2019年12月28日 星期六

108學年度高雄區公立高中聯合轉學考-升高二數學科詳解


高雄區公立高中 108 學年度聯合招考轉學生
升高二數學科試題詳解
一、單選題


3x5102x682x+28|2x+2|8{a=2b=8a+b=6(A)


POP=(2,3)2(x2)+3(y3)=02x+3y=13(A)



x=1+i2i=1+i+2i=1+3i(x1)2=(3i)2x22x+1=9x22x+10=0{a=2b=10a+b=8(C)


125712458123+1612246512+100=(127)12458123+1612246512+100=512458123+1612246512+100=(51258)123+1612246512+100=2123+1612246512+100=(212+16)122+46512+100=4012246512+100=(4012465)12+100=1512+100=180+100=280(B)



(2+323)2=2+32(2+3)(23)+23=421=22+323=2log4(2+323)=log42=12log222log22=14(D)




{pH=2.5pH=6.5{logH+=2.5logH+=6.5{H+=102.5H+=106.5102.5106.5=104(E)



f(x)=ax2+bx+c,{a1=1a2=4=a1+f(0)=1+f(0)f(0)=3a3=7=a2+f(1)=4+f(1)f(1)=3a4=12=a3+f(2)=7+f(2)f(2)=5{c=3a+b+c=34a+2b+c=5{a=1b=1c=3f(x)=x2x+3a5=a4+f(3)=12+323+3=21(A)


a1=2a2=11a1=1a3=11a2=12a4=11a3=2⇒<an>=2,1,12,2,={2n=3k+11k=3k+21/2n=3k,k=0,1,2,a2019+a2020+a2021=12+21=32(A)



::6661:1,5,7,964×4×1=1631631+16+16+16=49(D)


7O3XXX:a1Xa2Xa3Xa4,7Oa1,a2,a3,a4a1+a2+a3+a4=7,aiN{0};a2a3O{b1=a1b2=a2+1b3=a3+1b4=a4b1+b2+b3+b4=5H45=C85=56(B)



14861,862,86333C53=310141310=710(E)



{{50,100,x1,x2,,x46}{μ=70σ{80,70,x1,x2,,x46}{μ=70σ{148((5070)2+(10070)2+46i=1(xi70)2)=σ2148((8070)2+(7070)2+46i=1(xi70)2)=σ2{46i=1(xi70)2=48σ2130046i=1(xi70)2=48σ210048σ21300=48σ2100σ2=σ2+25σ>σ(σ+5)2=(σ2+25)+10σ=σ2+10σ(σ+5)2σ2σ+5σσσ5σ5σ<σ(B)

二、多重選擇題

156{|x+1|+|x5|66|x+1||x5|6(A)(C)(E)(B)(D)(BD)




{f(x)=a(x1)2+bf(3)>0f(4)<0{f(1)f(x)=03<x<4{f(1)>0f(1){b>0a<0f(x)x=1{f(1)=f(12)=f(1+2)=f(3)>0f(2)=f(13)=f(1+3)=f(4)<0f(x)1<x<3f(2)>f(3)>0f(2)=a+b>0(CDE)





{log2a=19log3b=12{a=219b=312ab=219312logab=19log2+12log3=19×0.301+12×0.4771=11.4442ab12,log2<0.4442<log3ab2{a=219b=312{loga=19log2=19×0.301=5.719logb=12log3=12×0.4771=5.7252a,b6,log5<0.719,0.7252<log6(0.4771+0.301=0.7784)ab5a+b7,1(BD)


(A):S>0(a1+a61)612>0a1+a61>0(B)×:a42+a60=2a1+100d=2a51<0(C):{a51<0S>0{a1>0d<0,(A)a1+a61>0a1+a61=a20+a42>0,d<0a20>a42a20>0(D):a1+a61>02(a1+30d)>02a31>0a31>0(E)×:(C)(ACD)




(A)×:7200=25×32×52(5+1)(2+1)(2+1)=54(B):(20+21+22+23+24+25)(30+31+32)(50+51+52)=(261)×13×31=25389(C):7200=2×(22)2×32×52(2+1)(1+1)(1+1)=12(D):(1+4+16)(1+9)(1+25)=5460(E)×:7200=22×23×32×521,81+8=9(BCD)



S=5R40Wn(S)=45!5!40!A:1R,4R40Wn(A)=44!4!40!B:10R,4R40Wn(B)=44!4!40!AB:1R10R,3R40Wn(AB)=43!3!40!(A):P(A)=n(A)n(S)=44!4!40!×5!40!45!=545=19(B)×:P(B)=n(B)n(S)=P(A)=19(C):P(AB)=n(AB)n(S)=43!3!40!×5!40!45!=545=199P(BA)=P(AB)P(A)=1/991/9=111(D)×:P(AB)=199,(C)(E)×:P(A)×P(B)=19×19=181111=P(AB)A,B(AC)


(A)×:n(S)=6×6×6=216(B)×:n(B)=3×6×6=108(C)×:A5=(1,1,3),(1,3,1),(3,1,1),(1,2,2),(2,1,2),(2,2,1)A5B=(2,1,2),(2,2,1)n(A5B)=2(D)×:A17=(6,6,5),(6,5,6),(5,6,6)n(A17)=3n(A5)P(A5)P(A17)(E):A3=(1,1,1)A3B=(CE)




(A):ρ=Cov(x,y)σXσY=0σXσY=0(B)×:b1=Cov(x,y)σ2X=0σ2X=0(C)×::y=b1x+b0(ˉx,ˉy),{ˉx=11515i=1xi=165/15=11ˉy=11515i=1yi=315/15=21(y21)=0(x11)y=21y21(D):(C)(E):(100,21)(ADE)


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