115年南科實中教甄-數學詳解

 國立南科國際實驗高級中學115學年度第1次教師甄選

第一部份、 填充題 70% 

解答:$$(\log x)(\log 5x) \left( \log{x\over 6} \right)-(\log 3x)^2+2= (\log x)(\log x+ \log 5) (\log x-\log 6)-(\log x+\log 3)^2+2 \\= (\log x)^3+ \left( \log{1\over 12} \right)(\log x)^2 -[(\log 5)(\log 6)+\log 9] \log x+ (\log 3)^2+2=0 \\ \Rightarrow \log \alpha+\log \beta+ \log \gamma=-\log {1\over 12} \Rightarrow \log \alpha \beta \gamma= \log 12 \Rightarrow \alpha \beta\gamma = \bbox[red, 2pt]{12}$$

解答:$$\cases{A(1,4) \\ B(3,2)} \Rightarrow \cases{圓心O=(A+B)/2=(2,3) \\ 圓半徑r= \overline{AB}/2= \sqrt 2} \Rightarrow 圓方程式: (x-2)^2+ (y-3)^2 =2\\ \Rightarrow P(\sqrt 2\cos \theta+2, \sqrt 2\sin \theta+3) \Rightarrow xy-3x-2y\\=(\sqrt 2\cos \theta+2)(\sqrt 2\sin \theta+3)-3(\sqrt 2 \cos \theta+2)-2(\sqrt 2\sin \theta+3) =2 \cos \theta \sin \theta-6 \\= \sin 2\theta-6 \Rightarrow 最大值為1-6=\bbox[red, 2pt]{-5}$$

解答:

$$f(x)= \sin x-3\cos x= \sqrt{10} \sin (x-\alpha), 其中 \cos \alpha = \frac{1}{\sqrt{10}},\sin \alpha = \frac{3}{\sqrt{10}}\\  \Rightarrow f(x)極大值為\sqrt{10},此時f'(x)=\cos x+3\sin x=0 \Rightarrow \tan x=-{1\over 3} \Rightarrow {\pi\over 2} \lt x\lt \pi 符合要求\\ \Rightarrow  f(x)  的圖形走勢:從點 (0, f(0)=-3) 開始遞增，上升至最高點 (x_{max}, \sqrt{10})，\\接著再遞減至右端點 (\pi, f(\pi)=3) \Rightarrow \cases{k\lt 3:只有一交點\\ \bbox[red, 2pt]{3\le k\lt \sqrt{10}}: 有兩個交點\\ k= \sqrt{10}:只有一切點}  $$

解答:$$假設\cases{首項a\\ 公比r} \Rightarrow \cases{a_2= ar=2 \\ a_6 =ar^5=1/8} \Rightarrow {ar^5\over ar}={1/8\over 2} \Rightarrow r^4={1\over 16} \Rightarrow r={1\over 2} \Rightarrow a=4 \\ \Rightarrow a_1a_2+ a_2a_3+ \cdots+ a_{99} a_{100} =a^2r+a^2r^3+a^2r^5+ \cdots+a^2 r^{197} =a^2r \left( 1 +r^2+\cdots+r^{196} \right) \\=a^2 r \cdot {1-r^{198}\over 1-r^2} =8\cdot {(2^{198}-1)/2^{198}\over 3/4}= {2^5\over 3}-{1\over 3\cdot 2^{193}} = \bbox[red, 2pt]{{32\over 3} \left( 1-4^{-99} \right)}$$

解答:$$\text{TENNESSEE}:連續兩個字母的組合有\text{TE, EN, NN, NE, ES, SS, SE, EE},共8個\\ \qquad \Rightarrow 若信件來自 \text{TENNESSEE}，看到「SS」的機率為{1\over 8} \\ \text{MISSISSIPPI}:同理,連續兩個字母的組合有10種, 其中\text{「SS」}有2種 \\ \qquad \Rightarrow 若信件來自 \text{MISSISSIPPI}，看到「SS」的機率為{2\over 10}={1\over 5}\\ 因此欲求之機率={1/5\over 1/5+1/8}= \bbox[red, 2pt]{8\over 13}$$

解答:$$橢圓\Gamma: {x^2\over a^2}+{y^2 \over b^2}=1 \Rightarrow c^2=a^2-b^2 \Rightarrow \cases{F_1(c,0)\\ F_2(-c,0)} \\ P(x_0,y_0) \in \Gamma \Rightarrow 切線L:  \frac{x_0 x}{a^2} + \frac{y_0 y}{b^2} = 1 \Rightarrow  (b^2 x_0)x + (a^2 y_0)y - a^2 b^2 = 0 \\ \Rightarrow \overline{F_1M}   = \frac{\vert{}(b^2 x_0)(c) + (a^2 y_0)(0) - a^2 b^2\vert{}}{\sqrt{(b^2 x_0)^2 + (a^2 y_0)^2}} = \frac{\vert{}b^2 c x_0 - a^2 b^2\vert{}}{\sqrt{b^4 x_0^2 + a^4 y_0^2}}  \\ \Rightarrow \overline{F_2N}  = \frac{\vert{}(b^2 x_0)(-c) + (a^2 y_0)(0) - a^2 b^2\vert{}}{\sqrt{(b^2 x_0)^2 + (a^2 y_0)^2}} = \frac{\vert{}-b^2 c x_0 - a^2 b^2\vert{}}{\sqrt{b^4 x_0^2 + a^4 y_0^2}} = \frac{\vert{}b^2 c x_0 + a^2 b^2\vert{}}{\sqrt{b^4 x_0^2 + a^4 y_0^2}}  \\ \Rightarrow \overline{F_1M} \cdot \overline{F_2N} =  \frac{\vert{}b^2 c x_0 - a^2 b^2\vert{} \cdot \vert{}b^2 c x_0 + a^2 b^2\vert{}}{b^4 x_0^2 + a^4 y_0^2} = \frac{\vert{}(b^2 c x_0)^2 - (a^2 b^2)^2\vert{}}{b^4 x_0^2 + a^4 y_0^2} = \frac{\vert{}b^4 c^2 x_0^2 - a^4 b^4\vert{}}{b^4 x_0^2 + a^4 y_0^2} \\=  \frac{b^4 \vert{}c^2 x_0^2 - a^4\vert{}}{b^4 x_0^2 + a^4 y_0^2} \cdots(1) \\ P\in \Gamma \Rightarrow  \frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} = 1 \Rightarrow b^2 x_0^2 + a^2 y_0^2 = a^2 b^2 \Rightarrow  a^4 y_0^2    = a^4 b^2 - a^2 b^2 x_0^2\; 代入(1) \\ \Rightarrow \overline{F_1M} \cdot \overline{F_2N} =\frac{b^4 \vert{}c^2 x_0^2 - a^4\vert{}}{b^4 x_0^2 + a^4 b^2 - a^2 b^2 x_0^2} =\frac{b^4 \vert{}c^2 x_0^2 - a^4\vert{}}{b^2  [a^4-x_0^2(a^2-b^2)]} =\frac{b^4 \vert{}c^2 x_0^2 - a^4\vert{}}{b^2  [a^4-x_0^2c^2]} =\bbox[red, 2pt]{b^2} \\ 註:-a\le x_0\le a \Rightarrow c^2x_0^2 \le c^2a^2 \lt a^4$$

解答:$$\overleftrightarrow{BC}:{x+6\over 1}={y-5\over -1} ={z+4\over -2} \Rightarrow \cases{方向向量 \vec v_1=(1,-1,-2) \\ P(-6,5,-4)在直線上} \\ \overleftrightarrow{FH}:{x-2 \over 1} ={y+3\over 5}={z-1\over 1} \Rightarrow 方向向量\vec v_2=(1,5,1) \\ 平面EFGH的法向量\vec n= \vec v_1\times \vec v_2= (9,-3,6) ,取\vec n=(3,-1,2) \\\Rightarrow 平面EFGH的方程式: 3(x-2)-(y+3)+2 (z-1)=0 \Rightarrow 3x-y+2z-11=0 \\ \Rightarrow 長方體的高h=d(P,EFGH)= \frac{\vert{}3(-6) - 5 + 2(-4) - 11\vert{}}{\sqrt{3^2 + (-1)^2 + 2^2}} = 3\sqrt{14} = \overline{CG} \\ 又\overline{CH} =d(H,\overleftrightarrow{BC}) ={ | \overrightarrow{PH}\times v_1| \over |v_1|} ={|(21,21,0)|\over \sqrt 6} =7\sqrt 3 \\ \Rightarrow \overline{GH} = \sqrt{\overline{CH}^2 - \overline{CG}^2} = \sqrt{(7\sqrt{3})^2 - (3\sqrt{14})^2} = \sqrt{147 - 126} = \sqrt{21} \\ 又\cases{ \overleftrightarrow{FH} \parallel \vec v_2 \Rightarrow \overleftrightarrow{FH} =c(1,5,1) \\ \overleftrightarrow{FG} \parallel \vec v_1 \Rightarrow \overleftrightarrow{FG} =a(1,-1,-2)} \Rightarrow \overleftrightarrow{GH} =\overleftrightarrow{FH}-\overleftrightarrow{FG} =(c-a,5c+a,c+2a)\\ \overleftrightarrow{GH} \bot \overleftrightarrow{FG} \Rightarrow \overleftrightarrow{GH} \cdot \vec v_1=0 \Rightarrow -6c-6a=0 \Rightarrow c=-a \\\overline{FH}^2 = \overline{FG}^2 + \overline{GH}^2 \Rightarrow 27c^2=6a^2+21 \Rightarrow c^2=1 \Rightarrow \overline{FG}= \sqrt{6a^2} = \sqrt{6c^2} = \sqrt 6\\ \Rightarrow 體積V = \overline{FG} \times \overline{GH} \times \overline{CG} = \sqrt{6} \times \sqrt{21} \times 3\sqrt{14} = 3 \times 42 = \bbox[red, 2pt]{126 }$$

解答:$$假設兩組的平均都是\mu \Rightarrow a+b+c+d+e+f= 6\mu \Rightarrow 6\mu+8+11+12+17 =10\mu \\ \Rightarrow 48=4\mu \Rightarrow \mu=12 \\假設兩組的變異數都是 \sigma^2 \Rightarrow   (a-12)^2 + (b-12)^2 + (c-12)^2 + (d-12)^2 + (e-12)^2 + (f-12)^2 = 6\sigma^2 \\ \Rightarrow  6\sigma^2 + (8-12)^2 + (11-12)^2 + (12-12)^2 + (17-12)^2=10\sigma^2 \Rightarrow \sigma^2={21\over 2}= \bbox[red, 2pt]{10.5}$$

解答:$$\lim_{n\to \infty} \left[ {n\over n^2}+{n\over (n+2)^2 }+{n\over (n+4)^2} + \cdots +{n\over (3n-2)^2} \right] = \lim_{n\to \infty} \sum_{k=0}^{n-1} {n\over (n+2k)^2} \\ = \lim_{n\to \infty} \sum_{k=0}^{n-1} {1\over n(+2(k/n))^2} = \int_0^1 {1\over (1+2x)^2} \,dx = \left. \left[ -{1\over 2(1+2x)} \right] \right|_0^1 =-{1\over 6}+{1\over 2}= \bbox[red, 2pt]{1\over 3}$$

解答:$$ \sum_{k=1}^{n}   {1\over k(k+1) (k+2)(k+3)} =  \frac{1}{3} \sum_{k=1}^{n}\left[ \frac{1}{k(k+1)(k+2)} - \frac{1}{(k+1)(k+2)(k+3)} \right] \\={1\over 3} \left[ \left( {1 \over 1\cdot 2\cdot 3} -{1\over 2\cdot 3\cdot 4} \right) + \left( { 1\over 2\cdot 3 \cdot 4} -{1\over 3\cdot 4\cdot 5} \right) + \cdots \right.\\+ \left.\left( {1 \over n\cdot (n+1)\cdot (n+2)} -{1\over (n+1)\cdot (n+2)\cdot (n+3)} \right)  \right ] \\={1\over 3} \left[ { 1\over 1\cdot 2 \cdot 3}- {1\over (n+1)\cdot (n+2)\cdot (n+3)}\right] ={119\over 2160} \Rightarrow {1\over (n+1)\cdot (n+2)\cdot (n+3)}={1 \over 720}\\ \Rightarrow n= \bbox[red, 2pt]7$$

解答:$$f(t) ={|0+0-15| \over \sqrt{9\cos^2 t+ 25\sin^2 t}} ={15\over \sqrt{9+16 \sin^2 t}} \Rightarrow 最小值為{15\over \sqrt{9+16}} ={15\over 5}= \bbox[red, 2pt]3$$

解答:$$  P = 113 \times 115 \times 117 \times 119 \times 121=    21892455585 \Rightarrow 147960^2 \lt P\lt 147961^2 \\ \sqrt{P+n}為正整數 \Rightarrow  n = 147961^2 - P = 21892457521 - 21892455585 = \bbox[red, 2pt]{1936 }$$

解答:$$假設投擲 10 次銅板，出現正面k次，反面為10-k次 \\ \Rightarrow 10次移動的總變動點數為S_{10}=5k+(-5)(10-k) =10k-50 \\ \Rightarrow 最終位置  X_{10}  相當於 S_{10}  除以 12 的餘數\\ \Rightarrow  X_{10} \equiv 12 + S_{10} \equiv 12 + 10k - 50 \equiv 10k - 38 \equiv -2k - 2 \pmod{12} \\ \Rightarrow \cases{k=0: X_{10} \equiv -2(0) - 2 = -2 \equiv 10 \pmod{12} \Rightarrow X_{10} = 10，機率為 \frac{1}{1024} \\k=1: X_{10} \equiv -2(1) - 2 = -4 \equiv 8 \pmod{12} \Rightarrow X_{10} = 8，機率為 \frac{10}{1024} \\k=2: X_{10} \equiv -2(2) - 2 = -6 \equiv 6 \pmod{12} \Rightarrow X_{10} = 6，機率為 \frac{45}{1024} \\k=3: X_{10} \equiv -2(3) - 2 = -8 \equiv 4 \pmod{12} \Rightarrow X_{10} = 4，機率為 \frac{120}{1024} \\k=4: X_{10} \equiv -2(4) - 2 = -10 \equiv 2 \pmod{12} \Rightarrow X_{10} = 2，機率為 \frac{210}{1024} \\k=5: X_{10} \equiv -2(5) - 2 = -12 \equiv 0 \pmod{12} \Rightarrow X_{10} = 12，機率為 \frac{252}{1024}\\ k=6: X_{10} \equiv -2(6) - 2 = -14 \equiv 10 \pmod{12} \Rightarrow X_{10} = 10，機率為 \frac{210}{1024} \\k=7: X_{10} \equiv -2(7) - 2 = -16 \equiv 8 \pmod{12} \Rightarrow X_{10} = 8，機率為 \frac{120}{1024} \\k=8 : X_{10} \equiv -2(8) - 2 = -18 \equiv 6 \pmod{12} \Rightarrow X_{10} = 6，機率為 \frac{45}{1024} \\k=9: X_{10} \equiv -2(9) - 2 = -20 \equiv 4 \pmod{12} \Rightarrow X_{10} = 4，機率為 \frac{10}{1024} \\k=10: X_{10} \equiv -2(10) - 2 = -22 \equiv 2 \pmod{12} \Rightarrow X_{10} = 2，機率為 \frac{1}{1024}} \\ \Rightarrow \cases{P(X_{10} = 12) = \frac{252}{1024} \\P(X_{10} = 10) = \frac{1 + 210}{1024} = \frac{211}{1024} \\P(X_{10} = 8) = \frac{10 + 120}{1024} = \frac{130}{1024} \\P(X_{10} = 6) = \frac{45 + 45}{1024} = \frac{90}{1024} \\P(X_{10} = 4) = \frac{120 + 10}{1024} = \frac{130}{1024} \\P(X_{10} = 2) = \frac{210 + 1}{1024} = \frac{211}{1024} } \\\Rightarrow E(X_{10}) = 12 \times \frac{252}{1024} + 10 \times \frac{211}{1024} + 8 \times \frac{130}{1024} + 6 \times \frac{90}{1024} + 4 \times \frac{130}{1024} + 2 \times \frac{211}{1024}={957\over 128} \\ \Rightarrow  E(X_{10} - 7) = E(X_{10}) - 7 = \frac{957}{128} - 7 =  \bbox[red, 2pt]{\frac{61}{128} }$$

解答:$$  \frac{1}{2} \times [a^2 + a] = 19a + 99 \Rightarrow [a^2+a]=38a+198 \in \mathbb N \Rightarrow 可假設38a=k\in \mathbb N \Rightarrow a={k\over 38} \\ x-1\lt [x] \le x \Rightarrow a^2 + a - 1 < [a^2 + a] \leq a^2 + a \Rightarrow  a^2 + a - 1 < 38a + 198 \leq a^2 + a \\  \Rightarrow   {37+\sqrt{2161} \over  2} \le a\lt {37+\sqrt{2165} \over 2} \Rightarrow 41.742 \le a\lt 41.765 \Rightarrow 1586.2 \le 38a \lt  1587.1\\ \Rightarrow 38a=1587 \Rightarrow a=\bbox[red, 2pt]{1587\over 38}$$

第二部份、 計算及證明題 30%

解答:$$S=a+b+c+d \Rightarrow  \frac{16d}{a+b+c} + \frac{25c}{a+b+d} + \frac{36b}{a+c+d} + \frac{49a}{b+c+d} \\ = \frac{16d}{S-d} + \frac{25c}{S-c} + \frac{36b}{S-b} + \frac{49a}{S-a}\\ =16\left(\frac{S}{S-d} - 1\right) + 25\left(\frac{S}{S-c} - 1\right) + 36\left(\frac{S}{S-b} - 1\right) + 49\left(\frac{S}{S-a} - 1\right) \\ = S \left( \frac{4^2}{S-d} + \frac{5^2}{S-c} + \frac{6^2}{S-b} + \frac{7^2}{S-a} \right) - (16 + 25 + 36 + 49) \\ = S \left( \frac{16}{S-d} + \frac{25}{S-c} + \frac{36}{S-b} + \frac{49}{S-a} \right) - 126 ,再利用柯西不等式\\  \left( (S-d) + (S-c) + (S-b) + (S-a) \right) \left( \frac{4^2}{S-d} + \frac{5^2}{S-c} + \frac{6^2}{S-b} + \frac{7^2}{S-a} \right) \geq (4 + 5 + 6 + 7)^2 \\ \Rightarrow  3S \left( \frac{16}{S-d} + \frac{25}{S-c} + \frac{36}{S-b} + \frac{49}{S-a} \right) \geq (22)^2 = 484 \\ \Rightarrow   S \left( \frac{16}{S-d} + \frac{25}{S-c} + \frac{36}{S-b} + \frac{49}{S-a} \right) \geq \frac{484}{3} \\ \Rightarrow S \left( \frac{16}{S-d} + \frac{25}{S-c} + \frac{36}{S-b} + \frac{49}{S-a} \right)-126 \geq \frac{484}{3} -126 = \bbox[red, 2pt]{106\over 3}$$

解答:$$  y = 3x  代入曲線方程式  (4 - 2\sin\theta)x^2 - (3\cos\theta)y = 0  \Rightarrow  x \left[ (4 - 2\sin\theta)x - 9\cos\theta \right] = 0  \\ \Rightarrow \cases{x_1=0\\ x_2=9\cos \theta/(4-2\sin \theta)} \Rightarrow 交點\cases{(0, 0) \\ (x_2,3x_2)} \Rightarrow 兩交點距離 d= \sqrt{x_2^2+9x_2^2} = \sqrt{10} |x_2|\\ 擬求|x_2|的最大值, 取f(\theta) ={9\cos \theta\over 4-2\sin \theta} \Rightarrow f'(\theta) =  \frac{18 - 36\sin\theta}{(4 - 2\sin\theta)^2}=0 \Rightarrow \sin \theta={1\over 2} \\ \Rightarrow \theta= \bbox[red, 2pt]{{\pi\over 6}, {5\pi\over 6}}$$

解答:$$\cases{\overline{AI}是\angle A的角平分線\Rightarrow \sin(A/2) = r/x \\\overline{BI}是\angle B的角平分線\Rightarrow \sin(B/2) = r/y \\\overline{CI}是\angle C的角平分線\Rightarrow \sin(C/2) = r/z } \Rightarrow \cases{x=r/\sin(A/2) \\ y=r/\sin (B/2) \\ z=r/\sin (C/2)} \\ \Rightarrow xyz ={r^3 \over \sin{A\over 2} \cdot \sin{B\over 2} \cdot \sin{C\over 2} } ={r^3 \over \sqrt{(s-b)(s-c)\over bc} \cdot \sqrt{(s-a)(s-c) \over ac} \cdot \sqrt{(s-a)(s-b) \over ab}} ={r^3 \over {(s-a)(s-b)(s-c) \over abc}} \\ ={r^3\cdot abc \over (s-a)(s-b)(s-c)}={r^3 \cdot abc \over r^2 s} ={r\cdot abc\over s} \Rightarrow xyzs=abcr \quad \bbox[red, 2pt]{故得證} \\ 註1: \triangle 面積=\sqrt{s(s-a)(s-b)(s-c)} =rs \Rightarrow (s-a)(s-b)(s-c)=r^2s \\ 註2: \cos A={b^2+c^2-a^2 \over 2bc} =1-2\sin^2{A\over 2} \Rightarrow 2\sin^2{A\over 2}={a^2-(b-c)^2\over 2bc} ={(a-b+c)(a+b-c) \over 2bc} \\\qquad ={2(s-b)\cdot 2(s-c) \over 2bc} \Rightarrow \sin^2{A\over 2}={(s-b)(s-c) \over bc}$$

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