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2022年7月28日 星期四

111年台南女中第2次教甄-數學詳解

國立臺南女中 111 學年度第 2 次教師甄選

填充題: (共 20 題,每題 5 分,共 100 分,答案請化為最簡分數或最簡根式)

解答{RBOC=θAOB=2θ{OAB=12R2sin2θ=100OBC=12R2sinθ=80sin2θsinθ=100802sinθcosθsinθ=54cosθ=58sinθ=398sin3θ=3sinθ4sin3θOACOBC=sin3θsinθ=34sin2θOAC=(34sin2θ)80=45
解答f(x)=(x+1)(x2+1)(x4+1)(x8+1)(x16+1)(x1)f(x)=(x21)(x2+1)(x4+1)(x8+1)(x16+1)=x321f(x)=x321x1f(4)=43213logf(4)=log(4321)log3log432log3=64log2log3=64×0.3010.4771=18.7869f(4)19
解答xP(k=x)2,121/363,112/364,103/365,94/366,85/3676/36,k50,k/2;=2000+50362(2+12+2(3+11)+3(4+10)+4(5+9)+5(6+8)+67)=2000+2536252=21756×2175=13050
解答Γ{O(0,0)A(8,0)B(0,6){C=(A+B)÷2=(4,3)R=5AB=6/8=34L:3x+4y+a=0d(C,L)=R|12+12+a|5=5|24+a|=25{a=1()a=493x+4y=49x=493
解答P=L¯ABBCPOAB=12¯BCׯBP¯BOׯAB=3ׯBP4ׯAB=12¯BP¯AB=23¯BP¯PA=21P=(B+2A)÷3=(4/3,4/3)L=CP:y=14x+1x4y+4=0(b,c)=(4,4)
解答{¯AF¯FG¯FG2=¯AG2¯AF2{¯AP¯DEGCDEFAPG=90¯AG2=¯AP2+¯PG2¯FG2=¯AP2+¯PG2¯AF2=(¯ADsin60)2+(202)2252=225+800625=20
解答3k=(d(O,L))2=(510)2=14
解答f(x)=ax3+bx2+cx+d(x+1)f(x)=ax4+(a+b)x3+(b+c)x2+(c+d)x+df(x)x25f(2)=58a+4b+2c+d=5(1)(x+1)f(x)=(x33)(ax+(a+b))+3x1{b+c=0(2)3a+3b+d=1(3)3a+c+d=3(4)(1)(4){a=1b=1c=1d=5a+b+c+d=4
解答f(x)=x0(at3+bt29)dt=14ax4+13bx39xf(x)=ax3+bx29x=3f(3)=027a+9b=93a+b=1(1)22f(x)dx=[120ax5+112bx492x2]|22=165a=32a=10(1)30+b=1b=292ab=20+29=49
解答f(x)=ax3+bx2+cx+dx=0y=x+5(=1){f(0)=1c=1f(0)=5d=5(1,8)=(b3a,f(b3a)){3a+b=0a+b+c+d=8a+b=2a=1
解答
{¯CD=x¯AB=¯AD=R¯AC=aABC=12aRsinθ=43aRsinθ=83cosCAD=cos(θ+45)=a2+R2x22aRx2=a2+R22aRcos(θ+45)=a2+R22aR(cos45cosθsin45sinθ)=a2+R22aR(cosθsinθ)=a2+R22aRcosθ+2aRsinθ=a2+R22aRcosθ+862a2R22aRcosθ+86=2aR2aRcosθ+86=283sinθ83sinθcosθ+86=86(2cosθsinθ+1)x286(2cosθsinθ+1)=86(k+1),k=2cosθsinθcosθ+ksinθ=2k2+1sin(θ+α)=2,{sinα=1/k2+1cosα=k/k2+1sin(θ+α)=2k2+112k2+1k1x286(1+1)=166x166=446
解答


x=x+{x}{x}=xxxxkx+3k=0xx=k(x3){Γ:y=xxL:k(x3)k;{Γ1(m,0)(m+1,1),mZ(m+1,1)LA(3,0)k{kABACkADAE{B(1,1)C(2,1)D(6,1)E(7,1){1<k<1214<k<13
解答CD=AC+AD=mAB+7ACAD=mAB+8AC¯BE¯EC=8m=aABEaACE8m=aABE1aABE=8mAD=tAEtAE=mAB+8ACAE=mtAB+8tACmt+8t=1t=m+8aACDaACE=t=m+8aACD=m+8aCDE=m+7aABE+aCDE=8m+m+728mm+7=42+7
解答

{A+D+G+F=15(1)B+D+E+G=16(2)C+E+F+G=17(3)G=5D+E+F=15(1)+(2)+(3)A+B+C+2(D+E+F)+3G=48A+B+C=4821535=3A+B+C+D+E+F+G=3+15+5=23Double doubles=4023=17
解答K(K=A,B,C,D,F,G,H)nEKn;Fn=Dn,Gn=Cn,Hn=Bn;{An=Bn1+Hn1=2Bn1(1)Bn=An1+Cn1(2)Cn=Bn1+Dn1(3)Dn=Cn1(4),{(2)(1)An=2An2+2Cn2(2)(4)(3)Cn=An2+Cn2+Cn2{An=2An2+2Cn2(5)Cn=An2+2Cn2(6){(5)(6)Cn=An2+An2An2=AnAn2(6)(5)An=2(Cn2Cn2)+2Cn2=2Cn2Cn2{An=2Cn2Cn2(7)Cn=AnAn2(8)(8)(7)An=4An22An4:{A1=A2=A3=0A4=2A=0A;A6=8A8=28a10=96a12=328a14=1120a16=3824
解答

(62)180÷6=1201,3,26163,32=1×615+3×615+2×315=4+235
解答
(X,Y)(13,12)XYX2Y2XY81164121889128114410810101001001001181216488129144811085050510510492b1=XYXYnX2(x)2n=4925050/5510502/5=45(ˉx,ˉy)=(50/5,50/5)=(10,10)y=45(x10)+10=45x+18(m,k)=(45,18)
解答
x264+y216{a=8b=4c=6416=43{F1(43,0)F2(43,0)¯F1F2=83{¯AF2=m¯AF1=n{m+n=2a=16m2+n2=¯F1F22=192{m=842n=8+42;{¯BF2=s¯BF1=t{s+t=2a=16(s+m)2+n2=t2(s+842)2+(8+42)2=t2{s=(24+42)/17t=(24842)/17=(¯AB+¯AF1¯BF1)÷2=(s+m+nt)÷2=(16+242172484217)÷2=24+4217

解答
{¯AF=¯FB=a¯AE=¯EC=ba2+b2=(¯BC2)2=92=81{¯BEm1=b/2a¯CFm2=2b/atan150=m2m11+m1m213=3b/2a1+b2/a2=3ba22(a2+b2)a=3ab162=ab54ab=183ABC=2ab=363GBC=13ABC=123

解答{¯BC=a¯AC=b¯AB=ctanA=15{sin15/4cosA=1/4:ABAC=bc/4{¯AD¯BD=12¯AD=13¯AB=c3¯AE¯CE=14¯AE=15¯AC=b5{¯AD=c/3¯AE=b/5APAB={¯ADׯAB=c3×c=c23(xAB+yAC)AB=xc2+ybc/44c2=12xc2+3ybc(1)APAC={¯AEׯAC=b5×b=b25(xAB+yAC)AC=xbc/4+yb24b2=5xbc+20yb2(2)(1)(2){12cx+3by=4c5cx+20by=4b{x=(80c12b)/225cy=(48b20c)/225b3x+5y=480bc36b2100c2225bc=480225(36b225c+100c225b)480225236225100225=480225260225=360225=85
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解題僅供參考,其他教甄試題及詳解


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