國立臺南女中 112 學年度第一次教師甄選數學科試題
一、填充題(每題 4 分，共 76 分)
解答：
$$\cos \angle AOB ={\overrightarrow{OA}\cdot \overrightarrow{OB} \over |\overrightarrow{OA}| |\overrightarrow{OB}|} ={1/2\over 1\cdot 1}={1\over 2} \Rightarrow \angle AOB=60^\circ \Rightarrow \triangle OAB為正\triangle\\ \overline{AB}中點P=(A+B)/2 = ({3\over 4},{1\over 4},{\sqrt 2\over 4}) \Rightarrow \overline{OP} = \overline{OB}\sin 60^\circ = 3\sqrt 3 \\ 假設C(a,b,c)\Rightarrow P= {3\sqrt 3\over 6}C \Rightarrow ({3\over 4},{1\over 4},{\sqrt 2\over 4})=({\sqrt 3\over 2}a,{\sqrt 3\over 2}b, {\sqrt 3\over 2}c)\\ \Rightarrow \cases{a=\sqrt 3/2\\ b=\sqrt 3/6\\ c=\sqrt 6/6} \Rightarrow C=\bbox[red,2pt]{\left({\sqrt 3\over 2},{\sqrt 3\over 6},{\sqrt 6\over 6} \right)}$$
解答：$$\cases{f(x)為二次式\\ y=f(x)通過原點} \Rightarrow f(x)=ax(x-b) \\ t=2 \Rightarrow f(4+t)=f(2-t) \Rightarrow f(6)=f(0) \Rightarrow 6a(6-b)=0 \Rightarrow \cases{b=6\\ a=0不合,違反f是二次式} \\ \Rightarrow f(x)=ax(x-6) =a(x-3)^2-9a\\ f(x-108)的最大值為12 \Rightarrow f(x)的最大值還是12 \Rightarrow -9a=12 \Rightarrow a=-{4\over 3} \\ \Rightarrow f(x)=-{4\over 3}x(x-6) \Rightarrow f(9)=-12\cdot 3 =\bbox[red, 2pt]{-36}$$
解答：$$\cases{\angle B=23^\circ \\ \angle C=37^\circ} \Rightarrow \angle A=180^\circ-\angle B-\angle C=120^\circ\\ 正弦定理:{\overline{BC} \over \sin A}=2R \Rightarrow {5\over \sin 120^\circ}={10\over \sqrt 3} =2R \Rightarrow R={5\over \sqrt 3}\\ 又\angle A=120^\circ \Rightarrow \angle BOC=240^\circ \Rightarrow \cos \angle BOC ={\overrightarrow{OB}\cdot \overrightarrow{OC} \over |\overrightarrow{OB}|\cdot |\overrightarrow{OC}|} = {\overrightarrow{OB}\cdot \overrightarrow{OC} \over R^2}\\ \Rightarrow -{1\over 2}={\overrightarrow{OB}\cdot \overrightarrow{OC} \over 25/3} \Rightarrow \overrightarrow{OB}\cdot \overrightarrow{OC}= \bbox[red,2pt]{-{25\over 6}}$$
解答：$$\omega={-1+\sqrt 3i\over 2} = \cos{2\pi\over 3}+i\sin{2\pi\over 3} \Rightarrow \omega^3=1 \Rightarrow 1+\omega+ \omega^2=0 \\\omega^2 \sum_{k=1}^{96}(-k)^{k+1}k \omega^{k-1} =\sum_{k=1}^{96}(-\omega)^{k+1}k = \\ \begin{array}{} \omega^2-2\omega^3+3\omega^4\\-4\omega^5+5 \omega^6-6\omega^7 \\+7\omega^8-8\omega^9+ 9\omega^{10} \\+ \cdots\\ -94\omega^{95}+ 95\omega^{96}-97\omega^{97}\end{array} =\color{blue}{ \begin{array}{} \omega^2-2 +3\omega\\-4\omega^2+5 -6\omega \\+7\omega^2-8+ 9\omega \\+ \cdots\\ -94\omega^{2}+ 95 -97\omega \end{array}} \\=(1-4+7-\cdots -94)\omega^2 +(-2+5-8+\cdots +95) +(3-6+9-\cdots-97)\omega \\=(-3-3-\cdots -3)\omega^2+(3+3+ \cdots +3)+(-3-3-\cdots -3)\omega \;(兩個一組合併計算)\\=(-3\times 16)\omega^2 +3\times 16+(-3\times 16)\omega =-48\omega^2+48-48\omega\\= -48(\omega^2+\omega)+48 = -48\cdot (-1)+48=\bbox[red, 2pt]{96}$$
解答：$$f(x)=\sqrt{10x-x^2}-\sqrt{16x-x^2-60} =\sqrt{5^2-(5-x)^2}-\sqrt{2^2-(8-x)^2}\\ \Rightarrow \cases{|5-x|\le 5\\ |8-x|\le 2} \Rightarrow \cases{0\le x\le 10\\ 6\le x\le 10}\Rightarrow 定義域:x\in [6,10] \\ 又\cases{5^2-(5-x)^2 遞減\\ x=6時,2^2-(8-x)=0為最小} \Rightarrow f(6)=\sqrt{24}-0= \bbox[red,2pt]{2\sqrt 6}為最大值$$
解答：$${假設|z|=a \in \mathbb N \Rightarrow z=ae^{i\theta} \Rightarrow {z-1\over z^2}={ae^{i\theta}-1\over a^2z^{2i\theta}} ={1\over a}e^{-i\theta}-{1\over a^2}e^{-2i\theta}} \in \mathbb R\\ \Rightarrow {1\over a}\sin(-\theta)={1\over a^2}\sin(-2\theta) \Rightarrow a\sin \theta =\sin (2\theta) =2\sin\theta \cos\theta \Rightarrow \sin\theta(a-2\cos\theta)=0\\ \Rightarrow \cases{\sin \theta =0,不合,因為z有虛部\\ \cos\theta =a/2} \Rightarrow \cos \theta= \cases{1/2\\ 2/2=1,不合,\theta=2k\pi,沒有虛部} \\ \Rightarrow \theta ={\pi\over 3},{5\pi\over 3} \Rightarrow z= \bbox[red,2pt]{{1\over 2}\pm {\sqrt 3\over 2}i}$$
解答：$$假設\overline{AC}=2R為直徑,因此取\cases{A(0,0)\\ C(2R,0)\\ B(R\cos\theta ,R\sin \theta)\\ D=(R\cos \phi, R\sin \phi)}\\ \overline{BD}=2\sqrt 5 \Rightarrow R^2(\cos \theta-\cos\phi)^2 +R^2( \sin \theta-\sin \phi)^2 = 20 \\ \Rightarrow R^2(2-2\cos\theta \cos \phi -2\sin\theta \sin \phi) =2R^2(1-\cos(\theta-\phi))=20 \cdots(1)\\ 再由\overrightarrow{AC}={3\over 2}\overrightarrow{AB} +{5\over 2}\overrightarrow{AD} \Rightarrow (2R,0)={3\over 2}(R\cos \theta,R\sin \theta) +{5\over 2} (R\cos \phi, R\sin \phi)\\ ({3\over 2}R\cos \theta +{5\over 2}R\cos \phi, {3\over 2}R\sin \theta+{5\over 2}R\sin \phi) \Rightarrow \cases{3\cos\theta +5\cos \phi=4\\ 3\sin \theta+ 5\sin \phi=0}\\ 兩式平方再相加\Rightarrow 34 +30(\cos\theta \cos \phi+ \sin \theta \sin \phi)=16 \\ \Rightarrow 30\cos(\theta-\phi)= -18 \Rightarrow \cos(\theta-\phi)=-{3\over 5} 代入(1) \Rightarrow {16\over 5}R^2=20 \\ \Rightarrow R^2={100\over 16} \Rightarrow R={10\over 4} \Rightarrow \overline{AC}=2R= \bbox[red, 2pt]5$$
解答：$$在n個球選r個不相鄰的方法數C^{n-r+1}_r\\9天都不吃吐司:每天有2種選擇,共2^9種\\9天中只有1天吃吐司:9天選一天吃吐司有C^9_1種選法,其它8天有2種選擇,共C^9_1\cdot 2^8種\\9天中有不相鄰2天吃吐司:9天選2天吃吐司有C^{9-2+1}_2= C^8_2種選法,其它7天有2種選擇,共C^8_2\cdot 2^7種\\ 9天中有不相鄰3天吃吐司:9天選2天吃吐司有C^{9-3+1}_3= C^7_3種選法,其它6天有2種選擇,共C^7_3\cdot 2^6種 \\9天中有不相鄰4天吃吐司:C^6_4\cdot 2^5種\\ 9天中有不相鄰5天吃吐司:C^5_5\cdot 2^4種\\總共有2^9+ C^9_12^8+ C^8_22^7+ C^7_32^6+ C^6_42^5 + C^5_52^4 =512+ 2304+3584+ 2240+480+ 16 \\= \bbox[red, 2pt]{9136}$$
解答：$$n^2+3n+43=k^2, k,n\in \mathbb N \Rightarrow 4n^2+12n+172=4k^2 \Rightarrow (2n+3)^2+ 163=(2k)^2\\ \Rightarrow (2k)^2-(2n+3)^2=163 \Rightarrow (2k+2n+3)(2k-2n-3)=163 \times 1\\ \Rightarrow \cases{2k+2n+3=163\\ 2k-2n-3=1},兩式相加\Rightarrow 4k=164 \Rightarrow k=41 \\ \Rightarrow n^2+3n+43=41^2 = \bbox[red,2pt]{1681}$$
解答：$$假設切線為y=ax+b \Rightarrow x^4-3x^2+2x+3=ax+b \\\Rightarrow x^4-3x^2+(2-a)x+3-b=0 恰有兩相異實根\alpha,\beta\\ 因此我們假設x^4-3x^2+(2-a)x+3-b=(x-\alpha)^2(x- \beta)^2 \\ \Rightarrow \cases{2\alpha+2\beta= 0 \\ \alpha^2+\beta^2+ 4\alpha\beta=-3\\ -2\alpha\beta( \alpha+\beta) =a-2\\ \alpha^2\beta^2 =3-b} \Rightarrow \cases{\alpha+\beta =0\\ \alpha\beta=-3/2} \Rightarrow \cases{a=2\\ b=3/4} \Rightarrow L: \bbox[red,2pt]{y=2x+{3\over 4} }$$
解答：$$由題意可知轉換矩陣T=\begin{bmatrix}5/8 & 3/8\\ 3/8& 5/8 \end{bmatrix},此題相當於求T^\infty\begin{bmatrix}1\\ 0 \end{bmatrix}= \begin{bmatrix}p_1\\ p_2 \end{bmatrix}中的p_1\\ 先將T對角化,即T=PDP^{-1} =\begin{bmatrix} -1 & 1\\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1/4 & 0\\ 0 & 1 \end{bmatrix} \begin{bmatrix} -1/2 & 1/2\\ 1/2 & 1/2 \end{bmatrix} \\\Rightarrow T^\infty=\begin{bmatrix} -1 & 1\\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1/4 & 0\\ 0 & 1 \end{bmatrix}^\infty \begin{bmatrix} -1/2 & 1/2\\ 1/2 & 1/2 \end{bmatrix} =\begin{bmatrix} -1 & 1\\ 1 & 1 \end{bmatrix} \begin{bmatrix} \color{blue}0 & 0\\ 0 & 1 \end{bmatrix} \begin{bmatrix} -1/2 & 1/2\\ 1/2 & 1/2 \end{bmatrix} \\= \begin{bmatrix} 1/2 & 1/2\\ 1/2 & 1/2 \end{bmatrix} \Rightarrow T^\infty\begin{bmatrix}1\\ 0 \end{bmatrix}= \begin{bmatrix}1/2\\ 1/2 \end{bmatrix} \Rightarrow p_1=\bbox[red, 2pt]{1\over 2}$$
解答：$$令\cases{P(0,0,0)\\ A(1,0,1)\\ B(1,1,0)\\ C(0,1,1)} \Rightarrow O=(A+B+C)/3=(2/3,2/3,2/3) \Rightarrow P=O/2=(1/3,1/3,1/3) \\ \Rightarrow \cases{\vec u=\overrightarrow{AP}= (-2/3,1/3,-2/3) \\ \vec v=\overrightarrow{BC}= (-1,0,1)} \Rightarrow \vec n= \vec u\times \vec v={1\over 3}(1,4,1) \\ \Rightarrow 過A法向量為\vec n的平面E:(x-1)+4y+(z-1)=0 \Rightarrow x+4y+z=2\\ \Rightarrow \cases{Q=E\cap \overline{PB}=(2/5,2/5,0)\\ R=E\cap \overline{PC}=(0,2/5,2/5)} \Rightarrow 上部份體積頂點為P,A,Q,R \\ \Rightarrow {1\over 6}\begin{Vmatrix} 0& 0 & 0 &1 \\1 &0 &1 &1 \\2/5 & 2/5 & 0 & 1 \\0 &2/5 &2/5 &1 \end{Vmatrix} ={1\over 6}\cdot {8\over 25}={4\over 75}=m\\ 原四面體體積={1\over 3} \Rightarrow 下部份體積={1\over 3}-{4\over 75}={21\over 75}=n\\ \Rightarrow {m\over n}=\bbox[red, 2pt]{4\over 21}$$
解答：$$第二張牌比第一張大的情形:(1,2-6),(2,3-6),(3,4-6),(4,5-6),(5,6),\\\quad 共5+4+\cdots+1=15種情形,機率=15/P^6_2=15/30={1\over 2}\\第三張牌比第1張大,也比第2張大的情形(不管第一,二張的大小):\\ \qquad (1,2,3-6): 1,2有2種排列,3-6有4種情形,此情形有2\times 4=8種;\\ \qquad 同理,(1,3,4-6),(1,4,5-6)...(4,5,6),機率=40/P^6_3 ={1\over 3}\\ 第四張比第1,2,3張都大的機率=90/P^6_4={1\over 4}\\ 第五張比前4張都大的機率={1\over 5},第六張比前5張都大的機率={1\over 6}\\ 因此期望值={1\over 2}+ {1\over 3}+ {1\over 4}+ {1\over 5}+ {1\over 6} ={174\over 120} =\bbox[red,2pt]{29\over 20}$$
解答：$$$$
解答：$$\angle C=90^\circ \Rightarrow \angle A=90^\circ \Rightarrow 直徑\overline{BD} = \sqrt{12^2+16^2}=20 \Rightarrow 半徑r=10\\ O為\overline{BD}中心點，也是圓心；圓周角\angle D=75^\circ \Rightarrow 圓心角\angle AOC=2\times 75^\circ =150^\circ\\ \Rightarrow \cos \angle AOC={r^2+r^2-\overline{AC}^2 \over 2r^2} \Rightarrow -{\sqrt 3\over 2} ={200-\overline{AC}^2\over 200} \Rightarrow \overline{AC}^2 =100(2+\sqrt 3) \\ \Rightarrow \overline{AC}= 10\sqrt{2+\sqrt 3} =5\sqrt{8+\sqrt{12}} =\bbox[red,2pt]{5(\sqrt 6+\sqrt 2)}$$
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解答：$$9\cos^2 x-3\sin x-7=0 \Rightarrow 9-9\sin^2x-3\sin x-7=0 \Rightarrow 9\sin^2 x+3\sin x-2=0\\ \Rightarrow (3\sin x-1)(3\sin x+2)=0 \Rightarrow \cases{\sin x={1\over 3}\\ \sin x=-{2\over 3} } \\\Rightarrow \cases{x= \theta_1,\pi-\theta_1,\theta_1+2\pi ,3\pi-\theta_1 \\ x=\theta_2,3\pi-\theta_2 , 2\pi+\theta_2 ,5\pi-\theta_2} \Rightarrow 所有根的和=6\pi+10\pi= \bbox[red,2pt]{16\pi}$$
解答：$$S_n=1+2+\cdots +n={n(n+1)\over 2} \Rightarrow S_n-1={n^2+n-2\over 2} ={(n+2)(n-1)\over 2} \\ \Rightarrow {S_n\over S_n-1}={n(n+1)\over (n-1)(n+2)} \Rightarrow T_n ={2\cdot 3\over 1\cdot 4} \times {3\cdot 4\over 2\cdot 5} \cdot {4\cdot 5\over 3\cdot 6}\cdots {n(n+1) \over (n-1)(n+2)} \\= {3\over 1}\cdot {n\over n+2} \Rightarrow T_{1998} ={3\cdot 1998\over 2000} =\bbox[red, 2pt]{2997\over 1000}$$
解答：$$假設5個正整數依序從小至大為x_1,x_2,x_3,x_4,x_5;由\cases{全距18\Rightarrow n_5=n_1+18\\ 中位數＝眾數=9 \Rightarrow x_3=9}\\ 因此5個正整數依序從小至大為x_1,x_2,9,x_4,x_1+18,\\再由平均數=13 \Rightarrow {2x_1+x_2+x_4+27\over 5}=13 \Rightarrow 2x_1+x_2+x_4=38\\ \begin{array}{} x_1 & x_2 & x_3 & x_4 & x_5 &\\\hline 1 & 9 & 9 & 19& 19 & \times:x_2+x_4\lt 36\\ 2 & 9 & 9 & 20 & 20& \times: x_2+x_4 \lt 34 \\ 3 & 9 & 9 & 21& 21& \times: x_2+x_4\lt 32 \\ 4& 9 & 9 & 21 & 22 & \bigcirc\\ \hdashline 5 &9 &9 &19 &23 & \bigcirc\\ 5 &8 &9 &20 &23 & \times:眾數不是9\\ 5 & 7 & 9 & 21 & 23 & \times:眾數不是9\\ 5 & 6 & 9 & 22 & 23& \times:眾數不是9\\ \hdashline 6 &9 & 9 &17 &24 & \bigcirc \\ 6 & 8& 9 &18&24 & \times:眾數不是9 \\ 6 & 7 & 9 & 19& 24 & \times:眾數不是9\\\hdashline 7 &9 & 9 &15 &25 & \bigcirc\\ 7 & 8 & 9 & 16 & 25 & \times:眾數不是9\\ \hdashline 8 & 9& 9 &13 & 26 & \bigcirc \\\hdashline 9 & 9 & 9 & 11&27 &\bigcirc\end{array}\\ 第二大的數字x_4=21,19,17,15,13,11,共有\bbox[red,2pt]6個$$
解答：$$假設L:y=ax+b \Rightarrow \cases{d(L,A(7,3))=6\\ d(L,B(6,6))=3} \Rightarrow \cases{|7a+b-3|=6\sqrt{a^2+1} \\ |6a+b-6| =3\sqrt{a^2+1}} \\ 先試7a+b-3=2(6a+b-6) \Rightarrow b=9-5a \Rightarrow |7a+9-5a-3|= |2a+6|=6\sqrt{a^2+1} \\ \Rightarrow (2a+6)^2 = 36(a^2+1) \Rightarrow 8a(4a-3)=0 \Rightarrow \cases{a=0 \Rightarrow b=9\\ a=3/4 \Rightarrow b=21/4} \\ \Rightarrow \cases{L:y=9 \Rightarrow d(L,C)=9\\ L:4y=3x+21 \Rightarrow d(L,C)=3} \Rightarrow d(L,C)=\bbox[ red, 2pt]{3或9}$$
解答：$$假設f(x)=ax^3+bx^2+cx+d \Rightarrow f'(x)=3ax^2 +2bx+c \Rightarrow f'(x)=6ax+2b \\ \Rightarrow f''(x)-3xf'(x)+9f(x) =3bx^2+(6a+c)x +(2b+9d)=0 \Rightarrow \cases{b=0\\ a=-c\\ d=0} \\ \Rightarrow f(x)=ax^3-ax,再將f(2)=6代入\Rightarrow 8a-2a=6 \Rightarrow a=1 \Rightarrow f(x)=x^3-x \\ 因此 g'(x)=f(x)=x^3-x=0 \Rightarrow \cases{x=0\\x=1} \Rightarrow \cases{g(0)= \int_0^0f(t)\,dt=0\\ g(1) = \int_0^1t^3-t\,dt = {1\over 4} -{1\over 2}=\bbox[red, 2pt]{-{1\over 4}}}$$
二、計算證明題(每題 6 分，共 24 分)
解答：$$假設有理根為{q\over p}(p,q互質) \Rightarrow a\cdot ({q\over p})^2+ b\cdot {q\over p}+c=0 \Rightarrow aq^2+ bpq+cp^2=0\\利用反證法,假設a,b,c都是奇數,由於p,q互質,因此\\ \text{Case I: }p,q都是奇數 \Rightarrow \cases{aq^2為奇數\\ bpq為奇數\\ cp^2為奇數} \Rightarrow 三奇數之和應為奇數,不可能為0\\ \text{Case II: }\cases{p為奇數\\ q為偶數}\Rightarrow \cases{aq^2為偶數\\ bpq為偶數\\ cp^2為奇數} \Rightarrow 二偶一奇之和應為奇數,不可能為0\\ \text{Case III: }\cases{p為偶數\\ q為奇數} \Rightarrow \cases{aq^2為奇數\\ bpq為偶數\\ cp^2為偶數} \Rightarrow 二偶一奇之和應為奇數,不可能為0\\ 因此a,b,c 三數不可能皆為奇數,至少有一個偶數,\bbox[red,2pt]{故得證}$$
解答：$$\cases{x^3y^2為六位數\\ \log x^3y^2不是整數} \Rightarrow 10^5 \lt x^3y^2 \lt 10^6 \Rightarrow 5\lt \log x^3y^2 \lt 6 \Rightarrow 5\lt \log x+2(\log x+\log y)\lt 6\cdots(1) \\ 又\cases{\log x\ge 0\\ \log y\ge 0} \Rightarrow 0\le 2(\log x+\log y)再加上(1) \Rightarrow 0\le 2(\log x+\log y) \lt 6\\ \Rightarrow 0\le \log x+\log y\lt 3,又\log x+\log y是整數,因此\log x+\log y=0,1,2\\ \text{Case II }\log x+\log y=0: 由於\cases{\log x\ge 0\\ \log y\ge 0} \Rightarrow \log x=\log y=0違反\log x,\log y都不是整數\\ \text{Case I }\log x+\log y=1: 由(1)得5 \lt \log x+2\lt 6 \Rightarrow 3\lt \log x\lt 4加上\log x+\log y=1\\ \qquad \Rightarrow -3\lt \log y\lt -2違反\log y\ge 0\\ 由\text{Case I & Case II}可知: \log x+\log y=2代入(1) \Rightarrow 1\lt \log x\lt 2 \Rightarrow 10\lt x\lt 100\\ \log x+\log y=\log xy=2 \Rightarrow xy=100 \Rightarrow 1\lt y\lt 10 \Rightarrow y=2,3,\dots,9 代入xy=100\\ \Rightarrow (x,y)=\bbox[red,2pt]{(50,2),(25,4),(20,5)}$$
解答：$$a_k =\sqrt{1+2+\cdots + k} =\sqrt{k(k+1)\over 2}\\ 由於k^2 \le k(k+1)\le (k+1)^2 \Rightarrow k \le \sqrt{k(k+1)}\le k+1 \Rightarrow {k\over \sqrt 2}\le a_k\le {k+1\over \sqrt 2} \\ \Rightarrow {1\over n^2}\sum_{k=1}^n {k\over \sqrt 2}\le {1\over n^2} \sum_{k=1}^n a_k\le {1\over n^2} \sum_{k=1}^n{k+1\over \sqrt 2} \Rightarrow {1\over n^2} {n(n+1)\over 2\sqrt 2}\le {1\over n^2} \sum_{k=1}^n a_k\le {1\over n^2} {n(n+3)\over 2\sqrt 2} \\ \Rightarrow \lim_{n\to \infty} \left({1\over n^2} {n(n+1)\over 2\sqrt 2} \right)\le \lim_{n\to \infty} \left({1\over n^2} \sum_{k=1}^n a_k \right) \le \lim_{n\to \infty}{1\over n^2} {n(n+3)\over 2\sqrt 2} \\ \Rightarrow {1\over 2\sqrt 2} \le \lim_{n\to \infty} \left( {1\over n^2} \sum_{k=1}^n a_k\right) \le {1\over 2\sqrt 2} \Rightarrow \lim_{n\to \infty} \left( {1\over n^2} \sum_{k=1}^n a_k\right) ={1\over 2\sqrt 2} =\bbox[red, 2pt]{\sqrt 2\over 4}$$
解答：$$假設兩焦點為F,F' 且\cases{\overline{FA}=p\\ \overline{FB}=q} \Rightarrow \cases{\overline{F'A}= 2a-p \\ \overline{F'B}=2a- q} \\ \Rightarrow \cases{\cos \angle AFF' = {p^2+ (2c)^2-(2a-p)^2 \over 2p(2c)} \\ \cos \angle BFF' ={q^2+(2c)^2-(2a-q^2) \over 2q(2c)}},\angle AFF'+\angle BFF'=180^\circ \\\Rightarrow {p^2+ (2c)^2-(2a-p)^2 \over 2p(2c)}=-{q^2+(2c)^2-(2a-q^2) \over 2q(2c)} \\ \Rightarrow p^2q+4c^2q-q(2a-p)^2 = -pq^2-4c^2p+p(2a-q)^2 \\ \Rightarrow pq(p+q)+4c^2(p+q)-q(4a^2-4ap+ p^2)-p(4a^2-4aq+q^2)=0\\ \Rightarrow pq(p+q)+4c^2(p+q)- 4a^2(p+q)-pq(p+q)+8apq=0\\ \Rightarrow (4c^2-4a^2)(p+q)+8apq \Rightarrow -4b^2(p+q)+8apq=0\\ \Rightarrow {p+q\over pq}={2a\over b^2} \Rightarrow {1\over p}+{1\over q}={2a\over b^2} \Rightarrow {1\over \overline{FA}} +{1\over \overline{FB}} ={2a\over b^2},\bbox[red,2pt]{故得證} $$
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解題僅供參考,其他教甄試題及詳解
解答：$$9\cos^2 x-3\sin x-7=0 \Rightarrow 9-9\sin^2x-3\sin x-7=0 \Rightarrow 9\sin^2 x+3\sin x-2=0\\ \Rightarrow (3\sin x-1)(3\sin x+2)=0 \Rightarrow \cases{\sin x={1\over 3}\\ \sin x=-{2\over 3} } \\\Rightarrow \cases{x= \theta_1,\pi-\theta_1,\theta_1+2\pi ,3\pi-\theta_1 \\ x=\theta_2,3\pi-\theta_2 , 2\pi+\theta_2 ,5\pi-\theta_2} \Rightarrow 所有根的和=6\pi+10\pi= \bbox[red,2pt]{16\pi}$$
解答：$$S_n=1+2+\cdots +n={n(n+1)\over 2} \Rightarrow S_n-1={n^2+n-2\over 2} ={(n+2)(n-1)\over 2} \\ \Rightarrow {S_n\over S_n-1}={n(n+1)\over (n-1)(n+2)} \Rightarrow T_n ={2\cdot 3\over 1\cdot 4} \times {3\cdot 4\over 2\cdot 5} \cdot {4\cdot 5\over 3\cdot 6}\cdots {n(n+1) \over (n-1)(n+2)} \\= {3\over 1}\cdot {n\over n+2} \Rightarrow T_{1998} ={3\cdot 1998\over 2000} =\bbox[red, 2pt]{2997\over 1000}$$
解答：$$假設5個正整數依序從小至大為x_1,x_2,x_3,x_4,x_5;由\cases{全距18\Rightarrow n_5=n_1+18\\ 中位數＝眾數=9 \Rightarrow x_3=9}\\ 因此5個正整數依序從小至大為x_1,x_2,9,x_4,x_1+18,\\再由平均數=13 \Rightarrow {2x_1+x_2+x_4+27\over 5}=13 \Rightarrow 2x_1+x_2+x_4=38\\ \begin{array}{} x_1 & x_2 & x_3 & x_4 & x_5 &\\\hline 1 & 9 & 9 & 19& 19 & \times:x_2+x_4\lt 36\\ 2 & 9 & 9 & 20 & 20& \times: x_2+x_4 \lt 34 \\ 3 & 9 & 9 & 21& 21& \times: x_2+x_4\lt 32 \\ 4& 9 & 9 & 21 & 22 & \bigcirc\\ \hdashline 5 &9 &9 &19 &23 & \bigcirc\\ 5 &8 &9 &20 &23 & \times:眾數不是9\\ 5 & 7 & 9 & 21 & 23 & \times:眾數不是9\\ 5 & 6 & 9 & 22 & 23& \times:眾數不是9\\ \hdashline 6 &9 & 9 &17 &24 & \bigcirc \\ 6 & 8& 9 &18&24 & \times:眾數不是9 \\ 6 & 7 & 9 & 19& 24 & \times:眾數不是9\\\hdashline 7 &9 & 9 &15 &25 & \bigcirc\\ 7 & 8 & 9 & 16 & 25 & \times:眾數不是9\\ \hdashline 8 & 9& 9 &13 & 26 & \bigcirc \\\hdashline 9 & 9 & 9 & 11&27 &\bigcirc\end{array}\\ 第二大的數字x_4=21,19,17,15,13,11,共有\bbox[red,2pt]6個$$
解答：$$假設L:y=ax+b \Rightarrow \cases{d(L,A(7,3))=6\\ d(L,B(6,6))=3} \Rightarrow \cases{|7a+b-3|=6\sqrt{a^2+1} \\ |6a+b-6| =3\sqrt{a^2+1}} \\ 先試7a+b-3=2(6a+b-6) \Rightarrow b=9-5a \Rightarrow |7a+9-5a-3|= |2a+6|=6\sqrt{a^2+1} \\ \Rightarrow (2a+6)^2 = 36(a^2+1) \Rightarrow 8a(4a-3)=0 \Rightarrow \cases{a=0 \Rightarrow b=9\\ a=3/4 \Rightarrow b=21/4} \\ \Rightarrow \cases{L:y=9 \Rightarrow d(L,C)=9\\ L:4y=3x+21 \Rightarrow d(L,C)=3} \Rightarrow d(L,C)=\bbox[ red, 2pt]{3或9}$$
解答：$$假設f(x)=ax^3+bx^2+cx+d \Rightarrow f'(x)=3ax^2 +2bx+c \Rightarrow f'(x)=6ax+2b \\ \Rightarrow f''(x)-3xf'(x)+9f(x) =3bx^2+(6a+c)x +(2b+9d)=0 \Rightarrow \cases{b=0\\ a=-c\\ d=0} \\ \Rightarrow f(x)=ax^3-ax,再將f(2)=6代入\Rightarrow 8a-2a=6 \Rightarrow a=1 \Rightarrow f(x)=x^3-x \\ 因此 g'(x)=f(x)=x^3-x=0 \Rightarrow \cases{x=0\\x=1} \Rightarrow \cases{g(0)= \int_0^0f(t)\,dt=0\\ g(1) = \int_0^1t^3-t\,dt = {1\over 4} -{1\over 2}=\bbox[red, 2pt]{-{1\over 4}}}$$
解答：$$假設有理根為{q\over p}(p,q互質) \Rightarrow a\cdot ({q\over p})^2+ b\cdot {q\over p}+c=0 \Rightarrow aq^2+ bpq+cp^2=0\\利用反證法,假設a,b,c都是奇數,由於p,q互質,因此\\ \text{Case I: }p,q都是奇數 \Rightarrow \cases{aq^2為奇數\\ bpq為奇數\\ cp^2為奇數} \Rightarrow 三奇數之和應為奇數,不可能為0\\ \text{Case II: }\cases{p為奇數\\ q為偶數}\Rightarrow \cases{aq^2為偶數\\ bpq為偶數\\ cp^2為奇數} \Rightarrow 二偶一奇之和應為奇數,不可能為0\\ \text{Case III: }\cases{p為偶數\\ q為奇數} \Rightarrow \cases{aq^2為奇數\\ bpq為偶數\\ cp^2為偶數} \Rightarrow 二偶一奇之和應為奇數,不可能為0\\ 因此a,b,c 三數不可能皆為奇數,至少有一個偶數,\bbox[red,2pt]{故得證}$$
解答：$$\cases{x^3y^2為六位數\\ \log x^3y^2不是整數} \Rightarrow 10^5 \lt x^3y^2 \lt 10^6 \Rightarrow 5\lt \log x^3y^2 \lt 6 \Rightarrow 5\lt \log x+2(\log x+\log y)\lt 6\cdots(1) \\ 又\cases{\log x\ge 0\\ \log y\ge 0} \Rightarrow 0\le 2(\log x+\log y)再加上(1) \Rightarrow 0\le 2(\log x+\log y) \lt 6\\ \Rightarrow 0\le \log x+\log y\lt 3,又\log x+\log y是整數,因此\log x+\log y=0,1,2\\ \text{Case II }\log x+\log y=0: 由於\cases{\log x\ge 0\\ \log y\ge 0} \Rightarrow \log x=\log y=0違反\log x,\log y都不是整數\\ \text{Case I }\log x+\log y=1: 由(1)得5 \lt \log x+2\lt 6 \Rightarrow 3\lt \log x\lt 4加上\log x+\log y=1\\ \qquad \Rightarrow -3\lt \log y\lt -2違反\log y\ge 0\\ 由\text{Case I & Case II}可知: \log x+\log y=2代入(1) \Rightarrow 1\lt \log x\lt 2 \Rightarrow 10\lt x\lt 100\\ \log x+\log y=\log xy=2 \Rightarrow xy=100 \Rightarrow 1\lt y\lt 10 \Rightarrow y=2,3,\dots,9 代入xy=100\\ \Rightarrow (x,y)=\bbox[red,2pt]{(50,2),(25,4),(20,5)}$$
解答：$$a_k =\sqrt{1+2+\cdots + k} =\sqrt{k(k+1)\over 2}\\ 由於k^2 \le k(k+1)\le (k+1)^2 \Rightarrow k \le \sqrt{k(k+1)}\le k+1 \Rightarrow {k\over \sqrt 2}\le a_k\le {k+1\over \sqrt 2} \\ \Rightarrow {1\over n^2}\sum_{k=1}^n {k\over \sqrt 2}\le {1\over n^2} \sum_{k=1}^n a_k\le {1\over n^2} \sum_{k=1}^n{k+1\over \sqrt 2} \Rightarrow {1\over n^2} {n(n+1)\over 2\sqrt 2}\le {1\over n^2} \sum_{k=1}^n a_k\le {1\over n^2} {n(n+3)\over 2\sqrt 2} \\ \Rightarrow \lim_{n\to \infty} \left({1\over n^2} {n(n+1)\over 2\sqrt 2} \right)\le \lim_{n\to \infty} \left({1\over n^2} \sum_{k=1}^n a_k \right) \le \lim_{n\to \infty}{1\over n^2} {n(n+3)\over 2\sqrt 2} \\ \Rightarrow {1\over 2\sqrt 2} \le \lim_{n\to \infty} \left( {1\over n^2} \sum_{k=1}^n a_k\right) \le {1\over 2\sqrt 2} \Rightarrow \lim_{n\to \infty} \left( {1\over n^2} \sum_{k=1}^n a_k\right) ={1\over 2\sqrt 2} =\bbox[red, 2pt]{\sqrt 2\over 4}$$
解答：$$假設兩焦點為F,F' 且\cases{\overline{FA}=p\\ \overline{FB}=q} \Rightarrow \cases{\overline{F'A}= 2a-p \\ \overline{F'B}=2a- q} \\ \Rightarrow \cases{\cos \angle AFF' = {p^2+ (2c)^2-(2a-p)^2 \over 2p(2c)} \\ \cos \angle BFF' ={q^2+(2c)^2-(2a-q^2) \over 2q(2c)}},\angle AFF'+\angle BFF'=180^\circ \\\Rightarrow {p^2+ (2c)^2-(2a-p)^2 \over 2p(2c)}=-{q^2+(2c)^2-(2a-q^2) \over 2q(2c)} \\ \Rightarrow p^2q+4c^2q-q(2a-p)^2 = -pq^2-4c^2p+p(2a-q)^2 \\ \Rightarrow pq(p+q)+4c^2(p+q)-q(4a^2-4ap+ p^2)-p(4a^2-4aq+q^2)=0\\ \Rightarrow pq(p+q)+4c^2(p+q)- 4a^2(p+q)-pq(p+q)+8apq=0\\ \Rightarrow (4c^2-4a^2)(p+q)+8apq \Rightarrow -4b^2(p+q)+8apq=0\\ \Rightarrow {p+q\over pq}={2a\over b^2} \Rightarrow {1\over p}+{1\over q}={2a\over b^2} \Rightarrow {1\over \overline{FA}} +{1\over \overline{FB}} ={2a\over b^2},\bbox[red,2pt]{故得證} $$
請問第10題為何可以直接假設恰好兩相異實根呢?不是也有兩相異實根+一組共軛虛根的可能嗎?
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