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

解： $$五座島最多可以蓋C^5_2=10條橋，只蓋4條橋有C^{10}_4=210種蓋法；其中不連通的方式:\\(1)4+1:4島相連，另一島不相連;4島互連可蓋C^4_2=6座橋，任取4座，共有C^6_4=15種情形;\\\qquad \qquad而五島選四島有C^5_4=5種選法，因此蓋4橋有5\times 15=75種蓋法;\\ (2)2+3: 2島互連，另3島互連，共有C^5_2=10種情形，每一種情形剛好蓋4座橋；\\因此蓋4橋可相連的方式有210-75-10=\bbox[red,2pt]{125}種。$$

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解： $$令三切點的x坐標分別為\alpha,\beta,\gamma \\\Rightarrow x^6-10x^5+29x^4-4x^3+ax^2-bx-c=(x-\alpha)^2 (x-\beta)^2(x-\gamma)^2\\ \Rightarrow \alpha+\beta+\gamma = 10\div 2= \bbox[red,2pt]{5}$$  ===========================================================

解： $$令\log_2 x=t \Rightarrow \log_{\sqrt 2}x =2t \Rightarrow {6\log_{\sqrt 2}x-8 \over 1+4(\log_2 x)^2} ={12t-8 \over 1+4t^2} \equiv g(t) \\ \Rightarrow g'(t)={12 \over 1+4t^2} -{8a(12t-8) \over (1+4t^2)^2} ={-4(6t+1)(2t-3) \over (1+4t^2)^2} = 0 \Rightarrow \cases{t=-1/6\\ t=3/2} \Rightarrow \cases{x=1/\sqrt[6]{2}\\ x=2\sqrt 2} \\ \Rightarrow \cases{f(1/\sqrt[6]{2}) =(-2-8)/(1+4/36)\\ f(2\sqrt 2)= (24\sqrt 2-8)/(1+32)} \Rightarrow f(2\sqrt 2) > f(1/\sqrt[6]{2}) \Rightarrow a=\bbox[red, 2pt]{2\sqrt 2}有最大值$$ ===========================================================

解： $$\log_a b=c \Rightarrow b=a^c，若\cases{a=2^x\\ b=2^y},其x,y\in\{1,2,...,25\}且x\ne y\\ \Rightarrow 2^y =(2^x)^c=2^{xc} \Rightarrow cx=y \Rightarrow x\mid y (x是y的因數)\\ \Rightarrow \begin{array}{}x& y &數量\\\hline 1& 2-25 & 24\\ 2 & 4,5,..,24& 11 \\ 3& 6,9,..,24& 7\\ 4 & 8,12,..,24 & 5\\ 5& 10,15,20,25 & 4\\ 6& 12,18,24 & 3\\ 7& 14,21 & 2\\ 8& 16,24 & 2\\ 9 & 18 & 1\\ 10& 20 & 1\\ 11& 22 & 1\\ 12& 24& 1\\\hline \end{array} \Rightarrow 24+11+\cdots + 1= 62 \\ \Rightarrow 機率為{62\over 25\times 24} =\bbox[red,2pt]{31\over 300}$$

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解： $$取3,3^2,\dots,3^{2n}的中間兩數3^n及3^{n+1}之間任一數，即可得最小之f(x),x\in[3^n,3^{n+1}]\\ \begin{aligned}f(3^n) &= (3^n-3)+(3^n-3^2)+\cdots +(3^n-3^n)+(3^{n+1}-3^n)+(3^{n+2}-3^n)+\cdots (3^{2n}-3^n)\\ &=n\cdot3^n-( 3+3^2+\cdots+ 3^n)+ (3^{n+1}+3^{n+2} +\cdots +3^{2n})-n\cdot 3^n \\ &=3^n(3+3^2+\cdots +3^n)-(3+3^2+\cdots +3^n) \\ &=(3^n-1)(3+3^2+\cdots +3^n) =(3^n-1){3^{n+1}-3 \over 3-1} =(3^n-1){3(3^n-1)\over 2} \\ &= \bbox[red,2pt]{{3\over 2}(3^n-1)^2}\end{aligned}$$

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解：

$$假設圓半徑為1\Rightarrow \overline{OA} =\overline{OB} =\overline{OC} =\overline{OD} =1\\ \triangle BCD: \angle DBC =180^\circ -45^\circ-75^\circ =60^\circ \Rightarrow \angle DOC= 2\angle DBC=120^\circ\\ \Rightarrow \angle ODC = \angle OCD = (180^\circ-120^\circ)\div 2=30^\circ\\\triangle OCD: {\overline{OC} \over \sin 30^\circ} ={\overline{CD} \over \sin 120^\circ} \Rightarrow {1\over 1/2} ={\overline{CD} \over \sqrt 3/2} \Rightarrow \overline{CD} =\sqrt 3 \\ \triangle BCD: {\overline{CD} \over \sin 60^\circ} = {\overline{BC} \over \sin 45^\circ}  \Rightarrow {\sqrt 3\over \sqrt 3/2} ={\overline{BC} \over 1/\sqrt 2} \Rightarrow \overline{BC}=\sqrt 2\\ \triangle BCD: \cos \angle DBC= \cos 60^\circ = {\overline{BD}^2+2-3\over 2\sqrt 2\times \overline{BD}} \Rightarrow \overline{BD} ={\sqrt 2+\sqrt 6\over 2}\\又\angle DAC = \angle DBC =60^\circ \Rightarrow \triangle OAD為一正\triangle  \Rightarrow \overline{AD}=1\\ 直角\triangle ABC: \cases{\overline{AC}=2 \\ \overline{BC}=\sqrt 2} \Rightarrow \overline{AB}=\sqrt 2\\ 令\cases{B為原點\\ \overrightarrow{BC}為x軸\\ \overrightarrow{BA}為y軸} \Rightarrow \cases{B(0,0)\\ C(\sqrt 2,0)\\ A(0,\sqrt 2)\\O(\sqrt 2/2,\sqrt 2/2)} ,又\overleftrightarrow{BD}斜率為\tan 60^\circ=\sqrt 3 \Rightarrow D(t,t\sqrt 3)\\ 由\overline{AD}=1 \Rightarrow t^2+(t\sqrt 3-\sqrt 2)^2=1 \Rightarrow t={\sqrt 2+\sqrt 6\over 4} \Rightarrow D({\sqrt 2+\sqrt 6\over 4},{3\sqrt 2+\sqrt 6 \over 4})\\ \overrightarrow{BD} = a\overrightarrow{OA} +b\overrightarrow{OB} \Rightarrow ({\sqrt 2+\sqrt 6\over 4},{3\sqrt 2+\sqrt 6 \over 4}) =a(-{\sqrt 2\over 2},{\sqrt 2\over 2}) +b(-{\sqrt 2\over 2},-{\sqrt 2\over 2})\\ \Rightarrow \cases{a+b=-{\sqrt 2+\sqrt 6 \over 2\sqrt 2} \\ a-b={\sqrt 6+3\sqrt 2\over 2\sqrt 2}} \Rightarrow (a,b)=\bbox[red,2pt]{({1\over 2},{-2-\sqrt 3\over 2})}\\...$$ =============================================================

解： $$P為橢圓上的切點，其切線斜率與L斜率相同;\\ 9x^2+4y^2 =36 \Rightarrow {d\over dx}(9x^2+4y^2)= {d\over dx}36 \Rightarrow 18x+8yy' = 0 \Rightarrow y'={18x\over -8y} = -{3\over 4}(L的斜率) \\ \Rightarrow y=3x 代回橢圓方程式\Rightarrow 9x^2+ 36x^2=36 \Rightarrow x^2={4\over 5} \Rightarrow x=-{2\over \sqrt 5} (x={2\over \sqrt 5}為最近的切點) \\ \Rightarrow y=3x = -{6\over \sqrt 5} \Rightarrow P坐標為\bbox[red,2pt]{(-{2\over \sqrt 5}, -{6\over \sqrt 5})}$$ ====================================================================

解： $$六面分別為1,1,1,2,2,3，其中有4個奇數，2個偶數 \Rightarrow P_1={2\over 6} ={1\over 3}\\又偶數=偶數+偶數或奇數+奇數 \Rightarrow P_n=P_{n-1}\times {1\over 3} +(1-P_{n-1})\times {2\over 3}= {2\over 3}-{1\over 3}P_{n-1}\\ \Rightarrow P_n=2({1\over 3}-{1\over 3^2} +{1\over 3^3}-\cdots +{1\over 3^n}(-1)^{n-1}) \cdots(1)\\ \Rightarrow {1\over 3}P_n=2({1\over 3^2}-{1\over 3^3} +{1\over 3^4}-\cdots +{1\over 3^{n+1}}(-1)^{n-1}) \cdots(2)\\ 式(1)+式(2) \Rightarrow {4\over 3}P_n=2({1\over 3}-{1\over 3^{n+1}}(-1)^{n-1}) \Rightarrow P_n={3\over 2}({1\over 3}-{1\over 3^{n+1}}(-1)^{n-1}) \\ \Rightarrow \lim_{n\to \infty} P_n ={3\over 2}({1\over 3}-0)=\bbox[red,2pt]{1\over 2}$$

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解： $${n+2\over n!+(n+1)!+(n+2)!} ={n+2\over n!(1+(n+1)+(n+2)(n+1))} ={n+2\over n!(n+2)^2} = {1\over n!(n+2)} \\ = {n+1\over (n+2)(n+1)n!} ={n+1\over (n+2)!} ={(n+2)-1\over (n+2)!} ={1\over (n+1)!}-{1\over (n+2)!} \\ \Rightarrow \sum_{n=1}^{2017} {n+2\over n!+(n+1)!+(n+2)!} =\sum_{n=1}^{2017}\left( {1\over (n+1)!}-{1\over (n+2)!} \right) \\=({1\over 2!}-{1\over 3!}) +({1\over 3!}-{1\over 4!}) +\cdots +({1\over 2018!}-{1\over 2019!})  = \bbox[red,2pt]{{1\over 2!}-{1\over 2019!}}$$ ===============================================================

解： $$假設切點為(a,f(a)) \Rightarrow 切線方程式: y-f(a)=f'(a)(x-a) \Rightarrow y-a^3+3a^2 -4=(3a^2-6a)(x-a)\\ 此切線過(p,4) \Rightarrow 4-a^3+3a^2 -4=(3a^2-6a)(p-a) \Rightarrow -a^3+3a^2-4 =3pa^2-3a^3-6ap+6a^2\\ \Rightarrow 2a^3+(-3-3p)a^2+6pa=0  \Rightarrow a(2a^2+(-3-3p)a+6p)=0;\\由於只有一切線，也就僅有一根a=0，且2a^2+(-3-3p)a+6p=0無實根，即判別式<0 \\ \Rightarrow (3+3p)^2-48p< 0 \Rightarrow 3p^2-10p+3 < 0 \Rightarrow (3p-1)(p-3)<0 \Rightarrow \bbox[red,2pt]{{1\over 3} < p < 3}$$ ========================================================================

解： $$x^2+ y^2=x+y= n \Rightarrow (x+y)^2=x^2+y^2+2xy \Rightarrow n^2=n+2xy \Rightarrow xy=(n^2-n)/2 \\ \Rightarrow (x-y)^2 = (x+y)^2-4xy =n^2-2(n^2-n)=2n-n^2 \Rightarrow x-y = \sqrt{2n-n^2}(負值不合,因為x>y)\\ \Rightarrow x^4-y^4 = (x^2+y^2)(x^2-y^2) = (x^2+y^2)(x+y)(x-y) =n^2\cdot \sqrt{2n-n^2} =\sqrt{2n^5-n^6}\\ 令f(n)=2n^5-n^6 \Rightarrow f'(n)=0 \Rightarrow 10n^4-6n^5=0 \Rightarrow 2n^4(5-3n)=0 \Rightarrow n=0,5/3\\ \Rightarrow f''(n)=40n^3-30n^4 \Rightarrow \cases{f''(0)=0\\ f''(5/3)< 0} \Rightarrow n=5/3時,f(n)有極大值\\\Rightarrow M= {5^2\over 3^2}\sqrt{{10\over 3}-{5^2\over 3^2}} = {25\sqrt 5\over 27} \Rightarrow (n,M)=\bbox[red,2pt]{({5\over 3},{25\sqrt 5\over 27})}$$ =======================================================================

解： $$\lim_{n\to \infty}{1\over n}\sqrt{1-{k^2\over 4n^2}} =2\lim_{n\to \infty}{1\over 2n}\sqrt{1-({k\over 2n})^2} =2 \int_0^{1/2} \sqrt{1-x^2}\;dx \\ =2\int_0^{\pi/6} \sqrt{1-\sin^2 u}\cos u \;du(其中\sin u =x) = \left. \left[ {1\over 2}\sin 2u + u \right]\right|_0^{\pi/6} =\bbox[red, 2pt]{{\sqrt 3\over 4}+{\pi \over 6}}$$ ==========================================================

解： $$\alpha,\beta為x^2-x+1=0的兩根\Rightarrow \cases{\alpha+\beta=1\\ \alpha\beta=1} \Rightarrow (\alpha^{n-1} +\beta^{n-1})(\alpha+\beta ) =\alpha^n+\beta^n+\alpha\beta(\alpha^{n-2}+\beta^{n-2})\\ \Rightarrow \alpha^n+\beta^n= \alpha^{n-1}+\beta^{n-1}-(\alpha^{n-2}+\beta^{n-2}) \Rightarrow \cases{S_n=S_{n-1}-S_{n-2},n\ge 3\\ S_1=\alpha+\beta=1\\ S_2=(\alpha+ \beta)^2-2\alpha\beta =-1},其中S_n =\alpha^n+\beta^n\\ \begin{array}{} n& S_n\\\hline 1 & 1 \\ 2 &  -1  \\ 3 & -1-1=-2 \\ 4 & -2-(-1)=-1\\ 5 & -1-(-2)=1 \\  6 & 1-(-1)=2 \\\hdashline 7 & 2-1=1 \\ 8& 1-2=-1\\ 9 & -1-1=-2\\ \cdots & \cdots\\\hline\end{array} \Rightarrow S_{n+6}= S_n,n\ge 1 \\ 因此由 S_1=S_5=1 \Rightarrow S_n =1,n=\cases{1\mod{6} \\ 5\mod{6}} \Rightarrow n=\cases{1,7,13,...,97(6\times 16+1)\\ 5,11,17,...,95(6\times 15+5)}\\ \Rightarrow 共有17+16=\bbox[red, 2pt]{33}個。$$  =======================================================================

解： $$\cases{P(x,y,z) \\A(1,-1,2) \\ B(1,1,0)\\ C(1,0,4)} \Rightarrow \overline{AP}^2+ \overline{BP}^2+\overline{CP}^2\\ = (x-1)^2 +(y+1)^2+(z-2)^2 +(x-1)^2+(y-1)^2+z^2 +(x-1)^2+y^2 +(z-4)^2\\ = 3((x-1)^2+y^2 +(z-2)^2) +10\\ 柯西不等式:((x-1)^2+y^2 +(z-2)^2) (1^2+1^2+1^2) \ge (x+y+z-3)^2 = (0-3)^2=9 \\ \Rightarrow (x-1)^2+y^2 +(z-2)^2\ge {9\over 3}=3\Rightarrow 3((x-1)^2+y^2 +(z-2)^2) +10 \ge  3\times 3+10= \bbox[red,2pt]{19}$$ ============================================================

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

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

解：

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

解：

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

解：

$$假設立方體邊長為a,A為原點及P(x,y,z),則\cases{A(0,0,0)\\ B(a,0,0)\\ C(a,a,0)\\ D(0,a,0)\\ E(0,0,a)\\ F(a,0,a)\\ G(a,a,a)\\ H(0,a,a)},如圖；\\ \cases{\overline{PA}=\sqrt 2\\ \overline{PB}= \overline{PD}=\sqrt 3\\ \overline{PE}=1} \Rightarrow \cases{x^2+y^2+z^2=2\cdots(1) \\ (x-a)^2+y^2+z^2 = 3 \cdots(2)\\ x^2+(y-a)^2+z^2 = 3 \cdots(3)\\ x^2+y^2+(z-a)^2=1 \cdots(4)},\\由(1)得\cases{y^2+z^2=2-x^2 代入(2) \Rightarrow (x-a)^2+2-x^2=3 \Rightarrow x=(a^2-1)/ 2a\\ x^2+z^2=2-y^2代入(3) \Rightarrow 2-y^2+(y-a)^2=3 \Rightarrow y=(a^2-1)/2a \\ x^2+y^2= 2-z^2代入(4) \Rightarrow 2-z^2+(z-a)^2=1 \Rightarrow z=(a^2+1)/2a}\\ 將P({a^2-1\over 2a},{a^2-1\over 2a},{a^2+1\over 2a})代回(1) \Rightarrow {(a^2-1)^2\over 4a^2} +{(a^2-1)^2\over 4a^2}+{(a^2+1)^2\over 4a^2}= 2 \\ \Rightarrow 3a^4-2a^2+3=8a^2 \Rightarrow 3a^4-10a^2+3=0 \Rightarrow (3a^2-1)(a^2-3)=0 \\ \Rightarrow a=\sqrt 3(a=1/\sqrt 3\Rightarrow x<0,y<0 \Rightarrow P在立方體外) \Rightarrow 體積=a^3 =(\sqrt 3)^3 = \bbox[red,2pt]{3\sqrt 3}$$

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解： $$(1) f(x+2y)=f(x)+g(y) \Rightarrow f'(x+2y)=f'(x) \Rightarrow f'(s)=f'(t)\;\forall s,t \Rightarrow f'(x)=c,c 為常數\\ (2)f'(x)=c\Rightarrow f(x)= cx+d, 其中c,d 為常數;又\cases{f(0)=1 \Rightarrow d=1\\ f'(0)=2 \Rightarrow c=2} \\ \Rightarrow f(x)=2x+1 \Rightarrow f(x+2y)=2x+4y+1 =f(x)+g(y)=2x+1+g(y) \Rightarrow g(y)=4y\\ \Rightarrow g(5)=\bbox[red,2pt]{20}$$

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