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2021年3月30日 星期二

105年全國高中教甄聯招-數學詳解

 教育部受託辦理105學年度公立高級中等學校教師甄選

一、單選題


:{45=455=565=675=795=9115=14+5+6+7+9+1=32{(A)125=2(B)135=3(C)145=4(D)155=5,(A)

(A)cos225sin225=cos50(B)1+sin3401sin340=1sin201+sin20<0(C)sin23cos112sin292sin67=sin23cos68+sin68cos23=sin(6823)=sin45=cos45(D)2tan67.51tan267.5=tan(2×67)=tan135=tan45=1(C)

x275y29=1c2=75+9=84x2t2+1+y27t=1t2+1=7t+c2=7t+84t2+t90=0(t9)(t+10)=0t=10(97t>0)(D)

2016=1214=a14+b14{a+b=12a<ba,bN(a,b)=(1,11)(5,6)5(x,y)=(114,11214),(2214,10214),,(5214,6214)(C)



f(x)=a(x2)(x3)(12)(13)+b(x1)(x3)(21)(23)+c(x1)(x2)(31)(32){f(1)=af(2)=bf(3)=cy=f(x)(1,a),(2,b),(3,b)abcd:{d=0:(a,b,c)=(1,1,1),(2,2,2),,(6,6,6)6d=1:(1,2,3),(2,3,4),,(4,5,6)4d=1:(6,5,4),(5,4,3),,(3,2,1)4d=2:(1,3,5),(2,4,6)2d=2:(6,4,2),(5,3,1)26+4+4+2+2=181863=112f(x)abcf(x)1112=1112(B)


ABCDECABDE:Day1Day2Day3Day4Day5AB?ACC?BD?DE?EDay2Day3Day4BC,D,EA3A,C,D,EB4A,C,ED3A,C,DE3CB,D,EA3A,D,EB3A,B,ED3A,B,DE3DB,C,EA3A,C,EB3A,B,C,ED4A,B,CE3EB,C,DA3A,C,DB3A,B,CD3A,B,C,DE4(3+4+3+3)×41=51(B)



α,β{α+β=4aαβ=4a24a3b+9αβ=2101,a,bN(α+β)24αβ=(αβ)216a24(4a24a3b+9)=40416a+12b=4404a+3b=110a=2,5,,26=2+3k,k=08(A)
:(x0,y0)ax+by=c{x=x0+bgcd(a,b)ty=y0agcd(a,b)t,tZ(x0,y0)2x+3y=kx0(x0+3t,y02t)tZx0x0300<x03(1)122yy0,y02,,y02×121y02×121>0y02×12202×122y0>2×121(2)(1)(2){x0=1,2,3y0=1,2k3×2=6(B)
(A)×:f(x)=3x3x22x{f(1)=0f(x)=9x22x2f(1)=922=5(1,f(1))=(1,0),5:y=5(x1)y=5x5(B):|10f(x)dx|=|[34x413x3x2]|10|=|712|=712(C):f(x)=018x2=0x=1/9f(1/9)=56243(D):f(x)=0x(x1)(3x+2)=0x=0,1,2/3(BCD)


(A):{A=42B=126C=54D=17{μA=42/6=7μB=126/6=21μC=54/6=9μD=17/6<3μB>μC>μA>μD(B)×:{Var(A)=(62+12+12+22+22+22)/6Var(C)=(62+12+12+22+22+22)/6σA=σC(C):{A,B,CD{r1,r2>0r3<0;{r1=(AiuA)(BiuB)(AiuA)2(BiuB)2=15050450=1r2=(AiuA)(CiuC)(AiuA)2(CiuC)2=505050=1r1=r2(D):σA=(62+12+12+22+22+22)/6=506=53=533(ACD)



f(x)=x52px4+x33px2+x2(A):f(1)=15p0,pZx1(B):f(1)=055p=0p=1f(2)=4444p=0(C):f(x)f(x)=0,x{±1,±2}{(A):f(1)0(B):f(1)=55pf(1)0p1f(2)0f(2)=4044p0pZf(x)(D)×:f(x)=(x+1)2(x3(p+2)x2+4px4)+(9p+10)x+(4p+5)(x+1)2{9p+10=04p+5=0(x+1)2(ABC)


(A)×:A=[4321]det(AλI)=0λ=1,2(AI)(A2I)=A23A+2I=0(B):A47A3+10A28A+3I=(A23A+2I)(A24A4I)12A+11I=12A+11I=[48362412]+[110011]=[37362423](C):B=[λ100λ2]=[1002](D)×:A=PBP1A10=PB10P1=P[100210]P1P1=[2311][2311](BC)

第二部份
一、 填充題
Γ:x2y2=9a=b=3{¯AF2¯AF1=2a=6¯BF2¯BF1=2a=6¯AF2+¯BF2(¯AF1+¯BF1)=12=¯AF2+¯BF2¯AB=12¯AF2+¯BF2=12+15=27ABF2=¯AB+¯AF2+¯BF2=15+27=42
{A(3,2,2)B(5,4,3)P(x,y,0){PE:z=0ABE¯PA+¯PB=(x+3)2+(y2)2+(2)2+(x+5)2+(y4)2+32ABE¯PA+¯PB=¯AB=22+22+52=33PL:ABEL:x+32=y22=z25(2t3,2t+2,5t+2)(2t3,2t+2,5t+2)=(x,y,0)5t+2=0t=25{x=2253=195y=225+2=14533(x,y)=(195,145)


{x+7y404x5y+1705x+2y200{A(2,5)B(3,1)C(4,0)f(x,y)=axy=k{f(C)f(A)f(C)f(B){4a2a54a3a1{a5/2a1/7a52


{a1=2an+1=2an1an=2an11=2(2an21)1=22an221==2n1a12n22n21=2n(2n11)=2n2n1+1=2n1+1>1000n=11(210=1024)


:

{A:B:{6A:16B:12A4B:44A2B:51+1+4+5=11

:

{cosPBA=4+a214a=a2+34acosPBC=4+a294a=a254a=sinPBA(a2+34a)2+(a254a)2=1a410a2+17=0=a2=10+322=5+22



{¯AP=a¯PQ=x{¯CQ=1a¯BP=2a¯BQ=2(1a)=1+a{cos60=12=(2a)2+(1+a)2x22(1+a)(2a)(1)cos45=22=(2a)2+x2(1+a)22(2a)x(2)(1)x2=3a23a+3(2)a=23(2+3,a<2)BPQABC=¯BPׯBQ¯BAׯBCBPQ=ABC×(2a)(1+a)4=3×(2a)(1+a)4=9334


{x=4x+7yy=6x+2y|x|3+|y|41=4×12×3×4=244752=27|4x+7y|3+|5x+2y|41=2427=89


[1232]5=45[1/42/43/42/4]5[1/42/43/42/4]1()[1/42/43/42/4]545[1/42/43/42/4]5[123321]=45(1+2+3+3+2+1)=45×12=1024×12=12288

(x6)2+(y7)2=10{C(6,7)r=10¯OC=62+72=85¯OA=¯OB=¯OC2r2=53COB=θ{sinθ=10/85cosθ=53/85sinAOP=sin2θ=2sinθcosθ=23017OAP:¯APsin2θ=¯OPsinOAP=¯OAsinAPO¯PO¯PA=sinOAPsin2θ=sinOAP230/17=kksinOAP=1k=17230=173060

計算題






{B=P0C=Pn{|AP0|=|AB|=1P0Pk=knP0Pn=knBC,k=0nSn=n1k=0APkAPk+1=n1k=0(AP0+P0Pk)(AP0+P0Pk+1)=n1k=0(|AP0|2+AP0P0Pk+1+P0PkAP0+P0PkP0Pk+1)=n1k=0(1+AP0P0Pk+1+P0PkAP0+k(k+1)n2)=n1k=0(1+AP0(P0Pk+P0Pk+1)+k(k+1)n2)=n1k=0(1+AB(2k+1nBC)+k(k+1)n2)=n1k=0(1+2k+1n(ABBC)+k(k+1)n2)=n1k=0(1+2k+1n(12)+k(k+1)n2)=n1k=0(12k+12n+k(k+1)n2)=nn2n+12n+n213nlimnSnn=limn(1n2n+12n2+n213n2)=112+13=56
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