114年景美女中教甄-數學詳解

 臺北市立景美女子高級中學 114 學年度第 1 次正式敎師甄選

一、填充題：84 分(每格 6 分) 

解答：$$\cases{10^n\lt p\lt q\lt 10^{n+1} \Rightarrow n\le \log p, \log q\lt n+1 \\ pq^2為13位數 \Rightarrow \log(pq^2) =\log p+2\log q=12} \Rightarrow 3n\le \log p+2\log q\lt 3n+3 \Rightarrow n=\bbox[red, 2pt]4$$

解答：

$$假設\cases{\overline{AC}=a, \overline{BC}=a+1, \overline{AB}=a+2\\ \angle B=\theta, \angle C=2\theta} \Rightarrow \cos \theta={(a+1)^2+(a+2)^2-a^2 \over 2(a+1)(a+2)} ={a+5\over 2(a+2)} \\ 正弦定理: {a\over \sin \theta} ={a+2\over \sin 2\theta} ={a+2\over 2\sin \theta \cos \theta} \Rightarrow a+2=2a\cos \theta =2a\cdot {a+5\over 2(a+2)} \\ \Rightarrow (a+2)^2=a(a+5) \Rightarrow a=4 \Rightarrow \cos \theta={3\over 4} \Rightarrow \sin \theta={\sqrt 7\over 4} \Rightarrow \triangle ABC={1\over 2}(a+1)(a+2) \sin \theta \\={1\over 2}\cdot 5\cdot 6\cdot {\sqrt 7\over 4} = \bbox[red, 2pt]{{15\over 4}\sqrt 7}$$

解答：$$由題意可知: P在\triangle ABC的投影點為\triangle ABC的外心\\\cases{a=\overline{BC}=7\\ b=\overline{AC}=5\\ c=\overline{AB}=3} \Rightarrow s=(a+b+c)/2=15/2 \Rightarrow a\triangle ABC =\sqrt{s(s-a)(s-b) (s-c)} \\ =\sqrt{{15\over 2}\cdot {1\over 2} \cdot {5\over 2 } \cdot {9\over 2}}   ={15\sqrt 3\over 4} ={abc\over 4R} ={105\over 4R}\Rightarrow 外接圓半徑R= {7\sqrt 3\over 3}\\ \Rightarrow 三角錐的高h= \sqrt{({25\sqrt 3\over 3})^2-({7\sqrt 3\over 3}^2)} =8\sqrt 3  \Rightarrow 體積={1\over 3}\cdot a\triangle \cdot h= \bbox[red, 2pt]{30}$$

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

解答：$$\triangle BPC:\triangle CPA:\triangle APB = 7:8:9 \Rightarrow 7\overrightarrow{PA}  +8\overrightarrow{PB}+ 9\overrightarrow{PC} =0 \Rightarrow 7\overrightarrow{AP} =   8\overrightarrow{PB} + 9\overrightarrow{PC} \\=8(\overrightarrow{PA} +\overrightarrow{AB}) +9(\overrightarrow{PA} +\overrightarrow{AC}) =17\overrightarrow{PA} +8\overrightarrow{AB} +9\overrightarrow{AC} \\ \Rightarrow 24\overrightarrow{AP}=8\overrightarrow{AB} +9\overrightarrow{AC} \Rightarrow \overrightarrow{AP} ={1\over 3}\overrightarrow{AB} +{3\over 8}\overrightarrow{AC} =a \overrightarrow{AB} +b\overrightarrow{AC} \Rightarrow (a,b)= \bbox[red, 2pt]{\left({1\over 3},{3\over 8} \right)}$$

解答：$$a_n={n-1\over n+1}a_{n-1} ={(n-1)(n-2)\over (n+1)n}a_{n-2} ={(n-1)! \over (n+1)!/2}a_1 ={2\over (n+1)n}a_1 =2\left( {1\over n}-{1\over n+1}\right) \\ \Rightarrow \sum_{n=1}^{100} a_n =2\sum_{n=1}^{100}\left( {1\over n}-{1\over n+1}\right) =2\left( 1-{1\over 101}\right) = \bbox[red, 2pt]{200\over 101}$$

解答：$${d\over dx} \int_{-x}^x f(t)\,dt =x^2+1 \Rightarrow g(x)=\int_{-x}^x f(t)\,dt =\int (x^2+1)\,dx ={1\over 3}x^3+x+ C \\ \Rightarrow g(0)= \int_0^0f(t)\,dt= 0=C \Rightarrow g(x) =\int_{-x}^x f(t)\,dt ={1\over 3}x^3+x \\\Rightarrow \int_{-1}^1 f(x)\,dx =g(1)={1\over 3}+1= \bbox[red, 2pt]{4\over 3}$$

解答：$$28\cdot 2-1=55= (10+1)(4+1) \Rightarrow n^2=2^{10} \cdot 3^4 \Rightarrow n=2^5\cdot 3^2= \bbox[red, 2pt]{288}$$

解答：$$\textbf{Case I }甲取4白入乙，再從乙任取5球入甲\Rightarrow 機率P_1= C^5_4/ C^7_4 =1/7\\ \textbf{Case II } 甲取3白1黑入乙，再從乙任取5球入甲\Rightarrow 機率P_2=C^5_3C^2_1/C^7_4 =4/7\\ \textbf{Case III }甲取2白2黑入乙，再從乙任取4白1黑球入甲\Rightarrow 機率P_3={C^5_2C^2_2\over C^7_4} \cdot {C^5_4\cdot C^2_1\over C^7_5} =20/147 \\\textbf{Case IV }甲取2白2黑入乙，再從乙任取3白2黑球入甲\Rightarrow 機率P_4={C^5_2C^2_2\over C^7_4} \cdot {C^5_3\cdot C^2_2\over C^7_5} =20/147 \\ \Rightarrow {P_3+P_4\over P_1+P_2 +P_3+P_4} =\bbox[red, 2pt]{8\over 29}$$

解答：

$$B,C為圓切點\Rightarrow \overline{AC}=\overline{AB}=1 , 因此假設\cases{A(0,0) \\B(1,0) \\ \angle BAC=\theta} \Rightarrow \cases{圓心O(1,r) \\ C(\cos \theta, \sin \theta) } \Rightarrow \overline{OC}^2=r^2 \\ \Rightarrow (\cos\theta-1)^2+(r-\sin\theta)^2=r^2 \Rightarrow r={1-\cos \theta\over \sin \theta}\\ \Rightarrow \cases{D=(A+B)/2=({1\over 2},0) \\E=(A+2C)/3 =({2\over 3}\cos \theta, {2\over 3}\sin \theta)} \Rightarrow \cases{L_1= \overleftrightarrow{BE}: y={\sin \theta\over \cos\theta-3/2}(x-1) \\ L_2=\overleftrightarrow{CD}: y={\sin \theta\over \cos \theta-1/2}(x-1/2)} \\ \Rightarrow P=L_1\cap L_2 =({1\over 4}+{1\over 2}\cos \theta, {1\over 2}\sin \theta) \Rightarrow \overline{OP}^2= r^2 \Rightarrow ({1\over 2}\cos \theta-{3\over 4})^2+({1\over 2}\sin \theta-r)^2= r^2 \\ \Rightarrow r={{13\over 16}-{3\over 4}\cos \theta\over \sin \theta} ={1-\cos \theta\over \sin \theta} \Rightarrow \cos \theta={3\over 4} \Rightarrow \overrightarrow{AB} \cdot \overrightarrow{AC} = |\overrightarrow{AB}| \cdot |\overrightarrow{AC}|\cos \theta=\cos \theta= \bbox[red, 2pt]{3\over 4}$$

解答：$$兩函數\cases{y=f(x)=a^x \\ y=g(x)=\log_a x}互為反函數 \Rightarrow 兩圖形對稱於直線L:y=x\\ 假設\cases{A(m,n) =(m,a^m)\\B(n,m)=(n, \log_a n)} \Rightarrow \cases{A(m,a^m) \\B(a^m,m)} \Rightarrow \overline{AB}=2\sqrt 6+2\sqrt 2 \Rightarrow 2(m-n)^2=(2\sqrt 6+2\sqrt 2)^2$$

解答：$$\cases{\triangle OAD為正\triangle  \Rightarrow \overline{OA} =\overline{AD} =2\\ \triangle OAB為正\triangle \Rightarrow \overline{AB}= \overline{OA}} \Rightarrow \overline{AB} =\overline{OB}=2 \\ 假設\cases{B(0,0,0) \\ A(0,2,0) \\C(4,0,0) \\ D(2,2,0) \\O(a,b,c)} \Rightarrow \cases{\overline{OB}=2\\ \overline{OA}=2 \\\overline{OD}=2} \Rightarrow \cases{a^2+b^2+c^2=4\\ a^2+(b-2)^2+c^2=4\\ (a-2)^2 +(b-2)^2+c^2=4} \Rightarrow \cases{a=1\\ b=1\\ c=\sqrt 2} \Rightarrow O(1,1,\sqrt 2) \\ \Rightarrow \cases{\overrightarrow{OA} =(-1,1,-\sqrt 2) \\ \overrightarrow{OC} =(3,-1,-\sqrt 2) \\ \overrightarrow{OD} =(1,1,-\sqrt 2)} \Rightarrow \cases{\vec u=\overrightarrow{OA} \times \overrightarrow{OD} =(0,-2\sqrt 2,-2) \parallel (0,\sqrt 2, 1) \\ \vec v= \overrightarrow{OD} \times \overrightarrow{OC} =(-2\sqrt 2,-2\sqrt 2,-4) \parallel (\sqrt 2,\sqrt 2, 2)} \\ \Rightarrow \theta為鈍角\Rightarrow \cos \theta =-{(0,\sqrt 2, 1)\cdot (\sqrt 2,\sqrt 2, 2) \over |(0,\sqrt 2, 1)|| (\sqrt 2,\sqrt 2, 2)|} =-{4\over \sqrt{24}} = \bbox[red, 2pt]{-{\sqrt 6\over 3}}$$

解答：$$取f(x)=2x^2+ax+b=g(x)+ cx+d \Rightarrow g(x)=2x^2+(a-c)x+b-d \\ \Rightarrow (f(x))^2=(g(x)+cx+d)^2= (g(x))^2+ 2(cx+d)g(x)+ (cx+d)^2\\ 由於(cx+d)^2= c^2x^2+ 2cdx +d^2 = {c^2\over 2}g(x)+ \left( 2cd-{c^2\over 2}(a-c)\right)x +\left( d^2-{c^2\over 2}(b-d)\right) \\ \Rightarrow \cases{ 2cd-{c^2\over 2}(a-c)=5\\ d^2-{c^2\over 2}(b-d)=3} \Rightarrow \cases{4cd-ac^2+c^3=10 \cdots(1)\\ 2d^2-bc^2+c^2d=6 \cdots(2)}\\ 同理,g(x)=f(x)-cx-d \Rightarrow (g(x))^2=(f(x))^2-2(cx+d)f(x)+(cx+d)^2\\ 由於(cx+d)^2=c^2x^2+2cdx+d^2 = {c^2\over 2}f(x)+\left( 2cd-{1\over 2}ac^2\right)x +\left(d^2-{1\over 2}bc^2 \right) \\ \Rightarrow \cases{2cd-{1\over 2}ac^2=1 \\d^2-{1\over 2}bc^2=1} \Rightarrow \cases{4cd-ac^2=2 \cdots(3)\\ 2d^2-bc^2=2 \cdots(4)} 代入(1)與(2) \Rightarrow \cases{c^3=8\\ c^2d=4} \Rightarrow \cases{c=2\\ d=1} \Rightarrow \cases{a=3/2\\ b=0} \\ \Rightarrow \cases{f(x)=2x^2+{3\over 2}x\\ g(x)=2x^2-{1\over 2}x-1} \Rightarrow f(x)+g(x)= \bbox[red, 2pt]{4x^2+x-1}$$

解答：

$$\cases{\overline{AR}=1/2 \\ \overline{OA}=1} \Rightarrow \overline{OR}=\sqrt 3/2 \Rightarrow \angle OAR={\pi\over 3} \Rightarrow \angle RAP =2\pi-{\pi\over 3}-{\pi \over 2}={7\over 6}\pi \\ \Rightarrow 梯形OAPQ+直角\triangle OAR+扇形APR面積={3\over 8}+{\sqrt 3\over 8}+{7\pi\over 48} \Rightarrow R=2\left( {3\over 8}+{\sqrt 3\over 8}+{7\pi\over 48}\right) \\= \bbox[red, 2pt]{18+6\sqrt 3+7\pi\over 24}$$

二、計算題：16 分(第 1 題 10 分，第 2 題 6 分) 

解答：$$已知\cases{E(X)=30\\ \sigma(X)=4\\ E(Y)=5\\ \sigma(Y)=5\\ \sigma(X+Y)=8} \Rightarrow Var(X+Y)=8^2= 4^2+5^2+2 Cov(X,Y) \\ \Rightarrow Cov(X,Y) ={23\over 2} \Rightarrow 相關係數\rho={Cov(X,Y) \over \sigma(X) \sigma(Y)} ={23/2\over 4\cdot 5} ={23\over 40}= \bbox[red, 2pt]{0.575}$$

解答：$$43x^2+14\sqrt 3xy+57y^2= [x,y] \begin{bmatrix}43& 7\sqrt 3\\ 7\sqrt 3& 57 \end{bmatrix}  \begin{bmatrix} x\\ y\end{bmatrix} \\令A= \begin{bmatrix}43& 7\sqrt 3\\ 7\sqrt 3& 57 \end{bmatrix}  \Rightarrow A的特徵值\lambda=36,64 \Rightarrow 43x^2+14\sqrt 3xy+57y^2=36x'^2+64y'^2=576\\ \Rightarrow {x'^2\over 16}+{y'^2\over 9}=1 \Rightarrow \cases{a=4\\ b=3} \Rightarrow 面積=ab\pi= \bbox[red, 2pt]{12\pi}$$

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