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2021年1月2日 星期六

106年中壢高中教甄-數學詳解

國立中央大學附屬中壢高級中學106學年度第1次教師甄選

一、填充題

A=[1ab01a001]=[100010001]+[0ab00a000]I+B{In=I,nNBk=0,k3,kZB2=[00a2000000]A10=(B+I)10=10k=0C10kBkI10k=2k=0C10kBkI10k=I+10B+45B2=[100010001]+[010a10b0010a000]+[0045a2000000]=[110a45a2+10b0110a001]=[1kapa2+qb01ka001]{k=10q=10p=45k+p+q=65
α,βx2+2x5=0{α+β=2αβ=5α2+β2=(α+β)22αβ=14(α+1)(α+2)(β+1)(β+2)(α3+β3)=[(α+1)(β+1)][(α+2)(β+2)](α3+β3)=(αβ+α+β+1)(αβ+2(α+β)+4)(α+β)(α2αβ+β2)=(52+1)(54+4)(2)(14+5)=(6)(5)(2)(19)=1140


z5+z1=0z5=1z|z5|=|1z||1z|=1{|z|=1|1z|=1{x2+y2=1(x1)2+y2=1z1,z2{z1=cosπ3+isinπ3z2=cosπ3+isinπ3z=12±32i

{448=4(112)=49=2232=62444488=44(1012)=4499=2211232=662444444888=444(10012)=444999=22111232=66622n44n88=n44(10012)=n44n99=22(11)232=(n66)2nk=1ak=6+66++n66=69(9+99++n99)=69(10+102++10nn)=69(10n+1109n)=22710n+169n2027=227[10n+19n10]a27[10n+1+bn+c](a,b,c)=(2,9,10)

{O(0,0)A(4,0)B(0,3)C(2,0){Γ:(x2)2+y2=4L1=AB:y=34x+3L2=BC:y=32x+3ΓL1P(3625,4825)L3=OP:y=43xL2L3Q(1817,2417)¯PQ=5485

36=22×32{a=2m13n1a=2m23n2c=2m33n3{(m1,m2,m3)=(2,2,2),(1,2,2),(0,2,2)(n1,n2,n3)=(2,2,2),(1,2,2),(0,2,2)(m1,m2,m3),(n1,n2,n3)1+3+3=77×7=49
y=x3+x+1(0,1)¯AB=¯BC{B=(0,1)ACLBL:y=mx+1A(a,ma+1){Ay=x3+x+1¯AB=5{ma+1=a3+a+1a2+(ma)2=5{m=a2+1(1)a2(1+m2)=5(2)(1)(2)a6+2a4+2a25=0a6+2a4+2a25=(a21)(a4+3a2+5)=0a2=1m=2L:y=2x+1
{32x2+13y2212x2y2=2xy123y2+32z2214y2z2=yz,32(x2+y2+z2)2xy+yz322xy+yz2xy+yz32
三中線為邊長的\triangle 面積={3\over 4}\triangle ABC(公式證明:\href{https://chu246.blogspot.com/2018/10/blog-post.html}{按這裡})={3\over 4}\times 3\sqrt{15}= {9\over 4}\sqrt{15}\\假設三中線長度為3,6,a,令s=(3+6+a)\div 2={a+9\over 2} \\\Rightarrow 三中線為邊長的\triangle 面積= \sqrt{s(s-3)(s-6)(s-a)} = \sqrt{{a+9\over 2} \cdot {a+3\over 2} \cdot {a-3\over 2} \cdot {9-a\over 2}} \\= {1\over 4}\sqrt{(81-a^2)(a^2-9)} ={9\over 4}\sqrt{15} \Rightarrow (81-a^2)(a^2-9)=81\times 15 \Rightarrow a^4-90a^2+1944=0 \\ \Rightarrow \cases{a^2=54\\ a^2=36} \Rightarrow 最大的a= \bbox[red,2pt]{3\sqrt 6}(\because 3\sqrt 6\gt 6)

由於圖形的對稱性,我們可以假設\cases{E[A_1\to A_6]=E_1\\ E[A_4\to A_6]=E_1\\ E[A_2\to A_6]=E_2 \\ E[A_3\to A_6]=E_3 \\E[A_5\to A_6]=E_3} ;\\因此\cases{E_3 = {1\over 4}((E[A_1\to A_6]+1) +(E[A_2\to A_6]+1) + (E[A_5\to A_6]+1)+ 1) \\ E_2= {1\over 4}((E[A_1\to A_6]+1) +(E[A_3\to A_6]+1) +(E[A_4\to A_6]+1) +(E[A_5\to A_6]+1)) \\ E_1={1\over 2}((E[A_2\to A_6]+1) +(E[A_3\to A_6]+1))} \\ \Rightarrow \cases{E_3={1\over 4}(E_1+E_2+E_3)+1\cdots(1) \\ E_2={1\over 4}(E_1+E_3+E_1+ E_3+4)={1\over 2}(E_1+E_3)+1 \cdots(2)\\ E_1={1\over 2}(E_2+ E_3)+1\cdots(3)}\\ 將(2)代入(1)及(3) \Rightarrow \cases{E_3={3\over 5}E_1+2\\ E_3=E_1-2} \Rightarrow {3\over 5}E_1+2 =E_1-2 \Rightarrow {2\over 5}E_1=4 \Rightarrow E_1= \bbox[red,2pt]{10}
(\sqrt 2+\sqrt 3)^{2016} =((\sqrt 2+\sqrt 3)^2)^{1008} = (5+2\sqrt 6)^{1008}\\ 令\cases{a=5+2\sqrt 6\\ b=5-2\sqrt 6} \Rightarrow \cases{a+b=10\\ ab=1} \Rightarrow a^2+b^2=(a+b)^2-2ab=98\\ a^n+b^n=(a+b)(a^{n-1}+b^{n-1})-ab(a^{n-2}+b^{n-2}) =10(a^{n-1}+b^{n-1})-(a^{n-2}+b^{n-2}) \\ \Rightarrow (a^n+b^n) \mod 10= -(a^{n-2}+b^{n-2}) \mod 10 \\ \Rightarrow \cases{(a+b) \mod 10=0\\ (a^2+b^2) \mod 10= -2=8\\ (a^3+b^3) \mod 10=0 \\ (a^4+b^4) \mod 10=-8 =2 \\ (a^5+b^5) \mod 10=0\\ \cdots } \Rightarrow \cases{(a^{4k}+b^{4k}) \mod 4=2\\ (a^{4k+1}+b^{4k+1}) \mod 4=0\\(a^{4k+2}+b^{4k+2}) \mod 4=8\\ (a^{4k+3}+b^{4k+3}) \mod 4=0},k=0,1,2,...\\ \Rightarrow (a^{1008}+b^{1008}) \mod 10 =(a^{4\cdot 252}+b^{4\cdot 252}) \mod 10 = 2 \\ \Rightarrow ((5+2\sqrt 6)^{1008} +(5-2\sqrt 6)^{1008}) \mod 10 =2 \\ 由於(5-2\sqrt 6)^{1008}近乎0,但不為0,因此(5+2\sqrt 6)^{1008}的個位數為1,小數點第1位為9\\,即(a,b)=\bbox[red,2pt]{(1,9)}
\cases{3x^2+y^2-3xy=3+2\sqrt 2 =(\sqrt 2+1)^2 \equiv a^2\\ y^2+ z^2-yz=9 +6 \sqrt 2= (\sqrt 6+\sqrt 3)^2\equiv b^2 \\ z^2+w^2+\sqrt 3zw =3+2\sqrt 2=(\sqrt 2+1)^2 \equiv a^2\\ w^2+3x^2+\sqrt 3wx =9+6\sqrt 2= (\sqrt 6+\sqrt 2)^2\equiv b^2} \\ \Rightarrow \cases{{(\sqrt 3x)^2+y^2-a^2 \over 2\sqrt 3xy} ={\sqrt 3\over 2} =\cos 30^\circ \\ { y^2+z^2-b^2 \over 2yz} ={1\over 2} =\cos 60^\circ \\ {z^2+w^2-a^2\over 2zw}=-{\sqrt 3\over 2} =\cos 150^\circ \\ {w^2+(\sqrt 3x)^2-b^2\over 2\sqrt 3wx}=-{1\over 2} =\cos 120^\circ},剛好\cases{30^\circ +60^\circ + 150^\circ + 120^\circ=360^\circ \\ 30^\circ + 150^\circ= 60^\circ+120^\circ =180^\circ}
因此我們可以假設有一圓內接矩形,其邊長分別為\sqrt 3x、y、z、w,兩對角線長為a及b;\\矩形各頂點角度就是30^\circ,60^\circ, 150^\circ,120^\circ ,如上圖;\\由托勒密定理(\href{https://chu246.blogspot.com/2020/11/ptolemys-theorem.html}{按這裡}): 圓內接矩形,兩對角線長之積 =兩對邊積之和;\\ 也就是\sqrt 3xz +yw = ab = (\sqrt 2+1)(\sqrt 6+\sqrt 3) = \bbox[red,2pt]{3\sqrt 3+2\sqrt 6}

二、 計算證明題

\log_{0.5}(x^2-2x-15) > \log_{0.25}(x+13)^2 = \log_{0.5}(x+13) \Rightarrow x^2-2x-15 < x+13 \\ \Rightarrow x^2-3x-28 < 0 \Rightarrow (x-7)(x+4) < 0 \Rightarrow -4 < x < 7 \cdots(1)\\ 又x^2-2x-15 > 0 \Rightarrow (x-5)(x+3) > 0 \Rightarrow x > 5或 x< -3 \cdots(2)\\再加上\cos 2x \ge 0 \Rightarrow -\pi/2 \pm 2k\pi \le 2x \le \pi/2 \pm 2k\pi \Rightarrow -\pi/4\pm k\pi \le x \le \pi/4 \pm k\pi\cdots(3)\\ 條件(1) \cap 條件(2) \cap 條件(3) \Rightarrow \bbox[red,2pt]{-5\pi/4 \le x < -3或 7\pi/4 \le x < 7}
(1)\cases{y^2=-x\\ y=k(x+1)},求交集\Rightarrow y^2=-({y\over k}-1) \Rightarrow y^2+{1\over k}y-1=0 \Rightarrow \cases{y_1+y_2= -1/k \\ y_1y_2=1}\\ \Rightarrow 兩交點為\cases{A(-y_1^2,y_1)\\ B(-y_2^2,y_2)} \Rightarrow \overrightarrow{OA} \cdot \overrightarrow{OB} =y_1^2y_2^2 +y_1y_2 =1-1=0 \Rightarrow \overline{OA} \bot \overline{OB},\bbox[red, 2pt]{故得證}(2)\triangle OAB = {1\over 2}\overline{OA}\cdot \overline{OB}= \sqrt{10} \Rightarrow \sqrt{y_1^4+y_1^2} \cdot \sqrt{y_2^4+y_2^2} =\sqrt{40} \\ \Rightarrow (y_1y_2)^4+ y_1^2y_2^2(y_1^2+y_2^2) +(y_1y_2)^2 =40 \Rightarrow 2+({1\over k^2}+2)=40 \Rightarrow {1\over k^2}=36 \\ \Rightarrow k=\bbox[red,2pt]{\pm {1\over 6}}\\ 註: y_1^2+y_2^2 =(y_1+y_2)^2-2y_1y_2= {1\over k^2}+2
將\cases{x+y+z=3 \\ x^3+y^3+z^3=3}\text{代入 } x^3+y^3+z^3-3xyz =(x+y+z)((x+y+z)^2 -3(xy+yz+zx))\\ \Rightarrow 3-3xyz= 3(9-3(xy+yz+zx)) \Rightarrow xyz-3(xy+yz+zx) =-8 \\ \Rightarrow xyz-3(xy+yz+zx)+ \color{blue}{9(x+y+z)-27} =-8 \Rightarrow (x-3)(y-3)(z-3)=-8\\ 因此我們有\cases{(x-3) +(y-3)+(z-3) =-6 \\ (x-3)(y-3)(z-3)=-8},由於x,y,z皆為整數,符合該二條件的解只有:\\ (x-3,y-3,z-3) =(-2,-2,-2) 或 (1,1,-8)的排列\\ \Rightarrow (x,y,z)=(4,4,-5)的排列 ((1,1,1)不合題意)\\ \Rightarrow x^2+y^2+z^2 =4^2+4^2 +5^2= \bbox[red,2pt]{57}

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