2019年12月24日 星期二

105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 105 學年度聯合轉學考招生考試
升高二數學科試題
一、單選題


{x=2+121=(2+1)2(21)(2+1)=3+22y=212+1=(21)2(21)(2+1)=322x+y=6(C)


{a3+a5=8a6+a10=32{(a1+2d)+(a1+4d)=8(a1+5d)+(a1+9d)=32{2a1+6d=82a1+14d=328d=24d=3(D)



6x×8y×9z=(2x×3x)×(23y)×(32z)=2x+3y×3x+2z=28×37{x+3y=8x+2z=7{x+3y=8x+2z=7x,y,z{x=5y=1z=1x+y+z=7(D)


2x4+3x32x2x2=(x1)(x+2)(2x2+x+1)(B)



|a+1|+|a2|=5{2a1=5a2a+1+2a=51a22a+1=5a1a=3,2b=3,2;a>b{a=3b=2ab=5(D)




{x2+8x7>0x20{(x7)(x1)<0x2{1<x<7x2x=3,4,5,6x=3log184,5,63(B)





xy4=10000log(xy4)=log(10000)logx+4logy=4(logx,logy)(0,1)(4,0)(A)


(A):{a3a8a8a3=a5a5a3=a2a3a2=a(B)×:{a=2b=2{a+b=0ab=2a,b(C)×:(B)(D)×:{a=2b=2{a+b=2+2b=2a+baba(E)×:{a=c=2b=d=2a+b,b+c,c+d,d+aa,b,c,d(A)


1410141024141410C147×C105(C)


f(1+i)=0x=1+i(x+1)2=i2x2+2x+2=0x2+2x+2f(x)f(x)=(x2+2x+2)(x+(a2))=x3+ax2+(2a2)x+(2a4){2a2=82a4=ba=5(x+(a2))=(x+3)f(x)f(3)=0(E)



=1%×0.91%×0.9+99%×0.01=0.0090.009+0.0099=1021(C)



x233444555566666y112312412351423z1211321432152433VVVVVVVxy=z15137715(B)


(D)




3(x,y,z)x=1C82x=2C72x=8C22x1×C92+2×C82++8×C22C92+C82++C22=330120=114(A)

二、多重選擇題



{a>0{>0g(x)(C)a<0{<0g(x)(B)(BC)


{an=C4n/24bn=C8n/28(A)×:a2=C42/24=6/16=3/8(B)×:b4=C84/28=70/256=35/128a2(C)×:b2=C82/28=28/256=7/642b2=7/32a2(D):b2=C82/28=C86/28=b6(E):{a3=C43/24=4/16=8/32b3=C83/28=56/256=7/32a3>b3(DE)




(A):S10=S15a1++a10=a1++a10+a11++a15a11++a15=0(a11+a15)×5÷2=0a11+a15=02a1+24d=0d=a1/12<0(B):a11++a15=0{a11+a15=0a12+a14=0a13=0{a12+a14=0a12>a14a12>0(C):d<0a14>a15a12+a15<a12+a14=0a12+a15<0(a1+11d)+(a1+14d)<0(a1+d)+(a1+24d)<0a2+a25<0(D):S14S11=a12+a13+a14=0(E)×:(A)2a1+24d=0a1+a25=0S25=0S26=S25+a26=a26<0(ABCD)



(1+2)4=C40+C412+C42(2)2+C43(2)3+C44(2)4{a4=C40+2C42+4C44b4=C41+2C43(A)×:a4=C40+2C42+4C44(B):b4=C41+2C43(C)×:a4+b4=17+12=2924(D):a242b24=1722×122=289288=1(E):(12)4=C40C412+C42(2)2C43(2)3+C44(2)4=(C40+2C42+4C44)(C41+2C43)2=a4b42(BDE)


(A):{σ(X+5)=σ(X)σ(Y+10)=σ(Y)Cov(X+5,Y+10)=Cov(X,y)ρ=Cov(X,Y)σXσY=Cov(X+5,Y+10)σX+5σY+10(B)×:{σ(2X)=2σ(X)σ(3Y)=3σ(Y)Cov(2X,3Y)=6Cov(X,y)ρ=Cov(2X,3Y)σ2Xσ3Y=6Cov(X,Y)6σXσY=Cov(X,Y)σXσY(C):{ρ=Cov(X,Y)σXσY=0.634β1=Cov(X,Y)σ2X=0.635β1ρ=σYσX=0.6350.634>1σY>σX(D)×:σXσY(E)×:(AC)




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