115年中正高中教甄-數學詳解

 臺北市立中正高級中學 115 學年度第 1 次專任教師甄選

一、填充題（每題 6 分）

解答:$$ a_7 = 1 \Rightarrow \cases{ a_1 = a_7 - 6d = 1 - 6d \\a_2 = a_7 - 5d = 1 - 5d\\ a_{12} = a_7 + 5d = 1 + 5d \\a_{13} = a_7 + 6d = 1 + 6d } \Rightarrow  \sum_{k=1}^{11} \frac{1}{a_k a_{k+1} a_{k+2}} = \frac{1}{2d} \sum_{k=1}^{11} \left( \frac{1}{a_k a_{k+1}} - \frac{1}{a_{k+1} a_{k+2}} \right) \\  = \frac{1}{2d} \left[ \left( \frac{1}{a_1 a_2} - \frac{1}{a_2 a_3} \right) + \left( \frac{1}{a_2 a_3} - \frac{1}{a_3 a_4} \right) + \cdots + \left( \frac{1}{a_{11} a_{12}} - \frac{1}{a_{12} a_{13}} \right) \right] \\ = \frac{1}{2d} \left( \frac{1}{a_1 a_2} - \frac{1}{a_{12} a_{13}} \right) ={1\over 2d} \left( \frac{1}{(1 - 6d)(1 - 5d)} - \frac{1}{(1 + 5d)(1 + 6d)}  \right) \\= \frac{1}{2d} \cdot \frac{22d}{(1 - 36d^2)(1 - 25d^2)} = \frac{11}{(1 - 36d^2)(1 - 25d^2)} =11 \\ \Rightarrow  (1 - 36d^2)(1 - 25d^2) = 1 \Rightarrow  900d^4 - 61d^2 = 0  \Rightarrow d^2 (900d^2-61)=0 \\ \Rightarrow  d^2 = \frac{61}{900} \Rightarrow d=\bbox[red, 2pt]{\sqrt{61} \over 30}$$

解答:$$-1\le x\le 1 \Rightarrow 取x=\cos \theta, 0\le \theta\le \pi \Rightarrow y=f(x)= {\sqrt{1-x^2} \over 5-x}= {\sin \theta\over 5-\cos \theta}  \Rightarrow y(5-\cos \theta)=\sin \theta\\ \Rightarrow \sin \theta+y\cos \theta=5y \Rightarrow \sqrt{1+y^2}\sin(\theta+\alpha)=5y \Rightarrow \sqrt{1+y^2}\ge |5y| \Rightarrow 1+y^2=25y^2 \\ \Rightarrow y^2\le {1\over 24} \Rightarrow  -\frac{1}{\sqrt{24}} \le y \le \frac{1}{\sqrt{24}} =\bbox[red, 2pt]{\sqrt 6\over 12}$$

解答:$$每個元素只有 3 種選擇:只屬於A、只屬於B、都不屬於 \Rightarrow 共有3^8種\\ A=\varnothing:每個元素只有 2 種選擇，共有2^8種 \\ B=\varnothing:每個元素只有 2 種選擇，共有2^8種 \\ A=B=\varnothing:每個元素只有 1 種選擇，共有1^8=1種 \\ 因此n(T)=3^8-2^8-2^8+1=6561-256-256+1 =\bbox[red, 2pt]{6050}$$

解答:$$ f(x) = 4x^3 - 6x^2 + 3x + 1 \Rightarrow f(1-x)=  4(1-x)^3 - 6(1-x)^2 + 3(1-x) + 1 \\ = -4x^3 + 6x^2 - 3x + 2 \Rightarrow f(x)+f(1-x)=3   \Rightarrow  S = \sum_{k=1}^{200} f\left(\frac{k}{201}\right)  \\= \left[ f\left(\frac{1}{201}\right) + f\left( \frac{200}{201}\right) \right] + \left[ f\left(\frac{2}{201} \right) + f\left( \frac{199}{201}\right) \right] + \cdots + \left[ f\left( \frac{100}{201} \right) + f\left(\frac{101}{201}\right) \right] \\=3+3+\cdots+3=3\times 100=\bbox[red, 2pt]{300}$$

解答:$$ \tan(45^\circ-\theta) = \frac{\tan 45^\circ - \tan \theta}{1 + \tan 45^\circ \tan \theta} = \frac{1 - \tan \theta}{1 + \tan \theta} \\ \Rightarrow  (1+\tan \theta)[1+\tan(45^\circ-\theta)] =  (1+\tan \theta) \left( 1 + \frac{1 - \tan \theta}{1 + \tan \theta} \right) = (1+\tan \theta) \left( \frac{2}{1 + \tan \theta} \right) =2 \\ \Rightarrow   P=(1+\tan1^\circ) \times (1+\tan2^\circ) \times \dots \times (1+\tan44^\circ) \times (1+\tan45^\circ)  \\= \left[ (1+\tan 1^\circ) (1+\tan 44^\circ) \right] \cdot \left[ (1+\tan 2^\circ) (1+\tan 43^\circ) \right] \cdots \left[ (1+\tan 22^\circ) (1+\tan 23^\circ) \right] \cdot (1+\tan 45^\circ)\\=2^{22}\cdot (1+\tan 45^\circ)=2^{22} \cdot (1+1)=2^{23} \Rightarrow 欲求\log_4 P =\log_4 2^{23} ={\log_2 2^{23} \over \log_2 4}= \bbox[red, 2pt]{23\over 2}$$

解答:$$設  \triangle ABC  的三邊長分別為  a-d 、 a 、 a+d   \Rightarrow  s = \frac{(a-d) + a + (a+d)}{2} = \frac{3a}{2} \\ \Rightarrow 三角形面積 \Delta = \sqrt{s(s - (a-d))(s - a)(s - (a+d))}  \\= \sqrt{\frac{3a}{2} \left(\frac{3a}{2} - a + d\right) \left(\frac{3a}{2} - a\right) \left(\frac{3a}{2} - a - d\right)}   = \frac{a}{4} \sqrt{3(a^2 - 4d^2)}\\  =r\cdot s= \frac{a}{4} \sqrt{3(a^2 - 4d^2)} = \frac{\sqrt{15}}{4} \cdot \frac{3a}{2} \Rightarrow  \sqrt{3(a^2 - 4d^2)} = \frac{3\sqrt{15}}{2}  \Rightarrow  a^2 - 4d^2 = \frac{45}{4} \\ \Rightarrow \Delta={a\over 4}\sqrt{3\cdot {45\over 4}} ={3a\sqrt{15}\over 8}\\ 又\Delta={三邊長乘積\over 4R}   = \frac{(a-d)a(a+d)}{4R} = \frac{a(a^2 - d^2)}{4R} = \frac{a(a^2 - d^2)}{4 \left(\frac{8\sqrt{15}}{15}\right)} ={3a\sqrt{15} \over 8} \\ \Rightarrow a^2-d^2=12 \Rightarrow \cases{a^2-4d^2=45/4\\ a^2-d^2=12} \Rightarrow d^2={1\over 4} \Rightarrow d= \bbox[red,2pt]{1\over 2}$$

解答:$$邊長為 2 的正八面體，兩平行面之間的距離公式為  \frac{\sqrt{6}}{3}\times 2 ={2\sqrt 6\over 3}  \Rightarrow B的z坐標為{2\sqrt 6\over 3} \\B在yz平面上\Rightarrow B的x坐標為0 \Rightarrow B=(0,y_B, {2\sqrt 6\over 3}) \\\Rightarrow B$ 到 y 軸的距離為  = \sqrt{0^2 + \left(\frac{2\sqrt{6}}{3}\right)^2} = \bbox[red, 2pt]{\frac{2\sqrt{6}}{3}}$$

解答:$$ S = \frac{1}{3} + \frac{4}{3^2} + \frac{9}{3^3} + \frac{16}{3^4} + \dots + \frac{n^2}{3^n} + \dots \\ \Rightarrow \frac{1}{3}S = \frac{1}{3^2} + \frac{4}{3^3} + \frac{9}{3^4} + \dots + \frac{(n-1)^2}{3^n} + \dots \\ \Rightarrow S-{1\over 3}S = \frac{2}{3}S = \frac{1}{3} + \frac{3}{3^2} + \frac{5}{3^3} + \frac{7}{3^4} + \dots + \frac{2n-1}{3^n} + \dots  \\ \Rightarrow {2\over 9}S = \frac{1}{3^2} + \frac{3}{3^3} + \frac{5}{3^4} + \dots + \frac{2n-3}{3^n} + \dots \\ \Rightarrow {2\over 3}S-{2\over 9}S=  \frac{4}{9}S = \frac{1}{3} + \frac{2}{3^2} + \frac{2}{3^3} + \frac{2}{3^4} + \dots ={1\over 3}+{2/9\over 1-1/3}={1\over 3}+{1\over 3}={2\over 3} \\ \Rightarrow S={2\over 3}\times {9\over 4}= \bbox[red, 2pt]{3\over 2}$$

解答:$$假設\cases{ P(x) = x^4 - 5x^3 + 8x^2 - 4x + 1 \\  Q(x) = x^5 - 6x^4 + 13x^3 - 12x^2 + 5x - 2\\  R(x) = x^2 - 2x + 2 } \Rightarrow F(x)=P^4(x)\cdot Q^3(x)\cdot R^7(x)\\ 取x=t+1 \Rightarrow \cases{ P(t+1) = (t+1)^4 - 5(t+1)^3 + 8(t+1)^2 - 4(t+1) + 1\\ \qquad=   1 + t - t^2 - t^3 + t^4 \\同理可得Q(t+1)   = -1 + t + t^2 - t^3 - t^4 + t^5 \\ 同理可得R(t+1) = t^2 + 1 } \\ 只考慮x^4以下的項次:\cases{P^4(t+1)  = 1 + 4t + 2t^2 - 12t^3 - 13t^4 + O(t^5) \\Q^3(t+1 )=   -1 + 3t + 0t^2 - 8t^3 + 3t^4 + O(t^5) \\ R^7(t+1)    = 1 + 7t^2 + 21t^4 + O(t^5) } \\ \Rightarrow F(t+1) \approx (1 + 4t + 2t^2 - 12t^3 - 13t^4) \cdot (-1 + 3t + 0t^2 - 8t^3 + 3t^4) \cdot (1 + 7t^2 + 21t^4) \\ \Rightarrow t^4係數為-3 \Rightarrow F^{(4)}(1)=4!\times(-3)=\bbox[red, 2pt]{-72}$$

解答:$$  f(x) = x^3 - x^2 + 3x + 31 =  (x-\alpha)(x-\beta) (x-\gamma) \\ \Rightarrow f(\alpha)= \alpha^3 - \alpha^2 + 3\alpha + 31 = 0 \Rightarrow  \alpha^3 = \alpha^2 - 3\alpha - 31 \\ \Rightarrow  \alpha^3 + 27 = \alpha^2 - 3\alpha - 4 =(\alpha-4)(\alpha+1) ,同理可得\cases{\beta^3+27=(\beta-4)(\beta+1) \\ \gamma^3+27=(\gamma -4) (\gamma+1)}\\ \Rightarrow 欲求(\alpha^3 + 27)(\beta^3 + 27)(\gamma^3 + 27) =  (\alpha - 4)(\alpha + 1)(\beta - 4)(\beta + 1)(\gamma - 4)(\gamma + 1) \\ = [(\alpha - 4)(\beta - 4)(\gamma - 4)] \cdot [(\alpha + 1)(\beta + 1)(\gamma + 1)]=(-f(4)) \cdot (-f(-1)) =f(4)\cdot f(-1) \\=\left(   4^3 - 4^2 + 3(4) + 31  \right) \cdot \left(    (-1)^3 - (-1)^2 + 3(-1) + 31  \right)= 91 \times 26 = \bbox[red, 2pt]{2366}$$

解答:$$A(2,-4)\cases{對直線  L_1: x + y - 2 = 0  的對稱點 A'=(6,0) \\ 對直線L_2:x-3y-6=0的對稱點A''=(2/5,4/5)}\\ \Rightarrow BC直線= \overleftrightarrow{A'A''}: \bbox[red, 2pt]{x+7y-6=0}$$

解答:$$10 天的訓練中，包含 5 個「高強度」(令為 H)與 5 個「修復」(令為 R) \Rightarrow 排列數為C^{10}_5=252\\ 第 1 天(4月10日)成為黃金孤立訓練日的機率:\\ \qquad 第 0 天為 R(已知)、第 1 天為 H、第 2 天為 R,將剩下的 4 個 H 與 4 個 R 排入剩餘的 8 天中，\\ \qquad 排列數為C^8_4=70 \Rightarrow 機率=70/252= 5/18\\ 第 10 天(4月19日)成為黃金孤立訓練日的機率:\\ \qquad 第 9 天為 R、第 10 天為 H、第 11 天為 R(已知),剩下的 8 天需排入 4 個 H 與 4 個 R，\\\qquad 排列數為 C^8_4=70 \Rightarrow 機率=70/252= 5/18 \\ 第 k 天(k = 2 \sim 9)成為黃金孤立訓練日的機率:\\ \qquad 第 k-1 天為 R、第 k 天為 H、第 k+1 天為 R,剩下的 7 天需排入剩下的 4 個 H 與 3 個 R, \\\qquad 排列數為C^7_4=35 \Rightarrow 機率=35/252=5/36\\ \Rightarrow 期望值 E = \frac{5}{18} + \frac{5}{18} + 8 \times \frac{5}{36} =\bbox[red, 2pt]{5\over 3}$$

二、計算題（除第 3 題 12 分外，其餘每題 8 分）

解答:$$欲證右半部:\sqrt[n]{2} \le 1+\frac{1}{n} \\ 白努利不等式: \left(1+\frac{1}{n}\right)^n \ge 1 + n\left(\frac{1}{n}\right) =2 \Rightarrow \sqrt[n]2\le 1+{1\over n}\\ 再證左半部: 1+\frac{1}{2n-1} \le \sqrt[n]{2} \\  1+\frac{1}{2n-1} = \frac{2n}{2n-1} = \frac{1}{\frac{2n-1}{2n}} = \frac{1}{1-\frac{1}{2n}} \cdots(1) \\再利用白努利不等式: \left(1-\frac{1}{2n}\right)^n \ge 1 + n\left(-\frac{1}{2n}\right)={1\over 2} \Rightarrow  \frac{1}{\left(1-\frac{1}{2n}\right)^n} \le 2 \\ 由(1)可得 \left(1+\frac{1}{2n-1}\right)^n \le 2 \Rightarrow  1+\frac{1}{2n-1} \le \sqrt[n]{2} \\ 因此 1+\frac{1}{2n-1} \le \sqrt[n]{2} \le 1+\frac{1}{n} \quad \bbox[red, 2pt]{故得證}$$

解答:$$\cases{A(1,2,-3) \\B(5,-4,1)} \Rightarrow \cases{M=\overline{AB}中點=(A+B)/2=(3,-1,-1)\\ \overline{AB}^2= 68}\\ 中線定理:  \overline{PA}^2 + \overline{PB}^2 = 2\overline{PM}^2 + \frac{1}{2}\overline{AB}^2  = 2\overline{PM}^2 + 34 \\ 由於\overline{PM}^2 的最小值= d^2(M,E) = \left( {2\over \sqrt{14}} \right)^2 ={2\over 7} \Rightarrow \overline{PA}^2 + \overline{PB}^2 最小值=  2\left(\frac{2}{7}\right) + 34 =  \bbox[red, 2pt]{\frac{242}{7} }$$

解答:$$\textbf{(1) }(x',y') \in \Gamma' \Rightarrow  \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \cos(-45^\circ) & -\sin(-45^\circ) \\ \sin(-45^\circ) & \cos(-45^\circ) \end{bmatrix} \begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ -\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{bmatrix} \begin{bmatrix} x' \\ y' \end{bmatrix} \\ \Rightarrow \cases{x=(x'+y')/\sqrt 2\\ y=(-x'+y')/\sqrt 2} \Rightarrow  \frac{\left[\frac{\sqrt{2}}{2}(x' + y')\right]^2}{160} + \frac{\left[\frac{\sqrt{2}}{2}(-x' + y')\right]^2}{40} = 1 \\ \Rightarrow  \frac{(x' + y')^2}{320} + \frac{(-x' + y')^2}{80} = 1 \Rightarrow  5x'^2 - 6x'y' + 5y'^2 = 320  \\ \Rightarrow \Gamma': \bbox[red, 2pt]{5x^2 - 6xy + 5y^2 = 320 } \\\textbf{(2) } \Gamma':5x^2 - 6xy + 5y^2 = 320 \Rightarrow  5y^2 - (6x)y + (5x^2 - 320) = 0 \\ \Rightarrow 判別式:  \Delta = (-6x)^2 - 4(5)(5x^2 - 320) \ge 0 \Rightarrow 64(100-x^2)\le 0 \Rightarrow x^2\le 100 \\ \Rightarrow -10\le x\le 10, 又64(100-x^2)必須是完全平方數, 因此x=0,\pm 6, \pm 8, \pm 10 \\ 找整數解\Rightarrow \cases{x=0 \Rightarrow  y=\pm 8\\ x=6\Rightarrow y=10 \\ x=-6 \Rightarrow y=-10 \\ x=8 \Rightarrow y=0 \\ x= -8 \Rightarrow y=0\\ x=10 \Rightarrow y=6\\ x=-10 \Rightarrow y=-6} \Rightarrow 格子點: \bbox[red, 2pt]{(0,\pm 8),(\pm 8,0), (6,10),(-6,-10), (10,6),(-10,-6)}$$

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