114年北一女教甄2-數學詳解

 臺北市立第一女子高級中學 114 學年度第二次正式教師甄選

一、填充題（每格 8 分，共 64 分）

解答:$$x^2+\alpha x+\beta=0 有重根r \Rightarrow x^2+\alpha x+\beta = (x-r)^2 \Rightarrow \cases{-2r=\alpha \\ r^2=\beta} \\ \alpha,\beta為x^2+kx-2025=0的兩根\Rightarrow \alpha \beta =-2025 \Rightarrow (-2r)r^2=-2025 \Rightarrow 2r^3=2025\\ \Rightarrow r= \sqrt[3]{2025\over 2}={1\over 2}\cdot \sqrt[3]{8100} = \bbox[red, 2pt]{{3\over 2}\sqrt[3]{300}}$$

解答:$$假設\vec b, \vec  c 夾角\theta, 又\vec a\bot \vec b, 因此\vec c, \vec a夾角(90^\circ-\theta)  \Rightarrow \cases{\cos \theta= \vec b\cdot \vec c/|\vec b||\vec c| \\ \cos (90^\circ-\theta) = \vec c\cdot \vec a/|\vec c||\vec a|} \\ \Rightarrow \cases{|\vec c| \cos \theta= 20|\vec a|/|\vec b| \\ |\vec c| \sin \theta=25|\vec b|/|\vec a|} \Rightarrow |\vec c|^2 ={400|\vec a|^2\over |\vec b|^2} +{625 |\vec b|^2 \over |\vec a|^2} \ge 2 \sqrt{{400|\vec a|^2\over |\vec b|^2} \cdot {625 |\vec b|^2 \over |\vec a|^2} }=1000\\ \Rightarrow |\vec c|\ge \sqrt{1000} = \bbox[red, 2pt]{10\sqrt{10}}$$

解答:$$假設\cases{R:紅鈕扣\\ G:綠鈕扣\\ Y:黃鈕扣}, 先將6R排成一列，產生7個間隔 \\ 由於同色不相鄰，必須將中間5個間隔 放入3G與3Y中的五個\\  \textbf{Case I }5個間隔放3G2Y且剩下1Y放頭或尾端: {5\choose 3}\cdot 2=20 \\\textbf{Case II }5個間隔放2G3Y且剩下1G放頭或尾端: {5\choose 3}\cdot 2=20 \\ \textbf{Case III }5個間隔放3G2Y且剩下1Y不放頭尾: \cases{G \cdot G \cdot Y \cdot Y \cdot \boxed{GY}排列數:\displaystyle {5!\over 2!2!} =30\\ G \cdot G \cdot Y \cdot Y \cdot \boxed{YG}排列數:\displaystyle {5!\over 2!2!} =30} \\ 以上合計:20+20+30+30= \bbox[red, 2pt]{100}$$

解答:$$\sum_{k=1}^7 {1\over a_k \cdot a_{k+1} \cdot a_{k+2}} = {1\over 2d} \sum_{k=1}^7 \left( {1\over a_k \cdot a_{k+1}  } -{1\over a_{k+1}\cdot a_{k+2}}\right) = {1\over 2d} \left( {1\over a_1a_2} -{1\over a_8a_9} \right) \\= {1\over 2d} \left( {1\over (1-4d) (1-3d)} -{1\over (1+3d)(1+4d)} \right) ={1\over 2d} \cdot {14d\over (1-9d^2) (1-16d^2)} \\={7\over (1-9d^2) (1-16d^2)} =7   \Rightarrow (1-9d^2) (1-16d^2)=1 \Rightarrow 144d^4-25d^2=0\\ \Rightarrow d^2(144d^2-25)=0 \Rightarrow d=\sqrt{25\over 144} = \bbox[red, 2pt]{5\over 12}$$

解答:$$f(x)=\max\{x,x^2,x^3-2x\} =\begin{cases} x^2,&-2\le x\le -1 \\ x^3-2x, &-1\le x\le 0\\ x, &0\le x\le 1\\ x^2,& 1\le x\le 2\end{cases} \\ \Rightarrow \int_{-2}^2 f(x)\,dx = \int_{-2}^{-1} x^2\,dx + \int_{-1}^0 (x^3-2x)\,dx +\int_0^1x\,dx +\int_1^2 x^2\,dx ={7\over 3}+{3\over 4}+{1\over 2} +{7\over 3} =\bbox[red, 2pt]{71\over 12}$$

解答:$$1- \left( {1\over 3} \cdot {1\over 4} \cdot {1\over 2}\right)12= 1-{1\over 2} = \bbox[red, 2pt]{1\over 2}$$

解答:$$2^x (1+\cos x)=2 \Rightarrow 1+\cos x=2^{1-x} \Rightarrow 取\cases{f(x)= 1+\cos x\\ g(x)=2^{1-x}}, 欲求兩圖形\cases{y=f(x) \\y=g(x)}的交點數\\ 其中\cases{0\le f(x) \le 2, 週期2\pi\\ g(x)\gt 0 且嚴格遞減} \\ \textbf{Case I }0\le x\le \pi: \cases{f(0)=g(0)=2 \\ f(\pi)=0 \lt g(\pi)} \Rightarrow 有2解\\ \textbf{Case II }\pi\le x\le 2\pi: \cases{f(\pi) \lt g(\pi) \\ f(2\pi) \gt g(2\pi)} \Rightarrow 有1解 \\ \textbf{Case III }2\pi \le x\le 2025\pi: \cases{2k\pi \le x\le (2k+1)\pi : \cases{f(2k\pi)\gt g(2k\pi) \\ f((2k+1)\pi) \lt g((2k+1)\pi)} \Rightarrow 有1解 \\ (2k+1)\pi \le x\le (2k+2)\pi : \cases{f((2k+1)\pi) \lt g((2k+1)\pi)\\ f((2k+2)\pi) \gt g((2k+2)\pi)} \Rightarrow 有1解} \\ \Rightarrow 共有2+1+ 2023= \bbox[red, 2pt]{2026}個解$$

解答:$$A= \begin{bmatrix} 0& 1/3 & 0 & 1/3 & 1/3 & 0 & 0 &0 \\ 1/3& 0 & 1/3& 0& 0& 1/3& 0& 0 \\ 0& 1/3& 0& 1/3& 0& 0& 1/3& 0\\ 1/3& 0& 1/3& 0& 0& 0& 0& 1/3\\ 1/3& 0& 0& 0& 0& 1/3& 0& 1/3\\ 0& 1/3& 0& 0& 1/3& 0& 1/3& 0\\ 0& 0& 1/3& 0& 0& 1/3& 0& 1/3\\ 0& 0& 0& 1/3& 1/3& 0& 1/3& 0\end{bmatrix} \\ \Rightarrow A^5 = \begin{bmatrix} 0 & \frac{61}{243} & 0 & \frac{61}{243} & \frac{61}{243} & 0 & \frac{20}{81} & 0 \\\frac{61}{243} & 0 & \frac{61}{243} & 0 & 0 & \frac{61}{243} & 0 & \frac{20}{81} \\0 & \frac{61}{243} & 0 & \frac{61}{243} & \frac{20}{81} & 0 & \frac{61}{243} & 0 \\\frac{61}{243} & 0 & \frac{61}{243} & 0 & 0 & \frac{20}{81} & 0 & \frac{61}{243} \\\frac{61}{243} & 0 & \frac{20}{81} & 0 & 0 & \frac{61}{243} & 0 & \frac{61}{243} \\0 & \frac{61}{243} & 0 & \frac{20}{81} & \frac{61}{243} & 0 & \frac{61}{243} & 0 \\\frac{20}{81} & 0 & \frac{61}{243} & 0 & 0 & \frac{61}{243} & 0 & \frac{61}{243} \\0 & \frac{20}{81} & 0 & \frac{61}{243} & \frac{61}{243} & 0 & \frac{61}{243} & 0\end{bmatrix} \Rightarrow P(A\to G)= \bbox[red, 2pt]{20\over 81}$$

二、計算證明題（共 86 分

解答:$$大體而言，第一個公式比第二個公式多了n-1個減法，當n很大時，第一種公式較費時；\\此外，當x_i與\mu 很接近時，第一種算法誤差較大$$

解答:$$x={5\over 4} \Rightarrow [x]+[3x] =1+3=4\ne 5 \Rightarrow 阿綠的答案顯然錯誤\\ 假設x=a+b, a\in \mathbb Z, 0\le b\lt 1 \Rightarrow 3x=3a+3b \\\Rightarrow \begin{cases}0\le b\lt 1/3 \Rightarrow [x]+3[x]= a+3a=4a =5,無解\\ 1/3\le b\lt 2/3 \Rightarrow [x]+[3x]= a+3a+1=5 \Rightarrow a=1 \Rightarrow 4/3\le x\lt 5/3\\ 2/3\le b\lt 1 \Rightarrow [x]+[3x]=a+3a+2=4a+2=5,無解  \end{cases} \\ \Rightarrow {4\over 3} \le x\lt {5\over 3}$$

解答:$$將圖形xy=k旋轉45^\circ, 即 \begin{bmatrix} x'\\ y'\end{bmatrix} = \begin{bmatrix} \sqrt 2/2 &-\sqrt 2/2\\ \sqrt 2/2 & \sqrt 2/2\end{bmatrix} \begin{bmatrix}x\\y \end{bmatrix} = \begin{bmatrix} {\sqrt 2\over 2}(x-y) \\ {\sqrt 2\over 2}(x+y)\end{bmatrix} \\ \Rightarrow (x')^2-(y')^2= {1\over 2}(4xy) ={1\over 2}\cdot 4k=2k \Rightarrow 旋轉後的圖形:(x')^2-(y')^2=2k 為雙曲線\\ \Rightarrow 旋轉前的圖形:xy=k也是雙曲線$$

解答:$$\cases{相異縱線間距為1 \Rightarrow 有n種 \\ 相異縱線間距為2 \Rightarrow 有n-1種 \\ \cdots \\相異縱線間距為n-1 \Rightarrow 有2種 \\ 相異縱線間距為n \Rightarrow 有1種 }\,,橫線也有相同的情形 \\ \Rightarrow 所有可能的矩形面積和= \left( \sum_{k=1}^n k(n-k+1)\right)^2\Rightarrow E(X_n)= {\left( \sum_{k=1}^n k(n-k+1)\right)^2\over {n+1\choose 2}^2}\\ \Rightarrow \lim_{n\to \infty }{E(X_n) \over n^2} = \lim_{n\to \infty }{\left( \sum_{k=1}^n k(n-k+1)\right)^2\over {n+1\choose 2}^2 n^2}  = \lim_{n\to \infty } \left({n(n+1)(2n+1)/6 \over n^2(n+1)/2} \right)^2 = \bbox[red, 2pt]{1\over 9}$$

解答:$$\cases{A(-1,2) \\ B(1,4) \\P(a,0)} \Rightarrow \cos \angle APB =f(a)={a^2+11a-4\over \sqrt{(a+1)^2+4} \cdot \sqrt{(a-1)^2+16}} \\ f'(a)=0 \Rightarrow 4a^3+36a^2+44a-84=0 \Rightarrow 4(a-1)(a+3)(a+7)=0 \Rightarrow a=1\\ \Rightarrow P= \bbox[red, 2pt]{(1,0)} \\\bbox[cyan, 2pt]{另解}:作A,B的外接圓且圓與x軸相切，切點即為P\\ \overline{AB}的中垂線L:x+y=3 \Rightarrow 圓心C=L\cap (x=a ) =(a,3-a) \Rightarrow \overline{AC}^2=\overline{CP}^2 \\ \Rightarrow (a+1)^2+(1-a)^2=(3-a)^2 \Rightarrow a^2+6a-7= 0 \Rightarrow (a-1)(a+7)=0 \\ \Rightarrow a=1 \Rightarrow P= \bbox[red, 2pt]{(1,0)}$$

====================== END ==========================
解題僅供參考，其他教甄試題及詳解