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2021年5月18日 星期二

110年警專40期數學科(甲組)詳解

臺灣警察專科學校110學年度專科警員班

第40期正期學生組新生入學考試甲組數學試題

壹、單選題

解答y=cosx2πy=14[2π,2π]44(D)

解答{L1:axy+b=0L2:cxy+d=0{L1=aL2=c{a>ca,c>0b<0d>0d>b(D)
解答3x54x14|x1|4|x+1|4|ax+1|b{a=1b=4(C)
解答{7x+3y=105x3y=2{x=1y=1(1,1,z),zRz(D)
解答41(C54C51)32(C53C52)23(C52C53)14(C51C54)5×5+10×10+10×10+5×5=250(A)
解答f(x)=x54x472x356x2+15x+8f(11)=11541147211356112+1511+8=(114)1147211356112+1511+8=71147211356112+1511+8=511356112+1511+8=112+1511+8=411+8=52(B)
解答y=ax2+bx+3ax=21{y(2)=1y(2)=0{4a2b+3a=1(1)4a+b=0(2)(2)b=4a(1)4a8a+3a=14a2a3=0(4a+3)(a1)=0a=34(a=1y)b=4a=3(D)
解答{a=21/2=12b=31/3=133c=41/4=21/2=a{a6=18b6=19a=c>b(C)
解答logx+logy=2log(xy)=2xy=102=100;k=4x+1y=x+4yxy=x+4y1004x+y24xy=40k40100=254x+1y25(D)
解答2log(7x)=log2+log(x3)log(7x)2=log2(x3)(7x)2=2(x3)x214x+49=2x6x216x+55=0(x5)(x11)=0x=5(x=117x>0)(B)
解答3i3+i(x(3i))(x(3+i))=x26x+10f(x)f(x)=(x26x+10)(x+a+6)+(6(a+6)6)x+1010(a+6)0a=5f(x)=(x26x+10)(x+1)f(x)=01(A)
解答{a4=42a10=27{a1+3d=42a1+9d=27{a1=99/2d=5/2Sn=n(2a1+(n1)d)÷2=12dn2+(a1d2)nddnSn=0dn+(a1d2)=0n=(d2a1)/d=12+995=20.320(C)
解答f(x)=(ax1x3)8=8k=0C8k(ax)k(1x3)8k=8k=0C8k(1)8kakx4k24k=6=C86(1)2a6=1792a6=1792÷28=64a=2(A)

解答:yμy=rσyσx(xμx)(63,68)6865=23σy2(6360)σy=3(C)
解答sinθcosθ=15(sinθcosθ)2=12sinθcosθ=1252sinθcosθ=2425(sinθ+cosθ)2=1+2sinθcosθ=1+2425=4925sinθ+cosθ=75{sinθcosθ=15sinθ+cosθ=75{sinθ=3/5cosθ=4/5tanθ=3/4(A)

解答


C=90¯BDA=90¯BD2=¯AB2+¯AD2=122+162=202=2r=¯BD=20r=10;ADC=60AOC=120:cosAOC=¯OA2+¯OC2¯AC22¯OA¯OC12=102+102¯AC221010¯AC2=300¯AC=103(B)
解答A=[acbd]{A[73]=[21]A[94]=[15]{[acbd][73]=[21][acbd][94]=[15]{{7a+3c=29a+4c=1{7b+3d=19b+4d=5{a=5b=11c=11d=26(B)
解答{O(1,2)P(3,5)OP=(4,3)P(4,3)L:4(x3)+3(y5)=0(0,9)L(A)
解答(sinθ+cosθ)2=1+2sinθcosθ=1+sin2θsinθ+cosθ=1+sin2θ{(A)sin12+cos12=1+sin24(B)sin32+cos32=1+sin64(C)sin52+cos52=1+sin104=1+sin86(D)sin72+cos72=1+sin144=1+sin36(C)(C)
解答=2/51/313/8=1/155/8=875(B)
解答a+b+c=0a+b=c|a+b|=|c|=5|a+b|2=25|a+b|2=(a+b)(a+b)=|a|2+2ab+|b|2=4+2ab+9=25ab=ab=6(1)|a+2b+3c|2=|a+2b+3(ab)|2=|2ab|2=|2a+b|2=(2a+b)(2a+b)=4|a|2+|b|2+4ab=16+9+24=49((1))|a+2b+3c|=49=7(C)
解答(x2+y2)(x2y2)=0x2y2=0(x+y)(xy)=0x+y=0xy=0(B)

解答|a||b|sinθ=|a×b|27sinθ=|(3,6,2)|14sinθ=32+62+(2)2=7sinθ=714=12(A)
解答ADB=90¯BD2=¯AB2¯AD2=242152=351CBD=90¯BC2=¯CD2¯BD2=202351=49¯BC=7sinθ=¯BC¯CD=720(A)
解答x2169+y2144{a=13b=12{M=a2=169m=b2=144(A)
解答z=(34i)(25i)(5+2i)(125i)|z|=|34i||25i||5+2i||125i|=5292913=513(B)
解答(2,0,0),(0,6,0),(0,0,4)EE:x2y6+z4=1E=114+126+116=1712=127(D)
解答(A)limn3n2+52n+7=(B)limn2n+5n4n+3n=(C)limnn2+n+1n=limnn2+n+1n2n2+n+1+n=limnn+1n2+n+1+n=limn1+1/n1+1/n+1/n2+1=1/2(D)limncos(nπ)=±1(C)
解答{x+2y+3z=6(1)2xy+z=2(2)3x+y+4z=k(3)=|123211314|=0{2×(2)+(1)5x+5z=10(2)+(3)5x+5z=k+2k+2=10k=8(D)
解答x=π2

貳、多重選題題

解答(A)×:[1420][a0]=[a2a](B):[0110][a0]=[0a](C)×:[1236][a0]=[a3a](D):[0321][a0]=[02a](E)×:[2110][a0]=[2aa](BD)
解答(A)×:{XB(10,0.5)YB(10,0.8){E(X)=np=5E(Y)=np=8E(Y)>E(X)(B):{Var(X)=np(1p)=100.50.5=2.5Var(Y)=100.80.2=1.6Var(X)>Var(Y)(C):P(X5)=5k=0C10k0.5k0.510k=0.510(C100+C101+C102+C103+C104+C105)=63810240.5(D)×:p=0.8P(Y5)<12(E):{XB(10,0.5)YB(10,0.8){XYP(Y>5)P(X>5)(BCE)
解答(B)×:(C)×:(D)×:(AE)
解答(B)×:2x>xx>2x(D)×:2x>x2|x|>|x|x(ACE)
解答{P(x,y)F1(3,2)F2(3,2)Γ:|¯PF1¯PF2|=k¯F1F2=60<k<6(AB)
解答(A)×:n=11=1+1+=(B)×:調n=11n=(C):n=112n=1/211/2=1(D):n=11n(n+1)=n=1(1n1n+1)=112+1213+=1(E)×:n=11n(n+2)=n=112(1n1n+2)=12(113+1214+1315+1416+)=12(1+12)=34(CD)
解答tanθ=4P(x,5)y=4xP(54,5){sinθ=4/17cosθ=1/17{sin(θ)=sinθ=4/17sin(90+θ)=cosθ=1/17=17/17cos(180θ)=cosθ=1/17=17/17(CDE)
解答(A):|k22k|0(k+2)(k2)0k2,2(B)×:k=2{2x+2y=52x+2y=5(C):k=2{2x+2y=32x2y=9{2x2y=32x2y=9(D)×:{2x2y=32x2y=9=622+22=322(E):2x+2y=51(ACE)

解答{an=C4n124bn=C8n128(A)×:a2=C42124=38(B)×:b4=C84128=35128a2(C):C82=C86b2=b6(D):{a3=C43124=1/4=8/32b3=C83128=56/256=7/32a3>b3(E)×:max{C8n,n[0,8]}=C84max{bn,n[0,8]}=b4b5(CD)

解答(A)×:ω=ei2π/7ω110=ei220π/7=ei(30+10π/7)=ei(10π/7)=ω51(B):ω=ei2π/7ω7=1ω71=0(ω1)(ω6+ω5++ω+1)=0ω=1+ω++ω6=01+ω++ω21=ω+ω7ω+ω14ω+ω21=ω21=1(C):ω7=1ω6=1ω=ˉω(D):f(x)=1+x2++x6=0ω,ω2,,ω6f(x)=(xω)(xω2)(xω6)f(1)=1+1++1=7=(1ω)(1ω2)(1ω5)(1ω6)(1ω)(1ω2)(1ω3)(1ω4)(1ω5)(1ω6)=7×:(E)(D)|(1ω)(1ω6)|=|(1ω)(1ω2)(1ω3)(1¯ω3)(1¯ω2)(1¯ω)|=|(1ω)(1ω2)(1ω3)||(1¯ω3)(1¯ω2)(1¯ω)|=|(1ω)(1ω2)(1ω3)|2=7|(1ω)(1ω2)(1ω3)|=|(1ω)||(1ω2)||(1ω3)|=7(BCD)
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