115年央大附中教甄-數學詳解

國立中央大學附屬中壢高級中學 115學年度第 1次教師甄選

ㄧ、填充題:每題 7 分，共 84 分。 

解答:

$$\cases{\angle HAO=\theta \\\angle HBO=2\theta\\ \angle HCO=3\theta \\ \overline{HO}=h\\ \overline{CO}=a} \Rightarrow \cases{\angle HBO= \angle HAO+\angle AHB \\ \angle HCO=\angle HBO+\angle BHC} \Rightarrow \cases{\angle AHB= \angle BHC=\theta \\ \overline{BH} =\overline{BA}=240}\\ \Rightarrow \overline{BH} 為\angle AHC的角平分線\Rightarrow \cases{\displaystyle {\overline{HA} \over \overline{HC}}= {\overline{AB} \over \overline{BC}}={240\over 90}={8\over 3} \\ \overline{HB}^2+ \overline{AB}\times\overline{BC} =\overline{HA} \times \overline{HC}} \Rightarrow \cases{\overline{HA}=8b\\ \overline{HC}=3b} \\ \Rightarrow 240^2+240\cdot 90=8b\cdot 3b \Rightarrow b^2=3300 \Rightarrow \cases{\triangle HCO: (3b)^2= a^2+h^2 \\ \triangle HBO:240^2=(90+a)^2+h^2} \\ \Rightarrow h^2=9b^2-a^2=240^2-(90+a)^2 \Rightarrow a=110 \Rightarrow h^2=240^2-200^2 =17600 \\ \Rightarrow h=\overline{OH} =\sqrt{17600} = \bbox[red, 2pt]{40\sqrt{11}}$$

解答:$$|z|=1 \Rightarrow z=\cos \theta+ i\sin \theta \Rightarrow |z^2+iz+1|= |\cos 2\theta-\sin \theta+1+i(\sin 2\theta +\cos \theta)| \\= \sqrt{(\cos2\theta-\sin \theta+1)^2+(\sin 2\theta+\cos \theta)^2} = \sqrt{3+2\cos 2\theta} \Rightarrow \cases{最大值a=\sqrt{5} \\ 最小值b=1} \\ \Rightarrow (a,b) = \bbox[red, 2pt]{(\sqrt 5,1)}$$

解答:$$a_n= S_n-S_{n-1} \Rightarrow \sqrt{S_n}+\sqrt{S_{n-1}} ={S_n-S_{n-1} \over 2n-1} ={(\sqrt{S_n}+\sqrt{S_{n-1}})(\sqrt{S_n}-\sqrt{S_{n-1}}) \over 2n-1} \\ \Rightarrow {\sqrt{S_n}-\sqrt{S_{n-1}} \over 2n-1}=1 \Rightarrow \sqrt{S_n}-\sqrt{S_{n-1}} =2n-1 \Rightarrow \cases{ \sqrt{S_2} -\sqrt{S_1}=1 \\ \sqrt{S_3}-\sqrt{S_2}=3\\ \cdots \cdots\\\sqrt{S_n}-\sqrt{S_{n-1}} =2n-1} \\ 全部相加\Rightarrow \sqrt{S_n}-\sqrt{S_1}=1+3+\cdots+2n-1 \Rightarrow \sqrt{S_n}-1=n^2-1 \Rightarrow S_n= \bbox[red, 2pt]{n^4}$$

解答:$$\cases{a=45+29\sqrt 2\\ b=45-29\sqrt 2} \Rightarrow \cases{a+b=90\\ ab=343=7^3} \Rightarrow (\sqrt[3]a+ \sqrt [3]b)^3=a+b+3\sqrt[3]{ab} (\sqrt[3]a+ \sqrt [3]b) =90+21(\sqrt[3]a+ \sqrt [3]b) \\ x=\sqrt[3]a+ \sqrt [3]b \Rightarrow x^3=90+21x \Rightarrow x^3-21x-90=0 \Rightarrow (x-6)(x^2+6x+15)=0 \\ \Rightarrow x=\sqrt[3]a+ \sqrt [3]b= \bbox[red, 2pt]6$$

解答:

$$假設\cases{圓心O(0,0)\\P(a,0) \\ \angle APO =\angle BPO=\theta} \Rightarrow \overline{AP} = \sqrt{\overline{OP}^2-\overline{OA}^2} =\sqrt{a^2-1} \Rightarrow  \cos\theta={\sqrt{a^2-1} \over a} \\ \Rightarrow \cos 2\theta=2\cos^2\theta-1 =1-{2\over a^2} \Rightarrow \overrightarrow{PA} \cdot  \overrightarrow{PB} =\overline{PA}\cdot \overline{PB} \cos 2\theta=(a^2-1)\cdot \left( 1-{2\over a^2} \right) \\=a^2+{2\over a^2}-3 \ge 2\sqrt{a^2\cdot {2\over a^2}}-3= \bbox[red, 2pt]{2\sqrt 2-3}$$

解答:$$25球任選兩球有C^{25}_2=300種, \\\cases{每行有5個號碼，任選兩個有C^5_2=10，共五行有5\times 10=50種\\每列有5個號碼，任選兩個有C^5_2=10，共五列有5\times 10=50種} \Rightarrow 中獎有100種\\ 每一行的每個數字都會與其它四個數字配對一次，也就是每行的數字被計算四次\\ 因此同行的組合獎金總和=(1+2+\cdots+ 25)\times 4=1300，同列獎金總和也是1300\\ 所有獎金總和=1300+1300=2600 \Rightarrow 期望值={100 \over 300} \times {2600\over 100}= \bbox[red, 2pt]{26\over 3}$$

解答:

$$f(x)為偶函數且週期為4 \Rightarrow I= \int_{-2}^6 f(x)\,dx\; 積分區間為兩個週期 \Rightarrow I=2\int_0^4 f(x)\,dx\\函數對稱半週期 \Rightarrow I =4\int_0^2f(x)\,dx  \Rightarrow 4 \left( \int_0^1 f(x)\,dx + \int_1^2 f(x)\,dx \right) \\函數對稱半週期 \Rightarrow 函數圖形對稱x=2 \Rightarrow f(x)=((4-x)-4)^2=x^2, x\in [1,2] \\ 也就是在區間[2,3]的圖形向左水平翻轉到[1,2],即1+\sqrt{1-x^2} 翻成x^2\\\Rightarrow I= 4 \left( \int_0^1 (1+\sqrt{1-x^2}) \,dx + \int_1^2 x^2 \,dx\right) =4 \left( 1+{\pi\over 4} +{7\over 3} \right)= \bbox[red, 2pt]{{40\over 3}+\pi}$$

解答:$$\lim_{n\to \infty} {n+n^2+\cdots +n^n \over 1^n+2^n+\cdots+n^n} =\lim_{n\to \infty} {(1/n^{n-1})+ (1/n^{n-2})+\cdots + 1 \over (1/n)^n+ (2/n)^n+\cdots+1^n} \\ 分子為等比級數\Rightarrow A_n=1+{1\over n}+\cdots+{1\over n^{n-1}} = {1-(1/n)^n\over 1-(1/n)} \Rightarrow \lim_{n\to \infty}A_n=1\\ 分母B_n= 1^n+ ({n-1\over n})^n+ \cdots+({2\over n})^n+ ({1\over n})^n =\sum_{k=0}^{n-1} \left( {n-k\over n} \right)^n =\sum_{k=0}^{n-1} \left( 1-{k\over n} \right)^n \\ \Rightarrow \lim_{n\to \infty} B_n= \sum_{k=0}^\infty e^{-k} ={1\over 1-e^{-1}}={e\over e-1} \\ \Rightarrow \lim_{n\to \infty} {A_n\over B_n} ={1\over {e\over e-1}} ={e-1\over e} = \bbox[red, 2pt]{1-{1\over e}}$$

解答:$$先求圓心高度變化，再求A相對圓心的高度變化，最後兩者相加\\圓心沿斜坡移動距離s={3\over 2} \times {14\over 9} \pi={7\over 3}\pi \Rightarrow 圓心高度變化\Delta h_c=s\sin \theta={7\over 3}\pi \times {3\over 5}= {7\over 5}\pi \\ 旋轉角度={s\over 圓半徑}={7\pi/3\over 1} ={7\over 3}\pi =2\pi+{\pi\over 3} \Rightarrow 轉了一圈又60^\circ\\ 假設\cases{圓心O(0,0)\\ A(0,-1)} \Rightarrow 順時針轉60^\circ 後，A的相對坐標為A'(-\sin 60^\circ,-\cos 60^\circ)=(-{\sqrt 3\over 2},-{1\over 2}) \\ \Rightarrow \overrightarrow{AA'}=(-{\sqrt 3\over 2},{1\over 2}) \Rightarrow 高度變化\Delta h_a=-{\sqrt 3\over 2}\sin \theta+ {1\over 2}\cos \theta={4-3\sqrt 3\over 10}\\ 總高度變化=\Delta h_c+\Delta h_a = \bbox[red, 2pt]{{1\over 10}(14\pi+4-3\sqrt 3)}$$

解答:$$\overline{P_1P_2} =\cdots =\overline{P_6P_7} =2 \Rightarrow \overline{P_1P_7} =6\times 2=12 \Rightarrow \cos \angle AP_7P_1={12^2+8^2-6^2\over 2\cdot 12\cdot 8} ={43\over 48} \\ \overrightarrow{AP_k} \cdot \overrightarrow{P_1P_7} = \left(\overrightarrow{AP_7}- \overrightarrow{P_kP_7}  \right) \cdot \overrightarrow{P_1P_7} =\overline{AP_7} \cdot \overline{P_1P_7} \cos \angle AP_7P_1 -\overline{P_kP_7}\cdot \overline{P_1P_7} \\=8\cdot 12\cdot {43\over 48}-12\overline{P_kP_7} =86 -12\overline{P_kP_7}\\\Rightarrow \sum_{k=1}^7 \overrightarrow{AP_k} \cdot \overrightarrow{P_1P_7} =7\times 86  -12(12+10+8+6+4+ 2+0)=602-12 \times 42= \bbox[red, 2pt]{98}$$

解答:$$假設\cases{C(0,0) \\A(0,2a) \\B(2b,0)} \Rightarrow M(b,0) \Rightarrow \cases{D=B以A為圓心旋轉90度 =(2a,2a+2b)\\E=C以A為圓心旋轉-90度 =(-2a,2a)} \\ \Rightarrow \cases{\overline{MD} =8 =\sqrt{(2a-b)^2+(2a+2b)^2} \\ \overline{ME}=6 =\sqrt{(2a+b)^2+4a^2}} \Rightarrow \cases{8a^2+4ab+5b^2=64\\ 8a^2+4ab+b^2=36} \Rightarrow 4b^2=28\Rightarrow b=\sqrt 7 \\ \Rightarrow \overline{BC}=2b= \bbox[red, 2pt]{2\sqrt 7}$$

解答:$$\cases{A(1,4,2) \\B(3,4,4)} \Rightarrow M=\overline{AB}中點=(2,4,3) \Rightarrow A,B的中垂面E_2的法向量=\overrightarrow{AB}=(2,0,2) \\ \Rightarrow E_2:2(x-2)+2(z-3)=0 \Rightarrow E_2:x+z=5 \\令 E_1:5x-2y+5z=3 \Rightarrow 球心O\in E_1\cap E_2 \Rightarrow O(t,11,5-t), t\in \mathbb R \Rightarrow 球半徑R =\overline{OA} \\ \Rightarrow R^2=(t-1)^2+(11-4)^2+(5-t-2)^2 =2t^2-8t+59 \\ \Rightarrow d(O,E)={|t+11+(5-t)-19|\over \sqrt{1^2+1^2+1^2}} = \sqrt 3為一常數 \Rightarrow d^2=3 \\ 設截圓半徑為r \Rightarrow r^2=R^2-d^2 \Rightarrow F(S)=\pi r^2 = \pi(2t^2-8t+56) =\pi(2(t-2)^2+48) \\ \Rightarrow 最小值= \bbox[red, 2pt]{48\pi}$$

二、計算證明題： 每題 8 分，共 16 分。

解答:$$\textbf{(1) } f(x)=(1+x)^n= \sum_{k=0}^n C^n_kx^k \Rightarrow f(1)=2^2=a_n \\ \Rightarrow f'(x)=n(1+x)^{n-1} = \sum_{k=0}^n kC^n_kx^{k-1} \Rightarrow b_n=f'(1)= \bbox[red, 2pt]{n2^{n-1}}  \\\textbf{(2) } g(x)=xf'(x)=nx(1+x)^{n-1} = \sum_{k=0}^n kC^n_kx^{k} \Rightarrow g'(x)= n(1+nx) (1+x)^{n-2} =\sum_{k=0}^n k^2C^n_kx^{k-1} \\ \Rightarrow d_n=g'(1)= \bbox[red, 2pt]{n(1+n)2^{n-2}}$$

解答:$$f(x)=x^{13}+x+90=(x^2-x+a) P(x) \Rightarrow \cases{f(0)=90= aP(0) \Rightarrow a是90的因數\\ f(1)=92=aP(1) \Rightarrow a是92的因數 \\f(-1)=88=(a+2)P(-1) \Rightarrow a+2是88的因數} \\ \text{gcd}(90,92)=2 \Rightarrow \cases{a=1 \Rightarrow a+2=3不是88的因數\\a=-2 \Rightarrow a+2=0不合 \\a=2 \Rightarrow a+2=4是88的因數\\ a=-1 \Rightarrow a+2=1是88的因數} \\ a=2 \Rightarrow x^2-x+2=0 \Rightarrow x^2=x-2 \Rightarrow x^4=-3x+2 \Rightarrow x^{12}=45x-46 \Rightarrow x^{13}=-x-90 \\\qquad \Rightarrow x^{13}+x+90=0 \Rightarrow x^2-x+2是f(x)的因式 \Rightarrow \bbox[red, 2pt]{a=2}$$

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