109年中正預校教甄-數學詳解

中正國防幹部預備學校109年教師甄選測驗試題

 

解答：

$$(x+1)^{26} =\sum_{k=0}^{26}C^{26}_kx^k = 1+\sum_{k=1}^{26}C^{26}_kx^k =1+8191= 2^{13}\\ \Rightarrow x=\sqrt 2e^{i{k\pi\over 13}}-1, k=0,1,\dots,25 \Rightarrow Re(x)=\sqrt 2\cos {k\pi\over 13}-1 \\若 Re(x) \gt 0,則\cos {k\pi\over 13} \gt {1\over \sqrt 2} =\cos {\pi\over 4} \Rightarrow 0 \le k \le 3 或23\le k\le 25\\ \Rightarrow Re(x) < 0 \Rightarrow 4\le k \le 22,共\bbox[red,2pt]{19}個$$

解答：

$$令\cases{E為\overline{BC}的中點\\ \overline{EF}\bot \overline{OA}}\qquad，如上圖\Rightarrow \overline{AE}= \overline{AC} \sin 60^\circ = 4\times {\sqrt 3\over 2} =2\sqrt 3\\依題意\overline{EF}=\sqrt 3 \Rightarrow \overline{AF}^2=\overline{AE}^2- \overline{EF}^2 = 12-3=9 \Rightarrow \overline{AF}=3 \\ 直角\triangle OCE: \overline{OE}^2 = \overline{OC}^2 -\overline{EC}^2 = a^2-2^2=a^2-4\\ 直角\triangle OEF: \overline{OE}^2 = (\overline{OA}-\overline{AF})^2 +\overline{EF}^2 \Rightarrow a^2-4=  (a-3)^2+3 \Rightarrow -4=-6a+12 \\ \Rightarrow a={16\over 6} =\bbox[red,2pt]{8\over 3}$$

解答： $$\cases{a_1+a_2 +a_3+\cdots +a_{2020}= 2021\\ {1\over a_1}+{1\over a_2}+{1\over a_3}+\cdots +{1\over a_{2020}}= 2021} \Rightarrow \cases{a_2 +a_3+\cdots +a_{2020} =2021-a_1\\ {1\over a_2}+{1\over a_3}+\cdots +{1\over a_{2020}}=2021-{1\over a_1}}\\ 由於(a_2 +a_3+\cdots +a_{2020})\left({1\over a_2}+{1\over a_3}+\cdots +{1\over a_{2020}} \right) \ge (1+1+\cdots +1)^2 =2019^2\\ \Rightarrow (2021-a_1)(2021-{1\over a_1}) \ge 2019^2 \Rightarrow 2021^2-2021(a_1+{1\over a_1})+1 \ge 2019^2 \\ \Rightarrow a_1+{1\over a_1} \le {2021^2-2019^2+1 \over 2021} = \bbox[red,2pt]{8081 \over 2021}$$

解答： $$假設\cases{a+b+2c=x \\ a+2b+c= y\\ a+b+c=z} \Rightarrow \cases{a=-x-y+3z\\ b=y-z\\ c=x-z} \\\Rightarrow {2b-2c\over a+b+2c} +{2a+4c\over a+2b+c} +{b\over a+b+c} ={2y-2x\over x} +{2x-2y+ 2z \over y}+{ y-z\over z} \\={2y\over x}-2+{2x+2z\over y}-2+{y\over z}-1  =  \left({2y\over x} +{2x\over y}\right) +\left({2z\over y} +{y\over z}\right) -5\\\ge 2\sqrt{{2y\over x}\cdot {2x\over y}} +2\sqrt{{2z\over y}\cdot {y\over z}}-5 =\bbox[red,2pt]{\sqrt 8-1}$$

解答： $$\cases{a^2+b^2=4 可以視為一圓,其圓心O_1(0,0),半徑r_1=2\\(c-7)^2+(d-24)^2=36可以視為一圓,其圓心O_2(7,24),半徑r_2=6}\\ \begin{vmatrix} a & b\\ c& d\end{vmatrix}的最大值M=\sqrt{a^2+b^2} \times \sqrt{c^2+d^2} =r_1\times (\overline{O_1O_2}+r_2) = 2(25+6)=62\\ (a-c)^2+(b-d)^2 最小值m= (\overline{O_1O_2}-r_1-r_2)^2 =(25-2-6)^2 =17^2 =289\\ 因此M+m= 62+289= \bbox[red,2pt]{351}$$ 註:學校公布的答案是$\bbox[blue,2pt]{79}$

解答： $$第12階至第20階的走法:2x+3y=8 \Rightarrow (x,y)=(4,0),(1,2)\Rightarrow \cases{2222排列數1\\233排列數=3}\Rightarrow 共4種\\第1階至第12階的走法:2x+3y=12 \Rightarrow (x,y)=(6,0),(3,2),(0,4) \\ \Rightarrow \cases{222222經過第8階\\ 22233排列數10，需扣除經過第8階的\cases{33222\\ 32322\\ 23322}\quad 3種走法,剩下7種\\ 3333排列數1} \\\Rightarrow 第1階至第12階有7+1=8種走法；\\因此共有8\times 4=\bbox[red,2pt]{32}種走法。$$

解答：

$$球心O(0,0,0)至平面E:3x+2y+ 2\sqrt 3 z=5的距離d(O,E)={ 5\over \sqrt{9+4+12}} =1\\所求體積相當於半徑為2的圓形弓形部分繞x軸旋轉所得體積，見上圖斜線區域；\\ 體積= \int_1^2 y^2 \pi dx =\pi\int_1^2 (4-x^2)dx = \pi \left. \left[4x-{1\over 3}x^3 \right] \right|_1^2 =\pi(8-{8\over 3}-4+{1\over 3})=\bbox[red,2pt]{5\over 3}\pi$$

解答： $$A=\begin{bmatrix} 2 & 2 & 1\\ 1 & 3& 1\\ 1 & 2& 2\end{bmatrix} = \begin{bmatrix} 1 & 2 & 1\\ 1 & 2& 1\\ 1 & 2& 1\end{bmatrix} +\begin{bmatrix} 1 & 0 & 0\\ 0 & 1& 0\\ 0 & 0& 1\end{bmatrix} = B+I\\ B=\begin{bmatrix} 1 & 2 & 1\\ 1 & 2& 1\\ 1 & 2& 1\end{bmatrix} \Rightarrow B^n=2^{2n-2}B,n\in N \\ 若矩陣M=\begin{bmatrix} a & b & c\\ d & e& f\\g & h& i\end{bmatrix},定義g(M)=h,則g(B^n)=2^{2n-2}g(B)=2^{2n-2}\cdot 2= 2^{2n-1}\\   A^2=(B+I)^2 = B^2+2B+I \Rightarrow g(A^2)= g(B^2)+ 2g(B)+g(I) =8+2\cdot 2+0=12\\ A^3=(B+I)^3 =B^3+3B^2+3B+I \Rightarrow g(A^3)=32+24+ 6+0=62 \\ A^4=(B+I)^4 =B^4 +4B^3 +6B^2 +4B +I\Rightarrow g(A^4)=128+128+48+8+0=312\\A^5 =(B+I)^5 =B^5+5B^4+10B^3 + 10B^2+5B+I \Rightarrow g(A^5)=512 +640+320+80+10=1562\\ \Rightarrow g(f(A))= 1562-7\times 312+9\times 62+9\times 12-7\times 2+8\times 0=\bbox[red,2pt]{30}$$

解答： $$\alpha,\beta,\gamma為x^3-2x^2-6x+5=0之三根\Rightarrow \cases{\alpha+\beta+\gamma = 2\\ \alpha\beta +\beta\gamma +\gamma\alpha=-6\\ \alpha\beta\gamma = -5}\\ \Rightarrow (\alpha+\beta)(\beta+ \gamma)(\gamma+ \alpha)=(2-\beta)(2-\gamma)(2-\alpha)\\\qquad = 8-4(\alpha+\beta +\gamma)+2 (\alpha\beta +\beta\gamma +\gamma\alpha) -\alpha\beta\gamma =-7\\ 因此(\alpha+\beta+\gamma)^5 -(\alpha^5+\beta^5+\gamma^5) = 5(\alpha+ \beta)(\beta+\gamma)(\gamma+\alpha)((\alpha+\beta+\gamma)^2-(\alpha\beta+ \beta\gamma+\gamma\alpha) \\ \Rightarrow 2^5-(\alpha^5+\beta^5+\gamma^5) =5 \cdot (-7)(4+6) \Rightarrow \alpha^5+\beta^5+\gamma^5=32+350= \bbox[red,2pt]{382}$$

解答： $$y的最小值2，最大值6 \Rightarrow 2\le y\le 6 \Rightarrow (y-2)(y-6)\le 0 \Rightarrow y^2-8y+12 \le 0 \cdots (1)\\ 由於y={ax^2+bx+6 \over x^2+2} \Rightarrow (a-y)x^2+bx +6-2y=0 此函數有解 \Rightarrow b^2-4(a-y)(6-2y) \ge 0 \\ \Rightarrow 8y^2-(8a+24)y+24a-b^2 \le 0 \Rightarrow y^2-(a+3)y+3a-{b^2\over 8} \le 0 與式(1)相同\\  \Rightarrow \cases{a+3=8\\ 3a-{b^2\over 8}=12} \Rightarrow a=\bbox[red,2pt]{5 }$$

解答： $$公式:C^n_0+ C^n_2 +C^n_4-n = C^{12}_0+C^{12}_2 +C^{12}_4-12=1+66+495-12= \bbox[red,2pt]{550}$$ 註:公式來源:$\href{http://math1.ck.tp.edu.tw/%E9%80%9A%E8%A8%8A%E8%A7%A3%E9%A1%8C/doc/answer55.doc}{按這裡}$

解答：

$$作\overline{AE}\bot \overline{BD}及\overline{CF} \bot \overline{BD}，如上圖；則\triangle MEA \sim \triangle MFC (AAA) \Rightarrow {\overline{AE} \over {CF}}={\overline{AM} \over \overline{MC}} \\ \Rightarrow {\overline{AM} \over \overline{MC}} = {\triangle ABD \over \triangle CBD} ={{1\over 2}\cdot \overline{AB}\cdot \overline{AD}\cdot \sin \angle BAD \over {1\over 2}\cdot \overline{CB}\cdot \overline{CD}\cdot \sin \angle BCD} = {20\sin \angle BAD \over 16\sin (\pi-\angle BAD)} = \bbox[red,2pt]{5\over 4}$$

解答： $$(x+2)^2 +(y+3)^2 =(x+5)^2+(y+7)^2 =(x+11)^2+(y+k)^2 \\ \Rightarrow 4x+6y+13 = 10x+14y+74 = 22x+2ky+ 121+k^2\\ \Rightarrow \cases{6x +8y+61=0\\ 12x+(2k-14)y+47+k^2=0}無解 \Rightarrow {6\over 12} ={8\over 2k-14} \ne {61\over 47+k^2}\\ {6\over 12} ={8\over 2k-14} \Rightarrow k=15 \Rightarrow \cases{{8\over 2k-14}={1\over 2} \\ {61\over 47+k^2} ={61\over 272}} \Rightarrow {8\over 2k-14} \ne {61\over 47+k^2}；因此，k=\bbox[red,2pt]{15}無解$$

解答： $$a=1223334444\cdots 999\\ \Rightarrow \cases{奇位數字和= 9 \times 5+8\times 4 + (7 +6 + 5)\times 3+ 4\times 2 + 3+ 2+ 1 =145\\ 偶位數字和= (9+8+7)\times 4 + 6\times 3+ (5+4+3)\times 2+2 = 140}\\ \Rightarrow 奇位數字和-偶位數字和=145-140=5 \Rightarrow a-5是11的倍數 \Rightarrow a=11k +5,k\in N \\ \Rightarrow a^2= (11k+5)^2 = 11^2k^2 + 110k + 25(= 11\times 2+3) =11s+ 3, s\in N\\ \Rightarrow a \equiv  \bbox[red,2pt]{3} \mod 11$$

解答： $$每個座位的鄰居數量如下表:\\\begin{array}{|c|c|c|c|c|}\hline 3& 5& 5& 5&3\\\hline 5& 8& 8& 8&5\\\hline5& 8& 8& 8&5\\\hline5& 8& 8& 8&5\\\hline 3& 5 & 5 & 5 &3\\\hline \end{array} \Rightarrow 鄰居總數=3\times 4+5\times 12+8\times 9=144\\ 因此機率為{144\over 25\times 24} ={6\over 25} ={n\over m}\Rightarrow m+n=\bbox[red,2pt]{31}$$

解答：

$$\cases{s=\sqrt{29}=\sqrt{2^2+(2+3)^2} \\ t= \sqrt{37} =\sqrt{1^2+(1+2+3)^2} \\u=\sqrt{52} =\sqrt{(1+3)^2 +(1+2+3)^2}}\\將s,t,u以不同的直角三角形的斜邊繪製，如上圖；則此三角形面積=正方形扣除三個直角三角形\\ =6\times 6-{1\over 2}(5\times 2+1\times 6+ 4\times 6)= 36-20=\bbox[red,2pt]{16}$$

解答：

$$f(f(x))=-5 \Rightarrow \cases{f(x)=2 \Rightarrow 兩圖形\cases{y=f(x)\\ y=2}有4個交點(上圖藍色圈圈)\\ f(x)=-3 \Rightarrow 兩圖形\cases{y=f(x)\\ y=-3}有5個交點(上圖綠色圈圈)\\ f(x)=k (上圖紅點) \Rightarrow 兩圖形\cases{y=f(x)\\ y=k(k>5)}有1個交點(上圖棕色圈圈)}\\ \Rightarrow f(f(x))=-5 有4+5+1= 10個交點，即有\bbox[red,2pt]{10}個相異實根$$ 註:學校公布的答案是$\bbox[blue,2pt]{9}$

解答： $${1\over 7} ={a_1\over 9}+ {a_2\over 9^2}+ {a_3\over 9^3}+ \cdots +{a_n\over 9^n}+\cdots \Rightarrow   {9\over 7}=1+{2\over 7} =a_1+ {a_2\over 9}+ {a_3\over 9^2}+ \cdots +{a_n\over 9^{n-1}}+\cdots\\ 由於a_1 \in \{0,1,...,6\}，因此a_1=1 \Rightarrow {2\over 7} ={a_2\over 9}+ {a_3\over 9^2}+ \cdots +{a_n\over 9^{n-1}}+\cdots \\ \Rightarrow {18\over 7}=2+{4\over 7}= a_2+ {a_3\over 9} +{a_4\over 9^2}+ \cdots \Rightarrow a_2=2 \Rightarrow {4\over 7}={a_3\over 9} +{a_4\over 9^2}+ \cdots \\\Rightarrow {36\over 7}=5+{1\over 7}=a_3+{a_4\over 9}+\cdots\\ \Rightarrow a_3=5 \Rightarrow \langle a_n \rangle = 1,2,5,1,2,5,\cdots \Rightarrow a_{300}= \bbox[red,2pt]{5} (\because 300\equiv 0 \mod 3)$$

解答：

$$令\overline{BD}=a \Rightarrow \cases{\cos \angle A={3+1-a^2 \over 2\sqrt 3} \\ \cos \angle C={1+1-a^2\over 2}} \Rightarrow \cases{a^2= 4-2\sqrt 3\cos \angle A \\ a^2=2-2\cos \angle C} \Rightarrow 4-2\sqrt 3\cos \angle A= 2-2\cos \angle C \\ \Rightarrow \cos \angle C= \sqrt 3\cos \angle A-1 \cdots(1)\\ 又\cases{S=\triangle ABD ={1\over 2}\cdot \overline{AB}\cdot \overline{AD} \sin \angle A ={\sqrt 3\over 2}\sin \angle A\\ T=\triangle BCD ={1\over 2}\cdot \overline{CB}\cdot \overline{CD} \sin \angle C ={1\over 2}\sin \angle C} \Rightarrow S^2+T^2 ={3\over 4}\sin^2\angle A+ {1\over 4}\sin^2\angle C \cdots(2)\\ 將(1)代入(2)\Rightarrow S^2+T^2 ={3\over 4}(1-\cos^2\angle A)+ {1\over 4}(1-\cos^2\angle C)\\ ={3\over 4}(1-\cos^2\angle A)+ {1\over 4}(1-(\sqrt 3\cos \angle A-1)^2)= -{3\over 2}\cos^2 \angle A+{\sqrt 3\over 2}\cos \angle A+{3\over 4}\\ \Rightarrow 當\cos \angle A={\sqrt 3\over 6}時，S^2+T^2 有極大值\bbox[red,2pt]{7\over 8}$$

解答： $$\cases{a(4-b) = 4\cdots(1)\\ b(4-c)=4\cdots(2) \\ c(4-a)=4\cdots (3)}，由(1) \Rightarrow a={4\over 4-b} 代入(3) \Rightarrow c(4-{4\over 4-b})=4 \Rightarrow c=1+{1\over 3-b}代入(2)\\ \Rightarrow b(3-{1\over 3-b})=4 \Rightarrow b={12-4b \over 8-3b} \Rightarrow b^2-4b+4=0 \Rightarrow (b-2)^2=0 \Rightarrow b=2 \\將 \cases{b=2代入(1) \Rightarrow a=2 \\ b=2代入(2) \Rightarrow c=2} \Rightarrow a+b+c=2+2+2 = \bbox[red,2pt]{6}$$

解答：

$$令\cases{\overline{CD}=w \\ \overline{AD}=h \\\overline{AE}=a\\ \overline{CF}=b} \Rightarrow \cases{\overline{BE}=w-a\\ \overline{BF}= h-b}，依題意\cases{\triangle ADE=2 \\\triangle BEF=3 \\ \triangle CDF=4} \Rightarrow \cases{ah=4 \cdots(1)\\ (w-a)(h-b)=6 \cdots(2)\\ bw=8 \cdots(3)} \\ 由(2) \Rightarrow wh-wb-ah+ab=6 \Rightarrow wh-8-4+ab=6 \Rightarrow wh+ab=18\cdots(3)\\ \triangle ADE \times \triangle CDF = 2\times 4 \Rightarrow {1\over 2}ah \times {1\over 2}bw = 8 \Rightarrow abhw=32 \cdots(4)\\ (3) \Rightarrow ab=18-wh 代入(4) \Rightarrow (18-wh)(wh)=32 \Rightarrow (wh)^2-18(wh)+32=0 \\ \Rightarrow (wh-2)(wh-16)=0 \Rightarrow wh=16 (2不合,\because 矩形ABCD \gt \triangle \triangle CDF)\\ \Rightarrow \triangle DEF = wh-\triangle ADE-\triangle BEF-\triangle CDF =16-2-3-4= \bbox[red,2pt]{7}$$

解答：

$$\cases{z_1=-3-\sqrt 3 i\\ z_2=\sqrt 3+ i\\ z=\sqrt 3\sin \theta+ i(\sqrt 3\cos \theta+2)} \Rightarrow \cases{A(z_1)=(-3,-\sqrt 3)\\ B(z_2)=(\sqrt 3,1)\\ \Gamma(z)=x^2+(y-2)^2 = 3} \Rightarrow L=\overleftrightarrow{AB}:x=\sqrt 3 y\\ 圓心(0,2)至L距離={2\sqrt 3\over \sqrt{3+1}} =\sqrt 3=圓半徑 \Rightarrow L為圓之切線 \\\Rightarrow |z-z_1|+ |z-z_2|的最小值=\overline{AB}= \sqrt{(3+\sqrt 3)^2 +(\sqrt 3+1)^2} =  \sqrt{16+2\sqrt{48}} \\=\sqrt{12}+\sqrt 4 = \bbox[red,2pt]{2+2\sqrt 3}$$

解答： $$a,b,c為x^3-3x^2+6x-1=0的三根\Rightarrow \cases{a+b+c=3 \\ ab+bc+ca = 6 \\ abc=1}\\ \Rightarrow a^2+b^2+c^2 =(a+b+c)^2-2(ab+ bc+ca)= 3^2-2\times 6= -3;\\ \begin{vmatrix} b^2+c^2 & ab & ac \\ ab & c^2+a^2 & bc\\ ac & bc & a^2+b^2 \end{vmatrix} = abc\begin{vmatrix} {b^2+c^2 \over a} & b & c \\ a & {c^2+a^2 \over b}& c\\ a & b & {a^2+b^2 \over c} \end{vmatrix} =\begin{vmatrix} b^2+c^2 & b^2 & c^2 \\ a^2 & c^2+a^2 & c^2\\ a^2 & b^2 & a^2+b^2 \end{vmatrix} \\ \underrightarrow{c_2+c_1,c_3+c_2}\begin{vmatrix} a^2+b^2+c^2 & a^2+b^2+c^2 & 2c^2 \\ 2a^2 & a^2+b^2+c^2 & a^2+b^2+c^2\\ a^2 & b^2 & a^2+b^2 \end{vmatrix} =\begin{vmatrix} -3 & -3 & 2c^2 \\ 2a^2 & -3 & -3\\ a^2 & b^2 & -3-c^2 \end{vmatrix} \\ = 9(-3-c^2)+4a^2b^2c^2 +9a^2 +6a^2c^2+6a^2(-3-c^2)-9b^2\\ =-23-9(a^2+b^2+c^2) =-23+27 = \bbox[red,2pt]{4}$$

解答：

111111+4444444444-66666=4444488889，再手算開根號，如上圖；因此 x=66667，各位數字和=6+6+6+6+7= $\bbox[red,2pt]{31}$

解答：

$$920 \div 2= 460 = 120+210+130 \\\Rightarrow 一個月的電費=120\times 1.63+210\times 2.38+130\times 3.52 = 1153元\\ \Rightarrow 兩個月的電費=1153\times 2= \bbox[red,2pt]{2306}元$$