107年臺中文華高中教甄-數學詳解

臺中市立文華高級中等學校 107 學年度第 1 次教師甄選
數學科專業知能試題本(填充題公告)

一、填充題

解： $$a_1=1 \Rightarrow a_2= \sqrt{S_1} +\sqrt{S_2}=1+\sqrt{1+a_2} \Rightarrow a_2=3 \Rightarrow a_3=\sqrt{S_2}+\sqrt{S_2+a_3} =2+\sqrt{4+a_3} \\ \Rightarrow a_3=5 \Rightarrow \cdots \Rightarrow a_n=2n-1 \Rightarrow a_{107} =2\times 107-1= \bbox[red,2pt]{213}$$

 解： $$f(x)=(x^2-x+3)^5 = \sum_{n=0}^5 C^5_nx^{2n} (3-x)^{5-n} \Rightarrow x^7係數= C^5_2(-1)^3 +C^5_3(-6)\\ =-10-60= \bbox[red,2pt]{-70}$$

解： $$此題相當於x+y+z=7,x,y,\in Z，再乘上旗子的排列數\\，即H^3_7 \times {7!\over 3!2!2!} =36\times 210 = \bbox[red,2pt]{7560}$$

解： $$很顯然\sqrt 6為x^3-3\sqrt 6x^2+17x-5\sqrt 6=0之一根，再利用長除法可得:\\x^3-3\sqrt 6x^2+17x-5\sqrt 6= (x-\sqrt 6)(x^2-2\sqrt 6x-5) =(x-\sqrt 6)(x+1-\sqrt 6)( x-1-\sqrt 6)\\ \Rightarrow 三根為\sqrt 6,-1+\sqrt 6,1+\sqrt 6;\\令 2S=三根之和=3\sqrt 6 \Rightarrow S={3\over 2}\sqrt 6 \\ \Rightarrow \triangle 面積=\sqrt{s(s-\sqrt 6)(s+1-\sqrt 6)(s-1-\sqrt 6)} =\sqrt{{3\sqrt 6\over 2} \cdot {\sqrt 6\over 2} \cdot (1+{\sqrt 6\over 2}) \cdot (-1+{\sqrt 6\over 2})} \\ =\sqrt{{9\over 2} \cdot {1\over 2}} =\bbox[red,2pt]{3\over 2}$$

解：

$$令\overline{CD}=a \Rightarrow {\triangle ABD \over \triangle ACD} ={\overline{BD} \over \overline{CD}} ={1\over a}\cdots (1);\\ 又\cases{\triangle ABD = {1\over 2}\overline{AD}\cdot \overline{AB}\sin 30^\circ \\ \triangle ACD ={1\over 2}\overline{AD}\cdot \overline{AC}} \Rightarrow {\triangle ABD \over \triangle ACD} = {\overline{AB}/2 \over \overline{AC} } = {1\over 2}\overline{AB} \cdots(2) \\ (1)=(2) \Rightarrow {1\over a}={\overline{AB} \over 2} \Rightarrow \overline{AB}={2\over a} \Rightarrow \cos \angle BAC = \cfrac{\overline{AC}^2 +\overline{AB}^2 -\overline{BC}^2}{2\overline{AB} \cdot \overline{AC}} \\\Rightarrow -{1\over 2}={1+4/a^2-(a+1)^2 \over 4/a} \Rightarrow a^4+2a^3-2a-4=0 \Rightarrow a^3(a+2)-2(a+2)=0 \\\Rightarrow (a^3-2)(a+2)=0 \Rightarrow a^3=2 \Rightarrow a=\bbox[red,2pt]{\sqrt[3]{2}}$$

解： $$令\cases{A(1,2,3)\\ B(-2,-1,2)} \Rightarrow \cases{|\overrightarrow{OA}|=\sqrt{14} \\ |\overrightarrow{OB}|=3 \\ \overrightarrow{OP} = (x-2y,2x-y, 3x+2y) = x\overrightarrow{OA} +y\overrightarrow{OB}} \\ \Rightarrow \cos \angle AOB = \cfrac{\overrightarrow{OA} \cdot \overrightarrow{OB}}{|\overrightarrow{OA}  ||\overrightarrow{OB}|} = \cfrac{2}{3\sqrt{14}} =\cfrac{\sqrt {14}}{21} \Rightarrow \sin \angle AOB= \cfrac{\sqrt{427}}{21} \\ \Rightarrow 所求面積|\overrightarrow{OA}||\overrightarrow{OB}|\sin \angle AOB \times (2-1)(1-(-1))=3\sqrt{14}\times \cfrac{\sqrt{427}}{21} \times 2= \bbox[red,2pt]{2\sqrt{122}}$$

解：

$$\cases{A(3,4,1)\\ B(1,-2,5) \\ E:2x-y+2z+9=0\\ 令f(x,y,z)=2x-y+2z+9} \Rightarrow \cases{f(A) >0 \\ f(B) >0 \\ d(E,A)=13/3 =\overline{AG}\\ d(E,B)=23/3 =\overline{BF}\\ \overline{AB}=\sqrt{56}}  \Rightarrow A,B 在E的同側，見上圖；\\ 因此\overline{BC} ={23\over 3}-{13\over 3}={10\over 3} \Rightarrow \overline{AC} =\sqrt{56-({10\over 3})^2} ={2\over 3}\sqrt{101}; \\ 令\overline{PG} =a \Rightarrow \overline{PF}={2\over 3}\sqrt{101}-a \Rightarrow \overline{PA}^2 +\overline{PB}^2= \overline{AG}^2 +\overline{PG}^2 + \overline{PF}^2+ \overline{BF}^2 \\ ={169\over 9} +a^2 + ({2\over 3}\sqrt{101}-a)^2+ {529\over 9} =2a^2 -{4\over 3}\sqrt{101}a+ {1102\over 9} \\ \Rightarrow 當a={\sqrt{101}\over 3}時，\overline{PA}^2 +\overline{PB}^2有最小值{202\over 9}-{404\over 9}+ {1102\over 9}={900\over 9}= \bbox[red,2pt]{100}$$

解： $$\cases{A(x_a,y_a)\\ (1,-4)=(A+B)/2} \Rightarrow B(2-x_a,-8-y_a); \\A,B皆在橢圓上，因此\cases{25x_a^2+4y_a^2=100 \cdots(1) \\ 25(2-x_a)^2 + 4(-8-y_a)^2 =100 \cdots(2)}\\ (2) \Rightarrow 25x_a^2-100x_a+100+4y_a^2 +64y_a+256=100 \Rightarrow -100x_a+ 64y_a+356=0 \\\Rightarrow x_a={16y_a+89\over 25} 代回(1) \Rightarrow  25\left( {16y_a+89\over 25}\right)^2+4y_a^2=100 \Rightarrow y_a=-4\pm {5\over 178}\sqrt{979}\\ \Rightarrow \overleftrightarrow{AB}斜率m={2y_a+8\over 2x_a-2} = \cfrac{2y_a+8}{2\cdot {16y_a+89\over 25}-2} = {50y_a+200\over 32y_a+128} =\cfrac{{250\over 178}\sqrt{979}}{{160\over 178}\sqrt{979}} ={25\over 16} \\ \Rightarrow \overleftrightarrow{AB}方程式: y+4={25\over 16}(x-1) \Rightarrow \bbox[red,2pt]{25x-16y=89}$$

解：

$$f(x)與g(x)互為反函數，因此兩者對稱於直線y=x；\\兩圖形y=f(x)與y=g(x)所圍面積=2\times (y=f(x)與直線y=x所圍面積)\\ y=f(x)=x^3+x^2+x = x \Rightarrow x^3+x^2=0 \Rightarrow x^2(x+1)=0 \Rightarrow x=-1,0 \\ \Rightarrow y=f(x)與直線y=x所圍面積 =\int_{-1}^0 x^3+x^2\;dx =\left. \left[ {1\over 4}x^4 +{1\over 3}x^3 \right] \right|_{-1}^0 = {1\over 12} \\ \Rightarrow 所求面積=2\times{1\over 12} =\bbox[red,2pt]{1\over 6}$$

解：

$$z_1= (2\cos \theta+3)+i(2\sin \theta+5) \Rightarrow \cases{x=2\cos\theta +3\\ y=2\sin \theta+5} \Rightarrow (x-3)^2 +(y-5)^2 =2^2 \\ \Rightarrow z_1 為一圓，圓心為O(3,5)，半徑r=2;\\ 又|z_2-z_1|=1，表示z_1與z_2的距離為1\Rightarrow z_2與z_1為同心圓，半徑為1及3(兩個圓的聯集);\\ z_3= k\omega+2 = (2-\sqrt 3k)+ki \Rightarrow \cases{x=2-\sqrt 3k\\ y=k} \Rightarrow x=2-\sqrt 3y為一直線；\\ |z_2-z_3|=z_2與z_3的距離=直線與兩圓的距離，其最小值=圓心至直線的距離再減去大圓半徑 \\ =d(O,x+\sqrt 3y=2)-3 =\cfrac{3+5\sqrt 3-2}{\sqrt{1+3}}-3 =\bbox[red,2pt]{\cfrac{-5+5\sqrt 3}{2}}$$

解： $$\begin{bmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\end{bmatrix} \begin{bmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3\end{bmatrix} =\begin{bmatrix} 16 & \alpha & 10\\ \alpha & 9 & \beta \\ 10 & \beta & 25 \end{bmatrix} \Rightarrow \cases{|\vec a|^2 =16 \\ |\vec b|^2 =9 \\ |\vec c|^2 = 25\\ \vec a\cdot \vec b=\alpha \\ \vec b\cdot \vec c=\beta \\ \vec a\cdot \vec c=10}\\ 令A=\begin{bmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\end{bmatrix} \Rightarrow AA^T=\begin{bmatrix} 16 & \alpha & 10\\ \alpha & 9 & \beta \\ 10 & \beta & 25 \end{bmatrix} \Rightarrow \det(A)^2 =\det(\begin{bmatrix} 16 & \alpha & 10\\ \alpha & 9 & \beta \\ 10 & \beta & 25 \end{bmatrix}) \\ \Rightarrow (20\sqrt 3)^2= -25\alpha^2+20\alpha\beta -16\beta^2+2700 \Rightarrow 25\alpha^2-20\alpha \beta+16\beta^2-1500=0\\ 又(\vec a+\vec c)\cdot \vec b = \vec a\cdot \vec b+ \vec b\cdot \vec c =\alpha+\beta \\ 因此我們有\cases{f(\alpha,\beta)=\alpha +\beta \\ g(\alpha,\beta) =25\alpha^2-20\alpha \beta+16\beta^2-1500}，\text{利用Lagrange 算子來求 }f\text{ 的極值} \\ \Rightarrow \cases{f_\alpha +\lambda g_\alpha=0\\ f_\beta +\lambda g_\beta \\ g=0} \Rightarrow \cases{1+\lambda(50\alpha-20\beta) =0 \\ 1+\lambda(-20\alpha+32\beta)=0 \\ g=0} \Rightarrow 50\alpha-20\beta= -20\alpha+32\beta \Rightarrow \alpha={26\over 35}\beta \\ 將\alpha={26\over 35}\beta 代入g=0 \Rightarrow \beta ={35\sqrt{305} \over 61} \Rightarrow \alpha ={26\sqrt{305} \over 61} \Rightarrow \alpha+\beta = \bbox[red,2pt]{\sqrt{305}}$$

解： $$\cases{E_n = E_{n-1}+({5\over 6})^{n-2}\\ E_2=2} \Rightarrow E_n=E_2+{5\over 6}+({5\over 6})^2+\cdots +({5\over 6})^{n-2} =2+ \cfrac{5/6-(5/6)^{n-1}}{1-5/6} \\ =2+5-6(5/6)^{n-1} = \bbox[red,2pt]{7-6\left( {5\over 6}\right)^{n-1}}$$

解： $$S(n)= a_1+a_2+\cdots +a_n=n^3-2n \\\Rightarrow a_n= S(n)-S(n-1)=n^3-2n-((n-1)^3-2(n-1)) = 3n^2-3n-1 \\ \Rightarrow \cases{ a_{2n}= 3(2n)^2-3(2n)-1= 12n^2-6n -1\\ a_{3n}=3(3n)^2-3(3n)-1 =27n^2-9n-1} \\ \Rightarrow \lim_{n\to \infty} \cfrac{\sqrt[3]{a_3+a_6+\cdots +a_{3n}} -\sqrt[3]{a_2+a_4+\cdots +a_{2n}}}{n} \\ =\lim_{n\to \infty} \cfrac{\sqrt[3]{\sum_{k=1}^n a_{3k}} -\sqrt[3]{\sum_{k=1}^n a_{2k}}}{n} \\= \lim_{n\to \infty} \cfrac{\sqrt[3]{\sum_{k=1}^n (27k^2- 9k-1)} -\sqrt[3]{\sum_{k=1}^n (12k^2-6k-1)}}{n}\\ = \lim_{n\to \infty} \cfrac{\sqrt[3]{{27\over 6}n(n+1)(2n+1)-{9\over 2}n(n+1)-n} -\sqrt[3]{2n(n+1)(2n+1)-3n(n+1)-n}}{n} \\ =\sqrt[3]{{27\over 3}} -\sqrt[3]{4} =\bbox[red,2pt]{\sqrt[3]{9}-\sqrt[3]{4}}$$

解：

$$\lim_{n\to \infty} \sum_{k=1}^n \cfrac{\sqrt{(3n+k)(n-k)}}{n^2} = \lim_{n\to \infty} \sum_{k=1}^n \cfrac{\sqrt{3n^2-2nk -k^2}}{n^2}\\ =\lim_{n\to \infty} {1\over n}\sum_{k=1}^n \sqrt{3-2{k\over n}-({k\over n})^2} = \int_0^1 \sqrt{3-2x-x^2}\;dx \\ 令y=\sqrt{3-2x-x^2 } \Rightarrow x^2+2x+y^2=3 \Rightarrow (x+1)^2+y^2 =2^2 \Rightarrow \cases{圓心O(-1,0)\\ 半徑r=2}\\ 因此積分區域(上圖著色區)為扇形區域扣除\triangle OAB={1\over 6}\cdot 2^2\pi -{1\over 2}\cdot 1\cdot \sqrt 3=\bbox[red, 2pt]{{2\pi\over 3}-{\sqrt 3\over 2}}$$

解： $$f(x)={x\over 1-x^2} =x(1+x^2+x^4+x^6+ \cdots) = x+x^3+x^5+x^7+\cdots\\ \Rightarrow f^{[7]} 的常數項為7!，即f^{[7]}(0)=7!= \bbox[red,2pt]{5040}$$

解： $$a_n=依題意所求之著色數\Rightarrow \cases{a_n+a_{n-1}=k(k-1)^{n-1} \\a_2=k(k-1) \\ a_3=k(k-1)(k-2)}，依題意\cases{k=5\\n=7} \\ \Rightarrow \cases{a_2= 5\times 4=20\\ a_3= 5\times 4\times 3 =60} \Rightarrow a_4=5\times 4^3-60 =260 \Rightarrow a_5=5\times 4^4-260 =1020 \\ \Rightarrow a_6= 5\times 4^5-1020 =4100 \Rightarrow a_7= 5\times 4^6-4100 =\bbox[red,2pt]{16380}$$

解： $$f(x)=\lim_{n\to \infty}\cfrac{x^{2n-1}+ax^2+bx}{x^{2n}+1} = \begin{cases} ax^2+bx & |x| \lt 1 \\ {1\over x} & |x| \gt 1\end{cases};\\ \cases{\lim_{x\to 1^+}f(x) =1\\ \lim_{x\to 1^{-}}f(x)=a+b\\ f(1)=\lim_{n\to \infty}\cfrac{1^{2n-1}+a+b}{1^{2n}+1}=\cfrac{a+b+1}{2}} \Rightarrow 1=a+b ={a+b\over 2} \Rightarrow a+b=1\cdots(1);\\ 同理\cases{\lim_{x\to -1^+}f(x) =a-b \\ \lim_{x\to -1^{-}}f(x)=-1\\ f(-1)=\lim_{n\to \infty}\cfrac{(-1)^{2n-1}+a-b}{(-1)^{2n}+1}=\cfrac{a-b-1}{2}} \Rightarrow a-b=-1=(a-b-1)/2 \\ \Rightarrow a-b=-1 \cdots(2)\\ 由(1)及(2)可得(a,b)=\bbox[red,2pt]{(0,1)}$$

證：

$$(x+3)(y-2)=-3 將中心點(-3,2)平移至原點(0,0)，則方程式成為xy=-3；\\此題則轉變為切線與坐標軸所圍三角形面積為定值；\\ xy=-3 \Rightarrow y+xy'=0 \Rightarrow 切點P(x_0,y_0)的切線L斜率m_L= y'(x_0)= -{y_0\over x_0} \\ \Rightarrow L: y=-{y_0\over x_0}(x-x_0)+ y_0 \Rightarrow \cases{L與y軸的交點A(0,2y_0)\\ L與x軸的交點B(2x_0,0)} \\\Rightarrow \triangle ABO面積= {1\over 2}|2x_0|\times |2y_0| =2|x_0y_0| =2|-3| =6，為定值，\bbox[red,2pt]{故得證}$$