112年台北市高中教甄聯招-數學詳解

 臺北市 112 學年度市立普通型暨技術型高級中等學校

正式教師聯合甄選

壹、選擇題：佔 20 分（共 4 題，每題 5 分）
單選題

解答: $$假設\cases{A(0,0,0) \\ B(1,0,1/2) \\ C(1,1,2/3)\\ D(0,1,a)} \Rightarrow \cases{\overrightarrow{AB} =(1,0,1/2) \\ \overrightarrow{AC}= (1,1,2/3)} \Rightarrow \vec n= \overrightarrow{AB}\times \overrightarrow{AC} =(-1/2,-1/6,1) \\ \Rightarrow \triangle ABC構成的平面E:-{1\over 2}x-{1\over 6}y+z=0\\ D在E上\Rightarrow 0-{1\over 6}+a=0 \Rightarrow a={1\over 6}，故選\bbox[red, 2pt]{(A)}$$

解答：$$此題相當於求兩圖形\cases{y=f(x)=\sin(\pi x) \\y=g(x)={1\over 2x}}的交點數 \\由於f(0)\ne g(0),僅需考慮x\in (0,3],且\cases{0\le |f(x)|\le 1 \\g(x)\gt 0 }\\ \Rightarrow 只需考慮f(x)\gt 0的範圍,即\cases{0\lt \pi x\lt \pi \\ 2\pi \lt \pi x\lt 3\pi} \Rightarrow \cases{0\lt x\lt 1\\ 2\lt x\lt 3}\\g(x)為遞減且明顯解:f(1/2)=g(1/2)=1為y=f(x)之最高點\\\Rightarrow 還有三個交點,分別位於(1/2,1),(2,5/2),(5/2,3) \Rightarrow 共四點，故選\bbox[red,2pt]{(D)}$$

解答：$$\begin{array}{c|cccccccccc} 0 & 1 & 2& 3& 4 & 5& 6 &7&8 &9 &10\\\hdashline 1 & 6 & 1 & 7 & 2& 8& 3& 9 & 4& 10 & 5\\ 2 & 3 & 6 & 9 & 1 & 4 & 7& 10 & 2 & 5 & 8 \\ 3& 7 & 3& 10 & 6 & 2 & 9 & 5 & 1 & 8 & \color{blue}4 \\ 4 & 9 & 7 & 5 & 3& 1& 10& 8 &6 & 4& 2\\ 5 & 10 & 9 & 8 & 7 & 6 & 5 & 4& 3& 2& 1\\\hline\end{array}\\ 由上表可知:經過5次洗牌後，數列剛好反序，因此經過10次洗牌後回到原序列\\ 洗了2023次牌相當於洗了3次牌,第10張是4,故選\bbox[red, 2pt]{(B)}$$

解答：$$ \sum_{k=1}^\infty k\cdot p^{k}(1-p) \ge 10 \Rightarrow \sum_{k=1}^\infty kp^k -\sum_{k=1}^\infty kp^{k+1} \ge 10 \\ \Rightarrow {p\over (1-p)^2}-{p^2\over (1-p)^2} \ge 10 \Rightarrow {p\over 1-p}\ge 10 \Rightarrow {1\over 1-p}-1\ge 10 \Rightarrow {1\over 1-p}\ge 11\\ \Rightarrow 1-p\le {1\over 11} \Rightarrow p\ge {10\over 11}, 故選\bbox[red,2pt]{(C)}$$

貳、非選擇題：佔 40 分（共 8 題，每題 5 分）

解答：

$$由題意:\cases{建仔身高\overline{AE}=1.8\\ 伊森身高\overline{BF}=1.6\\ 伊森前進距離\overline{AB}=45 \\ \angle CEH=\alpha\\ \angle CFG=\beta},並假設\cases{大樓高度\overline{CD}=h\\ \overline{BD}=w}\\則\cases{\tan \alpha={\overline{CH} \over \overline{EH}} \\ \tan \beta ={\overline{CG} \over \overline{FG}}} \Rightarrow \cases{{h-1.8\over w+45}={1\over 6} \\ {h-1.6\over w}={1\over 3}} \Rightarrow \cases{w=6h-10.8-45\\ w=3h-4.8} \\ \Rightarrow 6h-55.8=3h-4.8 \Rightarrow 3h=51 \Rightarrow h=17 \Rightarrow 教學大樓高度為\bbox[red,2pt]{17}公尺$$

解答：

$$P,Q皆在直線y=x上 \Rightarrow 假設\cases{Q(k,k)\\ P(k/2,k/2)} \Rightarrow \cases{k=\log_a k\\ {1\over 2}k = \log_a {k\over 2}} \\ \Rightarrow \cases{k=4\\ a=\sqrt 2} \Rightarrow Q(4,4) \Rightarrow  \overline{OQ}^2 =\bbox[red, 2pt]{32}$$

解答：$$\cases{F_1(-4,0)\\ F_2(4,0)\\ 頂點(0,3)} \Rightarrow \cases{c=4\\ b=3} \Rightarrow a=5 \Rightarrow \Gamma:{x^2\over 25}+ {y^2\over 9}=1\\  又\overline{PF_1}+\overline{F_1A} \ge \overline{PA} \Rightarrow \overline{PF_1}+ \overline{PF_2}+\overline{F_1A} \ge \overline{PA} +\overline{PF_2} \\ \Rightarrow \overline{PA} +\overline{PF_2} 的最大值=\overline{PF_1}+ \overline{PF_2}+\overline{F_1A} =2a+\sqrt{5^2+1} =\bbox[red, 2pt]{10+\sqrt{26}}$$

解答：$$\vec a\parallel \vec b \Rightarrow {6\over 8}={\sqrt{1-\sin \theta}\over \sqrt {\sin \theta}} \Rightarrow {9\over 16} ={1-\sin \theta \over \sin \theta} \Rightarrow \sin\theta ={16\over 25} \\ \Rightarrow \vec a\cdot \vec b最大值 =6 \sqrt{1-\sin\theta}+8 \sqrt{\sin \theta} =6\cdot \sqrt{9\over 25}+8 \cdot \sqrt{16\over 25} ={18+32\over 5}= \bbox[red, 2pt]{10}$$

解答：$$\cases{A(0,0,0) \\B(1,0,0)\\ C(0,1,0)\\ D(1,1,1)} \Rightarrow \cases{E_1=\triangle ABC:z=0 \\ E_2=\triangle ACD: x-z=0\\ E_3=\triangle ABD: -y+z=0 \\ E_4= \triangle BCD:x+y-z=1}\\ 假設球心P(a,b,c) \Rightarrow \cases{d(P,E_1)=c\\ d(P,E_2)=|a-c|/\sqrt 2 \\ d(P,E_3)=|-b+c|/\sqrt 2 \\ d(P,E_4)=|a+b-c-1|/\sqrt 3}\\ \triangle OAB為等腰直角 \Rightarrow a=b ，\\若a\lt c \Rightarrow d(P,E_2)={c-a\over \sqrt 2}=c \Rightarrow a=(1-\sqrt 2)c \lt 0 不合 \Rightarrow a\ge c \Rightarrow a=(\sqrt 2+1)c\\ 再考慮 d(P,E_4)={|(2\sqrt 2+1)c-1|\over \sqrt 3} =c \\ 若c\ge {1\over 2\sqrt 2+1} \Rightarrow c={1\over 2\sqrt 2-\sqrt 3+1} \Rightarrow a={\sqrt 2+1\over 2\sqrt 2-\sqrt 3+1} \gt 1 \Rightarrow P在四面體外側，不合\\ 因此c\le {1\over 2\sqrt 2+1} \Rightarrow c={1\over 2\sqrt 2+\sqrt 3+1} \Rightarrow a= \bbox[red,2pt]{\sqrt 2+1\over 2\sqrt 2+\sqrt 3+1}$$

解答：$$f(n)+n=2023 \Rightarrow n\lt 2023 \Rightarrow \max\{f(n),n\lt 2023\} =f(1999)=28\\ \Rightarrow \begin{array}{} n &  f(n) & f(n)+n \\\hline 2015 & 8 & \color{blue}{2023}\\\cdots\\2001 & 3 & 2004\\ 2000 & 2 & 2002\\ \color{red}{1999} & 28 & 2027 \\ 1998 & 27 &2025 \\ 1997 & 26 & \color{blue} {2023} \\ & & 越來越小\end{array} \Rightarrow 只有兩個解n=2015及1997\\ \Rightarrow 所有解的和＝2015+ 1997 = \bbox[red,2pt]{4012}$$

解答：$$\triangle ABC 面積=6={3x+4y+5z\over 2} \Rightarrow 3x+4y+5z=12\\ 利用\text{Lagrange multiplier}求解：假設\cases{f=3x^2+ y^2+2yz+ 2z^2 \\ g=3x+4y+5z-12} \\ \Rightarrow \cases{f_x=\lambda g_x \\f_y=\lambda g_y \\f_z=\lambda g_z \\g=0 } \Rightarrow \cases{6x=3\lambda \cdots(1)\\ 2y+2z=4\lambda \cdots(2)\\ 2y+4z=5\lambda\cdots(3)\\ 3x+4y+5z=12 \cdots(4)} ,由(1),(2),(3)\Rightarrow \cases{{3x\over y+z}={3\over 4} \\ {y+z\over y+2z} ={4\over 5} \\ {3x\over y+2z}={3\over 5}} \\ \Rightarrow \cases{x=z\\ y=3z}代入(4)\Rightarrow z={3\over 5} \Rightarrow \cases{x=z={3\over 5}\\ y={9\over 5}} \Rightarrow f({3\over 5}, {3\over 5},{9\over 5}) = \bbox[red,2pt]{36\over 5}為最小值$$

解答：$$假設X不含112，例X=\{a,b,c,d,a\gt b\gt c\gt d\} \Rightarrow f(X)=a-b+c-d\\ 又f(X\cup \{112\}) =112-a+b-c+d,因此f(X)+f(X\cup \{112\})=112\\ \cases{X不含112的有2^{111}個\\ X含112的也有2^{111}個} \Rightarrow \sum f(X) = \bbox[red,2pt]{112\cdot 2^{111}}$$

參、計算題：佔 40 分（共 4 題，每題 10 分）

解答：$$\mathbf{(1)}\; 行星E(400\cos \theta,400\sin \theta)每天繞行角度為{2\pi\over 500}={\pi \over 250} \\ \Rightarrow 第t天時，E坐標為(400\cos {t\pi\over 250}, 400\sin {t\pi\over 250})\\ 若E為原點，衞星M(\cos \theta, \sin \theta)每天繞行角度為{2\pi \over 50}={\pi\over 25}\\ \Rightarrow 第t天時，M坐標為(\cos {t\pi\over 25},\sin {t\pi\over 25})\\ 因此E坐標為(400\cos {t\pi\over 250}, 400\sin {t\pi\over 250})時，M坐標為\bbox[red,2pt]{(\cos {t\pi\over 25} +400\cos{t\pi\over 250},\sin {t\pi\over 25}+ 400\sin{t\pi\over 250})}\\ \mathbf{(2)}\;第t天時,\cases{\vec u=\overrightarrow{ES}=(-400\cos {t\pi\over 250}, -400\sin {t\pi\over 250}) \\ \vec v= \overrightarrow{EM}=(\cos{t\pi\over 25}, \sin{t\pi\over 25})} \\ \angle SEM=180^\circ \Rightarrow \cos \angle SEM=-1\Rightarrow \cos \angle SEM = {\vec u\cdot \vec v\over |\vec u||\vec v|} \\={-400\cos{t\pi\over 250}\cos {t\pi \over 5}- 400\sin{t\pi\over 250} \sin{t\pi\over 5} \over 400\cdot 1} = -\cos{t\pi\over 250}\cos {t\pi \over 5}-\sin{t\pi\over 250} \sin{t\pi\over 5} \\ =-\cos({t\pi\over 5}-{t\pi\over 250}) =-1 \Rightarrow  \cos({t\pi\over 5}-{t\pi\over 250})=1\Rightarrow {t\pi\over 5}-{t\pi \over 250} =\pi \\ \Rightarrow 49t\pi = 250\pi  \Rightarrow t={250\over 49} \approx 5.1 \Rightarrow 第\bbox[red, 2pt]{6}天$$

解答：$$\mathbf{(1)}(x,y)逆時鐘旋轉\theta後變為(X,Y) \Rightarrow \begin{bmatrix}X \\Y \end{bmatrix}=\begin{bmatrix} \cos \theta & -\sin \theta \\\sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix}x \\y \end{bmatrix} =\begin{bmatrix}x\cos\theta-y\sin \theta \\ x\sin\theta+y \cos\theta \end{bmatrix} \\ \Rightarrow 旋轉後的橢圓方程式\Gamma': (x\cos\theta-y\sin \theta)^2 -(x\cos\theta-y \sin \theta)( x\sin \theta+ y\cos \theta)+ (x\sin \theta+ y\cos \theta)^2=3 \\ \Rightarrow (\cos^2\theta -\cos\theta \sin\theta+\sin^2 \theta)x^2 +(-2\cos\theta \sin\theta-\cos^2\theta+\sin ^2\theta +2\cos\theta \sin\theta)xy \\\qquad +(\sin^2 \theta+\cos\theta \sin\theta+ \cos^2 \theta)y^2=3\\ \Rightarrow \bbox[red, 2pt]{(1-\cos\theta \sin\theta)x^2 + (\sin^2\theta -\cos^2\theta)xy +(1+\sin\theta \cos\theta)y^2 =3 }\\ \mathbf{(2)}\sin^2 \theta -\cos^2\theta =0 \Rightarrow \theta=45^\circ \Rightarrow \Gamma':(1-{1\over 2})x^2+0xy+ (1+{1\over 2})y^2=3 \\ \Rightarrow {1\over 2}x^2+{3\over 2}y^2=3 \Rightarrow {1\over 6}x^2+ {1\over 2}y^2=1 \Rightarrow \bbox[red, 2pt]{\cases{A=1/6\\ B=1/2}}$$

解答：$$\mathbf{(1)}\; 梯形面積={\overline{BC}+ \overline{AD} \over 2}\cdot \overline{BH} ={2+(2+2a)\over 2} \cdot b =\bbox[red, 2pt]{(a+2)b}\\\mathbf{(2)}\;a^2+b^2=1 \Rightarrow \cases{a= \cos \theta\\ b=\sin \theta} \Rightarrow (a+2)b= (\cos \theta +2)\sin \theta \equiv f(\theta) \\ \Rightarrow f(\theta) =2\sin\theta +{1\over 2}\sin 2\theta  \Rightarrow f'(\theta) =0 \Rightarrow 2\cos\theta+ \cos 2\theta = 2\cos \theta +2\cos^2\theta -1=0 \\\Rightarrow \cos \theta ={\sqrt 3-1\over 2}({-\sqrt 3-1\over 2}\lt 0, 不合) \Rightarrow a=\bbox[red, 2pt]{\sqrt 3-1\over 2}$$

解答：$$\mathbf{(1)}\; \bbox[red,2pt]{P在\overline{BC}上}，使得\overline{PA}+ \overline{PB}+ \overline{PC}+ \overline{PD} =\overline{AB}+2\overline{BC} +\overline{CD} =\overline{AD}+\overline{BC} 最小 \\\mathbf{(2)}\;假設\cases{\overline{AB}=\overline{CD}=m\\ \overline{BC}=2n}\;,並令\overline{BC}的中點O為原點,即\cases{A(-m-n,0)\\ B(-n,0)\\ C(n,0)\\ D(m+n,0)\\ P(a,b),b\ne 0} \\ 因此\cases{ \overline{PA}^2 = (a+m+n)^2+b^2\\ \overline{PD}^2 =(a-m-n)^2+b^2 \\ \overline{PB}^2 =(a+n)^2+b^2\\ \overline{PC}^2 =(a-n)^2+b^2} \Rightarrow \cases{\overline{PA}^2+ \overline{PB}^2 = 2a^2+2b^2 +2(m+n)^2 \\ \overline{PB}^2 +\overline{PC}^2 = 2a^2+ 2b^2+2n^2} \\ \Rightarrow \overline{PA}^2+ \overline{PB}^2\gt \overline{PB}^2 +\overline{PC}^2 \Rightarrow \overline{PA}+ \overline{PD} \gt\overline{PB} +\overline{PC},\bbox[red,2pt]{故得證}$$

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