國立嘉科實驗高級中學11 4學年度高中部教師甄選
一、 填充題（每題 7 分，共 56 分）
解答:$${1\over bc}+{ 1\over ac}+{1\over ab} ={a+b+c \over abc} ={1\over 7} \Rightarrow 7(a+b+c) =abc \Rightarrow abc 是7的倍數 \Rightarrow a,b,c其中之一為7 \\ 假設a=7 \Rightarrow 7+b+c=bc \Rightarrow bc-b-c+1=8 \Rightarrow (b-1)(c-1)=8 \\ \Rightarrow \cases{\cases{b-1=1\\ c-1=8} \Rightarrow \cases{b=2\\ c=9不合} \\ \cases{b-1=2\\ c-1=4} \Rightarrow \cases{b=3\\ c=5}} \Rightarrow (a,b,c)=(7,3,5)及其排列 \Rightarrow a^2+b^2+c^2\\=49 +9+25= \bbox[red, 2pt]{83}$$
解答:
$$\int (x^2-kx+k) \,dx = {1\over 3}x^3-{1\over 2}kx^2+kx +C \Rightarrow 取F(x) ={1\over 3}x^3-{1\over 2}kx^2+kx \\ \Rightarrow \cases{R_1= F(a)-F(0) \\R_2= F(a)-F(b)} \Rightarrow R_1=R_2 \Rightarrow F(b) =F(0) \Rightarrow {1\over 3}b^3-{1\over 2}kb^2+ kb=0 \\ \Rightarrow 2b^2-3kb+6k=0 \cdots(1) ,又 f(b)=0 \Rightarrow b^2-kb+k=0 \cdots(2) \\ (1)-(2)\times 2 \Rightarrow -kb+4k=0 \Rightarrow k(b-4)=0 \Rightarrow b=4 \;(k=f(0) \ne 0) \\ x^2-kx+k =(x-a)(x-b) \Rightarrow a+b=k=ab \Rightarrow a+4=4a \Rightarrow a={4\over 3} \Rightarrow k={16\over 3} \\ \Rightarrow (k,a,b)= \bbox[red, 2pt]{ \left( {16\over 3},{4\over 3},4 \right)}$$
解答:$$\tan \alpha+\log_3(3\tan \alpha+6) =2 \Rightarrow \tan \alpha+1+\log_3(\tan \alpha+2)=2 \Rightarrow \log_3(\tan \alpha+2) =1-\tan \alpha \\ \Rightarrow \tan \alpha +2 =3^{1-\tan \alpha} \Rightarrow (1-\tan \alpha)+3^{1-\tan \alpha}=3 \cdots(1) \\ \tan \beta+3^{\tan \beta-1}=4 \Rightarrow (\tan \beta-1)+3^{\tan \beta-1} =3 \cdots(2) \\取f(x)=x+3^x 為嚴格遞增函數 \Rightarrow f(x)=3 有唯一解\Rightarrow f(1-\tan \alpha)=f(\tan \beta-1) =3\\ \Rightarrow 1-\tan \alpha=\tan \beta-1 \Rightarrow \tan \alpha+\tan \beta= \bbox[red, 2pt]2$$
解答:$$\cases{d_1=d(O \to A_1)=1 \\d_2= d(A_1\to A_2)=-{2\over 3} \\d_3= d(A_2\to A_3) ={4\over 9} =(-{2\over 3})^2\\d_4= d(A_3\to A_4)=-{8\over 27} =(-{2\over 3})^3 } \Rightarrow d_1+d_2+ \cdots+d_{101} = \sum_{k=0}^{100} \left( -{2\over 3} \right)^k ={1-(-2/3)^{101} \over 1-(-2/3)} \\={3\over 5} \left( 1+{2^{101}\over 3^{101}} \right) ={3\over 5} \left( 1+({3-1\over 3})^{101} \right) \Rightarrow (p,q,r)= \bbox[red, 2pt]{(5,3,101)}$$
解答:$$\left| {z^2+1\over z+i}\right|+\left| {z^2+4i-3\over z-i+2}\right| =\left| {(z+i)(z-i)\over z+i}\right|+\left| {(z+i-2)(z-i+2)\over z-i+2}\right| = |z-i|+|z+i-2|=4 \\ \Rightarrow \overline{PF_1} +\overline{PF_2}=4, 其中\cases{P(x,y) \\ F_1(0,1)\\ F_2(2,-1)} \Rightarrow 橢圓\cases{中心點O(1,0)\\ a=2} \Rightarrow \cases{c=1\\ b=\sqrt 2} \\ \Rightarrow {(x-1)^2\over 4}+{y^2\over 2}=1 \Rightarrow P(2\cos \theta+1, \sqrt 2\sin \theta) \Rightarrow |z-1|= \overline{PO} =\sqrt{4\cos^2\theta+2\sin^2\theta} \\=\sqrt{2+2\cos^2\theta} \ge \bbox[red, 2pt]{\sqrt 2}$$
解答:
$$可星勝一局代表向右走一步，予熹勝一局代表向上走一步，\\比賽停止時位置(x,y)需符合\cases{x=y+2且x+y\le 8\\ y=x+2 且x+y\le 8 \\x+y=8}\;, 也就是上圖藍色端點位置\\ 到達每一端點(x,y)的機率為p^xq^y,方法(路徑)數為圖上紅字, 局數為p+q, \\因此局數的期望值 =2(p^2+q^2) +4(4p^3q+ 4pq^3) +6(4p^4q^2+ 4p^2q^4)+8(8p^3q^5+ 16p^4q^4+ 8 p^5q^3) \\= 2 \left( {9\over 16}+{1\over 16} \right) +16 \left( {27\over 256}+{1\over 256} \right) +24 \left( {3^4\over 4^6}+{1\over 4^6} \right) +64 \left( {3^3\over 4^8}+2 \cdot {3^4\over 4^8} +{3^5\over 4^8} \right) = \bbox[red, 2pt]{803\over 256}$$
解答:
$$假設O為原點及\cases{\vec a= \overrightarrow{OA} \\\vec b= \overrightarrow{OB} \\\vec c= \overrightarrow{OC} } \Rightarrow \cases{\vec a-\vec c= \overrightarrow{OA}-\overrightarrow{OC} =\overrightarrow{OA}+ \overrightarrow{CO} = \overrightarrow{CA} \\ \vec c-\vec b= \overrightarrow{OC}-\overrightarrow{OB} = \overrightarrow{OC}+ \overrightarrow{BO} =\overrightarrow {BC}} \\ \Rightarrow \overrightarrow{CA}與\overrightarrow{BC} 的夾角為{\pi\over 3} \Rightarrow \overrightarrow{CA}與\overrightarrow{CB} 的夾角為{2\pi\over 3} \Rightarrow \angle ACB=120^\circ \\\Rightarrow \vec a\cdot \vec b =|\vec a||\vec b| \cos \angle OAB =1\cdot 1\cdot \cos \angle OAB ={1\over 2} \Rightarrow \angle OAB=60^\circ \Rightarrow \triangle OAB為一正三角形\\ \Rightarrow \cases{當C=P=\triangle ABC的重心時, \overline{OC}最短\\ 當C=Q =P對稱於\overline{AB}的對稱點時, \overline{OC}最長} \Rightarrow \cases{\overline{OP} =\displaystyle {\sqrt 3\over 2} \cdot {2\over 3} \\ \overline{OQ} =\displaystyle {\sqrt 3\over 2}\cdot {4\over 3}} \\\Rightarrow \overline{OP}+ \overline{OQ} ={\sqrt 3\over 2} \cdot \left( {2\over 3}+{4\over 3} \right ) ={\sqrt 3\over 2}\cdot 2= \bbox[red, 2pt]{\sqrt 3}$$
解答:$$\left( 1+{1\over 2} \right)\left( 1+{1\over 3} \right) \left( 1+{1\over 7} \right) \cdots \left( 1+{1\over 25} \right)-1 ={3\over 2}\cdot {4\over 3}\cdot {8\over 7} \cdot {10\over 9}\cdot {14\over 13} \cdot {18\over 17} \cdot {26\over 25}-1 \\={256\over 85}-1= \bbox[red, 2pt]{171\over 85}$$
二、 計算證明題（每題 10 分，共 20 分）
解答:
$$\cases{E,S 互為\overline{CD}的對稱點 \Rightarrow \overline{CE} =\overline{CS}\\ E,Q互為\overline{BC}的對稱點 \Rightarrow \overline{CE} =\overline{CQ}} \Rightarrow S,E,Q皆在以C為圓心, 半徑為\overline{CE}的圓上\\ 同理可得\cases{E,S,T 皆在以D為圓心, 半徑為\overline{DE}的圓上 \\E,P,T 皆在以A為圓心, 半徑為\overline{AE}的圓上 \\E,P,Q 皆在以B為圓心, 半徑為\overline{BE}的圓上 } \\令\cases{\angle BEQ= \alpha\\ \angle AET=\beta} \Rightarrow \cases{\angle QEC=90^\circ-\alpha\\ \angle DET=90^\circ-\beta} \\ \cases{在圓C上對\stackrel{\large \frown}{EQ} 的圓周角與弦切角相等\Rightarrow \angle ESQ=\alpha \\在圓D上對\stackrel{\large \frown}{ET} 的圓周角與弦切角相等\Rightarrow \angle EST=\beta \\在圓A上對\stackrel{\large \frown}{ET} 的圓周角與弦切角相等\Rightarrow \angle EPT=90^\circ-\beta \\在圓B上對\stackrel{\large \frown}{EQ} 的圓周角與弦切角相等\Rightarrow \angle EPQ=90^\circ-\alpha } \Rightarrow \angle S+\angle P=180^\circ \\ \Rightarrow PQST共圓 \quad \bbox[red, 2pt]{QED.}$$
解答:$$取u=x+y \Rightarrow x^2+xy+y^2=(x+y)^2-xy=6 \Rightarrow xy=u^2-6 \\ \Rightarrow x^2y+xy^2-x^2-2xy-y^2+ x+y =xy(x+y)-(x+y)^2+x+y=(u^2-6)u-u^2+u \\=u^3-u^2-5u \Rightarrow 取f(u)=u^3-u^2-5u \Rightarrow f'(u)=3u^2-2u-5 = (3u-5)(u+1)=0 \\ \Rightarrow \cases{u=5/3 \Rightarrow f(5/3)= -175/27\\ u=-1 \Rightarrow f(-1) =3} \\ 考慮邊界問題: x^2+xy+y^2=6 \Rightarrow \cases{(x+y)^2-xy=6 \Rightarrow (x+y)^2=6+xy\ge 0\\ (x-y)^2+3xy =6 \Rightarrow (x-y)^2=6-3xy \ge 0} \\ \Rightarrow -6\le xy \le 2 \Rightarrow -6\le u^2-6\le 2 \Rightarrow 0\le u^2\le 8 \Rightarrow -2\sqrt 2\le u\le 2\sqrt 2 \Rightarrow \cases{f(2\sqrt 2)= -8+6\sqrt 2\\ f(-2\sqrt 2) =-8-6\sqrt 2} \\ \Rightarrow \bbox[red, 2pt]{\cases{最大值3\\ 最小值-8-6\sqrt 2}}$$
三、申論題（共 24 分）
$$\bbox[cyan,2pt]{略}$$
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