113年鳳山高中教甄-數學詳解

113 年國立鳳山高級中學教師甄選初試筆試試題【數學科】

一、 填充題 配分：每題 6 分，共 72 分。全對才給分。 答題： 請標註題號並依題序作答

解答: $$\sqrt{2023}=17\sqrt 7=\sqrt m+\sqrt n \Rightarrow \cases{m =a\sqrt 7 \\ n=b\sqrt 7 \\ a+b=17,a,b\in \mathbb N} \\ \Rightarrow (a,b)=(1,16), (2,15),\dots, (15,2),(16,1), 共\bbox[red, 2pt]{16}組解$$

解答: $$\left|\det(4\overrightarrow{OA}; 4\overrightarrow{OB}; 4\overrightarrow{OC})\right| -\left|\det(2\overrightarrow{OA}; 2\overrightarrow{OB}; 2\overrightarrow{OC})\right| =(4^3-2^3)\times 20=  \bbox[red, 2pt]{1120}$$

解答: $$A=\begin{bmatrix}0 & 0 & -2 \\1 & 2 & 1 \\ 1 & 0 & 3 \end{bmatrix} = \begin{bmatrix}-2 & 0 & -1 \\ 1 & 1 & 0 \\1 & 0 & 1 \end{bmatrix} \begin{bmatrix}1 & 0 & 0 \\0 & 2 & 0 \\0 & 0 & 2 \end{bmatrix} \begin{bmatrix}-1 & 0 & -1 \\1 & 1 & 1 \\1 & 0 & 2 \end{bmatrix} \\ \Rightarrow A^n= \begin{bmatrix}-2 & 0 & -1 \\ 1 & 1 & 0 \\1 & 0 & 1 \end{bmatrix} \begin{bmatrix}1 & 0 & 0 \\0 & 2^n & 0 \\0 & 0 & 2^n \end{bmatrix} \begin{bmatrix}-1 & 0 & -1 \\1 & 1 & 1 \\1 & 0 & 2 \end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix}-2^n+2 & 0 & -2^{n+1}+2 \\ 2^n-1 & 2^n & 2^n-1 \\2^n-1 & 0 & 2^{n+1}-1 \end{bmatrix}}$$

解答: $$地球是單位球,地球上每個點座標剛好就是單位向量\\\cases{A(1,0,0)\\ B(1/2,1/2,\sqrt 2/2) } \Rightarrow C=\overline{AB}中點=(3/4,1/4,\sqrt 2/4) \Rightarrow {\overrightarrow{OC} \over | \overrightarrow{OC}| } ={2\over \sqrt 3}({3\over 4},{1\over 4}, {\sqrt 2\over 4}) \\=\bbox[red, 2pt]{\left( {\sqrt 3\over 2}, {\sqrt 3\over 6}, {\sqrt 6\over 6} \right)}$$

解答: $$依題意\cases{a=a_1, a_n=a+(n-1)d\\ b=b_1, b_n=br^{n-1}} \Rightarrow \cases{a+b=1 \\ a+d+br=4\\ a+2d+br^2 =15\\ a+3d+br^3=2} \Rightarrow \cases{d+b(r-1)=4-1=3\\ d+br(r-1)=15-4=11\\ d+br^2(r-1)=2-15=-13} \\ \Rightarrow \cases{b(r-1)^2=11-3=8\\ br(r-1)^2=-13-11=-24} \Rightarrow {b(r-1)^2\over br(r-1)^2} ={8\over -24} \Rightarrow {1\over r}=-{1\over 3} \Rightarrow r=-3\\ \Rightarrow b(-3-1)^2=8 \Rightarrow b={1\over 2} \Rightarrow a=1-{1\over 2}={1\over 2} \Rightarrow d=3-b(r-1)= 5 \\ \Rightarrow c_6=a+5d+br^5= {1\over 2}+25+{1\over 2}(-243)= \bbox[red, 2pt]{-96}$$

解答: $$f(k)=\int_0^k \int_0^{\sqrt y} 1\,dxdy + \int_k^9 \int_{\sqrt y}^3 1\,dx dy={2\over 3}k^{3/2} +9-3k+{2\over 3}k^{3/2} =9+{4\over 3}k^{3/2}-3k \\ \Rightarrow f'(k)=2\sqrt k-3=0 \Rightarrow k={9\over 4} \Rightarrow f({9\over 4})={27\over 4} \Rightarrow \cases{\alpha=9/4\\ \beta=27/4} \Rightarrow \alpha+\beta= \bbox[red, 2pt]9$$

解答: $$\left\lfloor {x\over 2} \right\rfloor-\left\lfloor {x\over 3} \right\rfloor={x\over 7} 為一整數,因此x為7的倍數\\ 又{x\over 2}-{x\over 3}-1\lt {x\over 7} \lt {x\over 2}-{x\over 3}+1 \Rightarrow {x\over 6}-1\lt {x\over 7}\lt {x\over 6}+1 \Rightarrow -1\lt -{x\over 42}\lt 1\\ \Rightarrow -42\lt x\lt 42 \Rightarrow x=-35,-28,\dots,0,\dots, 35代回原式檢驗可得\\ x=-28,-14,-7,0,7,21 \Rightarrow \sum x=\bbox[red, 2pt]{-21}$$

解答: $$本題\bbox[red, 2pt]{送分}$$

解答: $$\cases{A(2,2)\\ B(0,8)\\ C(-16,16)} \Rightarrow \cases{\overline{AB}=2\sqrt{10}\\ \overline{BC}= 8\sqrt 5\\ \overline{CA}=2 \sqrt{130} \\重心M= (14/3,26/3)} \Rightarrow \cos B={40+320-520\over 160\sqrt 2} =-{1\over \sqrt 2} \\\Rightarrow \sin \angle B={1\over \sqrt 2} \Rightarrow \triangle ABC= {1\over 2}\cdot 2\sqrt {10} \cdot 8\sqrt 5 \cdot {1\over \sqrt 2}=40\\ \Rightarrow 重心繞x軸旋轉體積=(2\pi \cdot {26\over 3} \cdot 40 = \bbox[red, 2pt]{{2080\over 3}\pi}$$

解答: $$\begin{array}{} 連休三天&休休休營\bigcirc\bigcirc & 4\\ & 營休休休營\bigcirc & 2\\ & \bigcirc營休休休營& 2\\ & \bigcirc \bigcirc 休休休營& 4\\\hdashline 連休四天&休休休休營\bigcirc & 2\\ & 營休休休休營& 1\\ & \bigcirc 營休休休休& 2\\\hdashline 連休五天& 休休休休休營& 1\\ & 營休休休休休& 1\\\hdashline 連休六天& 休休休休休休& 1\\\hline & 合計&20 \end{array} \Rightarrow 開店數=2^6-20=\bbox[red, 2pt]{44}$$

解答:

$$\cases{原點O(0,0)\\ 圓心P(12,5)} \Rightarrow \overline{OP}與圓的交點即為所求之C且\overline{OC}=\overline{OP}-2=11\\ C與直線y=\sqrt 3x的對稱點C',與x軸的對稱點為C'',\\則\overline{C'C''}與直線y=\sqrt 3x的交點即為A,與x軸交點即為B \\ 因此\cases{\overline{AC} =\overline{AC'} \\ \overline{BC}= \overline{BC''} } \Rightarrow \triangle ABC周長= \overline{AC'} +\overline{AB}+ \overline{BC''}=\overline{C'C''}\\ 假設\cases{\angle AOC=\theta_1\\ \angle COB= \theta_2} ,直線y=\sqrt 3x與x軸的夾角為60^\circ, 即\theta_1+\theta_2 =60^\circ \\ 又\cases{C, C''對稱x軸\\ C,C'對稱直線y=\sqrt 3x} \Rightarrow \cases{\angle BOC''= \theta_1且\overline{OC''}= \overline{OC}=11\\ \angle AOC'=\theta_2 且\overline{OC'}=\overline{OC}=11 } \Rightarrow \angle C'OC''= 2(\theta_1+\theta_2)=120^\circ \\ \triangle OC'C'': \cos \angle C'OC'' ={11^2+11^2- 2\overline{C'C''}\over 2\cdot 11\cdot 11} \Rightarrow -{1\over 2}={242-\overline{C'C''}^2 \over 242} \Rightarrow \overline{C'C''}= \bbox[red, 2pt]{11\sqrt 3}$$

解答: $$條件1:所有數字都超過1:使用2,3任排,有2^4=16\\ 條件2:恰有1數字不超過2,其它三個數字都超過2,即X333任排,X=0,1,2, 有12種\\ 條件1\cap 條件2:1個2,3個3的排列,有4種\\ 因此共有 4^4-2^4-12+4= \bbox[red, 2pt]{232}$$

二、計算證明題 答題：請標註題號並依題序作答，切勿跳題回答，並完整說明。

解答: $$公式1:\tan(a+b)={\tan a+\tan b\over 1-\tan a\tan b} \Rightarrow \tan a\tan b\tan (a+b)= \tan(a+b)-\tan a-\tan b\\ 公式2: 若a+b=90^\circ \Rightarrow \tan a\cdot \tan b=1\\\begin{bmatrix}\tan 40^\circ & \tan 10^\circ &\tan 50^\circ \\ \tan 20^\circ & \tan 50^\circ &\tan 70^\circ\\\tan 10^\circ & \tan 70^\circ &\tan 80^\circ \end{bmatrix} =\begin{bmatrix}\tan 40^\circ & \tan 10^\circ &\tan (10^\circ+40^\circ) \\ \tan 20^\circ & \tan 50^\circ &\tan (20^\circ+ 50^\circ) \\\tan 10^\circ & \tan 70^\circ &\tan (10^\circ+ 70^\circ) \end{bmatrix} \\ \xrightarrow{-C_1-C_2+C_3 \to C_3} \begin{bmatrix}\tan 40^\circ & \tan 10^\circ & -\tan 10^\circ -\tan 40^\circ +\tan (10^\circ+40^\circ) \\ \tan 20^\circ & \tan 50^\circ & -\tan 20^\circ- \tan 50^\circ+ \tan (20^\circ+ 50^\circ) \\\tan 10^\circ & \tan 70^\circ & -\tan 10^\circ- \tan 70^\circ+\tan (10^\circ +70^\circ) \end{bmatrix} \\= \begin{bmatrix}\tan 40^\circ & \tan 10^\circ & \tan 10^\circ \cdot  \tan 40^\circ \cdot \tan (10^\circ+40^\circ) \\ \tan 20^\circ & \tan 50^\circ & \tan 20^\circ \cdot  \tan 50^\circ\cdot \tan (20^\circ+ 50^\circ) \\\tan 10^\circ & \tan 70^\circ & \tan 10^\circ\cdot  \tan 70^\circ \cdot \tan (10^\circ +70^\circ) \end{bmatrix} \\=  \begin{bmatrix}\tan 40^\circ & \tan 10^\circ & \tan 10^\circ   \\ \tan 20^\circ & \tan 50^\circ &   \tan 50^\circ   \\\tan 10^\circ & \tan 70^\circ &   \tan 70^\circ  \end{bmatrix} =\bbox[red, 2pt] 0$$

解答: $$x^3+y^4=z^5 \Rightarrow 2^{60n+24} +y^4=2^{60n+25} \Rightarrow y^4=2^{60n+25}-2^{60n+24}=2^{60n+24}(2-1)=2^{60n+24}\\ \Rightarrow \cases{\log_2 x=20n+8\\ \log_2y =15n+6\\ \log_2 z=12n+5}\\ \Rightarrow 6\log_2 x-32\log_2 y+30\log_2 z=120n+48-480n-192+360n+150 =\bbox[red, 2pt] 6$$

解答: $$f(x)=\sqrt{x-27}+\sqrt{40-x} +\sqrt x \Rightarrow f'(x)={1\over 2\sqrt{x-27}}- {1 \over 2\sqrt {40-x}}+ {1\over 2\sqrt{x}}\\ f'(x)=0 \Rightarrow {1\over \sqrt{40- x}} ={1\over \sqrt{x-27}} +{1\over \sqrt{x}} \Rightarrow x=36\; (註:{1\over 2}={1\over 3}+{1\over 6})\\ f(36)=\sqrt{9}+ \sqrt{4}+ \sqrt{36}=3+2+6=11,即最大值為\bbox[red,2pt]{11}$$
 

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解題僅供參考，其他歷年試題及詳解