Loading [MathJax]/jax/element/mml/optable/MathOperators.js

2021年1月22日 星期五

109年竹科實中教甄數學詳解


國立科學工業園區實驗高級中學 109 學年度
第一次教師甄選-數學科試題
填充題


解:
987657=3×329219=3×11×29929=3×11×1732173





解:
=abcde=a×104+b×103+c×102+d×10+ea>b>c>d>eabcde43210115C54=566C64=15217C74=35568321015784C43=46185C53=107186C63=209187210192873C32=395874C42=6101875C52=10111111875431108754210987541



解:


f(x)=x5x4x3x2x3f(x)=5x44x33x22x1f(x)f(x)=k=0ak+bk+ck+dk+ekxk+1;():f(x)f(x)=5x+1x2+3x3+7x4+15x5+41x6+a5+b5+c5+d5+e5=41:




解:


(abc)×8=(pqr)a=1(a2,)b2;(1bc)×d112×89=9968


解:
f(x)=ax2+bx+cf(x2)=f(x2)a(x2)2+b(x2)+c=a(x2)2+b(x2)+cax2+(b4a)x+4a2b+c=ax2+(4ab)x+4a2b+cb4a=4abb=4af(x)=ax2+4ax+c=a(x+2)2+c4a;x=22f(x)=02±2f(x)=a(x+2+2)(x+22);y1f(0)=1a(2+2)(22)=1a=12f(x)=12(x+2+2)(x+22)=12x2+2x+1



解:
1+1n2+1(n+1)2=1+(n+1)2+n2n2(n+1)2=1+2n2+2n+1n2(n+1)2=1+2n2+2nn2(n+1)2+1n2(n+1)2=1+2n(n+1)+1n2(n+1)2=(1+1n(n+1))2=(1+1n1n+1)2p=2019k=1(1+1n1n+1)=2019+2019k=1(1n1n+1)=2019+112020=202012020p2020


解:
log2(kx2)+x=2|x|+xlog2(kx2)=2|x|{y=log2(kx2)y=2|x|y{A(a,log2(ka2)+a)B(a,log2(k(a)2)a)¯AB=4a2+4a2=82a=4x=a=4log2(k42)+4=2|4|+4log2(16k)=16k=4096


解:
f(x)=739+x223xf(x)=07x39+x2=237x=29+x249x2=4(9+x2)45x2=36x=3645(


解:
令f(x)=g(x)+ax+b \Rightarrow \cases{f^2=(g+(ax+b))^2 = g^2+2(ax+b)g+(ax+b)^2 \\ g^2=(f-(ax+b))^2 = f^2-2(ax+b)f+ (ax+b)^2};\\ 又\cases{f^2除以g的餘式為4x+1\\ g^2除以f的餘式為5x+3} \Rightarrow (ax+b)^2 =\cases{a^2g(x)+4x+1 \\ a^2f(x)+5x+3 } \\ \Rightarrow a^2g(x)+4x+1=a^2f(x)+5x+3 \Rightarrow a^2(g-f)=x+2 \\\Rightarrow a^2(-ax-b)=x+2 \Rightarrow -a^3x-a^2b= x+2 \Rightarrow \cases{a=-1\\b=-2} 代回(ax+b)^2 =a^2g(x)+4x+1 \\\Rightarrow (-x-2)^2=g(x)+4x+1  \Rightarrow x^2+4x+4=g(x)+4x+1 \Rightarrow g(x) =\bbox[red,2pt]{x^2+3}


解:


P為\overline{BD}中點\Rightarrow P((2+4)/2,(4+2)/2,0)=(3,3,0);\\ 假設平面E與\overline{BC}交於Q點,則{四面體A-PQB體積\over 四面體A-BCD體積} ={\triangle BPQ\over \triangle BCD} ={\overline{BQ}\cdot \overline{BP} \over \overline{BC}\cdot \overline{BD}}={1\over 3}\\ \Rightarrow {\overline{BQ}\cdot \overline{BD}/2 \over 2\sqrt 5\cdot \overline{BD}}={1\over 3} \Rightarrow \overline{BQ}={4\over 3}\sqrt 5 \Rightarrow {\overline{BQ} \over \overline{BC}} = {4\sqrt 5/3\over 2\sqrt 5} ={2\over 3} \Rightarrow Q(2/3,4/3,0)\\ 因此\cases{\overrightarrow{AP} =(3,3,-1)\\ \overrightarrow{AQ}=(2/3,4/3, -1)} \Rightarrow 平面E 法向量\vec n= \overrightarrow{AP} \times \overrightarrow{AQ} =(-5/3,7/3,2) \\ \Rightarrow 平面E方程式: -{5 \over 3}x+{7\over 3}y+2(z-1)=0 \Rightarrow \bbox[red,2pt]{-5x+7y+6z=6}


-- END   (未公告計算題題目,填充題詳解僅供參考)  --





沒有留言:

張貼留言