113年全國高中教甄聯招-數學詳解

教育部受託辦理113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:$$\vec u\parallel \vec v \Rightarrow {b\over \cos B}={3a-c\over \cos C} \Rightarrow b\cos C=3a\cos B-c\cos B \\ \Rightarrow b\cos C+c\cos B=a=3a\cos B\Rightarrow \cos B={a\over 3a}={1\over 3} ,故選\bbox[red, 2pt]{(D)}$$

解答:$$令t=\log_5 4 \Rightarrow {1\over t}=\log_4 5 \Rightarrow f(\log_4 t)+f(\log_4 {1\over t}) =2^{\log_4 t}-2^{-\log_4 t} +2^{\log_4 {1\over t}}-2^{-\log_4{1\over t}} \\=2^{\log_4 t}-2^{\log_4{1\over t}}+2^{ \log_4{1\over t}}-2^{\log_4 t} =0, 故選\bbox[red, 2pt]{(D)}$$

解答:$$(a,b)=(1-9,1-9)共有9\times 9=81個格子點,其中需滿足b\gt a\cdot a^2-b\cdot a \\ \Rightarrow b\gt a^3-ab \Rightarrow ab+b\gt a^3\\ \begin{array}{}a& b \\\hline 1& 2b\gt 1 \Rightarrow b=1-9\\ 2& 3b\gt 8 \Rightarrow b=3-9\\ 3&4b\gt 27 \Rightarrow b=7-9\\ 4& \times\\\hline \end{array} \Rightarrow 共有9+7+3=19個 \Rightarrow 機率={19\over 81},故選\bbox[red, 2pt]{(C)}$$

解答:$$(a+3i)(b+i)=ab-3+(a+3b)i = 12+14i \Rightarrow \cases{ab=15\\ a+3b=14} \Rightarrow \cases{a=5\\ b=3}, a,b\in \mathbb Z \\ \Rightarrow \cases{\tan \alpha=3/5\\ \tan \beta=1/3} \Rightarrow \tan(\alpha+\beta) =\tan \theta={\tan \alpha+ \tan \beta\over 1-\tan \alpha \tan \beta } ={7\over 6} \\ \Rightarrow (\tan 45^\circ) 1\lt \tan \theta \lt \sqrt 3(\tan 60^\circ) \Rightarrow 45^\circ \lt \theta \lt 60^\circ,故選\bbox[red, 2pt]{(B)}$$

解答:$$假設\cases{\int_0^1 g(x)\,dx =a\\ \int_0^2 f(x)\,dx =b } \Rightarrow \cases{f(x)=x+3+a\\ g(x)=2x-9+b} \Rightarrow \cases{\int_0^2 f(x) \,dx = \int_0^2( x+3+a)\,dx \\ \int_0^1 g(x)\,dx = \int_0^1 (2x-9+b)\,dx} \\ \Rightarrow \cases{b= \left.\left[ {1\over 2}x^2+(3+a)x \right] \right|_0^2 =8+2a \\ a= \left.\left[  x^2+(b-9)x \right] \right|_0^1 =b-8} \Rightarrow \cases{2a-b=-8\\ a-b=-8} \Rightarrow \cases{a=0\\ b=8} \\ \Rightarrow f(x)=x+3+0 \Rightarrow f(3)=6,故選\bbox[red, 2pt]{(D)}$$

解答:$$x^2-2x+\sin C+\cos C =0 \Rightarrow (x-1)^2+\sin C+\cos C-1=0 \\ \Rightarrow \cases{重根1=\log_a b\\ \sin C+\cos C-1=0} \Rightarrow \cases{a=b\\ (\sin C+\cos C)^2=1+\sin 2C=1} \\ \Rightarrow \sin 2C=0 \Rightarrow 2C=180^\circ \Rightarrow C=90^\circ \Rightarrow 等腰直角三角形,故選\bbox[red, 2pt]{(D)}$$

解答:$$2^{1+\lfloor\log_2(N-1)\rfloor}-N=19 \Rightarrow  \lfloor\log_2(N-1)\rfloor=\log_2(N+19)-1\\ \lfloor\log_2(N-1)\rfloor是整數\Rightarrow \log_2(N+19)-1也是整數 \Rightarrow N+19是2的次方\\ \Rightarrow N+19=64,128 \Rightarrow N=45,109 \Rightarrow 45+109=154,故選\bbox[red, 2pt]{(C)}$$

解答:$${x^2\over 4}-{y^2\over 5}=1 \Rightarrow \cases{a=2\\ b=\sqrt 5} \Rightarrow c=3\Rightarrow \cases{F(3,0)\\ F'(-3,0)} \\ 過F(3,0)且斜率為1的直線L=\overleftrightarrow{PQ}: y=x-3 \Rightarrow \cases{P(x_1,x_1-3)\\ Q(x_2,x_2-3)}\\ 且{x^2\over 4}-{(x-3)^2\over 5}=1 \Rightarrow x^2+24x-56=0 \Rightarrow x=-12\pm 10\sqrt 2\\ \Rightarrow \overline{PQ}=\sqrt{2}|x_1-x_2| =\sqrt2 \cdot 20\sqrt 2=40\\ \cases{\overline{PF'}-\overline{PF}=2a=4\\ \overline{QF'}-\overline{QF}=2a=4} 兩式相加\Rightarrow \overline{PF'} +\overline{QF'} =\overline{PF} +\overline{QF} +8=40+8=48,故選\bbox[red, 2pt]{(D)},本題\bbox[cyan,2pt]{送分}$$

二、複選題（每題 4分，共 16分）

解答:$$(A)\bigcirc：f'(1)是f'(x)的極值 \Rightarrow f''(1)=0\\ (B)\bigcirc: 切線過原點及(1,0) \Rightarrow 切線:y＝0\Rightarrow f'(1)=0,因此我們有\cases{f(1)=0\\ f'(1)=0 \\ f''(1)=0} \\ \qquad \Rightarrow \cases{1+a+b+c=0\\ 3+2a+b=0\\ 6+2a=0} \Rightarrow \cases{a=-3\\ b=3\\ c=-1 } \Rightarrow f(x)=x^3-3x^2+3x-1 \\ f'(x)=0 \Rightarrow 3x^2-6x+3=3(x-1)^2=0 \Rightarrow x=1 \Rightarrow f''(1)=0無極值 \\(C)\bigcirc: 切線為y=0,即X軸 \\(D)\times: f'(x)=3(x-1)^2 \ge 0 \Rightarrow y=f(x)圖形遞增 \Rightarrow \cases{y=f(x)\\y=1}只有一個交點\\,故選\bbox[red, 2pt]{(ABC)}$$

解答:$$A(n)=A(n-1)+d_A(n), 其中\cases{A(0)=0 \\ d_A(0)=0\\ d_A(1)=1\\ d_A(n)=d_A(n-1)/2} \\ \Rightarrow A(n)=A(0)+d_A(1)+d_A(2) +\cdots +d_A(n)  =0+1+{1\over 2}+{1\over 4}+\cdots {1\over 2^{n-1}} =2-{1\over 2^{n-1}} \\ B(n)=B(n-1)-d_B(n), 其中\cases{B(0)=8 \\ d_B(0)=0\\ d_B(1)=4\\ d_B(n)= d_B(n-1)/3} \\ \Rightarrow B(n)=B(0)-(d_B(1)+d_B(2)+ \cdots +d_B(n))= 8-4(1+ 1/3+\cdots +1/3^{n-1})=2+{2\over 3^{n-1}}\\ C(n)={1\over 2}(A(n+B(n))= {1\over 2}\left( 2-{1\over 2^{n-1}}+2+{2\over 3^{n-1}}\right) =2-{1\over 2^n}+{1\over 3^{n-1}} \\(A) \bigcirc: C(1)=2-{1\over 2}+1={5\over 2} \\(B) \times:C(2)=2-{1\over 4}+{1\over 3} =2+{1\over 12} \lt c_1 \\(C)\times: {C(n+1)\over C(n)}非常數 \\(D)\bigcirc: \lim_{n\to \infty} \left( 2-{1\over 2^n}+{1\over 3^{n-1}}\right) =2\\,故選\bbox[red, 2pt]{(AD)}$$

解答:$$(p+q)^5 = q^5+C^5_1pq^4 +C^5_2 p^2q^3+ C^5_3p^3q^2+ C^5_4p^4q+ p^5\\(A)\times: P(X=1)=C^5_1pq^4=5\cdot {3\over 4}\cdot ({1\over 4})^4 ={15\over 1024}\\ (B)\bigcirc: \cases{P(X=2) =C^5_2p^2q^3\\ P(X=3)=C^5_3p^3q^2} \Rightarrow P(X=3)\gt P(x=2) \;(\because C^5_3=C^5_2,且p\gt q) \\(C)\bigcirc: f(p)=(p+q)^5 \Rightarrow E(X)= pf'(p)=5p(p+q)^4 =5\cdot {3\over 4}\cdot  ={15\over 4}  \\(D)\bigcirc: X\sim B(n,p) \Rightarrow \sigma(X)= \sqrt{npq} =\sqrt{5\cdot {3\over 4}\cdot {1\over 4}} ={\sqrt{15}\over 4}\\,故選\bbox[red, 2pt]{(BCD)}$$

解答:$$(A)\times:C_1繞x軸旋轉體積剛好為一單位球，體積={4\over 3}\pi \\(B)\bigcirc:理由同(A)\\ (C)\times: \cases{C_2圓心(0,1)繞x軸一圈距離=2\pi \\C_2圓面積=\pi} \Rightarrow C_2繞一圈體積＝2\pi\cdot \pi=2\pi^2\\ (D)\bigcirc: 理由同(C)\\,故選\bbox[red, 2pt]{(BD)}$$

第二部分：綜 合題 共 60分

一、填充 題（每 格 4 分，共 36分）

解答:$$|U_n-L_n|= {|f(3)-f(1)\over n/(3-1)} ={16\over n}\lt {1\over 10000} \Rightarrow n\gt 160000 \Rightarrow n=\bbox[red, 2pt]{160001}$$

解答:$$(4,-3)與(-3,4)對稱於直線x=-y \Rightarrow (1,2\sqrt 6)的對稱點為 \bbox[red, 2pt]{(-2\sqrt 6,1)}$$

解答:$$O為外心\Rightarrow \cases{\overrightarrow{AO} \cdot \overrightarrow{AB}= {1\over 2} |\overrightarrow{AB}|^2 \\ \overrightarrow{AO} \cdot \overrightarrow{AC}= {1\over 2} |\overrightarrow{AC}|^2 } \Rightarrow \cases{ (\overrightarrow{AB} +3 \overrightarrow{AC}) \cdot \overrightarrow{AB} = |\overrightarrow{AB}|^2+ 3\overrightarrow{AB} \cdot \overrightarrow{AC}={1\over 2} |\overrightarrow{AB}|^2 \\(\overrightarrow{AB} +3 \overrightarrow{AC}) \cdot \overrightarrow{AC} = 3|\overrightarrow{AC}|^2+ \overrightarrow{AB} \cdot \overrightarrow{AC}={1\over 2} |\overrightarrow{AC}|^2 } \\ \Rightarrow \overrightarrow{AB} \cdot \overrightarrow{AC} =-{1\over 6} \overline{AB}^2=-{5\over 2}\overline{AC}^2 \Rightarrow \overline{AB}= \sqrt{15}\cdot  \overline{AC} \\ \cos \angle BAC ={\overrightarrow{AB} \cdot \overrightarrow{AC} \over \overline{AB}\cdot \overline{AC}} ={-{5\over 2} \overline{AC}^2 \over \sqrt{15}\cdot \overline{AC}\cdot \overline{AC}} =-{5\over 2\sqrt{15}} \\ \Rightarrow \sin \angle BAC=\sqrt{1-\cos^2 \angle BAC} =\sqrt{1-{25\over 60}} =\bbox[red, 2pt]{\sqrt{21}\over 6}$$

解答:$$令g(x)=x^6-1 \Rightarrow f(1)=g(1)=0 \Rightarrow 1+a+b+c=0 \cdots(1)\\ \lim_{x\to 1} {f(x)\over g(x)} =\lim_{x\to 1} {f'(x)\over g'(x)} = \lim_{x\to 1} {3x^2+2ax+b\over 6x^5} = {3+2a+b\over 6}=2 \Rightarrow 2a+b=9 \cdots(2)\\ 對於(2)而言,滿足a\gt b\gt 0,a,b均為整數,只有\cases{a=4\\b=1} \Rightarrow 1＋4＋1＋c＝0 \Rightarrow c=-6\\ \Rightarrow \bbox[red, 2pt]{\cases{a=4\\b=1\\ c=-6}}$$

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:$$E(X)={1\over n}(a+(a+d)+\cdots+ (a+(n-1)d))=a+{n-1\over 2}d\\ E(X^2)={1\over n} \sum_{k=0}^{n-1}(a+kd)^2 ={1\over n} \left(\sum_{k=0}^{n-1}a^2+ 2ad \sum_{k=0}^{n-1}k+ d^2\sum_{k=0}^{n-1}k^2 \right)\\\qquad ={1\over n}\left( na^2+adn(n-1)+d^2\cdot {(n-1)n(2n-1)} \right) =a^2+ad(n-1)+d^2{(n-1)(2n-1)\over 6} \\ \Rightarrow Var(X)=E(X^2)-(E(X))^2 = d^2\left( {(n-1) (2n-1)\over 6} -{(n-1)^2\over 4}\right)^2 ={d^2(n^2-1)\over 12}=260 \\\Rightarrow  {13\over 4}\cdot {n^2-1\over 12}=260 \Rightarrow n^2=961 \Rightarrow n= \bbox[red, 2pt]{31}$$

解答:

$$對同弧的圓周角相等\Rightarrow \cases{\angle ABD=\angle ACD\\ \angle BAC= \angle BDC} \Rightarrow \triangle PAB\sim \triangle PCD (AA) \\ \Rightarrow {\overline{BP} \over \overline{PD}} ={\overline{AB}\over \overline{CD}} ={1\over 3} \Rightarrow \cases{\overline{BP}=1\\ \overline{PD}=3} \Rightarrow \triangle  ABP為正\triangle \Rightarrow  \triangle PCD也是正\triangle \\ \Rightarrow  \overline{PC}=3  \Rightarrow \triangle BCP面積={1\over 2} \cdot \overline{BP} \cdot \overline{PC} \cdot \sin \angle BPC={1\over 2}\cdot 1\cdot 3\cdot \sin 120^\circ =\bbox[red, 2pt]{3\sqrt 3\over 4}$$

解答:$$1- P(3顆 CPU 以上LAG) > 1- P(2顆 CPU 以上LAG) \\ \Rightarrow  1- ( C^4_3 \cdot p^3(1-p) + C^4_4\cdot p^4 ) > 1 -  C^2_2p^2  \Rightarrow 3p^4-4p^3+p^2\gt 0 \\ \Rightarrow p^2(3p-1)(p-1)\gt 0 \Rightarrow \bbox[red, 2pt]{0\lt p\lt {1\over 3}}$$

解答:$$f(x)\ge 0 \Rightarrow (x-2)^2-1\ge 0 \Rightarrow \cases{x-2\ge 1\\ x-2\le -1} \Rightarrow x\ge 3或x\le 1 \\ \Rightarrow f(|x|)=|f(x)| \Rightarrow (x\ge 3或x\le 1)\cap (x\ge 0) \Rightarrow \bbox[red, 2pt]{0\le x\le 1或x\ge 3}$$

二、 計算 證明 題（每題 8分，共 24分）

解答:$$利用柯西不等式: \left(\sum_{i=1}^n(x_i-\mu_x)^2 \right)\cdot \left(\sum_{i=1}^n(y_i-\mu_y)^2 \right) \ge \left(\sum_{i=1}^n(x_i-\mu_x) (y_i-\mu_y) \right)^2 \\ \Rightarrow \cfrac{\left(\sum_{i=1}^n(x_i-\mu_x) (y_i-\mu_y) \right)^2}{\left(\sum_{i=1}^n(x_i-\mu_x)^2 \right)\cdot \left(\sum_{i=1}^n(y_i-\mu_y)^2 \right)} \le 1 \\ \Rightarrow -1 \le \cfrac{ \sum_{i= 1}^n(x_i-\mu_x) (y_i-\mu_y)  }{\sqrt{\sum_{i=1}^n(x_i-\mu_x)^2}  \cdot \sqrt{\sum_{i= 1}^n(y_i-\mu_y)^2 }} \le 1 \\ \Rightarrow -1\le r\le 1 \Rightarrow |r| \le 1. \;\bbox[red, 2pt]{QED}$$

解答:$$假設Q(x,y)在直線L上,則\overline{PQ}=\sqrt{(x-x_0)^2 +(y-y_0)^2} = \sqrt{{(ax-ax_0)^2 \over a^2}+{(by-by_0)^2 \over b^2}}  \\ \Rightarrow \overline{PQ}^2= {(ax-ax_0)^2 \over a^2}+{(by-by_0)^2 \over b^2}\\ 利用柯西不等式:\overline{PQ}^2\cdot (a^2+b^2) \ge ((ax-ax_0)+(by-by_0))^2 =(ax_0+by_0+c)^2 \\ \Rightarrow \overline{PQ} \ge  {|ax_0 +by_0+c| \over \sqrt{a^2+b^2}} \Rightarrow \overline{PQ}的最短距離=d(P,L)=   {|ax_0 +by_0+c| \over \sqrt{a^2+b^2}}. \;\bbox[red, 2pt]{QED}$$

解答:$$f(x)=x^{18}+4x^{11}+1 =(x-x_1)(x-x_2)\cdots (x-x_{18})\\ \Rightarrow f(1)=1+4+1=6=(x_1-1) (x_2-1) \cdots (x_{18}-1)\\ x^6-1=(x^2-1)(x^4+x^2+1)=0的六根為\omega,\omega^2, \dots ,\omega^6,其中\omega=e^{\pi i/3} \\ \Rightarrow x^4+x^2+1=0的四根為\omega, \omega^2, \omega^4, \omega^5 ,且\omega^6=1,\omega^3=-1\\\Rightarrow x^4+x^2+1=(x-\omega)(x-\omega^2) (x- \omega^4)( x-\omega^5)\\ 因此\prod_{i=1}^{18} (x_i^4+x_i^2+1) = \prod_{i=1}^{18}(x_i-\omega)(x_i-\omega^2)(x_-\omega^4) (x_-\omega^5) \\=\left( \prod_{i=1}^{18}(x_i-\omega) \right) \left( \prod_{i=1}^{18}(x_i-\omega^2) \right) \left(\prod_{i=1}^{18}(x_i-\omega^3) \right) \left( \prod_{i=1}^{18}(x_i-\omega^4) \right) \\=f(\omega) f(\omega^2) f(\omega^4) f(\omega^5) \\=(\omega^{18}+4\omega ^{11}+1) (\omega^{36}+4\omega ^{22}+1) (\omega^{72}+4\omega ^{44}+1) (\omega^{90}+4\omega ^{55}+1) \\= (4\omega ^{5}+2) (4\omega ^{4}+2) ( 4\omega ^{2}+2) (4\omega +2) = 16(2\omega ^{5}+1) (2\omega ^{4}+1) ( 2\omega ^{2}+1) (2\omega +1)  \\=16(2\omega^5+ 2\omega^4+ 4\omega^3+1)(4\omega^3+ 2\omega^2+2\omega +1)=16(-2\omega^2- 2\omega-3)(2\omega^2+2\omega -3) \\=16((-3)^2-(2\omega^2+2\omega)^2) =16(9-4\omega^4-8\omega^3-4\omega^2) \\=16(13-4(\omega^4 + \omega^2+ 1) -8\omega^3 ) =16(13-0+8)=16\cdot 21=\bbox[red, 2pt]{336}$$

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解題僅供參考，其他歷年試題及詳解

註:數學科選擇第8題原公告答案是D不是C!!