pi i/11}, k=1,2,\dots,10 \Rightarrow f(z)=1+z+z^2+ \cdots+z^{10} = \prod_{k=1}^{10} (z-e^{2k\pi i/11}) \\ \Rightarrow f(1)=11= \prod_{k=1}^{10} (1-e^{2k\pi i/11}) =\prod_{k=1}^{10} \left|1-e^{2k\pi i/11} \right| =\prod_{k=1}^{10} \left|e^{k\pi i/11} \right|\cdot \left| e^{-k\pi i/11} -e^{k\pi i/11} \right| \\ =\prod_{k=1}^{10} 1\cdot \left| -2i\sin {k\pi\over 11} \right| =\prod_{k=1}^{10} 2 \left( \sin {k\pi\over 11} \right) =2^{10}\prod_{k=1}^{10} \sin{k\pi \over 11} =2^{10} \left( \prod_{k=1}^{5} \sin{k\pi \over 11} \right)^2 \\ \Rightarrow \left( \prod_{k=1}^{5} \sin{k\pi \over 11} \right)^2={11\over 2^{10}} \Rightarrow \prod_{k=1}^{5} \sin{k\pi \over 11} =\sqrt{11\over 2^{10}} =\bbox[red, 2pt]{\sqrt{11} \over 32}$$
解答:$$\cases{z_1=1+i\\ z_{n+1} =(i/2)z_n} \Rightarrow \langle z_n \rangle 為等比數列，首項為1+i, 公比為{i\over 2} \\ |z_{k+1} -z_k|= \left|{i\over 2}z_k-z_k \right| =\left|{i\over 2}-1 \right| \cdot |z_k|= {\sqrt{5} \over 2} \cdot |z_1|\cdot \left( \left|{i\over 2} \right| \right)^{k-1} ={\sqrt 5\over 2}\cdot \sqrt 2\cdot {1\over 2^{k-1}}={\sqrt{10} \over 2}\cdot{1\over 2^{k-1}} \\ \Rightarrow S_n= \sum_{k=1}^n |z_{k+1}-z_k|= {{\sqrt {10}\over 2} (1-{1\over 2^n}) \over 1-{1\over 2}} =\sqrt{10} \left( 1-{1\over 2^n} \right) \Rightarrow S=\lim_{n\to \infty }S_n= \sqrt{10} \\ \Rightarrow S-S_n=\sqrt{10}-\sqrt{10} \left( 1-{1\over 2^n} \right) =\sqrt{10}\cdot {1\over 2^n} \lt 10^{-20} \Rightarrow {1\over 2}-n\log 2\lt -20 \\ \Rightarrow n\gt {20.5\over \log 2} \approx 68.1 \Rightarrow n=\bbox[red, 2pt]{69}$$
解答:$$假設四位數為abcd，需滿足9倍數也是11的倍數，只有a+c=9且b+d=9\\ 相加為9共有4組:\{1,8\},\{2,7\}, \{3,6\},\{4,5\} \Rightarrow (a,c)有4種選擇，再乘上排列數2\Rightarrow 有8種排法\\ (b,d)有3種選擇，再乘上排列數2\Rightarrow 有6種排法；因此符合要求的數字共有8\times 6=48種\\ \Rightarrow 中獎機率={48\over P^9_4} ={48\over 3024}={1\over 64}\\ 任何一個中獎數字(例:1287)，將它前後對調(8712)也是中獎數字，兩這個數字相加一定是9999\\ 也就是說獎金的平均值為{9999\over 2} \Rightarrow 期望值={1\over 64}\cdot {9999\over 2}=\bbox[red, 2pt]{1111\over 14}$$
解答:$$取\cases{\vec u= \vec a\times \vec b =(7,1,-3) \\ \vec v= \vec b\times \vec c=(3,5,1) \\ \vec w= \vec c\times \vec a =(-1,-7,5)} \Rightarrow V= |\vec a\cdot (\vec b\times \vec c)| \Rightarrow V^2= \vec u\cdot (\vec v \times \vec w) =(7,1,-3)\cdot (32,-16,-16) \\=256 \Rightarrow V= \bbox[red, 2pt]{16}$$
解答:$$\det(A-\lambda I)=0 \Rightarrow \lambda =5,-1 \Rightarrow f(x)=x^{115}=(x-5)(x+1)Q(x)+ax+b \\ \Rightarrow \cases{f(5)=5^{115}= 5a+b\\ f(-1)=-1=-a+b} \Rightarrow \cases{a=(5^{115}+1)/6\\ b=(5^{115}-5)/6} \\ \Rightarrow A^{115}={5^{115}+1\over 6}A+{5^{115}-5\over 6} = \bbox[red, 2pt]{\begin{bmatrix}\displaystyle {5^{115}-2\over 3}& \displaystyle {5^{115}+1\over 3} \\ \displaystyle {2\cdot 5^{115}+2\over 3}& \displaystyle {2\cdot 5^{115}-1\over 3} \end{bmatrix}}$$
解答:$$S=\sum_{k=1}^\infty {1\over k^2} \Rightarrow \sum_{k=1}^\infty {1\over (2k-1)^2} = \sum_{k=1}^\infty{1\over k^2} -\sum_{k=1}^\infty{1\over (2k)^2} \;(奇數項=全部-偶數項) \\=S-{1\over 4}S={{3\over 4}}S \Rightarrow \lim_{n\to \infty} {\sum_{k=1}^\infty {1\over (2k-1)^2} \over \sum_{k=1}^\infty {1\over k^2} } = \bbox[red, 2pt]{3\over 4}$$
解答:$$f(z)=az^2+bz+c \Rightarrow f(e^{i\theta}) = (a\cos 2\theta+ b\cos \theta+c) +i(a\sin 2\theta+ b\sin \theta) \\ |f(z)|\le 1 \Rightarrow |f(z)|^2\le 1 \Rightarrow |f(e^{i\theta})|^2=(a\cos 2\theta+ b\cos \theta+c)^2 + (a\sin 2\theta+ b\sin \theta)^2 \\ = a^2+b^2+ c^2+ 2b(a+c)\cos \theta+ 2ac \cos 2\theta \le 1\\ 取\cos \theta={1\over \sqrt 3} \Rightarrow \cos 2\theta=-{1\over 3}代入上式\Rightarrow a^2+b^2 +c^2-{2\over 3}ac+{2\over \sqrt 3}ab +{2\over \sqrt 3}bc \le 1 \\ \Rightarrow \left( a+{1\over \sqrt 3} b-{1\over 3}c \right)^2+{2\over 3}b^2+{8\over 9}c^2+{8\over 3\sqrt 3}bc \le 1 \Rightarrow {2\over 3}b^2+{8\over 9}c^2+{8\over 3\sqrt 3}bc \le 1\\ 算幾不等式:{2\over 3}b^2+{8\over 9}c^2\ge 2\sqrt{{2\over 3}b^2 \cdot{8\over 9}c^2} ={8\over 3\sqrt 3}bc\; 代回上式\Rightarrow {8\over 3\sqrt 3}bc+{8\over 3\sqrt 3}bc\le 1 \\ \Rightarrow {16\over 3\sqrt 3}bc \le 1 \Rightarrow bc \le {3\sqrt 3\over 16} \Rightarrow bc的最大值為 \bbox[red, 2pt]{3\sqrt 3\over 16}$$
二、 計算證明題(共 15 分，需詳列計算或證明過程)
解答:
$$\cases{\angle BFH=\angle BDH=90^\circ \Rightarrow BDHF共圓(\overline{BH}為直徑) \Rightarrow \angle FBH=\angle FDH (對同弧) \\ \angle CEH=\angle CDH=90^\circ \Rightarrow CDHE共圓(\overline{CH}為直徑) \Rightarrow \angle HDE= \angle HCE \\ \angle BFC=\angle BEC=90^\circ \Rightarrow BCEF共圓(\overline{BC}為直徑) \Rightarrow \angle FBH=\angle HCE} \\ \Rightarrow \angle FDH= \angle FBH=\angle HCE= \angle HDE \Rightarrow \angle FDH=\angle HDE \Rightarrow \overline{DH} 為\angle FDE 的角平分線\\ 同理可得\cases{\overline{EH} 為\angle FED 的角平分線\\ \overline{FH} 為\angle DFE的角平分線} \Rightarrow H為\triangle DEF的內心, \bbox[red, 2pt]{故得證}$$
解答:$$x=2代入可得 10^2+11^2+12^2 =365= 13^2+14^2 \Rightarrow x=2為一實數解 \\ 10^x+11^x+12^x=13^x+14^x \Rightarrow \left( {10\over 12} \right)^x + \left( {11\over 12} \right)^x +1= \left( {13\over 12} \right)^x + \left( {14\over 12} \right)^x \\ 取f(x)=\left( {13\over 12} \right)^x + \left( {14\over 12} \right)^x -\left( {10\over 12} \right)^x - \left( {11\over 12} \right)^x,欲求f(x)=1 \\ 由於\cases{\left( {13\over 12} \right)^x 為嚴格遞增函數\\ \left( {14\over 12} \right)^x 為嚴格遞增函數\\-\left( {10\over 12} \right)^x 為嚴格遞增函數\\ - \left( {11\over 12} \right)^x 為嚴格遞增函數} \Rightarrow f(x)為嚴格遞增函數 \Rightarrow y=f(x)與y=1至多一個交點 \\ 又f(2)=1 \Rightarrow 10^x+11^x+12^x=13^x+14^x恰有一實數解，此解為x=2 \; \bbox[red, 2pt]{故得證}$$
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