臺中市立文華高級中等學校 109 學年度
第2次教師甄選-數學專業知能試題
解：
$$\cases{f(2x) < 0 共有11個整數解\\ f(2x+5) < 0 共有16個整數解} \Rightarrow \cases{f(x)< 0 共有11個偶數解\\ f(x) < 0 共有16個奇數解} \\ \Rightarrow f(x) < 0 有16+11=\bbox[red, 2pt]{27}個整數解$$
解：
$$\cases{3^{\alpha/2} =\sqrt 3-\alpha \\ \log_3 \beta = 2\sqrt 3-2\beta} \Rightarrow \cases{\sqrt 3^{\alpha} =\sqrt 3-\alpha \\ \log_{\sqrt 3} \beta = \sqrt 3-\beta}\\ 令A(a,\sqrt 3-a)為兩圖形\cases{y=\sqrt 3^x \\ y=\sqrt 3-x}的交點，及B(b,\sqrt 3-b)為兩圖形\cases{y=\log_{\sqrt 3}x \\ y=\sqrt 3-x}的交點;\\由於此二圖形\cases{y=\sqrt 3^x \\ y=\log_{\sqrt 3}x} 對稱於y=x \Rightarrow \cases{a=\sqrt 3-b \\ \sqrt 3-a =b} \Rightarrow a+b=\sqrt 3;\\ 因此(\alpha +\beta)^2-2(\sqrt 3)^\alpha -\log_3\beta = (\sqrt 3)^2-2(\sqrt 3-\alpha)-2(\sqrt 3-\beta) =3-4\sqrt 3+2(\alpha+\beta) \\ =3-4\sqrt 3+2\sqrt 3= \bbox[red,2pt]{3-2\sqrt 3}$$
解：
$$\cases{S=\{a,b,c,d,e,f\} \\ A=\{a,b\}} \Rightarrow P(A)={\#(A)\over \#(S)} ={2\over 6}={1\over 3}\\ A、B獨立\Rightarrow P(A\cap B)=P(A)P(B) = {1\over 3}P(B) \\ \Rightarrow \begin{array}{}P(A\cap B) & P(B) & B & \#(B)\\ \hline 0 & 0 & \emptyset & 1\\ 1/6 & 1/2 & \{a,\bigcirc,\triangle\} & C^4_2 \\ & & \{b,\bigcirc,\triangle\} & C^4_2 \\ & & \triangle,\bigcirc \in \{c,d,e,f\}\\ 2/6 & 1 & S & 1\\ \hdashline3/6 & 3/2 \not \lt 1 & 不存在 & 0 \\ 4/6,5/6,6/6& \not \lt 1 & 不存在 & 0\\\hline\end{array} \\ \Rightarrow \#(B) =1+C^4_2+C^4_2 +1= \bbox[red,2pt]{14}$$
解：
$$直線\overline{BE}交\overline{AC}於F,見上圖；又\cases{\angle CAD= \angle ADB-\angle ACB = 75^\circ -45^\circ=30^\circ \\ \angle BED=180^\circ -\angle AEB = 180^\circ- 120^\circ=60^\circ \\ \angle EBD=\angle AEB-\angle EDB=120^\circ -75^\circ =45^\circ}\\ \triangle BDE \Rightarrow {\overline{ED} \over \sin \angle EBD} ={ \overline{BD} \over \sin\angle BED} ={\overline{BE} \over \sin\angle BDE}\Rightarrow {4\sqrt 2\over \sin 45^\circ} ={\overline{BD} \over \sin 60^\circ}= {\overline{BE} \over \sin 75^\circ}\\\Rightarrow \cases{\overline{BD}=4\sqrt 3 \\ \overline{BE}=8\sin 75^\circ =8\sin(45^\circ +30^\circ)= 2(\sqrt 2+\sqrt 6)}\\ \angle FCB=\angle FBC=45^\circ \Rightarrow \overline{FB}= \overline{BC} \div \sqrt 2 = {6+4\sqrt 3 \over \sqrt 2} =3\sqrt 2+2\sqrt 6 \\\Rightarrow \overline{EF}= \overline{FB} -\overline{BE} =3\sqrt 2+2\sqrt 6-2\sqrt 2-2\sqrt 6=\sqrt 2 \Rightarrow \overline{AE}= 2\overline{EF}= 2\sqrt 2 \\ \Rightarrow \cos \angle AEB= {\overline{AE}^2 +\overline{BE}^2-\overline{AB}^2 \over 2\overline{AE} \times \overline{BE}} \Rightarrow \cos 120^\circ ={8+4(\sqrt 2+\sqrt 6)^2 -\overline{AB}^2 \over 8\sqrt 2(\sqrt 2+\sqrt 6)} \\ \Rightarrow -{1\over 2} = {40+16\sqrt 3-\overline{AB}^2 \over 16+16\sqrt 3} \Rightarrow \overline{AB}^2= 48+24\sqrt 3 \Rightarrow \overline{AB}= \bbox[red,2pt]{6+2\sqrt 3}$$
$$2\vec a\times \vec b+3\vec c\times \vec b+7\vec c\times \vec a = (12,-4,11) \Rightarrow \vec a\cdot (2\vec a\times \vec b+3\vec c\times \vec b+7\vec c\times \vec a ) = \vec a \cdot (12,-4,11)\\ \Rightarrow \vec 0-3\vec a\cdot(\vec b\times \vec c) + \vec 0 = (3,1,5)\cdot (12,-4,11)=87 \Rightarrow \vec a\cdot(\vec b\times \vec c)=-29 \\ \Rightarrow \begin{vmatrix} 3& 1& 5 \\ b_1& b_2 & b_3\\ c_1 & c_2 & c_3\end{vmatrix} =\vec a\cdot (\vec b\times \vec c) =\bbox[red, 2pt]{-29}$$
$$\cases{x+ay+a^2z =2a^3 \\x+by+b^2z =2b^3 \\x+cy+c^2z =2c^3 } 經過(-4,-2,4) \Rightarrow \cases{-4-2a+4a^2 =2a^3 \\-4-2b+4b^2 =2b^3 \\-3-2c+4c^2 =2c^3 }\\ \Rightarrow \cases{ a^3-2a^2+a+2=0 \\b^3-2b^2+b+2=0 \\c^3-2c^2+c+2=0 } \Rightarrow a,b,c 為x^3-2x^2+x+2=0 之三根 \\ \Rightarrow abc = 三根之積=\bbox[red,2pt]{-2}$$
$$令\cases{P({a^2\over 4},a) \\ A(2,1) \\ B(1,0)}\Rightarrow M= \overline{PA}+\overline{PB};\\P點軌跡為4x=y^2，為一拋物線，其焦點為F(1,0)=B，準線L:x=-1 \Rightarrow \overline{PB} = d(P,L) \\ \Rightarrow M 的最小值=d(A,L) = \bbox[red,2pt]{3}$$
$$N: 取3次號碼最大的數\\\begin{array}{clc} N & (a,b,c) & 個數 \\\hline 1 & (1,1,1) & 1\\ \hdashline 2 & (2,2,2) & 1\\ & (2,1,2) & 3 \\ & (2,1,1) & 3\\\hdashline 3 & (3,3,3) & 1\\ & (3,1-2,3) & 6 \\ & (3,1-2,1-2) & 12 \\\hdashline 4 & (4,4,4) & 1\\ & (4,1-3,4) & 9 \\ & (4,1-3,1-3) & 27\\\hdashline 5 & (5,5,5) & 1\\ & (5,1-4,5) & 12\\ & (5,1-4,1-4) & 48\\\hline \end{array}\\ \Rightarrow 期望值=(1\times 1+ 2(1+3+3) + 3(1+6+12) +4(1+9+27) +5(1+12+48)) \div 5^3 \\= {525 \over 125}= \bbox[red,2pt]{21 \over 5} $$
$$z=(2\sqrt 2+\sqrt 2i)x^2-\sqrt 2ixy-2\sqrt 2iy^2 +3\sqrt 2y+(3+4i)a -(2+i)b-(11\sqrt 2+i)=0\\ \Rightarrow \cases{Re(z)=0 \\ Im(R)=0} \Rightarrow \cases{2\sqrt 2x^2 +3\sqrt 2y+3a-2b-11\sqrt 2=0\\ \sqrt 2x^2-\sqrt 2xy-2\sqrt 2y^2+4a-b-1=0}\\ 由於a,b為有理數，a與b皆不含\sqrt 2 \Rightarrow \cases{\cases{2x^2+3y=11\\ 3a=2b}\\ \cases{x^2-xy-2y^2=(x+y)(x-2y)=0 \\ 4a-b=1}} \\\Rightarrow \cases{x=2\\y=1 \\ a=2/5\\ b=3/5} \Rightarrow x+y+a+b=\bbox[red,2pt]{4}$$
$$\begin{array}{} n & \lfloor \sqrt n \rfloor & \text{數量} &\text{累積和} \\\hline 1,3 & 1 & 2 & 2 \\\hdashline 5,7 & 2 & 2 & 6 \\\hdashline 9,11,15 & 3 & 3 & 15\\\hdashline 17,19,21,23 & 4 & 4 & 31\\\hdashline 25,27,29,31 & 5 & 6 & 61 \\ 33,35 \\\hdashline 37,39,41,43 & 6 & 6 & 97 \\ 45,47\\\hdashline 49,51,53,55 & 7 & 8 & 153\\ 57,59,61,63\\\hdashline 65,67,..,79 & 8 & 8 & 217 \\\hdashline 81,..,99 & 9& 10 & 307\\\hdashline 101,..,119 & 10 & 10 & 407 \\\hdashline 121,..,143 &11 & 12 & 539\\\hdashline 145,..,167 & 12 & 12 & 683\\\hdashline 169,..,195 & 13 & 14 & 865 \\\hdashline 197,..,223 & 14 & 14 & 1061\\ \hdashline 225,..,255 & 15 & 16 & 1301 \\\hdashline 257,..,287& 16 & 16 & 1557 \\ \hdashline 289,..,323 & 17 & 18 & 1863 \\\hdashline 325,..,\color{blue}{339} & 18 & 8 & 2007\\ 341 & 18 & 1 & 2025\\\hline\end{array}\\ \Rightarrow n的最大值為\bbox[red, 2pt]{339}$$
解：
解：
$$f(1-x)=1-f(x) \Rightarrow f(1)=f(1-0)=1-f(0)=1 \Rightarrow f(1/3)=f(1)/2=1/2 \\又 f(1/2)=f(1-1/2) =1-f(1/2) \Rightarrow 2f(1/2)=1 \Rightarrow f(1/2)=1/2\\ 綜合以上及題意: \cases{0\le x_1 < x_2 \le 1 \Rightarrow f(x_1)\le f(x_2)\\ f(1/2)=f(1/3)=1/2} \Rightarrow f(x)={1\over 2} \;\forall x\in[1/3,1/2];\\ f({109\over 2020}) =f({327\over 2020}\div 3) = {1\over 2} f({327\over 2020}) ={1\over 2} f({981\over 2020} \div 3) = {1\over 4}f({981\over 2020});\\ 由於{1\over 3} < {981\over 2020} <{1\over 2} \Rightarrow f({981\over 2020})={1\over 2} \Rightarrow {109\over 2020}= {1\over 4}\times {1\over 2} =\bbox[red, 2pt]{1\over 8}$$
解：
$$a_n= 7\left({1\over 10^n} + {1\over 10^{n+1}} + \cdots +{1\over 10^{2n-1}} \right) ={7\over 10^n} \left( 1+{1\over 10} +\cdots +{1\over 10^{n-1}}\right) ={7\over 9\times 10^{n-1}}(1-{1\over 10^n})\\ \Rightarrow S_n= \sum_{k=1}^n a_k = \sum_{k=1}^n {7\over 9\times 10^{k-1}}(1-{1\over 10^k}) ={7\over 9}\sum_{k=1}^n \left({1\over 10^{k-1}} -{1\over 10^{2k-1}} \right) \\ ={7\over 9} \left({1-{1\over 10^n}\over 1-10^{-1}} -{{1\over 10}-{1\over 10^{2n+1}}\over 1-10^{-2}} \right) ={7\over 9} \left( {100\over 99} -{1\over 9\times 10^{n-1}} +{{1\over 99\times 10^{2n-1}}}\right)\\ \Rightarrow \lim_{n\to \infty} S_n ={7\over 9}\times {100\over 99} = \bbox[red,2pt]{700 \over 891}$$
$$假設連續n個邊，及剩下的109-n邊，可以分別組合成一個n+1多邊形及一個110-n個多邊形，\\這兩個多邊形的面積和等於原正109邊形面積，也就是10；\\當n=2時，產生1組面積和為10的三角形及108邊形，共有109組，面積總和為1090；\\當n=3時，也有109組，面積總和也是1090；\\ n=54時為最大值，也就是n=2-54，共53種；因此總面積和=1090\times 53= \bbox[red,2pt]{57770}$$
解：
$$\cases{k=3\\ n=10}代公式:(k-1)(-1)^n+(k-1)^n = 2+2^{10}= \bbox[red,2pt]{1026}$$
解：$$\alpha,\beta,\gamma為x^3-2kx^2+(k^2+11)x-96=0之三根\Rightarrow \alpha+\beta +\gamma=2k \Rightarrow k={1\over 2}(\alpha+\beta +\gamma) \\ \Rightarrow \triangle面積=\sqrt{k(k-\alpha)(k-\beta)(k-\gamma)} =\sqrt{k(k^3-2k^3+(k^2+11)k-96)} =3\sqrt 3 \\ \Rightarrow k(11k-96)=27 \Rightarrow 11k^2-96k-27=0 \Rightarrow (11k+3)(k-9)=0 \\\Rightarrow k=\bbox[red, 2pt]{9}(k為三邊長之和的一半，必為正值)$$
解：
解：
解：$$球S:x^2+y^2+z^2-2x-4y+8z+18=0 \Rightarrow (x-1)^2+(y-2)^2 +(z+4)^2 =3 \\ \Rightarrow \cases{球心A(1,2,-4)\\ 球半徑R=\sqrt 3} \Rightarrow d(A,E)= {|1+4+8-16|\over \sqrt{1+4+4}} =1 = \overline{AB} (B為圓C之圓心)\\又P為切點\Rightarrow \angle APR=90^\circ = \angle BPR + \angle APB = \angle BPR+\angle BRP \\ \Rightarrow \angle APB= \angle BPR，再加上\angle ABP=\angle PBR = 90^\circ \Rightarrow \triangle ABP \sim \triangle PBR (AAA) \\ \Rightarrow {\overline {AB} \over \overline{PB}} ={\overline {PB} \over \overline{BR}} \Rightarrow {1\over \sqrt 2} = {\sqrt 2\over \overline{BR}} \Rightarrow \overline{BR} =2 \Rightarrow \overline{AR}=1+2=3;\\ 平面E的法向量(1,2,-2)即為直線\overline{AR}的方向向量\Rightarrow \overline{AR}的方程式: {x-1\over 1} = {y-2\over 2} ={x+4\over -2} \\ \Rightarrow B(t+1,2t+2,-2t-4)，B在E上 \Rightarrow t+1+4t+4+4t+8-16=0 \Rightarrow t=1/3 \\ \Rightarrow B(4/3,8/3,-14/3) \Rightarrow \overrightarrow{AB} =(1/3,2/3,-2/3)，再由3\overrightarrow{AB}=\overrightarrow{AR} \Rightarrow \bbox[red,2pt]{R(2,4,-6)}$$
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