109年台北市西松高中教甄數學詳解

臺北市立西松高級中學 109 學年度

教師甄試數學科試題

解：

$$A,B在x^2=4y 上 \Rightarrow \cases{A(2a,a^2)\\ B(2b,b^2)}，並假設\cases{L_1:過A之切線\\ L_2:過B之切線};\\ 又x^2=4y \Rightarrow y'={1\over 2}x \Rightarrow \cases{L_1線斜率={1\over 2}(2a)=a \\ L_2斜率={1\over 2}(2b)=b} \Rightarrow \cases{L_1: y-a^2=a(x-2a)\cdots(1)\\ L_2: y-b^2=b(x-2b) \cdots(2)} \\ (1)-(2)\Rightarrow -a^2+b^2 = (a-b)x-2(a^2-b^2) \Rightarrow x=a+b代回(1) \Rightarrow y-a^2=a(b-a) \\ \Rightarrow y=ab \Rightarrow L_1與L_2的交點C(a+b,ab)\\由題意知\overline{AB}^2= 4 = 4(a-b)^2+(a^2-b^2)^2 =4(a-b)^2+ (a+b)^2(a-b)^2 \\ =(a-b)^2((a+b)^2+4) = (x^2-4y)(x^2+4) \Rightarrow x^2-4y= {4\over x^2+4} \Rightarrow x^2-{4\over x^2+4} =4y\\ \Rightarrow {x^2(x^2+4)-4\over x^2+4}=4y \Rightarrow \bbox[red, 2pt]{y={x^4+ 4x^2-4\over 4(x^2+4)}}$$

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2. 設$A=\{(x,y)\mid (x-2)^2+2y^2 \le 4\},B=\{(x,y)\mid x\le 2y^2\},C= A\cap B$，若區域C繞x軸旋轉得一旋轉體，則此旋轉體體積為？

解：

$$此題相當於求上圖填色區域繞X軸旋轉的體積；先求兩圖形交點:\\ \cases{(x-2)^2+2y^2=4 \\ x=2y^2} \Rightarrow \cases{y^2=2-(x-2)^2/2\\ y^2=x/2} \Rightarrow 4-(x-2)^2=x \Rightarrow x^2-3x=0 \\ \Rightarrow x(x-3)=0 \Rightarrow \cases{x=0\\ x=3} \Rightarrow 交點\cases{O(0,0)\\ A(3,\sqrt{3/2})\\ B(3,-\sqrt{3/2})} \\ \Rightarrow 旋轉體積 \int_0^3 (2-{(x-2)^2\over 2})\pi\;dx - \int_0^3 {x\over 2}\pi \;dx={27\over 6}\pi -{9\over 4}\pi = \bbox[red, 2pt]{{9\over 4}\pi}$$

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解：

$$P在{x^2\over 25}+{y^2\over 16}=1上\Rightarrow \cases{P(5\cos \theta,4\sin \theta);\\ {2x\over 25}+{yy'\over 8}=0 } \Rightarrow {10\cos \theta \over 25} +{4 y'\sin \theta\over 8}=0 \Rightarrow y'=-{4\cos \theta\over 5\sin \theta} \\ \Rightarrow 過P的切線方程式為L:y-4\sin \theta =-{4\cos \theta\over 5\sin \theta}(x-5\cos\theta) \Rightarrow L:{\sin \theta \over 4}y+ {\cos \theta \over 5}x=1\\ \Rightarrow \cases{A({5\over \cos \theta},0) \\ B(0,{4\over \sin \theta})} \Rightarrow \overline{AB}^2 =({5\over \cos\theta})^2 +({4\over \sin \theta})^2 \Rightarrow \left(({5\over \cos\theta})^2 +({4\over \sin \theta})^2\right) (\cos^2\theta +\sin^2\theta) \ge (5+4)^2 \\ \Rightarrow \overline{AB}^2 \ge 9^2 \Rightarrow \overline{AB}最小值出現在{4\cos \theta\over \sin \theta} ={5\sin \theta\over \cos \theta} \Rightarrow 4\cos^2\theta= 5-5\cos^2 \theta \Rightarrow \cos\theta ={\sqrt 5\over 3} \\ \Rightarrow \sin \theta ={2\over 3} \Rightarrow L:{1\over 6}y+{\sqrt 5\over 15}x=1 \Rightarrow y=-{2\sqrt 5\over 5}x+6 \Rightarrow (a,b)= \bbox[red, 2pt]{(-{2\sqrt 5\over 5},6)}$$

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解：

$$\cases{f(x)=x^4 +ax^3+bx^2+cx +d\\ f(1)=2020\\ f(2)=4040\\ f(3)=6060},取g(x)=f(x)-2020x \Rightarrow g(1)=g(2)=g(3)=0\\ \Rightarrow g(x)=(x-1)(x-2)(x-3)(x+k) \Rightarrow \cases{f(7)=g(7)+2020\times 7=120(k+7)+2020\times 7\\ f(-3)= g(-3)-2020\times 3= -120(k-3)-2020\times 3}\\ \Rightarrow f(7)+f(-3)= 1200+8080  = \bbox[red, 2pt]{9280}$$

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解：

$$R+G+Y=20,0\le R,G,Y \le 12\xrightarrow{每色至少一球} R+G+Y=17, 0\le R, G, Y\le 11\\ \Rightarrow \begin{array}{c|c|c}R & G & Y &數量\\\hline  0 & 11 & 6 &\\ & 10& 7\\ & \cdots & \cdots & \\ &6 & 11 & 6\\\hline 1 & 11 & 5 & \\ & \cdots & \cdots &\\ & 5 & 11 & 7\\\hline \cdots & \cdots & \cdots &\\\hline 6& 11 & 0 & \\ & \cdots & \cdots & \\ & 0 & 11 & 12\\\hline 7 & 10 & 0 \\ & \cdots & \cdots \\ & 0 & 10 & 11\\\hline \cdots & \cdots & \cdots &\\\hline 11 & 6 & 0 &\\ & \cdots & \cdots & \\ & 0 & 6 & 7\\\hline\end{array} \\ \Rightarrow 共有(6+7+\cdots +12) +(11+\cdots +7)=63+45= \bbox[red,2pt]{108}種$$

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解：

$$f(x)=\sqrt{x^4-x^2-4x+5} -\sqrt{x^4-3x^2+4} \\=\sqrt{(x-2)^2+(x^2-1)^2}-\sqrt{(x-0)^2+(x^2-2)^2}\\ 令\cases{P(x,x^2)\\ A(2,1)\\ B(0,2)} \Rightarrow f(x)=\overline{PA} -\overline{PB} \le \overline{AB}=\sqrt{2^2+1^2} =\bbox[red, 2pt]{\sqrt 5}$$

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解：

$$\cases{f(x)=x^3+3x^2 +6x-8 \\ g(x)=x^3-6x^2+15x-2} \Rightarrow \cases{f'(x)=3x^2+6x+6 \\ g'(x)= 3x^2-12x+15} \Rightarrow \cases{判別式:36-78< 0\\ 判別式:144-12\cdot 15< 0} \\ \Rightarrow \cases{f(x)=0只有一實根\alpha\\ g(x)=0只有一實根\beta} ,令a=\alpha + \beta \Rightarrow \alpha = a-\beta\Rightarrow f(\alpha)=(a-\beta)^3+3(a-\beta)^2+6(a-\beta)-8=0 \\ \Rightarrow \beta^3-(3a+3)\beta^2+(3a^2+6a+6)\beta-(a^2+3a^2+6a-8)=0 \equiv g(\beta)=0 \\\Rightarrow \cases{3a+3=6\\ 3a^2+6a+6=15 \\ a^2+3a^2+6a-8=2}\Rightarrow \bbox[red, 2pt]{a=1}$$

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解：

$$令S=任二數相乘所得之總和; (5+7+\cdots +23)^2 = 5^2+7^2+\cdots +23^2 +2S\\ \Rightarrow 2S= (5+7+\cdots +23)^2 - (5^2+7^2+\cdots +23^2) =\left( \sum_{k=1}^{10} 2k+3 \right)^2 -\sum_{k=1}^{10}(2k+3)^2\\ =(2\times 55+30)^2- \left( 4\times {10\times 11\times 21 \over 6} +12\times{10\times 11\over 2} +90\right) =140^2-(1540+660+90)\\ =17310 \Rightarrow S=17310\div 2= \bbox[red, 2pt]{8655}$$

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解：

$$\cases{\vec a,\vec b皆為單位向量\\ \vec a,\vec b夾角60^\circ} \Rightarrow 假設\cases{\vec a=(1,0,0)\\ \vec b=({1\over 2},{\sqrt 3\over 2},0)} \Rightarrow \vec c=\vec a\times \vec b=(0,0,{\sqrt 3\over 2});\\ 令\vec p=(x,y,z) \Rightarrow \cos \theta = {\vec p\cdot \vec a\over |\vec p||\vec a|} = {\vec p\cdot \vec b\over |\vec p||\vec b|} = {\vec p\cdot \vec c\over |\vec p||\vec c|} \Rightarrow \cases{x=z\\y={1\over \sqrt 3}x}\\ \cos \theta ={\vec p\cdot \vec a\over |\vec p||\vec a|} = {x\over \sqrt{x^2+y^2+z^2}} = {x\over \sqrt{x^2+x^2/3+x^2}} = \sqrt{3\over 7} \\\Rightarrow \cos 2\theta =2\cos^2 \theta-1 ={6\over 7}-1= \bbox[red,2pt]{-{1\over 7}}$$

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解：

$$\cases{a_1=1\\ a_{n+1}=2a_n+n-1,n\in N} \Rightarrow a_{n+1}+(n+1)=2(a_n+n)\\ \Rightarrow a_n+n= 2(a_{n-1}+n-1) = 2(2(a_{n-2}+n-2))=2^2 (a_{n-2}+n-2)\\ = 2^3(a_{n-3}+n-3) = \cdots =2^{n-1}(a_1+1)=2^{n-1}a_1+2^{n-1} =2^{n-1}+2^{n-1} =2^{n} \\ \Rightarrow \bbox[red, 2pt]{a_n=2^n-n,n\in N}$$

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解：

(1) $$令內切圓圓心I及外接圓圓心O ;\\ \cases{\triangle IQR面積={1\over 2}r^2\sin \angle QIR = {1\over 2}r^2\sin ( 180^\circ -\angle C) ={1\over 2}r^2\sin \angle C\\ \triangle IPQ面積={1\over 2}r^2\sin \angle PIQ = {1\over 2}r^2\sin ( 180^\circ -\angle A) ={1\over 2}r^2\sin \angle A\\\triangle IPR面積={1\over 2}r^2\sin \angle PIR = {1\over 2}r^2\sin ( 180^\circ -\angle B) ={1\over 2}r^2\sin \angle B} \\ \Rightarrow \triangle PQR面積= \triangle IQR+\triangle  +IPQ +\triangle IPR ={1\over 2}r^2(\sin \angle A+\sin \angle B+\angle C)\\ ={1\over 2}r^2({\overline{BC}\over 2R}+{\overline{CA}\over 2R}+ {\overline{AB}\over 2R}) = {r^2\over 4R}(\overline{AB} +\overline{BC}+\overline{CA})\cdots (1)\\ \triangle ABC面積={1\over 2}r(\overline{AB} +\overline{BC}+\overline{CA}) \cdots(2) \Rightarrow {\triangle PQR面積 \over \triangle ABC面積} ={(1)\over (2)} = \bbox[red,2pt]{r \over 2R}$$ (2) $$尤拉定理: \overline{IO}^2 = R^2-2Rr \ge 0 \Rightarrow R \ge 2r \Rightarrow \bbox[red, 2pt]{{r\over 2R} \le {1\over 4}}$$

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解：

(1) $$\overline{PF}={1\over 2}d(P,L) \Rightarrow \sqrt{(x-3)^2+y^2} ={1\over 2}|x-1| \Rightarrow (x-3)^2+y^2={1\over 4}(x-1)^2 \\ \Rightarrow 4x^2-24x+36+4y^2=x^2-2x+1 \Rightarrow \bbox[red,2pt]{ 3x^2-22x+4y^2+35=0}$$ (2) $$3x^2-22x+4y^2+35=0 \Rightarrow 3(x^2-{22\over 3}x+({11\over 3})^2)+4y^2+35 = {121\over 3} \\ \Rightarrow 3(x-{11\over 3})^2+4y^2= {16\over 3} \Rightarrow {(x-11/3)^2 \over (4/3)^2} + {y^2 \over (2/ \sqrt 3)^2}=1 \Rightarrow \cases{a=4/3  \\ b= 2/\sqrt 3} \Rightarrow c=2/3\\ \Rightarrow {c\over a}= {2/3\over 4/3} ={1\over 2}= {\overline{PF} \over d(P,L)}，故得證。\\另，{(x-1)^2 \over 25} +{(y+2)^2\over 16} =1 \Rightarrow \cases{a=5\\b=4\\ 中心點O(1,-2)} \Rightarrow c=3 \Rightarrow e={c\over a}={3\over 5}  \\ \Rightarrow 準線: x-1=\pm {a^2\over c}= \pm {25\over 3} \Rightarrow 準線:x=1\pm {25\over 3}={28\over 3},-{22\over 3} \Rightarrow \bbox[red, 2pt]{ \cases{x=28/3\\ x=-22/3}}$$

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