110年竹北高中第1次教甄-數學詳解

國立竹北高中 110 年 第 1 次教師甄試

第一部分：填充題

解答： $$不請假\to 1\\請假1天\to C^5_1=5\\ 請假連續2天1次\to 4((1,2),(2,3),(3,4),(4,5))\\ 請假2天不連續\to 6((1,3),(1,4),(1,5), (2,4),(2,5), (3,5))\\ 請假3天完全不連續\to 1((1,3,5))\\ 請假3天:2天連續,1天不連續\to 6((1,2,4),(1,2,5), (2,3,5),(1,3,4), (1,4,5), (2,4,5))\\ 請假4天\to 1((1,2,4,5))\\ 共有1+5+4+6 +1+6+1 =\bbox[red, 2pt]{24}種$$

解答： $$S_8= a_1+a_2+ \cdots +a_8=2，由於a_k\in \{1,-1\},k\in\{1,\dots,8\}，\\因此a_1a_2\cdots a_8 為5個1及3個-1的排列，排列數為{8!\over 5! 3!} =56 \\ \Rightarrow 機率為\left({1\over 3} \right)^5\cdot \left({2 \over 3} \right)^3 \cdot 56 ={8\times 56 \over 3^8} =\bbox[red, 2pt]{448 \over 6561}$$

解答：

$$由上圖可知\cases{P=(C+D+G+H)/4 = (1/2,1,1/2)\\ Q=(A+B+C+D)/4 = (1/2,1/2,0)} \Rightarrow \overline{PQ} ={1\over \sqrt 2} \\ \Rightarrow 八面體體積={\sqrt 2\over 3}\cdot \left({1\over \sqrt 2}\right)^3 ={1\over 6} \Rightarrow {a\over b} ={1/6\over 1} =\bbox[red, 2pt]{1\over 6}$$

解答：

$$x^5=i = e^{i(90^\circ+360^\circ\cdot k)} \Rightarrow \omega_k =e^{i(18^\circ +72^\circ\cdot k)},k=0,1,2,3,4 \\\Rightarrow A(\omega_0), B(\omega_1), C(\omega_2), D(\omega_3),E(\omega_4),在單位圓上，且ABCDE為正五邊形，見上圖；\\顯然|1-\omega|的最大值為\overline{PC} = \overline{PQ}\cos \angle QPC = 2\cos 9^\circ = \bbox[red, 2pt]{2\sin 81^\circ}\\，其中\overline{PQ}為直徑，\angle QPC = (180^\circ -\angle COP)\div 2= (180^\circ-(18^\circ+72^\circ\times 2))\div 2=9^\circ$$

解答： $$\cases{\log_3 x+x-3=0\\ 3^x+x-3=0} \Rightarrow \log_3 x=3-x=3^x\\ 令\cases{f(x)=\log_3 x\\ g(x)=3^x}，則f(x)與g(x)互為反函數，因此兩圖形\cases{y=f(x)\\ y=g(x)}對稱於直線y=x\\ \Rightarrow 兩圖形的交點為(a,a)=(a,3-a) \Rightarrow a=3/2 \Rightarrow \log_3\alpha +3^\beta = {3\over 2}+{3\over 2}= \bbox[red, 2pt]{3}$$

解答： $$a,b,c為x^3-10x^2 +44x-14=0的三根 \Rightarrow \cases{a+b+c=10 \\ ab+bc+ca= 44 \\ abc=14} \\ 令s=(a+b+c)/2 = 5，則\triangle ABC面積= \sqrt{s(s-a)(s-b)(s-c)} \\= \sqrt{5(5-a)(5-b)(5-c)} =\sqrt{ 5(125-25(a+b+c) +5(ab+bc+ca) -abc)} \\ =\sqrt{5(125-250 +220-14)} = \sqrt{5\cdot 81}= \bbox[red, 2pt]{9\sqrt 5}$$

解答：

$$令B在原點，則\cases{A(0,8)\\ B(0,0)\\ C(6,0)\\ D(x,y)}；\\\angle B=90^\circ \Rightarrow \overline{AC}= \sqrt{6^2+8^2}=10，又\angle D=90^\circ \Rightarrow \overline{AD}=\sqrt{10^2-5^2}=5 \sqrt 3\\ \cases{\overline{AD}= 5\sqrt 3\\ \overline{CD}=5} \Rightarrow \cases{x^2+(y-8)^2 = 75\\ (x-6)^2+ y^2=25} \Rightarrow \cases{x=(9+4\sqrt 3)/2\\ y=(4+3\sqrt 3)/2} \\ \Rightarrow \cases{\overrightarrow{BC}=(6,0)\\ \overrightarrow{AD} =({9+4\sqrt 3\over 2}, {-12+3\sqrt 3\over 2})} \Rightarrow \overrightarrow{BC} \cdot \overrightarrow{AD}= \bbox[red, 2pt]{27+12\sqrt 3}$$

解答： $$22個不同數字任取3個，共有C^{22}_3組合，每一種組合的機率都是{3\over 66}\times {3\over 65} \times {3\over 64}\\ 因此機率為C^{22}_3 \times {3\over 66}\times {3\over 65} \times {3\over 64} =\bbox[red, 2pt]{63\over 416}$$

解答： $$由A(1,0)\to A'\left({24\over 25},{7\over 25} \right) \Rightarrow \cases{\sin \theta =7/25\\ \cos\theta = 24/25} \Rightarrow 旋轉(-\theta)矩陣R= \begin{bmatrix}\cos(-\theta) & -\sin(-\theta)\\ \sin(-\theta) & \cos(-\theta) \end{bmatrix} \\ =\begin{bmatrix}\cos(\theta) & \sin(\theta)\\ -\sin(\theta) & \cos(\theta) \end{bmatrix} = \begin{bmatrix}24/25 & 7/25\\ -7/25 & 24/25 \end{bmatrix}\\ 令M=\begin{bmatrix}a & b\\ c & d \end{bmatrix}，由題意知\cases{MRA'=A\\ MRC'=C} \Rightarrow \cases{\begin{bmatrix}a & b\\ c & d \end{bmatrix} \begin{bmatrix}24/25 & 7/25\\ -7/25 & 24/25 \end{bmatrix} \begin{bmatrix}24/25 \\ 7/25  \end{bmatrix} =\begin{bmatrix}1 \\ 0  \end{bmatrix} \\ \begin{bmatrix}a & b\\ c & d \end{bmatrix} \begin{bmatrix}24/25 & 7/25\\ -7/25 & 24/25 \end{bmatrix} \begin{bmatrix}-1/7 \\ 1  \end{bmatrix} =\begin{bmatrix}0 \\ 1  \end{bmatrix}} \\ \Rightarrow \cases{\begin{bmatrix}a & b\\ c & d \end{bmatrix}  \begin{bmatrix}1 \\ 0  \end{bmatrix} =\begin{bmatrix}1 \\ 0  \end{bmatrix} \\ \begin{bmatrix}a & b\\ c & d \end{bmatrix}  \begin{bmatrix}1/7 \\ 1 \end{bmatrix} =\begin{bmatrix}0 \\ 1  \end{bmatrix}} \Rightarrow \cases{a=1\\ c=0\\ a/7+b=0\\ c/7+d=1} \Rightarrow \cases{a=1\\ b=-1/7\\ c=0\\ d=1} \Rightarrow M= \bbox[red, 2pt]{\begin{bmatrix}1 & -1/7\\ 0 & 1 \end{bmatrix}}$$

解答： $$a_n = 2a_{n-1} +2^n = 2(2a_{n-2}+2^{n-1}) + 2^n = 2^2a_{n-2} +2\cdot 2^n\\ = 2^2(2a_{n-3} + 2^{n-2}) +2\cdot 2^n =2^3a_{n-3} + 3\cdot 2^n = \dots \\ =2^{n-1}a_1 + (n-1)2^n =2^{n-1} + (n-1)2^n =2^{n-1}(1+(n-1)\cdot 2) \\ = \bbox[red, 2pt]{(2n-1)2^{n-1}}$$

解答： $$1,2,\dots,8分成2組(不分組別)共有C^8_4\div 2=35\\ 1+2+\cdots+ 8=36 \Rightarrow 號碼和=18時，代表平手\\ 平手的情形有8721,8631, 8541,8532四種情形(不分組別)\\ 因此號碼和分出高低的有35-4=\bbox[red, 2pt]{31}種$$

解答：

$$x-my=n與x軸交於B(n,0)，且圓心C在\overline{OB}的中垂線上，因此假設C(n/2,k)，如上圖；\\\cases{\overline{CO}=10 \\ \overline{CA}=10} \Rightarrow \cases{{n^2\over 4}+ k^2 = 100 \cdots(1)\\ ({n\over 2}-5\sqrt 3)^2+ (k-5)^2=100 \cdots(2)}，將(1)代入(2) \Rightarrow 5\sqrt 3n+10k =100 \\\Rightarrow k=10-{\sqrt 3\over 2}n 代回(1) \Rightarrow {n^2 \over 4} +100-10\sqrt 3n+{3\over 4}n^2=100 \Rightarrow n^2-10\sqrt 3n=0\\ \Rightarrow n=\bbox[red, 2pt]{10\sqrt 3}$$

解答：

$$P、Q互為對稱點，且\cases{\overline{AB}與\overline{PQ}交於R\\\overline{AB}與\overline{OQ}交於S }，見上圖；\\則\cases{R=(P+Q)/2 = ((2\cos\theta+1)/2,\sin\theta)\\ S(1,\tan\theta)} \Rightarrow \cases{\overrightarrow{PQ} = (2\cos\theta-1,2\sin \theta) \\ \overrightarrow{SR} =(\cos\theta -1/2,\sin\theta -\tan\theta)} \\ 由於 \overrightarrow{PQ} \bot \overrightarrow{SR} \Rightarrow  \overrightarrow{PQ} \cdot \overrightarrow{SR}=0 \Rightarrow (2\cos\theta-1,2\sin \theta) \cdot (\cos\theta -1/2,\sin\theta -\tan\theta)=0\\ \Rightarrow {5\over 2}-2\cos\theta -2\sin\theta \tan\theta=0 \Rightarrow \cos\theta +{\sin^2\theta \over \cos \theta} ={5\over 4} \Rightarrow \cos\theta ={4\over 5} \\ \Rightarrow \cases{R(13/10,3/5)\\ S(1,3/4)} \Rightarrow \overleftrightarrow{AB} =\overleftrightarrow{RS}: y=-{1\over 2}(x-1)+{3\over 4} \Rightarrow \bbox[red, 2pt]{2x+4y=5}$$

解答：

$$假設地球為一單位球，小塊體積相當於上圖著色區域繞x軸旋轉所得體積\\小塊體積= \pi\int_{\sqrt 3/2}^1 (1-x^2)\;dx =({2\over 3}-{3\over 8}\sqrt 3)\pi；及球體積={4\over 3}\pi； \\因此, 小塊:大塊= ({2\over 3}-{3\over 8}\sqrt 3)\pi:{4\over 3}\pi-({2\over 3}-{3\over 8}\sqrt 3)\pi =1:{{2\over 3}+{3\over 8}\sqrt 3 \over {2\over 3}-{3\over 8}\sqrt 3} =1:R\\ \Rightarrow R={({2\over 3}+{3\over 8}\sqrt 3)^2 \over ({2\over 3})^2-({3\over 8}\sqrt 3)^2} ={({2\over 3}+{3\over 8}\sqrt 3)^2 \over {13/ 24^2}} ={(16+9\sqrt 3)^2 \over 13} ={(a+b\sqrt 3)^2\over c}\\ \Rightarrow (a,b,c)=\bbox[red, 2pt]{(16,9,13)}$$

解答： $$\cases{(p+q)^{50} =\sum_{k=0}^{50}C^{50}_kp^kq^{50-k} \\(p-q)^{50} =\sum_{k=0}^{50}C^{50}_kp^k(-q)^{50-k} } \Rightarrow \sum_{k=0}^{25}C^{50}_{2k}p^{2k}q^{50-2k} ={1\over 2}\left((p+q)^{50} +(p-q)^{50}\right) \\ = {1\over 2}\left(({2\over 3}+ {1\over 3})^{50} +({2\over 3}-{1\over 3})^{50}\right) ={1\over 2}(1+{1\over 3^{50}}) \equiv {1\over a}(b+{1\over c^d}) \\ \Rightarrow (a,b,c,d)=\bbox[red, 2pt]{(2,1,3,50)}$$

解答： $$令\cases{O(0,0,0)\\ A(5,0,0)\\ B(0,5,0)\\ C(0,0,5)}及地面E過O \Rightarrow E;x+ay+bz=0，又\cases{\text{dist}(A,E) =\overline{AA'} = 3 \\\text{dist}(B,E) =\overline{BB'} = 2} \\ \Rightarrow \cases{{5\over \sqrt{1+a^2+b^2}}=3 \cdots(1)\\ {{5a\over \sqrt{1+a^2+b^2}}}=2 \cdots(2)} ,由(1) \Rightarrow \sqrt{1+a^2+b^2}={5\over 3} 代入(2) \Rightarrow a={2\over 3} \Rightarrow \sqrt{1+{4\over 9}+b^2 }= {5\over 3} \\ \Rightarrow b^2={12\over 9} \Rightarrow b={2\sqrt 3\over 3} \Rightarrow \text{dist}(C,E) = \overline{CC'}= {5b\over \sqrt{1+a^2+b^2}} ={10\sqrt 3/3 \over 5/3} =\bbox[red, 2pt]{2\sqrt 3}$$

解答： $$令F(x)=f(x)\cdot g(x)，則F'(x) = \lim_{h\to 0}{F(x+h)-F(x)\over h} = \lim_{h\to 0}{f(x+h)g(x+h)-f(x)g(x)\over h} \\ = \lim_{h\to 0}{f(x+h)g(x+h)-f(x+h)g(x) +f(x+h)g(x)-f(x)g(x)\over h}\\ = \lim_{h\to 0}{f(x+h)(g(x+h)-g(x) )+g(x)(f(x+h) -f(x) \over h}\\ = \lim_{h\to 0}{f(x+h)(g(x+h)-g(x) ) \over h}+ \lim_{h\to 0}{g(x)(f(x+h) -f(x)) \over h} \\ =\lim_{h\to 0}f(x +h) \cdot \lim_{h\to 0}{ g(x+h)-g(x) \over h}+ \lim_{h\to 0}g(x) \cdot \lim_{h\to 0}{ f(x+h) -f(x) \over h} \\ = f(x) \cdot \lim_{h\to 0}{ g(x+h)-g(x) \over h}+ g(x) \cdot \lim_{h\to 0}{ f(x+h) -f(x) \over h}\\ =f(x)g'(x)+ g(x)f'(x)，\bbox[red,2pt]{故得證}$$

解答： $$(x,y)經旋轉\theta 後，變為(x',y')\\，即 (x,y)=(r\cos \alpha,r\sin \alpha) \Rightarrow (x',y')=(r\cos(\alpha+\theta),r\sin(\alpha +\theta) \\ =(r(\cos \alpha \cos\theta- \sin(\alpha)\sin \theta), r(\sin \alpha\cos \theta + \sin\theta \cos \alpha)) \\ =(x\cos \theta- y\sin \theta,y\cos \theta + x\sin\theta) = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix}\\ \Rightarrow \begin{bmatrix} x'\\ y' \end{bmatrix} =\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{bmatrix}\begin{bmatrix} x\\ y \end{bmatrix} \Rightarrow A= \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{bmatrix}，\bbox[red,2pt]{故得證}$$

解答：

(1) $$f(x)=x^3 -3x^2+(3+a^2)x-a^2 \\\Rightarrow g(x)=f(x+k)-f(x)= 3kx^2 +(3k^2-6k)x + k^3-3k^2+(3+a^2)k \\ \Rightarrow 判別式: (3k^2-6k)^2-12k(k^3-3k^2+(3+a^2)k) = -12a^2k^2-3k^4 \le 0\\ \Rightarrow g(x) \ge 0 ,即f(x+k) \ge f(x), \text{for }k \ge 0 \Rightarrow f(x)為遞增函數$$ (2) $$f(x)= x^3-3x^2 +(3+a^2)x-a^2 \Rightarrow f'(x)=3x^2-6x + (3+a^2)\\ \Rightarrow \cases{f'(x)為二次式且首項係數3 \gt 0\\ 判別式36-12(3+a^2) =-12a^2 \le 0 } \Rightarrow f'(x) \ge 0 \Rightarrow f(x)為遞增函數\\ \bbox[red, 2pt]{故得證}$$