114年全國高中教甄聯招-數學詳解

 教育部受託辦理114學年度

公立高級中等學校教師甄選

第一部分：選擇題（ 共40分）
一、單選題（ 每題2分， 共22分）

解答:$$\cases{a_1+a_2+ \cdots +a_{10} =60\\ a_2+a_4+ \cdots+ a_{10} =40} \Rightarrow  \cases{ \displaystyle a_1\cdot {1-r^{10} \over 1-r} =60 \cdots (1) \\ \displaystyle a_1r{1-r^{10}\over 1-r^2} =40 \cdots(2)} \\  \Rightarrow {(1) \over (2)} ={1+r\over r} ={3\over 2} \Rightarrow r=2 \Rightarrow a_1\cdot (2^{10} -1) =60 \Rightarrow a_1={60\over 1023} \approx 0.059，故選\bbox[red, 2pt]{(A)}$$

解答:$$乘積=\cases{1:(1,1,1,1) \Rightarrow 排列數=1\\ 8:\cases{(1,1,2,4) \Rightarrow 排列數=12\\ (1,2,2,2) \Rightarrow 排列數=4} \\27:(1,3,3,3) \Rightarrow 排列數=4 \\ 64:\cases{(1,4,4,4) \Rightarrow 排列數=4\\ (2,2,4,4)\Rightarrow 排列數=6}\\ 125:(1,5,5,5)\Rightarrow 排列數=4\\ 216:\cases{(1,6,6,6)\Rightarrow 排列數=4 \\ (2,3,6,6) \Rightarrow 排列數=12\\ (3,3,4,6) \Rightarrow 排列數=12\\ }} \Rightarrow 合計63 \Rightarrow 機率={63\over 6^4} \\ \Rightarrow X\sim Geo(p={63\over 6^4}) \Rightarrow E(X)={1\over p} ={6^4\over 63} ={144\over 7}\approx 20.57，故選\bbox[red, 2pt]{(C)}$$

解答:$$f(x) = \cos(2x)-\sqrt 3\sin(2x) =2\left( \sin{\pi\over 6} \cos(-2x) + \cos{\pi\over 6} \sin(-2x)\right) =2\sin(-2x+{\pi\over 6}) \\ \Rightarrow f(x+{5\pi\over 6}) =2\sin(-2x+\pi) =2\sin(2x) \Rightarrow g(x) = 2\sin(2x)-|x|\\ y=g(x)與x軸交點數相當於求兩圖形\cases{\Gamma_1:y=2\sin(2x)\\ \Gamma_2: y=|x|}的交點數\\ 除了(0,0)外，由於{\pi\over 2}\lt 2\lt {3\pi\over 4},因此在區間({\pi\over 4},{\pi\over 2})有另一交點，因此共有兩交點，故選\bbox[red, 2pt]{(B)}$$

解答:$${所取2張都是紅心且遺失的也是紅心\over 所取2張都是紅心且遺失的也是紅心+所取2張都是紅心且遺失的不是紅心} \\={C^{12}_2/C^{51}_2\times{1\over 4} \over C^{12}_2/C^{51}_2\times{1\over 4}+ C^{13}_2/C^{51}_2 \times{3\over 4}} ={11/850 \over 11/850 +39/850} ={11\over 50}，故選\bbox[red, 2pt]{(B)}$$

解答:$$f(x) =x(x^2+1)(x^3+x+2) \Rightarrow f'(x)= 6x^5+8x^3+6x^2+2x+2 \\\Rightarrow f''(x)=30x^4+24x^2 +12x+2 \gt 0  \Rightarrow f'(x) 遞增,又\cases{f'(x)為5次式 \\ f'(0)=2}, \\因此f'(x)=0恰有一解,假設為k,即f'(k)=0\\ \Rightarrow \int_0^a f'(x)=0 \Rightarrow a=0, a=m\lt k \left( -\int_m^k f'(x)\,dx =\int_k^0 f'(x)\,dx\right) \Rightarrow a有兩個，故選\bbox[red, 2pt]{(B)}$$

解答:$$f(x)=\left( x+{1\over \sqrt[5]x}\right)^n \Rightarrow 係數和為4096=f(1)=2^n \Rightarrow n=12 \\ \Rightarrow f(x)= \left( x+{1\over \sqrt[5]x}\right)^{12} =\sum_{k=0}^{12} C^{12}_k x^{12-k}x^{-k/5}  =\sum_{k=0}^{12} C^{12}_k x^{12-6k/5} \\ 12-{6\over 5}k 為整數\Rightarrow k=0,5,10 \Rightarrow \cases{整數有3個\\ 非整數有10個} \\ 先將10個非整數任排(10!)，在11個間隔中插入3個整數有C^{11}_3\cdot 3!排列數\\ \Rightarrow 機率為{10!\times C^{11}_3\cdot 3!\over 13!}={15\over 26}，故選\bbox[red, 2pt]{(D)}$$

解答:$$\left|{x\over 3} +{y\over 4}-{114\over 5} \right| ={1\over 60}|20x+15y-1368| \\ x,y\in \mathbb Z \Rightarrow 20x+15y為5的倍數 \Rightarrow 20x+15y-1368 可能為 2,3,8 \Rightarrow 取最小值2 \\ \Rightarrow 20x+15y-1368=2 \Rightarrow 20x+15y=1370 \Rightarrow 4x+3y=274 \Rightarrow 有解\cases{x=1\\ y=90} \\ \Rightarrow \left|{x\over 3} +{y\over 4}-{114\over 5} \right|最小值={1\over 60}\cdot 2={1\over 30}，故選\bbox[red, 2pt]{(C)}$$

解答:$$rref\left( \left[\begin{matrix}1 & 2 & 3 & 14\\2 & 3 & 1 & 11\\3 & 1 & 2 & 11\end{matrix}\right]\right) = \left[\begin{matrix}1 & 0 & 0 & 1\\0 & 1 & 0 & 2\\0 & 0 & 1 & 3\end{matrix}\right] \Rightarrow \cases{x=1\\ y=2\\ z=3}\\ \left[\begin{matrix}1 & 1 & 0 & \alpha\\0 & 1 & 1 & \beta\\1 & 0 & 1 & \gamma\end{matrix}\right] \Rightarrow \cases{x+y= \alpha\\ y+z=\beta\\ x+z=\gamma} \Rightarrow \cases{\alpha= 3\\ \beta=5\\ \gamma=4}，故選\bbox[red, 2pt]{(B)}$$

解答:$$a_n={2^3+1\over 2^3-1} \times {3^3+1\over 3^3-1} \times {4^3+1\over 4^3-1} \times \cdots \times {n^3+1\over n^3-1}  \\\qquad = {3\cdot 3\over 1\cdot 7} \times {4\cdot 7 \over 2\cdot 13} \times {5\cdot 13\over 3\cdot 21} \times \cdots \times {(n+1) (n^2-n+1)\over (n-1)(n^2+n+1)}  \\\qquad ={3\cdot (n+1)!/2 \over (n-1)!(n^2+n+1)} ={3\over 2}\cdot {(n+1)n\over n^2+n+1} \\ \Rightarrow \lim_{n\to \infty} a_n={3\over 2}，故選\bbox[red, 2pt]{(C)}$$

解答:$$a_n=3a_{n-1}-2(-1)^{n-1} \Rightarrow a_n+ k(-1)^{n} =3(a_{n-1}+k(-1)^{n-1}) \Rightarrow a_n =3a_{n-1} +3k(-1)^{n-1}-k(-1)^n \\ \Rightarrow -2(-1)^{n-1}= 3k(-1)^{n-1} -k(-1)^n =4k(-1)^{n-1} \Rightarrow k=-{1\over 2} \\ \Rightarrow a_n-{1\over 2}(-1)^n =3(a_{n-1}-{1\over 2}(-1)^{n-1}) \Rightarrow 取b_n= a_n-{1\over 2}(-1)^n \Rightarrow b_n=3b_{n-1},b_1=1  \\ \Rightarrow b_n=3^{n-1} \Rightarrow b_{114} =3^{113} =a_{114}-{1\over 2}(-1)^{114} \Rightarrow a_{114} =3^{113}+{1\over 2} \\ \Rightarrow \log(3^{113}) =113\log 3=113\cdot 0.4771= 53.9123 \Rightarrow a_{113} 是54位數，故選\bbox[red, 2pt]{(A)}$$

解答:

$$此題圖形簡單，可直接手繪求解，故選\bbox[red, 2pt]{(A)}$$

二、複選題（每題3分，共18分， 全對才給分）

解答:$$(A)\bigcirc: 1\gt \tan A\tan B\gt 0\Rightarrow \cases{\tan A\gt 0\\ \tan B\gt 0} \Rightarrow \tan C=\tan(\pi-(A+B))=-\tan(A+B) \\\qquad =-{\tan A+ \tan B\over 1-\tan A\tan B}\lt 0 \Rightarrow C為鈍角\\ (B)\bigcirc: \sin A+\cos A=\sqrt 2(\sin A\cos 45^\circ+\sin 45^\circ \cos A) =\sqrt 2\sin(45^\circ+A) ={1\over 4} \Rightarrow \sin(A+45^\circ) ={1\over 4\sqrt 2} \\\qquad \Rightarrow A+45^\circ \gt 135^\circ \Rightarrow A\gt 90^\circ \Rightarrow A為鈍角\\ (C) \bigcirc: \cases{若C為鈍角\Rightarrow \cos C=-\cos(A+B) \gt 0矛盾 \\ 若A為鈍角 \Rightarrow \sin C= \sin(A+B) \lt 0 矛盾} \Rightarrow \triangle ABC為銳角\triangle\\(D)\times: 若A為鈍角,仍符合\cases{\sin A=5/6\\ \sin B=4/5\\ \cos A=-\sqrt{11}/6\\ \cos B=3/5} 且\cases{\cos C=-\cos(A+B) \gt 0\\ \sin C=\sin(A+B)\gt 0}\\，故選\bbox[red, 2pt]{(ABC)}$$

解答:$$(A)\bigcirc: 若\lim_{x\to 0} f(x)=L\lt \infty 存在 \Rightarrow \lim_{x\to 0}\left(f(x)+{|x|\over x} \right) =L+\lim_{x\to 0}{|x|\over x} 不存在，矛盾\\ (B)\times: f(x)=x \Rightarrow \lim_{x\to 0}\left(f(x) \cdot {|x|\over x} \right) =\lim_{x\to 0} |x|=0存在,但 \lim_{x\to 0} f(x) =0亦存在\\ (C)\bigcirc: \lim_{x\to 0}\left(f(x) \cdot {|x|\over x} \right)=L\lt \infty 存在\Rightarrow \lim_{x\to 0}\left(f(x) \cdot {|x|\over x} \right)^2 =L^2 \Rightarrow  \lim_{x\to 0}\left((f(x))^2 \cdot {|x|^2\over x^2} \right) \\\qquad =\lim_{x\to 0} (f(x))^2 =L^2存在 \\ (D)\times: 令f(x)=\begin{cases}0 & x\lt 0\\ x+1& x \gt 0 \end{cases},則\lim_{x\to 0}f(x)不存在, 但\lim_{x\to 0}\left( f(x)\cdot {[x]\over x}\right) =0存在\\，故選\bbox[red, 2pt]{(AC)}$$

解答:$$\cases{333\cdots3=3+3\cdot 10+3\cdot 10^2+\cdots+3\cdot 10^{98} ={1\over 3}(10^{99}-1)\\ 666\dots6 = 6+ 6\cdot 10+\cdots+6\cdot 10^{98} ={2\over 3}(10^{99}-1)} \\\Rightarrow \overbrace{333\cdots3}^{99個} \times \overbrace{666\cdots 6}^{99個}={2\over 9}(10^{99}-1)^2 =2\times \overbrace{11\cdots 1}^{99個} \times \overbrace{99\cdots 9}^{99個} =2\times \overbrace{11\cdots 1}^{99個} \times (1\overbrace{00\cdots 0}^{99個} -1) \\=  \overbrace{22\cdots 2}^{99個}  \overbrace{00\cdots 0}^{99個} -\overbrace{22\cdots 2}^{99個} = \overbrace{22\cdots 2}^{98個} 1 \overbrace{77\cdots 7}^{97個}8 \\(A)\times: n=99\times 2=198\\ (B)\bigcirc\\ (C)\times:最高位數字為2\\ (D)\bigcirc\\，故選\bbox[red, 2pt]{(BD)}$$

解答:$$y=x^3+ax^2+a \Rightarrow y' =3x^2+2ax\\ 假設切點P(x_0,y_0)=(x_0, x_0^3+ax_0^2+a) \Rightarrow 切線斜率= 3x_0^2+2ax_0 \\ 假設原點O(0,0) \Rightarrow \overline{OP}斜率=切線斜率 \Rightarrow {x_0^3+ax_0^2+a\over x_0} =3x_0^2+2ax_0 \\ \Rightarrow 2x_0^3+ax_0^2-a=0有三相異實數解,因此取f(x)=2x^3+ax^2-a \Rightarrow f'(x)=6x^2+2ax =0 \\\Rightarrow 2x(3x+a)=0 \Rightarrow x=0, -a/3 \Rightarrow f(0)f(-a/3)\lt 0 \Rightarrow (-a)({a^3\over 9}-a) \lt 0 \\ \Rightarrow (-a^2)({a^2\over 9}-1)\lt 0 \Rightarrow {a^2\over 9}-1\gt 0 \Rightarrow a\gt 3,a\lt -3 \\ \Rightarrow \cases{(A)\bigcirc: \pi\approx 3.14\gt 3\\ (B) \bigcirc: \sqrt{2025} =45 \gt 3 \\(C) \times: \log 114\lt \log 1000=3 \\(D) \bigcirc: \displaystyle {2025\over 114} \gt {342\over 114}=3}，故選\bbox[red, 2pt]{(ABD)}$$

解答:$$(A)\bigcirc:a_n的個位數不可能為5，而b_n的個位數皆是5,因此\langle c_n\rangle 各項均異\\ (B)\times: 若a_{30} =c_{45}代表c_1-c_{45}中有30個a_n,15個b_n,而\cases{b_{15}= 5^{15} \Rightarrow \log b_{15} =15(1-0.301)= 10.485 \\ a_{30} =2^{30} \Rightarrow \log a_{30} =30\times 0.301= 9.03}\\ \qquad  \Rightarrow a_{30} \lt b_{15} \quad 矛盾 \\(C) \times: 若b_{10} =c_{30}代表c_1-c_{30}中有10個b_n,20個a_n,而\cases{b_{10} =5^{10} \Rightarrow \log b_{10} =6.99\\a_{20}=2^{20} \Rightarrow \log a_{20} =6.02 \\a_{21} =2^{21} \Rightarrow \log a_{21}=6.321\\ a_{22}=2^{22} \Rightarrow \log a_{22}=6.622 \\ a_{23}= 2^{23} \Rightarrow \log a_{23} =6.923} \\\qquad \Rightarrow 比b_{10}小的a_n有23個，所以b_{10} 在\langle c_n\rangle排名不是第30個,應該是c_{33} \\(D) \bigcirc: \cases{c_k=a_{20} =2^{20} \\c_{k+h}=a_{30} =2^{30}} \Rightarrow \cases{c_1-c_k 中有k-20個b_n\\ c_1-c_{k+h}中有k+h-30個b_n} \Rightarrow \cases{\log 5^{k-20} \lt \log 2^{20} \\ \log 5^{k+h-30} \lt \log 2^{30}} \\ \qquad \Rightarrow \cases{k-20 \lt 8.612 \\ k+h-30\lt 12.918} \Rightarrow h-10\lt 4.30 \Rightarrow h\lt 14.30 \Rightarrow h=14(可驗算)\\，故選\bbox[red, 2pt]{(AD)}$$

解答:

$$(A)\bigcirc: \overleftrightarrow{OB}是\angle B的角平分線 \Rightarrow 對稱點A'\in \overleftrightarrow{OB} \\(B)\times: 過A且與\overleftrightarrow{OB}垂直的直線\overleftrightarrow{AA'}: x-y=6 \Rightarrow \overleftrightarrow{OB}\cap \overleftrightarrow{AA'}=(4,-2)= (A+A')/2 \\ \qquad \Rightarrow A'=(6,0) \\(C) \times:假設A對\overleftrightarrow{OC}的對稱點為 \Rightarrow A''=({2\over 5}, {4\over 5}) \Rightarrow \overleftrightarrow{BC} =\overleftrightarrow{A'A''}: x+7y=6 \\ (D)\bigcirc: 圓心O =\overleftrightarrow{OB} \cap \overleftrightarrow{OC}=(3,-1) \Rightarrow  圓半徑r=d(O, \overleftrightarrow{BC} ) =\sqrt 2 \\ \qquad \Rightarrow 圓方程式: (x-3)^2+ (y+1)^2=2  \Rightarrow x^2+y^2-6x +2y+8=0\\，故選\bbox[red, 2pt]{(AD)}$$

第二部分： 綜合題（ 共60分）
一、 填充題（每題4分，共36分）

解答:$$平面E的法向量\vec n \parallel(\vec b\times \vec c) =(-10, 5,5) \Rightarrow \vec a在\vec n的正射影\vec u= (4,1,3) \cdot (-2,1,1)\cdot {(-2,1,1)\over (-2)^2+1^2+1^2} \\= -{2\over 3}(-2,1,1)= ({4\over 3},-{2\over 3}, -{2\over 3}) \Rightarrow x\vec b+ y\vec c=\vec a-\vec u \Rightarrow (2x+3y,3x+7y, x-y) =({8\over 3},{5\over 3},{11\over 3}) \\ \Rightarrow (x,y)= \bbox[red,2pt]{ \left({41\over 15},-{14\over 15} \right)}$$

解答:$${\triangle ABC\over \triangle BCD} ={3\over 1} \Rightarrow {\overline{CD} \over \overline{AD}} ={1\over 2} \Rightarrow {\tan A\over \tan C} ={\overline{BD}/\overline{AD} \over \overline{BD}/\overline{CD}} ={1\over 2} \Rightarrow \cases{\tan A=k\\ \tan C=2k} \\ \Rightarrow \tan B=-\tan(A+C)   ={3k\over 2k^2-1} \Rightarrow {2\over \tan A}+{1\over \tan B}+{3\over \tan C} ={2\over k}+{2k^2-1\over 3k}+ {3\over 2k} \\={ 4k^2+19\over 6k} ={2k\over 3}+{19\over 6k} \ge 2\sqrt{{2k\over 3}\times{19\over 6k}} = \bbox[red, 2pt]{{2\over 3}\sqrt{19}}$$

解答:$$f(k) = \cos{k\pi \over 11} \Rightarrow f(k)=\begin{cases} 0 & 1\le k\le 5\\ -1 & 6\le k\le 16\\ 0 & 17\le k\le 21\\ 1& k=22\end{cases} \Rightarrow \sum_{k=1}^{11} f(k)= -10 \\ \Rightarrow \sum_{k=1}^{114=22\cdot 5+4} f(k) = -10\cdot 5+0= \bbox[red, 2pt]{-50}$$

解答:$$\log_{n+1} a_n= {\log a_n\over \log (n+1)} =1+{1\over (n+1)\log(n+1)} = {(n+1)\log(n+1)+1\over (n+1)\log(n+1)} \\ \Rightarrow \log a_n= {(n+1)\log(n+1)+1\over (n+1) }= \log(n+1)+{1\over n+1} =\log ((n+1)\cdot 10^{1/n+1})\\ \Rightarrow a_n=(n+1)10^{1/n+1} \Rightarrow {a_n\over n+1} =10^{1/n+1}\lt 1.2 \Rightarrow {1\over  n+1}\lt \log 1.2 \approx 0.079 \approx {1\over 12.6} \\ \Rightarrow n= \bbox[red, 2pt]{12}$$

解答:

$$4\times {2\pi\over 12}(\cos 0+ \cos{2\pi\over 12}+ \cos{4\pi\over 12}) ={2\pi\over 3}(1+{\sqrt 3\over 2}+{1\over 2}) = \bbox[red, 2pt]{{\pi\over 3}(3+\sqrt 3)}$$

解答:$$\cases{\overrightarrow{OA} \times \overrightarrow{OB} =\overrightarrow{OC} \\ \overrightarrow{OA} \times \overrightarrow{OC} =\overrightarrow{OD}} \Rightarrow \cases{\overrightarrow{OA} \bot \overrightarrow{OC} \\\overrightarrow{OA}\bot \overrightarrow{OD} \\\overrightarrow{OC} \bot \overrightarrow{OD} \\ \overrightarrow{OB} \bot \overrightarrow{OC}} \Rightarrow \cases{A(k,0,0)\\ C(0,k,0)\\ D(0,0, m)\\ B(a,0,b)} \Rightarrow \overrightarrow{OA} \times \overrightarrow{OC} =(0,0,k^2) =(0,0,m) \Rightarrow m=k^2 \\ \Rightarrow \overrightarrow{OA} \times \overrightarrow{OB} =(k,0,0)\times (a,0,b) =(0,-bk,0) =(0,k,0) \Rightarrow b=-1 \Rightarrow \overline{BD} =\sqrt{a^2+(b-m)^2} \\=\sqrt{a^2+b^2-2bm+m^2} =\sqrt{k^2+2m+m^2} =\sqrt{k^2+2k^2 +k^4} = \bbox[red, 2pt]{k\sqrt{k^2+3}}$$

解答:$$(\sqrt 5+ \sqrt 6+\sqrt 7)( \sqrt 6+\sqrt 7-\sqrt 5) (\sqrt 5+\sqrt 7-\sqrt 6) (\sqrt 5+\sqrt 6-\sqrt 7) \\= ( (\sqrt 6+\sqrt 7)+\sqrt 5)( (\sqrt 6+\sqrt 7)-\sqrt 5) (\sqrt 5+(\sqrt 7-\sqrt 6)) (\sqrt 5 -(\sqrt 7-\sqrt 6)) \\=((\sqrt 6+\sqrt 7)^2-5) (5-(\sqrt 7-\sqrt 6)^2) =(8+2\sqrt{42}) (-8+2\sqrt{42}) =(2\sqrt{42})^2-8^2 \\=168-64= \bbox[red, 2pt]{104}$$

解答:$$分組的方法數:C^5_1C^4_2=30\\ 10張椅子5人入座後剩下5張椅子，剩下5張椅子中有6個間隔取3個給各組有C^6_3，\\組內排列數:2!\cdot 2!=4及3組排列數:3\\ 因此共有30\times C^6_3\times 4\times 3=\bbox[red, 2pt]{7200}坐法$$

解答:$$\sum_{k=1}^{114} (k!\times k)= \sum_{k=1}^{114} ((k+1)!-k!) =(2!-1!)+(3!-2!) +\cdots (115!-114!) =115!-1\\ 又115!是2025的倍數,因此(115!-1)除以2025的餘數=2025-1= \bbox[red, 2pt]{2024}$$

二、計算題（每題8分，共24分）

解答:

$$令\cases{\vec u= \overrightarrow{AB'}=\overrightarrow{AB}/|\overrightarrow{AB}| \\ \vec v=\overrightarrow{AD'}=\overrightarrow{AD}/|\overrightarrow{AD}| \\ \vec w = \overrightarrow{AC'} = \sqrt 3 \overrightarrow{AC}/|\overrightarrow{AC}| } \Rightarrow \cases{\vec u+\vec v=\sqrt 3 \vec w \\ |\vec u| =| \vec v|=  1\\ |\vec w|=\sqrt 3} \Rightarrow \cos \angle AD'C' = {|\vec v|^2+ |\vec u|^2-(\sqrt 3|\vec w|)^2 \over 2\cdot |\vec u| |\vec v|} \\={1+1- 3\over 2} =-{1\over 2} \Rightarrow \angle AD'C'=120^\circ \Rightarrow \angle DAC=\angle CAB=30^\circ \\ \Rightarrow \cases{ \overrightarrow{AC} = \overrightarrow{AB}逆時針旋轉30^\circ \times s = s \begin{bmatrix} \cos 30^\circ & -\sin 30^\circ\\ \sin 30^\circ & \cos 30^\circ \end{bmatrix} \begin{bmatrix} 5\\ 2\sqrt 3 \end{bmatrix} =s({3\over 2}\sqrt 3, {11\over 2} )\\ \overrightarrow{AD} = \overrightarrow{AB} 逆時針旋轉60^\circ \times t =t \begin{bmatrix}\cos 60^\circ & -\sin 60^\circ\\ \sin 60^\circ & \cos 60^\circ \end{bmatrix}\begin{bmatrix} 5\\ 2\sqrt 3 \end{bmatrix} =t(-{1\over 2}, {7\over 2}\sqrt 3)} \\ \Rightarrow \overrightarrow{CD} =\overrightarrow{AD}-\overrightarrow{AC} \Rightarrow (-7,-{\sqrt 3\over 3}) =(-{1\over 2}t-{3\over 2}\sqrt 3s, {7\over 2}\sqrt 3t-{11\over 2}s) \Rightarrow \cases{s =4\sqrt 3/3 \\t=2} \\ \Rightarrow \cases{\overrightarrow{AC} =(6, 22\sqrt 3 /3)\\ \overrightarrow{AD} = (-1,7\sqrt 3)} \Rightarrow \cases{|\overrightarrow{AB}|= \sqrt{37}\\ |\overrightarrow{AC}|= 4 \sqrt{111} /3 \\ |\overrightarrow{AD}|= 2\sqrt{37}} \Rightarrow ABCD面積=\triangle ABC+ \triangle ACD \\= {1\over 2} \sin 30^\circ (|\overrightarrow{AB}|\cdot |\overrightarrow{AC}| + |\overrightarrow{AD}|\cdot |\overrightarrow{AC}|) = \bbox[red, 2pt]{37\sqrt 3}$$

解答:$$\lim_{x\to 1}{1\over x-1}\left( {1-x^{100} \over 1-x}-100\right) =\lim_{x\to 1}{1\over x-1}\left( {(1-x)(1+x+x^2+ \cdots+ x^{99}) \over 1-x}-100\right) \\= \lim_{x\to 1}{1\over x-1}\left(  1+x+x^2 +\cdots+x^{99}-100\right) = \lim_{x\to 1} {(x+x^2+ \cdots+ x^{99}-99)'\over (x-1)'} \\=\lim_{x\to 1} (99x^{98} +98x^{97}+ \cdots +2x+ 1) =99+98+ \cdots +1 =50\times 99= \bbox[red, 2pt]{4950}$$

$$y={\cos x+2\sin x\over 2+\cos x} \Rightarrow  2y=(1-y)\cos x+2\sin x \\=\sqrt{(1-y)^2+2^2} \left( {1-y\over \sqrt{(1-y)^2+2^2}} \cos x+ {2\over \sqrt{(1-y)^2+2^2}} \sin x\right) =\sqrt{y^2-2y+5} \sin(x+\theta) \\ \Rightarrow \sin(x+\theta) ={2y\over \sqrt{y^2-2y+5}} \Rightarrow \left| {2y\over \sqrt{y^2-2y+5}}\right| \le 1 \Rightarrow 4y^2 \le y^2-2y+5 \\ \Rightarrow 3y^2+2y-5\le 0  \Rightarrow (3y+5) (y-1)\le 0 \Rightarrow -{5\over 3}\le y\le 1 \Rightarrow \bbox[red, 2pt]{\cases{最大值1\\ 最小值-{5\over3}}}$$

$$\bbox[cyan, 2pt]{試題疑義申復更正答案}$$

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解題僅供參考，其他教甄試題及詳解