115年松山家商教甄2-數學詳解

 臺北市立松山高級商業家事職業學校115學年度第2次教師甄選

壹、填充題（20 題，每題 5 分，合計 100 分）

解答:$$假設扇形\cases{半徑r\\ 弧長s} \Rightarrow 2r+s=20 \Rightarrow 算幾不等式:{2r+s} \ge 2\sqrt{2rs} \Rightarrow 10\ge \sqrt{2rs} \\ \Rightarrow rs\le 50\Rightarrow 面積{1\over 2}rs\le \bbox[red, 2pt]{25}$$

解答:$$x\ge {1\over 2} \Rightarrow x^2-2x+1-1=x^2-2x=x(x-2)\lt 0 \Rightarrow 0\lt x\lt 2 \Rightarrow {1\over 2}\le x\lt 2\\ x\lt {1\over 2} \Rightarrow x^2+2x-1-1=x^2+2x-2 =(x+1)^2-3\lt 0 \Rightarrow -\sqrt 3-1\lt x\lt \sqrt 3-1 \\兩條件取聯集\Rightarrow \bbox[red, 2pt]{-\sqrt 3-1\lt x\lt 2}$$

解答:$$y={x^2-5x+6\over x^2+5x+4} = {(x-3)(x-2) \over (x+1)(x+4)}\\ y\gt 0 \Rightarrow (x-3)(x-2)(x+1)(x+4)\gt 0 \Rightarrow x\gt 3, -1\lt x\lt 2, x\lt -4\\ y\lt 1 \Rightarrow {x^2-5x+6\over x^2+5x+4} -1\lt 0 \Rightarrow {-10x+2\over x^2+5x+4}\lt 0 \Rightarrow (-10x+2)(x+4)(x+1)\lt 0 \\ \qquad \Rightarrow 2(5x-1)(x+1)(x+4  )\gt 0 \Rightarrow x\gt {1\over 5}, -4\lt x\lt -1 \\ 兩條件\cases{x\gt 3, -1\lt x\lt 2, x\lt -4\\ x\gt {1\over 5}, -4\lt x\lt -1} \;取交集 \Rightarrow  \bbox[red, 2pt]{{1\over 5}\lt x\lt 2, x\gt 3}$$

解答:$$\begin{vmatrix} a^2+1 & ab & ac \\ ab & b^2+1 & bc \\ ac & bc & c^2+1 \end{vmatrix}=  (a^2+1) \begin{vmatrix} b^2+1 & bc \\ bc & c^2+1 \end{vmatrix} - ab \begin{vmatrix} ab & bc \\ ac & c^2+1 \end{vmatrix} + ac \begin{vmatrix} ab & b^2+1 \\ ac & bc \end{vmatrix}\\=(a^2b^2+ a^2c^2+ a^2+b^2+c^2+1)-a^2b^2 -a^2c^2= \bbox[red, 2pt]{a^2+b^2+c^2+1}$$

解答:

$$拋物線\Gamma: y^2=4x \Rightarrow c=1 \Rightarrow 焦點F(1,0) \\ A(-2,6)沿著水平方向(y=6)前進遇到\Gamma上一點P\Rightarrow P(k,6) 代入\Gamma \Rightarrow k=9 \Rightarrow P(9,6)\\  拋物線光學性質:平行於對稱軸(x 軸)的入射光線，經反射後必通過焦點F \\ \Rightarrow 反射光線沿著向量\overrightarrow{PF}=(-8,-6)前進\Rightarrow |\overrightarrow{PF}|=10 \Rightarrow \vec e={\overrightarrow{PF} \over |\overrightarrow{PF}|} = \left( -{4\over 5},-{3\over 5} \right)\\ \Rightarrow B=P-5\vec e=(9,6)-(-4,-3)= \bbox[red, 2pt]{(13,9)}$$

解答:$$x^5=1 \Rightarrow 1-x^5=0 \Rightarrow (1-x) (1+x+x^2+ x^3+x^4)=0 \\ \Rightarrow f(x)=1+x+x^2+ x^3+x^4 =(x-\omega_1) (x-\omega_2) (x-\omega_3) (x-\omega_4) \\ \Rightarrow f(3)={1-3^5\over 1-3}= {242\over 2}= \bbox[red, 2pt]{121}$$

解答:$$\cases{A(4,-4,6) \\B(2,0,2) \\C(4,-1,3)} \Rightarrow \cases{\overrightarrow{AC}=(0,3,-3) \\ \overrightarrow{AB} =(-2,4,-4)} \Rightarrow \overrightarrow{AH} =  \left(\frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AC}|^2}\right)\overrightarrow{AC} =(0,4,-4) \\ \Rightarrow H= A+ \overrightarrow{AH} =(4,-4,6)+(0,4,-4)= \bbox[red, 2pt]{(4,0,2)}$$

解答:$$\cases{A(3,0,0) \\ B(0,4,0) \\C(0,0,k)} \Rightarrow E:{x\over 3}+{y\over 4}+{z\over k}=1 \Rightarrow E的法向量\vec n_1= \left( {1\over 3},{1\over 4},{1\over k} \right)\\ xy平面:z=0 \Rightarrow 法向量\vec n_2=(0,0,1) \Rightarrow \cos 45^\circ = {\vec n_1\cdot \vec n_2\over |\vec n_1|| \vec n_2|} \Rightarrow {1\over 2}={1/k^2\over 1/9+1/16+1/k^2} \\ \Rightarrow k^2={144\over 25} \Rightarrow k=\bbox[red, 2pt]{12\over 5}$$

解答:$$L= \overleftrightarrow{PQ} 為兩平面\cases{x+y+z=1\\ 2x+2y+z=1}之交集 \Rightarrow L:(t,-t,1),t\in \mathbb R\\ L在x^2+y^2+z^2=4上\Rightarrow t^2+(-t)^2+1=4 \Rightarrow t=\pm \sqrt{3\over 2}\\ \Rightarrow \overline{PQ}=  \left( \sqrt{3/2}-(-\sqrt{3/2}) \right)\cdot \sqrt{1^2+(-1)^2+0} =2\sqrt{3\over 2}\cdot \sqrt 2= \bbox[red, 2pt]{2\sqrt 3}$$

解答:$$15^n=3^n\times 5^n \Rightarrow  a_n = (1 + 3 + 3^2 + \dots + 3^n)(1 + 5 + 5^2 + \dots + 5^n)  = \left(\frac{3^{n+1}-1}{3-1}\right) \left(\frac{5^{n+1}-1}{5-1}\right) \\= \frac{(3^{n+1}-1)(5^{n+1}-1)}{8} \Rightarrow  \lim_{n \to \infty} \frac{a_n}{15^n} = \lim_{n \to \infty} \frac{\frac{(3^{n+1}-1)(5^{n+1}-1)}{8}}{3^n \cdot 5^n}  = \frac{1}{8} \lim_{n \to \infty} \left( \frac{3^{n+1}-1}{3^n} \cdot \frac{5^{n+1}-1}{5^n} \right)\\= \frac{1}{8} \lim_{n \to \infty} \left( 3 - \frac{1}{3^n} \right) \left( 5 - \frac{1}{5^n} \right) ={1\over 8}\cdot 3\cdot 5= \bbox[red, 2pt]{15\over 8}$$

解答:$$每次操作後，濃度會變成前一次的  \frac{80}{100} = 0.8  倍 \\ 假設原濃度a_0，操作 n 次後的濃度為a_n= a_0\times 0.8^n \lt {1\over 10}a_0 \Rightarrow 0.8^n\lt 0.1 \\ \Rightarrow n\log 0.8 \lt \log 0.1=-1 \Rightarrow n(3\log 2-1)\lt -1 \Rightarrow n\gt {1\over 1-3\cdot 0.301} \approx 10.3 \Rightarrow n= \bbox[red, 2pt]{11}$$

解答:$$\log 3^{200} =200\log 3 \approx 200\cdot 0.4771 =95.42 \Rightarrow n=96 \\ \cases{3^1 \equiv 3\text{ mod 10} \\3^2 \equiv 9\text{ mod 10} \\3^3 \equiv 7\text{ mod 10} \\3^4 \equiv 1\text{ mod 10} \\3^5 \equiv 3\text{ mod 10} } \Rightarrow 循環數4 \Rightarrow 3^{200} =(3^4)^{25} \equiv 3^4 \equiv \text{ mod 10} =1 \Rightarrow a=1 \\ \log 3^{200}=95.42 \Rightarrow \log 2\lt 0.42\lt \log 3 \Rightarrow 最高位數字為2\Rightarrow b=2 \Rightarrow (n,a,b)= \bbox[red, 2pt]{(96,1,2)}$$

解答:$$\angle DAB= \angle BCD=90^\circ \Rightarrow ABCD為圓內接四邊形\\ 直角\triangle BAD: \overline{BD}= \sqrt{\overline{AB}^2+ \overline{AD}^2}=\sqrt{1+49} =5\sqrt 2\\ 依\href{https://chu246.blogspot.com/2020/11/ptolemys-theorem.html}{托勒密定理 \text{(Ptolemy's Theorem)}}:  \overline{AC} \times \overline{BD} = \overline{AB} \times \overline{CD} + \overline{BC} \times \overline{DA} \\ \Rightarrow \overline{AC}\cdot 5\sqrt 2= 1\cdot 5+5\cdot 7   \Rightarrow \overline{AC}={40\over 5\sqrt 2}= \bbox[red, 2pt]{4\sqrt 2}$$

解答:$$一個根為  1-2i \Rightarrow 另一個根為 1+2i \Rightarrow (x-(1-2i))(x-(1+2i))=x^2-2x+5 =(x-1)^2+4\gt 0\\ 再利用長除法可得 f(x)=(x^2-2x+5)(x^2-x-2) \lt 0 \Rightarrow x^2-x-2\lt 0\\ \Rightarrow (x-2)(x+1)\lt 0 \Rightarrow \bbox[red, 2pt]{-1\lt x\lt 2}$$

解答:$$\lim_{x\to 0} {f(3x)-f(\sin x) \over x} =\lim_{x\to 0} {{d\over dx}(f(3x)-f(\sin x)) \over {d\over dx}x} = \lim_{x\to 0}  \left( 3f'(3x)-(\cos x) f'(\sin x) \right) \\=3a-a= \bbox[red, 2pt]{2a}$$

解答:$$\textbf{Case I }1,2在同一列:在第1列或第2列，有2種情況;一列有3個格子選出2個排列1和2，\\\qquad 有P^3_2= 6種，因此共有2\times 6=12種\\ \textbf{Case II }1,2在同一行:共有3行，每行上下只能1,2或2,1，有2種排法，共3\times 2=6種 \\因此1 和 2 在同一列或同一行總共有：12+6=18種排法，剩下4格要排3,4,5,6，有4!=24種\\ 總共有18\times 24= \bbox[red, 2pt]{432}種$$

解答:$$\int_0^\pi {\sin x+\tan x\over \sec x} \,dx =\int_0^\pi (\sin x\cos x+\sin x) \,dx = \int_0^\pi ({1\over 2}\sin 2x+\sin x) \,dx \\= \left. \left[ -{1\over 4}\cos 2x -\cos x\right] \right|_0^\pi = \left( -{1\over 4}+1 \right) - \left( -{1\over 4}-1 \right) =\bbox[red, 2pt] 2$$

解答:$$\cases{ 1-\cos x= 2\sin^2(x/2) \\ 1+\cos x=2\cos^2(x/2)} \Rightarrow \int_0^{\pi/3} {1-\cos x\over 1+\cos x}\,dx = \int_0^{\pi/3} \tan^2 {x\over 2}\,dx =  \int_0^{\pi/3} \left( \sec^2 {x\over 2}-1 \right)\,dx  \\ = \left. \left[  2\tan{x\over 2}-x \right] \right|_0^{\pi/3} = \bbox[red, 2pt]{{2\sqrt 3\over 3}-{\pi\over 3}}$$

解答:$$y=x^2-{27\over x^2} \Rightarrow y'=2x+{54\over x^3} \Rightarrow y''=2-{162\over x^4}\\ y''=0 \Rightarrow x^4=81\Rightarrow x=\pm 3\Rightarrow y''在x=\pm 3處凹向性確實發生改變，故這兩處皆為反曲點\\ \Rightarrow \cases{y(3)=6\\ y(-3)=6} \Rightarrow 反曲點: \bbox[red, 2pt]{(3,6),(-3,6)}$$

解答:$$\lim_{n\to \infty} n^{-3/2} \sum_{k=1}^n \sqrt k =\lim_{n\to \infty} {1\over n} \sum_{k=1}^n \sqrt {k\over n} = \int_0^1 \sqrt x\,dx = \left. \left[{2\over 3} x^{3/2} \right] \right|_0^1= \bbox[red, 2pt]{2\over 3}$$

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