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2016年5月31日 星期二

99學年四技二專統測--數學(C)詳解

試題來源:技專校院入學測驗中心

L:y=14x+7=14x=287+4×728(B)





x=4y2+8y(y+1)2=14(x+4)(4,1)(B)




(A)sin885°=sin(885°360×2)=sin165°=sin15°(B)cos(430°)=cos(430°+360×2)=cos290°=cos70°(C)tan131°=tan49°<1(D)sin(2010°)=sin(2010°+360×6)=sin150°=sin30°(C)




A=(0,0)B=(4,8)D=(1,4)C=(4+1,8+4)=(5,12)|AC|+|BD|=52+122+32+42=13+5=18(B)




{L1:x+3y2=0L2:3x+y+2=0{3x+9y6=03x+y+2=08y8=0y=1,x=1L3:xy2=0=1A(1,1)1:xy+2=0(D)




C:(x+1)2+(y2)2=32(1,2),3L:3x+4y+5=0=|3+8+532+42|=2=bL=2=a(C)




log9(10x26x+5)log3x1=0log9(10x26x+5)log9x2log99=0log910x26x+59x2=010x26x+59x2=1x26x+5(x5)(x1)=0p+q=5+1=61p+q=16(A)




42C4231C3152C52C42×C31×C52=6×3×10=180(C)




a0.4a0.32a0.28a0.4a×0.50.32a×0.40.28a×0.60.2a+0.128a+0.168a=0.496a=0.496a=49.6%(D)





f(x)=g(x)x23x+5=2x+1(x4)(x1)=0x=1(a=1),x=4(b=4)f(x)=2x3,g(x)=2f(1)=1,g(x)=2m1+m2=21=1(D)




x+y10ABxy1BC(B)




(C)caf(x)dx=baf(x)dx+cbf(x)dx(C)




2313=10×23=203=40×13=403=603=20(B)




a,b,cs=(a+b+c)÷2r=abc4s(sa)(sb)(sc)=4×6×849×5×3×1=48135=1615=25615π(A)




(A)f(x)=(4x+5)(6x+7)f(x)=4(6x+7)+6(4x+5)=48x+58(B)f(x)=3x7+4x=x73+4xf(x)=73x43+4(C)f(x)=(4x+5)2f(x)=2(4x+5)×4=8(4x+5)(D)f(x)=4x+4x+1=4(x+1)x+1=4f(x)=0(B)




(A)limn3n2n5n=limn[(35)n(25)n]=0(C)limn0.01n5n1=0.015(D)limn(nn21)=limn((nn21)(n+n21)n+n21)=limn(1n+n21)=0(B)




f(x)=ax3+bx2+cx+d=p(x)(x21)=q(x)(x2)+6{f(1)=0a+b+c+d=0f(1)=0a+bc+d=0{b+d=0a+c=0f(2)=68a+4b+2c+d=66a+2(a+c)+3b+(b+d)=66a+3b=63(2a+b)=62a+b=2(C)




2(1+i)2+k(1+i)+6+2i=04i+k(1+i)+6+2i=06(1+i)+k(1+i)=0k=6(A)



S=1+13+122+133+124+135++122k+132k+1+=[1+122+124++122k+]+[13+133+135++132k+1+]=[11122]+[131132]=43+38=4124(A)




5r=4(340+352)25r4=(340+352)2log(5r4)=2log(340+352)=2log(235+352)=2log(5235)rlog52log2=2log(543)2log2=83log52log2r=83(A)



¯OA=2,¯OB=4¯ABD,使¯OD=3¯OB=4,¯OC=5O¯BC45,¯OA=2¯OC=5¯ACEF34ABCDEF6(C)





BDC=BAD+ABDABD=30¯AD=¯DB¯AD:¯DC=1:2¯AD=a,¯DC=2aDBC,¯BC2=¯DB2+¯DC22¯DB¯DCcosBDC=a2+4a24a2×12=3a2¯BC=3aDBC:a,2a,3aDCB=30(A)





A=(0,0),B=(xb,yb),C=(xc,yc),D=(xb+xc2,yb+yc2)ADBC=(xb+xc2,yb+yc2)(xcxb,ycyb)=12(xb+xc)(xcxb)+12(yb+yc)(ycyb)=12(x2cx2b)+12(y2cy2b)=12[(x2c+y2c)(x2b+y2b)]=12[¯AC2¯AB2]=12(2581)=28(A)




1+if1i1,1+i,1iff(x)=a(x1)(x2x+2)f(0)>0a×(1)×2>0a<0(A)f(2)=24a>0(B)f(2)=4a<0(C)f(4)=42a<0(D)f(6)=160a<0(C)


f(x)=(cosx+3sinx)(cosxsinx)=cos2x+2sinxcosx3sin2x=cos2x+2sinxcosx3(1cos2x)=4cos2x+2sinxcosx3=2(2cos2x1)+2sinxcosx1=2cos2x+sin2x1=5(25cos2x+15sin2x)1=5(sinαcos2x+cosαsin2x)1=5sin(α+2x)151f(x)51(D)


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