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2023年5月11日 星期四

112年鳳山高中教甄-數學詳解

國立鳳山高中 112 年教師甄試數學科試題

一、填充題(14 題,每題 5 分,占 70 分)

解答2022420233=(20231)420233=120233(20234420233+62023242023+1)=20234+62023420232+120233=20234=2019
解答{O(0,0,0)A(3,0,0)B(3,5,0)C(0,5,0)D(3,0,2)E(3,5,2)F(0,5,2)G(0,0,2){ab=ad=ae=9ac=af=ag=0bc=bf=25bd=9be=34bg=0cd=cg=0ce=cf=25de=13df=dg=4ef=29eg=fg=4,228228C72=22821=767
解答¯ADBAC¯AB¯AC=¯BC¯DC=31{¯AB=3k¯AC=k{¯AD=aBAD=DAC=θ{cosABD=(3k)2+a23223kacosDAC=k2+a2122ak9k2+a296ak=k2+a212ak3k2+a2+3a=3k23cosθ=k2+(3k23)12k3k23=2k213ksinθ=4k23kABC12(3ak+ak)sinθ=2aksinθ=23k23k4k23k=233k4+15k212=233(k252)2+274k2=5223274=3
解答Lagrange {f(x,y)=x+yg(x,y)=x+y4x+26y+1{fx=λgxfy=λgyg=0{1=λ(12x+2)1=λ(1+3y+1)2x+2=3y+1y=14(9x+14)y=14(9x+14)g=013x+144=4x+2+69x+184=13x+2(13x+14)2=522(x+2)169x22340x5212=0x=2340+89989122×169=21390+13641321313=90+323213
解答t=x2,(x2)6<x3x2+5x4t6<t3+5t2+13t+10t6t35t213t10<0(t+1)(t2)(t4+t3+3t2+4t+5)<01<t<21<x2<21<x<4{t2+t+1>02+4t+5>0,xRt4+t3+3t2+4t+5=t2(t2+t+1)+2t2+4t+5>0
解答z=eiθz3=ei3θ{3θθ=±60{θ=30θ=30=3303θθ=±60+360{θ=150θ=210(n,ϕ)=(4,330)
解答{A=(3+2)1000=(5+26)500=500k=0(500k)5500k2k6k/2B=(32)1000=(526)500=500k=0(500k)5500k(1)k2k6k/2A+B=2250k=0(5002k)55002k24k10:2×245004n{4mod10,n6mod10,nnN2×245002mod10A+B2,0<B<1A=(3+2)10001
解答349(1039)21247C492{2C102/C4921C101C391/C4922C102C492+C101C391C492=2049
解答
{A(1,0)B(2,0)C(x,y),B¯ACDDAB=DBAD¯ABD(12,a);¯AD¯AD¯DC=¯AB¯BC=3(x2)2+y2D=A¯BC+3C¯BC+312=(x2)2+y2+3x(x2)2+y2+36x3=3(x2)2+y2(2x1)2=(x2)2+y23x2y2=3x2=1+y23x>1(x=1C(1,0)A,B,C)Cx2y23=1,x>1
解答(logax)(logax2)=4(loga+logx)(loga+2logx)=42(logx)2+3logalogx+(loga)24=0logx=3loga±(loga)2+3241logx>03loga(loga)2+32>09(loga)2>(loga)2+32(loga)2>4{loga>2a>100()loga<20<a<1/100a>100,3loga<03loga(loga)2+320
解答x=3sinθdx=3cosθdθ03/213x2dx=0π/33cosθ3cos2θdθ=0π/31dθ=π3
解答
解答

29933
解答|a|=|b|=|a+b|=1ab120ab=12{ac=0bc<0ac90cb(36012090)=150cb=32{0va10vb1{0(xa+yc)a10(xa+yc)b1{0x1012x32y11232y323y13vc=(xa+yc)c=y=3
解答QLkQ=(2021t+k,2022t+k,2023t+k)E:2y=x+zdk=d(P,E)=|264|6=86=436

二、計算證明問答題(共 4 題,占 30 分)


解答

解答
解答


QP
解答¯Pn+2Pn:¯Pn+2Pn+1=3:1xn+2=14(3xn+1+xn)4xn3xn1xn2=04λ23λ1=0(4λ+1)(λ1)=0λ=14,1xn=C1(14)n+C2,{x1=1x2=2{1=14C1+C22=116C1+C2{C1=16/5C2=9/5xn=165(14)n+95limnxn=95
解答xyz=1{x=1yzy=1zxz=1xy13(xy+xy+yz)3x2yz=x(1)13(yz+yz+zx)3y2xz=y(2)13(zx+zx+xy)3z2xy=z(3)(1)+(2)+(3)xy+yz+zxx+y+z,

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解題僅供參考,其他教甄試題及詳解

3 則留言:

  1. 老師您好
    請問計算3是不是還沒補齊呢 呵呵
    再麻煩您了 謝謝

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