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2024年8月18日 星期日

113年彰化女中第2次教甄-數學詳解

 113 學年度 國立彰化女中 第二次教師甄選初試


解答:
ACD=θ{ADC=120+θB=60+θADC:1sinθ=3sin(120+θ)tanθ=37{sinθ=3/213cosθ=7/213BCD:¯BCsin(60θ)=2sin60=43¯BC=43sin(60θ)=613cosA=9+93613299=1113

解答:

|z1(3+3i)|=2P(z1)C1:(x3)2+(y3)2=22z2=a+biiz21=(1b)+ai|iz21|=1Q(z2)C2:x2+(y+1)2=1|z1z2|=¯PQ==32+42(2+1)=2


解答:{a=logx3b=logy2{2logx81=24logx3=24a=k3logy16=34logy2=34b=k{4a=log2k=logk/log24b=log3k=logk/log3{a=logk/4log2b=logk/4log3log3x+log2y=1a+1b=4log2logk+4log3logk=log(2434)logk=1k=2434=1296

解答:limn(1n2+2n+1n2+4n++1n2+2n2)=limnnk=11n2+2kn=limnnk=11n1+2kn=1011+2xdx=[1+2x]|10=31
解答:a2a=a(a1)=×=1000k=(125×8)ka=125p=8q±1{p=1,2,p=3a=125×3=375a+1=8×47=376a=376376×375=1000×141p=4,p=5a=125×5=625a1=624=8×78a=625625×624=1000×390p=6,7,8,a=376,625

解答:f(t)=(t1)2+(2t1)2+(2t3)2+(t2)2+(2t)2+(2t+1)2=9t218t+11+9t2+5f(t)=09t99t218t+11+9t9t2+5=03t210t+5=0t=5±103f(5103)<f(5+103)t=5103
解答:x2+y22x6y+8=0(x1)2+(y3)2=2y=32(x1)2(0,0)(2,2),=π20(32(x1)2)2dxπ20x2dx=π(4633π)83π=383π3π2

解答:

{()O1,r=1()O2,RC=¯O1Q¯AB¯BC=12¯AB=123{CO1B=60O1BC=30¯CQ=¯O1P¯PQ¯O1C=332tanQBC=¯CQ¯BC=23{QBC=15CQB=75¯O2B=¯O2Q=RBO2C=18075×2=30R=¯BC×2=3{AQBO2=16(3)2π=π2ABO2=1233sin60=334AQB=π2334,{APBO1=13πABO1=12sin120=3/4APB=π334=(π334)(π2334)=π6+32滿=(π6π+32)/π=32π16
解答:limn(13n2+13n2++13n2)<<limn(13n2+2n+13n2+2n++13n2+2n)limn(2n3n2)<<limn(2n3n2+2n)23<<23=233
解答:{C(0,0)A(1,1)B(1,1)P(12cosθ,12sinθ){AB=(2,0)BC=(1,1)AP=(12cosθ+1,12sinθ+1)AP=αAB+βBC(12cosθ+1,12sinθ+1)=(2αβ,β){α=1+14(sinθ+cosθ)β=12sinθ+14αβ=3+12sinθ+cosθ=3+52sin(θ+δ)=3+52
解答:a34a2=22034a2da=433

解答:{A(0,8)L1B(9,2)L4C(6,0)L2D(5,4)L3,{L1L2L3L4L1L3{L1:y=mx+a1a1=8y=mx+8L2:y=mx+a20=6m+a2y=mx6mL3:x+my=b15+4m=b1x+my=4m5L4:x+my=b29+2m=b2x+my=2m+9d(L1,L2)=d(L3,L4)|6m+8|m2+1=|2m14|m2+1(6m+8)2=(2m14)2{m=3/4d(L1,L2)=10m=11/2d(L1,L2)=25=102=100
解答:{(x1)20x2+2x+3=(x+1)2+2>0,(xk)(x6)0{k<6k<x<610x=5,4,,45k<4k>66<x<k9(x=1,10)x=7,8,,1515<k16{a=15b=16c=5d=4a+b+c+d=22
解答:{O(1,2)I(2,2)A(2,8){ABCΓ1:(x+1)2+(y2)2=45L=IA:x=2D=ΓL=(2,4)D,¯DI=6Γ2:(x2)2+(y+4)2=36Γ1Γ2(x+1)2+(y2)245=(x2)2+(y+4)236x2y=4:Γ1Γ2B,C
解答:


y=x2+bx+c=(xα)2β2(β>0){A(αβ,0)B(α+β,0)C(0,α2β2<0)α<βM(α,β2)SABM=122ββ2=β3():SACM=18SABMSABM+SACM=9SABM=β3=8β=2(2,5)5=(2α)24α=1y=(x1)24y=x22x3=x2+bx+c(b,c)=(2,3)
解答:{P(x,y)Q(0,a),¯PQ=23¯PBx2+(ya)2=49(x2+(y6)2)5x2+5y218ay+48y+9a2144=05166y(3a8)+9a2144=09a2646y(3a8)=0(3a8)(3a+86y)=0a=83Q(0,83)3¯PA+2¯PB=3¯PA+232¯PQ=3(¯PA+¯PQ)Q,P,A=3¯AQ=382+(8/3)2=36409=810
解答:{k=6n=7(k1)[(k1)n1+(1)n]k=5(561)6=13020

解答:{EAF=θ¯AB=¯AE=acosθ=a232asinθ=a4+10a292aSABC=12¯AB¯ACsinBAC=12asin(240θ)=12asin(60θ)=14(3cosθsinθ)=f(a)=18(3a233a4+10a29)f(a)=023a=2a3+10aa4+10a2916a6160a4+208a2=016a2(a410a2+10)=0a2=523f(a2=523)=18(3(523)33(523)2+10(523)9=18(238)=134
解答:{cos2A+cos2B+2sinAsinBcosC=158cos2B+cos2C+2sinBsinCcosA=149cos2A+2cos2B+cos2C+2sinB(sinAcosC+sinCcosA)=158+149cos2A+2cos2B+cos2C+2sinBsin(A+C)=cos2A+2cos2B+cos2C+2sin2B=24772cos2A+cos2C=247722cos2A+cos2C=10372(1)cos2Acos2C+2sinB(sinAcosCsinCcosA)=158149cos2Acos2C+2sin(A+C)sin(AC)=cos2Acos2C+cos(2C)cos(2A)=2372cos2Acos2C+(2cos2C1)(2cos2A1)=2372cos2A+cos2C=2372(2)(1)(2){cos2A=5/9cos2C=7/8cos2C+cos2A+2sinCsinAcosB=10372+2sinCsinA(cos(A+C))=10372+2sinCsinA(sinAsinCcosAcosC)=10372+2sin2Asin2C2sinCcosCsinAcosA=10372+24918±218782353=11172±43572
解答:



28()=128C95=129=79
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解題僅供參考,教甄歷年試題及詳解






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