新竹縣立湖口高中 115 年 第 2 次正式教師甄選
一、填充題(1 格 4 分，共 80 分)
解答:$$\cases{\alpha+\beta+ \gamma=0\\ \alpha\beta +\beta\gamma+ \gamma\alpha=-3\\ \alpha \beta\gamma=-1} \Rightarrow (\alpha+\beta+ \gamma)^2=\alpha^2+ \beta^2+ \gamma^2+2(\alpha\beta +\beta\gamma+ \gamma\alpha) \\ \Rightarrow 0= \alpha^2+ \beta^2 +\gamma^2-6 \Rightarrow \alpha^2+ \beta^2+\gamma^2=6 \\ 又\cases{\alpha^3-3\alpha+1 =0\\ \beta^3 -3\beta+1=0\\ \gamma^3-3\gamma+1=0} \Rightarrow \alpha^4+ \beta^4+\gamma^4=3(\alpha^2+\beta^2+\gamma^2)-(\alpha+\beta+\gamma) =3\cdot 6-0= \bbox[red, 2pt]{18}$$
解答:$$A= \begin{bmatrix}1& 1& 1\\0& 1& 1\\0& 0& 1 \end{bmatrix} = \begin{bmatrix}1&0&0\\0& 1&0\\0&0& 1 \end{bmatrix}+ \begin{bmatrix}0& 1& 1\\0& 0& 1\\0&0& 0 \end{bmatrix} =I+B \Rightarrow IB=BI=B \\B= \begin{bmatrix}0& 1& 1\\0& 0& 1\\0&0& 0 \end{bmatrix} \Rightarrow B^2 =\begin{bmatrix}0& 0& 1\\0& 0& 0 \\0&0& 0 \end{bmatrix} \Rightarrow B^3= \begin{bmatrix}0& 0& 0\\0& 0& 0 \\0&0& 0 \end{bmatrix} \Rightarrow B^n=0, n\ge 3 \\ A^n=(I+B)^n= \sum_{k=0}^n {n\choose k}B^k ={n\choose 0}I+ {n\choose 1}B +{n\choose 2}B^2+ 0+0\cdots+0 \\= \begin{bmatrix}1&0&0\\0& 1&0\\0&0& 1 \end{bmatrix}+ n\begin{bmatrix}0& 1& 1\\0& 0& 1\\0&0& 0 \end{bmatrix}+{n(n+1)\over 2}\begin{bmatrix}0& 0& 1\\0& 0& 0 \\0&0& 0 \end{bmatrix} = \begin{bmatrix}1& n& n(n+1)/2\\ 0& 1& n\\ 0&0& 1 \end{bmatrix} \\ \Rightarrow \sum_{n=1}^{100} A^n=\begin{bmatrix}\sum_{n=1}^{100}1& \sum_{n =1}^{100}n& \sum_{n=1}^{100}n(n+1)/2\\ 0& \sum_{n=1}^{100}1& \sum_{n =1}^{100}n \\ 0&0& \sum_{n=1}^{100}1 \end{bmatrix} \\ \Rightarrow 所有元素總和= 3\sum_{n=1}^{100}1+2 \sum_{n=1}^{100}+ \sum_{n=1}^{100}{n(n+1)\over 2} =3\cdot 100+2\cdot 5050+171700= \bbox[red, 2pt]{182100}$$
解答:$$假設\int_0^2 f(t)\,dt=C \Rightarrow f(x)=x^2+Cx \Rightarrow C=\int_0^2(t^2+Ct)\,dt = \left. \left[ {1\over 3}t^3+{1\over 2}Ct^2 \right] \right|_0^2={8\over 3}+2C\\ \Rightarrow C=-{8\over 3}\Rightarrow f(x)= \bbox[red, 2pt]{x^2-{8\over 3}x}$$
解答:$$由題意可知\cases{L_1的方向向量\vec u_1=(1,2,a)且通過P(1,-1,2) \\L_2的方向向量\vec u_2 =(1,1,-1) 且通過Q(-1,2,-1)}\\ \Rightarrow \cases{\overrightarrow{PQ}=(-2,3,-3) \\ \vec n=\vec u_1\times \vec u_2=(-a-2,a+1,-1) \Rightarrow |\vec n|=\sqrt{2a^2+6a+6}} \\\Rightarrow d(L_1,L_2) = {|\overrightarrow{PQ}\cdot \vec n| \over |\vec n|} ={|5a+10|\over \sqrt{2a^2+6a+6}} ={5\sqrt 6\over 6} \Rightarrow 2a^2+9a+9=0\\ \Rightarrow (2a+3)(a+3)=0 \Rightarrow a=\bbox[red, 2pt]{-3或-{3\over 2}}$$
解答:
$$s={1\over 2}(13+14+15)=21 \Rightarrow \triangle ABC=\sqrt{21(21-13)(21-14)(21-15)}=84 \\ \Rightarrow \triangle ABC= \triangle ABO+\triangle AOC \Rightarrow {1\over 2}(13r+15r)=84 \Rightarrow r={168\over 28} =\bbox[red, 2pt] 6$$
解答:$$\lim_{n\to \infty} \sum_{k=1}^{3n} {(n+k)^3\over n^4} =\lim_{n\to \infty} \sum_{k=1}^{3n} {(1+(k/n))^3\over n} = \int_0^3 (1+x)^3\,dx = \left. \left[ {1\over 4}(1+x)^4 \right] \right|_0^3 \\={1\over 4}(4^4-1) = \bbox[red, 2pt]{255\over 4}$$
解答:$$\cases{\cos (7\pi/8) =-\cos (\pi/8)\\ \cos(5\pi/8)=-\cos (3\pi/8) \\ \cos(3\pi/8)=\sin(\pi/8)} \Rightarrow \cos^6{\pi\over 8}+ \cos^6{3\pi\over 8}+ \cos^6{5\pi\over 8}+ \cos^6{7\pi\over 8} =2 \left( \cos^6{\pi\over 8}+ \cos^6{3\pi\over 8} \right) \\ =2 \left( \cos^6{\pi\over 8}+ \sin^6{\pi\over 8} \right) = 2 \left[ \left( \cos^2{\pi\over 8}+ \sin^2{\pi\over 8} \right)^3-3\cos^2{\pi\over 8} \sin^2{\pi\over 8}\left( \cos^2{\pi\over 8}+ \sin^2{\pi\over 8} \right) \right] \\=2 \left( 1-3 \cos^2{\pi\over 8} \sin^2{\pi\over 8}\right) =2 \left( 1-{3\over 4} \sin^2{ \pi\over 4}\right) =2 \left( 1-{3\over 8} \right) = \bbox[red, 2pt]{5\over 4}$$
解答:$$\sqrt{x^2-16} \Rightarrow x\ge 4或x\le -4 \Rightarrow \int_{-2\sqrt 3}^2 \sqrt{x^2-16}\,dx =虛數\\ 題目應\bbox[cyan,2pt]{更正}為 \int_{-2\sqrt 3}^2 (\sqrt{16-x^2}+2026xe^{x^2})\,dx\\ 取x=4\sin \theta \Rightarrow dx=4\cos \theta \,d\theta \Rightarrow I_1=\int_{-2\sqrt 3}^2 \sqrt{16-x^2}\,dx = \int_{-\pi/3}^{\pi/6} 16\cos^2\theta\,d\theta \\= 8\int_{-\pi/3}^{\pi/6} (1+\cos 2\theta)\,d\theta = 8 \left. \left[ \theta+{1\over 2}\sin 2\theta \right] \right|_{-\pi/3}^{\pi/6} =4\pi+4\sqrt 3 \\ 取u=x^2 \Rightarrow du=2x\,dx \Rightarrow I_2= \int_{-2\sqrt 3}^2 2026xe^{x^2}\,dx = \int_{12}^4 1013 e^u\,du=1013(e^4-e^{12}) \\ \Rightarrow I_1+I_2= \bbox[red, 2pt]{4\pi+4\sqrt 3+ 1013(e^4-e^{12})} \leftarrow 與公布的答案相同$$
解答:$${1\over x}+{1\over y}= {1\over2026} \Rightarrow 2026y+2026x =xy \Rightarrow xy-2026x-2026y+2026^2=2026^2\\ \Rightarrow (x-2026)(y-2026)=2026^2 =2^2\times 1013^2 \\ \Rightarrow \begin{array}{r|r}x-2026& y-2026 \\\hline 1& 2026^2\\ 2&2\times 1013^2\\ 4& 1013^2\\1013& 4\times 1013\\ 2\times 1013& 2\times 1013 \\\hline \end{array} \Rightarrow 共\bbox[red, 2pt]5組解$$
解答:$$P(z)=(r,0), r\gt 0 \Rightarrow \cases{Q(z^2(\cos 120^\circ+i\sin 120^\circ))=(-{1\over 2}r^2, {\sqrt 3\over 2}r^2) \\ R(z^3(\cos 240^\circ+ i\sin 240^\circ)) =(-{1\over 2}r^3,-{\sqrt 3\over 2}r^3)} \\ \Rightarrow \triangle PQR面積={1\over 2} \begin{Vmatrix} r&0& 1\\-{1\over 2}r^2& {\sqrt 3\over 2}r^2& 1\\ -{1\over 2}r^3& -{\sqrt 3\over 2}r^3& 1 \end{Vmatrix} = {\sqrt 3\over 4}r^3(1+r+r^2) ={\sqrt 3\over 2}\cdot {r^3+2\over r-1} \\ \Rightarrow r^3(r-1)(1+r+r^2)=2(r^3+2) \Rightarrow r^3(r^3-1)=2(r^3+2) \Rightarrow r^6-3r^3-4=0\\ \Rightarrow (r^3-4)(r^3+1)=0 \Rightarrow r^3=4 \Rightarrow r= \bbox[red, 2pt]{\sqrt[3]4}$$
解答:$$f(x)=x^3-3x^2+px+q \Rightarrow f'(x)=3x^2-6x+p \Rightarrow f''(x)=6x-6\\ f''(x)=0 \Rightarrow 6x-6=0 \Rightarrow x=1 \Rightarrow 對稱中心(1,f(1))=(1,p+q-1) 在y=2x-1上\\ \Rightarrow p+q-1=2-1 =1\Rightarrow q=3-p \Rightarrow f(x)= x^3-3x^2+px+(3-p)\\ x^3-3x^2+px+(3-p)=2x-1 \Rightarrow x^3-3x^2+(p-2)x+(4-p)=0\\ x=1為其中一解\Rightarrow (x-1)(x^2-2x+(p-4))=0 \Rightarrow x^2-2x+(p-4)=0有相異實數解\\ \Rightarrow \Delta=4-4(p-4) \gt 0 \Rightarrow \bbox[red, 2pt]{p\lt 5}$$
解答:$$f(x)={x^2+x+1\over x^2-x+1}=k \Rightarrow (k-1)x^2-(k+1)x+(k-1)=0有實數解\\ \Rightarrow (k+1)^2-4(k-1)^2 \ge 0 \Rightarrow (3k-1)(k-3)\le 0 \Rightarrow {1\over 3}\le k\le 3 \Rightarrow \cases{M=3\\ m=1/3} \\ \Rightarrow M-m=3-{1\over 3}= \bbox[red, 2pt]{8\over 3}$$
解答:$$z=\cos \theta+i\sin \theta \Rightarrow {1\over z}=\cos \theta-i\sin \theta \Rightarrow z+{1\over z}=2\cos \theta =2\cos 15^\circ \Rightarrow \theta=15^\circ \\ \Rightarrow z^{2026}+ {1 \over z^{2026}}=2\cos(2026\times 15^\circ) =2\cos 30390^\circ=2\cos 150^\circ = \bbox[red, 2pt]{-\sqrt 3}$$
解答:$$假設{{4-\sqrt 3\over 2}x^2-\sqrt 3x \over x^2+x+1}=k \Rightarrow \left( k-{4-\sqrt 3\over 2} \right)x^2+(k+\sqrt 3)x+k=0 \\ \Rightarrow \Delta=(k+\sqrt 3)^2-4 \left( k-{4-\sqrt 3\over 2} \right)k\ge 0 \Rightarrow 3k^2-8k-3\le 0 \Rightarrow (3k+1)(k-3)\le 0\\ \Rightarrow -{1\over 3}\le k\le 3 \Rightarrow y= \log_3 |k| \le 1 \Rightarrow \bbox[red, 2pt]{y\le 1}$$
解答:$$一扇門被改變狀態的次數，等於該門牌號碼的正因數個數。門要保持「打開」，必須被翻轉奇數次。\\只有完全平方數擁有奇數個正因數。在 1 到 100 中，完全平方數有 10 個，\\即1, 4, 9, 16, 25, 36, 49, 64, 81, 100。在沒有第 9 位同學搗蛋的情況下，\\這 10 扇門最後會是打開的，其餘 90 扇門是關閉的。\\而第 9 位同學負責處理所有9 的倍數的門。因此1-100的完全平方數扣除9的倍數，\\剩下1,4,16,25,49,64,100，共7扇門是保持打開。\\ 1-100中，9的倍數有\lfloor {100\over 9}\rfloor=11個，這個11個數中，有3個完全平方數:9,36,81 \\ 由於第9位同學的影響，11-3=8扇門保持開啟，因此共有7+8=\bbox[red, 2pt]{15}扇門開啟$$
解答:$$假設\cases{跨2階有a次\\ 跨1階有b次} \Rightarrow 2a+b=10\\ \cases{a=0:全部都是跨1階，排列只有1種\\ a=1\Rightarrow b=8\Rightarrow 排列數{9!\over 8!}=9\\ a=2 \Rightarrow b=6 \Rightarrow 在7個空隙中插入2個2階，有C^7_2=21種\\ a=3 \Rightarrow b=4 \Rightarrow 在5個空隙中插入3個2階，有C^5_3=10種\\ a=4 \Rightarrow b=2\Rightarrow 無法達成} \Rightarrow 合計:1+9+21+10=\bbox[red, 2pt]{41}$$
解答:$$取u=\sqrt{x-1} \Rightarrow x=u^2+1 \Rightarrow \sqrt{u^2+4-4u}+ \sqrt{u^2+9-6u} =\sqrt{(u-2)^2} +\sqrt{(u-3)^2}=1 \\ \Rightarrow |u-2|+|u-3|=1 \Rightarrow 2\le u\le 3 \Rightarrow 5\le u^2+1\le 10 \Rightarrow \bbox[red, 2pt]{5\le x\le 10}$$
$$h=d(O, \overline{PQ}) \Rightarrow \triangle OPQ={1\over 2}\cdot \overline{PQ}\cdot h \Rightarrow {1\over 2}\cdot 4\sqrt 2\cdot h=12\Rightarrow h=3\sqrt 2 ={|k|\over \sqrt 2} \Rightarrow |k|=6 \\ \Rightarrow \bbox[red, 2pt]{k=6} \quad(題意:k\gt 0) \Rightarrow L:x+y=6 \Rightarrow M=\overline{PQ}中點=兩直線\cases{x+y=6\\x=y}的交點(3,3) \\ P\in L \Rightarrow P(t,6-t) \Rightarrow \overline{MP}={1\over 2}\cdot 4\sqrt 2 \Rightarrow \sqrt{(t-3)^2+ (3-t)^2}=2\sqrt 2 \Rightarrow (t-3)^2=4 \\ \Rightarrow t=1\; (P在M的左邊\Rightarrow t\lt 3 \Rightarrow t=5不合) \Rightarrow P(1,5) \Rightarrow 5=a^1 \Rightarrow \bbox[red, 2pt]{a=5}$$
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解題僅供參考，其他教甄試題及詳解
解答:$$f(x)={2\sin x+2\cos x+4\over \sin x+\cos x+1}=2+{2\over \sin x+\cos x+1} = 2+{2\over \sqrt 2\sin(x+\pi/4)+1}\\ 已知{\pi\over 2}\le x\le {3\pi\over 4} \Rightarrow {3\pi\over 4}\le x+{\pi\over 4}\le \pi \Rightarrow 0\le \sin(x+{\pi\over 4}) \le {\sqrt 2\over 2}\\ \Rightarrow 1\le \sqrt 2 \sin(x+{\pi\over 4})+1\le 2 \Rightarrow 2+{2\over 2 }\le f(x)\le 2+{2\over 1} \Rightarrow \bbox[red, 2pt]{3\le f(x)\le 4}$$
二、計算證明題(共 20 分)
解答:$$X\sim B(n,p) \Rightarrow \cases{E(X)=np\\ Var(X)=np(1-p)} \Rightarrow E(X^2)=Var(X)+(E(X))^2= n^2p^2-np^2+np \\ E(X(X-1)(X-2)) =\sum_{x=3}^n x(x-1)(x-2){n\choose x}p^x(1-p)^{n-x} =\sum_{x=3}^n {n!\over (x-3)! (n-x)!}p^x(1-p)^{n-x} \\=n(n-1)(n-2)p^3 \sum_{x=3}^n{(n-3)!\over (x-3)!(n-x)!} p^{x-3}(1-p)^{n-x}\\ =n(n-1)(n-2)p^3 \sum_{k=0}^{n-3}{(n-3)!\over k!(n-3-k)!} p^k(1-p)^{n-3-k} =n(n-1)(n-2)p^3(p+(1-p))^{n-3} \\=n(n-1)(n-2)p^3 \Rightarrow E(X(X-1)(X-2))= E(X^3-3X^2+2X)= n(n-1)(n-2)p^3 \\ \Rightarrow E(X^3)=n(n-1)(n-2)p^3+3E(X^2)-2E(X) =n(n-1)(n-2)p^3+3(n^2p^2-np^2+np)-2np \\= \bbox[red, 2pt]{(n^3-3n^2+2n)p^3+(3n^2-3n)p^2+np}$$
解答:$$L_p=\left( {1+\sqrt 5\over 2} \right)^p+ \left( {1-\sqrt 5\over 2} \right)^p ={1\over 2^p} \left[ (1+\sqrt 5)^p+ (1-\sqrt 5)^p \right] ={N\over 2^p} \\ N=(1+\sqrt 5)^p+ (1-\sqrt 5)^p =\sum_{k=0}^p {p\choose k} \left[ ( \sqrt 5)^p+ ( -\sqrt 5)^p \right] =\begin{cases}0,& k為奇數\\ \sum_{k=0}^p {p\choose k} \left[ 2 ( \sqrt 5)^p \right],&k為偶數 \end{cases} \\ p 為奇質數，所以 k 的最大偶數值為 p-1，因此取偶數 k = 2m 代入上式 \\ \Rightarrow N= \sum_{m=0}^{(p-1)/2} {p\choose 2m} \cdot 2(\sqrt 5)^{2m} =2 \sum_{m=0}^{(p-1)/2} {p\choose 2m} \cdot 5^m \\已知 {p\choose k} \equiv 0 {\pmod p}, \text{ for }0\lt k\lt p \Rightarrow N \equiv 2\cdot {p\choose 0}\cdot 5^0 {\pmod p} \equiv 2 {\pmod p} \\ \Rightarrow N=2^pL_p \equiv 2 {\pmod p} \Rightarrow 2 L_p \equiv 2 {\pmod p} \Rightarrow L_p \equiv 1 {\pmod p} \quad \bbox[red, 2pt]{故得證}$$
解題僅供參考，其他教甄試題及詳解
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