Loading [MathJax]/extensions/TeX/mathchoice.js

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
=====================================================================
 

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
====================================================================


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
======================================================================



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
============================================================

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 ==========================================================


\alpha,\beta為x^2-x+1=0的兩根\Rightarrow \cases{\alpha+\beta=1\\ \alpha\beta=1} \Rightarrow (\alpha^{n-1} +\beta^{n-1})(\alpha+\beta ) =\alpha^n+\beta^n+\alpha\beta(\alpha^{n-2}+\beta^{n-2})\\ \Rightarrow \alpha^n+\beta^n= \alpha^{n-1}+\beta^{n-1}-(\alpha^{n-2}+\beta^{n-2}) \Rightarrow \cases{S_n=S_{n-1}-S_{n-2},n\ge 3\\ S_1=\alpha+\beta=1\\ S_2=(\alpha+ \beta)^2-2\alpha\beta =-1},其中S_n =\alpha^n+\beta^n\\ \begin{array}{} n& S_n\\\hline 1 & 1 \\ 2 &  -1  \\ 3 & -1-1=-2 \\ 4 & -2-(-1)=-1\\ 5 & -1-(-2)=1 \\  6 & 1-(-1)=2 \\\hdashline 7 & 2-1=1 \\ 8& 1-2=-1\\ 9 & -1-1=-2\\ \cdots & \cdots\\\hline\end{array} \Rightarrow S_{n+6}= S_n,n\ge 1 \\ 因此由 S_1=S_5=1 \Rightarrow S_n =1,n=\cases{1\mod{6} \\ 5\mod{6}} \Rightarrow n=\cases{1,7,13,...,97(6\times 16+1)\\ 5,11,17,...,95(6\times 15+5)}\\ \Rightarrow 共有17+16=\bbox[red, 2pt]{33}個。 =======================================================================
\cases{P(x,y,z) \\A(1,-1,2) \\ B(1,1,0)\\ C(1,0,4)} \Rightarrow \overline{AP}^2+ \overline{BP}^2+\overline{CP}^2\\ = (x-1)^2 +(y+1)^2+(z-2)^2 +(x-1)^2+(y-1)^2+z^2 +(x-1)^2+y^2 +(z-4)^2\\ = 3((x-1)^2+y^2 +(z-2)^2) +10\\ 柯西不等式:((x-1)^2+y^2 +(z-2)^2) (1^2+1^2+1^2) \ge (x+y+z-3)^2 = (0-3)^2=9 \\ \Rightarrow (x-1)^2+y^2 +(z-2)^2\ge {9\over 3}=3\Rightarrow 3((x-1)^2+y^2 +(z-2)^2) +10 \ge  3\times 3+10= \bbox[red,2pt]{19} ============================================================

n^3+108 = (n+11)(n^2-11n+121)-1223 \Rightarrow {n^3+108\over n+11}=(n^2-11n+121)-{1223\over n+11} \\ \Rightarrow n=1223-11= \bbox[red,2pt]{1212}為最大的整數使用得{n^3+108\over n+11}也是整數
============================================================

\\ 令\sin u={x-3\over 4} \Rightarrow \cos u\;du = {1\over 4} dx,則\int \sqrt{-x^2+6x+7}\;dx =  \int \sqrt{-(x-3)^2+4^2}\;dx \\=\int  4\sqrt{1-({x-3\over 4})^2} \;dx =\int 4\sqrt{1-\sin^2 u} \cdot 4 \cos u \;du  =16\int\sqrt{\cos ^2u}\cdot \cos u\;du = 16\int \cos^2 u\;du \\=8\int \cos 2u+1 \;du= 4\sin 2u +8u   =8\sin u\cos u+8u \\ = 8\cdot {x-3\over 4}\cdot {\sqrt{-x^2+6x+7}\over 4} +8\sin^{-1}{x-3\over 4} ={x-3\over 2}\cdot \sqrt{-x^2+6x+7} +8\sin^{-1}{x-3\over 4} \\ 因此\int_1^7 (-2+\sqrt{-x^2+6x+7})\;dx = \left. \left[ -2x+ {x-3\over 2}\cdot \sqrt{-x^2+6x+7} +8\sin^{-1}{x-3\over 4}\right] \right|_1^7\\ =(-14+0+8\sin^{-1}1) -(-2+(-1)\cdot 2\sqrt 3 +8\sin^{-1}{-1\over 2})\\ =-12+2\sqrt 3+ 8\cdot {\pi \over 2}-8\cdot {-\pi \over 6} = \bbox[red,2pt]{-12+2\sqrt 3+ {16\over 3}\pi} ============================================================



解:

正八邊形每一內角為\;(8-2)\times 180\div 8 = 135^\circ \Rightarrow \angle EFB= 180^\circ-135^\circ =45^\circ \\ \Rightarrow \angle BEF = \angle ABC-\angle AFE=60^\circ- 45^\circ=15^\circ  \Rightarrow \angle CED = 180^\circ -135^\circ-15^\circ = 30^\circ \\ \Rightarrow \angle CDE=30^\circ \Rightarrow \overline{CD}=\overline{CE}= 2;\\ \triangle CDE: \cos \angle CDE = \cos 120^\circ ={2^2+2^2-\overline{DE}^2 \over 2\times 2\times 2} \Rightarrow -{1\over 2}={8-\overline{DE}^2 \over 8} \Rightarrow \overline{DE}=2\sqrt 3;\\ \triangle BEF: {\overline{EF}\over \sin \angle EBF} = {\overline{EB}\over \sin \angle EFB} = {\overline{BF}\over \sin \angle BEF} \Rightarrow {2\sqrt 3\over \sin 120^\circ} = {\overline{EB}\over \sin 45^\circ} = {\overline{BF}\over \sin 15^\circ} \\\Rightarrow {2\sqrt 3\over \sqrt 3/2} = {\overline{EB}\over 1/\sqrt 2} = {\overline{BF}\over (\sqrt 6-\sqrt 2)/4} \Rightarrow \cases{\overline{EB}=2\sqrt 2\\ \overline{BF}=\sqrt 6-\sqrt 2} \\ \Rightarrow \overline{AF}= \overline{AB}+ \overline{BF} = \overline{CE}+ \overline{EB}+\overline{BF} =2+2\sqrt 2+\sqrt 6-\sqrt 2= \bbox[red,2pt]{2+\sqrt 6+\sqrt 2}===============================================================


假設\cases{O:外接圓圓心\\ P:內切圓圓心\\ R:外接圓半徑\\ r:內切圓半徑} \Rightarrow \overline{AG}=\overline{AO}\cos {\phi\over 2} =R\cos {\phi\over 2} \Rightarrow \overline{AB}=2\overline{AG} =2R\cos{\phi\over 2} \\ \Rightarrow \overline{AF}=\overline{AB} \cos{\phi \over 2} =2R\cos^2{\phi \over 2} \Rightarrow \overline{OP} =\overline{AF}-R-r =2R\cos^2{\phi \over 2}-R-r\\ \overline{GO}\parallel \overline{EP} \Rightarrow {\overline{GO} \over \overline{EP}} ={\overline{AO} \over \overline{AP}(=\overline{AO}+\overline{OP})} \Rightarrow {R\sin{\phi\over 2} \over r} ={R\over R+(2R\cos^2{\phi \over 2}-R-r)} ={R\over 2R(1-\sin^2{\phi \over 2})-r} \\ 令\cases{x=\sin{\phi\over 2} \\ r=k \\ R=(1+\sqrt 2) k},則上式為 {x\over 1}={1\over 2(1+\sqrt 2)(1-x^2)-1} \Rightarrow (2+2\sqrt 2)x^3-(1+2\sqrt 2)x+1=0 \\ \Rightarrow (x+1) ((2+2\sqrt 2)x^2-(2+2\sqrt 2)x+1)=0 \Rightarrow \cases{x=-1(不合,\because \phi/2 \ne 270^\circ)\\ x={(2+\sqrt 2)\pm 2 \over 2(2+2\sqrt 2)}= {\sqrt 2\over 2}或{2-\sqrt 2\over 2}} \\ \Rightarrow \sin{\phi\over 2}=x =\bbox[red,2pt]{{\sqrt 2\over 2},{2-\sqrt 2\over 2}} =============================================================



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}
=======================================================================


(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}
=================================================================
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。

    回覆刪除