Processing math: 100%

2020年2月12日 星期三

109年大學學測數學科詳解


109學年度學科能力測驗試題
數學考科
第壹部分:選擇題(占 6 0 分 )
一、單選題


{sinα=3/5=39/65sinβ=5/13=25/65sin30=1/2=32.5/65sinα>sin30>sinβ(2)



ABAC=ABADAB(ACAD)=0AB((CA+AD))=0AB((CD))=0ABCD=0(1)




(1)OP=OC+OEP=D(2)OP=14OC+12OEPODE(3)OP=14OC+12OEPOEF(4)OP=14OC12OEPOBC(5)OP=14OC12OEPOAB(2)


 A=[1134]A1=[4131]B=I+A+A1=[1001]+[1134]+[4131]=[6006]BA=[6006][1134]=[661824](5)



 {|x101|<5|x38|>3{5+101<x<5+101x>3+38x<3+38{4x15x10x310x15(3)



log(a2)+logb>1log(a2b)>log10a2b>10(a,b)=(2,36),(3,26),(4,16),(5,16),(6,16)4+5+6+6+6=2727/36=3/4(4)


f(x)=3x3f(x)=3x3=f(x)f(x)(cosθ,sinθ)Q=(cosθ,sinθ)=(cosθ,sin(θ))(4)

二、多選題


{滿A(1,3,1),(1,3,3),(1,3,5)滿B(1,3,2),(1,3,5),滿AB滿(1,3,1),(1,3,2),(1,3,3)1,2,3(1,2)




OPOQ=|OP||OQ|cosθ=2×2×cosθ=4cosθ={2θ=60,:P1OQ12θ=120,:Q1OQ24θ=180,:P1OQ22θ=240,:P1OP2(4,5)



(1):f(0)=4<0(2)×:{f(x)=f(x)y0<x<yf(x)<f(y)f(0)=4f(x)(3)×:(2)(4):01,01±2/3,f(±2/3)0(4):{f(0)=4f(1)=10f(0)×f(1)<0x[0,1],使f(x)=0(5)×:{f(1)=10f(2)=48+444>0f(1)×f(2)>012(1,4)



 
{loga=1.1logb=2.2logc=3.3{a=101.1b=102.2c=103.3(A)×:{a+c=101.1+103.3=102.2(101.1+101.1)2b=2×102.2a+c2b(B)×:a=101.1=10×100.1=10×1010>10(C):{log2000=log(2×103)=log2+log103=0.301+3=3.301log1000=33<3.3<3.3011000<c<2000(D)×:2a=2×101.1102.2(E):ba=101.1=cb(3,5)



(1)×:20132018(2)×:2015181.32016176.4(3):1070.9535.452011(4)×:49394049201865184.94972+78.8=150.8<184.9(5)×:79.469.1=10.3(1,3)




(1)×:{OAB¯OB=¯OAOAC¯OC=¯OA¯OB=¯OC,BOC=30OBC=180302=75OBC>BOC¯OC>¯BC(2):(1)(3)×:{OAB=12ׯOB2×sinAOBOBC=12ׯOB2×sinBOC,AOB=60>BOC=30OBCOAB(4):{¯OC=¯AC¯OB=¯AB¯OB=¯OBCABCOBCAB=COB=30(5)×:¯OAP使¯BP¯OA;¯BC=c,¯OC=¯OB=a,¯PB=¯PC=b,a>b;{cosBOC=2a2c22a2=1c2/2a2cosBPC=2b2c22b2=1c2/2b2c2=(1cosBOC)2a2=(1cosBPC)2b2(1cosBOC)a2=(1cosBPC)b21cosBOC<1cosBPCcosBPC<cosBOCBPC>BOC=3030(2,4)

第貳部分:選填題


:200×5=1000=1000200=8001調:800×0.5=4002調:400×0.5=2003調:200×0.5=100200+100=300



{:12×13×13=118:12×13×13=118118+118=19


 
{2x+y=10x2y+15=0A=(1,8)2x+y=10x2y=0B=(4,2)f(x,y)=3xy{f(A)=38=5f(B)=122=10c5





cosBAD=¯AD2+¯AB2¯BD22ׯADׯABcos135=4+2¯BD22×2×212=6¯BD242¯BD=10{¯OB=m¯OD=10m¯AO2={¯AB2¯OB2=2m2¯AD2¯OD2=4(10m)22m2=4(10m)2m=410¯AO2=¯AB2m2=21610=410¯AO=210¯AC=410=2105


 
{A(1,7,2)B(2,6,3)C(0,4,1)P{¯BC:x22=y+62=z32P=(t+2,t6,t+3),tRBC=(2,2,2)AP=(t+1,t13,t+1)APBC=0(t+1,t13,t+1)(2,2,2)=0t+1+t+13+t+1=03t+15=0t=5P=(5+2,56,5+3)=(3,1,2)


ABCD()O¯AB{A(2,0)B(2,0)C(3,14)D(3,14)y=a(x2)(x+2)D14=a×1×5a=145y=145(x24)x2=514y+4x2=4×556(y565)|556|=556





¯QT=23¯OQ=3¯PO¯OQ¯PO=¯PQ2¯QO2=43=1QPO=60QPT=120{PQST=120360ׯPQ2×π=43πPQT=¯QTׯPO÷2=3QRT=¯QO2×π÷2=32π=QRT(PQSTPQT)=32π(43π3)=16π+3{a=1/6b=3


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