114年基隆女中教甄-數學詳解

 國立基隆女中 114 學年第 1 次教師甄試 

一、填充題(每題 7 分，共 70 分)

解答:$$\lim_{n\to \infty}{5\over n^2}\left[ \sqrt{4n^2-(2\times 1^2)} +\sqrt{4n^2-(2\times 2^2)} +\cdots +\sqrt{4n^2-(2\times n^2)}  \right] \\= \lim_{n\to \infty} \sum_{k=1}^n {5\over n^2} \sqrt{4n^2-(2\times k^2)} = \lim_{n\to \infty} \sum_{k=1}^n {5\over n}\sqrt{4-(2\times (k/n)^2)} =\int_0^1 5\sqrt{4-2x^2}\,dx \\=\left. \left[ {5\over \sqrt 2}(x\sqrt{2-x^2}+2 \sin^{-1}{x\over \sqrt 2}) \right] \right|_0^1 = \bbox[red, 2pt]{5\sqrt 2(\pi +2) \over 4}$$

解答:$$f(x)={2x^2+4x-3\over x^2+1} =2+{4x-5\over x^2+1}\\ 取g(x)={4x-5\over x^2+1} \Rightarrow g'(x)={-4x^2+10x+4\over (x^2+1)^2} =0 \Rightarrow 2x^2-5x-2=0 \Rightarrow x= {5\pm \sqrt{41} \over 4} \\ \Rightarrow f((5+\sqrt{41})/5) =2+{8\sqrt{41}\over 41+5\sqrt{41}} =\bbox[red, 2pt]{\sqrt{41}-1\over 2}  $$

解答:$$\sqrt m+\sqrt n=\sqrt{2783} =11\sqrt {23} =a\sqrt{23}+b\sqrt{23} \Rightarrow a+b=11, a,b \in \mathbb N \\ \Rightarrow (a-1)+(b-1)=9\Rightarrow 共有H^2_{9} =C^{10}_9=\bbox[red, 2pt]{10}組$$

解答:$$\cos \angle AOB= {\overrightarrow{OA} \cdot \overrightarrow{OB} \over |\overrightarrow{OA}| |\overrightarrow{OB} |} ={1\over \sqrt{13} \cdot 5} ={13+5^2-\overline{AB}^2 \over 10\sqrt{13}} \Rightarrow \overline{AB} =6 \\ \cos \angle BOC= {\overrightarrow{OC} \cdot \overrightarrow{OB} \over |\overrightarrow{OC}| |\overrightarrow{OB} |} ={-11\over 25} ={50-\overline{BC}^2\over 50} \Rightarrow \overline{BC}=6\sqrt 2\\ \cos \angle AOC ={\overrightarrow{OA} \cdot \overrightarrow{OC} \over |\overrightarrow{OA}| |\overrightarrow{OC} |} ={1\over 5\sqrt{13}} ={38-\overline{AC}^2\over 10\sqrt{13}} \Rightarrow \overline{AC}=6\\ \Rightarrow \overline{BC}^2=\overline{AB}^2+\overline{AC}^2 \Rightarrow \angle BAC=90^\circ \Rightarrow a\triangle ABC={1\over 2} \cdot 6\cdot 6=18\\ 假設\cases{A(0,0,0) \\C(3\sqrt 2,3\sqrt 2,0) \\ B(-3\sqrt 2, 3\sqrt 2,0) \\ O(0,a,b)} \Rightarrow \cases{\overline{OA}^2=a^2+b^2=13\\ \overline{OB}^2 =18+(a-3\sqrt 2)^2+b^2=25} \Rightarrow \cases{a=2\sqrt 2\\ b=\sqrt 5} \\ \Rightarrow 四面體體積={1\over 3}\cdot 18\cdot \sqrt 5 = \bbox[red, 2pt]{6\sqrt 5}$$

解答:$$利用f(x)=x-\lfloor {x\over 3}\rfloor, 將x=525迭代,即f(\cdots f(f(535))\cdots), 可得535 \to 350\to 234 \to \cdots 4\to 3\\ 再倒推回去(想不出更好的方法), 可求得最後第三個人是位在475$$

解答:$$有兩個「大」相鄰:{6!\over 2!2!} -{5!\over 2!2!}=150種排列法\\有兩個「江」相鄰:{6!\over 3!2!} =60種排列法\\有兩個「海」相鄰:{6!\over 3!2!} =60種排列法\\有兩個「大」相鄰且有兩個「江」相鄰:{5!\over 2!}-{4!\over 2}=48 \\有兩個「大」相鄰且有兩個「海」相鄰:{5!\over 2!}-{4!\over 2}=48 \\有兩個「海」相鄰且有兩個「江」相鄰:{5!\over 3!} =20\\ 兩個「大」相鄰且兩個「江」相鄰且兩個「海」相鄰:4!-3!=18 \\ {7!\over 3!2!2!}-(150+60+60)+(48+48+20)-18= \bbox[red, 2pt]{38}$$

解答:$$z=\cos \theta+i\sin \theta \Rightarrow z^2-2z+5=\cos 2\theta+i\sin 2\theta-2(\cos \theta+i\sin \theta)+5 \\ =(\cos 2\theta-2\cos \theta+5)+ i(\sin 2\theta-2\sin \theta) \\\Rightarrow |z^2-2z+5|=|(\cos 2\theta-2\cos \theta+5)+ i(\sin 2\theta-2\sin \theta)| \\ =\sqrt{(\cos 2\theta-2\cos \theta+5)^2+ (\sin 2\theta-2\sin \theta)^2} =\sqrt{20\cos^2\theta-24\cos \theta+20} \\ 令g(\theta) = 20\cos^2\theta-24\cos \theta+20 \Rightarrow g'(\theta) =0 \Rightarrow \cos\theta={3\over 5} \Rightarrow \sin \theta=\pm {4\over 5} \Rightarrow z_0={3\over 5}\pm {4\over 5}i \\ \Rightarrow m=|z_0^2-2z_0+5| =\sqrt{20\cdot {9\over 25}-24\cdot {3\over 5}+20} =\sqrt{64\over 5} ={8\sqrt 5\over 5} \\\Rightarrow (z_0,m) = \bbox[red, 2pt]{\left( {3\over 5}\pm {4\over 5}i,{8\sqrt 5\over 5}\right)}$$

解答:$$代公式:p^2+6p+27p^2+4p^3-27\cdot {9\over 4}p^2 \gt 0\\ \Rightarrow 16p^3-131p^2+24p=p(16p-3)(p-8)\gt 0 \Rightarrow \bbox[red, 2pt]{p\gt 8 或0\lt p\lt {3\over 16}} \\ 公式來源:《\href{https://math.ntnu.edu.tw/~maco/macobook/arith/arith35.pdf}{按這裡}》$$

解答:$$\prod_{k=1}^n2^{\lfloor \log_{45}k \rfloor} =2^{\sum_{k=1}^n \lfloor \log_{45}k \rfloor} =1024^{1024} =2^{10240} \Rightarrow \sum_{k=1}^n \lfloor \log_{45}k \rfloor=10240\\ 由於 \lfloor \log_{45}k \rfloor =\begin{cases} 0& 1\le k\le 44\\ 1& 45\le k\le45^2-1=2024\\ 2& 2025\le k\le 45^3-1=91124\end{cases} \Rightarrow 10240=(2024-45+1)+ 2\times 4130 \\ \Rightarrow n=2025+4130-1=\bbox[red, 2pt]{6154}$$

解答:$$此題\bbox[cyan,2pt]{有疑義}, 利用電腦計算可求得比5大之值，應無最大值$$

二、計算證明題(每題 10 分，共 30 分)

解答:$$Q(a,b)={1\over n}\sum_{i=1}^n (y_i-(a+bx_i))^2 \Rightarrow \cases{Q_a =-{2\over n} \sum_{i=1}^n (y_i-a-bx_i) \\Q_b=-{2\over n} \sum_{i=1}^n (x_iy_i-ax_i-bx_i^2)}\\ Q_a=0 \Rightarrow {1\over n}\sum_{i=1}^n (y_i-a-bx_i)=0 \Rightarrow u_y-a-bu_x=0 \Rightarrow a=u_y-bu_x\\ Q_b=0 \Rightarrow {1\over n}\sum_{i=1}^n  (x_iy_i-ax_i-bx_i^2)=0 \Rightarrow {1\over n}\sum_{i=1}^nx_iy_i-au_x-{1\over n}\sum_{i=1}^n bx_i^2 =0 \\ \Rightarrow {1\over n}\sum_{i=1}^nx_iy_i-(u_y-bu_x)u_x-{1\over n}\sum_{i=1}^n bx_i^2 =0 \Rightarrow b={{1\over n}\sum_{i=1}^n (x_iy_i)-u_xu_y\over {1\over n}\sum_{i=1}^n x_i^2-u_x^2} \\ \Rightarrow b={\sum_{i=1}^n(x_i-u_x)(y_i-u_y)\over \sum_{i=1}^n(x_i-u_x)^2} ={\sum_{i=1}^n(x_i-u_x)(y_i-u_y)\over \sigma_x^2}\\ \textbf{(1) }迴歸直線y=a+bx=u_y-bu_x+bx=b(x-u_x)+u_y \Rightarrow 通過(u_x,u_y)\quad \bbox[red, 2pt]{QED}\\\textbf{(2) } b={\sum_{i=1}^n(x_i-u_x)(y_i-u_y)\over \sigma_x^2} ={\sum_{i=1}^n(x_i-u_x)(y_i-u_y)\over \sigma_x \cdot \sigma_y} \cdot {\sigma_y\over \sigma_x} =r\cdot {\sigma_y\over \sigma_x} \\\qquad \Rightarrow b=r\cdot {\sigma_y\over \sigma_x}\quad \bbox[red, 2pt]{QED}$$

解答:$$過P(1,1)之直線L: y=m(x-1)+1 \Rightarrow 弦\overline{A_1A_4}兩頂點\cases{A_1(x_1,mx_1-m+1) \\A_4 (x_4,mx_4-m+1)} \\ \Rightarrow \cases{\overline{PA_1} =\sqrt{(x_1-1)^2+(mx_1-m)^2} =\sqrt{(m^2+1) (x_1-1)^2}= |x_1-1| \sqrt{m^2+1}\\PA_4 =\sqrt{(x_4-1)^2+ (mx_4-m)^2} =|x_4-1|\sqrt{m^2+1}} \\ \Rightarrow \overline{PA_1}\cdot \overline{PA_2}=|x_1-1||x_4-1| {(m^2+1)} =|1-(x_1+x_4)+x_1x_4 |{(m^2+1)} \cdots(1) \\ 將y=m(x-1)+1代入\Gamma \Rightarrow 4x^2+9(mx-m+1)^2=36 \\ \Rightarrow (9m^2+4)x^2+ 18m(1-m)x+9(1-m)^2-36=0 \Rightarrow \cases{兩根之和=x_1+x_4={18m(m-1)\over (9m^2+4)} \\ 兩根之積=x_1x_4={9(1-m)^2-36\over (9m^2+4)}} \\ 代入(1) \Rightarrow \overline{PA_1}\cdot \overline{PA_2}= \left| 1-{18m(m-1)\over (9m^2+4)} +{9(1-m)^2-36\over (9m^2+4)}\right| (m^2+1) ={23\over 9m^2+4}\cdot(m^2+1) \\ \Rightarrow {1\over \overline{PA_1}\cdot \overline{PA_2}}= {9(m^2+1)-5\over 23(m^2+1)} ={9\over 23}-{5\over 23}\cdot {1\over m^2+1}  ={9\over 23}-{5\over 23}\cos^2 \theta \\ \Rightarrow {1\over \overline{PA_1} \cdot \overline{PA_2}} +{1\over \overline{PA_2}\cdot \overline{PA_5}} + {1\over \overline{PA_3}\cdot \overline{PA_6}} ={9\over 23}\cdot 3 -{5\over 23}(\cos^2 \theta +\cos^2(\theta+{\pi\over 3}) + \cos^2(\theta+{2\pi\over 3})) \\= {27\over 23}-{5\over 23}\cdot {3\over 2} = \bbox[red, 2pt]{39\over 46}$$

解答:$$$$

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解題僅供參考，其他教甄試題及詳解