115年建功高中教甄-數學詳解

 新竹市立建功高中 115 年第一次正式教師甄選

一、填充題：(每題 5 分，共 75 分)

解答:$${通過英檢的碩士人數 \over 通過英檢人數} ={40\% \times 60\% \over 50\%\times 30\%+40\% \times 60\%+ 10\%\times 100\%} =\bbox[red, 2pt]{24\over 49}$$

解答:$$從 7 位數學老師中選出 2 位，有C^7_2=21種 \Rightarrow 從8人選3人給第1組,有C^8_3=56種 \\ \Rightarrow 剩下5人選3位給第2組,有C^5_3=10種，因此共有21\times 56\times 10= \bbox[red, 2pt]{11760}種分組$$

解答:$$假設正四面體的 \cases{高h\\ 稜長a} \Rightarrow h={\sqrt 6a\over 3}=d(A,\triangle BCD) ={6+6+2\over \sqrt{2^2+3^2+6^2}} =2 \Rightarrow a=\sqrt 6 \\ \Rightarrow \triangle BCD面積={\sqrt 3\over 4}a^2= {3\over 2}\sqrt 3 \Rightarrow 體積={1\over 3}\cdot {3\over 2}\sqrt 3 \cdot 2= \bbox[red, 2pt]{\sqrt 3}$$

解答:

$$\cases{直線L:2x+y+2=0 \\A(-1,0) \\B(1,0)} \Rightarrow P在\overline{AB}中垂線上(y軸)上又P\in L\Rightarrow P(0,-2) \\ y=f(x)=a(x+1)(x-1) 通過P(0,-2) \Rightarrow a=2 \Rightarrow y=f(x)=2(x^2-1)\\ 直線L'\parallel L且通過B \Rightarrow L': 2x+y=2 \Rightarrow B'= L'\cap \Gamma =(-2,6) \\ B'(-2,6)平移至B(1,0) \Rightarrow P(0,-2)平移至Q   \Rightarrow \overline{PQ}= \overline{B'B}=\sqrt{3^2+6^2} = \bbox[red, 2pt]{3\sqrt 5}$$

解答:$$取g(x)=f(x)-3 \Rightarrow g(1-i)=g(2)=0 \Rightarrow x=1\pm i,2 為g(x)=0的三根\\ \Rightarrow g(x)=f(x)-3=k(x-2)(x^2-2x+2) \Rightarrow f(x)=k(x-2)(x^2-2x+2)+3 \\ \Rightarrow f(0)=-4k+3=-3 \Rightarrow k={3\over 2} \Rightarrow f(x)={3\over 2}(x-2)(x^2-2x+2)+3\\ 又x^2f(x)+12=3x^2+f(x) \Rightarrow (x^2-1)f(x)+12-3x^2=0 \\ \Rightarrow (x^2-1) \left( {3\over 2}(x-2)(x^2-2x+2)+3 \right)+12-3x^2 =0  \Rightarrow  x^5-4x^4+5x^3+ \cdots=0 \\ \Rightarrow \cases{a+b+c+d+e=4\\ ab+bc+ \cdots+ea=5} \Rightarrow (a+b+\cdots+e)^2=(a^2+\cdots+e^2)+2(ab+bc+\cdots+ea) \\ \Rightarrow 16=(a^2+\cdots+e^2)+10 \Rightarrow a^2+b^2+\cdots+e^2=\bbox[red, 2pt]6$$

解答:$$假設\cases{\overline{BC}=a\\ \overline{CA}=b\\ \overline{AB}=c} \Rightarrow \cases{ \overrightarrow{AB} \cdot \overrightarrow{AC} =bc \cos A= (b^2+c^2-a^2) /2 \\ \overrightarrow{BA} \cdot \overrightarrow{BC} =ac \cos B= (a^2+c^2-b^2) /2 \\\overrightarrow{CA} \cdot \overrightarrow{CB} =ab \cos C= (a^2+b^2-c^2) /2 } \\ 假設\overrightarrow{AB} \cdot \overrightarrow{AC} = 2\overrightarrow{BA} \cdot \overrightarrow{BC} = 3\overrightarrow{CA} \cdot \overrightarrow{CB} = 6K \Rightarrow \cases{b^2+c^2-a^2=12K\\ a^2+c^2-b^2=6K\\ a^2+b^2-c^2=4K} \\ \Rightarrow a^2+b^2+c^2=22K \Rightarrow \cases{a^2=5K\\ b^2=8K\\ c^2=9K} \Rightarrow c^2\gt b^2\gt a^2 \Rightarrow \angle C為最大角 \\ \Rightarrow \cos C={a^2+b^2-c^2 \over 2ab} ={5K+8K-9K\over 2\cdot \sqrt{5K} \cdot \sqrt{8K}} ={4\over 4\sqrt{10}} ={1\over \sqrt{10}} \Rightarrow \sin C={3\over \sqrt{10}} = \bbox[red, 2pt]{3\sqrt{10} \over 10}$$

解答:$$\log_a b=x \Rightarrow \log_a b+\log_b a =x+{1\over x}={5\over 2} \Rightarrow   2x^2-5x+2=0 \Rightarrow (2x-1)(x-2)=0 \\ \Rightarrow x={1\over 2} \;(a\gt b\gt 1 \Rightarrow \log_a b \lt 1) \Rightarrow \log_a b={1\over 2} \Rightarrow a=b^2 \\\Rightarrow {b\over a+4}={b\over b^2+4} ={1\over b+{4\over b}} \le \bbox[red, 2pt]{1\over 4}\;\left( b+{4\over b}\ge 2\sqrt{b\cdot {4\over b}}=4 \right)$$

解答:$$z_1= a+bi \Rightarrow z_2=a-bi \Rightarrow |z_1-z_2|=|2bi|=2|b|=4\sqrt 3 \Rightarrow |b|=2\sqrt 3 \Rightarrow \cases{z_1=a+2\sqrt 3i\\ z_2=a-2\sqrt 3i} \\ \Rightarrow z_1z_2=a^2+12 \Rightarrow {z_1\over z_2^2} ={z_1z_2\over z_2^3} ={a^2+12\over (a-2\sqrt 3i)^3} \in R \Rightarrow (a-2\sqrt 3i)^3 \in R \\ \Rightarrow (a^2-12)2\sqrt 3+4\sqrt 3a^2=0 \Rightarrow a^2=4 \Rightarrow |z_1|=\sqrt{a^2+12} =\bbox[red, 2pt]4 \\ (b=\pm 2\sqrt 3 \Rightarrow a^2=4)$$

解答:$$(a-b)^2 =a^2-2ab+b^2 \ge 0 \Rightarrow a^2+b^2 \ge 2ab \Rightarrow 2(a^2+b^2)\ge a^2+2ab+b^2=(a+b)^2 \\ 取\cases{a=5-x^2\\ b=x+3} \Rightarrow 2[(5-x^2)^2+(x+3)^2] \ge  (-x^2+x+8)^2 \\ \Rightarrow f(x) =\sqrt{2[(5-x^2)^2+(x+3)^2]} +(x^2-x+1)\ge |-x^2+x+8|+x^2-x+1 \\取t=x^2-x \Rightarrow f(x)\ge g(t)=|8-t|+t+1 \Rightarrow \cases{8-t\ge 0 \Rightarrow g(t)=9\\ 8-t\le 0 \Rightarrow g(t)=2t-7 \ge 9} \\ \Rightarrow f(x)的最小值為\bbox[red, 2pt]9$$

解答:$$令a_i =2b_i-1 \Rightarrow \cases{a_i=1 \Rightarrow b_i=1 \\a_i=-1 \Rightarrow b_i=0} \Rightarrow b_i \in \{0,1\} \\ f(2)=-53 =\sum_{i=0}^6(2b_i-1)2^i =2\sum_{i=0}^6 b_i- \sum_{i=0}^6 2^i= 2\sum_{i=0}^6 b_i-127 \Rightarrow \sum_{i=0}^6 b_i=(127-53)/2=37 \\37= 100101_2 \Rightarrow \cases{b_5=b_2=b_0=1 \\ b_k=0, k\ne0,2,5} \Rightarrow \sum_{i=0}^6 b_i=3\\ \Rightarrow f(1)= \sum_{i=0}^6 a_i= \sum_{i=0}^6(2b_i-1) =2 \sum_{i=0}^6b_i-7=2\cdot 3-7= \bbox[red, 2pt]{-1}$$

解答:$$x^3+2025x+1=0 \Rightarrow \cases{a+b+c=0\\ ab+bc+ca=2025\\ abc=-1} \\ \Rightarrow a^3b^2+a^2b^3+b^3c^2+b^2c^3 +c^3a^2+c^2a^3 =a^2b^2(a+b)+ b^2c^2(b+c)+ c^2a^2(c+a) \\=a^2b^2(-c)+b^2c^2(-a)+c^2a^2(-b) =-abc(ab+bc+ca)=1\cdot 2025= \bbox[red, 2pt]{2025}$$

解答:$${4^a+4^{a+k}+ \cdots+4^{a+mk} \over 2^a+2^{a+k}+ \cdots+ 2^{a+mk}} ={4^a(4^{(m+1)k}-1)/(4^k-1)\over 2^a(2^{(m+1)k}-1)/(2^k-1)} =2^a\cdot {2^{(m+1)k}+1\over 2^k+1} =1928 =2^3\cdot 241 \\ \Rightarrow a=3 \Rightarrow {2^{(m+1)k}+1\over 2^k+1}=241 \Rightarrow 2^{(m+1)k}=241\cdot 2^k+240 \\ \Rightarrow \cases{k=1 \Rightarrow 2^{m+1}=722\Rightarrow 不合\\ k=2 \Rightarrow 2^{2(m+1)} =1204 \Rightarrow 不合\\ k=4 \Rightarrow 2^{4(m+1)}=4096 \Rightarrow m=2} \Rightarrow (a,k,m)= \bbox[red, 2pt]{(3,4,2)}$$

解答:$$\lim_{n\to \infty} {4\over n} \sum \left( 5k-4\over n \right)^3 =4 \int_0^1 (5x)^3\,dx = 500\int_0^1 x^3\,dx =500\cdot {1\over 4}=\bbox[red, 2pt]{125}$$

解答:$$三角形面積={1\over 2}\cdot 7\cdot 9 \sin \theta \Rightarrow 兩個不同\triangle ,面積相等,代表\theta_1+\theta_2=180^\circ \\ 假設\cases{第1個\triangle 的第3邊邊長為a\\ 第2個\triangle 的第3邊邊長為b} \Rightarrow \cases{a^2=7^2+9^2-2\cdot 7\cdot 9\cdot\cos \theta\\ b^2=7^2+9^2-2\cdot 7\cdot 9\cdot \cos(180^\circ-\theta)} \\ \Rightarrow a^2+b^2=260,由於a,b\in \mathbb R且9-7=2\lt a,b\lt 9+7=16 \\\Rightarrow 從3^2,4^2, \dots,15^2中挑兩個數相加等於260 \Rightarrow 8^2+14^2=260 \Rightarrow a+b=8+14= \bbox[red, 2pt]{22}$$

解答:$${a_n-1\over n-1}={a_{n-1}+1\over n} \Rightarrow na_n-(n-1)a_{n-1}=2n-1 \\ \Rightarrow \cases{2a_2-a_1=2\cdot 2-1\\ 3a_3-2a_2=2\cdot 3-1\\4a_4-3a_3=2\cdot 4-1 \\ \cdots\\ na_n-(n-1)a_{n-1}=2\cdot n-1} \Rightarrow 全部相加可得:na_n-a_1= \sum_{k=2}^n(2k-1)=n^2-1\\ \Rightarrow a_n=(n^2-1+2)/n= n+{1\over n} \Rightarrow a_n^2=n^2+ 2+{1\over n^2}\\\Rightarrow \sum_{n=1}^{50}a_n^2 = \sum_{n=1}^{50}n^2 +2\sum_{n=1}^{50}1+\sum_{n=1}^{50}{1\over n^2}=43025+\sum_{n=1}^{50}{1\over n^2} \Rightarrow \left[  \sum_{n=1}^{50}a_n^2 \right]=43025+1= \bbox[red, 2pt]{43026}\\ \text{Fourier series: }f(x)=x^2= {\pi^2\over 3}+ \sum_{n=1}^\infty{4\over n^2} \Rightarrow f(\pi)=\pi^2+{\pi^2\over 3}+ \sum_{n=1}^\infty{4\over n^2} \Rightarrow  \sum_{n=1}^\infty{1\over n^2}={\pi^2\over 6} \approx 1.6$$

二、計算題：(共 25 分)

解答:$$\textbf{(1) }已知\bbox[red, 2pt]{L_1:y=1} \Rightarrow (x,1)\in L_1 \Rightarrow   \begin{bmatrix}a&1\\-1&a \end{bmatrix} \begin{bmatrix}x\\1 \end{bmatrix} = \begin{bmatrix}ax+1\\ -x+a \end{bmatrix} = \begin{bmatrix}x'\\ y' \end{bmatrix} \\ \quad \Rightarrow x'+ay'=a^2+1 \Rightarrow \bbox[red, 2pt]{L_2: x+ay=a^2+1} \\\textbf{(2) }P=L_1\cap L_2 =(a^2-a+1,1) \\ A^{-1} ={1\over a^2+1} \begin{bmatrix}a&-1\\1& a \end{bmatrix} \Rightarrow A^{-1} \begin{bmatrix}x\\1 \end{bmatrix} ={1\over a^2+1} \begin{bmatrix}ax-1\\x+a \end{bmatrix} = \begin{bmatrix}x''\\ y'' \end{bmatrix} \Rightarrow x''=ay''-1 \\\qquad \Rightarrow L_3: x-ay+ 1=0 \Rightarrow Q=L_1\cap L_3 =(a-1 ,1) \Rightarrow R=L_2\cap L_3=({a^2\over 2},{a^2+2\over 2a}) \\ \Rightarrow \bbox[red, 2pt]{P(a^2-a+1,1), Q(a-1,1), R \left( {a^2\over 2},{a^2+2\over 2a} \right)} \\\textbf{(3) }S(a)={1\over 2} \begin{Vmatrix} a^2-a+1& 1& 1\\ a-1& 1& 1\\ {a^2\over 2}& {a^2+2\over 2a}& 1 \end{Vmatrix} =\bbox[red, 2pt]{(a^2-2a+2)^2\over 4a} \\\textbf{(4) }{S(a) \over a} = \left( {a^2-2a+2\over 2a} \right)^2 = \left( {a\over 2}-1+{1\over a} \right)^2 \ge (\sqrt 2-1)^2= \bbox[red, 2pt]{3-2\sqrt 2} \\ 註:{a\over 2}+{1\over a} \ge 2\sqrt{{a\over 2}\cdot {1\over a}} =\sqrt 2$$

解答:$$\textbf{(1) }f(x)=x^3-3x^2+2x \Rightarrow f'(x)=3x^2-6x+2 \Rightarrow f'(k)=3k^2-6k+2 \\ \Rightarrow L_k: y-f(k)=f'(x)(x-k) \Rightarrow \bbox[red, 2pt]{L_k: y=(3k^2-6k+2)x-2k^3+3k^2} \\\textbf{(2) }f(x)=x^3-3x^2+2x =(3k^2-6k+2)x-2k^3+3k^2 \\\Rightarrow g(x)=x^3-3x^2+(6k-3k^2)x+2k^3-3k^2=0\\ 由於\cases{f(x)是3次式\\ P是切點\\ Q是交點}\Rightarrow g(x)=0的三根為k,k,q \Rightarrow k+k+q=3 \Rightarrow q=3-2k \\\qquad \Rightarrow Q的x座標為\bbox[red, 2pt]{3-2k} \\\textbf{(3) }公式: A(k)={1\over 12}|(3-2k)-k|^4 ={27\over 4}(1-k)^4 \\由於0\lt k\lt {1\over 2} \Rightarrow 最小值=A(1/2)= \bbox[red, 2pt]{27\over 64} \quad \href{https://www.youtube.com/watch?v=p5qszz_d__I}{公式來源}$$

$$\bbox[cyan,2pt]{學校未公告答案，解題僅供參考}$$

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