111年嘉義高中教甄-數學詳解

國立嘉義高級中學111學年度第1次教師甄選

填充題：共 20 題，每題 5 分，合計 100 分

解答： $$2^n是13位數\Rightarrow 12\le \log 2^n \lt 13 \Rightarrow 12\le n\log 2\lt 13 \Rightarrow {12\over 0.301}\le n \lt {13\over 0.301}\\ \Rightarrow 39.86\le n\lt 43.19 \Rightarrow n=\bbox[red, 2pt]{40,41,42,43}$$

解答： $$1^2-2^2+3^2-4^2+\cdots +39^2-40^2+41^2 = \sum_{k=1}^{21} (2k-1)^2 - \sum_{k=1}^{20} (2k)^2 \\ =\sum_{k=1}^{21}\left( 4k^2-4k+1\right)-\sum_{k=1}^{20}\left(4k^2 \right) =4\cdot 21^2-4\sum_{k=1}^{21}k +21 =4\cdot 441-4\cdot \cfrac{21\cdot 22  }{2}+21\\ =1764-924+21 =\bbox[red, 2pt]{861}$$

解答： $$\cases{n+11 =7k\\ n+7=11m} \Rightarrow \cases{n+18= 7k+7 =7k'\\ n+18=11m+11 =11 m'} \Rightarrow n+18是7的倍數也是11的倍數\\ \Rightarrow n+18是77的倍數\Rightarrow n+18 = 77t \Rightarrow \bbox[red, 2pt]{n=77t-18,t\in \mathbb{Z}}$$

解答： $$甲寄出\cases{遺失(p=0.2)\\ 乙收到(p=0.8) \cases{回信(p=0.6) \cases{遺失(p=0.2)\\ 甲收到(p=0.8)}\\ 不回信(p=0.4)}} \\ \Rightarrow {甲沒收到回信且乙有收到信\over 甲沒收到回信} ={0.8\cdot 0.4+ 0.8\cdot 0.6\cdot 0.2 \over 0.2 +0.8\cdot 0.4+ 0.8\cdot 0.6\cdot 0.2} ={52/125\over 77/125} =\bbox[red, 2pt]{52\over 77}$$

解答： $$x+2y+3=0 的方向向量為\vec n=(2,-1)，又\vec a及\vec b在\vec n上有相同的正射影 \Rightarrow \vec a\cdot \vec n= \vec b\cdot \vec n\\ \Rightarrow (1,m)\cdot (2,-1)= (n,2)\cdot (2,-1) \Rightarrow m+2n=4 \Rightarrow (m^2+n^2 )(1^2+2^2)\ge (m+2n)^2\\ \Rightarrow m^2+n^2 \ge 4^2/5={16\over 5} \Rightarrow |\vec a|^2 +|\vec b|^2 = m^2+n^2 +5 \ge {16\over 5}+5= \bbox[red, 2pt]{41\over 5}$$

解答： $$\cases{L_1:{x-a\over 2} ={y-3\over 1} ={z-2\over -1} \\ L_2: {x-1\over 1} ={y-2\over 1} ={z-a\over -2}} \Rightarrow \cases{L_1通過P(a,3,2)且方向向量為(2,1,-1)\\ L_2通過Q(1,2,a)且方向向量為(1,1,-2)} \\ \Rightarrow 平面的法向量為 (2,1,-1)\times (1,1,-2) =(-1,3,1) \\\Rightarrow 平面方程式: -x+3y+z=k \Rightarrow x-3y-z=d \Rightarrow \cases{b=-3\\ c=-1}\\ 再將P,Q代入平面方程式\Rightarrow \cases{a-9-2=d\\ 1-6-a=d} \Rightarrow a-11=-5-a \Rightarrow a=3 \Rightarrow d=-8 \\ \Rightarrow (a,b,c,d) =\bbox[red, 2pt]{(3,-3,-1,-8)}$$

解答： $$A=\begin{bmatrix} -3 & 1\\ 1 & -3\end{bmatrix} \Rightarrow \det(A-\lambda I)=0 \Rightarrow \lambda =-2,-4\\ \cases{\lambda=-2 \Rightarrow (A-\lambda I)X=0 \Rightarrow X=(t,t)\\ \lambda =-4 \Rightarrow (A-\lambda I)X=0 \Rightarrow X=(s,-s)} \Rightarrow P=\begin{bmatrix} t & s\\ t & -s\end{bmatrix} =-2st =-2 \\由於a,b,c,d\in \mathbb{Z}，因此 \cases{取s=t=1 \Rightarrow P=\begin{bmatrix} 1 & 1\\ 1 & -1\end{bmatrix} \\ 取s=t=-1  \Rightarrow P=\begin{bmatrix} -1 & -1\\ -1 & 1\end{bmatrix}} \Rightarrow P=\bbox[red,2pt]{\begin{bmatrix} 1 & 1\\ 1 & -1\end{bmatrix} 或\begin{bmatrix} -1 & -1\\ -1 & 1\end{bmatrix}}$$

解答： $$最大角為\theta \Rightarrow \cos \theta ={5+6-7\over 2\sqrt{30}} ={2\over \sqrt{30}} \Rightarrow \sin \theta =\sqrt{26\over 30} \Rightarrow \triangle ABC={1\over 2}\cdot \sqrt 5\cdot \sqrt 6\cdot \sqrt{26\over 30} ={1\over 2}\sqrt{26} \\ \Rightarrow \triangle A'B'C' =\triangle ABC \cdot \det(T)={1\over 2}\sqrt{26} \cdot \begin{vmatrix}1 & 2\\ -1 & 4 \end{vmatrix} ={1\over 2}\sqrt{26} \cdot 6 =\bbox[red, 2pt]{3\sqrt{26}}$$

解答： $$a_n= 1+3+5+\cdots +4n+1 = (4n+2)(2n+1)\div 2 = (2n+1)^2 \Rightarrow \lim_{n\to \infty}{a_n\over n^2} = \lim_{n\to \infty}{(2n+1)^2 \over n^2} \\= \bbox[red, 2pt]{4}$$

解答： $$\Gamma:9(x^2-2x+1)+ 16(y^2-4y+4)=144 \Rightarrow \cfrac{(x-1)^2}{16}+\cfrac{(y-2)^2}{9}=1\\ P\in \Gamma  \Rightarrow P(4\cos \theta+1,3\sin \theta+2) \Rightarrow d(P,L)= \left|\cfrac{8\cos\theta-18-15\sin \theta }{\sqrt{4+25}} \right| \\ =\cfrac{|17\sin(\alpha-\theta)-18|}{\sqrt{29}} \Rightarrow 最大值發生在8\cos\theta-15\sin \theta=17，即\cases{\cos\theta = 8/17 \\ \sin \theta =-15/17}\\ \Rightarrow P({32\over 17}+1,{-45\over 17}+2) =\bbox[red, 2pt]{({49\over 17},-{11\over 17})}$$

解答： $$\cases{a,b,f,g任排有4!排法\\ 在5個間隙插入c,d,e，有C^5_3插法\\ c,d,e任排有3!=6種排法} \Rightarrow c,d,e不相鄰有4!\times C^5_3\times 3=1440排法;\\ \cases{a,b相鄰,f,g任排有3!\times 2=12排法\\ 在4個間隙插入c,d,e，有C^4_3插法\\ c,d,e任排有3!=6種排法} \Rightarrow c,d,e不相鄰且a,b相鄰有12\times C^4_3\times 3=288排法;\\ 因此a,b不相鄰且c,d,e不相鄰有1440-288= \bbox[red, 2pt]{1152}種$$

解答：

$$過P作垂直線，分別交\overleftrightarrow{AD}與x軸交於R、S，並令\overline{PS}=h，見上圖;\\\overline{PS}=h \Rightarrow \overline{CS}= h/\sqrt 3 \Rightarrow \overline{SO}= 1-{h\over \sqrt 3}  =\overline{RD} \Rightarrow \cos \angle RAP = \\ \cases{\triangle ADP \sim \triangle QCP \Rightarrow \overline{DP}: \overline{PC} =\overline{AD}: \overline{QC}\\ \triangle CSP \sim \triangle DRP \Rightarrow \overline{DP}: \overline{PC} = \overline{RP}: \overline{PS}} \Rightarrow \cfrac{\overline{AD}}{ \overline{QC}} = \cfrac{\overline{RP}}{ \overline{PS}} \Rightarrow \cfrac{2}{ \overline{QC}} = \cfrac{\sqrt 3-h}{ h} \\ \Rightarrow \overline{QC} ={2h\over \sqrt 3-h} \Rightarrow \cases{\triangle ADP = \overline{AD}\times \overline{RP}\div 2 = \sqrt 3-h\\ \triangle CQP= \overline{QC}\times \overline{PS}\div 2= h^2/(\sqrt 3-h)} \\ \Rightarrow f(h)=\triangle ADP+\triangle CQP = (\sqrt 3-h)+{h^2 \over \sqrt 3-h}，則 f'(h)=0 \Rightarrow -1+{2h\over \sqrt 3-h}+ {h^2\over (\sqrt 3-h)^2}=0\\ \Rightarrow -(\sqrt 3-h)^2 +2h(\sqrt 3-h)+h^2=0 \Rightarrow 2h^2-4\sqrt 3h+3=0 \Rightarrow h=\sqrt 3-\sqrt 6/2\\ \Rightarrow f(\sqrt 3-\sqrt 6/2) = {\sqrt 6\over 2}+ \cfrac{\left({2\sqrt 3-\sqrt 6\over 2} \right)^2}{\sqrt 6/2} ={\sqrt 6\over 2}+ {9-6\sqrt 2\over \sqrt 6} =\bbox[red, 2pt]{2\sqrt 6-2\sqrt 3}$$

解答：

$$\cos \angle C={1\over 2}= {6^2+2^2-\overline{BD}^2\over 2 \cdot 2\cdot 6}={40-\overline{BD}^2 \over 24} \Rightarrow \overline{BD}=2\sqrt 7 \Rightarrow {2\over \sin \angle CBD} ={2\sqrt 7\over \sqrt 3/2}\\ \Rightarrow \sin \angle CBD = {\sqrt 3\over 2\sqrt 7} \Rightarrow \overline{CP} = \overline{BC}\cdot \sin \angle CBD = 6\cdot {\sqrt 3\over 2\sqrt 7} =3\sqrt{3\over 7}\\ \overline{DP}\parallel \overline{AQ} \Rightarrow \cfrac{\overline{CD}}{\overline{CA}} =\cfrac{\overline{CP}}{ \overline{CQ}} \Rightarrow {2\over 6} ={3\sqrt{3\over 7} \over \overline{CQ}} \Rightarrow \overline{CQ} =9\sqrt{3\over 7} =\bbox[red, 2pt]{{9\over 7}\sqrt{21}}$$

解答：

$$假設\overline{CE}與\overline{AB}交於O，並以\cases{O為空間的原點\\ \overleftrightarrow{AB}為y軸\\ \overleftrightarrow{CE}為x軸};\\ 由於\cases{\angle AQB=90^\circ \\ \overline{AB}=4\\ \overline{QB}=2} \Rightarrow \angle ABC=60^\circ \Rightarrow \overline{OE}=\overline{BE}\cdot \sin 60^\circ =3\sqrt 3 \Rightarrow 摺起後E(0,0,3\sqrt 3);\\又\cases{\overline{DP} =\overline{CD}\sin 30^\circ = 2\\ \overline{CP}= \overline{CD}\cos 60^\circ =2\sqrt 3 \Rightarrow \overline{OP} =\sqrt 3} \Rightarrow D(-\sqrt 3,2,0) \Rightarrow \overline{DE}= \sqrt{3+4+27} =\bbox[red, 2pt]{\sqrt{34}}$$

解答： $$E_1\parallel E_2 \Rightarrow 角柱高h= d(E_1,E_2) = {9\over \sqrt 3} =3\sqrt 3，又\cases{P= E_1\cap E_3\cap E_4 =(11/3, -1/3,-4/3)\\ Q=E_1\cap E_4\cap E_5 = (3,1,-2)\\ R= E_1\cap E_3\cap E_5 =(5/3,-1/3, 2/3)} \\ \Rightarrow \cases{\vec u= \overrightarrow{PQ} =(-2/3,4/3,-2/3)\\ \vec v= \overrightarrow{PR} =(-2,0,2)} \Rightarrow \triangle PQR ={1\over 2} \sqrt{|\vec u|^2 |\vec v|^2 -(\vec u\cdot \vec v)^2} ={1\over 2}\sqrt {64/3-0} =4/\sqrt 3\\ \Rightarrow 三角柱體積={4 \over \sqrt 3}\times 3\sqrt 3= \bbox[red, 2pt]{12}$$

解答：

$$通過Q(2,1)的直線L:y=m(x-2)+1 代入y=f(x)=x^2-x \Rightarrow x^2-x=m(x-2)+1\\ \Rightarrow x^2-(m+1)x+2m-1=0 \Rightarrow 判別式=0，即(m+1)^2-4(2m-1)=0 \Rightarrow m^2-6m+5=0\\ \Rightarrow (m-5)(m-1)=0 \Rightarrow \cases{m=5 \Rightarrow L_1: y=5x-9\\ m=1 \Rightarrow L_2:y=x-1} \Rightarrow \cases{A= \Gamma \cap L_2=(1,0)\\ B=\Gamma \cap L_1=(3,6) \\ C= (x=2)\cap \Gamma =(2,2)} \\ 所圍面積=\int_1^2 (x^2-x)-(x-1)\,dx+\int_2^3 (x^2-x)-(5x-9)\,dx \\=\int_1^2 x^2-2x+1\,dx+\int_2^3 x^2-6x+9\,dx =\left. \left[ {1\over 3}x^3-x^2+x \right] \right|_1^2 +\left. \left[ {1\over 3}x^3-3x^2+ 9x \right] \right|_2^3  \\ = {1\over 3}+{1\over 3}= \bbox[red, 2pt]{2\over 3}$$

解答： $$(2)\times: 假設兩圓\cases{圓B:(x-a)^2+(y-b)^2 =r^2 \\圓C:x^2+(y+2)^2=4}相切 \Rightarrow 兩圓心距離=兩半徑之和\\ \qquad \Rightarrow a^2 +(b+2)^2 =(r+2)^2 \cdots(1);又(0,4)在圓B上\Rightarrow a^2+(b-4)^2=r^2\cdots(2)\\ \qquad 由(1)及(2)可知圓心(a,b)軌跡為雙曲線\\ (3)\times: 假設兩圓\cases{圓B:(x-a)^2+(y-b)^2 =r^2 \\圓C:x^2+y^2=49}相切 \Rightarrow 兩圓心距離=兩半徑之和 \\ \qquad \Rightarrow \cases{a^2+b^2=(r+7)^2 \\a^2+(b-4)^2=r^2} \Rightarrow a^2+b^2 =(\sqrt{a^2+(b-4)^2}+7)^2 \Rightarrow a^2+{33\over 49}(b-2)^2={33\over 4}為一橢圓 \\故選\bbox[red, 2pt]{(145)}$$

解答： $$a_{n+2}\cdot a_n + 6a_{n+2}\cdot a_{n+1} =a_{n+1}\cdot a_n \Rightarrow  a_{n+2}( a_n + 6  a_{n+1})= a_{n+1}\cdot a_n \Rightarrow \cfrac{a_n + 6  a_{n+1} }{ a_{n+1}\cdot a_n} ={1\over a_{n+2}}\\ \Rightarrow {1\over a_{n+1}} +{6\over a_n} ={1\over a_{n+2}} \Rightarrow b_{n+2}= b_{n+1}+ 6b_n, 其中\,b_n={1\over a_n} \Rightarrow b_{n+2}-b_{n+1}-6b_n,初始值\cases{b_1=1\\ b_2=1/2}\\ \Rightarrow 特徵多項式\,\lambda^2-\lambda-6=0 \Rightarrow \lambda=3,-2 \Rightarrow b_n= C_1(-2)^n +C_2\cdot 3^n;\\ 再由初始值\cases{b_1=1 =-2C_1+ 3C_2\\ b_2=1/2 =4C_1+9C_2} \Rightarrow \cases{C_1=-1/4\\ C_2=1/6} \Rightarrow b_n= {1\over a_n}= -{1\over 4}(-2)^n -{1\over 6}\cdot 3^n \\ \Rightarrow a_n =\cfrac{1}{-{1\over 4}(-2)^n -{1\over 6}\cdot 3^n} =\cfrac{1}{{1\over 2}(-2)^{n-1} -{1\over 2}\cdot 3^{n-1}} = \bbox[red, 2pt]{\cfrac{2}{ (-2)^{n-1} - 3^{n-1}}  , n\in \mathbb{N}}$$

解答： $$\cases{\sqrt x\left(1+{1\over x+y}\right)=2 \\\sqrt y\left(1 -{1\over x+y}\right)= \sqrt 2 } \Rightarrow \cases{ 1+{1\over x+y} =2/\sqrt x \cdots(1) \\ 1 -{1\over x+y} = \sqrt 2 /\sqrt y \cdots(2)}\\ \Rightarrow (1)^2 -(2)^2 =\left(1+{1\over x+y}\right)^2- \left(1 -{1\over x+y}\right)^2={4\over x} -{2\over y} \Rightarrow  {4\over x+y}= {4y-2x\over xy}\\ \Rightarrow 4xy =(x+y)(4y-2x) \Rightarrow x^2+xy-2y^2=0 \Rightarrow (x+2y)(x-y)=0\\ \Rightarrow x=y (\because x,y\ge 0 \Rightarrow x\ne -2y)代入(1) \Rightarrow 1+{1\over 2x}={2\over \sqrt x} \Rightarrow {(2x+1)^2\over 4x^2} ={4\over x} \\\Rightarrow 4x^2-12x+1=0 \Rightarrow x={3+2\sqrt 2\over 2}(負值不合)  \Rightarrow (x,y)= \bbox[red, 2pt]{({3+2\sqrt 2\over 2}, {3+2\sqrt 2\over 2})}$$

解答： $$令q=1-p，則E(n)= \sum_{n=1}^\infty nq^{n-1}p =p\sum_{n=1}^\infty{d\over dq}q^n =p{d\over dq}\sum_{n=1}^\infty q^n =p{d\over dq}{q\over 1-q} ={p\over (1-q)^2 }= \color{blue}{1\over p}\\同理,E(n(n-1)) =\sum_{n=1}^\infty n(n-1) q^{n-1}p =pq \sum_{n=1}^\infty n(n-1) q^{n-2}  =pq \sum_{n=1}^\infty {d^2\over dq^2}q^n \\= pq {d^2\over dq^2}\sum_{n=1}^\infty q^n =pq {d^2\over dq^2} {q\over 1-q} =pq \cdot {2\over (1-q)^3} ={2(1-p)\over p^2} \\\Rightarrow E(n^2)= E(n(n-1)+n) =E(n(n-1)) +E(n) ={2(1-p)\over p^2} +{1\over p}= \color{blue}{2-p\over p^2}\\ E(n(n-1)(n-2)) = \sum_{n=1}^\infty n(n-1)(n-2) q^{n-1}p = pq^2\sum_{n=1}^\infty n(n-1)(n-2) q^{n-3 } \\=pq^2\sum_{n=1}^\infty {d^3\over dq^3} q^n =pq^2 {d^3\over dq^3}\sum_{n=1}^\infty  q^n =pq^2 {d^3\over dq^3}{q\over 1-q}= pq^2 \cdot {6\over (1-q)^4} =\color{blue} {6(1-p)^2\over p^3}\\ 最後E(n^3) = E(n(n-1)(n-2)) +3E(n^2)-2E(n) ={6(1-p)^2\over p^3}+3\cdot {2-p\over p^2}-2\cdot {1\over p} \\=\bbox[red, 2pt]{p^2-6p+6 \over p^3}\\註:如果你有背幾何分布的期望值及變異數，就可以很快算出答案$$

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