114年新北高中教甄聯招-數學詳解

 新北市公立高級中等學校 114 學年度教師聯合甄選

一、填充題： 共 10 題，每題 7 分。

解答: $$\cases{\log_2(x+1) +\log_2(y+1) =4\\ xy-x-y=-1} \Rightarrow \cases{\log_2(x+1)(y+1)=4 \\ xy-x-y+1=0} \Rightarrow \cases{(x+1)(y+1) =16 \cdots(1)\\ (x-1)(y-1)=0 \cdots(2)} \\ \text{eqn.}(2) \Rightarrow \cases{x=1 \Rightarrow y+1=8 \Rightarrow y=7\\ y=1 \Rightarrow x+1=8 \Rightarrow x=7} \Rightarrow \cases{(x,y)=(1,7) \\ (x,y)=(7,1)} \Rightarrow x+y=\bbox[red, 2pt]8$$

解答: $$物理系只能拆成兩組，座號為(9,11,13)與(10,12), 數學系只能拆成4+4或3+5\\數學系拆成4+4有C^8_4/2分法，再將物理系的兩組插入，共有{1\over 2}C^8_4\times 2=70種;\\數學系拆成3+5有C^8_3分法，再將物理系的兩組插入，共有C^8_3\times 2=112種\\ 因此共有70+112= \bbox[red, 2pt]{182}種分組方式$$

解答: $$假設三重根為a,另一根為b，則x^4-6x^2+\alpha x+(5-\alpha) =(x-a)^3(x-b) \\=x^4-(3a+b)x^3+ (3ab+3a^2)x^2- (a^3+3a^2b)x+ a^3b \\ \Rightarrow \cases{3a+b=0\\ 3ab+3a^2=-6} \Rightarrow b=-3a \Rightarrow a^2=1 \Rightarrow \cases{a=1 \Rightarrow b=-3 \Rightarrow \cases{\alpha=8\\ 5-\alpha=-3}\\ a=-1\Rightarrow b=3 \Rightarrow \cases{\alpha=-8 \\5-\alpha=-3}不合} \\ \Rightarrow \alpha= \bbox[red, 2pt]8$$

解答:

$$z=x+yi \Rightarrow 2\sqrt{x^2+y^2} \lt \sqrt{(x-1)^2+y^2} \Rightarrow x^2+{2\over 3}x+y^2\lt {1\over 3 }   \Rightarrow \left(x+{1\over 3} \right)^2+y^2\lt {4\over 9}\\ 圖形\Gamma: \left(x+{1\over 3} \right)^2 +y^2= {4\over 9} 為一圓\Rightarrow \cases {圓心A(0,-1/3) \\ 圓半徑r=2/3\\\Gamma與y軸交於B(0,\sqrt 3/3), C(0,-\sqrt 3/3)\\ 原點O(0,0)} \Rightarrow \angle OAB =60^\circ \\\Rightarrow \cases{\triangle ABC= \sqrt3 /9 \\ 扇形ABC =4\pi/27} \Rightarrow 欲求面積= \bbox[red, 2pt]{{4\pi\over 27} -{\sqrt 3\over 9}}$$

解答: $$\cases{兩次都抽到1號的機率=1/4^2 =1/16\\兩次都抽到2號的機率=((1/4)\times(1/2))^2=1/64 \\兩次都抽到3號的機率= ((1/4)\times(1/2))^2=1/64 \\ 兩次都抽到4號的機率= ((1/4)\times(1/2))^2=1/64 \\兩次都抽到5號的機率=((1/4)\times(1/2))^2=1/64 \\ 兩次都抽到6號的機率=1/4^2=1/16 } \\\Rightarrow 兩次抽到相同的期望值= {2\times 1\over 16}+ {4\times 1\over 64} ={3\over 16} \\ \Rightarrow 兩次抽到不同的期望值= 2-{3\over 16}= {29\over 16} = \bbox[red, 2pt]{1.8125}$$

解答: $$f(x)=x^5+x^2+1 =(x-r_1)(x-r_2)\cdots (x-r_5)\\ P(x) =x^2-2 =(x+\sqrt 2)(x-\sqrt 2) =(\sqrt 2-x) (-\sqrt 2-x) \\ \Rightarrow P(r_1)P(r_2)\cdots P(r_5) =(\sqrt 2-r_1)(-\sqrt 2-r_1) \times(\sqrt 2-r_2)(-\sqrt 2-r_2)\times \cdots \times(\sqrt 2-r_5)(-\sqrt 2-r_5) \\\qquad =[(\sqrt 2-r_1)(\sqrt 2-r_2)\cdots (\sqrt 2-r_5)] \times [(-\sqrt 2-r_1)(-\sqrt 2-r_2)\cdots (-\sqrt 2-r_5)] \\\qquad =f(\sqrt 2)\times f(-\sqrt 2) =(3+4\sqrt 2)(3-4\sqrt 2) =9-32= \bbox[red, 2pt]{-23}$$

解答: $$y={x+b\over x-2}=1+{ b+2\over x-2} \Rightarrow x={b+2\over y-1}+2= {2y+b\over y-1} \Rightarrow  f(y)={x+a\over 2x+1} ={(a+2)y+ b-a\over 5y+2b-1} \\ \Rightarrow \cases{\lim_{y\to \infty} f(y)=(a+2)/5=1 \Rightarrow a=3\\ \lim_{y \to 0} f(y)=(b-a)/(2b-1) =0 \Rightarrow a=b} \Rightarrow a=b=3 \Rightarrow a+b=\bbox[red, 2pt] 6$$

解答: $${n^5+2n^2+1\over n^2+3} =n^3-3n+2+{9n-5\over n^2+3} \Rightarrow {9n-5\over n^2+3} \in \mathbb Z\\ 僅需考慮n=-9,-8,\dots,0,1,\dots,9 \Rightarrow n=\bbox[red, 2pt]{1,8}$$

解答: $$先排奇數1,3,5,7，有4!=24種排法，而3個偶數插入奇數間隔中，僅有1種插法，所以共有\bbox[red, 2pt]{24}種$$

解答: $${1\over 2}\cos {\pi\over 3} = {1\over 2}e^{i\pi /3} 的實部 \Rightarrow 取z={1\over 2}e^{i\pi /3} \Rightarrow z^n={1\over 2^n}e^{n\pi i/3}  \\ \Rightarrow f(z) =\sum_{n=1}^\infty z^n ={z\over 1-z}\Rightarrow zf'(z)={z\over (1-z)^2} =\sum_{n=1}^\infty nz^{n}  =\sum_{n=1}^\infty {n\over 2^n} e^{n\pi i/3} \\z={1\over 2}e^{i\pi /3} ={1\over 4}(1+\sqrt 3i) \Rightarrow zf'(z) = {{1\over 4}(1+\sqrt 3 i) \over ({3\over 4}-{\sqrt 3i\over 4})^2}= {2\over 3}\cdot {1+\sqrt 3i\over 1-\sqrt 3i} =-{1\over 3}+{\sqrt 3\over 3}i \\ \Rightarrow \sum_{n=1}^\infty {n\over 2^n}\cos{n\pi \over 3} =zf'(z)的實部= \bbox[red, 2pt]{-{1\over 3}}$$

二、計算題： 共 3 題，每題 10 分。

解答: $$C^n_2-C^m_2 ={1\over 2}(n(n-1)-m(m-1)) =2025 \Rightarrow n^2-m^2-n+m=4050 \\ \Rightarrow (n-m)(n+m)-(n-m)= 4050 \Rightarrow (n-m)(n+m-1)=4050 =2\times 3^4\times5^2\\ n,m\in \mathbb N且n-m\lt n+m-1,欲求n-m之最大值，即將4050化成最相近的兩數相乘 \\4050=54\times 75 \Rightarrow n-m最大值為\bbox[red, 2pt]{54}$$

解答: $$a_1+2a_2+ \cdots+ na_n \lt 2025 +(a_1 +a_2^2+ \cdots+a_n^n) \\\Rightarrow (a_1-a_1^1)+ (2a_2-a_2^2)+  \cdots +(na_n-a_n^n)\lt 2025 \Rightarrow \sum_{k=1}^n(ka_k-a_k^k) \lt 2025\\因此我們取 f(x)=kx-x^k \Rightarrow f'(x)=k-kx^{k-1} \Rightarrow f''(x)=-k(k-1)x^{k-2} \\ f'(k)=0 \Rightarrow x=1 \Rightarrow f''(1)=-k(k-1) \lt 0, k=2,3, \dots \Rightarrow f(1)=k-1為極大值 \\ \Rightarrow \sum_{k=1}^n(k-1)= {n(n+1)\over 2}-n={n^2-n\over 2} \lt 2025 \Rightarrow n(n-1)\lt 4050 \Rightarrow 最大的n=\bbox[red, 2pt]{64}$$

解答:

$$已知\angle PAB=\angle PBC=\angle PCA=\theta又 \angle PDA=\angle PFA=90^\circ \Rightarrow PDAF共圓 \\ \Rightarrow 對同弧的圓周角相等:\cases{\angle FAP=\angle PDF =1 \\ \angle PAD= \angle PFD = \theta}\\ 同理, \cases{PDBE共圓 \Rightarrow \cases{\angle PBD=\angle PED=2\\ \angle PBE =\angle PDE =\theta} \\PECF共圓 \Rightarrow \cases{\angle PCE=\angle PFE=3 \\ \angle PCF =\angle PEF =\theta}} \Rightarrow \cases{\angle A=\theta+1=\angle EDF\\ \angle B=\theta+ 2=\angle DEF\\ \angle C=\theta+3 = \angle DFE} \\\Rightarrow \triangle ABC \sim \triangle  DEF\quad \bbox[red, 2pt]{QED}$$

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