113年竹東高中教甄-數學詳解

國立竹東高級中學 113 學年度第一次教師甄試數學科試卷

一、填充題(每題 7 分,共 56 分)

解答: $$x^2-3x+1=0 \Rightarrow x-3+{1\over x}=0 \Rightarrow x+{1\over x}=3 \Rightarrow \cases{\left( x+{1\over x} \right)^2 =3^2 \\\left( x+{1\over x} \right)^3 =3^3 }\\\Rightarrow \cases{x^2+{1\over x^2}+2=9\\ x^3+{1\over x^3}+ 3(x+{1\over x})=27} \Rightarrow \cases{x^2+{1\over x^2} =7 \\ x^3+{1\over x^3}=18} \Rightarrow \left( x^2+{1\over x^2} \right)^2=7^2 \\ \Rightarrow x^4+{1\over x^4}=47 \Rightarrow \left(x^4+{1\over x^4} \right) \left( x^3+{1\over x^3} \right)=47\times 18 \Rightarrow x^7+{1\over x^7}+x+{1\over x}=846 \\ \Rightarrow x^7+{1\over x^7}=746-3= \bbox[red, 2pt]{843}$$

解答: $$f(x)=x+\sqrt{k-x^2} \Rightarrow k-x^2\ge 0 \Rightarrow k\ge x^2\ge 0\\ f'(x)=0 \Rightarrow 1={x\over \sqrt{k-x^2}} \Rightarrow x^2={k\over 2} \Rightarrow x=\pm \sqrt{ k\over 2} \\\Rightarrow 最大值f(\sqrt{  k\over 2}) =\sqrt{ k\over 2}+ \sqrt {k\over 2} =2\sqrt{13} \Rightarrow k=26\\ f(x)最小值發生在k-x^2=0 \Rightarrow x^2=k \Rightarrow x=-\sqrt{26} \Rightarrow f(-\sqrt{26})= \bbox[red, 2pt]{ -\sqrt{26}}$$

解答: $$\cases{F_1:x+y+z=0\\ F_2:x+y+z=k} \Rightarrow d(F_1,F_2)={|k| \over \sqrt 3} \\ \cases{L_1= E_1\cap E_2 :x=y=z\\ L_2= E_2\cap E_3: x-4=y+4=z \\ L_3=E_1\cap E_3: x-8=y-4=z} \Rightarrow \cases{A=F_1\cap L_1=(0,0,0)\\ B=F_1\cap L_2= (4,-4,0)\\ C=F_1\cap L_3 =(4,0,-4)} \\ \Rightarrow \triangle ABC 面積={1\over 2}\Vert \overrightarrow{AB} \times \overrightarrow{AC} \Vert ={1\over 2}\cdot 16\sqrt 3=8\sqrt 3\\ \Rightarrow 三角柱體積=  8\sqrt 3\cdot {| k|\over \sqrt 3} =30 \Rightarrow k= \bbox[red, 2pt]{\pm {15\over 4}}$$

解答: $$\begin{bmatrix}x' \\y' \end{bmatrix} =\begin{bmatrix}0 & 1 \\1 & 0 \end{bmatrix} \begin{bmatrix}1 & 3 \\0 & 1 \end{bmatrix} \begin{bmatrix}x   \\y \end{bmatrix} = \begin{bmatrix}y   \\x+3y \end{bmatrix} \Rightarrow \begin{bmatrix}x   \\ y \end{bmatrix} =\begin{bmatrix}y'-3x'   \\ x' \end{bmatrix} \\ \Rightarrow x^2+y^2=1 \Rightarrow (y'-3x')^2+x'^2=1 \Rightarrow \Gamma: \bbox[red, 2pt]{(y-3x)^2+x^2=1}$$

解答: $$b_n=\sum_{k=1}^{2n}{1\over n} \cdot {k\over \sqrt{n^2+k^2}} \Rightarrow a_n={1\over 2}b_n , \text{ as }n\to \infty \\ \Rightarrow \lim_{n\to \infty} b_n=\int_0^2 {x\over\sqrt{1+x^2}}\,dx = \int_1^5 {1/2\over \sqrt u}\,du =\sqrt 5-1 \Rightarrow \lim_{n\to \infty} a_n ={1\over 2}\lim_{n\to \infty} b_n =\bbox[red, 2pt]{\sqrt 5-1\over 2}$$

解答: $$\bbox[cyan, 2pt]{此題不計分}$$

解答: $$\cases{{a\over 3}+{b\over 4}+{c\over 6}=1 \\{a\over 4}+{b\over 5}+{c\over 7}=1 \\{a\over 6}+{b\over 7}+{c\over 9}=1 } \Rightarrow \cases{4a+3b+2c=12 \\ 35a+28b +20c=140\\ 21a+18b+14c=126} \Rightarrow \begin{bmatrix}4 & 3 & 2 \\35 & 28 & 20 \\21 & 18 & 14\end{bmatrix} \begin{bmatrix}a\\ b \\c   \end{bmatrix}= \begin{bmatrix}12 \\140 \\ 126 \end{bmatrix} \\\Rightarrow A= \begin{bmatrix}4 & 3 & 2 \\35 & 28 & 20 \\21 & 18 & 14\end{bmatrix}  \Rightarrow A^{-1}=\begin{bmatrix}16 & -3 & 2 \\-35 & 7 & -5 \\21 & \frac{-9}{2} & \frac{7}{2}\end{bmatrix} \Rightarrow \begin{bmatrix}a\\ b \\c   \end{bmatrix}= A^{-1} \begin{bmatrix}12 \\140 \\ 126 \end{bmatrix} = \begin{bmatrix}24 \\-70\\ 63 \end{bmatrix} \\ \Rightarrow a+b+c= 24-70+63= \bbox[red, 2pt]{17}$$

解答: $$a_2=a_1 \Rightarrow a_3=a_2+a_1 =2a_2 \Rightarrow a_4=a_3+a_2+a_1=4a_2 \Rightarrow a_n=2^{n-2}a_2 \\ \Rightarrow {1\over a_1}+ {1\over a_2}+ \cdots+ {1\over a_n}+\cdots= {1\over 7}+ {1\over 7}+{1\over 7\cdot 2}+{ 1\over 7\cdot 2^2}+\cdots \\={1\over 7} +{1\over 7} (1+{1\over 2}+ {1\over2^2}+\cdots)={1\over 7} +{1\over 7}\cdot 2= \bbox[red, 2pt]{3\over 7}$$

二、計算證明題(共 44 分)

解答: $$由於\sum_{k=i}^n C^n_kC^k_i= C^n_i 2^{n-i}, 因此\sum_{k=3}^n C^n_kC^k_3= C^n_3 2^{n-3} =\bbox[red, 2pt]{{1\over 3}2^{n-4}(n-2)(n-1)n}\\可以想成有n個物品分成三群,各有3個,k-3個,及n-k個的方法數\\ 先將n個取3個有C^n_3,剩下n-3個分成2群有2^{n-3}種分法,因此有C^n_32^{n-3}種分法$$

解答: $$\textbf{(1)}\; \Gamma_3:y=\sqrt{(1-x^2)^3} \Rightarrow V_3= \int_{-1}^1 (1-x^2)^3\pi\,dx =2 \pi \int_{0}^1 (1-x^2)^3 \,dx \\\quad = 2\pi \left. \left[-{1\over 7}x^{7}+{3\over 5}x^{5}-x^3+x \right] \right|_0^1 =2\pi \cdot {16\over 35}=\bbox[red, 2pt]{32\pi\over 35}\\ \textbf{(2)}\; 假設I_n= \int_0^1 (1-x^2)^n \,dx, 取\cases{u=(1-x^2)^n \\ dv=dx} \Rightarrow \cases{u=-2nx(1-x^2)^{n-1} \\ v=x} \\\quad \Rightarrow I_n=\left. \left[ x(1-x^2)^n\right] \right|_0^1+2n \int_0^1 x^2(1-x^2)^{n-1}\,dx =0+2n \int_0^1 (1-(1-x^2))(1-x^2)^{n-1}\,dx \\\quad =  2n \int_0^1 (1-x^2)^{n-1}\,dx -2n \int_0^1 (1-x^2)^n\,dx   =2nI_{n-1}-2nI_n  \Rightarrow  I_n={2n \over 2n+1}  I_{n-1} \\ \Rightarrow {V_{n+1} \over V_n} ={{2n+2\over 2n+3}I_n\over I_n} ={2n+2\over 2n+3} \quad \bbox[red, 2pt]{QED}$$

解答: $$y={4-3\sin x\over 2+\cos x} \Rightarrow 3\sin x+y\cos x=4-2y \Rightarrow \sqrt{y^2+9} \sin(x+\alpha) =4-2y \\ \Rightarrow \sin(x+\alpha)={4-2y\over \sqrt{y^2+9}} \Rightarrow \left| {4-2y\over \sqrt{y^2+9}}\right|\le 1 \Rightarrow (4-2y)^2 \le y^2+9\\ \Rightarrow 3y^2-16y+7\le 0 \Rightarrow (y-{8+\sqrt{43}\over 3}) (y-{8-\sqrt{43}\over 3}) \le 0 \Rightarrow \bbox[red, 2pt]{{8-\sqrt{43}\over 3}\le y\le {8+\sqrt{43}\over 3}}$$

解答: $$假設A(a,b), 由於{1\over 3}(2A+B)=P(1,2) \Rightarrow B(3-2a,6-2b) \Rightarrow \cases{a^2+b^2=37\\ (3-2a)^2+(6-2b)^2=37} \\ \Rightarrow a+2b=13 \Rightarrow a=13-2b \Rightarrow (13-2b)^2+b^2=37 \Rightarrow 5b^2-52b+132=0 \\ \Rightarrow (b-6)(5b-22)=0 \Rightarrow \cases{b=6 \Rightarrow a=1\\ b=22/5 \Rightarrow a=21/5} \Rightarrow \cases{A(1,6), B(1,-6)\\ A(21/5,22/5), B(-27/5,-14/5)} \\\Rightarrow \overleftrightarrow{AB} : \bbox[red, 2pt]{x=1,3x-4y+5=0}$$
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