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2020年11月18日 星期三

108年臺中二中教甄-數學詳解

臺中市立臺中第二高級中等學校108學年度

第一次教師甄選數學科題目卷

一、填充題:10 題共 50 分

1(logn2n)1=1logn2+lognn1=1logn2=log2n2056n=2log2n=(1+1)+22=42+2+2+2+23=83++3++210=102410++10+911++11=(10k=1k2k)+9×11=18533

2. 設kZ,若 x2+y2=2k2xy=4k 的圖形不相交,則k=_____

{x2+y2=2k2xy=4k{y2=2k2x2y2=16k2x22k2x2=16k2x2x42k2x2+16k2=0x2=2k2±4k464k224k464k2<04k2(k216)<0{k2<16k0{4<k<4k0k=±1,±2,±3
a=1021+31021=a31063=(1021)3=(a3)310631021+3=(a3)3a=a39a2+27a27a=a29a+26+a27a10631021+3=a29a+26+a27a=a29a+26(0<a27a<1)100a=1021+33mod100(a29a+26)mod100(927+26)mod1008=08
A=90z3zz2z=k(cos90+isin90)z(z+1)(z1)z(z1)=kiz+1=kiz=1+ki|z|=1+k2=2k=±3z=1±3i
{tanA=1tanB=2tanC=tan(π(A+B))=tan(A+B)=tanA+tanB1tanAtanB=3{sinA=1/2sinB=25sinC=310{ha=csinBhb=asinChc=bsinAabchahbhc=1sinAsinBsinC=106=53
NN4,5,63N10N3331333/642224233/64221324!/2=1212/651112514/2=1212/65113414/2=1212/651122315!/4=3030/6610110=24/65+30/6636/65+30/66+6/64=24×6+3036×6+30+6×36=2977
nk=1k2+3k+1(k+2)!=nk=1(k+2)(k+1)1(k+2)!=nk=1(1k!1(k+2)!)=(113!)+(12!14!)+(13!15!)++(1(n1)!1(n+1)!)+(1n!1(n+2)!)=1+12!1(n+1)!1(n+2!)=321(n+1)!1(n+2)!
x3+3x2=(xa)(xb)(xc){a+b+c=0ab+bc+ca=3abc=2a3+3a=b3+3b=c3+3c=2a,bx3+3x2=0{a3+3a=2b3+3b=2a3+3a=b3+3ba3b3+3a3b=0(ab)(a2+ab+b2)+3(ab)=0(ab)(a2+ab+b2+3)=0a2+ab+b2=3(ab)2=33ab=3(ab+1)=3(2c+1)=3(c+2c){(ab)2=3(ab+1)=3(c+2c)(bc)2=3(bc+1)=3(a+2a)(ca)2=3(ca+1)=3(b+2b)=(ab)2+(bc)2+(ca)2=3(1+2c+1+2a+1+2b)=3(3+2(ab+bc+ca)abc)=3(3+232)=18;=(ab)2(bc)2+(bc)2(ca)2+(ca)2(ab)2=9((ab+1)(bc+1)+(bc+1)(ca+1)+(ca+1)(ab+1))=9((2b+ab+bc+1)+(2c+bc+ca+1)+(2a+ca+ab+1))=9(2(a+b+c)+2(ab+bc+ca)+3)=9(0+6+3)=81=(ab)2(bc)2(ca)2=3(c+2c)3(a+2a)3(b+2b)=27(a+2)(b+2)(c+2)abc=272(a+2)(b+2)(c+2)=272(f(2))=272f(2)=272(862)=216x3+18x2+81x+216
西:(x2+y2)(t2+(z)2)(xtyz)2{x2+y2=16z2+t2=25xtyz=20(x2+y2)(t2+(z)2)=16×25=400=202西xt=yz=45z=54yxz=54xyx2+y22x2y28xy81054xy10xz10
{OA=(1,3,2)OB=(1,2,1)OC=(1,1,0)OAxOByOC=(1+xy,32x+y,2x)|OAxOByOC|2=(1+xy)2+(32x+y)2+(2x)2西:((1+xy)2+(32x+y)2+(2x)2)(12+12+(1)2)(1+xy+32x+y+2+x)2=62=36|OAxOByOC|363=23=m|OAxOByOC|=m1+xy=32x+y=2+x{x=0y=1(x,y,m)=(0,1,23)

二、計算證明題:5 題共 50 分



{ADE:¯AE2=¯AD2+¯DE2=62+32=45ADB:¯AB2=¯AD2+¯DB2=62+62=72{¯AE=35¯AB=62AEB:cosAEB=45+972185=15sinAEB=25CDE:¯DEsinDCE=¯DCsinDEC31/2=¯CD2/5¯CD=625sinBDC=sin(180(45+DEC))=sin(45+DEC)=sin45cosDEC+sinDECcos45=12(15)+2512=110BCD=12¯CDׯBDsinBDC=126256110=185
f(x)=x3+ax+b=0α,β,γ{α+β+γ=0αβ+βγ+γα=aαβγ=b(f)=(αβ)2(βγ)2(γα)2=(4a3+27b2)f(x)(1),(2);(1)(f)0(0)(2)(f)<04a3+27b2>04a3+27b2>0
:12[f,f]f(x0)=12,x0[0,1]0abc1f(b)=|ba|+|cb|3=ca313<12f(b)<12{f(1)=((1a)+(1b)+(1c))÷3=1(a+b+c)/3f(0)=(a+b+c)/3f(1)+f(0)2=12max
4. 證明:n\in N,1+\cfrac{2}{3n-2}\le \sqrt[n]{3} \le 1+\cfrac{2}{n}
\cfrac{3+\overbrace{1+\cdots + 1}^{n-1個1}}{n} \ge \sqrt[n]{3\cdot 1} \Rightarrow \cfrac{n+2}{n} \ge \sqrt[n]{3} \Rightarrow \sqrt[n]{3} \le 1+\cfrac{2}{n} \cdots(1)\\ \cfrac{1/3+\overbrace{1+\cdots + 1}^{n-1個1}}{n} \ge \sqrt[n]{1 \over 3} \Rightarrow \cfrac{1/3+n-1}{n} \ge   \cfrac{1}{\sqrt[n]{3}} \Rightarrow \cfrac{3n-2 }{3n} \ge   \cfrac{1}{\sqrt[n]{3}} \Rightarrow \cfrac{3n}{3n-2}\le  \sqrt[n]{3} \\ \Rightarrow 1+\cfrac{2}{3n-2}\le \sqrt[n]{3} \cdots(2) \\ 由(1)及(2) \Rightarrow 1+\cfrac{2}{3n-2}\le \sqrt[n]{3} \le 1+\cfrac{2}{n},\bbox[red,2pt]{故得證}。

假設\cases{\triangle 三邊長為a,b,c \\ s=(a+b+c)\div 2},則\triangle 面積= rs = \sqrt{s(s-a)(s-b)(s-c)} ={abc\over 4R};\\ 由於三邊成等差,故令\cases{a=x-d\\ b=x\\ c=x+d } \Rightarrow s=\cfrac{3}{2}x\Rightarrow \triangle 面積= rs= {3\over 2}rx \\ \Rightarrow \cases{ {3\over 2}rx= {abc\over 4R} ={x(x^2-d^2)\over 4R} \Rightarrow x^2-d^2 =6Rr\\{3\over 2}rx= \sqrt{s(s-a)(s-b)(s-c)} =\sqrt{{3x^2\over 4} ({x^2\over 4}-d^2) } } \\ \Rightarrow \cases{ x^2=d^2+6Rr\\{9r^2x^2\over 4} ={3x^2\over 4} ({x^2\over 4}-d^2) \Rightarrow x^2=4(3r^2+d^2)} \Rightarrow d^2+6Rr=12r^2+4d^2 \\\Rightarrow 3d^2=6Rr-12r^2 \Rightarrow d^2=2Rr-4r^2 \Rightarrow d=\sqrt{2Rr-4r^2},\bbox[red,2pt]{故得證}。

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