朱式幸福: 114年文華高中教甄-數學詳解

臺中市立文華高級中等學校 114 學年度第 1 次教師甄選

一、填充題：請將正確答案填入正確的題格中，分式須化至最簡，根式須有理化，否則不予計分，全對才給分，不需計算過程。
第一部份：(每格 4 分)

解答: $$\cases{P=a^3-4a^2-6a+4\\ Q=a^2-3a+1} \Rightarrow P=Q(a-1)-10a+5 \Rightarrow (Q-10)a+(-P-Q+5)=0 \\ 由於Q-10及-P-Q+5皆為有理數，而a為無理數，因此\cases{Q-10=0\\ -P-Q+5=0} \Rightarrow \cases{P=-5\\ Q=10} \\ \Rightarrow 10=a^2-3a+1 \Rightarrow a^2-3a-9=0 \Rightarrow a= \bbox[red, 2pt]{3+3\sqrt{5}\over 2} \; (\because a\gt 0)$$

解答:

$$P=\overline{AC} \cap \overline{BD} \Rightarrow \cases{\triangle ADP \sim \triangle BCP \\ \triangle ABP \sim \triangle DCP} \Rightarrow \cases{\displaystyle {\overline{AP} \over \overline{BP} }={\overline{PD} \over \overline{PC}} ={\overline{AD} \over \overline{BC}} ={9\over 5} \\ \displaystyle { \overline{PB}\over \overline{PC}} ={\overline{AP} \over \overline{PD}} ={\overline{AB} \over \overline{CD}}={9\over 3}} \\ \Rightarrow \overline{AP}: \overline{BP}: \overline{CP}: \overline{DP}: 27:15:5:9 \Rightarrow \cases{\overrightarrow{CP}= {3\over 8} \overrightarrow{CB}+{5\over 8} \overrightarrow{CD} \\ \overrightarrow{CP}={5\over 32} \overrightarrow{CA}} \\ \Rightarrow \overrightarrow{CA} ={32\over 5} \overrightarrow{CP} ={32\over 5 } \left({3\over 8} \overrightarrow{CB}+{5\over 8} \overrightarrow{CD} \right) ={12\over 5} \overrightarrow{CB}+ 4 \overrightarrow{CD} \Rightarrow (x,y) = \bbox[red, 2pt]{\left({12\over 5},4 \right)}$$

解答: $$取g(x)=f(x)-7 \Rightarrow g(2-\sqrt 3)=g(1+\sqrt 5)=0 \Rightarrow 2\pm \sqrt 3,1\pm \sqrt 5為g(x)=0之四根 \\ \Rightarrow g(x)=(x^2-4x+1)(x^2-2x-4) \Rightarrow f(x)= (x^2-4x+1)(x^2-2x-4)+7 \\x^2=6代入f(x) \Rightarrow 餘式=(7-4x)(2-2x)+7 = 8x^2-22x+21 =48-22x+21= \bbox[red, 2pt]{-22x+69}$$

解答: $$令\cases{a=\log 2\\ b=\log 5} \Rightarrow a+b=1 \Rightarrow 原式L=5^{a^3}\times 8^{b^2} \times 5^{b^3} \Rightarrow \log L=a^3b+b^2\log 8+b^3b \\=(a^3+b^3)b+3ab^2 =((a+b)(a^2-ab+b^2))b+3ab^2 =a^2b-ab^2+b^3+3ab^2 \\=a^2b+2ab^2+b^3 =ab(a+b)+ab^2+b^3 =ab+ab^2+b^3=ab+b^2(a+b) =ab+b^2 \\=b(a+b)=b=\log 5 \Rightarrow L= \bbox[red, 2pt]5$$

解答:

$$假設原長方形在x-y平面上，沿著對角線\overline{BD}將C點摺起，摺起C點後各點坐標可假設為:\\ \cases{A(0,0,0)\\ B(10,0,0)\\C(a,b,c) \\D(0,5,0 )} \Rightarrow 折起後C點與xy平面的距離c=\overline{CE}\sin 60^\circ =2\sqrt 5\cdot {\sqrt 3\over 2}=\sqrt{15} \\ \Rightarrow 折起後\overline{CE}在xy平面投影長\overline{EF}=\overline{CE}\cos 60^\circ= \sqrt 5 \Rightarrow \cases{\overline{F'C} =\overline{FC}\cos \theta= 3 \\ \overline{FF'} =\overline{CF}\sin \theta=6} \\\Rightarrow \cases{a=10-3=7\\ b=5-6=-1} \Rightarrow 折起後C點坐標(7,-1,\sqrt{15}) \Rightarrow \overline{AC}=\sqrt{65}\\ \Rightarrow \cos \angle ADC={5^2+10^2-(\sqrt{65})^2 \over 2\cdot 5\cdot 10 } ={60\over 100} = \bbox[red, 2pt]{3\over 5}$$

解答: $$(1)紅球沒有被連續取出的方法:三顆黑球與四顆白球先任排，有8個空隙選2個插入紅球，\\\quad 共有C^8_2種\\ (2)紅球要最後取完:紅球放最後，前面有三顆黑球及四顆白球，共7個空隙(兩紅球不相鄰)\\\quad 讓紅球插入，共有C^7_1種選法\\ 因此機率={C^7_1 \over C^8_2} =\bbox[red, 2pt]{1\over 4}$$

解答: $$假設弦\overline{PQ}中點=(-5,4) \Rightarrow \cases{P(a,b) \\Q(c,d)\\ a+c=-10\\ b+d=8} \Rightarrow \cases{4a^2-9(b-2)^2=36\\ 4c^2-9(d-2)^2=36} \\ \Rightarrow 4(a^2-c^2)-9((b-2)^2-(d-2)^2)=0 \Rightarrow 4(a+c)(a-c)-9(b+d-4)(b-d) =0\\ \Rightarrow -40(a-c)-36(b-d)=0 \Rightarrow \overline{PQ}斜率={b-d\over a-c} =-{40\over 36} =-{10\over 9} \\ \Rightarrow 弦方程式: y=-{10\over 9}(x+5)+4 \Rightarrow \bbox[red, 2pt]{10x+ 9y=-14}$$

解答: $$\cases{P(\cos \theta, \sin \theta) \\Q(6,0) \\R(0,-2)} \Rightarrow |z-6|^2+|z+2i|^2 = \overline{PQ}^2+ \overline{PR}^2 =(\cos \theta-6)^2+\sin^2 \theta +\cos^2 \theta+ (\sin\theta +2)^2 \\=42-12\cos \theta+4\sin \theta=42+4\sqrt{10}\left({ 4\over 4\sqrt {10}}\sin \theta-{12\over 4\sqrt{10}} \cos \theta \right) =42+4\sqrt{10}\sin(\theta-\alpha) \\ \Rightarrow 最小值 \bbox[red, 2pt]{42-4\sqrt{10}}$$

第二部份：(每格 6 分)

解答: $$f(x)=g(x) \Rightarrow x^4+ax^2+(b-c)x+3-a= (x-\alpha)^2 (x-\beta)^2 \\=x^4-2(\alpha+\beta)x^3+ ((\alpha+ \beta)^2+2\alpha \beta)x^2-2(\alpha+\beta)\alpha \beta x+ \alpha^2\beta^2 \Rightarrow \alpha+\beta=0 \\ \Rightarrow \cases{2\alpha\beta=a\\ b-c=0\\ 3-a=\alpha^2 \beta^2} \Rightarrow 3-a={a^2\over 4} \Rightarrow a^2+4a-12=0 \Rightarrow (a+6)(a-2)=0 \Rightarrow a=\bbox[red, 2pt]{-6} \\若a=2 \Rightarrow \cases{\alpha \beta=1\\ \alpha+\beta =0} \Rightarrow \alpha,\beta非實根$$

解答: $$f(x)=x^2+4x+7=(x+2)^2+3 \Rightarrow 圖形y=f(x)對稱於x=-2 \\ \Rightarrow \cases{圖形\Gamma_1: y=3^{f(x)} 對稱於x=-2 \\ 圖形\Gamma_2: y=81\times \log_3 f(x)對稱於x=-2} \Rightarrow 兩圖形交點亦對稱x=-2 \\ \Rightarrow \alpha對稱於x=-2的對稱點為(-2-\alpha)=\beta \Rightarrow \alpha+\beta= \bbox[red, 2pt]{-4}$$

解答: $$邊長6的正四面體體積減去角落四個邊長2的正四面體體積={\sqrt 2\over 12}(6^3-2^3\times 4) = \bbox[red, 2pt]{{46\over 3}\sqrt 2}$$

解答: $$f(x)=3x^{20}+192 x^8+3 =3(x-\alpha_1)(x-\alpha_2) \cdots (x-\alpha_{20})\\ \Rightarrow (x-\alpha_1)(x-\alpha_2) \cdots (x-\alpha_{20})={1\over 3}f(x)\\ g(x)=x^3-3x^2+4x-2 =(x-1)(x^2-2x+2) =(x-1)(x-\omega)(x-\bar \omega),\\ 其中\omega,\bar \omega為x^2-2x+2=0的兩根 \Rightarrow \omega^2=2\omega-2 \Rightarrow \omega^4=4\omega^2-8\omega+4=4(2\omega-2)-8\omega+4 \\ \Rightarrow \omega^4=-4 \Rightarrow \omega^8=16 \Rightarrow \omega^{20}=-1024 \Rightarrow f(\omega) =3\cdot(-1024) +192\cdot (-4)+3=3\\同理, f(\bar \omega)=3及f(1)=3+192+3=198\\ g(\alpha_1)\times g(\alpha_2)\times \cdots\times g(\alpha_{20} ) =(\alpha_1-1)(\alpha_1-\omega)(\alpha_1-\bar \omega) \times (\alpha_2-1)(\alpha_2-\omega)(\alpha_2-\bar \omega) \\ \qquad \quad \times \cdots \times (\alpha_{20}-1)(\alpha_{20}-\omega)(\alpha_{20}-\bar \omega) \\\qquad ={1\over 3}f(1)\cdot {1\over 3}f(\omega) \cdot {1\over 3}f(\bar\omega) =66\cdot 1\cdot 1=\bbox[red, 2pt]{66}$$

解答: $$ax^2+bx+c \ge 0 \Rightarrow \cases{0\lt a\lt b\\ b^2-4ac\le 0} \Rightarrow \cases{1-a/b\gt 0\\ b/a\le 4c/b} \\ \Rightarrow {4a+3b+2c\over b-a} ={4(a/b) +3+2(c/b) \over 1-a/b} \ge {4(a/b)+3+(1/2)(b/a) \over 1-a/b} \\={4t+3 +1/2t\over 1-t} =k(取t= {a\over b} \in \mathbb R) \Rightarrow (2k+8)t^2+ (6-2k)t+1=0 \\ \Rightarrow (6-2k)^2-4(2k+8)\ge 0 \Rightarrow k^2-8k+1 \ge 0 \Rightarrow (k-(4+ \sqrt{15})) (k-(4-\sqrt{15})) \ge 0 \\ \Rightarrow k \ge 4+\sqrt{15} \Rightarrow \bbox[red, 2pt]{m\lt 4+\sqrt{15}}$$

解答: $$S_n=n^2+4n+3 \Rightarrow a_n =S_n-S_{n-1} = n^2+4n+3-((n-1)^2+4(n-1)+4)=2n+3 \\ \Rightarrow \mu_n={S_n\over n}=n+4+{3\over n} \Rightarrow \sigma_n^2={1\over n}\sum_{k=1}^n a_k^2 -\mu_n^2 ={1\over n}\sum_{k=1}^n (2k+3)^2 -\mu_n^2 \\={2\over 3}(n+1)(2n+1)+6(n+1)+9-(n+4+{3\over n})^2 \Rightarrow n^2係數={1\over 3} \\ \Rightarrow \mu_n^2的n^2係數=1 \Rightarrow \lim_{n\to \infty}{\sigma_n^2\over \mu_n^2}={1/3\over 1} =\bbox[red, 2pt]{1\over 3}$$

解答: $$假設\langle a_n \rangle 公差為d_a \Rightarrow T_k= {1\over 2}(2a_1+(k-1)d_a)k =(a_1-{1\over 2}d_a)k+{1\over 2}d_ak^2 \\ \Rightarrow \sum_{k=1}^n T_k= (a_1-{1\over 2}d_a)\sum_{k=1}^n k+ {1\over 2}d_a\sum_{k=1}^nk^2 ={1\over 2}(a_1-{1\over 2}d_a)n(n+1) +{1\over 12}d_an(n+1)(2n+1) \\={1\over 2}n(n+1)a_1+ \left({1 \over 12}n(n+1)(2n+1)-{1\over 4}n(n+1) \right)d_a ={1\over 2}n(n+1)a_1+ {1\over 6}n(n-1)(n+1)d_a\\ 同理可得\sum_{k=1}^n S_k= {1\over 2}n(n+1)b_1+ {1\over 6}n(n-1)(n+1)d_b \\ \Rightarrow { \sum_{k=1}^n T_k\over \sum_{k=1}^n S_k} ={{1\over 2}a_1+{1\over 6}(n-1)d_a \over {1\over 2} b_1+{1\over 6}(n-1)d_b} ={3a_1+(n-1)d_a\over 3b_1+(n-1)d_b} ={2n+3\over 3n+4} \Rightarrow \cases{d_a=2 \\ d_b=3} \\ \Rightarrow {2n+3a_1-2\over 3n+3b_1-3}={2n+3\over 3n+4} \Rightarrow \cases{a_1= 5/3\\ b_1=7/3} \Rightarrow {a_7 \over b_{10}} ={a_1+ 6d_a\over b_1+9d_b} ={5/3+12\over 7/3+27} = \bbox[red, 2pt]{41\over 88}$$

解答:

$$假設\cases{\Gamma_1: x^2+y^2=1\\ \Gamma_2: (x-1)^2+y^2=1\\ \Gamma_3: x^2+(y-1)^2=1\\ \Gamma_4: (x-1)^2+ (y-1)^2=1} \Rightarrow \cases{P=\Gamma_2 \cap \Gamma_4 =(1-\displaystyle {\sqrt 3\over 2},{1\over 2}) \\Q=\Gamma_1\cap \Gamma_2 =(\displaystyle {1\over 2},1-{\sqrt 3\over 2})} \\ \Rightarrow \cases{S= 2\int_0^{1 /2} 1-\sqrt{1-x^2}\,dx =1-{\pi\over 6}-{\sqrt 3\over 4} \\T= \int_{1/2}^{ \sqrt3/2} (\sqrt{1-x^2}-{1\over 2})\,dx ={1\over 4}-{\sqrt 3\over 4}+{\pi\over 12}} \Rightarrow 4S+4T= \bbox[red, 2pt]{5-2\sqrt 3-{\pi\over 3}}\\ 說明:r=1 \Rightarrow (\overline{PA}-1)(\overline{PB}-1)(\overline{PC}-1)(\overline{PD}-1) \ge 0 簡寫成\alpha\beta \gamma \delta \ge 0 \\ \Rightarrow 四項\alpha\beta \gamma \delta中,\cases{ 四項皆\le 0, 也就是T區域\\ 其中二項\le 0 也就是S區域} \Rightarrow 欲求之機率=4S+4T的面積$$