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:
五位數=abcde=a×104+b×103+c×102+d×10+e,需符合a>b>c>d>eabcde數量小計43210115−−−−C54=566−−−−C64=15217−−−−C74=35568321015784−−−C43=46185−−−C53=107186−−−C63=209187210192873−−C32=395874−−C42=6101875−−C52=10111⇒第111位是87543、第110位是87542、第109位是87541。
解:
f(x)=x5−x4−x3−x2−x−3⇒f′(x)=5x4−4x3−3x2−2x−1⇒f′(x)f(x)=∞∑k=0ak+bk+ck+dk+ekxk+1;由長除法(見上圖)可知:f′(x)f(x)=5x+1x2+3x3+7x4+15x5+41x6+⋯,因此a5+b5+c5+d5+e5=41公式來源:按這裡
解:
(abc)×8=(pqr)⇒a=1(若a≥2,乘積為四位數)且b≤2;又(1bc)×d為四位數,因此可得112×89=9968
令f(x)=ax2+bx+c,由f(x−2)=f(−x−2)⇒a(x−2)2+b(x−2)+c=a(−x−2)2+b(−x−2)+c⇒ax2+(b−4a)x+4a−2b+c=ax2+(4a−b)x+4a−2b+c⇒b−4a=4a−b⇒b=4a⇒f(x)=ax2+4ax+c=a(x+2)2+c−4a;在x軸截得線段長=2√2⇒f(x)=0的兩根為−2±√2⇒f(x)=a(x+2+√2)(x+2−√2);再由y截距為1,即f(0)=1⇒a(2+√2)(2−√2)=1⇒a=12⇒f(x)=12(x+2+√2)(x+2−√2)=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+1n−1n+1)2⇒p=2019∑k=1(1+1n−1n+1)=2019+2019∑k=1(1n−1n+1)=2019+1−12020=2020−12020⇒p最接近之正整數為2020
解:
log2(kx2)+x=2|x|+x⇒log2(kx2)=2|x|,而兩圖形{y=log2(kx2)y=2|x|皆對稱y軸,因此兩交點可設為{A(a,log2(ka2)+a)B(−a,log2(k(−a)2)−a),再由¯AB=√4a2+4a2=8√2⇒a=4;將x=a=4代回原式,可得log2(k⋅42)+4=2|4|+4⇒log2(16k)=16⇒k=4096
解:
f(x)=73√9+x2−23x⇒f′(x)=0⇒7x3√9+x2=23⇒7x=2√9+x2⇒49x2=4(9+x2)⇒45x2=36⇒x=√3645(∵
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}
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}
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