115年鳳山高中教甄-數學詳解

 115 學年度國立鳳山高級中學教師甄選

一、 填充題，每題 6 分，共 72 分

解答:$$m^3+8n^3+18mn =27 \Rightarrow m^3+(2n)^3+(-3)^3-3(m\cdot 2n\cdot (-3)) \\=(m+2n-3)(m^2+4n^2+9- 2mn +6n+4m)=0 \Rightarrow m+2n=3 \\ 算幾不等式:{(m/3)+ (m/3)+ (m/3)+2n\over 4} \ge \sqrt[4]{{m^3\over 27}\cdot 2n} \Rightarrow {3\over 4} \ge \sqrt[4]{{m^3\over 27}\cdot 2n} \\ \Rightarrow {81\over 256 } \ge {2m^3n\over 27} \Rightarrow m^3n \le {81\cdot 27\over 256\cdot 2} = \bbox[red, 2pt]{2187\over 512}$$

解答:$$A^{-1} 不存在\Rightarrow \det(A)=ad-bc =0 \Rightarrow ad=bc \\ \textbf{Case I }ad= bc=0 \Rightarrow ad=0的情形:(a,d)=(0,0),(0,\pm 1),(0,-2),(\pm 1, 0),(-2,0)，共7種\\ \qquad \Rightarrow bc=0也有7種，因此共有7\times 7=49種\\ \textbf{Case II }ad=bc=1 \Rightarrow (a,d)=(1,1),(-1,-1)，共2種;bc=1也有2種，共4 種\\ \textbf{Case III }ad=bc=-1 \Rightarrow (a,d)=(1,-1),(-1,1)，共2種;bc=-1也有2種，共4種\\ \textbf{Case IV }ad=bc=2 \Rightarrow (a,d)=(-1,-2),(-2,-1)，共2種;bc=2也有2種，共4種\\ \textbf{Case V }ad=bc=-2 \Rightarrow (a,d)=(-2,1),(1,-2)，共2種;bc=-2 也有2種，共4種\\\textbf{Case VI }ad=bc=4 \Rightarrow (a,d)=(-2,-2)，共1種;bc=4也有1種，共1種\\ 總共有49+4\times 4+1=66種,機率為{66\over 4^4} =\bbox[red ,2pt]{33\over 128}$$

解答:$$A,D,E在平面z=0上，構成了邊長為2的正三角形\\ 正八面體的高h=\sqrt {2\over 3} \cdot 2={2\sqrt 6\over 3},而B位於\triangle ADE的上方,即B到y軸距離=h= \bbox[red, 2pt]{2\sqrt 6\over 3}$$

解答:

$$\Gamma:{x^2\over 36}+{y^2\over 27}=1 \Rightarrow \cases{a=6\\ b=3\sqrt 3} \Rightarrow c=3 \Rightarrow F(3,0)為右焦點 \Rightarrow 離心率e={c\over a}={1\over 2} \\ \Rightarrow 右準線方程式:x={a^2\over c} =12 \Rightarrow L:x=12剛好為右準線 \\ \Rightarrow {\overline{FA_k} \over d_k}=e={1\over 2} \Rightarrow \overline{FA_k} ={1\over 2}d_k\\ 假設 \cases{A_k(x_k,y_k) \\ \overline{FA_k}與正x軸的夾角為\theta_k}\Rightarrow x_k-3=\overline{FA_k} \cos \theta_k \Rightarrow x_k=\overline{FA_k} \cos \theta_k+3 \\ \Rightarrow d_k=12-x_k=12-(\overline{FA_k} \cos \theta_k+3) =9-\overline{FA_k} \cos \theta_k =9-{1\over 2}d_k\cos \theta_k \\ \Rightarrow d_k \left( 1+{1\over 2}\cos \theta_k \right) =9 \Rightarrow {1\over d_k}={2+\cos \theta_k\over 18} \Rightarrow \sum_{k=0}^{11} {1\over d_k} =\sum_{k=0}^{11}{2+\cos \theta_k\over 18} \\=\sum_{k=0}^{11}{1\over 9}+{1\over 18}\sum_{k=0}^{11} \cos \theta_k ={12\over 9}+0= \bbox[red, 2pt]{4\over 3}$$

解答:

$$L:{x-5\over 3} ={y-5\over 2}={z-4\over -2} \Rightarrow \cases{L通過P(5,5,4)\\ L方向向量\vec v=(3,2,-2)}\\ 假設L_1=E_1\cap E_2 \Rightarrow \cases{L_1通過O(0,0,0) \\ L_1 \parallel L} \Rightarrow d(L,L_1)= {|\overrightarrow{OP} \times \vec v| \over |\vec v|} ={|(-18,22,-5)| \over \sqrt{9+4+4}}=7 \\ 作一個垂直於L_1 的截平面，在此截面上\cases{L_1投影為一點O'\\ L投影為一點P',已知\overline{O'P'}=7\\ E_1與E_2投影為兩條通過O'的直線} \\ \Rightarrow \sin {\theta\over 2}={2\over 7} \Rightarrow \cos {\theta\over 2}={3\sqrt 5\over 7} \Rightarrow \sin \theta=2\sin {\theta\over 2} \cos {\theta\over 2}= \bbox[red, 2pt]{12\sqrt 5\over 49}$$

解答:$$g(x)={x-3\over x+1} \Rightarrow g(g(x))={x+3\over 1-x} \Rightarrow g(g(g(x))) =x \Rightarrow 循環數3\\  \Rightarrow \begin{cases} f(x) + 2f\left(\frac{x-3}{x+1}\right) = x & \cdots (1) \\f\left(\frac{x-3}{x+1}\right) + 2f\left(\frac{x+3}{1-x}\right) = \frac{x-3}{x+1} & \cdots (2) \\f\left(\frac{x+3}{1-x}\right) + 2f(x) = \frac{x+3}{1-x} & \cdots (3) \end{cases} \quad 式(3)\Rightarrow  f\left(\frac{x+3}{1-x}\right)={x+3\over 1-x}-2f(x) 代入(2) \\ \Rightarrow f\left(\frac{x-3}{x+1}\right)={x-3\over x+1}-{2(x+3) \over 1-x}+4f(x) 代入(1) \Rightarrow 9f(x)=x-{2(x-3) \over x+1}-{4(x+3) \over x-1} \\ \Rightarrow f(x)= \bbox[red, 2pt]{x^3-6x^2-9x-19\over 9(x^2-1)}$$

解答:$$X\sim Geo(p=0.1) \Rightarrow P(X=k)=0.9^{k-1}\cdot 0.1 \Rightarrow P(X\ge c) = \sum_{k=c}^\infty 0.9^{k-1}\cdot 0.1=0.1\cdot {0.9^{c-1} \over 1-0.9} \\=0.9^{c-1}\le 0.1 \Rightarrow (c-1)\log 0.9\le \log(0.1) \Rightarrow (c-1)(2\log 3-1)\le -1 \Rightarrow c-1\ge {-1\over 0.0458} \\ \Rightarrow c\ge 22.85 \Rightarrow 拒絕域 \bbox[red, 2pt]{\{X\ge 23\}}$$

解答:$$\cos^2 x+\cos^2 y= {1\over 2}(\cos 2x+1)+{1\over 2}(\cos 2y+1) =1+{1\over 2}(\cos 2x+\cos 2y) \\=1+{1\over 2}\cdot 2\cos(x+y)\cos(x-y)=1+\cos(x+y) \cos 60^\circ=1+{1\over 2}\cos(x+y) \\ \Rightarrow \cases{最大值=1+1/2 =3/2\\ 最小值=1-1/2 =1/2} \Rightarrow (M,m) = \bbox[red, 2pt]{\left( {3\over 2},{1\over 2} \right)}$$

解答:$$x^2-(3a+1)x-2a+2=0有虛根\Rightarrow \Delta=(3a+1)^2-4(-2a+2)\lt0 \Rightarrow 9a^2+14a-7\lt 0 \\ z及\bar z為其兩根\Rightarrow \begin{cases}z + \bar{z} = 3a+1\\z\bar{z} = -2a+2 \end{cases} \\已知z^3為實數\Rightarrow z^3=\bar z^3 \Rightarrow z^3-\bar z^3=(z-\bar z)(z^2+ z \bar z+ \bar z^2)=0 \\ z為虛根\Rightarrow z\ne \bar z \Rightarrow z^2+ z \bar z+ \bar z^2=0 \Rightarrow (z+\bar z)^2-z\bar z=0 \Rightarrow (3a+1)^2-(-2a+2)=0 \\ \Rightarrow 9a^2+8a-1=0 \Rightarrow (9a-1)(a+1)=0 \Rightarrow a= \bbox[red, 2pt]{-1或{1\over 9}}$$

解答:$$\cases{A(2,4) \\B(8,13) \\C(1,1) \\ P(x,y)} \Rightarrow \cases{\overrightarrow{AP}= (x-2,y-4) \\ \overrightarrow{BP} =(x-8,y-13) \\ \overrightarrow{CP}=(x-1,y-1)} \Rightarrow \cases{\overrightarrow{AP} \cdot \overrightarrow{BP} =x^2-10x+16+y^2-17y+52 \\ |\overrightarrow{CP}|^2 =x^2+y^2-2x-2y+2} \\ \Rightarrow \overrightarrow{AP} \cdot \overrightarrow{BP} = |\overrightarrow{CP}|^2 \Rightarrow 8x+15y-66=0 為一直線，設為L\\ 在\overline{AB}上取一點M,滿足2\overrightarrow{MA}+ \overrightarrow{MB}=0 \Rightarrow M=(2A+B)/3=(4,7) \\ 欲求|2\overrightarrow{AP} +\overrightarrow{BP}| =|2(\overrightarrow{MP}-\overrightarrow{MA})+(\overrightarrow{MP}-\overrightarrow{MB})|= 3|\overrightarrow{MP}| \\ |\overrightarrow{MP}|的最小值=d(M,L)={71\over 17} \Rightarrow |2\overrightarrow{AP} +\overrightarrow{BP}|最小值={3\cdot 71\over 17} =\bbox[red, 2pt]{213\over 17}$$

解答:$$\vec a=(x,y,z) \Rightarrow \text{proj}_{(1,2,2)} \vec a=(1,2,2) \Rightarrow {\vec a\cdot (1,2,2) \over |(1,2,2)|} \cdot {(1,2,2)\over |(1,2,2)|} =(1,2,2) \\ \Rightarrow  {\vec a\cdot (1,2,2) \over |(1,2,2)|^2}=1 \Rightarrow {x+2y+2z\over 9}=1 \Rightarrow x+2y+2z=9 \cdots(1) \\ 同理, \cases{ {\vec a\cdot (2,3,6)\over |(2,3,6)|^2}=1 \\{\vec a\cdot (3,1,4)\over |(3,1,4)|^2}=1} \Rightarrow \cases{2x+3y+6z=49 \cdots(2) \\ 3x+y+4z=26 \cdots(3)} \;由(1),(2),(3)可得\cases{x=-7\\y=-5\\ z=13} \\\Rightarrow \vec a= \bbox[red, 2pt]{(-7,-5,13)}$$

解答:$$f(x)=x^3-{3\over 2}x^2-{1\over 4}x+1 \Rightarrow f(1-x)=-x^3+{3\over 2}x^2+{1\over 4}x +{1\over 4} \\ \Rightarrow f(x)+ f(1-x)= {5\over 4} \Rightarrow f \left( {k\over 2026} \right)+f \left( {2026-k\over 2026} \right) ={5\over 4} \\ \Rightarrow 2 \sum_{k=0}^{2026} f\left( {k\over 2026} \right) =2027\times {5\over 4} \Rightarrow \sum_{k=0}^{2026} f\left( {k\over 2026} \right) =2027\times {5\over 8} ={10135\over 8} \\ \Rightarrow \sum_{k=1}^{2026} f\left( {k\over 2026} \right) ={10135\over 8}-f(0)={10135\over 8}-1= \bbox[red, 2pt]{10127\over 8}$$

二、 計算證明題，共 28 分

解答:$$取R=\sqrt{a^2+b^2} \Rightarrow a\cos x+b\sin x=c \Rightarrow R\cos (x-\phi)=c \Rightarrow \cos(x-\phi)={c\over R}, 其中\cases{\cos \phi=a/R\\ \sin \phi=b/R}\\ \alpha,\beta為相異兩根\Rightarrow \cases{\cos (\alpha-\phi) =c/R \cdots(1)\\ \cos(\beta-\phi)=c/R \cdots(2)} \Rightarrow (1)-(2)=\cos(\alpha-\phi)-\cos(\beta-\phi)=0 \\ \Rightarrow -2\sin {\alpha+\beta-2\phi\over 2} \sin{\alpha-\beta\over 2}=0 \Rightarrow \sin {\alpha+\beta-2\phi\over 2}=0 \;(\alpha-\beta \ne k\pi \Rightarrow \sin(\alpha-\beta)/2 \ne 0) \\ \Rightarrow \cos^2 {\alpha+\beta-2\phi\over 2} =1-\sin^2{\alpha+\beta-2\phi\over 2}=1\\ (1)+(2)=\cos(\alpha-\phi)+ \cos(\beta-\phi)={2c\over R} \Rightarrow 2\cos {\alpha+\beta-2\phi \over 2} \cos {\alpha-\beta\over 2}={2c\over R} \\ \Rightarrow \cos {\alpha+\beta-2\phi \over 2} \cos {\alpha-\beta\over 2}={ c\over R} \Rightarrow \cos^2 {\alpha+\beta-2\phi \over 2} \cos^2 {\alpha-\beta\over 2}={ c^2\over R^2} \\ \Rightarrow 1\cdot \cos^2{\alpha-\beta\over 2} ={c^2\over R^2} \Rightarrow \cos^2{\alpha- \beta\over 2}={c^2\over a^2+b^2},\bbox[red, 2pt]{故得證}$$

解答:$$\textbf{法一 }標準式:已知\cases{對稱軸平行y軸\\ 焦點F(2,-1)}  \Rightarrow 對稱軸:x=2 \Rightarrow \cases{頂點V(2,k)\\ 焦距c} \Rightarrow -1=k+c \\ \Rightarrow k=-1-c \Rightarrow V(2,-1-c) \Rightarrow 拋物線標準式:(x-h)^2=4c(y-k) \\ \Rightarrow (x-2)^2=4c(y+1+c)通過P(-1,3) \Rightarrow (-1-2)^2=4c(3+1+c) \Rightarrow 4c^2+16c-9=9 \\ \Rightarrow (2c-1)(2c+9)=0 \Rightarrow \cases{c=1/2\Rightarrow  (x-2)^2=2(y+3/2) \\c=-9/2 \Rightarrow (x-2)^2=-18(y-7/2)}  \\\textbf{法二 }拋物線定義: 對稱軸平行y軸\Rightarrow 準線L:y=d\\ 已知\cases{P(-1,3) \\F(2,-1)} \Rightarrow \overline{PF}=5 \Rightarrow d(P,L)=5 \Rightarrow |3-d|=5 \Rightarrow d=-2或8 \\ \Rightarrow \cases{d=-2 \Rightarrow \sqrt{(x-2)^2+(y+1)^2}=|y+2| \Rightarrow (x-2)^2=2y+3 \\d=8 \Rightarrow \sqrt{(x-2)^2+(y+1)^2}=|y-8| \Rightarrow (x-2)^2=-18y+63}\\ \bbox[red, 2pt]{方法二比較好},利用定義可理解曲線幾何意義,省去背公式$$

解答:$$\textbf{(1) }x^2+(a-2)x+(a^2-a-2)=0有兩實根b,c \Rightarrow \Delta=(a-2)^2-4(a^2-a-2)\ge 0 \\ \Rightarrow a^2\le 4\Rightarrow (a-2)(a+2)\le 0 \Rightarrow \bbox[red, 2pt]{-2\le a\le 2} \\ \textbf{(2) }\cases{兩根之和=b+c=2-a\\ 兩根之積=bc=a^2-a-2} \Rightarrow b^3+c^3=(b+c)((b+c)^2-3bc) \\=(2-a)((2-a)^2-3(a^2-a-2)) = 2a^3-3a^2-12a+20 \\ \Rightarrow f(a)=(a+1)^3+b^3+c^2=(a^3+3a^2+3a+1)+( 2a^3-3a^2-12a+20) \\ =3a^3-9a+21 \Rightarrow f'(a)=9a^2-9=0 \Rightarrow a=\pm 1 \Rightarrow \cases{f(2)= 27\\ f(1)=15\\ f(-1)=27\\ f(-2)=15} \Rightarrow 最小值:\bbox[red, 2pt]{15}$$

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解題僅供參考，其他教甄試題及詳解