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2024年4月3日 星期三

113年台師大附中教甄-數學詳解

國立臺灣師範大學附屬高級中學 113 學年度
第 1 次專任教師甄選數學科筆試

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:x=8log2x9log3x4log2x+log0.50.25=2log2x33log3x22log2x2+2=x3x2x2+2x32x2x+2=0(x21)(x2)=0x=1,2(x=1,x>0)=3
解答:(12x)5(1+4x2)5(1+2x)5=(14x2)5(1+4x2)5=(116x4)5=5k=0C5k(16)kx4k{k=3a=C53(16)3=10163k=4b=C54(16)4=5164ba=5×16410×163=8
解答:E(X)=(a1++a12)÷12=13a1++a12=13×12=156Var(X)=E(X2)(E(X))252=E(X2)132E(X2)=194a21++a212=19412=232812i=1aibi=12i=1ai(ai135)=1512i=1(a2i13ai)=15(a21++a212)135(a1++a12)=152328135156=3005=60

解答:

x2+y2=10r=10=¯OP¯AO=62+72=85¯AP=8510=53PAO=θ{sinθ=10/85cosθ=53/85sin2θ=2sinθcosθ=23017APB:¯BPsin2θ=¯BAsinAPB¯BA¯BP=sinAPBsin2θ=sinAPB230/17=kksinAPB=1,k=17230=173060
解答::m=6,n=6,:(n1)m+(1)m(n1)=56+5=15630=30×521k=521
解答:
{A(1,7)B(7,1)O(0,0)P(x,y){PA=(x1,y7)PB=(x7,y+1)PAPB=(x1)(x7)+(y7)(y+1)=19(x4)2+(y3)2=6,Q(4,3),r=6¯OP=¯OQ+r=5+6
解答:[(3a2b+1)2+(2a+b2)2+(4a5b3)2][(2)2+12+12][2(3a2b+1)+(2a+b2)+(4a5b3)]2[(3a2b+1)2+(2a+b2)2+(4a5b3)2]6(7)2[(3a2b+1)2+(2a+b2)2+(4a5b3)2]496=496
解答:PLP(t+1,t+2,t){¯PA=(t+1)2+(t1)2(t3)2¯PB=t2+(t1)2+(t+2)2{¯PA=3(t1)2+8/3¯PB=3(t+1)2+2/3¯PA+¯PB=3(¯QC+¯QD),{Q(t,0)C(1,26/3)D(1,6/3)¯QC+¯QD=¯CD=10¯PA+¯PB=310=30

解答:H93(4!2!2!+5!2!3!+6!2!4!)H44H93=51155775=3135:35,,3,44:H44:,3593,H93=H44H93=57752:224!2!2!93H933:235!2!3!93H934:246!2!4!93H93H93(4!2!2!+5!2!3!+6!2!4!)H44H93
解答:α,α,β,β,γ,γ,cosα+cosβ+cosγ=16i=1cosθi=26i=12cosθi=46i=2cosθi=42cos120=5
解答:limx(5x5+3x4+4x3+3x3x3+3x2+4x+1)=limx(x(51+3x+4x2+3x4)x(31+3x+4x2+1x3))=limx51+3x+4x2+3x431+3x+4x2+1x31x=limx(51+3x+4x2+3x431+3x+4x2+1x3)(1x)=limx(15(1+3x+4x2+3x4)4/5(3+8x+12x3)13(1+3x+4x2+1x3)2/3(3+8x+3x2))=351=25
解答:(x1)f(x)=4x1f(t)dtddx((x1)f(x))=ddx(4x1f(t)dt)f(x)+(x1)f(x)=4f(x)f(x)3x1f(x)=0,I(x)=e3x1dx=1(x1)3I(x)f(x)3x1I(x)f(x)=01(x1)3f(x)3(x1)4f(x)=0(1(x1)3f(x))=01(x1)3f(x)=Cf(x)=C(x1)3f(0)=C=2C=2f(x)=2(x1)3f(5)=243=128

解答:ω=t(3+4i)+(22t)i=3t+(2+2t)iL:2x+3y=23 z8=12+32i=cos(2π3+2kπ)+isin(2π3+2kπ)zi=cos(3k+112π)+isin(3k+112π),i=0,1,,78Pi(cos(3k+112π),sin(3k+112π)),i=07,LL,,i01{P0(cos(π/12),sin(π/12))=((6+2)/4,(62)/4P1(cos(π/3),sin(π/3))=(1/2,3/2)|(6+52)/423|>|5/223|=d(P1,L)=|5/223|7=235/27=4215714
解答:13x210xy+13y26x42y27=[x,y][135513][xy]+[6,42][xy]27A=[135513]=[2/22/22/22/2][80018][2/22/22/22/2][6,42][2/22/22/22/2]=[242,182]8x2+18y2242x182y27=08(x3/2)2+18(y1/2)2=72(x3/2)29+(y1/2)24=1{a=3b=22b2a=83
解答:P(X=k)=0.8k1×0.2{P(X=10)=0.890.2>0.02P(X=11)=0.8100.2>0.02P(X=12)=0.8110.20.017k=12log(0.8100.2)=10log0.8+log0.2=10(3log21)+log21=31log211=31×0.30111=1.669log0.02=log22=0.3012=1.699log(0.8100.2)>log0.020.8100.2>0.02
解答:{x(y+zx)=392x2y(z+xy)=522y2z(x+yz)=782z2{xy+xz=39x2yz+xy=52y2xz+yz=78z22(xy+yz+zx)=169(x2+y2+z2)x2+y2+z22(xy+yz+zx)=(x+y+z)2=169x+y+z=13{y+zx=132xz+xy=132yx+yz=132z{13x2x2=392x213y2y2=522y213z2z2=782z2{x=3y=4z=6abc=346=72
解答:
{A(0,a)B(3a,0)C(0,0)P(cosθ,sinθ){¯PA2=cos2θ+(sinθa)2=2¯PB2=(cosθ3a)2+sin2θ=10{2asinθ+a2=123acosθ+3a2=9{sinθ=(a21)/2acosθ=(3a29)/23asin2θ+cos2θ=(a21)24a2+(3a29)212a2=13(a21)2+(3a29)2=12a2a46a2+7=0a2=3+2ABC=12¯CA¯CB=123a2=33+62


解答:

|a|=1A(1,0);|ab|=12BA,1/2|5ac|=1CO1(5,0),1adπ4DL:xy¯BD+¯CD,L,O2O1,¯AO2AB,O2CC¯BD+¯CD=¯AO2¯O2C¯AB=26112=32+26

二、 證明題:(共 10 分。請用黑色或藍色原子筆寫在作答卷上,須詳細過程,否則酌予扣分)

解答:(1a3(b+c)+1b3(a+c)+1c3(a+b))(a(b+c)+b(a+c)+c(a+b))(1a+1b+1c)2(1a3(b+c)+1b3(a+c)+1c3(a+b))(2(ab+bc+ca)(ab+bc+caabc)21a3(b+c)+1b3(a+c)+1c3(a+b)ab+bc+ca232(
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解題僅供參考,其他歷年試題及詳解




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