Processing math: 85%

2023年7月30日 星期日

112年新竹中學教甄-數學詳解

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)

解答滿nf(n)n=3:101f(3)=1n=4:1011,1101f(4)=2n=5:10101,11011f(5)=201011f(n)=f(n2)+f(n3),n6n678910111213141516f(n)34579121621283749f(16)=49
解答:(1)(2)(3)(4)(5)(6)(7)(8)(1)(6)(1)(8)14(2,3,6,7)10(4)(5)12924!=2492×24=2208
解答=+=0.001×0.99+0.999×0.02=0.001×0.99×0.990.001×0.99×0.99+0.999×0.02×0.02=363511
解答

COQPR=60QOR=120¯QRsinQOR=33sin120=6=2rr=3¯OR=¯OQ=3CORCOQ(SSS)COQ=COR=120cosCOQ=12=¯OC2+32722¯CO3¯OC=5¯CP=4COP=θ{sinθ=4/5cosθ=3/5{POR=120θPOQ=120+θPOR+POQ=1232(sin(120θ)+sin(120+θ))=923cosθ=27103OPR=1232sin120=943PQR=27103+943=99203
解答ω=cos(2π/7)+isin(2π/7)=e2πi/7ω6=e12πi/7=e2πi/7cos(2π/7)=(ω+ω6)/2cos(4π/7)=(ω2+ω5)/2,cos(6π/7)=(ω3+ω4)/2x3+ax2+bx+c=0{α=ω+ω62β=ω2+ω52γ=ω3+ω42{a=α+β+γb=αβ+βγ+γαc=αβγα+β+γ=12(ω+ω2++ω6)=12(1)=12a=12αβ+βγ+γα=142(ω+ω2++ω6)=12b=12αβγ=18(2+ω+ω2+s+ω6)=18c=18abc=132
解答z1,z2,z3,z4P(z)=0{z1+z2+z3+z4=4z1(z2+z3+z4)+z2(z3+z4)+z3z4=3z1z2z3+z2z3z4+z3z4z1+z1z2z4=2z1z2z3z4=1{f(z1)=4iˉz1f(z2)=4iˉz2f(z3)=4iˉz3f(z4)=4iˉz4Q(z)=0{B=(4i)2×3=48D=(4i)4×1=256B+D=25648=208
解答

¯BD=¯AC=2+1+3=6BDD():¯BD2=¯BD2+¯DD2=(¯BY+¯DX)2+¯XY262=(¯BY+¯DX)2+12¯BY+¯DX=35=ABC+ADC=12¯AC(¯BY+¯DX)=12635=335
解答L:y=tanθxP(a,b)LL:y=1tanθ(xa)+bP=LL=(a+btanθ1+tan2θ,atanθ+btan2θ1+tan2θ)=[11+tan2θtanθ1+tan2θtanθ1+tan2θtan2θ1+tan2θ][ab]A=[11+tan2θtanθ1+tan2θtanθ1+tan2θtan2θ1+tan2θ]A2=AA112=A=[11+tan2θtanθ1+tan2θtanθ1+tan2θtan2θ1+tan2θ]
解答
cosC=42+6228246=12C=60{B(0,0)C(4,0)A(1,33)¯ACQ¯BQ=23Q(3,3){L1=BQ:3y=3xL2=CP:x+3y=4P=L1L2=(2,23/3){AP=(1,73/3)AB=(1,33)AC=(3,33)(α,β)=(13,49)


解答:y=ax+b,(x,y)=(4,5),(6,5),(8,7),(10,9),(12,9)XYX2XY451620653630876456109100901291441084035360304{ˉx=40/5=8ˉy=35/5=7a=n(xy)(x)(y)nx2(x)2=5×30440×355×360402=35(ˉx,ˉy)y=35(x8)+7y=35x+115(a,b)=(35,115)
解答f(x)=anx2+an1xn1++a1x+a0pxqp,qpanqa0:f(x)pxqf(q/p)=0an(qp)n+an1(qp)n1++a1(qp)+a0=0anqn+an1pqn1++a1pn1q+a0pn=0{p(an1qn1++a1pn2q+a0pn1)=anqnpanqnq(anqn1+an1pqn2++a1pn1)=a0pnqa0pnp,q{panqa0
解答an=2an1+n=2(2an2+(n1))+n=22an2+2(n1)+n=22(2an3+(n2))+2(n1)+n=23an3+22(n2)+2(n1)+n=2n1a1+2n22+2n33++22(n2)+2(n1)+n=2n11+2n22+2n33++22(n2)+2(n1)+20n2an=2n+2n12+2n23++23(n2)+22(n1)+2n2anan=an=2n+2n1+2n2++22+2n=2n+1n2
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}

=============== END ============

學校未公告答案,解題僅供參考,其他教甄試題及詳解

3 則留言:

  1. 第一題是2019AMC改題,https://www.youtube.com/watch?v=1XEV-SufWng&list=PLmRNZ9_6PAB25t4F3XbO7eJ_RTaMaFUHE&index=4

    回覆刪除
    回覆
    1. 謝謝告知,這份考卷還有其他的AMC試題!

      刪除