115年永春高中教甄4-數學詳解

 臺北市立永春高級中學 115 學年度第四次正式教師甄選

一、填充題（每格 7 分，共 70 分）

解答:$$z=a+bi \Rightarrow \bar z=a-bi \Rightarrow z^{2025}=\bar z \Rightarrow |z|^{2025}=|z| \Rightarrow |z|(|z|^{2024}-1)=0 \\ \textbf{Cases I }|z|=0 \Rightarrow z=0 \Rightarrow (a,b)=(0,0) \\ \textbf{Case II }|z|=1 \Rightarrow z\cdot \bar z =|z|^2 =1 \Rightarrow \bar z={1\over z} \Rightarrow z^{2025}={1\over z} \Rightarrow z^{2026}=1 有2026組相異解\\ 共有1+2026= \bbox[red, 2pt]{2027}組解$$

解答:$$\cases{P(紅球比黑球晚取完): \frac{2}{2+4} = \frac{1}{3} \\   P(紅球比白球晚取完): \frac{2}{2+3} = \frac{2}{5} \\   P(紅球最後才被取完):\frac{2}{9}} \\\Rightarrow 欲求機率 = 1 - P(\text{紅球比黑球晚取完}) - P(\text{紅球比白球晚取完}) + P(\text{紅球最後才被取完})\\ =1-{1\over 3}-{2\over 5}+{2\over 9}= \bbox[red, 2pt]{22\over 45}$$

解答:$$ \sin \angle A = \sin(180^\circ - (\angle B + \angle C)) = \sin(\angle B + \angle C) = \frac{\sqrt{15}}{4} \Rightarrow  \cos^2 A = 1 - \sin^2 A =     \frac{1}{16} \\ \Rightarrow  \vert{}\cos A\vert{} = \frac{1}{4} \Rightarrow \cases{ \overline{AE} = \overline{AB} \cdot \vert{}\cos A\vert{} \\ \overline{AD} = \overline{AC} \cdot \vert{}\cos A\vert{} } \Rightarrow  \frac{\overline{AE}}{\overline{AB}} = \frac{\overline{AD}}{\overline{AC}} = \vert{}\cos A\vert{}  \Rightarrow  \triangle ADE \sim \triangle ABC \\ \Rightarrow  \frac{\overline{DE}}{\overline{BC}} = \vert{}\cos A\vert{} \Rightarrow  \overline{DE} = \overline{BC} \times \vert{}\cos A\vert{} = 7 \times \frac{1}{4} = \bbox[red, 2pt]{\frac{7}{4} }$$

解答:$$假設在 5 次投擲中，出現x次正面及y次反面 \Rightarrow x+y=5 \Rightarrow z_5=1\cdot 2^x\cdot i^y =2 \\ \Rightarrow 2^x\cdot i^y=2 \Rightarrow  |2^x\cdot i^y|= |2| \Rightarrow 2^x=2 \Rightarrow  x=1\Rightarrow y=4 \\ \Rightarrow P=C^5_1 \left( {1\over 2} \right)^x \left( {1\over 2} \right)^y  = \binom{5}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^4 =\bbox[red, 2pt]{ \frac{5}{32}}$$

解答:$$ \frac{x-3}{x+1} = 2 \Rightarrow x=-5 \Rightarrow  f\left(\frac{-5-3}{-5+1}\right) + f\left(\frac{3+(-5)}{1-(-5)}\right) = -5 \Rightarrow  f(2) + f\left(-\frac{1}{3}\right) = -5 \cdots(1) \\ \frac{3+x}{1-x} = 2 \Rightarrow x=-{1\over 3} \Rightarrow  f\left(\frac{-\frac{1}{3}-3}{-\frac{1}{3}+1}\right) + f\left(\frac{3+(-\frac{1}{3})}{1-(-\frac{1}{3})}\right) = -\frac{1}{3} \Rightarrow  f(-5) + f(2) = -\frac{1}{3} \cdots(2) \\\frac{x-3}{x+1} = -\frac{1}{3} \Rightarrow x=2 \Rightarrow  f\left(\frac{2-3}{2+1}\right) + f\left(\frac{3+2}{1-2}\right) = 2 \Rightarrow  f\left(-\frac{1}{3}\right) + f(-5) = 2 \cdots(3) \\ \Rightarrow (1)+(2)+(3) \Rightarrow  f(2) + f\left(-\frac{1}{3}\right) + f(-5) = -\frac{5}{3}  \cdots(4) \Rightarrow  (4)-(3) \Rightarrow  f(2) = \bbox[red, 2pt]{-{11\over 3}}$$

解答:

$$ Q_1, Q_2  為切點\Rightarrow \cases{ \angle OQ_1P = \angle OQ_2P = 90^\circ \\ \overline{OQ_1} = \overline{OQ_2} = 圓半徑=1} \Rightarrow 四邊形 OQ_1PQ_2$ 的面積 A = \triangle OQ_1P +\triangle OQ_2P \\  = 2 \times \left( \frac{1}{2} \times \overline{OQ_1} \times \overline{PQ_1} \right) = \overline{OQ_1} \times \overline{PQ_1} = 1 \times \overline{PQ_1} = \overline{PQ_1} \\ 直角\triangle OQ_1P:  \overline{OP}^2 = \overline{OQ_1}^2 + \overline{PQ_1}^2 = 1^2 + \overline{PQ_1}^2  \Rightarrow  \overline{PQ_1} = \sqrt{\overline{OP}^2 - 1}  \Rightarrow A=\sqrt{\overline{OP}^2 - 1} \\ 假設P(\sqrt 2\cos \theta, \sqrt 3\sin \theta) \Rightarrow \overline{OP}^2 = x^2 + y^2 = (\sqrt{2}\cos\theta)^2 + (\sqrt{3}\sin\theta)^2 = 2\cos^2\theta + 3\sin^2\theta  \\= 2(1 - \sin^2\theta) + 3\sin^2\theta = 2 + \sin^2\theta 最大值為3\Rightarrow A的最大值=\sqrt{3-1}= \bbox[red, 2pt]{\sqrt 2}$$

解答:$$\overline{AB}^2 = (a-1)^2 + (\sqrt{1-a^2} - 0)^2 + (0-0)^2 =  2 - 2a>0 \Rightarrow a<1 \\C(x,y,z ) \Rightarrow x^2 +y^2 +z^2=1\\ \triangle ABC 為正三角形\Rightarrow \overline{AC}^2 = \overline{BC}^2 = \overline{AB}^2 = 2 - 2a\\\overline{AC}^2 = (x-1)^2 + y^2 + z^2 =  2 - 2x=2-2a \Rightarrow x=a \Rightarrow \overline{BC}^2 = (x-a)^2 + (y-\sqrt{1-a^2})^2 + z^2 \\= 0 + y^2 - 2y\sqrt{1-a^2} + (1-a^2) + z^2 = y^2 - 2y\sqrt{1-a^2} + 1 - a^2 + 1 - a^2 - y^2\\ = 2 - 2a^2 - 2y\sqrt{1-a^2} =2-2a \Rightarrow 2a - 2a^2 = 2y\sqrt{1-a^2} \Rightarrow  y\sqrt{1-a^2} = a(1-a) \\ \Rightarrow y = \frac{a(1-a)}{\sqrt{1-a^2}} = \frac{a(1-a)}{\sqrt{(1-a)(1+a)}} = a\sqrt{\frac{1-a}{1+a}} \Rightarrow z^2 = 1 - x^2 - y^2 = 1 - a^2 - a^2\left(\frac{1-a}{1+a}\right)\\=   1 - a^2\left(1 + \frac{1-a}{1+a}\right) = 1 - a^2\left(\frac{1+a+1-a}{1+a}\right) = 1 - \frac{2a^2}{1+a} \ge 0\Rightarrow   2a^2 - a - 1 \le 0 \\ \Rightarrow (2a+1)(a-1) \le 0 \Rightarrow \bbox[red,2pt] {-\frac{1}{2} \le a \le 1}$$

解答:$$令y=x^2 \Rightarrow  y^{1013} - y^{1012} - y^{1011} - \dots - y - 2 = 0 \Rightarrow y^{1013} - (y^{1012} + y^{1011} + \dots + y + 1) - 1 = 0 \\ \Rightarrow  y^{1013} = (y^{1012} + y^{1011} + \dots + y + 1) + 1  ={y^{1013}-1\over y-1}+1 \Rightarrow y=2為其中一根 \\ 也就是f(2)=0, 其中f(y) =  y^{1013} - y^{1012} - y^{1011} - \dots - y - 2 \\依據笛卡兒符號法則(\href{https://en.wikipedia.org/wiki/Descartes%27s_rule_of_signs}{\text{Descartes' Rule of Signs})}, f(y)=0恰有 1 個正實根\\ 因此y=2是唯一正實根, 即x^2=2 \Rightarrow x=\pm \sqrt 2 \Rightarrow x_1^2+ x_2^2=2+2= \bbox[red, 2pt]{4}$$

解答:

$$拋物線\Gamma: y=x^2 \Rightarrow c={1\over 4}\Rightarrow \cases{焦點F(0,1/4)\\ 準線L: y=-1/4} \\ P(t,t^2) \in \Gamma \Rightarrow a=\overline{PF} =d(P,L) =t^2+{1\over 4}\\ 假設P'(t',t'^2)=\Gamma \cap \overleftrightarrow{PF} \Rightarrow \overline{PP'}為一焦弦 \Rightarrow t\cdot t'=-4c^2=-{1\over 4} \Rightarrow t'=-{1\over 4t}\\ \Rightarrow Q的x坐標=-t' ={1\over 4t} \Rightarrow b= \overline{QF}=d(Q,L)=  \left( {1\over 4t} \right)^2- \left( -{1\over 4} \right)={1\over 16t^2}+{1\over 4} \\ \Rightarrow  a + 4b = (t^2 + \frac{1}{4}) + 4(\frac{1}{16t^2} + \frac{1}{4}) =  t^2 + \frac{1}{4t^2} + \frac{5}{4} \ge 2 \sqrt{t^2\cdot {1\over 4t^2}}+{5\over 4} =1+{5\over 4}= \bbox[red, 2pt]{9\over 4}$$

解答:

$$假設\cases{  \overline{OA} = a\\ \overline{OB} = b\\ \overline{OM} = m\\ \overline{ON} = n} \Rightarrow \cases{ \overrightarrow{OA} \cdot \overrightarrow{OB} = ab\cos(3\theta) = k \\\overrightarrow{OM} \cdot \overrightarrow{OB} = mb\cos(2\theta) = 2k \\ \overrightarrow{ON} \cdot \overrightarrow{OB} = nb\cos(\theta) = 3k } \Rightarrow \cases{1/a=b\cos (3\theta) /k\\ 1/m=b\cos(2\theta)/2k \\ 1/n =b\cos \theta/3k} \\ S_{\triangle OAN} = S_{\triangle OAM} + S_{\triangle OMN} \Rightarrow   \frac{1}{2}an\sin(2\theta) = \frac{1}{2}am\sin(\theta) + \frac{1}{2}mn\sin(\theta) \\\Rightarrow  2an \cos(\theta) = am + mn \Rightarrow  \frac{2\cos(\theta)}{m} = \frac{1}{n} + \frac{1}{a} \Rightarrow  2\cos(\theta) \cdot \left(\frac{b\cos(2\theta)}{2k}\right) = \frac{b\cos(\theta)}{3k} + \frac{b\cos(3\theta)}{k} \\ \Rightarrow  3\cos(\theta)\cos(2\theta) = \cos(\theta) + 3\cos(3\theta) \Rightarrow  3\cos(\theta)(2\cos^2(\theta) - 1) = \cos(\theta) + 3(4\cos^3(\theta) - 3\cos(\theta)) \\ \Rightarrow  6\cos^3(\theta) - 5\cos(\theta) = 0 \Rightarrow \cos \theta(6\cos^2\theta-5) =0 \Rightarrow 6\cos^2\theta=5 \Rightarrow \cos \theta =\bbox[red, 2pt]{\sqrt{30} \over 6}$$

二、計算題（每題 15 分，共 30 分）

解答:

$$\textbf{(a) }假設切點P(t,f(t)) \Rightarrow \cases{f(t)=g(t) \\f'(t)=g'(t)} \Rightarrow \cases{n\ln t=at^n \\ n/t=ant^{n-1}} \Rightarrow at^n=1=n\ln t \Rightarrow t=e^{1/n} \\ \quad \Rightarrow a \left( e^{1/n} \right)^n=1 \Rightarrow ae=1 \Rightarrow a= \bbox[red, 2pt]{1\over e} \\ \textbf{(b) } \cases{f(x)=0 \Rightarrow x=1\\ g(x)=0 \Rightarrow x=0} \Rightarrow S_n =  \int_0^{e^{1/n}} g(x) \, dx - \int_1^{e^{1/n}} f(x) \, dx = \int_0^{e^{1/n}} \frac{1}{e} x^n \, dx  - \int_1^{e^{1/n}} n \ln x \, dx \\ \quad = {1\over e} \left. \left[ {1\over n+1}x^{n+1} \right] \right|_0^{e^{1/n}}- \left. \left[ nx\ln x- nx\right] \right|_1^{e^{1/n}}= {e^{1/n} \over n+1} - \left( n-(n-1)e^{1/n}  \right) = \bbox[red, 2pt]{\frac{n^2}{n+1} e^{1/n} - n } \\\textbf{(c) } \lim_{n \to \infty} \left( \frac{n^2}{n+1} e^{1/n} - n \right)  = \lim_{n \to \infty} \frac{n^2 (e^{1/n} - 1) - n}{n+1} =\lim_{n \to \infty} \frac{n^2 \left( 1 + \frac{1}{n} + \frac{1}{2n^2} + O(\frac{1}{n^3}) - 1 \right) - n}{n+1}  \\ \quad  = \lim_{n \to \infty} \frac{\frac{1}{2} + O(\frac{1}{n})}{n+1} = \bbox[red, 2pt]0$$

解答:$$\textbf{(a) }\cases{d(p^k)=k+1 \\d(p^{k+1}) =k+2} \Rightarrow \cases{f(p^k)=(k+1)/p^{k/2} \\f(p^{k+1}) = (k+2)/ p^{(k+1)/2}} \Rightarrow  \frac{k+1}{p^{k/2}} \le \frac{k+2}{p^{(k+1)/2}}  \\ \qquad \Rightarrow  (k+1)p^{1/2} \le k+2  \Rightarrow p^{1/2} \le {k+2\over k+1} =1+{1\over k+1} \le 1+{1\over 2} \; (\because k是正整數) \\\qquad \Rightarrow \sqrt p\le 1.5 \Rightarrow p\le 1.5^2=2.25 \Rightarrow p=2 \Rightarrow \sqrt 2\le 1+{1\over k+1} \Rightarrow k+1\le \sqrt 2+1 \approx 2.414\\ \qquad \Rightarrow k+1=2 \Rightarrow k=1 \Rightarrow (p,k) = \bbox[red, 2pt]{(2,1)} \\\textbf{(b) }   n = p_1^{a_1} p_2^{a_2} \cdots p_m^{a_m} \Rightarrow  f(n) = \frac{(a_1+1)(a_2+1)\cdots(a_m+1)}{\sqrt{p_1^{a_1} p_2^{a_2} \cdots p_m^{a_m}}} = \prod_{i=1}^m \frac{a_i+1}{p_i^{a_i/2}} = \prod_{i=1}^m f(p_i^{a_i})  \\ \quad p=2 \Rightarrow \cases{ a=1: f(2^1) = \frac{2}{\sqrt{2}} = \sqrt{2} \approx 1.414 \\  a=2: f(2^2) = \frac{3}{\sqrt{4}} = 1.5 \\ a=3: f(2^3) = \frac{4}{\sqrt{8}} = \sqrt{2} \approx 1.414 } \Rightarrow 最大值f(2^2)=1.5 \\ \quad p=3\Rightarrow \cases{ a=1: f(3^1) = \frac{2}{\sqrt{3}} \approx 1.154 \\a=2: f(3^2) = \frac{3}{\sqrt{9}} = 1 } \Rightarrow 最大值f(3^1)={2\over \sqrt 3} \\ \quad p\ge 5\Rightarrow  a=1: f(p^1) = \frac{2}{\sqrt{p}} \le \frac{2}{\sqrt{5}} \approx 0.894 < 1 \\ 要使 f(n)  達到最大，只能取用   f(p^a) > 1  的部分乘積,\\即 \max f(n) = f(2^2) \times f(3^1) = \frac{3}{2} \times \frac{2}{\sqrt{3}} = \sqrt{3} ,此時n=2^2\times 3^1=12 \\ \Rightarrow \bbox[red, 2pt]{f(n) 的最大值為 \sqrt{3}，此時 n = 12}$$

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解題僅供參考，其他教甄試題及詳解