114年家齊高中教甄-數學詳解

 國立臺南家齊高級中等學校 114 學年度第一次教師甄選

一、 填充題(1~4 題每題 5 分， 5~10 題每題 6 分， 11~13 題每題 8 分)

解答:$$\cases{f(a,b,c,d)=d\\ g(a,b,c,d)=a-2b+3c-4d-6\\ h(a,b,c,d)=a^2+4b^2+9c^2+16d^2-12} \Rightarrow \cases{f_a= \lambda g_a+\mu h_a \\ f_b= \lambda g_b+ \mu h_b\\ f_c=\lambda g_c+ \mu h_c\\ f_d=\lambda g_d+ \mu h_d\\ g=0\\ h=0} \Rightarrow \cases{0=\lambda+\mu(2a)\\ 0=-2\lambda +\mu(8b) \\ 0= 3\lambda +\mu(18c) \\1=-4\lambda+\mu(32d) \\a-2b+3c-4d=6\\ a^2+4b^2+9c^2+16d^2=12} \\ \Rightarrow a=-2b=3c \Rightarrow \cases{b=-a/2 \\c=a/3} \Rightarrow \cases{a-2b+3c-4d=3a-4d=6 \\ 3a^2+16d^2=12} \Rightarrow 3a^2+(3a-6)^2=12 \\ \Rightarrow a^2-3a+2=0 \Rightarrow (a-2)(a-1)=0 \Rightarrow \cases{a=1 \Rightarrow d=(3a-6)/4=-3/4 \\a=2\Rightarrow d=(3a-6)/4 =0}  \\ \Rightarrow d的最小值= \bbox[red, 2pt]{-{3\over 4}}$$

解答:$$連續五個自然數相加:x+(x+1)+\cdots+(x+4)=5(x+2) \Rightarrow A=\{5(x+2) \mid 1\le x \le 2021 \}\\ 連續7個自然數相加:y+(y+1)+\cdots +(y+6)=7(y+3) \Rightarrow B=\{7(y+3) \mid1\le y\le 2019\} \\ A\cap B \Rightarrow 5(x+2)=7(y+3) \Rightarrow x=5(y=2),12(y=7), \dots,2021(y=1442) \\ \Rightarrow x=5,5+7,\cdots,5+7\cdot 288 \Rightarrow 共\bbox[red, 2pt]{289}個$$

解答:$$\cases{x=[x]+a\\ y=[y]+b} \Rightarrow \cases{(x+[y])^2=([x]+[y]+a)^2 =2025.114\\ ([x]+y)^2 =([x]+[y]+b)^2=2025.1911} \Rightarrow \cases{[x]+[y]=45\\ b\gt a}\\ \Rightarrow k=[x-y]=[[x]+a-[y]-b]最大值發生在\cases{[x]=45\\ [y]=0},此時k=[45+a-b]= \bbox[red, 2pt]{44}$$

解答:$$x^3=1 \Rightarrow \cases{1+\omega+\omega^2=0 \\ \omega^3 =1} \Rightarrow \omega^a+\omega^b +\omega^c=0 \Rightarrow (a,b,c)=(1,2,3),(4,2,3), (1,5,3), (1,2,6),\\(4,5,3), (4,2,6),(1,5,6),(4,5,6) ，共8種，加上排列數為8\times 3!=48 \Rightarrow 機率為{48\over 6^3}= \bbox[red, 2pt]{2\over 9}$$

解答:

$$假設\cases{重心G\\ 外接圓半徑r\\ \overline{CF}=a},由於H,O,G在一直線上且平行於\overline{BC} \Rightarrow {\overline{AH} \over \overline{HF}}={\overline{AG} \over \overline{GM}} ={2\over 1} \Rightarrow \overline{AH}=5\\ \cases{直角\triangle OMB: \overline{OB}^2=r^2= \overline{OM}^2+ \overline{BM}^2=2.5^2+(5.5+a)^2 \\直角\triangle AHO: \overline{OA}^2 =r^2= \overline{AH}^2+ \overline{OH}^2 =5^2+5.5^2} \Rightarrow 5^2+5.5^2=2.5^2+(5.5+a)^2 \\ \Rightarrow a^2+11a-18.75=0 \Rightarrow (a-1.5)(a+12.5)=0 \Rightarrow a=1.5 \Rightarrow \overline{BC}=2(5.5+a)=\bbox[red, 2pt]{14}$$

解答:

$$\cases{A(1,2) \\B(4,-7)} \Rightarrow 直線L'=\overleftrightarrow{AB}:y=-3x+5 \Rightarrow Q=L'\cap L =(0,5) \\ 假設A,B,P在一圓上，且P為切點，依割線定理:\overline{PQ}^2 =\overline{QA}\times \overline{QB} \\ P\in L \Rightarrow P(10-2t,t) \Rightarrow (10-2t)^2+(5-t)^2=\sqrt{10}\times 4\sqrt{10} =40 \\ \Rightarrow t^2-10t+17=0 \Rightarrow t=5\pm 2\sqrt 2 \Rightarrow P \bbox[red, 2pt]{(4\sqrt 2,5-2\sqrt 2)} \quad (P在第一象限, t=5+2\sqrt 2不合)$$

解答:

$$\alpha^2+ \beta^2+\gamma^2-2\alpha-4\beta-6\gamma+13=0 \Rightarrow (\alpha-1)^2+ (\beta-2)^2+(\gamma-3)^2 =1\\ (\alpha-(\delta+8))^2 +(\beta-(2\delta+9))^2 +(\gamma-(3\delta+10))^2 = \overline{PQ}^2, \\\qquad 其中\cases{P\in 球:(x-1)^2+ (y-2)^2+( z-3)^2=1\\ Q\in 直線L:{x-8}={y-9\over 2} ={z-10\over 3} \Rightarrow Q=(t+8,2t+9,3t+10), t\in \mathbb R} \\ \Rightarrow \overline{PQ} =d(球心,L)-球半徑=\sqrt{(t+7)^2+ (2t+7)^2+ (3t+7)^2} -1 \\=\sqrt{14(t+3)^2+21}-1 \Rightarrow \overline{PQ}最小值=\sqrt{21}-1 \Rightarrow \overline{PQ}^2最小值= \bbox[red, 2pt]{22-2\sqrt{21}}$$

解答:$$y=f(x)=x^4+ax^3+ax^2+x+1 \Rightarrow f'(x)=4x^3+3ax^2+2ax+1 \Rightarrow f'(0)=1\\ \Rightarrow 過(0,1)切線L的斜率為1 \Rightarrow L: y=x+1代回曲線可得x^4+ax^3+ax^2+x+1=x+1 \\ \Rightarrow x^4+ax^3+ax^2=0 \Rightarrow x^2(x^2+ax+a)=0 \Rightarrow x^2+ax+a=0有重根\\ \Rightarrow 判別式a^2-4a=0 \Rightarrow a(a-4) =0\Rightarrow a=\bbox[red, 2pt]4 (a=0 \Rightarrow 只有在x=0有切點,不符題意)$$

解答:$$\cases{f(x,y)= x^2+y^2\\ g(x,y)= x^2+xy+y^2-6} \Rightarrow \cases{f_x= \lambda g_x\\ f_y =\lambda g_y\\ g=0} \Rightarrow \cases{2x=\lambda(2x+y) \\ 2y = \lambda(x+2y) \\x^2+xy+y^2=6} \Rightarrow {x\over y} ={2x+y\over x+2y} \\ \Rightarrow x^2=y^2 \Rightarrow \cases{x=y \Rightarrow 3y^2=6 \Rightarrow y=\pm \sqrt 2\\ x=-y \Rightarrow y^2=6 \Rightarrow y=\pm \sqrt 6} \Rightarrow (x,y)=(\pm \sqrt 2, \pm\sqrt 2), (\pm \sqrt 6,\mp\sqrt 6) \\ \Rightarrow x^2+y^2=4, 12 \Rightarrow (M,m)= \bbox[red, 2pt]{(12,4)}$$

解答:$$P(x,y)與Q對稱於直線y=2x+1 \Rightarrow Q({-3x+4y-4\over 5},{4x+3y+2\over 5}) \\ \Rightarrow \begin{bmatrix} x'\\ y'\end{bmatrix}= \begin{bmatrix} {\sqrt 2\over 2} &-{\sqrt 2\over 2}\\{\sqrt 2\over 2}& {\sqrt 2\over 2}\end{bmatrix} \begin{bmatrix} {-3x+4y-4\over 5}\\{4x+ 3y+2 \over 5}\end{bmatrix} =\begin{bmatrix} {\sqrt 2\over 10}(-7x+y-6)\\ {\sqrt 2\over 10}(x+7y-2) \end{bmatrix} = {\sqrt 2\over 10}\begin{bmatrix} -7& 1 \\ 1& 7\end{bmatrix} \begin{bmatrix} x\\ y\end{bmatrix} + {\sqrt 2\over 10}\begin{bmatrix} -6\\ -2\end{bmatrix} \\ \Rightarrow (a,b,c)= \bbox[red, 2pt]{ \left(-{7\over 10}\sqrt 2, {1\over 10}\sqrt 2,  {1\over 10}\sqrt 2\right)}$$

解答:$$假設期望值為E \Rightarrow \cases{出現反\Rightarrow E+1\\ 出現正正\Rightarrow E+1\\ 出現正反反 \Rightarrow E+3} \Rightarrow E={2\over 3}(E+1)+{1\over 3^2}(E+1)+ {2^2\over 3^3}(E+3) \\ \Rightarrow 27E=18E+18+3E+3+4E+12 \Rightarrow 2E=33 \Rightarrow E= \bbox[red, 2pt]{33\over 2}$$

解答:$$1\le a_k\le 6 \Rightarrow {\sum_{k=1}^n (6k-6)^5\over n^6}\le {S_n \over n^6} \lt {\sum_{k=1}^n (6k)^5\over n^6} \\ \Rightarrow \lim_{n \to \infty} {\sum_{k=1}^n (6k-6)^5\over n^6} \le \lim_{n \to \infty} {S_n \over n^6} \lt \lim_{n \to \infty}{\sum_{k=1}^n (6k)^5\over n^6} \\ \Rightarrow 左極限=右極限=\int_0^1 6^5x^5\,dx =6^4= \bbox[red, 2pt]{1296}$$

解答:$$E:y-z=2 \Rightarrow z=y-2代入S \Rightarrow x^2+y^2+(y-2)^2=4 \Rightarrow x^2+2(y-1)^2=2\\ \Rightarrow {x^2\over 2}+(y-1)^2=1 在xy平面的投影: \bbox[red, 2pt]{\cases{{x^2\over 2}+(y-1)^2=1\\z=0}}$$

二、 計算證明題(共 20 分)

解答:$$\left| a-2b^2\right|+ \left| b-2a^2\right| \ge \left| a-2b^2+b-2a^2\right| =\left| (a+b)-2(a^2+b^2)\right| =\left| 2(a^2+b^2)-(a+b)\right| \\\ge  2(a^2+b^2)-(a+b) = (a^2+2ab+b^2)+(a^2+b^2-2ab)-(a+b) \\=(a+b)^2+(a-b)^2-(a+b) \ge (a+b)^2-(a+b)\ge 3^2-3=6 \\ \Rightarrow \left| a-2b^2\right|+ \left| b-2a^2\right|\ge 6\quad \bbox[red, 2pt]{故得證}$$

解答:$$P\in \Gamma \Rightarrow P=(4\cos \theta, 3\sin \theta)\\ {x^2\over 16}+{y^2\over 9}=1 \Rightarrow {{x\over 8}+{2yy'\over 9}}=0 \Rightarrow y'=-{9x\over 16y} \Rightarrow y'(P)=-{9\over 16} \cdot{4\cos \theta\over 3\sin \theta} =-{3\cos \theta\over 4\sin \theta} \\ \Rightarrow L: y=-{3\cos \theta\over 4\sin \theta} (x-4\cos \theta)+3\sin \theta \Rightarrow \cases{\overline{ON} =d(O,L) = {12\over \sqrt{9+7\sin^2 \theta}}\\ \overline{OP} =\sqrt{9+7\cos^2\theta}} \\ \Rightarrow \overline{NP}^2= \overline{OP}^2-\overline{ON}^2 =9+7\cos^2\theta-{144\over 9+ 7\sin^2\theta} ={49 \cos^2\theta \sin^2\theta\over 9+ 7\sin^2\theta} \Rightarrow \overline{NP}={7\cos \theta \sin \theta\over \sqrt{9+7\sin^2\theta}} \\ \Rightarrow 直角\triangle OPN面積={1\over 2} \overline{NO} \cdot \overline{NP} =f(\theta)={42\cos \theta \sin \theta\over 9+7\sin^2\theta} ={21\sin 2\theta\over 9+7\sin^2\theta} \\ \Rightarrow f'(\theta)=0 \Rightarrow 42\cos 2\theta(9 +7\sin^2\theta) =147\sin^2 2\theta \Rightarrow \cos 2\theta={7\over 25} \Rightarrow \cos \theta= \pm {4\over 5} \Rightarrow \sin \theta=\pm {3\over 5} \\ \Rightarrow P= \bbox[red, 2pt]{({16\over 5},{9\over 5}), ({16\over 5},-{9\over 5}), (-{16\over 5},{9\over 5}), (-{16\over 5},-{9\over 5})}$$

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