Processing math: 100%

網頁

2020年9月27日 星期日

109年台中文華高中第1次教甄-數學科詳解

 臺中市立文華高級中等學校 109 學年度
第1次教師甄選-數學科試題



解:
Γ1:x233+y249=1{a1=7b1=33a21=b21+c21c1=4{F1(0,4)F2(0,4)PΓ1¯PF1+¯PF2=2a1=14;:cosF1PF2=¯PF12+¯PF22¯F1F222¯PF1ׯPF2cos60=1422¯PF1ׯPF2822¯PF1ׯPF212=1322¯PF1ׯPF22¯PF1ׯPF2¯PF1ׯPF2=44(¯PF1¯PF2)2=(¯PF1+¯PF2)24¯PF1ׯPF2=1424×44=20|¯PF1¯PF2|=20=25=2a2a2=5c22=c21=a22+b2216=5+b22b2=11=2b2=211


解:
|z13+4i|=1{C(3,4)r=1;z2z1=1+3i=2(cos60+isin60){|z2|=2|z1|AOB=60OAB=12¯OAׯOBsinAOB=12¯OA×(2¯OA)sin60=32¯OA2;¯OA=¯OC+r=5+1=6OAB=32×62=183



解:
1(3)(10.2)3=0.83n0.83n10.83n>0.9990.001>0.83nlog0.001>log0.83n3>3n(3log21)3>3n(3×0.3011)=0.291nn>30.29110.3n=11



解:
(x0,y0)f(x)=a(xx0)3+b(xx0)+y0;y=f(x)=a(x+2)3+b(x+2)+5f(x)=3a(x+2)2+bx=1y=7x+1{f(1)=7f(1)=8{27a+b=727a+3b+5=8{a=1/3b=2y=13(x+2)32(x+2)+5



解:
{x+y+z=1(1)x2+y2+z2=3(2)x3+y3+z3=5(3)(x+y+z)2=(x2+y2+z2)+2(xy+yz+zx)1=3+2(xy+yz+zx)xy+yz+zx=1(1)×(2)=(x+y+z)(x2+y2+z2)=(x3+y3+z3)+xy(x+y)+yz(y+z)+zx(z+x)3=5+xy(1z)+yz(1x)+zx(1y)=5+(xy+yz+zx)3xyz=513xyzxyz=1/3(1)×(3)=(x+y+z)(x3+y3+z3)=x4+y4+z4+xy(x2+y2)+yz(y2+z2)+zx(z2+x2)5=x4+y4+z4+xy(3z2)+yz(3x2)+zx(3y2)=x4+y4+z4+3(xy+yz+zx)xyz3x2yzxy2z=x4+y4+z43xyz(x+y+z)=x4+y4+z431/3x4+y4+z4=5+3+1/3=253



解:
f(x)=x3+(1a)x2+(a+5)x+2a+5f(1)=01;f(x)=01,1+d,1+2d{=3+3d=a1=(1+d)(1+2d)=(2a+5){a=3d2(1)2d23d4=2a(2)(1)(2)2d23d4=6d4d(2d9)=0d=9/2(d=02)a=3d2=27/22=232



解:
f(x+2)=1+f(x)1f(x)f(3)=1+f(1)1f(1)=222=21f(5)=222=2+1f(7)=2+22=21f(9)=22+2=12=f(1):f(2k1)={12if (kmod4)=121if (kmod4)=22+1if (kmod4)=312if (kmod4)=0{1021=2×5111(k=511mod4=3)2025=2×10131(k=1013mod4=1){f(1021)=2+1f(2025)=12f(1021)+f(2025)=2

解:
A(0,0)B(6,5)6()5();62(11);1(,);311212;51,5;2×5=10;x,yx+y=5,x,yNH23=C43=4;310×4=40





解:
{k0f(x)g(x)dx=25k5k4+k2πk0g2(x)f2(x)dx=(49k9+k8+23k6165k5+83k3)π{ddkk0f(x)g(x)dx=ddk(25k5k4+k2)ddkk0g2(x)f2(x)dx=ddk(49k9+k8+23k6165k5+83k3){f(k)g(k)=2k44k3+2k(1)g2(k)f2(k)=4k8+8k7+4k516k4+8k2g2(k)f2(k)f(k)g(k)=4k8+8k7+4k516k4+8k22k44k3+2k=2k4+4kf(k)+g(k)=2k44k(2)(2)(1)2g(k)=4k36kg(k)=2k33kg(x)=2x33x



解:


D{P¯GH¯BQ:¯QC=2:1R¯AB{P(2,0,2)Q(4,1,0)R(a,3,0);¯EQ:x4=y32=z22S(4t,2t+3,2t+2){PS=(4t2,2t+3,2t)PR=(a2,3,2)PSPR2t+33=2t2t=3/5S(125,95,45){¯SP=425+8125+3625=115¯SE=14425+3625+3625=665¯EP=4+9=13cosPSE=¯SE2+¯SP2¯EP22ׯSEׯSP=12252×66256=666




解:


LAB{A(14y21,y1)B(14y22,y2)LmL=y1y2(y21y22)/4=4y1+y2L:yy1=4y1+y2(x14y21)x=(yy1)(y1+y2)4+14y21=14(y1+y2)y14y1y2=98=y1y214(y1+y2)y14y1y214y2dy=[18(y1+y2)y214y1y2y112y3]|y1y2=(18(y1+y2)y2114y21y2112y31)(18(y1+y2)y2214y1y22112y32)=124(y31y32)18(y21y2y1y22)=124(y1y2)3(y1y2)3=27y1y2=3(y1y2)2=9y2y2=12(y21+y229)P¯ABP(18(y21+y22),12(y1+y2))(x,y)y2=14(y1+y2)2=14(y21+y22)+12y1y2=14(y21+y22)+1212(y21+y229)=12(y21+y22)94=4x94y2=4x94



解:
f(x)=x3kx2+4kf(x)=3x22kx=0x=0,2k/3;f(0)f(2k/3)<04k(4k427k3)<016k2(1127k2)<0k2>27k>33k<33f(x)=0f(0)<04k<0k<0k<33


解:
{yx>2zy>3{yx+3zy+4xyz1481585915711151259157710141528111518k=112k(k+1)=1201 15x,y,z滿xyzH153=C173=680120680=317



A=[3113]12A=[3/21/21/23/2]=[cos(π/6)sin(π/6)sin(π/6)cos(π/6)](12A)30=[cos(5π)sin(5π)sin(5π)cos(5π)]=[1001]A30=[23000230]Cn2r=Cn2r2Cn2r=Cn2n2r=Cn2r2n2r=r2r/n=1/2A30B=[23000230][1/21/2]=[229229]






{x2+(y1)2=1y=x+2{A(0,2)D(1,1){x2+(y1)2=1y=x{C(0,0)B(1,1)xV1=π01(x+2)2(11x2)2dx=π012x2+4x+2+21x2dx=π[23x3+2x2+2x+(1x2+sin1x)]|01=π(1(2312π))=12π2+53πxV2=π10(1+1x2)2x2dx=π1022x2+21x2dx=π[2x23x3+1x2+sin1x]+|10=π(43+12π1)=12π2+13πV1+V2=π2+2π



{O(0,0,0)A(3,1,2)B(1,1,4)C(2,3,1){0x+y+z6{A(18,6,12)B(6,6,24)C(12,18,6)0x+y+z3{A(9,3,6)B(3,3,12)C(6,9,3){OABC=1618612662412186OABC=169363312693=1618612662412186169363312693=16(6333)312114231=16×189×14=441


-- END   (解題僅供參考,學校未提供計算題資料)  --





沒有留言:

張貼留言