2021年3月15日 星期一

106年台中文華高中教甄-數學詳解

臺中市立文華高級中等學校106學年度第2次教師甄選

一、填充題

解答f(x)=(x2+2x+3)(x2+2x2)+5x2+10x+4=(x2+2x)2+(x2+2x)6+5(x2+2x)+4=(x2+2x)2+6(x2+2x)2f(x)=2(x2+2x)(2x+2)+6(2x+2)f(x)=0x=1f(1)=162=7(α,m)=(1,7)

解答log2x+logx86=0logxlog2+3log2logx6=0(logx)26log2(logx)+3(log2)2=0{logα+logβ=6log2logαlogβ=3(log2)2(logα)2+(logβ)2=(logα+logβ)22logαlogβ=30(log2)2logαβ+logβα=logβlogα+logαlogβ=(logα)2+(logβ)2logαlogβ=30(log2)23(log2)2=10

解答logB=I+fIflogA=(I1)+(f+log5)A=10I+f+log51=B10log51=B×510=B2AB+152=B2+152=0B=15A=152A+B=152+15=452

解答(m2+1)x24mx+2=0x=2m±2m22m2+1{α=2m+2m22m2+1β=2m2m22m2+1{α3β=4m+42m22m2+12αβ=4m2+1:2αβ=α3βm+2m22=12m22=(m+1)2=m2+2m+1m22m3=0m=3(m=12x2+4x+2=0x=1)

解答(1+2x)n=nk=0Cnk2kxkx3Cn323=4n(n1)(n2)31an=341n(n1)(n2)=38(1(n1)(n2)1n(n1))100k=31an=38(121132+132143++19998110099)=38(1219900)=3849499900=494926400

解答{:34+3=84:4×32×4=144:4×3×4=48:42×2=3284+144+48+32=308

解答x+y+z+w=20,1x,y,z,w6x+y+z+w=16,0x,y,z,w5(x,y,z,w)(5,5,5,1)4!/3!=4(5,5,4,2)4!/2=12(5,5,3,3)4!/(2!2!)=6(5,4,4,3)4!/2!=12(4,4,4,4)14+12+6+12+1=35

解答5a1a5滿{a3=6(a1++a5)÷5=6a1a2a3a4a512610111369111468119102368119101567118102467118102567108934671089356794567816666=16C52×C52=425
解答f(x)=|x+22220172x+2018201920172201823x+20192|f(x)x36abc=f(0)6f(0)=|222201720182019201722018220192|=2|111201720182019201722018220192|=2|111012040358072|=2|111010040352|=2(20)=4abc=46=23:3x+20193x+20192

解答y=3sin2x+43sinxcosxcos2x=34cos2x+23sin(2x)=32(cos(2x)+1)+23sin(2x)=23sin(2x)2cos(2x)+1=4(32sin(2x)12cos(2x))+1=4sin(2xπ6)+12xπ6=3π2x=56πy4+1=3(a,m)=(56,3)

解答
sinθ=45cosθ=35cos2θ=cos2θsin2θ=725POX=α{XOP1=α,P1P¯OXP2OY=θα,P2P¯OYP1OP2=(θα)+θ+α=2θcosP1OP2=102+102¯P1P222×10×10¯P1P2=1022cos2θ=102+1425=106425=16PQR=¯PR+¯PQ+¯QR=¯P2R+¯P1Q+¯QR=¯P1P2=16



解答

C:(x7)2+(y+4)2=5{O(7,4)r=5A(5,2)yAA=(5,2)¯AOyPCQ¯PA+¯PQ=¯PA+¯PQ=¯AOr=122+625=55

解答
{PA+2PB+3PC=CA=CP+PAQA+2QB+3QC=2AB=2AQ+2QBRA+2RB+3RC=3BC=3BR+3RC{PB+2PC=0QA+QC=0RA+5RB=0{|PB|=2|PC||QA|=|QC||RA|=5|RB|{AQR:ABC=15:26=5:12BPR:ABC=12:63=1:9CPQ:ABC=11:23=1:6ABCPQR=115121916=112536=3611

解答
{0απ/3π/2β3π/4{0sinα3/21/2cosβ0{x=5sinα+cosβy=4cosβ+sinα{sinα=(4xy)/19cosβ=(5yx)/19{0(4xy)/193/21/2(5yx)/190{L1:4x=yL2:4xy=193/2L3:x=5yL4:5yx=19/2{d=dist(L1,L2)=193217L1,L2L3,L4P(0,0),Q(1/2),R,S¯PQ=172PQRS=dׯPQ=193217×172=1964



解答
13{3/463/21233/4234×620+32×1220+334×220=9203

解答f(x)=x42x3+3x(x2(3t37t2+5t1)dt)6{f(2)=1616+606=0f(x)=4x36x2+3(3x37x2+5x1)f(2)=11(5)=6limh0f(2+3h)4h=limh0f(2+3h)f(2)4h=limh034f(2+3h)f(2)3h=34f(2)=346=92

解答f(x)=3x+5x4f(x)=123x+525x4=53x5x423x5x4f(x)=053x=5x47525x=5x4x=7930f(7930)=37930+7964=1130+556=1130+ 51130=61130=3305

解答
ABC{¯BC=x¯AC=y¯AB=z{¯AF=5¯BE=6¯CD=10x,y,z滿{x=y225+z225y=x236+z236z=x2100+y2100ABC=125x=126y=1210z=k{x=2k/5y=k/3z=k/5s=(x+y+z)/2=7k/15ABC=s(sx)(sy)(xz)=7k15k152k154k15=214152k2=kk=152214x+2y+4z=2k5+2k3+4k5=2815k=2815152214=1514



解答A=[1551]=P[6004]P1P=[1111],P1=[1/21/21/21/2]An=[1111][6n00(4)n][1/21/21/21/2]=[6n(4)n6n(4)n][1/21/21/21/2]=[(6n+(4)n)/2(6n+(4)n)/2(6n+(4)n)/2(6n+(4)n)/2]=[abcd]b=6n+(4)n2

解答¯AB=92+122=15¯BP=a¯PA=5415a=39aABP=s(sa)(s39+a)(s15),s=54/2=27ABP=27(27a)(a12)(12)f(a)=(27a)(a12)=a2+39a324f(39/2)ABP=27(27392)(39212)(12)=2715215212=18×152=135

解答a2|x|8x2+5|x|1yα+β=0α2β=α+2α=3α=34α=14a1/212+54=1a=14a=116




3 則留言:

  1. 您好:請問第8題的第一組答案1,2,6,7,8是不是不符合?謝謝

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    回覆
    1. 把1,2,6,7,8換成4,5,6,7,8就對了,謝謝提醒!! 已修訂

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