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2023年6月23日 星期五

112年屏東高中教甄-數學詳解

國立屏東高級中學112學年度正式教師甄試

一、填充題(共10題,每題6,共60分)

解答:$$\begin{array}{ccl} a& 種類&數量 \\\hline 兩位數& 93,84,\dots,39 & 7\\\hdashline 三位數 & 9cd& H^2_3=4\\ & 8cd& H^2_4=5\\ & \cdots& \cdots \\ & 3cd& H^2_9=10\\ & 2cd,需扣除(c,d)= (10,0),(0,10)& H^2_{10} -2=9\\ & 1cd,需扣除(c,d)=(11,0), (0,11),(10,1),(1,10)& H^2_{11}-4 = 8 \\\hdashline 四位數 & 9bcd & H^3_3=10 \\ & 8bcd & H^3_4=15\\ & \cdots & \cdots\\ & 3bcd & H^3_9= 55\\ & 2bcd,需扣除(10,0,0)的排列數& H^3_{10}-3=63 \\ & 1bcd,需扣除(11,0,0),(10,1,0)的排列數9 & H^3_{11}-9=69 \\\hdashline 五位數 & 9abcd & H^4_3=20 \\ & 8abcd& H^4_4= 35\\ & \cdots & \cdots \\ & 3abcd & H^4_9=220\\ & 2abcd,需扣除(a,b,c,d)=(10,0,0,0)的排列數4 & H^4_{10}-4 =282 \\ & 1abcd,需扣除(a,b,c,d)= (11,0,0,0),(10,1,0,0)的排列數16 & H^4_{11}-16=348\\\hline  \end{array}\\ 合計 7+(4+5+6+7+8+9+10+9+8) +(10+15+ 21+28+ 36+45+55+ 63+69)\\ +(20+35+ 56+84+120+165+220+282+348)= 7+ 66+ 342+1330=\bbox[red,2pt]{1745}$$
解答:$$\cases{\overrightarrow{AB}=(-1,1,-1) \\ \overrightarrow{AC}= (1,-1,-1)} \Rightarrow \vec n = \overrightarrow{AB} \times \overrightarrow{AC} =(-2,-2,0) \Rightarrow 平面E=\triangle ABC: x+y=1\\假設垂心H(a,b,c) \Rightarrow \cases{\overrightarrow{AH} \bot \overrightarrow{BC} \\\overrightarrow{BH} \bot \overrightarrow{AC} \\\overrightarrow{CH} \bot \overrightarrow{AB} } \Rightarrow \cases{(a,b-1,c-2) \cdot (2,-2,0)=0 \\ (a+1,b-2,c-1) \cdot (1,-1,-1)=0\\ (a-1,b,c-1) \cdot (-1,1,-1)=0} \Rightarrow H(a,a+1,3)\\ 又\\H需在\triangle ABC平面上 \Rightarrow a+a+1=1 \Rightarrow a=0 \Rightarrow H\bbox[red, 2pt]{(0,1,3)}$$
解答
$$\angle A=\angle C=90^\circ \Rightarrow ABCD共圓,假設此外接圓圓心O,半徑=a,如上圖\\ 因此\angle ABO=45^\circ \Rightarrow \angle OBC=60^\circ-45^\circ =15 \Rightarrow \cases{\overline{BC}=2a\cos 15^\circ \\ \overline{CD} =2a\sin 15^\circ} \\ \Rightarrow \overline{BC}+\overline{CD}=2a(\cos 15^\circ+ \sin 15^\circ) =2a\cdot {\sqrt 6\over 2} =\sqrt 6a = k \Rightarrow a={k\over \sqrt 6} \\ \Rightarrow ABCD周長= 2\sqrt 2a+ k = {2\sqrt 2 \over \sqrt 6}k +k =\bbox[red, 2pt]{\left( {3+2\sqrt 3\over 3}\right)k}$$
解答:$$依學校公佈的答案為\color{blue}{43},可是不存在兩正整數a,b滿足a^2+b^2=43,題目有\bbox[red, 2pt]{疑義}$$
解答:$$$$
解答:$$$$
解答
$$由上圖可知:\cos \theta ={1\over 3} \Rightarrow \sin \theta={2\sqrt 2\over 3} \Rightarrow h=10+\sum_{k=0}^\infty \left({10\over 2^k} +{10\over 2^{k+1}}\right)\sin \theta \\ =10+10(2+1)\cdot {2 \sqrt 2\over 3} =\bbox[red, 2pt]{10+20\sqrt 2}$$
解答:$$第一個聯立方程式除了明顯解(0,0,0)外,還有一解(3,-1,2),代表有無限多組解\\ 第二個聯立方程式是第一個方程組三個平面的平移,顯然其解為過(6,2,0)方向向量為(3,-1,2)的直線\\ 即 {x-6\over 3} ={y-2\over -1}={z\over 2},也就是\bbox[red,2pt]{(3t+6,-t+2,2t), t\in \mathbb R}$$
解答:$$迴歸直線y=28+{2\over 5}x一定通過(\mu_x,\mu_y) \Rightarrow \mu_y=28+{2\over 5}\cdot 60=52\\ 又\cases{p=-x/2+8\\ q=y/4-7} \Rightarrow \cases{\mu_p=-60/2+8=-22\\\sigma_p=\sigma_x/2\\ \mu_q= 52/4-7= 6\\ \sigma_q=\sigma_y/4} \Rightarrow q,p迴歸直線斜率b=-{2\over 5}\times {\sigma_y/4\over \sigma_x/2} =-{1\over 5} \\ \Rightarrow q,p迴歸直線: q=-{1\over 5}(p+22)+6 \Rightarrow -{1\over 5}p+{8\over 5} \Rightarrow (a,b)=\bbox[red, 2pt]{\left({8\over 5},-{1\over 5}\right)}$$
解答:$$f(x)=x^3+9x^2+8x+5 \Rightarrow f''(x)=6x+18=0 \Rightarrow x=-3 \\ \Rightarrow y=f(x)的對稱中心P(-3,f(-3)=35) \\ 而(f(s)+f(t))\div 2=(42+28)\div 2=35 =f(-3) \Rightarrow (s+t)\div 2=-3 \Rightarrow s+t=\bbox[red, 2pt]{-6}$$
解答:$$令f(x)={x^2-x+1\over x^2+x+1} \Rightarrow f'(x)={x^2-1\over (x^2+x+1)^2} =0 \Rightarrow x=\pm 1 \Rightarrow \cases{f(1)=1/3\\ f(-1)=3} \\ \Rightarrow \cases{M=\log_2 3 \\ m=-\log 3} \Rightarrow M-m=2\log_2 3 =\log_2 9 \Rightarrow 8^{M-m} =8^{\log_2 9} =2^{\log_2 9^3} \\=9^3 = \bbox[red, 2pt]{729}$$
解答:$$令a_3=a_4=a \Rightarrow a_5=2a \Rightarrow a_3^2+a_4^2+a_5^2=6a^2\lt 2344 \Rightarrow a  \le 19\\ 由於a_5=2a 是完全平方數,取a=18\Rightarrow a_3=a_4=18, a_5=36 \\ \Rightarrow a_1^2+a_2^2 =2344-18^2-18^2-36^2=400 \cdots(1)\\ 又\sum_{i=1}^5 a_i =a_1+a_2+18+18+36=a_1+a_2+72=k^2 \cdots(2)\\ 由(1)及(2) \Rightarrow \cases{a_2=16\\ a_1=12} \Rightarrow a_1+a_5=12+36=\bbox[red, 2pt]{48}$$
解答:$$x+{2\over x-1}=x-1+{2\over x-1}+1=(x-1)+{x+1\over x-1} \ge 2\sqrt{(x-1)\cdot {x+1\over x-1}} =2\sqrt{x+1}\\  同理可得y+{2\over y-1}=(y+1)+{y+1\over y-1} \ge 2\sqrt{y+1} \\ 因此我們有x+{2\over x-1}+y+{2\over y-1}\ge 2(\sqrt{x+1}+\sqrt{y+1}),依題意,等號成立\\ \Rightarrow \cases{x-1={x+1\over x-1}\\ y-1={y+1\over y-1}} \Rightarrow \cases{x^2-3x=0\\ y^2-3y=0} \Rightarrow \bbox[red, 2pt]{\cases{x=3\\y=3}}$$
解答:$$令a=x+y,其中\cases{x\in \mathbb z\\ 0\le y\lt 1} \Rightarrow \left[ a+{19\over 100}\right] + \left[ a+{20\over 100}\right] +\cdots + \left[ a+{85\over 100}\right] \\= 67x +\left[ y+{19\over 100} \right] + \left[ y+{20\over 100}\right] +\cdots + \left[ y+{85\over 100}\right] =500\\ \Rightarrow x=7 且\left[ y+{19\over 100} \right] + \left[ y+{20\over 100}\right] +\cdots + \left[ y+{85\over 100}\right]=500-67\times 7=31 \\ \Rightarrow \cases{\left[ y+{19\over 100} \right] = \left[ y+{20\over 100}\right] =\cdots = \left[ y+{54\over 100}\right]=0 \\\left[ y+{55\over 100} \right] = \left[ y+{56\over 100}\right] =\cdots + \left[ y+{85\over 100}\right]=1} \Rightarrow \cases{y+0.54\lt 1\\ y+0.55\ge 1} \\ \Rightarrow 0.45\le y\lt 0.46 \Rightarrow [100a] =[100x+100y] =\bbox[red, 2pt]{745}$$
解答:$$\cases{O(0,0)\\A(-4,2)} \Rightarrow \cases{\overline{OA}=2\sqrt 5\\ \overline{OA}的中垂線L:2x-y+5=0},由於C\in L \Rightarrow C(a,2a+5)\\ 正五邊形每一內角為{(5-2)\times 180\over 5}=108^\circ =60^\circ+45^\circ \Rightarrow \cos \angle B={(2\sqrt 5)^2+(2\sqrt 5)^2-\overline{AC}^2\over 2\cdot 2\sqrt 5\cdot 2\sqrt 5} \\ \Rightarrow {\sqrt 2-\sqrt 6\over 4} ={40-\overline{AC}^2\over 80} \Rightarrow \overline{AC}^2=40-20(\sqrt 2-\sqrt 6) =\overline{OC}^2=a^2+(2a+5)^2 \\ \Rightarrow a^2+4a-3+4(\sqrt 2-\sqrt 6)=0 \Rightarrow a=$$
解答:$$f(x)=(3x+2)^3 (7-2x)^2 \\\Rightarrow f'(x)=9(3x+2)^2(7-2x)^2 -4(3x+2)^3(7-2x) = 5(3x+2)^2(7-2x)(-6x+11) \\\Rightarrow f''(x)=30(3x+2)(7-2x)(-6x+11) -10(3x+2)^2(-6x+11)-30(3x+2)^2 (7-2x)\\ 若f'(x)=0 \Rightarrow \cases{x=-2/3\\ x=7/2\\ x=11/6} \Rightarrow \cases{f''(-2/3)=0\\ f''(7/2)\gt 0\\ f''(11/6) \lt 0}\\ \Rightarrow f(11/6)= \left( {11\over 2}+2\right)^3\left( 7-{11\over 3} \right)^2 =\bbox[red, 2pt]{9375\over 2}為範圍內的最大值$$

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