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2020年5月4日 星期一

109學年度四技二專統測--數學(C)詳解


109學年度科技校院四年制與專科學校二年制
統一入學測驗試題本數學(C)詳解


22<0limx2x20(B)


{a=tan480=tan120<0b=sec135<0c=cos(60)>0{(b,c)=(,+)(a,b)=(,)(c,a)=(+,)(c,b)=(+,)(A)



f(x)=p(x)(x1)(x2+x+1)+3x2+5x2=p(x)(x1)(x2+x+1)+3(x2+x+1)+2x5=(x2+x+1)(p(x)(x1)+3)+2x52x5(B)



(A)×:10020=10020=80(B):39206070(C)×:(80)(D)×:39×90%=35.1PR904(B)



6×14=84(D)


cos21=cos(8059)=cos80cos59+sin80sin59=b1a2+a1b2(A)


(x4)225+(y+2)2144=1{C(4,2)a=12b=5(A)×:a2=b2+c2144=25+c2c=119=(4,2±119)(B)×:(4,2±12)(C)×:(4±5,2)(D):=2b2a=2×2512=256(D)


(3+i)z=23+2iz=23+2i3+i=(23+2i)(3i)(3+i)(3i)=4+43i4=1+3i=2(12+32i)=2(cos2π3+sin2π3)2π3(B)



{a5=41a13=73{a1+4d=41a1+12d=73{a1=25d=4{a19=25+18×4=97a20=25+19×4=1014203011100(B)



{A(2,2)B(9,3/2)C(9,1)D(2,8)ABCD=BCDEABE=(¯BC+¯DE)ׯBE2¯AEׯBE2=(5/2+19/2)×727/2×72=42494=1194(A)


cos3x12cos3x121=12cos3x1=0cos3x=123x=π3+nπ,n=0,1,2,...x=π9+nπ3,n=0,1,2,...x[0,2π]x=π9,49π,79π,119π,149π,179π6{a=6b=1ab=6(A)


10(1+3%)r=20(1+3%)r=21.03r=2rlog1.03=log2r=log2log1.03=0.3010.012823.520+23.5=43.544(C)


f(x)=x33x224x+32f(x)=3x26x24f(x)=6x6f(x)=03x26x24=0x22x8=0(x4)(x+2)=0f(4),f(2);{f(4)=246>0f(4)=48f(2)=126<0f(2)=60;{2[3,3]f(3)=50f(3)=40{f(2)=60=mf(3)=40=nmn=100(C)




xyz36123621BG4×4!2(x,x,y,z)=4×12=484644864=127(C)註:(1,1,1,2)或(2,2,2,1)不是BG,要重新擲骰子!


Pk+k2+24k+k1=0k22k+1=0(k1)2=0k=1k=1x2+y2+2x4y+11=0(x+1)2+(y2)2=5(A)


log10x2(x2+3x+2){10x2>010x21x2+3x+2>0{10>x>10x±3(x+32)214>0{x=3,2,1,0,1,2,3x±3x+32>12,x+32<12{x=2,1,0,1,2x>1,x<2x=0,1,23(A)



f(x)={2x1x>2x22x+3x2f(x)={2x>22x2x22=2×22f(2)=2(B)


α,βx2+5x+k=0{α+β=5(1)αβ=k(2){f(x)xα1f(x)xβ2{f(α)=1f(β)=2{2α2+7α+5=12β2+7β+5=2{2α2+7α+6=02β2+7β+3=0{(2α+3)(α+2)=0(2β+1)(β+3)=0{α=3/2,2β=1/2,3(1){α=2β=3(2):k=αβ=(2)×(3)=6(C)



{AaBb5a+3b=11070(1){A1.5aB1.5b(5+3)×1.5a+(3+9)×1.5b=421202a+3b=7020(2);(1)(2)3a=4050a=1350(A)



|a12b13c1a12c1a1a22b23c2a22c2a2a32b33c3a32c3a3|C2C3=|a12b13c12c1a1a22b23c22c2a2a32b33c32c3a3|=2|a12b13c1c1a1a22b23c2c2a2a32b33c3c3a3|=2|a12b13c1a1c1a22b23c2a2c2a32b33c3a3c3|C1C2+3C3=2|2b1a1c12b2a2c22b3a3c3|=4|b1a1c1b2a2c2b3a3c3|=4|a1b1c1a2b2c2a3b3c3||a12b13c1a12c1a1a22b23c2a22c2a2a32b33c3a32c3a3|=8=4|a1b1c1a2b2c2a3b3c3||a1b1c1a2b2c2a3b3c3|=2(C)


{A(1,1)B(5,2)C(5,2)DC=(x,y){AB=(4,3)AC=(4,1)DC=(x,y)AD=ACAB|AB|×AB|AB|=(4,1)(4,3)42+32×(4,3)42+32=1325(4,3)D=1325(4,3)+(1,1)=(7725,1425)DC=(57725,2+1425)=(4825,6425)=(x,y)x+y=4825+645=11225(D)


(ax+1)4=a4x4+4a3x3+6a2x2+4ax+1{4a3>a44a3>6a24a3>4a4a3>1{a3(a4)<02a2(32a)<04a(1a2)<0a3>1/4{0<a<43/2<a1>a>0,1<a<0,a>1a>1/343/2<a<4(D)


:cosA=¯AB2+¯AC2¯BC22ׯABׯACcos120=32+62¯BC22×3×612=45¯BC236¯BC2=45+18=63¯BC=37;¯AD¯BD:¯DC=¯AB:¯AC=3:6=1:2¯DC=23¯BC=23×37=27(C)



{D(1,2)G(3,2)DG(A)f(x,y)=3x+4y+2f(D)f(G)>0(B)f(x,y)=3x+4y6f(D)f(G)>0(C)f(x,y)=6x+8y+3f(D)f(G)>0(D)f(x,y)=6x+8y2f(D)f(G)=8×(4)<0(D)




{11f(x)dx=+=5=12×1×12=14=32×3×12=9411(f(x)g(x))dx=+=514+94=7(D)

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