111年全國教甄聯招-數學詳解

教育部受託辦理111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3分，共24分）

解答： $$f(x,y)= 3x-2y+k \Rightarrow f(P)f(Q)\lt 0 \Rightarrow (k+18)(k-13)\lt 0 \Rightarrow -18\lt k\lt 13 \\ \Rightarrow k=-17,-16,\dots,12 \Rightarrow 共有17+13=30個整數k，故選\bbox[red,2pt]{(C)}$$

解答： $$y=px^3-3px^2+ (3p+q)x-p-q+6 \Rightarrow y'=3px^2-6px+(3p+q) \Rightarrow y''=6px-6p\\ 若y''=0 \Rightarrow 6px-6p=0 \Rightarrow x=1 \Rightarrow 對稱中心坐標(1,y(1)) =(1,6)，故選\bbox[red,2pt]{(C)}$$

解答：

$$令\cases{A(4,3)\\ B(b,0)\\ C(a,a)}，則\cases{B在x軸上\\ C在直線y=x上}\\ 而\sqrt{(4-b)^2+ 3^2} +\sqrt{(a-b)^2+ a^2} +\sqrt{(4-a)^2 +(3-a)^2} = \overline{AB} +\overline{BC} +\overline{CA}\\ 因此取\cases{A對稱y=x的對稱點A'(3,4)\\ A對稱x軸的對稱點A''(4,-3)}，則\overline{A'A''}與x軸的交點即為B、與y=x的交點即為C\\ \Rightarrow \overline{AB} +\overline{BC} +\overline{CA} =\overline{A'A''} =\sqrt{1+7^2} =  {5\sqrt 2}為最小值，故選\bbox[red,2pt]{(B)}$$

解答：

$$令\cases{\overline{CD}=a\\ \angle BCD = \alpha} \Rightarrow \cases{\tan \alpha = \overline{BD}/\overline{CD}=1/2a \\ \tan(\theta +\alpha)= 2.5/a = 5/2a}\\ \tan(\theta+\alpha) = {\tan\theta +\tan \alpha\over 1-\tan \theta \tan \alpha} \Rightarrow {5\over 2a} ={\tan\theta +1/2a\over 1-\tan\theta/2a} \Rightarrow (2a+{5\over 2a})\tan \theta = 4\\ \Rightarrow \tan \theta = f(a)={8a\over 4a^2+5} \Rightarrow f'(a)= {8\over 4a^2+5}-{64a^2\over (4a^2+5)^2} ={40-32a^2 \over (4a^2+5)^2} =0 \Rightarrow a={\sqrt 5\over 2}\\，故選\bbox[red,2pt]{(B)}$$

解答：

$$由上圖可知：截平面為五邊形，故選\bbox[red,2pt]{(B)}$$

解答： $$令\cases{f(x)=10-x\\ g(x)=10^x\\ h(x)=\log x}，則\cases{f(x)=g(x) \Rightarrow x=\alpha\\ f(x)=h(x) \Rightarrow x=\beta\\ g(x)= h^{-1}(x)} \\\Rightarrow \cases{y=f(x)與y=g(x)的交點為(\alpha, 10^\alpha) \Rightarrow f(\alpha)=g(\alpha)\Rightarrow 10^\alpha=10-\alpha \cdots(1)\\y=f(x)與y=h(x)的交點為(\beta,\log \beta) \Rightarrow (\log \beta,\beta)=(\alpha,10^\alpha) \Rightarrow \log\beta =\alpha \cdots(2)} \\   \Rightarrow 10^\alpha+\log \beta = (10-\alpha)+\alpha = 10，故選\bbox[red,2pt]{(C)}$$

解答： $$\lim_{n\to \infty} {1\over n^3}\sum_{k=1}^{3n} (4n+k)^2 =\lim_{n\to \infty} {1\over n^3}\sum_{k=1}^{3n} (16n^2+8nk +k^2) \\=\lim_{n\to \infty} {1\over n^3}\left(16n^2\sum_{k=1}^{3n} 1+ 8n\sum_{k=1}^{3n} k +\sum_{k=1}^{3n} k^2 \right) \\=\lim_{n\to \infty} {1\over n^3}\left(16n^2\cdot 3n+ 8n\cdot {3n(3n+1)\over 2} +{3n(3n+1)(6n+1)\over 6} \right) =48+ 36+9 =93，故選\bbox[red,2pt]{(C)}$$

解答： $$3087是9的倍數 \Rightarrow 3087\times a也是9的倍數 \Rightarrow 9+5+3+2+1+0 =20 需再加7才是9的倍數\\，故選\bbox[red,2pt]{(D)}$$

二、複選題（每題4分，共16分，全對才給分）

解答： $$(A)\times: P\in L_2 \Rightarrow P(t+1,2t-4,2t-2) \Rightarrow d(P,E)={2t+2-2t+4-6\over \sqrt 5} =0\\\qquad \Rightarrow L_2在E上，重疊非平行\\(B)\bigcirc:\cases{L_1方向向量\vec u=(2,3,2)\\ L_2方向向量\vec v=(1,2,2)} \Rightarrow \vec n= \vec u\times \vec v=(2,-2,1)；取\cases{A\in L_1\\ D=\overline{BC}中點\in L_2} \\ \qquad \Rightarrow \cases{A(2t+3,3t+5,2t+3)\\ D(s+1,2s-4,2s-2)},t,s\in \mathbb{R} \Rightarrow \overrightarrow{AD} =(s-2t-2,2s-3t-9,2s-2t-5)\\ \qquad，且\overrightarrow{AD}\parallel \vec n \Rightarrow {s-2t-2\over 2} ={2s-3t-9\over -2} ={2s-2t-5\over 1} \Rightarrow \cases{t=-1\\ s=2}\Rightarrow A=(1,2,1)\\(C)\times: H=D=(3,0,2)\\ (D)\bigcirc: \overline{AH}= \sqrt{2^2+2^2+1^2}=3 \Rightarrow \overline{BC}={3\over \sqrt 3}\times 2 \Rightarrow \triangle ABC={1\over 2}\times {3\over \sqrt 3}\times 2\times 3 =3\sqrt 3\\，故選\bbox[red,2pt]{(BD)}$$

解答： $$(A)\bigcirc: g(x)=f(x^2-4x+3)= f((x-3)(x-1)) \Rightarrow g(3)=f(0)=-28\\ (B)\times: x=2+\sqrt 2 \Rightarrow x^2=4x-2 \Rightarrow x^3=4x^2-2x \Rightarrow f(2+\sqrt 2)= (4x^2-2x)-9x^2+26x-28\\ \qquad =-5x^2+24x-28 = -5(4x-2)+24x-28 = 4x-18 = 4(2+\sqrt 2)-18 = 4\sqrt 2-10\\ (C)\bigcirc: f'(x)=3x^2-18x+26 \Rightarrow f'(2)=2 \Rightarrow L:y=2(x-2)-4 = 2x-8 \\ \qquad \Rightarrow L與坐標軸的交點\cases{A(0,-8)\\ B(4,0)} \Rightarrow \triangle OAB面積={1\over 2}\times 8\times 4=16 \\ (D)\times: f(x)\lt 2 \Rightarrow x^3-9x^2+26x-30\lt 0 \Rightarrow (x-5)(x^2-4x+6)\lt 0 \Rightarrow x\lt 5\\\qquad \Rightarrow 有無限多個整數解\\\\，故選\bbox[red,2pt]{(AC)}$$

解答： $$a= 三次皆為紅球或三次皆為黑球的機率=({10\over 15})^3 +({5\over 15})^3 = {1\over 3}\\ b=連續三次皆為黑球的機率=({5\over 15})^3 ={1\over 27}\\ c=連續三次皆為紅球的機率= ({10\over 15})^3 ={8\over 27}\\ 因此a\gt c\gt b且a=b+c，故選\bbox[red,2pt]{(BCD)}$$

解答： $$假設正八邊形頂點依序為A_1,A_2,\dots, A_8\\(A)\bigcirc: 四條不同的直徑，每條直徑左右各有三個頂點，可決定6個直角\triangle，因此共有4\times 6=24個直角\triangle\\(B) \bigcirc:  \cases{ \angle A_{i-1}A_iA_{i+1} = 135^\circ \\\angle A_{i-1}A_iA_{i+2}= 112.5^\circ \\ \angle A_{i-2}A_iA_{i+1} = 112.5^\circ},i=1-8 \Rightarrow  共有8\times 3=24個鈍角\triangle \\(C)\bigcirc: 共有C^8_3=56個\triangle ，扣除24個直角\triangle、24個鈍角\triangle ，剩下56-24-24=8個銳角\triangle\\ (D)\bigcirc: C^8_4= 70個四邊形\\，故選\bbox[red,2pt]{(ABCD)}$$

第二部分：綜合題（共60分）

一、填充題（每題4分，共36分）

解答： $$取\cases{f(a,b)= 2^a+ 4^b\\ g(a,b)=a+b-13}，\text{依 Lagrange 算子}\cases{f=\lambda g\\ g=0} \Rightarrow \cases{f_a=\lambda g_a\\ f_b=\lambda g_b\\ g=0} \Rightarrow \cases{\ln 2\cdot 2^a =\lambda \cdots(1)\\ \ln 4\cdot 4^b =\lambda \cdots(2)\\ a+b-13=0 \cdots(3)}\\ {(1)\over (2)} \Rightarrow {2^a\over 2\cdot 4^b}=1 \Rightarrow 2^a=2^{2b+1} \Rightarrow a=2b+1 代入(3) \Rightarrow 3b=12 \Rightarrow b=4\Rightarrow a=9 \\\Rightarrow f(9,4) =2^9+ 4^4= 512+ 256 = \bbox[red, 2pt]{768}$$

解答： $$假設\cases{R(20,0)\\ P(12,16)\\ \angle POR=\theta} \Rightarrow \theta = \tan^{-1}{4\over 3} \Rightarrow \tan{\theta\over 2} = {1\over 2} \Rightarrow \cases{\sin (\theta/2) =1/\sqrt 5\\ \cos (\theta/2) =2/\sqrt 5}\\ 甲繞圓心2\pi +\theta \Rightarrow 乙繞圓心(2\pi+\theta)\div 2 = \pi +{1\over 2}\theta (第三象限)\\ \Rightarrow 乙的位置=(-20\cos{\theta\over 2},-20\sin{\theta\over 2}) = \bbox[red, 2pt]{(-8\sqrt 5,-4\sqrt 5)}$$

解答： $$\begin{array}{} B & A & 數量\\\hline 20 & 1-17 & 17 \\ 19 & 1-16 & 16 \\ 18 & 1-15 & 15\\ 17 & 1-14,20 & 15\\ 16 & 1-13,19-20 & 15\\ \cdots & \cdots & 15\\ 4 & 1,7-20 & 15 \\ 3 & 6-20 & 15\\ 2& 5-20 & 16 \\ 1 & 4-20& 17\\\hline \end{array} \Rightarrow 機率={1\over 20}\cdot {1\over 19}(17+16+ 15\cdot 16+ 16+17) =\bbox[red, 2pt]{153\over 190}$$

解答： $$x^6-8= (x^2)^3-2^3= (x^2-2)(x^4+2x^2+ 4) =0 \Rightarrow x^6=8的六根扣除x^2=2的二根即為\\x^4+2x^2+ 4=0的四根；即\cases{A(\sqrt 2/2,\sqrt 6/2) \\ B(-\sqrt 2/2, \sqrt 6/2)\\ C(-\sqrt 2/2,-\sqrt 6/2)\\ D(\sqrt 2/2,-\sqrt 6/2)}\\ \Rightarrow 長方形ABCD面積= \overline{AB}\times \overline{BC} =\sqrt 2\times \sqrt 6 = \bbox[red,2pt]{2\sqrt 3}$$

解答： $$a\gt 1 \Rightarrow 3\log_a x \gt 2\log_a x\gt \log_a x \Rightarrow A在B的右方且C在\overline{AB}的上方；\\因此假設A(\alpha,\beta) \Rightarrow B(\alpha-12,\beta) \Rightarrow C(\alpha-12,\beta+12)\Rightarrow \log_a \alpha= 2\log_a (\alpha-12) \Rightarrow (\alpha-12)^2 =\alpha \\ \Rightarrow \alpha^2-25\alpha +144=0 \Rightarrow (\alpha-16)(\alpha-9)=0 \Rightarrow \alpha=16 (\alpha=9不合, \because \alpha-12 \lt 0)\\ \Rightarrow \beta = \log_a 16 \Rightarrow \cases{A(16,\log_a 16)\\ B(4,\log_a 16) \\ C(4, \log_a 16+12)}  \Rightarrow 3\log_a 4=\log_a 16+12 = 2\log_a 4+ 12 \\ \Rightarrow \log_a 4 = 12 \Rightarrow a^{12}=4 \Rightarrow a^3= \sqrt 2 \Rightarrow a^3+a^{-3} =\sqrt 2+{1\over \sqrt 2} = \bbox[red, 2pt]{3\sqrt 2\over 2}$$

解答： $$\alpha,\beta,\gamma 為f(x)=0之三根 \Rightarrow \cases{\alpha +\beta +\gamma = -3\\ \alpha\beta +\beta\gamma +\gamma\alpha = -4\\ \alpha\beta \gamma = 2} 且f(\alpha)=f(\beta) = f(\gamma)=0\\ \cases{f(x)= x^3+3x^2 -4x-2\\ g(x)=x^4+6x^3 +5x^2-16x-2} \Rightarrow g(x)=f(x)(x+3)+(-2x+4) \\ \Rightarrow \cases{g(\alpha) = f(\alpha)(\alpha+3)+(-2\alpha+4) = -2\alpha+4 \\ g(\beta)= f(\beta)(\beta+3) +(-2\beta+4) = -2\beta+4 \\ g(\gamma) = f(\gamma)(\gamma+3) +(-2\gamma+4) = -2\gamma+4}  \\ \Rightarrow  {1\over g(\alpha)} +{1\over g(\beta)} +{1\over g(\gamma)} =-{1\over 2}\left({1\over \alpha-2} +{1\over \beta-2} +{1\over \gamma-2}  \right)\\ =-{1\over 2}\cdot { \alpha\beta +\beta\gamma +\gamma\alpha-4(\alpha+\beta +\gamma)+12\over \alpha\beta \gamma -2(\alpha\beta +\beta\gamma +\gamma\alpha) +4(\alpha+ \beta +\gamma)-8} =-{1\over 2}\cdot{ -4-4\cdot (-3)+12\over 2-2\cdot (-4)+4\cdot(-3)-8} =\bbox[red, 2pt]{1}$$

解答： $$10位數:9876543210\to 只有一個\\ 9位數:將9876543210拿掉任一個數字，剩下就符合條件的9位數，因此有C^{10}_1個\\8位數:將9876543210拿掉任2個數字，剩下就符合條件的8位數，因此有C^{10}_2個\\ \dots\\3位數:將9876543210拿掉任7個數字，剩下就符合條件的3位數，因此有C^{10}_3個\\ 總共有C^{10}_0+ C^{10}_1 +C^{10}_2 + \cdots +C^{10}_7 =\sum_{k=0}^{10} C^{10}_k -C^{10}_8-C^{10}_9-C^{10}_{10} =2^{10} -45-10-1 = \bbox[red, 2pt]{968}$$

解答：

$$x^2+y^2=|x|+|y| \Rightarrow \begin{cases} (x-1/2)^2+(y-1/2)^2 =1/2& \text {if }x,y\ge 0 \\  (x+1/2)^2+(y-1/2)^2 =1/2 & \text {if }x\le 0 ,y\ge 0 \\ (x-1/2)^2+(y+1/2)^2 =1/2 & \text {if }x\ge 0,y\le 0 \\ (x+1/2)^2+(y+1/2)^2 =1/2 & \text {if }x,y\le 0 \end{cases}\\ 所圍面積= 邊長為\sqrt 2的正方形ABCD及半徑為1/\sqrt 2的四個半圓=(\sqrt 2)^2+ 2\times ({1\over \sqrt 2})^2\pi \\ =\bbox[red, 2pt]{2 +\pi}$$

解答： $$y=ax+b=x^2 \Rightarrow x^2-ax-b=0恰有一根\Rightarrow a^2+4b=0 \cdots(1)\\ y=ax+b= (x-2)^2+12 \Rightarrow x^2-(a+4)x +16-b=0 恰有一根 \\\Rightarrow (a+4)^2-64+4b=0 \Rightarrow a^2+8a+16-64+4b=0\cdots (2) \\將(1)代入(2) \Rightarrow 8a-48=0 \Rightarrow a=6 代回(1) \Rightarrow b=-9 \Rightarrow a-b= \bbox[red,2pt]{15}$$

二、計算證明題（每題8分，共24分）

解答： $$等差數列\langle a_n\rangle ，其中\cases{a_1=4\\ d=8} \Rightarrow a_1+a_2+\cdots +a_n = {(2a_1+(n-1)d)n\over 2} ={(8+(n-1)8)n \over 2} =4n^2\\ 因此\cases{a_2+ a_4+ \cdots +a_{2n} =(a_2+ a_{2n})n\div 2 = 8n^2+4n\\ a_1+a_3+ \cdots +a_{2n-1} =(a_1+ a_{2n-1})n\div 2= 8n^2-4n} \\ \Rightarrow \lim_{n\to \infty} \left(\sqrt{a_2+ a_4+ \cdots +a_{2n}}-\sqrt{a_1+a_3+ \cdots +a_{2n-1}} \right)\\ =\lim_{n\to \infty}\left( \sqrt{8n^2+4n}-\sqrt{8n^2-4n}\right)=\lim_{n\to \infty} \cfrac{8n}{\sqrt{8n^2+4n}+\sqrt{8n^2-4n}} =\cfrac{8}{2\sqrt 2+2\sqrt 2} =\bbox[red, 2pt]{\sqrt 2}$$

解答：

(1) $$y=\ln x \Rightarrow x= e^y \Rightarrow {d\over dx}x ={d\over dx}e^y \Rightarrow 1=y'e^y =y'x \Rightarrow y'={1\over x} \Rightarrow \bbox[red,2pt]{{d\over dx}\ln x={1\over x}}$$

(2) $$S_n= \sum_{k=1}^n {1\over (2k)(2k-1)} = \sum_{k=1}^n \left({1\over  2k-1}-{1\over 2k} \right) = \sum_{k=1}^n \left({1\over  2k-1}-({1\over k}-{1\over 2k}) \right) \\ = \sum_{k=1}^n \left({1\over  2k-1}+{1\over 2k} \right) -\sum_{k=1}^n {1\over k} =\sum_{k=1}^{2n} {1\over k}-\sum_{k=1}^n {1\over k} =\sum_{k=n+1}^{2n} {1\over k} \\ =\sum_{k= 1}^{n}{1\over n+k} =\sum_{k= 1}^{n}{1\over n} \cdot {1\over 1+k/n} \Rightarrow \lim_{n\to \infty} S_n =\int_0^1 {1\over 1+x}\,dx = \ln 2-\ln 1= \bbox[red, 2pt]{\ln 2}$$

解答： $$(\log 6x) (\log 3x)+ k= (\log 2+\log 3x)(\log 3x)+ k= (\log 3x)^2 +\log 2(\log 3x) +k= 0 \\ 有相異二正根 \Rightarrow (\log 2)^2 -4k \gt 0 \Rightarrow \bbox[red,2pt]{k\lt {1\over 4}(\log 2)^2 }\\ 假設\alpha,\beta 為(\log 3x)^2 +\log 2(\log 3x) +k= 0的二根\Rightarrow  \log 3\alpha+ \log 3\beta =-\log 2 \Rightarrow \log 9\alpha\beta = \log {1\over 2}\\  \Rightarrow 9\alpha\beta ={1\over 2} \Rightarrow 兩根之積=\alpha\beta =\bbox[red,2pt]{1\over 18}$$
 

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解題僅供參考，其他教甄試題及詳解