2019年12月10日 星期二

108學年度臺北市聯合轉學考-高中升高二-數學科詳解


臺北市高級中等學校 108 學年度聯合轉學考招生考試
高中數學科試題(升高二)
一、單選題

(A)(B)=(52)/5>0(A)>(B)(C)(D)=(52)/7>0(C)>(D){(22+35)/5=(142+2110)/35(32+45)/7=(152+2010)/35(142+2110)(152+2010)=102>0(A)>(C){(22+35)/5=(42+610)/10(2+5)/2=(52+55)/10(42+610)(52+55)=62>0(A)>(E)(A)



f(x)=(x4+2x23x+5)(x2+3x2)+x3x2+2x4=(x4+2x23x+5)(x2+3x2)+(x2+3x2)(x4)+16x12=(x4+2x22x+1)(x2+3x2)+16x12x4+2x22x+1(D)


1<x<4(x4)(x+1)<0x23x4<0x2+3x+4>0{a=1b=4bx2+2ax12<04x22x12<02(2x+3)(x2)<032<x<2(B)


2x517x4+52x368x2+35x6=(2x1)(x48x3+22x223x+6)=(2x1)(x2)(x36x2+10x3)=(2x1)(x2)(x3)(x23x+1)=01/2+2+3=11/2(E)



1024<2019<2048log21024<log22019<log2204810<log22019<1120<2log22019<2220<log220192<2221(D)




x=3132×0.0154103×0.0112logx=13(log132+log0.0154)12(log103+log0.0112)=13(log(1.32×100)+log1.54100)12(log(1.03×100)+log1.12100)=13(log1.32+2+log1.542)12(log1.03+2+log1.122)=13(log1.32+log1.54)12(log1.03+log1.12)=13(0.1206+0.1875)12(0.0128+0.0492)=13×0.308112×0.062=0.10270.031=0.0717logx=0.0717x1.18(E)




(甲乙不同船,其他人任坐) - (甲乙不同船而且超載)
=(甲有3種選法、乙有兩種選法,其他人都有3種選法=3×2×34) - (甲有3種選法、乙有兩種選法,其他人同坐甲船或同坐乙船(3×2×2) ,因此共有3×2×343×2×2=48612=474安全過渡法, 故選(A)



+=1×0.9991×0.999+99999×0.001=0.999100.9981100(D)


a<bc(a,b,c)(6,5,51)5(6,4,41)4(6,1,1)1(5,4,41)4(5,3,31)3(5,1,1)1(2,1,1)1(5+4++1)+(4+3++1)++(1)=5+4×2+3×3+2×4+1×5=35356×6×6=35216(B)


0.9n<25nlog0.9<log2log5n(2log31)<log2(1log2)n(2×0.47711)<2×0.30110.0458n<0.398n>0.3980.0458=8.69n=9(B)



{a1+a3+a5=12S13=208{a1+(a1+2d)+(a1+4d)=12(2a1+12d)×13/2=208{a1+2d=4a1+6d=16{a1=2d=3S8=(2a1+7d)×8÷2=(4+21)×4=68(E)


4×2×3=24a1a2a3a4b1b2c1c2c36205304503203602003002101501000(1,1,1),(1,1,2),(1,1,3)(1,2,1),(1,2,2)(2,1,1),(2,1,2),(2,1,3)(2,2,1)(3,1,1),(3,1,2)1112411=13(C)



x+y+z=7H37=C97(C)




{EX=5EX2(EX)2=42EX2=16+52=41E(3X25)=3EX25=3×415=118(D)但公佈的答案是(C),隨便舉個例子:{x1=1x2=9{=(1+9)÷2=5=((15)2+(95)2)÷2=4{3x215=23x225=238=(2+238)÷2=118110

二、多重選擇題



(A)×:{log73<1log74<1(log73<1)(log74<1)<11(B)×:log212)(log32)=log12log2×log2log3=log12log3=log3124(C):log23=log3log2=(1/2)log3(1/2)log2=log3log2=log23(D):5log53=alog55log53=log5alog53=log5aa=3=log327(E):y=0.2x0.2a>0.2bb>a(CDE)


(A)×:f(1)×f(2)=(2)×3<012(B):f()>0f(3)×f()<03(C):f()>0f(1)×f()<01(,1),(1,2),(2,3),(3,)4(D):(E)×:(x(1+2)(x2)=01+2212(BCD)



(A)×:E(1.5X+10)=1.5E(X)+101.510(B)×:σ(1.5X+10)=1.5σ(X)1.5(C):(D):(E)×:(CD)


  
(A):/=3/7(B):+=37×46+47×36=27+27=47(C)×:+=37×47+47×37=2449(D):(B)(E)×:()()()()(4×3×2+3×2×4+4×3×3+3×4×3)÷(7×6×5)=47(ABDE)


<bi>=2,4,6,<2i>,i=1,2,a1=b1a2=b2+b3a3=b4+b5+b6a4=b7+b8+b9+b10{1b1nbkn+1bk+nnn(A):a5=22+24+26+28+30=130(B)×:nbk,k=1+n1i=1i=n(n1)2+1bk=2k=n(n1)+2(C):nbk,k=ni=1i=n(n+1)2bk=2k=n(n+1)(D):an=(n(n1)+2)+(n(n1)+4)++n(n+1)=(n2n+2)+(n2n+4)++(n2n+2n)=n(n2n)+2(1+2++n)=n(n2n)+n(n+1)=n3+n(E)×:ni=1an=2+4+6++n(n+1)=2(1+2+3++n(n+1)2)=(n(n+1)2+1)n(n+1)2=n(n+1)(n2+n+2)4(ACD)




解題僅供參考

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