臺北市立松山高級中學114學年度第1次正式教師甄選
一、 填充題（答案請寫在答案欄，每題4分，共20分）
解答:$$ S= \sum_{n=1}^\infty {2^{n-1} \over 1+4^{2^n}} = {1\over 2}\sum_{n=1}^\infty {2^{n} \over 1+4^{2^n}} ={1\over 2 } \left( \sum_{n=0}^\infty {2^{n} \over 1+4^{2^n}} -{2^0\over 1+4^{2^0}}\right) ={1\over 2 } \left( \sum_{n=0}^\infty {2^{n} \over 1+4^{2^n}} -{1\over 5}\right)\\={1\over 2} \left( {1\over 3}-{1\over 5} \right) = \bbox[red, 2pt]{1\over 15}\\ 補充證明: f(x)={1\over x-1} =\sum_{k=0}^\infty {2^k\over x^{2^k}+1} \quad \text{for }|x|\gt 1 \\ 由於{1\over y-1}-{1\over y+1}={2\over y^2-1} \Rightarrow {1\over y+1}={1\over y-1}-{2\over y^2-1} \\ 因此{2^k\over x^{2^k}+1} =2^k \left( {1\over x^{2^k}-1}-{2\over x^{2^{k+1}}-1} \right) ={2^k\over x^{2^k}-1}-{2^{k+1}\over x^{2^{k+1}}-1} \\ \Rightarrow S_N= \sum_{k=0}^N{2^k\over x^{2^k}+1}= \sum_{k=0}^N \left( {2^k\over x^{2^k}-1}-{2^{k+1}\over x^{2^{k+1}}-1} \right) \\= \left( {1\over x-1}-{2\over x^2-1} \right) + \left( {2\over x^2-1}-{4\over x^4-1} \right) + \cdots+ \left( {2^N\over x^{2^N}-1}-{2^{N+1}\over x^{2^{N+1}}-1} \right) \\={1\over x-1}-{2^{N+1}\over x^{2^{N+1}}-1} \Rightarrow f(x)={1\over x-1} =\sum_{k=0}^\infty {2^k\over x^{2^k}+1} \quad \text{for }|x|\gt 1\\ 因此\sum_{n=0}^\infty {2^{n} \over 1+4^{2^n}}= f(4)={1\over 3}$$
解答:
$$(x-4)^2+y^2=r^2 \Rightarrow \cases{圓心P(4,0)\\ 圓半徑r} \\ 相交四相異點\Rightarrow 圓半徑尺寸介於拋物線內切圓與外切圓之間 \\ y^2=x 代入圓 \Rightarrow (x-4)^2+x=r^2 \Rightarrow x^2-7x+16-r^2=0有實根\\ \Rightarrow 49-4(16-r^2)\gt 0\Rightarrow r^2\gt {15\over 4} \Rightarrow r\gt {\sqrt{15}\over 2} \\ 又外切圓半徑r=4 \Rightarrow \bbox[red, 2pt]{{\sqrt{15}\over 2} r\lt 4}$$
解答:
$$d=\sqrt{2a^2-6a+5} +\sqrt{b^2-4b+5}+ \sqrt{2a^2-2ab+b^2} \\=\sqrt{(a-2)^2+ (a-1)^2} +\sqrt{(b-2)^2 +(0-1)^2} +\sqrt{(a-b)^2 +(a-0)^2} \\= \overline{AP}+ \overline{BP}+ \overline{AB},其中\cases{A(a,a)\in L:y=x\\ B(b,0) \in x軸\\P(2,1)} \\ 作\cases{P'與P對稱於直線L\\ P''與P對稱於x軸} \Rightarrow \cases{P'(1,2) \\ P''(2,-1)} \Rightarrow d的最小值=\overline{P'P''} =\bbox[red, 2pt]{\sqrt{10}},此時\cases{A=L\cap \overline{P'P''} \\B=x軸\cap \overline{P'P''}}$$
解答:
$$10個點任取4個共有C^{10}_4=210種取法, 四點共面的情況有:\\\textbf{Case I }四面體每個面有6個點, 任取4個皆共面, 有4\times C^6_4=60種\\ \textbf{Case II }將四面體均分的平面, \{P,Q,T,U\}, \{Q,R,S,U\}, \{P,R,S,T\},共3種 \\ \textbf{Case III }稜邊與對邊中點構成的平面,如:\{A,P,B, T\}, 6條稜邊有6種\\ 因此不共面有210-60-3-6=141種,機率為{141\over 210} = \bbox[red, 2pt]{47\over 70}$$
解答:
$$令\cases{A(1,0,0) \\B(0,1,1)\\ C(0,0,1)} \Rightarrow 平面E=\triangle ABC:x+z=1\\ P\in \triangle ABC \Rightarrow P= C+ u\overrightarrow{CA}+ v\overrightarrow{CB} =(0,0,1)+u(1,0,-1)+v(0,1,0) =(u,v,1-u), \\其中u,v \ge 0, 且u+v\le 1 \\z=1-u \Rightarrow P(1-z,y,z), 0\le y\le z \Rightarrow 旋轉半徑r=d(P,z軸)=\sqrt{(1-z)^2+y^2}\\ 繞z軸旋轉圖形為空心的, 內部是圓角錐,假設內半徑r \Rightarrow r^2= (1-z)^2\\ 外半徑R \Rightarrow R^2=(1-z)^2+z^2=2z^2-2z+1 \\ \Rightarrow 旋轉體積V=\int_0^1 \pi(R^2-r^2)\,dz =\int_0^1 \pi z^2\,dz= \bbox[red, 2pt]{\pi\over 3}$$
二、 計算題（需詳列計算過程，每題 8 分，共 80 分）
解答:
$$假設大扇形半徑為2a, 由於B為\overline{AC}中點, 小扇形半徑為a, 即\cases{\overline{BA} = \overline{BD} =\overline{BC}=a\\ \overline{CE}=2a} \\ \triangle BAD: {\overline{AD} \over \sin 120^\circ}=2a \Rightarrow \overline{AD}= \sqrt 3 a, 又\cases{\angle ABD= \angle ACE\\ \overline{AB}:\overline{AC}= \overline{BD}: \overline{CE}} \\\Rightarrow A,D,E在一直線上, 且\overline{DE}= \overline{AD} = \sqrt 3a\\\cases{B為小扇形圓心\\ \angle ABD=120^\circ } \Rightarrow \angle APD=120^\circ \Rightarrow \cos \angle APD=-{1\over 2}={3^2+ \overline{PD}^2 -3a^2 \over 2\cdot 3\cdot \overline{PD}} \cdots(1) \\ \overline{PD}為\triangle PAE的中線 \Rightarrow 3^2+7^2=2(\overline{PD}^2+3a^2) \cdots(2)\\ 由(1)及(2) \Rightarrow 3a^2= \overline{PD}^2+3\overline{PD}+9=29-\overline{PD}^2 \Rightarrow 2\overline{PD}^2+ 3\overline{PD} -20=0 \\ \Rightarrow (\overline{PD} +4)(2 \overline{PD} -5)=0 \Rightarrow \overline{PD}={5\over 2} \Rightarrow a^2={91\over 12} \Rightarrow 小扇形面積={1\over 3}a^2\pi= \bbox[red, 2pt]{91\pi\over 36}$$
解答:$$\cases{xy=z-x-y\\ xz=y-x-z\\ yz=x-y-z} \Rightarrow \cases{xy+x+y+1=z+1\\ xz+x+z+1=y+1\\ yz+y+z+1=x+1} \Rightarrow \cases{(x+1) (y+1)=z+1\\ (x+1)(z+1)= y+1\\ (y+1)(z+1)= x+1} \\ 取\cases{a=x+1\\ b=y+1\\ c=z+1} \Rightarrow \cases{ab= c\\ ac=b\\ bc=a} \Rightarrow 三式相乘:(abc)^2=abc \\ \textbf{Case I }abc=0 \Rightarrow a=b=c=0 \\ \textbf{Case II }abc=1 \Rightarrow c\cdot c=1 \Rightarrow c=\pm 1 \Rightarrow \cases{(a,b,c)=(1,1,1) \\(a,b,c)= (1,-1,-1) \\(a,b,c)= (-1,1,-1)\\(a,b,c)=(-1,-1,1)} \\ 由\textbf{Case I &II }可得(x,y,z)=\bbox[red, 2pt]{(-1,-1,-1), (0,0,0), (0,-2,-2), (-2,0,-2), (-2,-2,0)}$$
解答:$$\textbf{(1) }取x=y=0 \Rightarrow f(0)f(0)=f(0)+0 \Rightarrow f(0)(f(0)-1)=0 \Rightarrow f(0)=0或1\\ \quad 若f(0)=0 \cases{取x=0,y=1 \Rightarrow f(0)f(1)=f(0+1)+ 0\cdot 1 \Rightarrow f(1)=0 \\ 取x=1,y=-1 \Rightarrow f(1)f(-1)= f(0)-1 \Rightarrow f(1)f(-1)=-1\ne 0} \Rightarrow 矛盾 \\\quad \Rightarrow \bbox[red, 2pt]{f(0)=1} \\\textbf{(2) } f(x)f(-x)=f(0)-x^2 =1-x^2 \Rightarrow f(1)f(-1)=1-1=0 \\ \Rightarrow \cases{f(1)=0 \Rightarrow f(x)f(1)=f(x+1)+x \Rightarrow f(x+1)=-x \Rightarrow f(x)=1-x\\ f(-1)=0 \Rightarrow f(x)f(-1)=f(x-1)-x \Rightarrow f(x-1)=x \Rightarrow f(x)=1+x} \\ \Rightarrow \bbox[red, 2pt]{f(x)=1-x 或1+x}$$
解答:$$\lim_{n\to \infty} \left( 2b_n+{6n-30\over n} \right) \lt \lim_{n\to \infty} a_n \Rightarrow 2\lim_{n \to \infty} b_n +6\lt 12 \Rightarrow \lim_{n \to \infty} b_n\lt {12-6\over 2}=3 \cdots(1)\\ \lim_{n\to \infty} a_n\lt \lim_{n\to \infty} 4b_n \Rightarrow 12\lt 4\lim_{n\to \infty} b_n \Rightarrow 3\lt\lim_{n\to \infty} b_n \cdots(2) \\ 由(1)及(2) \Rightarrow 3\lt \lim_{n\to \infty} b_n\lt 3 \Rightarrow 由夾擠定理可得\bbox[red, 2pt]{\langle b_n \rangle 收斂且\lim_{n\to \infty} b_n=3}$$
解答:$$\cases{A(0,0)\\ B(2,0) \\C(3\cos 60^\circ, 3\sin 60^\circ)=(3/2,3\sqrt 3/2)} \Rightarrow D(\cos \theta+{3\over 2}, \sin \theta+{3\sqrt 3\over 2}) \\ \Rightarrow \left| \overrightarrow{AB} +\overrightarrow{AC} +\overrightarrow{CD}\right| =\left| (2,0)+ ({3\over 2}, {3\sqrt 3\over 2})+(\cos \theta, \sin \theta)\right| =\sqrt{ \left( {7\over 2}+ \cos \theta \right)^2+ \left( {3\sqrt 3\over 2}+ \sin \theta \right)^2} \\=\sqrt{7\cos \theta+3\sqrt 3\sin \theta+20} =\sqrt{\sqrt{76} \sin(\theta+ \alpha)+20} \Rightarrow 最小值=\sqrt{20-\sqrt{76}} =\sqrt{20-2\sqrt{19}} \\= \bbox[red, 2pt]{\sqrt{19} -1}$$
解答:$$\overline{AC}為直徑\Rightarrow \angle ADC=90^\circ \Rightarrow \overline{AD}^2+\overline{DC}^2= \overline{AC}^2=27 \\ \Rightarrow 周長S=\overline{AD}+ \overline{DC}+ \overline{AB}+ \overline{BC} =\overline{AD}+ \overline{DC}+ 9\\ 當\overline{AD} =\overline{DC}時, S有最大值, 此時2\overline{AD}^2=27 \Rightarrow \overline{AD}={3\sqrt 6\over 2} \Rightarrow S= \bbox[red, 2pt]{3\sqrt 6+9}$$
解答:
$$\textbf{(1) }假設E, F分別為\Gamma在\overline{AC}, \overline{AB}的切點, 並令\cases{\overline{AE}= \overline{AF}=a\\ \overline{BD} =\overline{BF}=b\\ \overline{CD}=\overline{CE}=c} \Rightarrow 2(a+b+c) =5+6 +7=18\\ \quad \Rightarrow a+b+c=9\Rightarrow \cases{a=9- \overline{BC}=2 \\b=9-\overline{AC}=3\\ c=9-\overline{AB}= 4} \Rightarrow b:c=3:4 \Rightarrow \overrightarrow{AD}={{4\over 7}} \overrightarrow{AB}+{3\over 7} \overrightarrow{AC} \\\quad \Rightarrow (s,t)= \bbox[red, 2pt]{\left( {4\over 7},{3\over 7} \right)} \\\textbf{(2) } \overrightarrow{AD}={{4\over 7}} \overrightarrow{AB}+ {3\over 7} \overrightarrow{AC} \Rightarrow \overline{AD}^2= \left( {{4\over 7}} \overrightarrow{AB}+ {3\over 7} \overrightarrow{AC} \right) \cdot \left( {{4\over 7}} \overrightarrow{AB}+{3\over 7} \overrightarrow{AC} \right) \\ \quad ={16\over 49}\cdot 5^2+{9\over 49}\cdot 6^2 +{24\over 49}\overrightarrow{AB}\cdot \overrightarrow{AC} ={724\over 49}+{24\over 49} \overline{AB}\cdot \overline{AC} \cos A \\\quad ={724\over 49}+{720\over 49}\cdot {5^2+6^2-7^2 \over 2\cdot 5\cdot 6} = {724\over 49}+{144\over 49} ={868\over 49} \Rightarrow \overline{AD}= {2\sqrt{217} \over 7}\\ \quad 依圓冪性質:\overline{AF}^2 =\overline{AD}\times \overline{AP} \Rightarrow 4= {2\sqrt{217} \over 7}\times \overline{AP} \Rightarrow \overline{AP}= {2\sqrt{217}\over 31} \\ \quad \Rightarrow {\overline{AP} \over \overline{AD}} ={7\over 31} \Rightarrow \overrightarrow{AP}={7\over 31} \overrightarrow{AD} ={7\over 31} \left( {{4\over 7}} \overrightarrow{AB} +{3\over 7} \overrightarrow{AC} \right) ={4\over 31} \overrightarrow{AB}+ {3\over 31}\overrightarrow{AC} \\ \quad \Rightarrow (p,q)= \bbox[red, 2pt]{\left( {4\over 31},{3\over 31} \right)}$$
解答:$$\overline{FG} \bot \overline{FA} \Rightarrow \angle AFG=90^\circ \Rightarrow \overline{AG}^2=\overline{AF}^2+ \overline{FG}^2=26^2+32^2=1700\\ \cases{A在平面CDEF的投影點H\\ G在平面CDEF} \Rightarrow \overline{AH}\bot \overline{GH} \Rightarrow \angle AHG=90^\circ \Rightarrow \overline{AG}^2= \overline{AH}^2+ \overline{GH}^2 \\ \Rightarrow \overline{GH}^2 =1700-(\overline{AD}\sin 30^\circ)^2= 1700-100 =1600 \Rightarrow |\overrightarrow{GH}|= \overline{GH}= \sqrt{1600} =\bbox[red, 2pt]{40}$$
解答:$$L_1:{x-7\over 4} ={y-2\over -1}= {z-4\over 2} \Rightarrow L_1方向向量\vec u=(4,-1,2) \Rightarrow P(7,2,4)\in L_1 \\ L_2:\cases{2x+y-1=0\\ 3y-2z+11=0} \Rightarrow L_2方向向量\vec v=(2,1,0) \times (0,3,-2) =(-2,4,6) \Rightarrow Q(0,1,7)\in L_2 \\ \Rightarrow \vec n= \vec u\times \vec v=(-14,-28,14) \Rightarrow d= \left|{ \overrightarrow{PQ} \cdot \vec n \over |\vec n|} \right|=2 \sqrt 6 \Rightarrow 稜長= \sqrt 2 d=4\sqrt 3 \\ \Rightarrow 體積V={\sqrt 2\over 12}\cdot \left( 4\sqrt 3 \right)^3= \bbox[red, 2pt]{16\sqrt 6}, L_1與L_2關係:\bbox[red, 2pt]{歪斜}$$
解答:$$\bbox[cyan,2pt]{略}$$
解題僅供參考，其他教甄試題及詳解
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