$$\log(y-6)+\log(y-4)-\log(x^2+8)=0 \Rightarrow \cases{y-6\gt 0\\ y-4\gt 0} \Rightarrow y\gt 6\\ \log((y-6)(y-4))=\log(x^2+8) \Rightarrow (y-6)(y-4)=x^2+8 \Rightarrow (y-5)^2-x^2=9 \\ \Rightarrow \Omega:{(y-5)^2\over 9}-{x^2\over 9}=1, y\gt 6, 僅雙曲線的上半部\\ C:x^2+(y-3)^2=4 \Rightarrow \cases{\text{center: }O_c(0,3) \\ \text{radius }r=2} \\ \text{Let }P(x,y)\in \Omega \Rightarrow d=d(O_c,P) =\sqrt{x^2+(y-3)^2} \Rightarrow d^2=x^2+(y-3)^2 = \left[ (y-5)^2-9 \right]+(y-3)^2 \\=2y^2-16y+25 \Rightarrow d_{min}^2= 2\cdot 8^2-16\cdot 8+25= 25 \Rightarrow d_{min}=5\; ((0,8)為雙曲線頂點)\\ \Rightarrow \overline{PQ}_{min} =5-r= \bbox[red, 2pt]3$$
解答:$$假設x=10^{100} \Rightarrow a={x^{150} \over x+3}= Q(x)+{3^{150}\over x+3},其中Q(x)={x^{150}-3^{150}\over x+3},餘數{3^{150} \over 10^{100}+3} \lt 1 \\ \Rightarrow Q(10^{100})= \bbox[red, 2pt]{10^{15000}-3^{150} \over 10^{100}+3}$$
解答:$$假設事件 A:右手攤開為人頭，事件B:左手攤開為人頭\\箱中共有 7 枚硬幣，每枚有 2 面，總共有 14 面,其中人頭的面數為:1 \times 2+5 \times 1 = 7 面 \\ \Rightarrow P(A)={7\over 14}={1\over 2} \\考慮抽出的 2 枚硬幣的組合:\cases{抽到「雙面人頭」與「一頭一字」 \Rightarrow 機率=5/42\\ 抽到兩枚「一頭一字」 \Rightarrow 機率=5/42} \\ \Rightarrow P(A\cap B)={5\over 42}+{5\over 42} ={5\over 21} \Rightarrow P(B\mid A)={P(A\cap B) \over P(A)}={5/21\over 1/2}= \bbox[red, 2pt]{10\over 21}$$
解答:$$f(n) = \begin{cases} n-3,& n\ge 1000\\ f(f(n+7)),& n\lt 1000\end{cases} \Rightarrow \cases{f(999)=f( f(1006)) =f(1003)=1000\\ f(998) = f(f(1005)) = f(1002) = 999 \\f(997) = f(f(1004)) = f(1001) = 998\\ f(996) = f(f(1003)) = f(1000) = 997 \\ f(995) = f(f(1002)) = f(999)=1000\\ \cdots} \\ \Rightarrow 循環數4 \Rightarrow f(n) = \begin{cases} 1000,& n\equiv 3 \text{ (mod 4)} \\ 999,& n\equiv 2 \text{ (mod 4)} \\ 998,& n\equiv 1 \text{ (mod 4)} \\ 997,& n\equiv 0 \text{ (mod 4)} \\\end{cases} \Rightarrow f(n)=997 + (n \text{ mod 4}), n\lt 1000 \\ 90=22\times 4+2 \Rightarrow f(90)=997+2= \bbox[red, 2pt]{999}$$
解答:$$\text{Let $A$ be the $2 \times 2$ matrix corresponding to the transformation $f$.} \\ \cases{(f\circ f)(\vec u)=f(\vec v) \\ (f\circ f)(\vec v)=f(\vec u)} \Rightarrow \cases{A^2 \vec u=A\vec v\\ A^2\vec v=A\vec u} \Rightarrow \cases{A^3\vec u=A^2\vec v =A\vec u \Rightarrow (A^3-A)\vec u=0 \\A^3\vec v=A^2\vec u=A\vec v \Rightarrow (A^3-A)\vec v=0} \\ \Rightarrow A^3-A= A(A-I)(A+I) =0 \Rightarrow 特徵值\lambda=0,1,-1 \\A = \begin{bmatrix}a& b\\c& d \end{bmatrix} \xrightarrow{條件(2)} \begin{bmatrix}a& b\\c& d \end{bmatrix} \begin{bmatrix}1\\1 \end{bmatrix} = \begin{bmatrix}a+b\\ c+d \end{bmatrix}= \begin{bmatrix}1\\0 \end{bmatrix} \Rightarrow \cases{a+b=1\\ c+d= 0} \Rightarrow \cases{b=1-a\\ d=-c} \\ \Rightarrow A= \begin{bmatrix}a& 1-a\\ c& -c \end{bmatrix} \Rightarrow \cases{tr(A)=a-c\\ \det(A)=-ac-c(1-a)=-c}\\ A為2\times 2矩陣,但特徵值有3種可能0,1,-1，因此有以下可能: \\ \textbf{Case I }\cases{\lambda_1=1\\ \lambda_2=0} \Rightarrow \cases{tr(A)=1=a-c\\ \det(A)=0 =-c} \Rightarrow \cases{a=1\\ c=0} \Rightarrow A_1 = \begin{bmatrix}1& 0\\0& 0 \end{bmatrix} \\\textbf{Case II }\cases{\lambda_1= -1\\ \lambda_2=0} \Rightarrow \cases{tr(A)= -1=a-c\\ \det(A)=0 =-c} \Rightarrow \cases{a=-1\\ c=0} \Rightarrow A_2 = \begin{bmatrix}-1& 2\\ 0& 0 \end{bmatrix} \\\textbf{Case III }\cases{\lambda_1=1\\ \lambda_2= -1} \Rightarrow \cases{tr(A)=0 =a-c\\ \det(A)= -1 =-c} \Rightarrow \cases{a=1\\ c=1} \Rightarrow A_3= \begin{bmatrix}1&0\\1& -1 \end{bmatrix} \\ \Rightarrow A_1+A_2+A_3 = \bbox[red, 2pt]{\begin{bmatrix}1& 2\\1& -1 \end{bmatrix}}$$
解答:$$\cases{E_1: x+y+z=5，法向量 \vec{n}_1 = (1, 1, 1) \\ E_2: x+y+z=11，法向量 \vec{n}_2 = (1, 1, 1) \\E_3: x-2y+z=3，法向量 \vec{n}_3 = (1, -2, 1) \\ E_4: x-z=5，法向量 $\vec{n}_4 = (1, 0, -1) \\E_5: x-y=2，法向量 \vec{n}_5 = (1, -1, 0)} \Rightarrow E_1 \parallel E_2 \Rightarrow E_1與E_2為三角柱的底面\\ \Rightarrow \vec n_3\cdot \vec n_1=\vec n_4\cdot \vec n_1= \vec n_5 \cdot \vec n_1=0 \Rightarrow 直角三角柱 \Rightarrow 三角柱的高h=d(E_1,E_2) ={11-5\over \sqrt 3}=2\sqrt 3 \\ \Rightarrow \cases{A= E_3\cap E_4\cap E_1 =(14/3,2/3,-1/3) \\B=E_4\cap E_5\cap E_1 =(4,2,-1) \\C=E_3\cap E_5\cap E_1=(8/3,2/3,5/3)} \Rightarrow 底面積={1\over 2} |\overrightarrow{AB} \times \overrightarrow{AC}| ={4\sqrt 3\over 3} \\ \Rightarrow V=底面積\times 高={4\sqrt 3\over 3}\times 2\sqrt 3= \bbox[red, 2pt]8$$
解答:
$$ACED共圓(\Gamma_1) \Rightarrow \cases{\angle DEC= 180^\circ-\angle A=120^\circ \Rightarrow \angle BED=60^\circ\\ \angle ADE=180^\circ-\angle C=120^\circ \Rightarrow \angle BDE=60^\circ} \Rightarrow \triangle BDE為正\triangle \\ 假設\overline{AD}=x \Rightarrow \overline{BD}=3-x \Rightarrow \Gamma_2半徑R_2={3-x\over \sqrt 3} \\ \triangle ACD: \overline{CD}^2=3^2+x^2-2\cdot 3\cdot x\cos 60^\circ =x^2-3x+9 \\ \Rightarrow 2R_1={\overline{CD} \over \sin A}= {\sqrt{x^2-3x+9} \over \sqrt 3/2} \Rightarrow \Gamma_1半徑R_1= \sqrt{x^2-3x+9\over 2}\\ 兩圓面積和S= \pi R_1^2+ \pi R_2^2= \pi \left[ \left(\frac{\sqrt{x^2 - 3x + 9}}{\sqrt{3}}\right)^2 + \left(\frac{3-x}{\sqrt{3}}\right)^2 \right] =f(x)={\pi\over 3}(2x^2-9x+18) \\ \Rightarrow f(9/4)= \bbox[red, 2pt]{21\pi\over 8}$$
解答:$$先計算「不考慮首尾限制」的直線排列數:\\
從 n 個排成一列的箱子中，選取 k 個不相鄰的箱子放球，其方法數=C^{n-k+1}_k= C^{10-4+1}_4=35 種\\扣除「首尾同時放球」的違規情形: C^{6-2+1}_2=10種\\ 合法的方法數35-10= \bbox[red, 2pt]{25}種$$
從 n 個排成一列的箱子中，選取 k 個不相鄰的箱子放球，其方法數=C^{n-k+1}_k= C^{10-4+1}_4=35 種\\扣除「首尾同時放球」的違規情形: C^{6-2+1}_2=10種\\ 合法的方法數35-10= \bbox[red, 2pt]{25}種$$
解答:$$n^2+8n-1 \equiv 0 \text{ (mod 13)} \Rightarrow (n+4)^2-17\equiv 0 \text{ (mod 13)} \Rightarrow (n+4)^2 \equiv 4\text{ (mod 13)} \\ \Rightarrow \cases{n+4 \equiv 2 \text{ (mod 13)} \Rightarrow n \equiv 11 \text{ (mod 13)} \Rightarrow n=13k+11\\ n+4 \equiv 11 \text{ (mod 13)} \Rightarrow n \equiv 7 \text{ (mod 13)} \Rightarrow n=13k+7} \\ \Rightarrow \cases{1\le 13k+11\le 2026 \Rightarrow k=0,1,2,\dots,155，共156個\\ 1\le 13k+7\le 2026 \Rightarrow k=0,1,2,\dots, 155，共156個} \Rightarrow 合計 \bbox[red, 2pt]{312}個$$
解答:$$設 \omega 為方程式 x^2 - x + 1 = 0 的根，即 \omega^2 - \omega + 1 = 0 \Rightarrow (\omega+1)(\omega^2-\omega+1)= \omega^3+1=0 \\ \Rightarrow \omega^3=-1 \Rightarrow f(\omega)= \omega^{2026}+a \omega^{100}+b= (-\omega)+a(-\omega)+b=0 \Rightarrow (-1-a)\omega +b=0\\ \Rightarrow \cases{-1-a=0 \Rightarrow a=-1\\ b=0} \Rightarrow (a,b)= \bbox[red, 2pt]{(-1,0)}$$
解答:$$f(n)=k \in \mathbb Z \Rightarrow k\le \log_2 n\lt k+1 \Rightarrow 2^k\le n\lt 2^{k+1} \\ 在區間[2^k,2^{k+1}-1]包含2^{k+1}-2^k=2^k個整數\\ 2^{10} \le 2026\lt 2^{11} \Rightarrow f(2026)=10 \Rightarrow k=0-9為完整組別 \\ \Rightarrow \sum_{k=0}^9k\cdot 2^k = 8194, 接著計算k=10 (n=1024-1026) 的項數為2026-1024+1=1003 \\ \Rightarrow \sum_{n=1}^{2026}f(n)=8194+1003\times 10= \bbox[red, 2pt]{18224}$$
解答:$$\omega^7 =1 \Rightarrow \omega^{7-k} ={\omega^7\over \omega^k}={1\over \omega^k} \Rightarrow {1\over 1+\omega^k}+ {1\over 1+\omega^{7-k}} = {1\over 1+\omega^k}+ {1\over 1+1/\omega^{k}} \\ = {1\over 1+\omega^k}+ {\omega^k \over \omega^{k}+1}=1 \Rightarrow {1\over \omega}+{1\over 1+\omega^6} = {1\over \omega^2}+{1\over 1+\omega^5} = {1\over \omega^3}+{1\over 1+\omega^4} =1 \\ \Rightarrow \sum_{k=1}^6 {1\over 1+\omega^k}= \bbox[red, 2pt]3$$
解答:$$f(x)={x\over \ln x} \Rightarrow f'(x)= {\ln x-1\over (\ln x)^2} =0 \Rightarrow \ln x=1 \Rightarrow x=e \\ \Rightarrow f(e)={e\over 1} =\bbox[red, 2pt]e$$
解答:$$V= \int 2\pi xf(x)\,dx = \int_0^{\pi} 2\pi x\sin x\,dx =2\pi \left. \left[ -x\cos x+\sin x \right] \right|_0^\pi =2\pi \cdot \pi = \bbox[red, 2pt]{2\pi^2}$$
二、計算題： 6 題，每題 5 分（共 30 分）
解答:$$利用柯西不等式的\text{Titu }引理\text{(Titu's Lemma)}: {x_1^2\over y_1} + {x_2^2\over y_2} +\cdots+ {x_n^2\over y_n} \ge {(x_1+x_2+ \cdots+x_n)^2 \over y_1+y_2 + \cdots+ y_n} \\ 取\cases{x_i=p_i\\ y_i=p_i+ q_i + r_i} 代入上式 \Rightarrow \sum_{i=1}^n {p_i^2\over p_i+ q_i +r_i} \ge {(\sum_{i=1}^n p_i)^2 \over \sum_{i=1}^n (p_i+ q_i+ r_i)} ={(\sum_{i=1}^n p_i)^2 \over 3\sum_{i=1}^n p_i } ={\sum_{i=1}^n p_i \over 3 } \\ \Rightarrow \sum_{i=1}^n {p_i^2\over p_i+ q_i +r_i} \ge{1\over 3}\sum_{i=1}^n p_i, \bbox[red, 2pt]{故得證}$$
解答:$$令 S_k = \sum_{i=1}^n a_i^k 為根的 k 次方和 \Rightarrow \sum_{i=1}^n (a_i^{n-1} - a_i^n) = \sum_{i=1}^n a_i^{n-1} - \sum_{i=1}^n a_i^n = S_{n-1} - S_n \cdots(1)\\ a_i是x^n+x+1=0的根\Rightarrow a_i^n=-a_i-1 \Rightarrow S_n= \sum_{i=1}^n (-a_i-1)=-S_1-n \\a_i^n+a_i+1=0 \Rightarrow a_i^{n-1}+1+{1\over a_i}=0 \Rightarrow a_{i}^{n-1}=-1-{1\over a_i} \Rightarrow S_{n-1}= \sum_{i=1}^n \left( -1-{1\over a_i}\right) \\=-n-\sum_{i=1}^n{1\over a_i} =-n- \left( -{一次項係數\over 常數項} \right) =-n-(-1)=1-n\\ 將S_n及S_{n-1}代入(1)可得: S_{n-1}-S_n=(1-n)-(-S_1-n)=1+S_1 (所有根的和)=1+0=1 \\ \Rightarrow \sum_{i=1}^n (a_i^{n-1} - a_i^n) = 1, \bbox[red, 2pt]{故得證} $$
解答:$$對於任意多項式P(x)及兩個變數 a 和 b，恆有 (a - b) \mid (P(a) - P(b)) \\ \cases{P(z)=f(z)\\a=f(z)\\ b=0} \Rightarrow f(z) -0\mid f(f(z))-f(0) \Rightarrow f(z)可以整除 f(f(z))-f(0) \\ 又\cases{a=f(f(z)) \\b=f(0)} \Rightarrow f(f(z))-f(0) \mid f(f(f(z)))-f(f(0)) \Rightarrow f(z) \mid f(f(f(z)))-f(f(0)) \cdots(1)\\ 假設f(z)=0的所有根為r_1,r_2,\dots, r_n \Rightarrow f(z)=(z-r_1)(z-r_2)\cdots (z-r_n) \\ \Rightarrow f(0)=(-1)^n(r_1r_2\cdots r_n) \xrightarrow{條件(2)} f(0)=(-1)^n \cdot 1=(-1)^n \Rightarrow f(0) \in\{1,-1\} \\ 條件1: z^2-1可以整除f(z) \Rightarrow 1和-1都是f(z)=0的根 \Rightarrow f(1)=f(-1)=0 \\ \Rightarrow f(f(0))=f(1或-1)=0 代回(1) \Rightarrow f(z) \mid f(f(f(z)))-0 \Rightarrow f(z)可以整除 f \circ f\circ f(z) \; \bbox[red, 2pt]{故得證}$$
解答:$$利用數學歸納法證明\\ n=1 \Rightarrow \cases{a_1=1\\ \sqrt{2\cdot 1-1}=1} \Rightarrow a_1\ge \sqrt{2\cdot 1-1} \Rightarrow 不等式成立\\ 假設n=k時不等式成立, 即a_k\ge \sqrt{2k-1} \\ n=k+1 \Rightarrow a_{k+1}=a_k+{1\over a_k} \Rightarrow a_{k+1}^2= a_k^2+2+{1\over a_k^2}\ge a_k^2+2 \ge (\sqrt{2k-1})^2+2= 2k+1 \\ \qquad \Rightarrow a_{k+1}\ge \sqrt {2k+1} =\sqrt{2(k+1)-1} \Rightarrow 不等式成立\\ 由數學歸納法可知，對所有正整數 n ， a_n \geq \sqrt{2n - 1} 恆成立, \bbox[red, 2pt]{故得證}$$
解答:
$$\overline{AE}為外接圓直徑\Rightarrow \cases{\overline{BE} \bot \overline{AB} \\ \overline{CE} \bot \overline{AC}}, 又H為垂心\Rightarrow \cases{\overline{CH} \bot \overline{AB} \\ \overline{BH} \bot \overline{AC}} \Rightarrow \cases{\overline{BE} \parallel \overline{CH} \\ \overline{CE} \parallel \overline{BH}} \\ \Rightarrow HBEC為平行四邊形\; \bbox[red, 2pt]{故得證}$$
解答:$$k\ge 2, k\in \mathbb N \Rightarrow k^2 \gt k(k-1) \Rightarrow {1\over k^2} \lt {1\over k(k-1)}={1\over k-1}-{1\over k} \\ \Rightarrow {1\over 2^2}+{1\over 3^2}+\cdots+{1\over n^2} \lt \left( {1\over 1}-{1\over 2} \right)+ \left( {1\over 1}-{1\over 3} \right) +\cdots+ \left( {1\over n-1}-{1\over n} \right)=1-{1\over n} \\ \Rightarrow {1\over 2^2}+{1\over 3^2}+\cdots+{1\over n^2} \lt1-{1\over n}, \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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