Processing math: 100%

2022年7月7日 星期四

111年台中市國中聯合教甄-數學詳解

111 學年度臺中市政府教育局受託辦理本市立國民中學

(含本市立高級中等學校附設國中部)教師甄選

選擇題(共 50 題,每題 2分,共 100 分)

解答f(x)=xsin(x)+ex{f(2h)=2hsin(2h)+e2hf(0)=1f(2h)=2hsin(2h)+e2hlimh0f(2h)2f(0)+f(2h)h2=limh04hsin(2h)+e2h+e2h2h2=limh04sin(2h)+8hcos(2h)+2e2h2e2h2h by l'Hospital rule=limh016cos(2h)16hsin(2h)+4e2h+4e2h2 by l'Hospital rule=242=12(D)
解答x2+3xy+y2+xy=10x2x+3y+3xy+2yy+1y=0y=2x+3y+13x+2y1|(1,2)=3232(1,2):y=32(x1)+2y=32x+72(B)
解答f(x)=ddx(cosx11+t4dt)=1+(cosx)4(sinx)f(π4)=1+14(22)=52(22)=104(C)
解答k=1(x+1)kk2kan=(x+1)nn2nr=limn|an+1an|=limn|x+12nn+1|<1|x+12|<12<x+1<23<x<1[3,1)(x=3){a=3b=1a+4b=3+4=1(C)
解答LA(2,1,3)(1,3,2)(E:x3y+2z=4)L:x21=y13=z32;PLP(t+2,3t+1,2t+3),tRd(P,E)=|t+2+9t3+4t+641+9+4|=14t+114t=114P(2714,1714,207)EP=(A+Q)/2QAQ=2PA=(137,107,197)=(α,β,γ)β+γ=297(B)
解答A=[0112]=[1110][1101][0111]=PDP1An=P[1n01]P1A111=P[111101]P1=[11101111][0111]=[110111111112]=[abcd]a+c+d=110+111+112=113(D)
解答2,4,6,855434×5×5×4×3=1200(A)
解答10n2>9nlog10n2>log9nn2>2nlog3=2n×0.47710.0458n>2n>43.67n=44(C)

解答{a=32+3b=323x=abx3=(ab)3=(a3b3)3ab(ab)=2331xx3+3x=23x3+3x+1=23+1(A)
解答x=1i3i=(3+1)(31)i4=12(6+24624i)=12(cos(15)isin(15))x10=132(cos(150)isin(150))=132(3212i)64x10=3i(A)
解答(A):661(mod13)661111(mod13)66111113(B)×:29995111837(mod13){88(mod13)8212(mod13)835(mod13)841(mod13)858(mod13)48378(mod13)2999+513(C)×:6833333331111(mod13)2999+113(D):1850=(182)25(131)25(mod13)1850+113(C)
解答p(x)=1+x+x2+x3+x4,(1+x+x2+x3+x4)(x+x2++x10)2=p(x)(xp(x)+x6p(x))2=p(x)(x6+x)2p(x)2=p3(x)(x2+2x7+x12){p3(x)x3=10p3(x)x8=15x10=2×10+15=35(C):p(x)=1,1,1,1,1p2(x)1,2,3,4,5,4,3,2,1p3(x)1,3,6,10,15,18,19,18,15,10,6,3,1
解答logx+2logy=1logxy2=1xy2=103x+2y226xy2=603x+2y2260=4153x+2y2415(C)
解答3±5=6=9cosθcosθ=2312sin2θ2=23sin2θ2=16sinθ2=66(A)
解答{1+2xx2+x33+x+x2+2x31+x2x2+2x302x+x2+2x3A=[1211311211220212 ]rref(A)=[1003010300140000]Rank(A)=3(C)
解答{3x+3yz=104xy3z=2mnx4y2z=m2[331413n42][xyz]=[102mm2]Ax=bdet([341413n42])=01010n=0n=1{3x+3yz=10(1)4xy3z=2m(2)x4y2z=m2(3){3×(3)(1)15y5z=3m164×(3)(2)15y5z=2m83m16=2m8m=8mn=81=7(D)
解答
|(z1)(z+12)|=|z1||z+12|=¯PAׯPB,{PA(1,0)B(1/2,0){cosPOA=cosθ=¯OA2+¯OP2¯PA22¯OA¯OP=2¯PA22¯PA2=22cosθcosPOB=cosθ=¯OB2+¯OP2¯PB22¯OB¯OP=54¯PB2¯PB2=cosθ+54α2=f(θ)=¯PA2ׯPB2=(22cosθ)(cosθ+54)=2cos2θ12cosθ+52α2=f(18)=264+116+52=8132(B)
解答limx0sinxxx=limx0cosx11=111=0(C)
解答f(x)=cos(sin(x))f(x)=sin(sin(x))cos(x)f(0)=0(B)

解答y=x2y=2x|(1,1)=2L2L:y=2(x1)+1y=2x1x(12,0)x=12(D)
解答f(x,y)=x2+y2{fx=2xfy=2yf(x,y)=(2x,2y)f(1,1)=(2,2)(C)
解答u=x2du=2xdxπ0xcosx2dx=π012cosudu=[12sinu]|π0=0(B)
解答limn1nnk=0(kn)4=10x4dx=15(D)
解答42(l'Hospital rule):limsπsπcos3xdxsπ=limsπcos3s1=1(D)
解答xk=1+1k2>1f(xk)=1(D)
解答14+(14)2+(14)3+=1/411/4=13(C)
解答(D)(D)
解答sinx=xx33!+x55!x77!+(B)
解答y=k=0xkk!y=k=1xk1(k1)!=k=0xkk!=yy=y(B)
解答y=k=0kxkk!y=k=1k2xk1k!y=k=2k2(k1)xk2k!y=k=3k2(k1)(k2)xk3k!y(0)=9213!=3(C)
解答(2×3)(m×n)(5×8){m=3n=5(C)
解答|3468|=3864=00(C)
解答AX=0{x+y+z=0x+2y+3z=0(B)
解答A=[1111]/2=[1/21/21/21/2]=[cos45sin45sin45cos45](C)
解答A=[112134112]rref(A)=[101011000]Rank(A)=2Null(A)=1(D)
解答2×3=6(D)
解答{u=x2du=2xdxdv=cosxdxv=sinxx2cosxdx=x2sinx2xsinxdx{u=xdu=dxdv=sinxdxv=cosxx2cosxdx=x2sinx2(xcosx+sinx)π0x2cosxdx=[x2sinx+2xcosx2sinx]|π0=2π(D)
解答E(X+1)=E(X)+1=3E(X)=2;E[(X+1)2]=E(X2+2X+1)=E(X2)+2E(X)+1=E(X2)+4+1=20E(X2)=15Var(2X1)=E[(2X1)2](E(2X1))2=E(4X24X+1)(2E(X)1)2=4E(X2)4E(X)+1(221)2=608+19=44(C)
解答{u=f(x)du=f(x)dxdv=sinxdxv=cosxf(x)sinxdx=cosxf(x)+f(x)cosxdx{u=f(x)df=f(x)dxdv=cosxdxv=sinxf(x)cosxdx=f(x)sinxf(x)sinxdxf(x)sinxdx=cosxf(x)+f(x)sinxf(x)sinxdxπ0(f(x)+f(x))sinxdx=[cosxf(x)+f(x)sinx]|π0=f(π)12=0f(π)=12(B)
解答5:(1,4),(4,1),(2,3),(3,2)p=436=19P(X=1)=p,P(X=2)=(1p)p,..,P(X=n)=(1p)n1p,nNXGeometric(p)E(X)=1pE(5X1)=5E(X)1=5p1=451=44(D)
解答f(x)=cos22x+4cos2x4f(x)=4cos2xsin2x8cosxsinx=4cos2xsin2x4sin2x=4sin2x(cos2x+1)=0{sin2x=0x=kπ/2,kZcos2x=1x=(2k1)π/2,kZcos2x=1cos2x=0,f(x)f(π/2)=1+04=3(B)
解答limx0x32x0t21+t3dt=limx02x0t21+t3dtx3=limx04x21+8x323x2=limx08x23x2+24x5=limx016x6x+120x4=limx0166+480x3=166=83(D)
解答x+1x=2cosθ(x+1x)2=x2+1x2+2=4cos2θx2+1x2=4cos2θ2x3+1x3=(x+1x)(x21+1x2)=2cosθ(4cos2θ3)=2(4cos3θ3cosθ)=2cos3θ(A)
解答a+b4=a3+a3+a3+b44a3b2714a3b27a3b27loga3blog273loga+logb3log3(D)
解答38+36+34+32+134+33+32+3+1=131013213513=310182351=35+14=61(B)
解答x=p{x2+kx+1=(xp)(xq)x2+x+k=(xp)(xr),x2+kx+1x2+x+k=xqxrx3+(kr)x2+(1kr)xr=x3+(1q)x2+(kq)xkq{kr=1q(1)1kr=kq(2)r=kq(3),(3)(1)(2){kkq=1qq=1(4)1k2q=kq(5)(4)(5)1k2=k1k2+k2=0(k+2)(k1)=0k=1,212=1(A)
解答|ab||a+b|ab05050550504140231221122143231133113441050012(D)
解答d=32ai=32×50=1600>1504d32d=16ai=16×50=800<1504(1504800)÷16=44a1=a2==a6=16,a7=a8==a50=32滿16ai,i=150d=16(C)
解答


{y24x20(y+2x)(y2x)0|x|22x2;=8×2=16(B)
解答

=BPC=112(2a)2=a22+2a1(C)
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解題僅供參考,教甄歷屆試題及詳解


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