112年新竹中學數學科代理教師甄選
一、填充題(共60分,每題6分)
解答:假設滿足條件且長度為n的數量為f(n)n=3:只有101一種⇒f(3)=1n=4:有兩種1011,1101⇒f(4)=2n=5:也是兩種10101,11011⇒f(5)=2只要在合乎條件的字串中後面加上01或011讓長度加長且合乎條件因此f(n)=f(n−2)+f(n−3),n≥6⇒n678910111213141516f(n)34579121621283749⇒f(16)=49參考資料解答:甲乙相鄰且甲不靠窗:乙甲DEFGHK⋯(1)AC甲乙FGHK⋯(2)AC乙甲FGHK⋯(3)ACD甲乙GHK⋯(4)ACD乙甲GHK⋯(5)ACDE甲乙HK⋯(6)ACDE乙甲HK⋯(7)ACDEFG甲乙⋯(8)接下來計算(1)−(6)丙丁不相鄰且皆不靠窗(1)與(8)各有14種,(2,3,6,7)各有10種,(4)及(5)各有12種,合計92種最後將剩下的四人任排有4!=24種,一共92×24=2208種
解答:第1次檢查有心臟病的比率=有心臟病被正確診斷+無心臟病被誤診=0.001×0.99+0.999×0.02第1次檢查有心臟病且第2次檢查也有心臟病而真的有心臟病第1次檢查有心臟病且第2次檢查也有心臟病=0.001×0.99×0.990.001×0.99×0.99+0.999×0.02×0.02=363511
解答:
假設半圓心C,小圓圓心O,如上圖小圓圓周角∠QPR=60∘⇒圓心角∠QOR=120∘⇒¯QRsin∠QOR=3√3sin120∘=6=2r小圓半徑r=3⇒¯OR=¯OQ=3⇒△COR≅△COQ(SSS)⇒∠COQ=∠COR=120∘⇒cos∠COQ=−12=¯OC2+32−722⋅¯CO⋅3⇒¯OC=5⇒¯CP=4令∠COP=θ⇒{sinθ=4/5cosθ=3/5且{∠POR=120∘−θ∠POQ=120∘+θ⇒△POR+△POQ=12⋅32(sin(120∘−θ)+sin(120∘+θ))=92√3cosθ=2710√3又△OPR=12⋅32sin120∘=94√3⇒△PQR=2710√3+94√3=9920√3
解答:假設ω=cos(2π/7)+isin(2π/7)=e2πi/7⇒ω6=e12πi/7=e−2πi/7⇒cos(2π/7)=(ω+ω6)/2同理,cos(4π/7)=(ω2+ω5)/2,cos(6π/7)=(ω3+ω4)/2因此x3+ax2+bx+c=0的三根為{α=ω+ω62β=ω2+ω52γ=ω3+ω42⇒{−a=α+β+γb=αβ+βγ+γα−c=αβγα+β+γ=12(ω+ω2+⋯+ω6)=12(−1)=−12⇒a=12αβ+βγ+γα=14⋅2(ω+ω2+⋯+ω6)=−12⇒b=−12αβγ=18(2+ω+ω2+⋅s+ω6)=18⇒c=−18因此abc=132
解答:z1,z2,z3,z4為P(z)=0的四根⇒{z1+z2+z3+z4=−4z1(z2+z3+z4)+z2(z3+z4)+z3z4=3z1z2z3+z2z3z4+z3z4z1+z1z2z4=−2z1z2z3z4=1又{f(z1)=4iˉz1f(z2)=4iˉz2f(z3)=4iˉz3f(z4)=4iˉz4為Q(z)=0的四根⇒{B=(4i)2×3=−48D=(4i)4×1=256⇒B+D=256−48=208
解答:
解答:假設ω=cos(2π/7)+isin(2π/7)=e2πi/7⇒ω6=e12πi/7=e−2πi/7⇒cos(2π/7)=(ω+ω6)/2同理,cos(4π/7)=(ω2+ω5)/2,cos(6π/7)=(ω3+ω4)/2因此x3+ax2+bx+c=0的三根為{α=ω+ω62β=ω2+ω52γ=ω3+ω42⇒{−a=α+β+γb=αβ+βγ+γα−c=αβγα+β+γ=12(ω+ω2+⋯+ω6)=12(−1)=−12⇒a=12αβ+βγ+γα=14⋅2(ω+ω2+⋯+ω6)=−12⇒b=−12αβγ=18(2+ω+ω2+⋅s+ω6)=18⇒c=−18因此abc=132
解答:z1,z2,z3,z4為P(z)=0的四根⇒{z1+z2+z3+z4=−4z1(z2+z3+z4)+z2(z3+z4)+z3z4=3z1z2z3+z2z3z4+z3z4z1+z1z2z4=−2z1z2z3z4=1又{f(z1)=4iˉz1f(z2)=4iˉz2f(z3)=4iˉz3f(z4)=4iˉz4為Q(z)=0的四根⇒{B=(4i)2×3=−48D=(4i)4×1=256⇒B+D=256−48=208
解答:
等腰梯形對角線等長⇒¯BD=¯AC=2+1+3=6直角△BDD′(見上圖):¯BD2=¯BD′2+¯DD′2=(¯BY+¯DX)2+¯XY2⇒62=(¯BY+¯DX)2+12⇒¯BY+¯DX=√35⇒梯形面積=△ABC+△ADC=12⋅¯AC(¯BY+¯DX)=12⋅6⋅√35=3√35
解答:L:y=tanθ⋅x⇒過P(a,b)且與L垂直的直線L′:y=−1tanθ(x−a)+b⇒P′=L∩L′=(a+btanθ1+tan2θ,atanθ+btan2θ1+tan2θ)=[11+tan2θtanθ1+tan2θtanθ1+tan2θtan2θ1+tan2θ][ab]⇒A=[11+tan2θtanθ1+tan2θtanθ1+tan2θtan2θ1+tan2θ]⇒A2=A⇒A112=A=[11+tan2θtanθ1+tan2θtanθ1+tan2θtan2θ1+tan2θ]
解答:
解答:L:y=tanθ⋅x⇒過P(a,b)且與L垂直的直線L′:y=−1tanθ(x−a)+b⇒P′=L∩L′=(a+btanθ1+tan2θ,atanθ+btan2θ1+tan2θ)=[11+tan2θtanθ1+tan2θtanθ1+tan2θtan2θ1+tan2θ][ab]⇒A=[11+tan2θtanθ1+tan2θtanθ1+tan2θtan2θ1+tan2θ]⇒A2=A⇒A112=A=[11+tan2θtanθ1+tan2θtanθ1+tan2θtan2θ1+tan2θ]
解答:
cosC=42+62−282⋅4⋅6=12⇒∠C=60∘,因此令{B(0,0)C(4,0)⇒A(1,3√3)假設¯AC上的垂足為Q,則¯BQ=2√3⇒Q(3,√3)⇒{L1=↔BQ:3y=√3xL2=↔CP:x+√3y=4⇒P=L1∩L2=(2,2√3/3)⇒{→AP=(1,−7√3/3)→AB=(−1,−3√3)→AC=(3,−3√3)⇒(α,β)=(13,49)
解答:本題相當於求迴歸直線:y=ax+b,樣本點為(x,y)=(4,5),(6,5),(8,7),(10,9),(12,9)XYX2XY451620653630876456109100901291441084035360304⇒{ˉx=40/5=8ˉy=35/5=7⇒a=n∑(xy)−(∑x)(∑y)n∑x2−(∑x)2=5×304−40×355×360−402=35又迴歸直線必經(ˉx,ˉy)⇒y=35(x−8)+7⇒y=35x+115⇒(a,b)=(35,115)
解答:敘述:若整係數多項式f(x)=anx2+an−1xn−1+⋯+a1x+a0有一次因式px−q,其中p,q為整數且互質,則p∣an且q∣a0證明:f(x)有一次因式px−q⇒f(q/p)=0⇒an(qp)n+an−1(qp)n−1+⋯+a1(qp)+a0=0⇒anqn+an−1pqn−1+⋯+a1pn−1q+a0pn=0⇒{p(an−1qn−1+⋯+a1pn−2q+a0pn−1)=−anqn⇒p∣−anqnq(anqn−1+an−1pqn−2+⋯+a1pn−1)=−a0pn⇒q∣−a0pn由於p,q互質,因此{p∣anq∣a0,故得證。
解答:an=2an−1+n=2(2an−2+(n−1))+n=22an−2+2(n−1)+n=22(2an−3+(n−2))+2(n−1)+n=23an−3+22(n−2)+2(n−1)+n=2n−1a1+2n−2⋅2+2n−3⋅3+⋯+22(n−2)+2(n−1)+n=2n−1⋅1+2n−2⋅2+2n−3⋅3+⋯+22(n−2)+2(n−1)+20⋅n⇒2an=2n+2n−1⋅2+2n−2⋅3+⋯+23(n−2)+22(n−1)+2⋅n⇒2an−an=an=2n+2n−1+2n−2+⋯+22+2−n=2n+1−n−2
解答:\mathbf{(1)}\; 假設三狀態,分別為S_1\cases{甲:1白1紅\\ 乙:1白2紅},S_2\cases{甲:2紅\\ 乙:2白1紅},S_3\cases{甲:2白\\ 乙:3紅}\\ P(S_1\to S_1)=P(甲抽白球至乙,乙抽白球至甲)+P(甲抽紅球至乙,乙抽紅球至甲)={5\over 8}\\ P(S_1\to S_2)=P(甲抽白球至乙,乙抽紅球至甲)={1\over 4} \\ P(S_1\to S_3)=P(甲抽紅球至乙,乙抽白球至甲) ={1\over 8}\\ 同理P(S_2\to S_1)={1\over 2}, P(S_2\to S_2)={1\over 2};P(S_3\to S_3)= {1\over 4}, P(S_3\to S_1)= {3\over 4}\\因此狀態轉換矩陣A=\begin{bmatrix} 5/8 & 1/2 & 3/4 \\ 1/4 & 1/2 & 0\\ 1/8 & 0 & 1/4\end{bmatrix} \Rightarrow A^2 \begin{bmatrix} 1\\ 0 \\0 \end{bmatrix} = \left[\begin{matrix}\frac{39}{64} & \frac{9}{16} & \frac{21}{32} \\\frac{9}{32} & \frac{3}{8} & \frac{3}{16} \\\frac{7}{64} & \frac{1}{16} & \frac{5}{32} \end{matrix}\right]\begin{bmatrix} 1\\ 0 \\0 \end{bmatrix} =\begin{bmatrix} 1\\ 0 \\0 \end{bmatrix} =\begin{bmatrix}\frac{39}{64} \\\frac{9}{32} \\ \frac{7}{64}\end{bmatrix}\\ \Rightarrow 2紅球在甲箱的機率(S_2) =\bbox[red, 2pt]{9\over 32} \\ \mathbf{(2)}\; A=\left[\begin{matrix} -2 & 1 & 6 \\ 1 & -2 & 3 \\1 & 1 & 1 \end{matrix}\right] \left[\begin{matrix} 0 & 0 & 0 \\0 & \frac{3}{8} & 0 \\0 & 0 & 1\end{matrix}\right]\left[\begin{matrix}\frac{-1}{6} & \frac{1}{6} & \frac{1}{2} \\\frac{1}{15} & \frac{-4}{15} & \frac{2}{5} \\\frac{1}{10} & \frac{1}{10} & \frac{1}{10}\end{matrix}\right] \\\Rightarrow A^\infty =\left[\begin{matrix} -2 & 1 & 6 \\ 1 & -2 & 3 \\1 & 1 & 1 \end{matrix}\right] \left[\begin{matrix} 0 & 0 & 0 \\0 & \color{blue}0 & 0 \\0 & 0 & 1\end{matrix}\right]\left[\begin{matrix}\frac{-1}{6} & \frac{1}{6} & \frac{1}{2} \\\frac{1}{15} & \frac{-4}{15} & \frac{2}{5} \\\frac{1}{10} & \frac{1}{10} & \frac{1}{10}\end{matrix}\right] = \left[\begin{matrix} \frac{3}{5} & \frac{3}{5} & \frac{3}{5} \\\frac{3}{10} & \frac{3}{10} & \frac{3}{10} \\\frac{1}{10} & \frac{1}{10} & \frac{1}{10}\end{matrix} \right] \\ \Rightarrow A^{\infty} \begin{bmatrix} 1\\ 0\\ 0\end{bmatrix}= \left[\begin{matrix}\frac{3}{5} \\\frac{3}{10} \\\frac{1}{10}\end{matrix}\right] \Rightarrow P(S_2)= \bbox[red, 2pt]{3\over 10}
解答:敘述:若整係數多項式f(x)=anx2+an−1xn−1+⋯+a1x+a0有一次因式px−q,其中p,q為整數且互質,則p∣an且q∣a0證明:f(x)有一次因式px−q⇒f(q/p)=0⇒an(qp)n+an−1(qp)n−1+⋯+a1(qp)+a0=0⇒anqn+an−1pqn−1+⋯+a1pn−1q+a0pn=0⇒{p(an−1qn−1+⋯+a1pn−2q+a0pn−1)=−anqn⇒p∣−anqnq(anqn−1+an−1pqn−2+⋯+a1pn−1)=−a0pn⇒q∣−a0pn由於p,q互質,因此{p∣anq∣a0,故得證。
解答:an=2an−1+n=2(2an−2+(n−1))+n=22an−2+2(n−1)+n=22(2an−3+(n−2))+2(n−1)+n=23an−3+22(n−2)+2(n−1)+n=2n−1a1+2n−2⋅2+2n−3⋅3+⋯+22(n−2)+2(n−1)+n=2n−1⋅1+2n−2⋅2+2n−3⋅3+⋯+22(n−2)+2(n−1)+20⋅n⇒2an=2n+2n−1⋅2+2n−2⋅3+⋯+23(n−2)+22(n−1)+2⋅n⇒2an−an=an=2n+2n−1+2n−2+⋯+22+2−n=2n+1−n−2
解答:\mathbf{(1)}\; 假設三狀態,分別為S_1\cases{甲:1白1紅\\ 乙:1白2紅},S_2\cases{甲:2紅\\ 乙:2白1紅},S_3\cases{甲:2白\\ 乙:3紅}\\ P(S_1\to S_1)=P(甲抽白球至乙,乙抽白球至甲)+P(甲抽紅球至乙,乙抽紅球至甲)={5\over 8}\\ P(S_1\to S_2)=P(甲抽白球至乙,乙抽紅球至甲)={1\over 4} \\ P(S_1\to S_3)=P(甲抽紅球至乙,乙抽白球至甲) ={1\over 8}\\ 同理P(S_2\to S_1)={1\over 2}, P(S_2\to S_2)={1\over 2};P(S_3\to S_3)= {1\over 4}, P(S_3\to S_1)= {3\over 4}\\因此狀態轉換矩陣A=\begin{bmatrix} 5/8 & 1/2 & 3/4 \\ 1/4 & 1/2 & 0\\ 1/8 & 0 & 1/4\end{bmatrix} \Rightarrow A^2 \begin{bmatrix} 1\\ 0 \\0 \end{bmatrix} = \left[\begin{matrix}\frac{39}{64} & \frac{9}{16} & \frac{21}{32} \\\frac{9}{32} & \frac{3}{8} & \frac{3}{16} \\\frac{7}{64} & \frac{1}{16} & \frac{5}{32} \end{matrix}\right]\begin{bmatrix} 1\\ 0 \\0 \end{bmatrix} =\begin{bmatrix} 1\\ 0 \\0 \end{bmatrix} =\begin{bmatrix}\frac{39}{64} \\\frac{9}{32} \\ \frac{7}{64}\end{bmatrix}\\ \Rightarrow 2紅球在甲箱的機率(S_2) =\bbox[red, 2pt]{9\over 32} \\ \mathbf{(2)}\; A=\left[\begin{matrix} -2 & 1 & 6 \\ 1 & -2 & 3 \\1 & 1 & 1 \end{matrix}\right] \left[\begin{matrix} 0 & 0 & 0 \\0 & \frac{3}{8} & 0 \\0 & 0 & 1\end{matrix}\right]\left[\begin{matrix}\frac{-1}{6} & \frac{1}{6} & \frac{1}{2} \\\frac{1}{15} & \frac{-4}{15} & \frac{2}{5} \\\frac{1}{10} & \frac{1}{10} & \frac{1}{10}\end{matrix}\right] \\\Rightarrow A^\infty =\left[\begin{matrix} -2 & 1 & 6 \\ 1 & -2 & 3 \\1 & 1 & 1 \end{matrix}\right] \left[\begin{matrix} 0 & 0 & 0 \\0 & \color{blue}0 & 0 \\0 & 0 & 1\end{matrix}\right]\left[\begin{matrix}\frac{-1}{6} & \frac{1}{6} & \frac{1}{2} \\\frac{1}{15} & \frac{-4}{15} & \frac{2}{5} \\\frac{1}{10} & \frac{1}{10} & \frac{1}{10}\end{matrix}\right] = \left[\begin{matrix} \frac{3}{5} & \frac{3}{5} & \frac{3}{5} \\\frac{3}{10} & \frac{3}{10} & \frac{3}{10} \\\frac{1}{10} & \frac{1}{10} & \frac{1}{10}\end{matrix} \right] \\ \Rightarrow A^{\infty} \begin{bmatrix} 1\\ 0\\ 0\end{bmatrix}= \left[\begin{matrix}\frac{3}{5} \\\frac{3}{10} \\\frac{1}{10}\end{matrix}\right] \Rightarrow P(S_2)= \bbox[red, 2pt]{3\over 10}
解答:\mathbf{(1)}\;\cases{L垂直平面N\\ \overleftrightarrow {FJ}垂直平面N} \Rightarrow L\parallel \overleftrightarrow {FJ}再加上\angle ACF =\angle CAJ=90^\circ \Rightarrow ACFJ為矩形\\ 同理ABEJ也是矩形,因此\cases{\overline{AC}=\overline{FJ} \\\overline{IE} =\overline{AB}= \overline{AC}/2} \Rightarrow {\overline{IE} \over \overline{FJ}}={1\over 2} ={\overline{DI} \over \overline{DJ}} \Rightarrow I為\overline{DJ}中點\\ 三角形中線定: \overline{AJ}^2 +\overline{AD}^2= 2(\overline{AI}^2+ \overline{IJ}^2) \Rightarrow 100=2(25+ \overline{IJ}^2) \Rightarrow \overline{IJ}= \bbox[red, 2pt]5 \\\mathbf{(2)}\; \overline{IJ}=5 \Rightarrow \overline{DJ}=10 \Rightarrow \overline{DJ}^2 = \overline{AJ}^2 +\overline{AD}^2 \Rightarrow \angle JAD=90^\circ\\ d(L,M)= A至底邊\overline{DJ}的高 = 8\times 6\div 10= \bbox[red,2pt]{24\over 5}
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第一題是2019AMC改題,https://www.youtube.com/watch?v=1XEV-SufWng&list=PLmRNZ9_6PAB25t4F3XbO7eJ_RTaMaFUHE&index=4
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