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
解:A=[1ab01a001]=[100010001]+[0ab00a000]≡I+B⇒{In=I,n∈NBk=0,k≥3,k∈ZB2=[00a2000000]⇒A10=(B+I)10=10∑k=0C10kBkI10−k=2∑k=0C10kBkI10−k=I+10B+45B2=[100010001]+[010a10b0010a000]+[0045a2000000]=[110a45a2+10b0110a001]=[1kapa2+qb01ka001]⇒{k=10q=10p=45⇒k+p+q=65解:α,β為x2+2x−5=0的兩根⇒{α+β=−2αβ=−5⇒α2+β2=(α+β)2−2αβ=14因此(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)(α2−αβ+β2)=(−5−2+1)(−5−4+4)(−2)(14+5)=(−6)(−5)(−2)(19)=−1140
解:
z5+z−1=0⇒z5=1−z⇒|z5|=|1−z|⇒|1−z|=1兩式{|z|=1|1−z|=1之交集,相當於兩圓{x2+y2=1(x−1)2+y2=1之交集,即上圖之z1,z2⇒{z1=cosπ3+isinπ3z2=cos−π3+isin−π3⇒z=12±√32i
解:{44−8=4(11−2)=4⋅9=22⋅32=624444−88=44(101−2)=44⋅99=22⋅112⋅32=662444444−888=444(1001−2)=444⋅999=22⋅1112⋅32=6662⋯2n個⏞4⋯4−n個⏞8⋯8=n個⏞4⋯4(10⋯01−2)=n個⏞4⋯4⋅n個⏞9⋯9=22⋅(1⋯1)2⋅32=(n個⏞6⋯6)2⇒n∑k=1ak=6+66+⋯+n個⏞6⋯6=69(9+99+⋯+n個⏞9⋯9)=69(10+102+⋯+10n−n)=69(10n+1−109−n)=22710n+1−69n−2027=227[10n+1−9n−10]≡a27[10n+1+bn+c]⇒(a,b,c)=(2,−9,−10)
解:{O(0,0)A(4,0)B(0,3)C(2,0)⇒{圓Γ:(x−2)2+y2=4直線L1=↔AB:y=−34x+3直線L2=↔BC:y=−32x+3⇒Γ與L1相交於P(3625,4825)⇒L3=↔OP:y=43x⇒L2與L3交於Q(1817,2417)⇒¯PQ=5485
解:圖形y=x3+x+1的對稱點為(0,1),又¯AB=¯BC⇒{B=(0,1)A、C互為對稱點⇒L經過B⇒L:y=mx+1⇒A(a,ma+1)⇒{A在y=x3+x+1上¯AB=√5⇒{ma+1=a3+a+1√a2+(ma)2=√5⇒{m=a2+1⋯(1)a2(1+m2)=5⋯(2),將(1)代入(2)⇒a6+2a4+2a2−5=0⇒a6+2a4+2a2−5=(a2−1)(a4+3a2+5)=0⇒a2=1⇒m=2⇒L:y=2x+1
解:{√32x2+1√3y2≥2√12x2y2=√2xy12√3y2+√32z2≥2√14y2z2=yz,兩式相加⇒√32(x2+y2+z2)≥√2xy+yz⇒√32≥√2xy+yz⇒√2xy+yz最大值為√32
解: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)
解: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}
解:(\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
解:\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}解:
(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
解: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}
解: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}
你好:請問a=6為什麼不符合兩邊和大於第三邊呢?
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