臺北市立建國高級中學 115 學年度第1次正式教師甄選
一、填充題（每題 6 分，共計 72 分）
解答:$${z\over w}的主幅角為{\pi\over 3} \Rightarrow z與w的夾角為60^\circ \Rightarrow z與2w的夾角為60^\circ \\\Rightarrow |z-2w|^2 =|z|^2+|2w|^2-2|z||2w| \cos 60^\circ \Rightarrow 9=|z|^2+4|w|^2-2|z||w|\\ 令\cases{|z|=x\\ |w|=y} \Rightarrow 在已知x^2-2xy+4y^2=9的條件下求x+y的最大值\\ 利用\text{Lagrange Multiplier }求極值, \cases{f(x,y)=x+y\\ g(x,y)=x^2-2xy+4y^2-9} \Rightarrow \cases{f_x=\lambda g_x\\ f_y= \lambda g_y\\ g=0} \\ \Rightarrow \cases{1=\lambda(2x-2y) \\1=\lambda(-2x+8y) \\ x^2-2xy+4y^2=9} \Rightarrow x={5\over 2}y \Rightarrow {25\over 4}y^2-5y^2+4y^2=9 \Rightarrow y^2={36\over 21} \Rightarrow y={6\over \sqrt{21}} \\ \Rightarrow x={5\over 2}\cdot {6\over \sqrt{21}} ={15\over \sqrt{21}} \Rightarrow x+y={21\over \sqrt{21}} \Rightarrow 最大值為 \bbox[red, 2pt]{\sqrt{21}}$$
解答:$$公式:a^3+b^3+c^3-3abc =(a+b+c) (a^2+b^2+c^2-ab-bc-ca) \\ 取\cases{a=x\\ b=y\\c=-1} \Rightarrow x^3+y^3-1+3xy =(x+y-1)(x^2+y^2-xy+x+y+1)=0 \\ 由於x^2+y^2-xy+x+y+1 =(x-{1\over 2}y)^2+{3\over 4}y^2+x+y+1 \gt 0, \forall x,y \gt 0 \\ 因此x+y-1=0 \Rightarrow 在x+y=1的條件下求x^2y的最大值 \\ 算幾不等式: {x/2+x/2+y\over 3} \ge \sqrt[3]{x^2y\over 4} \Rightarrow {1\over 3}\ge \sqrt[3]{x^2y\over 4} \Rightarrow {1\over 27}\ge {x^2y\over 4} \Rightarrow x^2y\le \bbox[red, 2pt]{4\over 27}$$
解答:$${a_{n+2}+a_n\over a_{n+1}+2}=2 \Rightarrow a_{n+2}-2a_{n+1}+a_n=4 \Rightarrow (a_{n+2}-a_{n+1}-(a_{n+1}-a_n))=4 \\ \Rightarrow b_{n+1}-b_n=4, 其中b_n=a_{n+1}-a_n \Rightarrow \langle b_n\rangle 為公差為4的等差數列 \\ \Rightarrow b_1=a_2-a_1=4 \Rightarrow b_n=4+4(n-1)=4n \Rightarrow a_n= a_1+\sum_{k=1}^{n-1}b_k = a_1+\sum_{k=1}^{n-1}4k \\=1+4\times {n(n-1)\over 2} \Rightarrow a_n=2n^2-2n+1 = \left( \sqrt 2k-{\sqrt 2\over 2} \right)^2+{1\over 2} \\ \Rightarrow \left( \sqrt 2k-{\sqrt 2\over 2} \right)^2\lt 2k^2-2k+1=a_k \le(\sqrt 2k)^2=2k^2 \Rightarrow \sqrt 2k-{\sqrt 2\over 2}\lt \sqrt{a_k}\le \sqrt 2k \\ \Rightarrow \sum_{k=1}^n \left( \sqrt 2k-{\sqrt 2\over 2} \right) \lt \sum_{k=1}^n \sqrt{a_k} \le \sum_{k=1}^n \sqrt 2k \Rightarrow {\sqrt 2\over 2}n^2\lt \sum_{k=1}^n \sqrt{a_k} \le{\sqrt 2\over 2}(n^2+n) \\ \Rightarrow {\sqrt 2\over 2} \lt {1\over n^2}\sum_{k=1}^n \sqrt{a_k} \le{\sqrt 2\over 2}+{\sqrt 2\over 2n} \\ \lim_{n\to \infty} {\sqrt 2\over 2}=\lim_{n\to \infty} \left( {\sqrt 2\over 2}+{\sqrt 2\over 2n} \right) ={\sqrt 2\over 2},由夾擠定理可知 \lim_{n\to \infty} {1\over n^2}\sum_{k=1}^n \sqrt{a_k} = \bbox[red, 2pt]{\sqrt 2\over 2}$$
解答:$$已知{\cos A\over \cos B}={b\over a}={4\over 3}, 加上正弦定理 {\sin A\over \sin B}={a\over b} \Rightarrow \cases{\sin A={3\over 4}\sin B\\ \cos A={4\over 3} \cos B} \\ \Rightarrow \sin^2 A+\cos^2 A=1={9\over 16}\sin^2 B+{16\over 9}\cos^2 B ={9\over 16}(1-\cos^2 B)+{16\over 9}\cos^2 B \\ \Rightarrow \cos^2 B={9\over 25} \Rightarrow \cos B={3\over 5} (若\cos B\lt 0 \Rightarrow \cos A\lt 0, 三角形兩內角不可能同時為鈍角) \\ \Rightarrow \cos A={3\over 5} \cdot {4\over 3} ={4\over 5 } \Rightarrow \cases{\sin A=3/5\\ \sin B=4/5} \Rightarrow \cos C=-\cos(A+B)=-(\cos A\cos B-\sin A\sin B)=0\\ \Rightarrow \angle C=90^\circ \Rightarrow \cases{a=3\\ b=4} \\ 取\cases{C(0,0) \\ A(0,4) \\ B(3,0)} \Rightarrow 內切圓半徑r={a+b-c\over 2}=1 \Rightarrow 內切圓圓心I(1,1) \\ P在內切圓上 \Rightarrow P(1+\cos \theta, 1+\sin \theta) \Rightarrow d=\overline{PA}^2+ \overline{PB}^2 +\overline{PC}^2 \\= (1+\cos \theta)^2+(\sin \theta-3)^2+(\cos\theta-2)^2+(1+\sin \theta)^2+(1+\cos \theta)^2 +(1+\sin \theta)^2 \\=20-2\sin \theta \Rightarrow \cases{d_{max}=22\\ d_{min}=18} \Rightarrow d_{max}+d_{min} = \bbox[red, 2pt]{40}$$
解答:$$z=x+iy \Rightarrow {z+1\over z-1} ={x+1+iy\over x-1+iy} ={(x+1+iy)(x-1-iy)\over (x-1+iy)(x-1-iy)} 為純虛數 \\ \Rightarrow 分母實部=(x^2-1)+y^2=0 \Rightarrow x^2+y^2=1 \Rightarrow z在單位圓上 \Rightarrow z=\cos \theta +i\sin \theta \\ \Rightarrow |z^2-2z+3|^2 =(z^2-2z+3)(\bar z^2-2 \bar z+3) \stackrel{z\bar z=1}{=} 3(z^2+\bar z^2)-8(z+\bar z)+14\\ =12\cos^2\theta-16\cos \theta+8 \quad (z+\bar z=2\cos \theta , z^2+\bar z^2 =2\cos 2\theta=4\cos^2\theta-2) \\=12(\cos\theta-{2\over 3})^2+{8\over 3} \Rightarrow |z^2-2z+3|^2最小值為{8\over 3} \Rightarrow |z^2-2z+3| 2最小值為 \bbox[red, 2pt]{2\sqrt 6\over 3}$$
解答:$$依\text{Lifting The Exponent Lemma (LTE)},質數p=2,x,y皆為奇數, 指數N為偶數時,\\ v_2(x^N-y^N) =v_2(x-y)+v_2(x+y)+v_2(N)-1, 其中v_2(K)代表數字K含有質因數2的數量\\ 將本題數字\cases{x=3\\ y=1\\ N=2^{2026}} 代入上式\Rightarrow v_2(3^{2^{2026}}-1) =v_2(3-1)+v_2(3+1)+v_2(2^{2026})-1 \\ \Rightarrow v_2(2)+v_2(4)+2026-1= \bbox[red, 2pt]{2028} \\ \href{https://pregatirematematicaolimpiadejuniori.wordpress.com/wp-content/uploads/2016/07/lte.pdf}{LTE參考資料}$$
解答:$$k \in [-1,1] \Rightarrow kx的最小值是-x, \text{ for }x\gt 0 \Rightarrow 5\ln x+x^2-8x+a \le -x\\ \Rightarrow a\le -5\ln x-x^2+7x =f(x) \Rightarrow f'(x)=-{5\over x}-2x+7=0 \Rightarrow 2x^2-7x+5=0 \\ \Rightarrow (2x-5)(x-1)=0 \Rightarrow x=1,5/2 \Rightarrow \cases{f(1)=6\\ f(5/2) =11.25-5\ln 2.5 \approx 6.7\\ f(4)=12-5\ln 4 \approx 5.1} \\ \Rightarrow 最小值發生在x=4 \Rightarrow \max(a)=f(4)=12-5\ln 4= \bbox[red, 2pt]{12-10\ln 2}$$
解答:$$x\cdot [x\cdot [x \cdot[x]]] \approx x^4=114 \Rightarrow x\in [3,4) 或x\in [-4,-3) \\\textbf{Case I: }x\in [3,4) \Rightarrow [x]=3 \Rightarrow x\cdot [x\cdot [3x]]=114 \\\Rightarrow \cases{[3x]=9 \Rightarrow 3\le x\lt 10/3 \Rightarrow x\cdot [9x]=114 \Rightarrow 27\le 9x\lt 30\Rightarrow x\gt 10/3不合 \\ [3x]=10 \Rightarrow 10/3\le x \lt 11/3 \Rightarrow x\cdot[10x]=114,取N=[10x] \Rightarrow x=114/N \\\qquad \Rightarrow [10\cdot (114/N)] =N \Rightarrow [1140/N]=N \Rightarrow N^2\le 1140\lt N(N+1)無解\\ [3x]=11 \Rightarrow 11/3\le x\lt 4 \Rightarrow x[11x]=114, 取N=[11x] \\\qquad \Rightarrow x=114/N \le 114/40=2.85 \lt 11/3不合} \\ \textbf{Case II: }x\in [-4,-3) \Rightarrow 12\lt -4x\le 16 \Rightarrow k=[-4x]可能值為12,13,\dots,16 \\ \qquad \Rightarrow N=[12x]=-36時符合題意，即x= \bbox[red, 2pt]{114/(-37)} \\ 只能一個一個試,沒什麼好方法!!!$$
解答:$$t=\sqrt{x-1}\Rightarrow x=t^2+1 \\\Rightarrow f(t)= (t^2+1)^2-2(t^2+1)-2-16t+(t^2+1) \sqrt{(t^2+1)^2-4(t^2+1)-32t+76} \\\qquad =t^4-16t-3+(t^2+1)\sqrt{t^4-2t^2-32t+73} \\ \Rightarrow f'(t)=0 \Rightarrow (要算很久) t={1+\sqrt{17}\over 4} \Rightarrow x_0=t^2+1 \Rightarrow x_0={17+\sqrt{17} \over 8} \\ \Rightarrow m=f(x_0)=-6 \Rightarrow (x_0,m) = \bbox[red, 2pt]{\left( {17+\sqrt{17}\over 8} ,-6\right)} \\ \bbox[cyan,2pt]{另解}: 取\cases{\vec u=(x-2,2\sqrt{x-1}) \\ \vec v=(x-4, 2\sqrt{x-1}-8)} \Rightarrow \cases{|\vec u|=x\\ |\vec v|=\sqrt{x^2-4x-32\sqrt{x-1}+76} \\ \vec u\cdot \vec v=x^2-2x+4-16\sqrt{x-1}} \\ \Rightarrow f(x)=\vec u\cdot \vec v-6+|\vec u||\vec v| \ge -6 \;(\vec u\cdot \vec v\ge -|\vec u||\vec v|) \Rightarrow m=-6\\ 最小值發生在\vec v=k\vec u \Rightarrow {x-4\over x-2}={2\sqrt{x-1}-8\over 2\sqrt{x-1}} =k \Rightarrow \sqrt{x-1}=2(x-2)\; (x必須大於等於2)\\ \Rightarrow 4x^2-17x+17=0 \Rightarrow x_0={17+\sqrt{17} \over 8} \;({17-\sqrt{17}\over 8}\lt 2不合)\\\Rightarrow (x_0,m) = \bbox[red, 2pt]{\left( {17+\sqrt{17}\over 8} ,-6\right)}$$
解答:$$令 S 為樣本空間，A 為乘積 a 是完全平方數的事件，B 為總和 b 是完全平方數的事件\\ |S|= {13\choose 3}=286\\ 事件A:a=x\cdot y\cdot z \Rightarrow (x,y,z) \in \{(1,4,9), (1,2,8), (4,2,8), (9,2,8), (1,3,12), (4,3,12), (9,3,12),\\ (2,6,12), (8,6,3), (2,10,5), (8,10,5),(2,3,6),(6,8,12)\}，\Rightarrow |A|=13\\ 事件B:\cases{b=9:(1,2,6),(1,3,5),(2,3,4)3種\\ b=16:(1,2,13), (1,3,12), (1,4,11), (1,5,10), (1,6,9), (1,7,8),\\ \qquad (2,3,11), (2,4,10), (2,5,9), (2,6,8),(3,4,9), (3,5,8), \\\qquad(3,6,7),(4,5,7) 14種\\ b=25 : (1,11,13), (2,10,13), (2,11,12), (3,9,13), (3,10,12),\\\qquad (4,8,13), (4,9,12), (4,10,11), (5,7,13), (5,8,12), (5,9,11), \\\qquad(6,7,12), (6,8,11), (6,9,10), (7,8,10) 15 種\\ b=36:(11,12,13)} \Rightarrow |B|=33 \\ A\cap B =(1,3,12) \Rightarrow |A\cap B|=1 \\ \Rightarrow P(A\cup B) ={|A|+|B|-|A\cap B| \over |S|}={13+33-1\over 286} = \bbox[red, 2pt]{45\over 286}$$
解答:$$取\cases{A(0,0,0) \\B(1,0,0) \\C(1/2,\sqrt 3/2,0)} \Rightarrow \cases{A_1(0,0,1) \\B_1(1,0,1) \\C_1(1/2,\sqrt 3/2,1)} \Rightarrow \cases{\overrightarrow{AB_1} =(1,0,1) \\ \overrightarrow{C_1B} =(1/2,-\sqrt 3/2,-1) \\ \overrightarrow{AB}=(1,0,0)} \\ \Rightarrow \vec n= \overrightarrow{AB_1} \times \overrightarrow{C_1B} =({\sqrt 3\over 2},{3\over 2},-{\sqrt 3\over 2}) \Rightarrow d(\overline{AB_1}, \overline{C_1B}) = {\overrightarrow{AB}\cdot \vec n\over |\vec n|} ={\sqrt 3/2 \over \sqrt{15}/2} = \bbox[red, 2pt]{\sqrt 5\over 5}$$
二、計算題（各題配分列於題後，全部 2 大題，共計 28 分）
解答:$$\textbf{(a) }使用反證法，假設a\ge 0 \\|x-1|+3|x-115|=|x-1|+|115-x|+2|x-115 |\ge |(x-1)+(115-x)|+2\cdot 0\\\qquad =114 \Rightarrow f(x)=|x-1|+3|x-115|+a|x-b| \ge 114與f(x)最小值為12矛盾\\ 因此a\gt 0, \bbox[red, 2pt]{故得證} \\\textbf{(b) } x\gt 115 \Rightarrow f(x)=(x-1)+a(x-b)+3(x-115)=(a+4)x-(ab+346)有最小值 \\\qquad \Rightarrow a+4\ge 0 \Rightarrow a=-1,-2,-3,-4\\ f(x)的最小值發生在x=1,b,或115\\ x=1 \Rightarrow f(1)=a|1-b|+342 \Rightarrow \cases{a=-3\\ b=111} \Rightarrow f(1)=12,其他不合\\ x=b \Rightarrow f(b)=344-2b=12 \Rightarrow b=166 \not \lt 115\\ x=115 \Rightarrow 114+a(115-b)=12 \Rightarrow \cases{a=-1 \Rightarrow b=13\\ a=-2 \Rightarrow b=64\\ a=-3\Rightarrow b=81\\ a=-4\Rightarrow b=89.5} \\ \Rightarrow 符合條件的(a,b)= \bbox[red, 2pt]{(-1,13),(-2,64),(-3,81),(-3,111)}$$
解答:$$a_n={n^3\over (n-1)(n-2)} =n+3-{1\over n-1}+{8\over n-2} \\ \Rightarrow S= \sum_{n=3}^{p-1} (n+3)-\sum_{n=3}^{p-1}{1\over n-1}+8\sum_{n=3}^{p-1}{1\over n-2} \\ 定義調和級數H_k=1+{1\over 2}+{1\over 3}+\cdots+{1\over k} \Rightarrow S={1\over 2}(p^2+5p-24)-(H_{p-2}-1)+8H_{p-3} \\ \Rightarrow S={1\over 2}(p^2+5p-22)-H_{p-2}+8H_{p-3}\\={1\over 2}(p^2+5p-22)-H_{p-1}+{1\over p-1}+8 ( H_{p-3}-{1\over p-1}-{1\over p-2})\\={1\over 2}(p^2+5p-22)+7H_{p-1}-{7\over p-1}-{8\over p-2}\\ 由於\cases{{1\over 2}(p^2+5p-22) \equiv -{22\over 2}\equiv -11 {\mod p} \\ H_{p-1} \equiv 0{\mod p} \\ -{7\over p-1} \equiv 7 {\mod p}\\ -{8\over p-2} \equiv 4{\mod p}} \Rightarrow S\equiv -11+0+7+4\equiv 0 {\mod p} \\ 因為 S={b\over a} \Rightarrow p必定為b的因數 \quad \bbox[red, 2pt]{故得證}$$
解題僅供參考，其他教甄試題及詳解
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