110年台南女中第1次教甄-數學詳解

國立臺南女中 110 學年度第一次教師甄選

一、填充題

解答：$$ 由於k\in \mathbb{N}，因此 若f(k)是最大值\Rightarrow \cases{f(k)\ge f(k+1)\\ f(k)\ge f(k-1)} \Rightarrow \cases{f(k)/f(k+1) \ge 1\\ f(k)/f(k-1)\ge 1} \\\Rightarrow \cases{({k\over k+1})^2\cdot 1.05^2 \ge 1 \\ ({k\over k-1})^2 \cdot {1\over 1.05^2} \ge 1} \Rightarrow \cases{(1-{1\over k+1}) \cdot 1.05 \ge 1\\ (1+{1\over k-1})\cdot {1\over 1.05} \ge 1} \Rightarrow \cases{k\ge 20\\ 1\le k\le 21} \Rightarrow k=\bbox[red, 2pt]{20或21}$$

解答：$$y=x^2 \Rightarrow y'=2x \Rightarrow y'(1)=2 \Rightarrow \overleftrightarrow{Q_1P_2}:y=2x-1 \Rightarrow Q_2({1\over 2},0) \Rightarrow P_2({1\over 2},{1\over 4})\\ 同理可得\cases{P_3({1\over 4},0),Q_3({1\over 4},{1\over 16}) \\ P_4({1\over 8},0), Q_4({1\over 8},{1\over 64}) \\ \cdots} \Rightarrow \cases{\overline{P_1Q_1}=1 \\ \overline{P_2Q_2}=1/4 \\ \overline{P_3Q_3}=1/4^2 \\ \overline{P_4Q_4}=1/4^3 \\\cdots } \Rightarrow \sum_{k=1}^\infty \overline{P_kQ_k} ={1\over 1-1/4} =\bbox[red, 2pt]{4\over 3}$$

解答：

$$令\overline{AC}與\overline{OB}的交點為D，見上圖；由於\overline{CP}\parallel \overline{OB} \parallel \overline{AQ}，\\因此{\overline{AD}\over \overline{DC}} ={OAQB面積\over OBPC}={40\over 12} ={10\over 3} ={\triangle OAD\over \triangle OCD} \Rightarrow \cases{\triangle OAD={10\over 13}\triangle OAC={180 \over 13} \\ \triangle OCD={3\over 13}\triangle OAC={54 \over 13}} \\ \Rightarrow \triangle ABD ={1\over 2}OAQB-\triangle OAD = 20-{180\over 13}={80\over 13};\\ 又{\overline{OD} \over \overline{DB}} ={\triangle OAD\over \triangle ABD} ={180/13 \over 80/13} ={9\over 4} \Rightarrow \overrightarrow{OB}= {13\over 4}\overrightarrow{OD} ={13\over 9}\left( {3\over 13}\overrightarrow{OA} +{10\over 13} \overrightarrow{OC}\right) = {1\over 3}\overrightarrow{OA} +{10\over 9} \overrightarrow{OC}\\ \Rightarrow (a,b)=\bbox[red, 2pt]{({1\over 3},{10\over 9} )}$$

解答：$$令f(x)=(x-a_1)(x-a_2)\cdots (x-a_{2011})，則\cases{x^{2020}的係數=-\sum a_i\\ x^{2019}係數=\sum a_ia_j\\ x^{2018}係數=-\sum a_ia_ja_k\\ \dots \\ 常數項=-a_1a_2\cdots a_{2011}}\\ \Rightarrow b_{2010} =1-f(1)=1-(1-{1\over 11^2})(1-{1\over 12^2})(1-{1\over 13^2})\cdots (1-{1\over 2021^2}) \\ =1- {10\cdot 12\over 11^2} \times {11\cdot 13\over 12^2} \times {12\cdot 14\over 13^2} \times \cdots \times {2020\times 2022\over 2021^2} \\ =1- {10\over 11} \times {2022\over 2021} =1-{20220\over 22231} = \bbox[red, 2pt]{2011\over 22231}$$

解答：$$\cases{7x-{1\over y}=16\\ xy+{1\over xy}=30} \Rightarrow \left(7x-{1\over y} \right)\left(y-{7\over x} \right) = 7(xy+{1\over xy})-50 \Rightarrow 16\left(y-{7\over x} \right) = 7\cdot 30-50\\ \Rightarrow y-{7\over x}=10 \Rightarrow 2xy-20x+9 = 2x(y-10)+9 = 2x(y-(y-{7\over x}))+9 =14+9= \bbox[red, 2pt]{23}$$

解答：$$r_n及r_{n+1}為x^2-a_nx+\left({1\over 5}\right)^n=0之二根\Rightarrow \cases{a_n= r_n+r_{n+1}\\ r_nr_{n+1}= {1\over 5^n}} \\ 由\cases{r_1=2 \\ r_nr_{n+1}={1\over 5^n}}\Rightarrow \cases{r_1=2\\r_{2k+1}={2\over 5^k}\\ r_{2k}={1\over 2\cdot 5^k}},k\in \mathbb{N} \Rightarrow \sum_{n=1}^\infty a_n = r_1+2(r_2 +r_3+\cdots +r_n+\cdots)\\ =2+ 2\sum_{k=1}^\infty \left( {1\over 2\cdot 5^k} +{2\over 5^k}\right) =2+ 2\cdot {5\over 2}\sum_{k=1}^\infty {1\over 5^k} =2+5 \times {1\over 4} =\bbox[red, 2pt]{13\over 4}$$

解答：$$令\cases{N=轉向次數\\ 右:向右走\\ 上:向上走\\ x_i:向右走的之數\\ y_i向上走的次數}，則\cases{捷徑數相當於5個右4個上的排列數=C^9_4=126 \\ \sum x_i=5\\ \sum y_i=4 \\ x_i,y_j \in \mathbb{N}}\\ \begin{array}{}N & 排列 -------------------& 數量\\\hline 1 & (右)^5(上)^4 & 1\\ & (上)^4(右)^5 & 1\\\hdashline 2 & (右)^{x_1}(上)^4(右)^{x_2} & H^2_3=4 \\ & (上)^{y_1}(右)^5(上)^{y_2} & H^2_2=3 \\\hdashline 3  & (右)^{x_1}(上)^{y_1}(右)^{x_2}(上)^{y_2} & H^2_3\times H^2_2=12 \\ & (上)^{y_1}(右)^{x_1}(上)^{y_2}(右)^{x_2} & H^2_2\times H^2_3=12 \\\hdashline 4 & (右)^{x_1}(上)^{y_1}(右)^{x_2}(上)^{y_2}(右)^{x_3} & H^3_2\times H^2_2=18 \\ & (上)^{y_1}(右)^{x_1}(上)^{y_2}(右)^{x_2}(上)^{y_3} & H^3_1\times H^2_3=12 \\\hdashline 5 & (右)^{x_1}(上)^{y_1}(右)^{x_2}(上)^{y_2}(右)^{x_3}(上)^{y_3} & H^3_2\times H^3_1=18 \\ & (上)^{y_1}(右)^{x_1}(上)^{y_2}(右)^{x_2}(上)^{y_3} (右)^{x_3} & H^3_2\times H^3_1=18 \\\hdashline 6 &  (右)^{x_1}(上)^{y_1}(右)^{x_2}(上)^{y_2}(右)^{x_3}(上)^{y_3} (右)^{x_4}& H^4_1\times H^3_1=12 \\ & (上)^{y_1}(右)^{x_1}(上)^{y_2}(右)^{x_2}(上)^{y_3} (右)^{x_3}(上)^{y_4} & H^3_2=6 \\\hdashline 7 &  (右)^{x_1}(上)^{y_1}(右)^{x_2}(上)^{y_2}(右)^{x_3}(上)^{y_3} (右)^{x_4} (上)^{y_4}& H^4_1 =4 \qquad\\ & (上)^{y_1}(右)^{x_1}(上)^{y_2}(右)^{x_2}(上)^{y_3} (右)^{x_3}(上)^{y_4}(右)^{x_4} & H^4_1=4 \qquad\\\hdashline 8 & 右上右上右上右上右 & 1\\\hline\end{array}\\ \Rightarrow 轉向次數的期望值= (1\times 2 +2\times 7 + 3\times 24 + 4\times 30  +5\times 36  +6\times 18  +7\times 8 +8)\div 126\\ ={560\over 126} =\bbox[red, 2pt]{40\over 9}$$

解答：$$\cases{|z_1|=2\\ |z_2|=3} \Rightarrow \cases{z_1=2(\cos A+i\sin A)\\ z_2= 3(\cos B+i \sin B)}，\\依題意3z_1-2z_2 ={3\over 2}-i \Rightarrow \cases{6(\cos A -\cos B) ={3\over 2}\\6(\sin A -\sin B)=-1} \Rightarrow \cases{-12\sin {A+B\over 2} \sin {A-B\over 2} ={3\over 2} \cdots(1)\\ 12\cos {A+B\over 2}\sin {A-B\over 2}=-1 \cdots(2)}\\ {(1)\over (2)} \Rightarrow -\tan {A+B\over 2}=-{3\over 2} \Rightarrow \tan{A+B\over 2}={3\over 2} \Rightarrow \cases{\sin{A+B\over 2}= {3\over \sqrt{13}} \\\cos{A+B\over 2} ={2\over \sqrt{13}}} \\\Rightarrow \cases{\sin(A+B)= 2\sin{A+B\over 2}\cos {A+B\over 2} ={12\over 13} \\ \cos(A+B) = \cos^2{A+B\over 2}-\sin^2{A+B\over 2} =-{5\over 13}}\\ \Rightarrow z_1z_2= 6(\cos(A+B)+i\sin (A+B)) =\bbox[red,2pt]{-{30\over 13}+{72\over 13}i}$$

解答：$$假設\overline{PB}=a \Rightarrow \cases{\cos \angle A={4-a^2\over 2\sqrt 3} \Rightarrow a^2=4-2\sqrt 3\cos \angle A\\ \cos \angle Q={2-a^2 \over 2} \Rightarrow a^2=2-2\cos \angle Q} \\\Rightarrow 4-2\sqrt 3\cos \angle A=2-2\cos \angle Q \Rightarrow  \cos \angle Q=\sqrt 3\cos \angle A-1\\又\cases{S= {1\over 2}\sqrt 3\sin \angle A\\ T={1\over 2}\sin \angle Q} \Rightarrow S^2+T^2 = {3\over 4}\sin^2 \angle A+{1\over 4}\sin^2 \angle Q = {3\over 4}(1-\cos^2 \angle A) +{1\over 4}(1-\cos^2 \angle Q) \\ =1-{3\over 4}\cos^2 \angle A-{1\over 4}(\sqrt 3\cos \angle A-1)^2= -{3\over 2}\cos^2 \angle A+{\sqrt 3\over 2}\cos \angle A+{3\over 4}\\ \Rightarrow 當\cos \angle A={\sqrt 3\over 6}時，S^2+T^2有極大值\bbox[red, 2pt]{7\over 8}$$

解答：

$$欲求之體積=三角錐J-ABK - 三角錐J-DGH - 三角錐K-BCI\\,其中\cases{G,H,I為平面H與圖形的交點\\ J為\overleftrightarrow{HI}與z軸的交點\\ K為\overleftrightarrow{HI}與xy平面的交點}\\\cases{\cases{\overrightarrow{AB}=(2,1,0)\\ \overrightarrow{AK}=(0,3,0)\\ \overrightarrow{AJ}=(0,0,3)}\Rightarrow 三角錐J-ABK體積={1\over 6}\begin{Vmatrix} 2& 1 & 0\\ 0 & 3 &0\\ 0 & 0 & 3\end{Vmatrix}=3 \\\cases{\overrightarrow{JD}=(0,0,-1)\\ \overrightarrow{AK}=(2/3,1/3,-1)\\ \overrightarrow{JH}=(0,1,-1)}\Rightarrow 三角錐J-DGH體積={1\over 6}\begin{Vmatrix} 0& 0 & -1\\ 2/3 & 1/3 &-1\\ 0 & 1 & -1\end{Vmatrix}={1\over 9} \\ \cases{\overrightarrow{KI}=(0,-1,1)\\ \overrightarrow{KC}=(0,-1,0)\\ \overrightarrow{KB}=(2,-2,0)}\Rightarrow 三角錐K-BCI體積={1\over 6}\begin{Vmatrix} 0& -1 & 1\\ 0 & -1 &0\\ 2 & -2 & 0\end{Vmatrix}={1\over 3}}\\ \Rightarrow 欲求之體積=3-{1\over 9}-{1\over 3}=\bbox[red, 2pt]{23\over 9}$$

解答：

    由題意知: 每抓一次，箱子(唉!題目的箱子就是盒子啦!)就少一個白球或2個黑球；也就是說箱子內的黑球永遠都是偶數，如果最後剩下一球，那一定是白球，所以最後一球是黑色的機率為\(\bbox[red, 2pt]{0}\)

解答：$$利用因式分解公式: a^3+b^3+c^3-3abc =(a+b+c)(a^2+b^2+c^2-(ab+bc+ca))\\ ={1\over 2}(a+b+c)^2((a-b)^2 +(b-c)^2+(c-a)^2)\\ 現在m^3+n^3+99mn = 33^3 \Rightarrow m^3+n^3+ (-33)^3-3mn(-33)=0\\ \Rightarrow (m+n-33)(m^2+n^2+(-33)^2-(mn-33n-33m))=0 \\ \Rightarrow {1\over 2}(m+n-33)((m-n)^2+(n+33)^2 + (m+33)^2)=0\\ \Rightarrow \cases{m+n=33\\ m=n=-33}；由題意知m\cdot n \ge 0且m,n\in \mathbb{Z}\\ \Rightarrow (m,n)=(33,0),(32,1),\dots,(1,32),(0,32),(-33,-33)，共有\bbox[red, 2pt]{35}組解$$

解答：$$除了完全平方數外，其它數字的因數是偶數個，因此\cases{編號是完全平數的門是打開的\\編號不是完全平數的門是關閉的}\\1-100的完全平方數:1^2, 2^2, \dots,10^2，共10個；\\6號同學沒來，有16(100/6)扇門的狀態被改變，即\cases{完全平方數36改成關閉\\ 其它15個門改成打開}\\ 總結以上，打開門的有10-1+15= \bbox[red, 2pt]{24}扇$$

解答：$$f(x) =\int_0^x (t+a)(t+2)^3\;dt \Rightarrow  f'(x)=(x+a)(x+2)^3 \Rightarrow f''(x)=(x+2)^3+ 3(x+a)(x+2)^2\\y=f(x)在x=2有水平切線，即f'(2)=0 \Rightarrow (2+a)\cdot 4^2=0 \Rightarrow a=-2\\\Rightarrow f''(x)=(x+2)^3+3(x-2)(x+2)^2 = 4(x+2)^2 (x-1)，因此若 f''(x)=0 \Rightarrow x=1,-2\\ \Rightarrow 反曲點為(1,f(1))，而f''(-2^+)f''(-2^-) \gt 0\Rightarrow (-2,f(-2))不是反曲點；\\ \Rightarrow f(x)=\int_0^x (t-2)(t+2)^3\;dt = {1\over 5}x^5+x^4-8x^2-16x \Rightarrow f(1)={1\over 5}+1-8-16=-{114\over 5} \\ \Rightarrow 反曲點坐標為 \bbox[red, 2pt]{(1,-{114\over 5})}$$

解答：

$$依題意:兩圖形\cases{y=f(x)=\left| \log x\right|\\ y=g(x)=ax+b} 有三個相圖交點，分別是k,2k,3k,k\in \mathbb{R^+}\\ 由y=f(x)圖形可知\cases{f(k) \lt 0\\f(2k),f(3k)\gt 0} \Rightarrow \cases{-\log k = ak+b \cdots(1)\\ \log 2k= 2ak +b \cdots(2) \\ \log 3k= 3ak+b \cdots(3)}\\ (2)-(1) \Rightarrow \log 2k +\log k=ak \Rightarrow a={\log 2k^2 \over k} 代回(1) \Rightarrow -\log k=\log 2k^2 +b \Rightarrow b=-\log 2k^3\\ 將\cases{a={\log 2k^2 \over k}\\ b=-\log 2k^3} 代入(3) \Rightarrow \log 3k = 3\log 2k^2-\log 2k^3 =\log{8k^6\over 2k^3} =\log 4k^3\\ \Rightarrow 3k=4k^3 \Rightarrow k(4k^2-3)=0 \Rightarrow k={\sqrt 3\over 2} \Rightarrow (a,b)=({2\over \sqrt 3}\log{3\over 2},-\log {3\sqrt 3\over 4})\\ =\bbox[red, 2pt]{\left({2\sqrt 3\over 3}\log {3\over 2},\log {4\sqrt 3\over 9}\right)}$$

解答：

$$拋物線\Gamma:y^2=4x的焦點F(1,0)\Rightarrow \cases{焦點F(1,0)\\ 準線L:x=-1}\\P在\Gamma上，可表示成P(t^2,2t),t\in R，且\overline{PF} =\text{dist}(P,L)=t^2+1\\因此k=\left|t^2-\sqrt{(t^2-5)^2 +(2t-3)^2}\right| =\left|\overline{PF}-1-\overline{PQ} \right|,其中Q(5,3)\\ k的最大值發生在\overline{PF}-\overline{PQ}=-\overline{FQ}，即k=|-1-\overline{FQ}|=|-1-5|= \bbox[red, 2pt]{6}，\\此時P=A，其中A,B為\overleftrightarrow{QF}與\Gamma 的交點，見上圖；$$

解答：$$令\cases{x'=3x+y\\ y'=2x-7y} \Rightarrow {x'^2\over 5}+{y'^2\over 4}\le 1為一橢圓封閉區域，面積為\sqrt 5\cdot 2\pi\\ 又\begin{Vmatrix}{\partial \over \partial x}x' & {\partial \over \partial y}x' \\{\partial \over \partial x}y' &{\partial \over \partial x}y'\end{Vmatrix} =\begin{Vmatrix}3 & 1\\ 2 & -7 \end{Vmatrix} =23 \Rightarrow {(3x+y)^2 \over 5} +{(2x-7y)^2 \over 4}\le 1所圍面積=\bbox[red, 2pt]{2\sqrt 5\pi\over 23}$$

解答：$$P在L:{x\over 1} ={y-3 \over -1} ={z-1\over 2}，P可表示成P(t,-t+3,2t+1),t\in \mathbb{R}\\ \Rightarrow \overline{PA}+ \overline{PB} = \sqrt{(t-1)^2 +(-t-1)^2+ 4t^2} +\sqrt{(t-3)^2 +(-t+1)^2+ (2t-7)^2} \\ =\sqrt{6t^2+2} +\sqrt{6t^2-36t+59} =\sqrt 6\left( \sqrt{t^2+{1\over 3}} +\sqrt{t^2-6t+{59\over 6}}\right) \\ =\sqrt 6\left( \sqrt{(t-0)^2+(0-{1\over \sqrt 3})^2} +\sqrt{(t-3)^2+(0+\sqrt{5\over 6})^2}\right)\\ 相當於(t,0)至A(0,{1\over \sqrt 3})加上(t,0)至B(3,-\sqrt{5\over 6})的距離和，最小值即為\overline{AB}\\此時(t,0)即為\overline{AB}與x軸的交點；\\由A及B可得\overleftrightarrow{AB}:y=-{\sqrt 5+\sqrt 2\over 3\sqrt 6}x+{1\over \sqrt 3}，與x軸的交點為({3\sqrt 2\over \sqrt 5+\sqrt 2},0)，即t={3\sqrt 2\over \sqrt 5+\sqrt 2}\\ \Rightarrow P之x坐標即為t值，也就是{3\sqrt 2\over \sqrt 5+\sqrt 2}=\sqrt 2(\sqrt 5-\sqrt 2)=\bbox[red, 2pt]{\sqrt{10}-2}$$

解答：$$三向量\vec a,\vec b,\vec c所展開的四面體體積=V \Rightarrow 其展開的六面體體積=6V\\ 將\alpha,\beta,\gamma 視為立體空間的三坐標，則\cases{|\alpha| \le 1\\ |\alpha+\beta|\le 1\\ |\alpha+ \beta+\gamma | \le 1}所圍的體積為8\\ \Rightarrow S的體積=6V\times 8= \bbox[red, 2pt]{48}V\\註:\cases{|x|\le 1\\ |x+y|\le 1\\ |x+y+z| \le 1} \Rightarrow \int_{-1}^1\int_{-1-x}^{1-x} \int_{-1-x-y}^{1-x-y}1\;dz\;dy\;dx=8$$

解答：$$a_{n+3} =a_{n+2}-a_{n+1}+a_n \Rightarrow a_{n+3}+a_{n+1} =a_{n+2}+a_n \Rightarrow \cases{a_4+a_2=a_3+a_1 \\ a_5+a_3= a_4+a_2} \\ \Rightarrow a_5+a_3= a_3+a_1 \Rightarrow a_5=a_1 ，同理可得\cases{a_1=a_5\\ a_2=a_6\\ a_3=a_7\\ a_4=a_8\\ \cdots}，即 a_m=a_{(m \mod 4)}\\ 因此 \sum_{k=-1}^{102}(-1)^ka_k = (-a_1+a_2-a_3+a_4) + (-a_5+a_6-a_7+a_7)+ \cdots \\\qquad +(-a_{97}+ a_{98}- a_{99}+a_{100})-a_{101}+a_{102} = 0 + 0 +\cdots +0 -a_{101}+a_{102}\\ =-a_1+a_2 =a_3-a_4 = a_{75}-a_{24} =13-71= \bbox[red, 2pt]{-58}$$

解答：$$令u=25-x^2,則du=-2xdx；因此\int_3^5 x\sqrt{25-x^2}\;dx = \int_{16}^0 -{1\over 2}u^{1/2}\;du = \left. \left[ -{1\over 3} u^{3/2} \right] \right|_{16}^0\\ ={64\over 3} \Rightarrow f(x)在[3,5]的函數值平均為{\int_3^5f(x)\;dx \over 5-3} ={64\over 3} \times {1\over 2} =\bbox[red, 2pt]{32\over 3}$$

解答：

$$3\overline{AD} =2\overline{DE}=\overline{EB} \Rightarrow \overline{AD} :\overline{DE} :\overline{EB} =2:3:6 \\\Rightarrow \triangle CAD: \triangle CAD: \triangle CDE: \triangle CEB: \triangle ABC=2:3:6:11 \\ \Rightarrow {1\over 2}ab \sin \alpha: {1\over 2}bc \sin \beta: {1\over 2}cd \sin \gamma: {1\over 2}ad =2:3:6:11，其中\cases{\overline{CA}= a\\ \overline{CD}= b\\ \overline{CE}= c\\ \overline{CB}= d} \\ \Rightarrow ab \sin \alpha: bc \sin \beta: cd \sin \gamma: ad =2:3:6:11 \Rightarrow \cases{\sin\alpha = {2\over ab}k \\\sin\beta = {3\over bc}k \\\sin\gamma = {6\over cd}k \\ad=11k}\\ \Rightarrow {\sin \alpha\cdot \sin \gamma \over \sin \beta}={{12\over abcd}k^2 \over {3\over bc}k} ={4k\over ad} ={4k\over 11k}= \bbox[red, 2pt]{4\over 11}$$

二、計算題

解答：$$\cases{A(a,a^3) \\ B(b,b^3)\\ C(c,c^3)} \Rightarrow {1\over 3}(\overrightarrow{OA} +\overrightarrow{OB} +\overrightarrow{OC} ) = \overrightarrow{OG}  \Rightarrow {1\over 3}(a+b+c,a^3+b^3+c^3) =(0,-2)\\ \Rightarrow \cases{a+b+c =0 \\ a^3+b^3+c^3 =-6} \Rightarrow \cases{c=-a-b\\ a^3+b^3-(a+b)^3+6 =0} \\ 令\cases{f(a,b)=-a-b\\ g(a,b)=a^3+b^3-(a+b)^3+6 =0}，利用 \text{Lagrange 乘數法} \Rightarrow \cases{{\partial \over \partial a}f+ \lambda {\partial \over \partial a}g=0 \\ {\partial \over \partial b}f+ \lambda {\partial \over \partial b}g=0 \\ g=0} \\ \Rightarrow \cases{-1 +\lambda (3a^2-3(a+b)^2) =0 \cdots(1)\\ -1+ \lambda(3b^2-3(a+b)^2)=0 \cdots(2)\\ a^3+b^3-(a+b)^3+6 =0 \cdots(3)},由(1)及(2)可得 \lambda = {1\over -6ab-3b^2} = {1\over -3a^2-6ab} \\ \Rightarrow \cases{a=b\\ a=-b} 代入(3) \Rightarrow \cases{2a^3-8a^3+6=0\\ 0-0+6=0} \Rightarrow \cases{a=1=b\\ 6=0矛盾,不合} \Rightarrow c=f(1,1)=\bbox[red, 2pt]{-2}為極值(最大值)$$

解答：$$\cases{n是奇數\Rightarrow 3n^2+n+1 = 奇+奇+奇=奇數\\ n是偶數\Rightarrow 3n^2+n+1= 偶+偶+奇=奇數} \qquad\Rightarrow 3n^2+n+1是奇數\\利用反證法:\\假設存在正整數n，使得f(n)=2；由於3n^2+n+1為奇數且各數字和=2\\，因此3n^2+n+1只能首尾數字是1，其它皆是0 \Rightarrow 3n^2+n+1=10^k+1,k\in \mathbb{N}\\ \Rightarrow n(3n+1)= 2^k\cdot 5^k \Rightarrow \cases{n=2^k\cdots(1)\\ 3n+1=5^k \cdots(2)}，將(1)代入(2) \Rightarrow 3\cdot 2^k+1=5^k \\ \Rightarrow \cases{3\cdot 2^k+1 \gt 5^k,\text{ if }k=1\\ 3\cdot 2^k+1 \lt 5^k,\text{ if }k\ge 2} \Rightarrow 3\cdot 2^k+1\ne 5^k, \forall k\in \mathbb{N}\Rightarrow 矛盾\\ \Rightarrow f(n)\ne 2, \forall n\in \mathbb{N}$$

解答：$$學校公布的作法:\\2^{234} =\sum_{k=0}^{99} \lfloor 2^kx\rfloor \le \sum_{k=0}^{99} 2^kx =(2^{100}-1)x \Rightarrow x \ge {  2^{234} \over 2^{100}-1} =2^{134}+ 2^{34} +{2^{34}\over 2^{100}-1}\\ 設最小的x=2^{134} +2^{34}+ {y\over 99} \Rightarrow \sum_{k=0}^{99}\lfloor 2^k \cdot {y\over 2^{99}}\rfloor =2^{34};\\ 取y=2^{33+1}時，可得最小的x=\bbox[red, 2pt]{2^{134}+ 2^{34} +{2^{33}+1\over 2^{99}}}$$

解答：

學校公布的作法:

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解答僅供參考

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