115年台師大附中教甄-數學詳解

國立臺灣師範大學附屬高級中學115學年度第1次專任教師甄選

一、選填題：（每題 5 分，共 80 分。填在答案卷上， 分數或根式須以最簡形式回答，否則不予計分）

解答:$$21!是9的倍數也是11的倍數\Rightarrow \cases{5+1+0+9+\cdots+a+b= 55+a+b \\ 奇數位數:5+0+0+4+1+1+0+a=11+a\\ 偶數位數:1+9+9+2+7+7+9+b=44+b} \\ \Rightarrow \cases{55+a+b是9的倍數\\ (11+a)-(44+b)=a-b-33是11的倍數} \Rightarrow \cases{a+b=8或17\\a-b=0} \Rightarrow (a,b)=\bbox[red, 2pt]{(4,4)}$$

解答:$$\cases{H_0: p=0.02\\ H_1: p\lt 0.02 \\ 顯著水準\alpha =0.01\\ 拒絕域 X\in [n, \infty)} \Rightarrow P(X\ge n)=(1-p)^{n-1} =0.98^{n-1} \le 0.01 \Rightarrow (n-1)\log 0.98 \le -2 \\ \Rightarrow n-1\ge {-2\over \log 0.98} \approx {-2\over -0.0088} \approx 227.3 \Rightarrow n\ge 228.3 \Rightarrow n=\bbox[red, 2pt]{229} \\ 註:\log 98=\log 2+2\log 7=0.301+2\times 0.8451=1.9912 \Rightarrow \log 0.98=1.9912-2 =-0.0088$$

解答:$$\lim_{n\to \infty} \sum_{k=1}^n {n\over n^2+k^2} = \lim_{n\to \infty} \sum_{k=1}^n {1/n\over 1+(k/n)^2} = \int_0^1 {1\over 1+x^2} \,dx = \left. \left[ \tan^{-1} x \right] \right|_0^1 = \bbox[red, 2pt]{\pi\over 4}$$

解答:$${x^2\over 16}-{y^2\over 9}=0 \Rightarrow  \left( {x\over 4}-{y\over 3} \right) \left( {x\over 4}+{y\over 3} \right) =0 \Rightarrow 漸近線\cases{L_1: y=3x/4\\ L_2: y=-3x/4} \\ 假設\cases{A\in L_1 \Rightarrow A(4s,3s) \Rightarrow \overline{OA} =5|s| \\B\in L_2 \Rightarrow B(4t,-3t) \Rightarrow \overline{OB} =5|t|} \Rightarrow \overline{OA} \times \overline{OB}=25|s||t|=150   \Rightarrow |st|=6 \\ P(x,y)={1\over 2}(A+B) =(2(s+t),{3\over 2}(s-t)) \Rightarrow \cases{s+t=x/2\\ s-t=2y/3} \\ \Rightarrow (s+t)^2-(s-t)^2=4st \Rightarrow {x^2\over 4}-{4y^2\over 9}=4st \Rightarrow {x^2\over 16}-{y^2\over 9}=st \\ \Rightarrow {x^2\over 16}-{y^2\over 9}=6或{x^2\over 16}-{y^2\over 9}=-6 \Rightarrow qr=6\times(-6)= \bbox[red, 2pt]{-36}$$

解答:$$假設\cases{A(0,0,0) \\B(3,0,3) \\C(3,3,0) \\D(0,3,3)} \Rightarrow \cases{P=(2B+A)/3=( 2,0,2) \\Q=(2C+A)/3 =(2,2,0) \\R=(2D+C)/3= (1,3,2) \\S=(2D+B)/3 =(1,2,3)} \Rightarrow \cases{\overrightarrow{PQ} \times \overrightarrow{PR}=(6,2,2) \\ \overrightarrow{BC} \times \overrightarrow{BD}=(9,9,9)} \\ \Rightarrow \cases{BCD法向量\vec n_1=(1,1,1) \\PQRS法向量\vec n_2=(3,1,1)} \Rightarrow \cos \theta={\vec n_1\cdot \vec n_2\over |\vec n_1|| \vec n_2|} ={5\over \sqrt{33}} \\ \Rightarrow 兩平面夾角的餘弦值= \bbox[red, 2pt]{\pm {5\sqrt{33} \over 33}}$$

解答:

$$\cases{A(2,1,-2) \\B(4,-1,-4)} \Rightarrow M=\overline{AB}中點=(3,0,-3) \\ 依\triangle PAB中線定理: \overline{PA}^2+ \overline{PB}^2= 2\overline{PM}^2+{1\over 2} \overline{AB}^2 \\\Rightarrow 求 \overline{PA}^2+ \overline{PB}^2的最小值,相當於求\overline{PM}^2的最小值\\ M(3,0,-3)在E的投影點M'(1,-2,-2) \Rightarrow \cases{\overline{MM'}=3 \\ \overline{OM'} =3}, \\又\angle MM'P=90^\circ \Rightarrow \overline{PM}^2= \overline{PM'}^2+ \overline{MM'}^2 \Rightarrow 最小的\overline{PM'}=\overline{OM'}-\overline{OP}=3-1=2 \\ \Rightarrow 此時\overline{PM}^2=2^2+3^2=13 \Rightarrow 最小的\overline{PA}^2+ \overline{PB}^2=2\cdot 13+6= \bbox[red, 2pt]{32}$$
 

解答:$$t=\sin x+\cos x =\sqrt 2 \sin(x+\pi/4) \in [-\sqrt 2, \sqrt 2] \Rightarrow t^2=1+2\sin x \cos x\Rightarrow \sin x\cos x ={t^2-1\over 2} \\ \Rightarrow \sin^4x+\cos^4 x=(\sin^2x+ \cos^2 x)^2-2\sin^2x \cos^2x =1-2 \left( {t^2-1\over 2} \right)^2 ={1\over 2}(-t^4+2t^2+1) \\ \Rightarrow \sin^3x+\cos^3 x=(\sin x+\cos x)(\sin^2x-\sin x\cos x +\cos^2x) =t(1-{t^2-1\over 2}) ={1\over 2}(-t^3+3t) \\ \Rightarrow f(x) =g(t)={1\over 2}(-t^4+2t^2+1)+{1\over 2}(-t^3+3t)+1 ={1\over 2}(-t^4-t^3+ 2t^2+ 3t+3) \\ \Rightarrow g'(t)=0 \Rightarrow -4t^3-3t^2+4t+3= (1-t)(1+t)(4t+3)=0 \Rightarrow t=1,-1,-{3\over 4} \\ \Rightarrow \cases{g(1)=3\\ g(-1)=1\\ g(-3/4)= 507/512} 還有端點\cases{g(\sqrt 2) =(3+\sqrt 2)/2\\ g(-\sqrt 2)=(3-\sqrt 2)/2} \Rightarrow \cases{M=3\\ m=(3-\sqrt 2)/2} \\\Rightarrow M-m=\bbox[red,2pt]{3+\sqrt 2\over 2}$$

解答:$${m+n\over 1+mn} ={1\over 64}\Rightarrow mn-64m-64n=-1 \Rightarrow (m-64)(n-64)=4095\\ 取\cases{x=m-64\\ y=n-64} \Rightarrow m+n=x+y+128 \Rightarrow 在xy=4095的條件下求x+y的最小值 \\ 由於m,n\in \mathbb N \Rightarrow m,n\ge 1\Rightarrow x,y\ge -63 \Rightarrow xy=(-63)\cdot (-63)=3969 \lt 4095 \Rightarrow x,y 皆為正數\\ 4095=64^2-1 =64\times 65 \Rightarrow 取\cases{x=63\\ y=65} \Rightarrow m+n=256 \Rightarrow \log_\sqrt 2 (m+n) =\log_\sqrt 2 256 =\bbox[red, 2pt]{16}$$

解答:

$$假設直徑\overline{AB} =a \Rightarrow \cases{\angle  ACB=90^\circ\\ \angle ADB=90^\circ} \Rightarrow \cases{\overline{AC} =\sqrt{a^2-4} \\ \overline{BD}= \sqrt{a^2-144}} \\ \href{https://chu246.blogspot.com/2020/11/ptolemys-theorem.html}{托勒密定理}:\overline{AC} \times \overline{BD} = \overline{AB}\times \overline{CD}+ \overline{BC}\times \overline{AD} \Rightarrow \sqrt{a^2-4} \times \sqrt{a^2-144} =9a+24 \\ \Rightarrow a^3-229a-432=0 \Rightarrow (a-16)(a^2+16a+27)=0 \Rightarrow a= \bbox[red, 2pt]{16}$$

解答:$$本題\bbox[cyan,2pt]{送分}$$

解答:

$$從 9 個格子中隨機選出 4 個標記，總方法數為C^9_4=126, 將9個格子分三類\\第一類:角落格子(編號:1,3,7,9)\\\qquad 這個格子要得分，必須將2個緊鄰格被標記，剩下2個標記由剩下的7的格子中選出\\\qquad 有C^7_2=21 種選法，這個格子得分機率={21\over 126}，共4個角落格，貢獻4\times{21\over 126}={84\over 126} \\第二類:邊界格(編號:2,4,6,8)\\ \qquad 在3個緊鄰格中恰有2個被標記，有C^3_2=3種;剩下2個標記由6(9-3)個格子選出\\\qquad 有C^6_2=15種，得分機率={3\times 15\over 126}，共4個邊界格，貢獻4\times{3\times 15\over 126}={180\over 126} \\ 第三類:中心格(編號5)\\ \qquad 在四個緊鄰格中恰有2個被標記，有C^4_2=6種; 剩下2個標記由5(9-4)個格子選出\\\qquad 有C^5_2=10種，得分機率={10\times 6\over 126}，只有1個中心格，貢獻{60\over 126}  \\ 期望值={84\over 126}+{180\over 126}+{60\over 126}={324\over 126}=\bbox[red, 2pt]{18\over 7}$$

解答:$$A(3,0,0) \to B(0,3,3) \Rightarrow \cases{-X方向(代號a)3\\ Y方向(代號b)3\\ Z方向(代號c)3} \Rightarrow 3個a,3個b,3個c 排列數={9!\over 3!3!3!}=1680種走法\\ 令f(x,y,z)=2x+2y+z-4 \Rightarrow \cases{f(A)=2\gt 0\\ f(B)=5\gt 0} \Rightarrow 路徑上的點P需符合f(P)\gt 0\\ 在3個a,3個b,3個c的排列中挑出不合條件的排列數\\aXXXXXXXX:A(3,0,0)\to (2,0,0) \Rightarrow f(2,0,0)=0 \Rightarrow 剩下 8 步排列數:{8!\over 2!3!3!} =560 \\ caaXXXXXX:剩下 6 步排列數:{6!\over 1!3!2!}=60 \\ baaXXXXXX:剩下 6 步排列數:{6!\over 1!2!3!}=60 \\ ccaaXXXXX:剩下 5 步排列數:{5!\over 1!3!1!}=40\\ \cases{bcaa\\ cbaa\\ baca\\ caba} \Rightarrow 剩下4步排列數{4!\over 2!2!}=6 \Rightarrow 總共4\times 6=24種 \\ \cases{cccaaa\\ ccacaa\\ caccaa} \Rightarrow  3\times {3!\over 3!}=3; \cases{bbaaa\\ babaa}\Rightarrow 2\times {4!\over 1!3!} =8 \\ \cases{1b2c2a+a} \Rightarrow 13\times {3!\over 2!1!}=39  \\ \Rightarrow 1680-(560+60+60 +40+24+3 +8+39) =\bbox[red, 2pt]{886}$$

解答:$$已知\cases{圓C_1: 圓心O_1(-5,0), 半徑r_1=4\\ 圓C_2: 圓心O_2(5,0), 半徑r_2= 2\\ 圓C_3: 圓心O_3(29,24), 半徑r_3=2}, 假設外切圓C:圓心O(x,y), 半徑r \\ \Rightarrow \cases{\overline{OO_1}=r+r_1=r+4\\ \overline{OO_2}=r+r_2=r+2\\ \overline{OO_3} =r+r_3=r+2} \Rightarrow \overline{OO_2} =\overline{OO_3} \Rightarrow O在\overline{O_2O_3}的中垂線L:x +y=29上    \\ 又\overline{OO_1} -\overline{OO_2}=2為一雙曲線\Gamma: \cases{O_1,O_2為焦點\Rightarrow c=5\\ 2a=2 \Rightarrow a=1} \Rightarrow b^2=5^2-1=24 \\ \Rightarrow \Gamma:{x^2-{y^2\over 24}=1}, 又 \overline{OO_1}\gt \overline{OO_2} \Rightarrow O在\Gamma的右半邊,即x\gt 0 \\ 由\cases{x+y=29\\ x^2-y^2/24=1 } \Rightarrow 23x^2+58x-865=0 \Rightarrow (x-5)(23x+173)=0 \Rightarrow x=5\\ \Rightarrow y=29-5=23 \Rightarrow 圓心坐標\bbox[red, 2pt]{(5,24)}$$

解答:$$f(x)=x^3-x \Rightarrow f'(x)=3x^2-1 \\\Rightarrow 在切點(t,t^3-t)斜率為f'(t)的切線方程式: y=(3t^2-1)(x-t)+t^3-t \Rightarrow y=(3t^2-1)x-2t^3\\ 同理,g(x)=(x-1)^3-(x-1)+2 \Rightarrow g'(x)=3(x-1)^2-1 \\ \Rightarrow 在切點(u+1,u^3-u+2)的切線方程式:y=(3u^2-1)(x-(u+1)) =u^3-u+2 \\ \Rightarrow y=(3u^2-1)x-2u^3-3u^2+3 \\ 兩條切線\cases{y=(3t^2-1)x-2t^3\\ y=(3u^2-1)x-2u^3-3u^2+3}需相等\Rightarrow \cases{t^2=u^2 \Rightarrow t=\pm u\\ -2t^3=-2u^3-3u^2+3}\\ u=t \Rightarrow -3t^2+3=0 \Rightarrow \cases{t=1\Rightarrow 切線方程式: y=2x- 2 \\t=-1 \Rightarrow 切線方程式: y=2x+2} \Rightarrow \cases{a=2\\ (b,c)=(2,-2)或(-2,2)} \\ \Rightarrow abc= \bbox[red, 2pt]{-8}$$

解答:

$$A為\Gamma_1切點\Rightarrow \angle DAC =\angle B \Rightarrow \triangle DAC\sim \triangle DBA \;(AAA) \Rightarrow {\overline{DA} \over \overline{DB}} = {\overline{DC} \over \overline{DA}} = {\overline{AC} \over \overline{AB}} ={6\over 8} \\ \Rightarrow {\overline{DC} \over \overline{DB}} ={\overline{DC} \over \overline{DA}} \times {\overline{DA} \over \overline{DB}} ={6\over 8}\times {6\over 8} ={9\over 16} \Rightarrow \overline{DC}=9 \\ 令P=\overleftrightarrow{AE} \cap \overline{CD} 及 \overline{DP}=a \Rightarrow \overline{PC}=9-a \Rightarrow \cases{\Gamma_1:\overline{PE}\times \overline{PA}= \overline{PC}\times \overline{PB} \\ \Gamma_2: \overline{PD}^2= \overline{PE}\times \overline{PA}} \Rightarrow \overline{PD}^2=\overline{PC}\times \overline{PB} \\ \Rightarrow a^2=(9-a)(16-a) \Rightarrow a={144\over 25} \Rightarrow \overline{PC} =9-a ={81\over 25} \Rightarrow \overline{PB}=16-a={256\over 25} \\ 又\cases{\triangle PEB \sim \triangle PCA (AAA) \Rightarrow {\overline{BE} \over \overline{CA}} ={\overline{BP} \over \overline{AP}} \cdots(1) \\ \triangle PEC \sim \triangle PBA(AAA) \Rightarrow {\overline{EC} \over \overline{AB}} ={\overline{PC} \over \overline{PA}} \cdots(2)} \Rightarrow {\overline{BE} \over \overline{EC}} ={\overline{CA} \over \overline{AB}} \times {\overline{BP}\over \overline{CP}} ={6\over 8} \times {256\over 81}= \bbox[red, 2pt]{64\over 27}$$

解答:$$a_{n+1} ={7a_n+ \sqrt{45a_n^2-36}\over 2} \Rightarrow \cases{n=0 \Rightarrow a_1=5\\ n=1 \Rightarrow a_2=34} \\ F_1=F_2=1 \Rightarrow F_3=2\Rightarrow F_4=3 \Rightarrow F_5=5 \Rightarrow F_6=8 \Rightarrow F_7=13 \Rightarrow F_8=21 \Rightarrow F_9=34 \\ \Rightarrow \cases{a_0=1=F_1\\ a_1=5 =F_5\\ a_2=34=F_9} \Rightarrow a_n=F_{4n+1} \Rightarrow 2025=4m+1 \Rightarrow \bbox[red, 2pt]{506}$$

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