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2020年10月28日 星期三

108年臺北市麗山高中教甄-數學詳解

臺北市立麗山高級中學108學年度第1次教師甄選
數學科試題卷

C52=104C104=210:(1)4+1:4;4C42=64C64=15;C54=545×15=75;(2)2+3:23C52=10442107510=125
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xα,β,γx610x5+29x44x3+ax2bxc=(xα)2(xβ)2(xγ)2α+β+γ=10÷2=5 ===========================================================


log2x=tlog2x=2t6log2x81+4(log2x)2=12t81+4t2g(t)g(t)=121+4t28a(12t8)(1+4t2)2=4(6t+1)(2t3)(1+4t2)2=0{t=1/6t=3/2{x=1/62x=22{f(1/62)=(28)/(1+4/36)f(22)=(2428)/(1+32)f(22)>f(1/62)a=22 ===========================================================

logab=cb=ac{a=2xb=2y,x,y{1,2,...,25}xy2y=(2x)c=2xccx=yxy(xy)xy12252424,5,..,241136,9,..,24748,12,..,245510,15,20,254612,18,243714,212816,242918110201112211224124+11++1=626225×24=31300
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3,32,,32n3n3n+1f(x),x[3n,3n+1]f(3n)=(3n3)+(3n32)++(3n3n)+(3n+13n)+(3n+23n)+(32n3n)=n3n(3+32++3n)+(3n+1+3n+2++32n)n3n=3n(3+32++3n)(3+32++3n)=(3n1)(3+32++3n)=(3n1)3n+1331=(3n1)3(3n1)2=32(3n1)2
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解:

1¯OA=¯OB=¯OC=¯OD=1BCD:DBC=1804575=60DOC=2DBC=120ODC=OCD=(180120)÷2=30OCD:¯OCsin30=¯CDsin12011/2=¯CD3/2¯CD=3BCD:¯CDsin60=¯BCsin4533/2=¯BC1/2¯BC=2BCD:cosDBC=cos60=¯BD2+2322ׯBD¯BD=2+62DAC=DBC=60OAD¯AD=1ABC:{¯AC=2¯BC=2¯AB=2{BBCxBAy{B(0,0)C(2,0)A(0,2)O(2/2,2/2),BDtan60=3D(t,t3)¯AD=1t2+(t32)2=1t=2+64D(2+64,32+64)BD=aOA+bOB(2+64,32+64)=a(22,22)+b(22,22){a+b=2+622ab=6+3222(a,b)=(12,232)... =============================================================
PL;9x2+4y2=36ddx(9x2+4y2)=ddx3618x+8yy=0y=18x8y=34(L)y=3x9x2+36x2=36x2=45x=25(x=25)y=3x=65P(25,65) ====================================================================


1,1,1,2,2,342P1=26=13=++Pn=Pn1×13+(1Pn1)×23=2313Pn1Pn=2(13132+133+13n(1)n1)(1)13Pn=2(132133+134+13n+1(1)n1)(2)(1)+(2)43Pn=2(1313n+1(1)n1)Pn=32(1313n+1(1)n1)limnPn=32(130)=12
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n+2n!+(n+1)!+(n+2)!=n+2n!(1+(n+1)+(n+2)(n+1))=n+2n!(n+2)2=1n!(n+2)=n+1(n+2)(n+1)n!=n+1(n+2)!=(n+2)1(n+2)!=1(n+1)!1(n+2)!2017n=1n+2n!+(n+1)!+(n+2)!=2017n=1(1(n+1)!1(n+2)!)=(12!13!)+(13!14!)++(12018!12019!)=12!12019! ===============================================================

(a,f(a)):yf(a)=f(a)(xa)ya3+3a24=(3a26a)(xa)(p,4)4a3+3a24=(3a26a)(pa)a3+3a24=3pa23a36ap+6a22a3+(33p)a2+6pa=0a(2a2+(33p)a+6p)=0;a=02a2+(33p)a+6p=0<0(3+3p)248p<03p210p+3<0(3p1)(p3)<013<p<3 ========================================================================



x2+y2=x+y=n(x+y)2=x2+y2+2xyn2=n+2xyxy=(n2n)/2(xy)2=(x+y)24xy=n22(n2n)=2nn2xy=2nn2(,x>y)x4y4=(x2+y2)(x2y2)=(x2+y2)(x+y)(xy)=n22nn2=2n5n6f(n)=2n5n6f(n)=010n46n5=02n4(53n)=0n=0,5/3f(n)=40n330n4{f(0)=0f(5/3)<0n=5/3,f(n)M=52321035232=25527(n,M)=(53,25527) =======================================================================


limn1n1k24n2=2limn12n1(k2n)2=21/201x2dx=2π/601sin2ucosudu(sinu=x)=[12sin2u+u]|π/60=34+π6 ==========================================================


α,βx2x+1=0{α+β=1αβ=1(αn1+βn1)(α+β)=αn+βn+αβ(αn2+βn2)αn+βn=αn1+βn1(αn2+βn2){Sn=Sn1Sn2,n3S1=α+β=1S2=(α+β)22αβ=1,Sn=αn+βnnSn1121311=242(1)=151(2)=161(1)=2721=1812=1911=2Sn+6=Sn,n1S1=S5=1Sn=1,n={1mod65mod6n={1,7,13,...,97(6×16+1)5,11,17,...,95(6×15+5)17+16=33 =======================================================================
{P(x,y,z)A(1,1,2)B(1,1,0)C(1,0,4)¯AP2+¯BP2+¯CP2=(x1)2+(y+1)2+(z2)2+(x1)2+(y1)2+z2+(x1)2+y2+(z4)2=3((x1)2+y2+(z2)2)+10西:((x1)2+y2+(z2)2)(12+12+12)(x+y+z3)2=(03)2=9(x1)2+y2+(z2)293=33((x1)2+y2+(z2)2)+103×3+10=19 ============================================================

n3+108=(n+11)(n211n+121)1223n3+108n+11=(n211n+121)1223n+11n=122311=1212使n3+108n+11
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sinu=x34cosudu=14dxx2+6x+7dx=(x3)2+42dx=41(x34)2dx=41sin2u4cosudu=16cos2ucosudu=16cos2udu=8cos2u+1du=4sin2u+8u=8sinucosu+8u=8x34x2+6x+74+8sin1x34=x32x2+6x+7+8sin1x3471(2+x2+6x+7)dx=[2x+x32x2+6x+7+8sin1x34]|71=(14+0+8sin11)(2+(1)23+8sin112)=12+23+8π28π6=12+23+163π ============================================================



解:

(82)×180÷8=135EFB=180135=45BEF=ABCAFE=6045=15CED=18013515=30CDE=30¯CD=¯CE=2;CDE:cosCDE=cos120=22+22¯DE22×2×212=8¯DE28¯DE=23;BEF:¯EFsinEBF=¯EBsinEFB=¯BFsinBEF23sin120=¯EBsin45=¯BFsin15233/2=¯EB1/2=¯BF(62)/4{¯EB=22¯BF=62¯AF=¯AB+¯BF=¯CE+¯EB+¯BF=2+22+62=2+6+2===============================================================


{O:P:R:r:¯AG=¯AOcosϕ2=Rcosϕ2¯AB=2¯AG=2Rcosϕ2¯AF=¯ABcosϕ2=2Rcos2ϕ2¯OP=¯AFRr=2Rcos2ϕ2Rr¯GO¯EP¯GO¯EP=¯AO¯AP(=¯AO+¯OP)Rsinϕ2r=RR+(2Rcos2ϕ2Rr)=R2R(1sin2ϕ2)r{x=sinϕ2r=kR=(1+2)kx1=12(1+2)(1x2)1(2+22)x3(1+22)x+1=0(x+1)((2+22)x2(2+22)x+1)=0{x=1(, =============================================================



假設立方體邊長為a,A為原點及P(x,y,z),則\cases{A(0,0,0)\\ B(a,0,0)\\ C(a,a,0)\\ D(0,a,0)\\ E(0,0,a)\\ F(a,0,a)\\ G(a,a,a)\\ H(0,a,a)},如圖;\\ \cases{\overline{PA}=\sqrt 2\\ \overline{PB}= \overline{PD}=\sqrt 3\\ \overline{PE}=1} \Rightarrow \cases{x^2+y^2+z^2=2\cdots(1) \\ (x-a)^2+y^2+z^2 = 3 \cdots(2)\\ x^2+(y-a)^2+z^2 = 3 \cdots(3)\\ x^2+y^2+(z-a)^2=1 \cdots(4)},\\由(1)得\cases{y^2+z^2=2-x^2 代入(2) \Rightarrow (x-a)^2+2-x^2=3 \Rightarrow x=(a^2-1)/ 2a\\ x^2+z^2=2-y^2代入(3) \Rightarrow 2-y^2+(y-a)^2=3 \Rightarrow y=(a^2-1)/2a \\ x^2+y^2= 2-z^2代入(4) \Rightarrow 2-z^2+(z-a)^2=1 \Rightarrow z=(a^2+1)/2a}\\ 將P({a^2-1\over 2a},{a^2-1\over 2a},{a^2+1\over 2a})代回(1) \Rightarrow {(a^2-1)^2\over 4a^2} +{(a^2-1)^2\over 4a^2}+{(a^2+1)^2\over 4a^2}= 2 \\ \Rightarrow 3a^4-2a^2+3=8a^2 \Rightarrow 3a^4-10a^2+3=0 \Rightarrow (3a^2-1)(a^2-3)=0 \\ \Rightarrow a=\sqrt 3(a=1/\sqrt 3\Rightarrow x<0,y<0 \Rightarrow P在立方體外) \Rightarrow 體積=a^3 =(\sqrt 3)^3 = \bbox[red,2pt]{3\sqrt 3}
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(1) f(x+2y)=f(x)+g(y) \Rightarrow f'(x+2y)=f'(x) \Rightarrow f'(s)=f'(t)\;\forall s,t \Rightarrow f'(x)=c,c 為常數\\ (2)f'(x)=c\Rightarrow f(x)= cx+d, 其中c,d 為常數;又\cases{f(0)=1 \Rightarrow d=1\\ f'(0)=2 \Rightarrow c=2} \\ \Rightarrow f(x)=2x+1 \Rightarrow f(x+2y)=2x+4y+1 =f(x)+g(y)=2x+1+g(y) \Rightarrow g(y)=4y\\ \Rightarrow g(5)=\bbox[red,2pt]{20}
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註:解題僅供參考

1 則留言:

  1. 第5題,3的k次冪,應是3*k,解法就合理。若是3的k次冪,好像x=3的2n次冪,才會使和為最小。

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