113年南湖高中教甄-數學詳解

 臺北市立南湖高級中學 113 學年度第 1 次正式教師甄選

一、填充題 (14 題，每題 5 分)

解答:$$\cases{L_1:3x-y=0\\ L_2:x+y=0} \Rightarrow \cases{\vec u=(3,-1)\\ \vec v=(1,1)} \Rightarrow \cos \theta= {\vec u\cdot \vec v\over |\vec u||\vec v|} ={1\over \sqrt 5} \Rightarrow \sin \theta={2\over \sqrt 5} \\ 假設平行四邊形邊長為\alpha,\beta \Rightarrow \alpha\beta \sin \theta= \sqrt 5 \Rightarrow \alpha \beta={5\over 2}\\ \Rightarrow 周長=2(\alpha+\beta)\ge 4\sqrt{\alpha \beta}=4\sqrt{5\over 2} = \bbox[red, 2pt]{2\sqrt{10}}$$

解答:$$令u=2^x \Rightarrow u^2=4^x \Rightarrow (u-4)^3+(u^2-8)^3=(u^2+u-12)^3 \\ \Rightarrow (u^2+u-12)^3-3(u-4)(u^2-8)(u^2+u-12)=(u^2+u-12)^3 \\ \Rightarrow (u-4)(u^2-8)(u^2+u-12)=0 \Rightarrow (u-4)(u^2-8)(u+4)(u-3)=0 \\ \Rightarrow (u-4) (u-2\sqrt 2)(u-3)=0 \Rightarrow \cases{2^x=4 =2^2\\ 2^x=2\sqrt 2=2^{3/2}\\ 2^x=3} \Rightarrow \cases{x=2 \\x=3/2\\ x=\log_2 3} \\ \Rightarrow 2+{3\over 2}+\log_2 3= \bbox[red, 2pt]{{7\over 2}+\log_2 3}$$

解答:$$a_n a_{n+1} a_{n+2}a_{n+3}= a_n+a_{n+1}+ a_{n+2}+ a_{n+3} \Rightarrow a_{n+3}={a_n+a_{n+1}+ a_{n+2} \over a_n a_{n+1} a_{n+2}-1} \\\Rightarrow a_4={1+2+1\over 1\cdot 2\cdot 1-1} =4 \Rightarrow a_5={2+ 1+ 4\over 2\cdot 1\cdot 4-1}=1 \Rightarrow a_6=2 \Rightarrow a_7=1 \\ \Rightarrow \langle a_n \rangle =1,2,1,4,1,2,1,4,\dots 循環數4 \Rightarrow \sum_{k=1}^{4\times 28+1} a_k= (1+2+1+4)\times 28+1= \bbox[red, 2pt]{225}$$

解答:$$若a+b=30^\circ \Rightarrow \tan 30^\circ ={1\over \sqrt 3}={\tan a^\circ+ \tan b^\circ \over 1-\tan a^\circ \tan b^\circ}\\ \Rightarrow \tan a^\circ + \tan b^\circ +\sqrt 3(\tan a^\circ+ \tan b^\circ)= 1 \Rightarrow (\tan a^\circ+ \sqrt 3)(\tan b^\circ+ \sqrt 3) \\=\tan a^\circ \tan b^\circ+ \sqrt 3(\tan a^\circ+\tan b^\circ) +3=1+3=4\\ \Rightarrow (\tan 15^\circ+ \sqrt 3)(\tan 15^\circ+ \sqrt 3)=4 \Rightarrow \tan 15^\circ+ \sqrt 3 =\sqrt 4=2\\ 因此\log_8 (\tan 1^\circ+\sqrt 3)(\tan 2^\circ+ \sqrt 3)\cdots (\tan 29^\circ+ \sqrt 3) \\=\log_8 [(\tan 1^\circ +\sqrt 3) (\tan 29^\circ+\sqrt 3)][(\tan 2^\circ+\sqrt 3) (\tan 28^\circ+\sqrt 3)] \\\qquad \cdots [(\tan 14^\circ+\sqrt 3) (\tan 16^\circ+\sqrt 3)] (\tan 15^\circ +\sqrt 3) \\=\log_8 (4^{14}\cdot 2) ={\log_2 2^{29}\over \log_2 8 }=\bbox[red, 2pt]{29\over 3}$$

解答:$$7^{12} = 1\text{ mod }13 \Rightarrow 7^{2024}=\left( 7^{12}\right)^{168} \cdot 7^8 =7^8 \text{ mod }13 =3 \text{ mod }13 \Rightarrow 餘數為\bbox[red, 2pt]3$$

解答:$$(x-1)f(x+1)=(x+2)f(x) \Rightarrow \cases{x=-2 \Rightarrow f(-1)=0\\ x=-1 \Rightarrow f(0)=0 \\ x=0 \Rightarrow f(1)=0}  \Rightarrow f(x)= (x-1)x(x+1) p(x) \\ \Rightarrow (x-1)x(x+1)(x+2) p(x+1) =(x+2)(x-1)x(x+1)p(x) \\ \Rightarrow p(x+1)=p(x) \Rightarrow p(x)=常數,又f(x)首項係數為1 \Rightarrow p(x)=1\\ \Rightarrow f(x)=(x-1)x(x+1)= \bbox[red, 2pt]{x^3-x}$$

解答:$$f'(x)={-2\over \sqrt{24-4x}} +{5\over 2\sqrt{5x+15}} =0 \Rightarrow 25(24-4x)=15(5x+15) \Rightarrow x=2\\ \Rightarrow \cases{f(2)= \sqrt{16}+ \sqrt{25}=9\\ f(-3)= 6\\ f(6)= \sqrt{45} =3\sqrt 5} \Rightarrow (M,m)= \bbox[red, 2pt]{(9,6)}$$

解答:$$\log(10^x+200) \gt {x\over 2}+1+\log 3 \Rightarrow 10^x+200 \gt 10^{x/2}\cdot 10\cdot 3 \Rightarrow (10^{x/2})^2-30\cdot 10^{x/2}+200 \gt 0 \\ \Rightarrow (10^{x/2}-10)(10^{x/2}-20)\gt 0 \Rightarrow \cases{10^{x/2} \gt 20  \Rightarrow {x/2} \gt 1+\log 2 \Rightarrow x\gt 2+2\log 2\\ 10^{x/2} \lt 10 \Rightarrow x/2\lt 1 \Rightarrow x\lt 2} \\ \Rightarrow \bbox[red, 2pt]{x\lt 2或x\gt 2+2\log 2}$$

解答:$${x^2\over 25} +{y^2\over 16}=1 \Rightarrow \cases{a=5\\ b=4} \Rightarrow c=3 \Rightarrow \cases{F_1(-3,0)\\ F_2(3,0)} \\ \Rightarrow F_2對稱於L的對稱點F_2'(10,7) \Rightarrow \overline{PF_1}+ \overline{PF_2} 最小值=\overline{F_1F_2'} =\sqrt{13^2+7^2} =\bbox[red, 2pt]{\sqrt{218}},P\in L$$

解答:$$\cases{A(2,0,3)\\ B(-1,0,6)\\ C(4,0,3)\\ D(3,-2,2)} \Rightarrow \cases{L_1=\overleftrightarrow{AB}: {x-2\over -3}={z-3\over 3},y=0\\ L_2=\overleftrightarrow{CD}: {x-4\over -1} ={y\over -2} ={z-3\over -1}} \Rightarrow \cases{L_1方向向量\vec u=(-3,0,3)\\ L_2方向向量\vec v=(-1,-2,-1) \\ L_1,L_2歪斜} \\ \Rightarrow \vec n=\vec u\times \vec v=(6,-6,6) \Rightarrow \overrightarrow{AC}=(2,0,0)在\vec n的投影長d={12\over 6\sqrt 3} ={2\over \sqrt 3} \\ \Rightarrow \triangle PAB面積={1\over 2}\overline{AB} \cdot d ={1\over 2} \cdot 3\sqrt{2} \cdot {2\over \sqrt 3}=\bbox[red, 2pt]{\sqrt 6}$$

解答:$$\angle ACB=90^\circ \Rightarrow 取\cases{C(0,0,0)\\ A(12,0,0)\\ B(0,b,0)\\ P(6,b/2,6)}, 又Q\in y軸 \Rightarrow Q(0,k,0), k\in \mathbb R \\ \Rightarrow \overline{PQ}=\sqrt{72+(b/2-k)^2}\Rightarrow 當k=b/2時, \overline{PQ}最小值=\sqrt{72} =\bbox[red, 2pt]{6\sqrt 2}$$

解答:$$假設甲隊獲勝，且甲隊第1名選手贏x_1場、第2名選手贏x_2場、\dots、第6名選手贏x_6場，\\ 即x_1+ x_2+ \cdots+ x_6=6 有H^6_6=462種賽程; 同理,乙隊獲勝也有462種賽程，共有\bbox[red, 2pt]{924}種賽程$$

解答:$$P(n)={2\over 3}P(n-1)+{1\over 3}(1-P(n-1)) ={1\over 3} P(n-1)+{1\over 3} = {1\over 3}({1\over 3}P(n-2)+{1\over 3})+{1\over 3} \\={1\over 3^2}P(n-2)+{1\over 3^2}+{1\over 3} =\cdot s={1\over 3^{n-1}} P(1)+{1\over 3^{n-1}} +{1\over 3^{n-2}}+\cdots +{1\over 3} \\={1\over 3^n}+{1\over 3^{n-2}}+\cdots +{1\over 3} ={1\over 2}-{1\over 2\cdot 3^n} \Rightarrow (a,b)= \bbox[red, 2pt]{({1\over 2},-{1\over 2})}$$

解答:$$此題相當於f(L)=L',其中L\bot L'且L\cap L'=(3,1)\\ 因此\begin{bmatrix}x'-3  \\y'-1 \end{bmatrix} = k\begin{bmatrix}\cos (\pi/2) & -\sin (\pi/2) \\\sin (\pi/2) & \cos (\pi/2) \end{bmatrix} \begin{bmatrix}x-3  \\y-1 \end{bmatrix} =k\begin{bmatrix}1-y  \\x-3 \end{bmatrix} \\ \Rightarrow \begin{bmatrix}x'  \\y' \end{bmatrix} = \begin{bmatrix}k-ky+3  \\kx-3k+1 \end{bmatrix} = \begin{bmatrix}2 & a \\0 & 1 \end{bmatrix} \begin{bmatrix}x \\y \end{bmatrix} =\begin{bmatrix}2x+ay  \\y \end{bmatrix} \\ \Rightarrow \cases{2x+(a+k)y=k+3 \\ kx-y=3k-1} \Rightarrow {2\over k}={a+k\over -1} ={k+3\over 3k-1} \Rightarrow 2(3k-1)=k(k+3)\\ \Rightarrow k^2-3k+2=0 \Rightarrow \cases{k=1 \Rightarrow 2={a+1\over-1} \Rightarrow a= -3\\k=2 \Rightarrow 1={a+2\over -1} \Rightarrow a=-3} \Rightarrow a=-3 \\ \Rightarrow \begin{bmatrix}x'  \\y' \end{bmatrix} =\begin{bmatrix}2x-3y  \\y \end{bmatrix}\\ 由於L: y-1=m(x-3) \Rightarrow L':y'-1=-{1\over m}(x'-3) \Rightarrow y-1=-{1\over m}(2x-3y-3) \\ \Rightarrow -m(y-1)=2x-3y-3 \Rightarrow (3-m)y+m=2x-3 \Rightarrow (3-m)(m(x-3)+1)+m=2x-3 \\ \Rightarrow (3-m)m=2 \Rightarrow m^2-3m+2=0 \Rightarrow m=2,1 \Rightarrow (a,m)=\bbox[red, 2pt]{(-3,1),(-3,2)}$$

============== END =============

解題僅供參考，教甄歷年試題及詳解