115年陽明交大附中教甄-數學詳解

國立陽明交大附中 115 學年度 第 1 次教師甄選

一、填充題(每題 6 分，共 78 分)

解答:

$$\cases{A(0,0) \\B(\sqrt 2,0) \\C(1+\sqrt 2,1) \\E(\sqrt 2,2+\sqrt 2) \\ G(-1,1+\sqrt 2)} \Rightarrow \cases{L_1= \overleftrightarrow{AC}:y={x\over 1+\sqrt 2} \\ L_2=\overleftrightarrow{AE}:y=(1+\sqrt 2)x \\ L_2=\overrightarrow{BG}:y=-x+\sqrt 2} \Rightarrow \cases{P= L_1\cap L_3=(1,{1\over 1+\sqrt 2}) \\Q=L_2\cap L_3=({1\over 1+\sqrt 2},1)} \\ \Rightarrow \triangle APQ= {1\over 2} \begin{Vmatrix}  1& {1\over 1+\sqrt 2}& 1\\ {1\over 1+\sqrt 2}& 1& 1\\ 0& 0& 1\end{Vmatrix} = {1\over 2} \left( 1-{1\over (1+\sqrt 2)^2} \right) =\bbox[red,2pt]{\sqrt 2-1}$$

解答:$$\sqrt{60+10\sqrt{35}} =\sqrt{60+2\sqrt{25\times 35}} = \sqrt{35}+\sqrt{25} =\sqrt{35}+5  \Rightarrow \sqrt{60-10\sqrt{35}} = \sqrt{35}-5 \\ \Rightarrow  \sqrt{(60+10\sqrt{35})^{3/2} -(60-10\sqrt{35})^{3/2}} =\sqrt{(\sqrt{35}+5)^3 -(\sqrt{35}-5)^3} \\= \sqrt{10(60+10\sqrt{35}+10+60-10\sqrt{35})} =\sqrt{1300} = \bbox[red, 2pt]{10\sqrt{13}}$$

解答:$$假設\cases{違規 A:甲、乙皆被選出，且甲、乙分開 \\違規 B:丙、丁皆被選出，且丙、丁相鄰} \\ \Rightarrow 總排列數-(違規A)-(違規B)+(違規A且違規B) \\ 總排列數=P^7_5= {7!\over 2!} =2520 \\ \textbf{違規A}:甲、乙確定入選，剩下5人再選3人，有C^5_3=10種選法; 先將此3人(不含甲乙)\\\qquad 排列，有3!=6種排法，再將甲、乙二人插入4個間隔中，有P^4_2=12種排法。\\\qquad 因此違規A有10\times 6\times 12=720種排法\\ \textbf{違規B}:丙、丁確定入選，剩下5人再選3人，有C^5_3=10種選法;丙丁綁在一起,可左右\\\qquad 互換有2種排列,此4人任排有4!=24種排法，違規B有10\times 2\times 24=480種\\ \textbf{違規A且違規B}:甲乙丙丁皆入選,剩下3人選1人(假設此人代號A)有C^3_1=3種選法;\\\qquad 丙丁綁在一起可左右互換有2種排列，丙丁與A有2種排列，產生3個間隔\\\qquad 讓甲、乙插入，有P^3_2=6種插入排列，因此總共有3\times 2\times 2\times 6=72種排列\\ 結果:2520-720-480+72=\bbox[red, 2pt]{1392}$$

解答:$$\cases{\overleftrightarrow{AC}:{x-2\over 2}={y-7\over 1} ={z+3\over 2} \Rightarrow方向向量\vec u_1=(2,1,2) \\\overleftrightarrow{BD}:{x-5\over 2}={4-y\over 2} =6-z \Rightarrow方向向量\vec u_2=(2,-2,-1) } \Rightarrow \vec u_1\cdot \vec u_2=0 \Rightarrow \overleftrightarrow{AC}\bot \overleftrightarrow{BD} \\ \Rightarrow \overleftrightarrow{AC}與\overleftrightarrow{BD}歪斜\Rightarrow \overline{CD} =兩歪斜線距離 \\ \vec n=\vec u_1\times \vec u_2=(3,6,-6) \Rightarrow 取\vec n=(1,2,-2) \\ \cases{P_1(2,7,-3) \in \overleftrightarrow{AC} \\ P_2(5,4,6) \in \overleftrightarrow{BD}} \Rightarrow \overrightarrow{P_1P_2} =(3,-3,9) \Rightarrow \overline{CD} = {|\overrightarrow{P_1P_2}\cdot \vec n| \over ||\vec n||} =7\\ \cases{\overrightarrow{AC}\bot \overrightarrow{BD} \\ \overrightarrow{AC} \bot \overrightarrow{CD} \\\overrightarrow{CD} \bot \overrightarrow{BD}} \Rightarrow   |\overrightarrow{AB}|^2= |\overrightarrow{AC} + \overrightarrow{CD} -\overrightarrow{BD}|^2 = (\overrightarrow{AC})^2 +(\overrightarrow{CD})^2+(\overrightarrow{BD})^2  \\ =4^2+7^2+4^2=81 \Rightarrow \overline{AB}= \sqrt{81} =\bbox[red, 2pt] 9$$

解答:

$$f(x)=3x^3-18x+4\sqrt 2 \Rightarrow f'(x)=0 \Rightarrow 9x^2-18=9(x^2-2)=0 \Rightarrow x=\pm \sqrt 2 \\ \Rightarrow f''(x)=18x \Rightarrow \cases{f''(\sqrt 2)\gt 0\\ f''(-\sqrt 2)\lt 0} \Rightarrow \cases{f(\sqrt 2)=-8\sqrt 2為相對極小值\\ f(-\sqrt 2)=16\sqrt 2為相對極大值} \\ \Rightarrow y=|f(x)|圖形為凹向上,中間有兩個波峰(-\sqrt2,16\sqrt 2), (\sqrt 2,8\sqrt 2)\\ \Rightarrow 與水平線y=8\sqrt 2恰有五個交點\Rightarrow 2^k=8\sqrt 2=2^{7/2} \Rightarrow k= \bbox[red, 2pt]{7\over 2}$$

解答:$$\lim_{n\to \infty} \left( \sum_{n=1}^n {1\over n}f(2+{2i\over n}) \right) =\lim_{n\to \infty} \left( \sum_{n=1}^n {1\over n}(2+{2i\over n})({2i\over n}) ({2i\over n}-2) \right) = \int_0^1 (2+2x)x(2x-2)\,dx  \\=8\int_0^1(x+1)x(x-1)\,dx = 8\int_0^1 (x^3-x)\,dx =8 \left. \left[ {1\over 4}x^4-{1\over 2}x^2 \right] \right|_0^1=8\cdot (-{1\over 4}) = \bbox[red, 2pt]{-2}$$

解答:$$直線直線L_1: {x-3\over 2}={y-2\over 1}={z-2\over 1} =\{(2t+3,t+2,t+2) \mid t\in \mathbb R\} 分別代入E_1, E_2\\ 可得\cases{t=-1 \Rightarrow A(1,1,1) \\t=0 \Rightarrow B(3,2,2)} ; \quad 又\cases{E_1法向量\vec n_1=(1,1,1) \\E_2法向量\vec n_2 =(1,-1,2)} \Rightarrow \vec v= \vec n_1\times \vec n_2 =(3,-1,-2) \\ \Rightarrow L_2=E_1\cap E_2 的方向向量=\vec v \Rightarrow 取C(4,-1,0)\in L_2 \\ \Rightarrow 兩歪斜線L_1,L_2的距離= {|\overrightarrow{CP}\cdot \vec v|\over |\vec v|}={12\over 5\sqrt 3} \\ \Rightarrow 最小面積={1\over 2}\cdot \overline{AB} \cdot d(L_1,L_2) ={1\over 2}\cdot \sqrt 6\cdot {12\over 5\sqrt 3} = \bbox[red, 2pt]{{6\over 5}\sqrt 2}$$

解答:

$$假設\cases{A(1,\sqrt 3) \\B(1,-\sqrt 3)}, P\in \overline{AB} \Rightarrow P=(1,t),-\sqrt 3\le t\le \sqrt 3 \Rightarrow \overline{OP}斜率=t \Rightarrow \overline{MN}斜率為-{1\over t} \\ \Rightarrow \overleftrightarrow{MN}: y=-{1\over t}(x-1)+t \Rightarrow t^2-yt+1-x=0 \Rightarrow 判別式:y^2-4(1-x)\ge 0 \\ \cases{x^2+y^2\le 4\\ y^2\ge 4(1-x)} 所圍面積=半圓-拋物線與y軸所圍區域={1\over 2}2^2\pi-\int_{-2}^2 (1-{y^2\over 4})\,dy= \bbox[red, 2pt]{2\pi-{8\over 3}}$$

解答:$$\cases{\omega_1=z_1/3+2z_2/3 \\ \omega_2=3z_1/4+z_2/4} \Rightarrow \cases{z_1= (8\omega_2-3\omega1)/5\\ z_2= (9\omega_1-4\omega_2)/5} \Rightarrow {z_2\over z_1} ={9\omega_1-4\omega_2\over 8\omega_2-3\omega_1} ={9-4(\omega_2/ \omega_1) \over -3+8(\omega_2/\omega_1)} \\={9-4(1+\sqrt 3i) \over -3+8(1+\sqrt 3i)} ={5-4\sqrt 3i\over 5+8\sqrt 3i} \Rightarrow \left|{z_2\over z_1} \right|= \left| {5-4\sqrt 3i\over 5+8\sqrt 3i} \right|= \bbox[red, 2pt]{\sqrt{73\over 217}}$$

解答:$$xf(x-1)=(x-5)f(x) \Rightarrow \cases{f(0)=0\\ f(1)= 1\cdot f(0)/(-4)=0\\ f(2)= 2f(1)/(-3)=0\\ f(3)=3f(2)/(-2)=0 \\f(4)=4f(3)/(-1)=0} \\ \Rightarrow f(x)=x(x-1)(x-2) (x-3)(x-4)P(x) 代回xf(x-1)=(x-5)f(x) \\ \Rightarrow P(x-1)=P(x) \Rightarrow P(x)為常數\Rightarrow P(x)=k  \Rightarrow f(x)= kx(x-1)(x-2) (x-3)(x-4) \\ \Rightarrow f(6)=1=k\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2 \Rightarrow k={1\over 720} \Rightarrow f({5\over 2})={1\over 720}\cdot {5\over 2}\cdot {3\over 2}\cdot {1\over 2}\cdot {-1\over 2} \cdot {-3\over 2} =\bbox[red, 2pt]{1\over 512}$$

解答:$$\cos \angle A=\cos 2\theta={4^2+5^2-\overline{BC}^2 \over 2\cdot 4\cdot 5} \Rightarrow \overline{BC}^2=41-40\cos 2\theta \Rightarrow \overline{BC} =\sqrt{41-40\cos 2\theta} \\又{\overline{BC} \over \sin 2\theta} =2R \Rightarrow R\cdot \theta= {\sqrt{41-40\cos 2\theta}\cdot \theta\over 2\sin 2\theta} \Rightarrow \lim_{\theta \to 0} R\cdot \theta =\lim_{\theta \to 0}  {(\sqrt{41-40\cos 2\theta}\cdot \theta)'\over (2\sin 2\theta)'} \\= \lim_{\theta \to 0}{\sqrt{41-40\cos 2\theta} +{40\sin 2\theta \cdot \theta\over \sqrt{41-40\cos 2\theta}}\over 4\cos 2\theta} =\bbox[red, 2pt]{1\over 4}$$

解答:$$假設\cases{平常成績X\\ 段考成績Y\\ 學期成績Z} , 已知\cases{\sigma_X=20\\ \sigma_Y=15\\ \sigma_Z=13\\Z=0.4X+0.6Y} \\ \Rightarrow Var(Z)= Var(0.4X+0.6Y) \Rightarrow 13^2= 0.4^2 \cdot20^2+ 0.6^2\cdot 15^2+2 \cdot 0.4\cdot 0.6\cdot r\cdot 20\cdot 15 \\ \Rightarrow r=\bbox[red, 2pt]{1\over 6}\\ 公式:Var(aX+ bY)= a^2Var(X)+b^2Var(Y)+ 2ab r\sigma_X\sigma_Y$$

解答:$$轉移矩陣T= \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\0.4 & 0 & 0.6 & 0 & 0 & 0 \\0 & 0.4 & 0 & 0.6 & 0 & 0 \\0 & 0 & 0.4 & 0 & 0.6 & 0 \\0 & 0 & 0 & 0.4 & 0 & 0.6 \\0 & 0 & 0 & 0 & 0 & 1\end{bmatrix} \Rightarrow \text{canonical form} \begin{bmatrix}I&0\\R&Q \end{bmatrix},其中\\ Q= \begin{bmatrix}0 & 0.6 & 0 & 0 \\0.4 & 0 & 0.6 & 0 \\0 & 0.4 & 0 & 0.6 \\0 & 0 & 0.4 & 0 \end{bmatrix}, R= \begin{bmatrix} 0.4 & 0 \\0 & 0 \\0 & 0 \\0 & 0.6\end{bmatrix}, I= \begin{bmatrix}1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1 \end{bmatrix} \\ \Rightarrow B=(I-Q)^{-1}R =\begin{pmatrix}P(1 \to 0) & P(1 \to 5) \\P(2 \to 0) & P(2 \to 5) \\P(3 \to 0) & P(3 \to 5) \\P(4 \to 0) & P(4 \to 5)\end{pmatrix} = \begin{bmatrix} \frac{130}{211} & \frac{81}{211} \\ \mathbf{\frac{76}{211}} & \mathbf{\frac{135}{211}} \\\frac{40}{211} & \frac{171}{211} \\\frac{16}{211} & \frac{195}{211}\end{bmatrix} \\ \Rightarrow P(2\to 5)= \bbox[red, 2pt]{135\over 211}$$

二、計算證明題： (22 分，須詳列出過程，否則不予計分)

解答:$$假設截圓C_h的半徑為r_h \Rightarrow {r_h\over 底面半徑} ={頂點到平面的距離\over 圓錐總高}\Rightarrow {r_h\over 2} ={6-h\over 6} \Rightarrow r_h={6-h\over 6}\\ R_h為正方形面積最大\Rightarrow R_h對角線長即為圓直徑 \Rightarrow 正方形邊長={2r_h\over \sqrt 2} \\ \Rightarrow R_h最大面積= \left( {2r_h\over \sqrt 2} \right)^2=2r_h^2 =2 \left( {6-h\over 6} \right)^2={2\over 9}(6-h)^2 \\ \Rightarrow 錐體體積V(h)={1\over 3}\times {2\over 9}(6-h)^2\times h ={2\over 27}h(6-h)^2 \Rightarrow V'(h) ={2\over 9}(6-h)(2-h)=0 \\ \Rightarrow h=2 (0\lt h\lt 6 \Rightarrow h\ne 6) \Rightarrow V(2)= \bbox[red, 2pt]{64\over 27}$$

解答:$$\cases{g(114) =f(114) \\g(115) =f(115) \\g(116) =f(116) } \Rightarrow g(x)-f(x)=(x-114) (x-115)(x-116) \\ \Rightarrow g(x)=f(x)+(x-114) (x-115)(x-116) \\\Rightarrow g(t+115)=f(t+115)+(t+1)t(t-1) =f(t+115)+t^3-t \\ f(x)= 123x^2+ 234x-345 \Rightarrow f(t+115)=123t^2+At+B, \cases{A=123\cdot 230+234\\ B=123\cdot 115^2+234\cdot 115-345} \\ \Rightarrow g(t+115)=t^3+123t^2+(A-1)t+B \Rightarrow h(t+115) =(A-1)t+B \\ \Rightarrow g(t+115)-h(t+115)=t^3+123t^2 \Rightarrow \text{area} =\int_{114}^{116} |g(x)-h(x)|\,dx =\int_{-1}^{1 }|t^3+123t^2|\,dt \\=\int_{-1}^1 t^3\,dt +123\int_{-1}^1t^2\,dt =0+ 246 \int_0^1 t^2\,dt =246\cdot {1\over 3}= \bbox[red, 2pt]{82}$$

解答:$$f(x)= k(x-a)(x-b)(x-c) \Rightarrow {f'(x)\over f(x)} ={1\over x-a}+{1\over x-b}+{1\over x-c} \\ \Rightarrow {d\over dx} \left( {f'(x)\over f(x)} \right) ={d\over dx} \left( {1\over x-a}+{1\over x-b}+{1\over x-c} \right) \\ \Rightarrow {f''(x)f(x)-[f'(x)]^2\over [f(x)]^2} =- \left( {1\over (x-a)^2}+{1\over (x-b)^2}+{1\over (x-c)^2}  \right) \\ \Rightarrow f''(x)f(x)-[f'(x)]^2=- [f(x)]^2  \left( {1\over (x-a)^2}+{1\over (x-b)^2}+{1\over (x-c)^2}  \right)\\ 狀況一: x\ne a 且x\ne b且x\ne c \Rightarrow \cases{[f(x)]^2 \gt 0\\  {1\over (x-a)^2}+{1\over (x-b)^2}+{1\over (x-c)^2}\gt 0} \Rightarrow f''(x)f(x)-[f'(x)]^2\lt 0 \\\qquad \Rightarrow [f'(x)]^2\gt f''(x)f(x) \\ 狀況二:x=a或x=b或x=c \Rightarrow f(x)=0 \Rightarrow f''(x)f(x)=0 \Rightarrow [f'(x)]^2\ge  f''(x)f(x) \\ 因此[f'(x)]^2\ge  f''(x)f(x)\quad \bbox[red, 2pt]{故得證}$$

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