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2020年6月15日 星期一

98年警專28期甲組數學科詳解


臺灣警察專科學校專科警員班二十八期(正期學生組)
新生入學考試甲組數學科試題
壹、單選題


(log23)(log37)(log78)=log3log2×log7log3×log8log7=log8log2=log28=3(C)


cos1590=cos(360×4+150)=cos150=cos30=32(D)


s=(¯AB+¯BC+¯AC)÷2=(5+6+7)÷2=9ABC=s(s¯AB)(s¯BC)(s¯AC)=9(95)(96)(97)=9×4×3×2=66(B)



{(A)log27>0(B)log37>0(C)log0.27=log7log0.2=log7log2log10=log7log21<0(D)log0.37=log7log31<0log27>log37(A)(A)



(0,0)(5)(B)(B)


S:x2+y2+z2=49{O(0,0,0)R=7a=dist(O,E)=|1222+12+(2)2|=4rR2=r2+a249=r2+16r2=33=r2π=33π(D)





A{A(0,0,0)B(2,0,0)D(0,3,0)E(0,0,3){AB=(2,0,0)AD=(0,3,0)EB=(2,0,3)ED=(0,3,3){u=AB×AD=(0,0,6)v=EB×ED=(9,6,6)cosθ=uv|u||v|=366×317=217sinθ=1317(D)



|4×23×3+642+(3)2|=55=1(A)



AP=xAB+yAC={2AD+yACxAB+32AE{2x+y=1x+32y=1{x=1/4y=1/2(C)




1=180π{sin1=sin180πsin57sin2=sin2×180πsin114=sin66sin3=sin3×180πsin171=sin9sin4=sin4×180πsin228<0sin2(B)



:{a1+2b1+3c1=d1a2+2b2+3c2=d2a3+2b3+3c3=d3{a14+2b14+3c14=4d1a24+2b24+3c24=4d2a34+2b34+3c34=4d3(4,4,4){a1x+2b1y+3c1z=4d1a2x+2b2y+3c2z=4d2a3x+2b3y+3c3z=4d3(D)


|123215a54|=0430+10a3a+1625=0a=5(D)





{y=3sinxy=x;Aky=3sinxAk=(π2+2kπ,3),k=0,1,2,¯OAkmk=3π2+2kπ{m0=6π>1m1=65π<1y=xm=1m1<m<m0133(A)




tanα,tanβx2+5x+2=0{tanα+tanβ=5tanαtanβ=2tan(α+β)=tanα+tanβ1tanαtanβ=51=5{sin(α+β)=526cos(α+β)=126sin2(α+β)+4sin(α+β)cos(α+β)+7cos2(α+β)=2526+2026+726=5226=2(B)



f(x)=3sinxcosx+1=2(32sinx12cosx)+1=2(cosysinxsinycosx)+1=2sin(xy)+11f(x)3(C)


a=2+2i1+3ia2=3ia4=223ia8=883ia12=(223i)(883i)=64(D)


{1=1010+6+4=125=610+6+4=31010=410+6+4=210=1×12+5×310+10×210=4(D)



{3=C33/C1233=C43/C1233=C53/C1233=(C33+C43+C53)/C123=15220=344(B)


log12100=100log12=100(log3+2log2)=100(0.4771+0.602)=107.91107+1=108(B)



西:(x2+y2+z2)(62+22+32)(6x+2y+3z)29×49(6x+2y+3z)2216x+2y+3z2121(C)


x2+2x+7=(x+1)2+66(x2+2x+7)(x+1)(x+2)<0(x+1)(x+2)<02<x<1(A)



A=[3512]det(A)=65=1A1=[2513]=[abcd]{c=1d=3c+d=1+3=2(A)


=0.2×0.40.4×0.6+0.4×0.2+0.2×0.4=0.080.4=0.2=15(B)



(1,0,1),(1,1,0),(0,1,1)(0,0,0)=381=37(C)




P=6=6×PAB=6×(r×32r×12)=332r2APO¯OP=¯OA2¯AP2=36r216×332r2×36r2=34r236r2=3436r4r6f(r)=36r4r6f(r)=144r36r5=6r3(24r2)r2=24f(r)r=24=26(D)


A=[100030002]A3=[130003300023]=[1000270008]b=27(D)



{u=AB=(1,0,3)v=7x+4y4z=0=(7,4,4)n=u×v=(1,0,3)×(7,4,4)=(12,17,4)E:nA(2,1,1)E:12(x2)+17(y1)4(z+1)=012x17y+4z3=0(B)



y=f(x)=x36x2+9x2f(x)=3x212x+9f(x)=6x12f(x)=06x12=0x=2(2,f(2))=(2,0)(C)



R=21f(x)dx=21x2dx=[13x3]|21=13(81)=73(C)


10f2(x)πdx=π10xdx=π2(A)

貳、多重選擇題


ω=cos2π5+isin2π5ω5=cos2π+isin2π=1(A)×:ω5=1ω51=0(ω1)(ω4+ω3+ω2+ω+1)=0ω4+ω3+ω2+ω+1=0ω4+ω3+ω2+ω=11(B):ω5=1ω10=1(C)×:(1ω)(1ω4)(1ω2)(1ω3)=(1ωω4+ω5)(1ω2ω3+ω5)=(2ωω4)(2ω2ω3)=42ω22ω32ω+ω3+ω42ω4+ω6+ω7=42ω2ω2ω3ω4+ω6+ω7=42ω2ω2ω3ω4+ω+ω2=4ωω2ω3ω4=4(1)=51(D):(1+ω)(1+ω4)(1+ω2)(1+ω3)=(1+ω+ω4+1)(1+ω2+ω3+1)=(2+ω+ω4)(2+ω2+ω3)=4+2ω+2ω2+3ω3+3ω4+ω6+ω7=4+3(ω+ω2+ω3+ω4)=4+3×(1)=1(E)×:11+ω+11+ω2+11+ω3+11+ω4=(1+ω2)(1+ω3)(1+ω4)+(1+ω)(1+ω3)(1+ω4)+(1+ω)(1+ω2)(1+ω4)+(1+ω)(1+ω2)(1+ω3)(1+ω)(1+ω2)(1+ω3)(1+ω4)=(1+ω2)(1+ω3)(1+ω4)+(1+ω)(1+ω3)(1+ω4)+(1+ω)(1+ω2)(1+ω4)+(1+ω)(1+ω2)(1+ω3)=(2+ω2+ω3)(1+ω4)+(2+ω+ω4)(1+ω3)+(2+ω+ω4)(1+ω2)+(2+ω2+ω3)(1+ω)=(2+ω2+ω3)(2+ω+ω4)+(2+ω+ω4)(2+ω2+ω3)=2(2+ω2+ω3)(2+ω+ω4)=2(4+3(ω+ω2+ω3+ω4))=2(43)=21(BD)





y=log2x(A):y=2y=log2x(4,2)(B):x+y=2(C)×:log2x=x+24×2x=x(D)×:2x>log2x(E):log0.5x=log2xX(1,0)(ABE)


(A)×:z=3(B)×:2x+3y=6(C):x32=y43=4z1(2t+3,3t+4,t+4)(D):(E):(26t,5,9+8t)(CDE)



(B)×:f(1)=limh0f(1+h)f(1)h=limh0|h|h=±1f(1)(ACDE)公布的答案是(ACE)



(A):|123102030306090|10r1+r2|123000306090|=0(B):|123456789|4r1+r2,7r1+r3|1230360612|2r2+r3|123036000|=0(C):|123234345|2r1+r2,3r1+r3|123012024|2r2+r3|123012000|=0(D)×:|111234223242|c1+c2,c1+c3|10021222512|2c2+c3|1002102252|=20(E):|693142034562333|=6|23171017562333|2c3+c1,3c3+c26|00141411712212233|c1+c26|00141017122033|=0(ABCE)




f(x)=x33x2+7f(x)=3x26xf(x)03x26x03x(x2)0x2x0(ABE)



(A)×:{32π<θ<2πcosθ=35sinθ=45(B)×:32π<θ<2π34π<θ/2<πcosθ2<0cosθ=cos(2θ2)=2cos2θ21=35cos2θ2=45cosθ2=25(C):sinθ=2sinθ2cosθ2=45sinθ2=45×12×(52)=15(D):cos2θ=2cos2θ1=2×9251=725(E):sin2θ=2sinθcosθ=2×(45)×35=2425(CDE)




(A):=4×4×2×12=16(B):f(x,y)=x+y{f(A)=2f(B)=4f(C)=2f(D)=44(C):g(x,y)=xy{g(A)=2g(B)=4g(C)=2g(D)=4=4(D):h(x,y)=x+3y{h(A)=6h(B)=4h(C)=6h(D)=4=6(E):p(x,y)=x3y{p(A)=6p(B)=4p(C)=6p(D)=4=6(ABCDE)


f(x)=2x3+6x2+13f(x)=6x2+12xf(x)=12x+12;f(x)=06x2+12x=06x(x+2)=0x=0,2{f(0)=120f(2)=120x=0,2(CE)


(A):{A2={2,4,6,,60}P(A2)=30/60=1/2A3={3,6,9,,60}P(A3)=20/60=1/3A2A3={6,12,18,,60}P(A2A3)=10/60=1/6P(A2A3)=P(A2)P(A3)(B)×:{A7={7,14,,56}P(A2)=8/60=2/15A3A7={21,42}P(A3A7)=2/60=1/30P(A3)P(A7)=2/451/30=P(A3A7)(C):A2A7={14,28,42,56}P(A2A7)=4/60=1/15=P(A2)P(A7)(D):{A5={5,10,,60}P(A5)=12/60=1/5A2A3A5={30,60}P(A2A3A5)=2/60=1/30P(A2)P(A3)P(A5)=12×13×15=130=P(A2A3A5)(E)×:A2A3A7={42}P(A2A3A7)=1/60P(A2)P(A3)P(A7)=12×13×215=1451/60(ACD)


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