110年桃園高中教甄-數學詳解

 桃園高中 110 學年度第 1 次教師甄選

一、填充題：共二十題，每題全對得５分

解答：$$\cases{({5\over 2})^{1/x}=10 \\ 4^{-1/y}=10} \Rightarrow \cases{{5\over 2}=10^x \\ {1\over 4}=10^y} \Rightarrow \cases{ x=\log{5\over 2} =1-2\log 2\\y =\log{1\over 4} =-2\log 2}\Rightarrow  x-y=1 \\ 又(x-y)^3= x^3-y^3-3xy(x-y) \Rightarrow 1= x^3-y^3-3xy \Rightarrow y^3-x^3+3xy = \bbox[red, 2pt]{-1}$$

解答：$$a_n={a_{n-1}\over 1+2a_{n-1}} \Rightarrow {1\over a_n} ={1+2a_{n-1} \over a_{n-1}} =2 +{1\over a_{n-1}} \Rightarrow b_n = b_{n-1}+2,\text{where }b_n={1\over a_n} \Rightarrow b_1=1\\ \Rightarrow b_n= b_{n-1}+2 = b_{n-2}+2\times 2= \cdots = b_1+2\times (n-1) =1+2n-2=2n-1\\ \Rightarrow a_n={1\over b_n} = \bbox[red, 2pt]{1\over 2n-1}$$

解答：$${n+2\over n!+(n+1)! +(n+2)!} = {n+2\over n!(1+(n+1) +(n+1)(n+2))} = {n+2 \over n! (n+2)^2} ={1\over n!(n+2)}\\ ={n+1 \over (n+2)!} ={(n+2)-1\over (n+2)!} ={1\over (n+1)!} -{1\over (n+2)!} \\ 因此S_n =\left({1\over 2!}-{1\over 3!} \right)+ \left({1\over 3!} -{1\over 4!}\right) + \cdot +\left( {1\over (n+1)!}-{1\over (n+2)!}\right) \\ = \bbox[red, 2pt]{{1\over 2}-{1\over (n+2)!}}$$

解答：$$(x+1)^n = C^n_0+ C^n_1x+ C^n_2x^2+\cdots +C^n_kx^k+ \cdots+C^n_nx^n\\ \Rightarrow \int_0^1(x+1)^n\;dx = \left. \left[ C^n_0x+{1\over 2} C^n_1x^2+ {1\over 3}C^n_2x^3+\cdots +{1\over k+1}C^n_kx^{k+1} \cdots+ {1\over n+1}C^n_nx^{n+1}  \right]\right|_0^1 \\ \Rightarrow \left. \left[{1\over n+1}(x+1)^{n+1}\right] \right|_0^1 =C^n_0 +{1\over 2}C^n_1 +{1\over 3}C^n_2+ \cdots +{1\over n+1}C^n_n ={4095\over n+1} \\ \Rightarrow {1\over n+1}2^{n+1}-{1\over n+1}={4095\over n+1} \Rightarrow 2^{n+1}-1=4095 \Rightarrow n=\bbox[red, 2pt]{11}$$

解答：$$1-20任取3數為a,b,c，則\\ \begin{array}{lll|c} a & b & c& 數量\\\hline 1 & 4 & 7-20 & 14\\ 1 & 5 & 8-20 & 13 \\ 1 & \cdots & \cdots \\ 1 & 17 & 20 & 1\\\hdashline 2 & 5 & 8-20 & 13\\ 2 & \cdots & \cdots \\ 2 & 17 & 20 & 1\\\hdashline \dots & \dots \\\hdashline 13 & 16 & 19-20 & 2\\ 13 & 17 & 20 & 1\\\hdashline 14 & 17 & 20 & 1\\\hline\end{array} \Rightarrow 共有\sum_{n=1}^{14}\sum_{k=1}^n k = {1\over 2}\sum_{n=1}^{14}n(n+1) =560\\ 機率為{560 \over C^{20}_3} ={560\over 1140} =\bbox[red, 2pt]{28\over 57}$$

解答：$$\lim_{x\to 1}{x^3f(1)-f(x^2) \over x-1} = \lim_{x\to 1}{ (x^3f(1)-f(x^2))' \over (x-1)'} = \lim_{x\to 1} (3x^2f(1)-2x f'(x^2))   = \bbox[red, 2pt]{3a-2b}$$

解答：$$某餐做飯、洗碗寫成(做飯,洗碗)，午晚兩餐可表達成(A,B)、(C,D)；\\若B=C，則A,B,C,D 有5\times 4\times 1\times 4=80種；\\若B \ne C，則A,B,C,D 有5\times 4\times 3\times 3= 180種；\\因此共有80+180 = \bbox[red, 2pt]{260}種情形$$

解答：$${45n-277\over n-1}=37 \Rightarrow n=30 \Rightarrow 29個數的總和=45n-277=1073，\\ 若28個數都是最小值1，則最大數為1073-28= \bbox[red, 2pt]{1045}$$

解答：$$1A3B的排列數:4個正確數字都被選到，這4數字中有1個位置正確，\\另3個數都不在自己的位置有2種排法\\，因此有C^{4}_4\times C^4_1\times 2排法，機率為{C^{4}_4\times C^4_1\times 2\over P^{10}_4}；\\ 1A3B\to 4A: 4個數字中有1個在正確的位置，另外3個都不在自己的位置有2種排法\\，機率為{1\over C^4_1\times 2}；\\因此第1次猜到1A3B且第2次猜到4A的機率為{C^{4}_4\times C^4_1\times 2\over P^{10}_4}\times {1\over C^4_1\times 2} ={1 \over P^{10}_4} =\bbox[red, 2pt]{1\over 5040}$$

解答：$$a_{n+1} \equiv 將n+1個圓盤由A柱移到B柱\\ 步驟1:將n個圓盤由A柱移到B柱，需要a_n步；\\ 步驟2:將第n+1個圓盤由A柱移到C柱，需要1步；\\步驟3:將n個圓盤由B柱移到A柱，需要a_n步；\\ 步驟4:將第n+1個圓盤由C柱移到B柱，需要1步；\\步驟5:將n個圓盤由A柱移到B柱，需要a_n步；\\因此，a_{n+1} =a_n+1+a_n+1+a_n = 3a_n +2 \Rightarrow (p,k)=\bbox[red, 2pt]{(3,2)}$$

解答：$$y=f(x)=x^3 +ax^2+x+1 \Rightarrow 切點P(t,f(t)) \Rightarrow 切點斜率為f'(t)=3t^2+2at+1 \\ \Rightarrow 切線方程式: y=(3t^2+2at+1)(x-t)+ t^3+at^2+t+1\\ 原點(0,0)在切線上，因此 0=(3t^2+2at+1)(-t)+ t^3+at^2+t+1 \\ \Rightarrow 2t^3+at^2-1=0 有三相異實數解；\\令g(t)=2t^3+at^2-1 \Rightarrow g'(t)=6t^2+2at=0 \Rightarrow t(3t+a)=0 \Rightarrow t=0,-a/3 \\ \Rightarrow g(0)(g(-a/3)) \lt 0 \Rightarrow -{2\over 27}a^3 +{1\over 9}a^3-1 \gt 0 \Rightarrow {1\over 27}a^3 \gt 1 \Rightarrow \bbox[red, 2pt]{a \gt 3}$$

解答：

$$令A為原點，則\cases{小圓C_1:x^2+(y-\sqrt 6)^2 =6\\ 大圓C_2:(x-3\sqrt 2)^2 +y^2 = 18} \Rightarrow \cases{圓心C_1(0, \sqrt 6),半徑r_1=\sqrt 6\\ 圓心C_2(3\sqrt 2,0),半徑r_2=3\sqrt 2}\\ \Rightarrow \cases{x^2+y^2 =2\sqrt 6y \\ x^2+y^2 = 6\sqrt 2 x} \Rightarrow 2 \sqrt 6y = 6\sqrt 2 x \Rightarrow y=\sqrt 3x\\ \Rightarrow \overleftrightarrow{AP}: y= \sqrt 3x代入C_1 \Rightarrow P({3\over 2}\sqrt 2,{3\over 2}\sqrt 6) \Rightarrow \overline{OP} =\sqrt{{18\over 4}+ {54\over 4}}=\sqrt{18}= 3\sqrt 2=r_2 \\ \Rightarrow \triangle APO_2為正\triangle  \Rightarrow \angle PO_2A=60^\circ；\\ 又\cos \angle PO_1A= {6+6- 18 \over 2\cdot \sqrt 6\cdot\sqrt 6}=-{1\over 2} \Rightarrow \angle PO_1A=120^\circ \\ \Rightarrow \cases{棕色面積= 扇形O_2AP - \triangle O_2AP ={1\over 6}\times 18\pi-{\sqrt 3\over 4}\times 18 =3\pi-{9\over 2}\sqrt 3\\ 藍色面積= 扇形O_1PA- \triangle O_1PA ={1\over 3}\times 6\pi -{1\over 2}\times 6 \times {\sqrt 3\over 2} =2\pi-{3\over 2}\sqrt 3} \\ \Rightarrow 著色面積=3\pi-{9\over 2}\sqrt 3+2\pi-{3\over 2}\sqrt 3 =\bbox[red, 2pt]{5\pi-6\sqrt 3}$$

解答：$$(z+z^2+z^4)(z^3+z^5+z^6) = z(1+z+z^3)z^3(1+z^2+z^3) \\=z^4(1+z+ z^2+z^3+z^4+z^5+z^6+ 2z^3) =z^4(0+2z^3)=2z^7=2\\ \Rightarrow (z+z^2+z^4)(z^3+z^5+z^6)=2 \equiv \alpha \beta =2\cdots(1)\\ 1+z+ z^2+z^3+z^4+z^5+z^6=0 \Rightarrow (z+z^2+z^4)+(z^3+z^5+z^6)=-1 \equiv \alpha +\beta=-1 \cdots(2)\\ 由(1)及(2) \Rightarrow \alpha,\beta 為x^2+x+2=0的兩根 \Rightarrow \alpha = z+z^2+z^4=\bbox[red, 2pt]{-1\pm \sqrt{7}i\over 2}$$

解答：$$\cases{(1+\sqrt 3)^n =   a_n+b_n\sqrt 3 \\(1-\sqrt 3)^n =   a_n-b_n\sqrt 3} \Rightarrow ((1+\sqrt 3)(1-\sqrt 3))^n = (a_n+ b_n\sqrt 3)(a_n-b_n\sqrt 3) \\ \Rightarrow (-2)^n =a_n^2-3b_n^2 \Rightarrow ({a_n\over b_n})^2= {(-2)^n\over b_n^2} +3 \Rightarrow \lim_{n\to \infty} {a_n\over b_n} =\lim_{n\to \infty} \sqrt{{(-2)^n\over b_n^2} +3} =\bbox[red, 2pt]{\sqrt 3}$$

解答：$$依題意\begin{cases}f(x)=a_{n-1}x^{n-1}, & {1\over n}\lt x\le {1\over n-1} \\ f(x)=a_nx^n, & {1\over n+1}\lt x\le {1\over n}\end{cases}；\\由於f(x)在(0,1]連續，因此 a_{n-1}({1\over n})^{n-1}=a_n({1\over n})^n \\\Rightarrow a_n=na_{n-1} =n\cdot (n-1)a_{n-2} =\cdots = n\cdot (n-1)\cdots 2\cdot a_1 =\bbox[red, 2pt]{n!}$$

解答：$$f(x)=-2(x+2)^3 -2(x+2)+3 \Rightarrow f(1.99)=-2\cdot 3.99^3-2\cdot 3.99 +3 \\=-127.042398-7.98+3 = \bbox[red, 2pt]{-132.022398}$$

解答：$$令\cases{a_n:格子數為n，上色要求符合題意，但最右格為紅色\\ b_n:格子數為n，上色要求符合題意，但最右格為綠色}\qquad \qquad \qquad，則\cases{a_1=1\\ b_1=1\\ c_{10}=a_{10}+b_{10}即為所求}\\ 因此\cases{a_1=1\\b_1=1} \Rightarrow \cases{a_2=1\\b_2=2} \Rightarrow \cases{a_3=2\\b_3=3} \Rightarrow \cases{a_4=3\\b_4=5} \Rightarrow \cases{a_5=5\\b_5=8} \Rightarrow \cases{a_6=8\\b_6=13} \\\Rightarrow \cases{a_7=13\\b_7=21} \Rightarrow \cases{a_8=21\\b_8=34} \Rightarrow \cases{a_9=34\\b_9=55} \Rightarrow \cases{a_{10}=55\\b_{10} =89} \Rightarrow c_{10}=55+89=\bbox[red, 2pt]{144}$$

解答：

$$取一白球相當於向右走一格，取一紅球相當於向上走一格，過程中白球數需大於等於紅球數，\\相當於行走在著色區域，見上圖；因此從A走到B有90種走法，機率為{90\over {10!\over 4! 6!}} =\bbox[red, 2pt]{3\over 7}$$

解答：$$考慮只有1組的情形：8件都是陰性，機率為0.8^8，只需要檢驗1次，期望值為0.8^8；\\8件中有一件是陽性，機率為1-0.8^8，需要檢驗9次，期望值為9(1-0.8^8)；\\總共有100組，期望值為100(0.8^8+9(1-0.8^8)) = \bbox[red, 2pt]{900-800\cdot 0.8^8}$$

解答：$$單位圓上兩點z_1,z_2,且\overline{z_1z_2}={\sqrt 6-\sqrt 2\over 2},則\cos \angle z_1Oz_2={1+1-({\sqrt 6-\sqrt 2\over 2})^2 \over 2} ={\sqrt 3\over 2} \\\Rightarrow \angle z_1Oz_2=30^\circ \Rightarrow |z_1-z_2| < {\sqrt 6-\sqrt 2\over 2} \Rightarrow \angle z_1Oz_2 < 30^\circ;\\現在正2020邊形z_1z_2...z_{2020},每一內角為{360^\circ \over 2020} \Rightarrow {360^\circ \over 2020}\times n < 30^\circ \Rightarrow n\le 168\\ \Rightarrow z_i的左邊有168個z_j(i\ne j),右邊也有168個z_j(i\ne j),使得|z_i-z_j|< {\sqrt 6-\sqrt 2\over 2}\\也就是說對任一z_i而言,在其它2019個z_j(i\ne j)有168\times 2=336個符合條件\\,其機率為{336\over 2019} =\bbox[red,2pt]{112\over 673}$$

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解題僅供參考，其他教甄試題及詳解