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2024年9月11日 星期三

113年台師大附中教甄2-數學詳解

 國立臺灣師範大學附屬高級中學113學年度第2次專任教師甄選

一、 選填題:(每題5分,共90分。填在答案卡上,分數或根式須以最簡形式回答,否則不予計分)

解答:1aanbn=anan1,n=2,3,bn=2,3,1,2,3,1,bn=bn+325n=1an=a1+(a1+b1)+(a1+b1+b2)+(a1+b1+b2+b3)+(a1+b1++b24)=25a1+24b1+23b2+22b3+21b1+20b2+19b3++b3=25a1+b1(24+21+18++3)+b2(23+20++2)+b3(22+19++1)=25a+22782+325822382=25a+4242024a64
解答:(1)f(x)=a(x1)2+bf(2)=a+b=5(2)y=g(x)=2(x1)+b=2x+b+2(3)h(x)=(a(x1)2+b)(2x+b+2)x3=2a|2a|=6a=3(f(x)a>0)b=2{f(x)=3(x1)2+3g(x)=2x+4h(x)=(3(x1)2+3)(2x+4)y=p(x)=h(x1)=(3(x2)2+3)(2x+6)=6x3+42x2102x+90h(x1)=18x2+84x102h(h1)=36x+84=0x=8436=73

解答:

PQRS{SMR=MRQ=α¯SR=¯PQ=6PMN=MNR=θ¯MN{¯MR=¯MP=43RMN=PMN=θMSR:6sinα=43sinS=sinQ=5/4sinα=158cosα=78cosα2=1542θ+α=180sinθ=sin(90α2)=cosα2=154RMN:¯MNsinα=43sinθ¯MN=43154158=23
解答:


yx5014(2)253(4)492,1(5)5220(5)5111(6)6132(6)6133(4)494(2)2550189


解答:Q(x)=2(xα)(xβ)f(x)=Q(x2+8x17)=2(x2+8x17α)(x2+8x17β)f(2)=02(3α)(3β)=0α=3β=3Case I α=β=3f(x)=2(x2+8x20)2=2(x2)2(x+10)2Case II α=3,β3f(x)=2(x2)(x+10)(x2+8x17β)x2+8x17β=064+4(17+β)=0β=33g(x)=Q(2x25x+9)=2(2x25x+9α)(2x25x+9β){Case I α=β=3g(1/2)=2(73)2=32Case II α=3,β=33g(1/2)=2(73)(7+33)=32032+320=352

解答:1,2,3,4,5,6,7,8S=14

解答:{A(1,2,3)B(3,4,5)P(x,y,0)¯PA2+¯PB2=(x1)2+(y2)2+9+(x3)2+(y4)2+25=2(x2+y24x6y+32)=2((x2)2+(y3)2+19){x=2y=3:2×19=38

解答:Cn2=36n=9,
解答:n=12:Cn0+Cn1+Cn2+Cn3=C120+C121+C122+C123=1+12+66+220=299,
解答:(1+x)16=C160+C161x+C162x2++C1616x16f(x)=x(1+x)16=C160x+C161x2+C162x3++C1616x17f(x)=(1+x)15(1+17x)=C160+2C161x+3C162x2++17C1616x16f(x)=(1+x)14(32+272x)=2C161+23C162x+34C163x2++1617C1616x15f(1)=214×304=2C161+23C162+34C163++1617C16161216(2C161+23C162+34C163++1617C1616)=214×304216=76
解答:I=limn2nk=14n2(2n+5k)3=limn2nk=14/n(2+5k/n)3=204(2+5x)3dxu=2+5xdu=5dxI=12245u3du=772
解答:x7=2772,ω,ω2,,ω6,ω=2e2π/7f(x)=x727=(x2)(xω)(xω2)(xω6)|f(x)|=|x727|=|x2||xω||xω2||xω6||f(1+i)|=|(1+i)727|=¯PA¯PB¯PC¯PG(1+i)2=2i(1+i)6=8i(1+i)7=88i|f(1+i)|=|88i27|=|1208i|=1202+82=8226


解答:{L1u=(3,6,53)L2v=(2,1,0)n=u×v=(53,103,15)nL2(0,0,0)E:3x+23y+3z=0L1(1,2,3)E=22=d(L1,L2)a=222=4=1122a3=1632

解答:Xi=iX=X1+X2++X6XiGeo(pi=7i6)E(Xi)=67iE(X)=E(X1+X2++X6)=E(X1)+E(X2)++E(X6)=1+65+64+63+62+61=14.7,

解答:Γ1=Γ229x383x+6k+6=0x312x+27k+27=0f(x)=x312x+27k+27f(x)=03x212=0x=±2f(x)=0f(2)f(2)<0(27k+11)(27k+43)<04327<k<1127k=1(k)f(x)=x312x=0x=0,±23Γ1Γ2=g(x)=29x383xg(x)2323|g(x)|dx=2230|g(x)|dx=2[236x443x2]|023=16
解答:{u=19x+13yv=25x+17y|(u,v)(x,y)|=19132517=2|u|3+|v|4=1=3×4×2=24|19x+13y|3+|25x+17y|4=1=242=12
解答:

Γ1:¯BC=10c=5b2=d(B,L)d(C,L)=255155=75b=53a=75+25=10Γ1:(x+1)2100+y275=1Γ2:c=5(x+1)2ky225k=1D(6,94)25k81/1625k=116k2881k+10000=0(16k625)(k16)=0{k=625/16a=25/4>c(5)k=16a=4Γ2:(x+1)216y29=1Γ1Γ2(x+1)2=100(1y275)=16(1+y29)y2=27|y|=33sABC=12¯BC|y|=121033=153=b2,
解答:x2+xy+y2=[x,y][11/21/21][xy]=[x,y][22222222][120032][22222222]T[xy][x,y]=[x,y][22222222]=[22(x+y),22(x+y)]{x=yx2y=x+y2{x2+xy+y2=x22+y22/3=1(1)x23xy2y2=x23xy2y2(2),(1){x=2cosθy=23sinθ(2)2cos2θ23sinθcosθ43sin2θ=53cos2θ3sin2θ+13=259+3+13=1+2133


解答:


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解題僅供參考,教甄歷年試題及詳解






2 則留言:

  1. 選填D y=0時, x=-5~5吧?從圖也可以看出來

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