107年台北市國中教甄聯招數學詳解

臺北市 107 學年度市立國民中學正式教師
聯合甄選-數學科題本

解：

$$正六邊形每個內角為(6-2)\times 180\div 6=120^\circ \Rightarrow 陰影面積=正六邊形面積扣除二個圓面積\\=6\sqrt 3-2\pi，故選\bbox[red, 2pt]{(B )}$$

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解：

$$a-3(a-y) < y-4 \Rightarrow y < a-2 \equiv y< 1 \Rightarrow a-2=1 \Rightarrow a=3，故選\bbox[red, 2pt]{(B )}$$ 本題試題有疑義，答案修正為A、B、D均給分

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解：

$$f(x) =x^3-x^2+1 \Rightarrow f'(x)=3x^2-2x \Rightarrow f'(1)=3-2=1 \\ \Rightarrow 切線斜率為f'(1)=1且過(1,1)，方程式為y-1=x-1 \Rightarrow y=x，故選\bbox[red, 2pt]{(C )}$$

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解：

$$x^2-x-1=0 \Rightarrow x^2=x+1 \Rightarrow x^4=(x+1)^2 =x^2+2x+1=3x+2 \\ \Rightarrow x^8 =(x^4)^2 =(3x+2)^2 =9x^2+12x +4 = 9x+9+12x+4 =21x+13 \\ \Rightarrow x^8+x+1 =21x+13+x+1=22x+14，故選\bbox[red, 2pt]{( D)}$$

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解：

$$\int_0^1 x^4\;dx = \left. \left[{1\over 5}x^5 \right] \right|_0^1 ={1\over 5}=0.2，故選\bbox[red, 2pt]{(B)}$$

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解：

$$只有f(x)=x^2-2同時經過(-1,-1)及(2,2)，其他(A)(C)(D)均不過(2,2)，故選\bbox[red, 2pt]{(B)}$$

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解：

$$x^2y+xy^2 +x+y=63 \Rightarrow xy(x+y) +x+y =(xy+1)(x+y)=63 \Rightarrow x+y ={63\over 6+1}=9\\ \Rightarrow x^2+y^2 = (x+y)^2-2xy =9^2-2\times 6 = 81-12=69，故選\bbox[red, 2pt]{( A)}$$

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解：

$$a^2-2018a+1=0 \Rightarrow a^2-2017a+{2018\over a^2+1} = a-1+{2018\over 2018a}=a-1+{1\over a} ={a^2+1\over a}-1\\ ={2018a\over a} -1 =2018-1=2017，故選\bbox[red, 2pt]{(C )}$$

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解：

$$以C為圓心\Rightarrow \cases{A逆時針旋轉60度成為E \\ B逆時針旋轉60度成為D} \Rightarrow \angle CED = \angle CAB=110^\circ \\\Rightarrow \angle AED =\angle CED-\angle CEA = 110^\circ-60^\circ =50^\circ，故選\bbox[red, 2pt]{( C)}$$

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解：

$$\cases{\overline{AB}=4 \Rightarrow f(x)=0的兩根為\alpha,\alpha+4 \Rightarrow \alpha+\alpha+4=-{b\over a}\\ x=3有極值\Rightarrow -{b\over 2a}=3 \Rightarrow -{b\over a}=6} \Rightarrow 2\alpha +4=6\Rightarrow \alpha=1\\ \Rightarrow f(x)=0的兩根為1及5 \Rightarrow f(x)=k(x-1)(x-5) \Rightarrow f(1)=a+b+c=0，故選\bbox[red, 2pt]{( B)}$$

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解：

$$\cases{l\to 12(偶) \to {12\over 2}+13 = 19 \to s\\ o\to 15(奇) \to {15+1\over 2} = 8 \to h\\ v\to 22(偶) \to {22\over 2}+13 = 24 \to x\\ e\to 5(奇) \to {5+1\over 2}  = 3 \to c } \Rightarrow shxc，故選\bbox[red, 2pt]{( B)}$$

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解：

$${N^2+7 \over N+4} =N-4 + {23\over N+4} \Rightarrow N+4=23k;\\ 1\le N\le 2018 \Rightarrow 1\le k \le 87 \Rightarrow 共有87個，故選\bbox[red, 2pt]{(C )}$$

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解：

$$\begin{array}{|c|c|c|} \hline 4 & a+1 & 2\\\hline  & 5 & \\\hline a& 1 & a-2 \\\hline\end{array} \Rightarrow 4+a+1+2(第1列)=a+1+a-2(第3列) \\\Rightarrow a=8 \Rightarrow P的位置:a-2=6，故選\bbox[red, 2pt]{( C)}$$

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解：

$$a={20182017\over 2017} = {20180000\over 2017} +1\\ b= {20172018 \over 2018} ={20170000\over 2018 }+1 \\ c={20172018 \over 2017} =10000+{2018\over 2017} \\ d={20182017\over 2018} =10000+{2017\over 2018} \\ \Rightarrow \cases{a>b \\ c> d\\ a> c\\ d> b} \Rightarrow a > c> d >b，故選\bbox[red, 2pt]{(B )}$$

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解：

$$正方形E的邊長為\sqrt{6^2+5^2}=\sqrt{61} \\\Rightarrow 直角\triangle G的短邊長= 正方形F的邊長=\sqrt{10^2-61}=\sqrt{39}\\ \Rightarrow 正方形D的邊長=\sqrt{39-5^2} =\sqrt{14}，故選\bbox[red, 2pt]{(B )}$$

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解：

$$\cases{a^2+b^2=m^2 \cdots(1)\\ (1-a)^2+1=m^2\cdots(2) \\(1-b)^2+1=m^2 \cdots(3)} \Rightarrow \cases{(2)及(3) \Rightarrow (1-a)^2=(1-b)^2 \Rightarrow a=b\cdots(3) \\(1)及(2) \Rightarrow a^2+b^2=(1-a)^2+1 \Rightarrow b^2=2-2a\cdots(4)}\\ 將(3)代入(4) \Rightarrow a^2+2a-2=0 \Rightarrow a=\sqrt 3-1 \Rightarrow m^2=a^2+b^2 =2(\sqrt 3-1)^2=8-4\sqrt 3 \\ \Rightarrow 正\triangle 面積={\sqrt 3\over 4}m^2 = {\sqrt 3\over 4}\times (8-4\sqrt 3)=2\sqrt 3-3，故選\bbox[red, 2pt]{(A )}$$

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解：

$$作\overline{BG}\bot \overline{CD}，見上圖；由於ABCD為等腰梯形，因此\overline{GC} = (\overline{CD}-\overline{AB})\div 2=(10-4)\div 2=3\\ \Rightarrow  \overline{DG}=\overline{CD}-\overline{CG}=10-3=7 ;又B為斜邊中點，即為\triangle DEF外接圓圓心\\\Rightarrow \overline{BF}=\overline{BD} \Rightarrow \overline{GF}= \overline{GD}=7 \Rightarrow \overline{CF}= \overline{GF}- \overline{GC} =7-3=4，故選\bbox[red, 2pt]{( C)}$$

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解：

$$(B)\times: \cases{\triangle CAP面積={1\over 2}\times \overline{AC}\times \overline{AP}\sin \angle CAP \\\triangle BAP面積={1\over 2}\times \overline{AP}\times \overline{AB}\sin \angle PAB} \Rightarrow \triangle CAP \ne \triangle BAP\\，故選\bbox[red, 2pt]{(B)}$$

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解：

$$\stackrel{\large{\frown}}{BC} =2 \stackrel{ \large{\frown}}{AB} \Rightarrow \cases{ \angle COB=120^\circ \\ \angle AOB=60^\circ};\\圓周長=2r\pi =4\pi \Rightarrow 75\pi \div 4\pi = 18+{3\over 4}，相當於滾了18又{3\over 4}圈，而{3\over 4}圈相當於270^\circ \\ \Rightarrow \stackrel{\large{\frown}}{DA}接觸地面，故選\bbox[red, 2pt]{( C)}$$

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解：

$$\cases{\angle POR=120^\circ \\ \angle POA=45^\circ} \Rightarrow \angle AOR = 120^\circ-45^\circ =75^\circ，故選\bbox[red, 2pt]{(D)}$$

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解：

$$54x \le 906 \Rightarrow x \le 16，由x從1至16試算，可得(x,y)=(2,38),(9,20),(16,2)，三組解，故選\bbox[red, 2pt]{(B )}$$

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解：

$$\cases{\overline{AB}=c \\ \overline{AC}=b \\ \angle A=180^\circ -60^\circ=120^\circ} \Rightarrow \cos \angle A= \cos  120^\circ={b^2+c^2-2^2 \over 2bc} =-{1\over 2} \Rightarrow b^2+bc+c^2 =4\\ \Rightarrow (b+c)^2 =bc+4 \Rightarrow {b+c\over 2} \ge \sqrt {bc} \Rightarrow (b+c)^2 \ge 4bc \Rightarrow bc+4 \ge 4bc \Rightarrow bc \le {4\over 3}\\ \triangle ABC 面積={1\over 2}bc \sin \angle A ={1\over 2}bc \times {\sqrt 3\over 2} = {\sqrt 3\over 4}bc \le {\sqrt 3\over 4}\times {4\over 3} ={\sqrt 3\over 3}={1\over \sqrt 3}，故選\bbox[red, 2pt]{(B)}$$

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解：

$$\cases{\overline{AE}=\overline{ED} \\ \overline{FC}=3\overline{BF}} \Rightarrow \cases{\triangle AOE 面積=\triangle EOD 面積=a \\ \overline{FC}=3\overline{BF} \\\triangle ABO+\triangle OBC = \triangle DOC+\triangle OBC} \Rightarrow \cases{\triangle AOE 面積=\triangle EOD 面積=a \\\triangle OFC 面積=3\triangle OBF 面積= 3b \\ \triangle ABO 面積=\triangle DOC  面積=c};\\ 又\cases{\overline{AE} =\overline{BF} \Rightarrow {\triangle AEO \over \triangle OBF} ={h_1\over h_2} \Rightarrow {a\over b} ={h_1\over h_2} \\ \triangle OAD \sim \triangle OCB \Rightarrow {\triangle ADO \over \triangle OBC}={h_1^2\over h_2^2}} \Rightarrow {2a\over 4b} ={h_1^2 \over h_2^2} ={a^2\over b^2} \Rightarrow b=2a;\\ 此外，{\triangle ACD \over \triangle ABC} = {\overline{AD} \over\overline{ BC} }\Rightarrow {2a+c \over 4b+c} ={2\over 4} \Rightarrow 4a+2c=4b+c=8a+c \Rightarrow c=4a; \\四邊形ABFOE面積=21 \Rightarrow a+b+c=21 \Rightarrow a+2a+4a=21 \Rightarrow a=3 \Rightarrow \cases{b=2a=6\\ c=4a=12} \\ \Rightarrow 梯形ABCD面積=2a+4b+2c= 6+24+24 =54，故選\bbox[red, 2pt]{(A)}$$

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解：

$$\cases{判別式>0 \\ 首項係數\ne 0} \Rightarrow \cases{(2m+1)^2 -4(m-2)^2 >0 \\ m-2\ne 0} \Rightarrow \cases{4m-3> 0\\ m\ne 2} \\ \Rightarrow m> {3\over 4} 且 m\ne 2，故選\bbox[red, 2pt]{(C )}$$

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解：

$$\cases{\angle A=\angle A\\ \angle ACB=\angle ADC=90^\circ} \Rightarrow \triangle ACB \sim \triangle ADC \Rightarrow {\overline{AC} \over \overline{AB}} ={\overline{AD} \over \overline{AC}} \Rightarrow {2\sqrt{10} \over 5k} ={4k \over 2\sqrt{10}} \\ \Rightarrow 20k^2=40 \Rightarrow k=\sqrt 2 \Rightarrow \overline{AD}= 4k=4\sqrt 2，故選\bbox[red, 2pt]{( A)}$$

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解：

$$\cases{\angle ABD=\angle DBC \\\angle ABC=30^\circ} \Rightarrow \angle ABD=15^\circ; 又\angle DAB=90^\circ \Rightarrow \angle ADB= 90^\circ-15^\circ =75^\circ，故選\bbox[red, 2pt]{(D)}$$

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解：

$$直角\triangle ABD \Rightarrow 10^2=8^2+\overline{BD}^2 \Rightarrow \overline{BD}=6;\\ 又\overline{OP} \parallel \overline{BD} \Rightarrow {\overline{AP} \over \overline{AD} }={\overline{AO} \over \overline{AB} } ={1\over 2} \Rightarrow \overline{AP}={1\over 2}\overline{AD}= 4;\\ 由於\cases{\angle CAO= \angle APO =90^\circ \\ \angle AOP=\angle AOP} \Rightarrow \triangle CAO \sim \triangle APO \Rightarrow {\overline{AC} \over \overline{AO}} ={\overline{AP} \over \overline{PO}} \Rightarrow {\overline{AC} \over 5} = {4\over 3}\\ \Rightarrow \overline{AC} ={20\over 3}，故選\bbox[red, 2pt]{( C)}$$

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解：

$$x=1有極大值\Rightarrow y'(1)=0 \Rightarrow 2a+b=0，故選\bbox[red, 2pt]{(D)}$$

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解：

$$x^3係數=- (C^3_3 +C^4_3+\cdots +C^9_3) =-\sum_{n=3}^9 C^n_3 =-\sum_{n=3}^9({1\over 6}n^3 -{1\over 2}n^2+{1\over 3}n) \\ =-\left({1\over 6}(45^2-1-8)-{1\over 2}({1\over 6}\times 9\times 10\times 19-1-4)+{1\over 3}(45-1-2) \right) =-210，故選\bbox[red, 2pt]{( D)}$$

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解：

$$a_3=a_2+a_1 \Rightarrow a_4=a_3+a_2=2a_2+a_1 \Rightarrow a_5=a_4+a_3=3a_2+2a_1 \\ \Rightarrow a_6=a_5+a_4=5a_2+3a_1 \Rightarrow a_7=a_6+a_5 = 8a_2+5a_1 \\ \Rightarrow a_8=a_7+a_6 =13a_2+8a_1;\\ 現在a_7=120 \Rightarrow 8a_2+5a_1=120 \Rightarrow \cases{a_1=8 \\ a_2=10} \Rightarrow a_8=13a_2+8a_1 =130+64=194，故選\bbox[red, 2pt]{(B )}$$

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解：

$$前3分鐘，每分鐘注水{20\over 3}公分；之後，每分鐘注水{5\over 3}公分；\\假設圓柱體底面積為A，則容器容積為50A；鐵塊體積為3\times ({20\over 3}-{5\over 3})A= 15A；\\ 因此鐵塊:容器=15A:50A = 3:10，故選\bbox[red, 2pt]{(A )}$$

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解：

$$x^2+{1\over x^2} = (x+{1\over x})^2-2 \Rightarrow x^3+{1\over x^3} =(x+{1\over x})(x^2+{1\over x^2}-1) =(x+{1\over x})((x+{1\over x})^2-3) \\ 現在x+{1\over x}= a+\sqrt{a^2-1}+  {1\over a+\sqrt{a^2-1}} =a+\sqrt{a^2-1}+a-\sqrt{a^2-1}=2a \\ \Rightarrow x^3+{1\over x^3} =(x+{1\over x})((x+{1\over x})^2-3)=(2a)(4a^2-3) =8a^3-6a，故選\bbox[red, 2pt]{(A )}$$

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解：

$$\sqrt{a_2+2a_1} +\sqrt{a_3+2a_2} +\sqrt{a_4+2a_3} +\dots+\sqrt{a_{10}+2a_9} =0\\ \Rightarrow \cases{a_2 +2a_1=0 \\a_3+2a_2 =0\\ a_4+2a_3=0 \\ \cdots \\ a_{10}+2a_9=0}，全部相加\Rightarrow 3\sum_{n=1}^{10}a_n-a_1-2a_{10}=0 \Rightarrow \sum_{n=1}^{10}a_n = {1\over 3}(a_1+2a_{10})\\ 又a_{10}=-2a_9 =-2(-2a_8)=2^2a_8 =2^2(-2a_7) =-2^3a_7 = \dots =-2^9a_1=-2^{10} =-1024\\ \Rightarrow \sum_{n=1}^{10}a_n = {1\over 3}(a_1+2a_{10})= {1\over 3}(2-2048) =-682，故選\bbox[red, 2pt]{(D )}$$

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解：

$$C(0,0) \Rightarrow \cases{A(-3,3\sqrt 3)\\ B(-6,0) \\ D(2,2\sqrt 3)\\ E(4,0)} \Rightarrow \cases{N={1\over 2}(A+E) = ({1\over 2},{3\over 2}\sqrt 3) \\  M={1\over 2}(B+D)= (-2,\sqrt 3)} \Rightarrow \cases{\overline{CM}=\sqrt 7\\ \overline{CN}=\sqrt 7 \\ \overline{MN}=\sqrt 7} \\ \Rightarrow \triangle MNC為一正\triangle \Rightarrow \triangle MNC 面積={\sqrt 3\over 4}(\sqrt 7)^2= {7\over 4}\sqrt 3，故選\bbox[red, 2pt]{(C )}$$

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解：

$$令正方形邊長為m \Rightarrow {\overline{DC} \over \overline{AF}} = {\overline{DE} \over \overline{EA}} \Rightarrow {b-m \over b } ={m \over a} \Rightarrow m={ab\over a+b} ; \\ {無陰影面積\over 陰影面積} ={\triangle EDC + \triangle CBF \over 正方形abcd} ={{1\over 2}m(b-m)+ {1\over 2}m(a-m) \over m^2} ={a+b-2m \over 2m} \\={a+b \over 2m}-1 ={(a+b)^2 \over 2ab}-1 ={a^2+b^2 \over 2ab} ={c^2\over 2ab}，故選\bbox[red, 2pt]{(D)}$$

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解： $$a_2=3a_1-1 \Rightarrow a_3=3a_2-1 =3^2a_1-3-1 \Rightarrow a_4= 3a_3-1 =3^3a_1-3^2-3-1 \\ \Rightarrow a_n =3^{n-1}a_1-3^{n-2}-3^{n-3}-\cdots - 1 =3^{n-1}-3^{n-2}-3^{n-3}-\cdots - 1 \\ =3^{n-1}-(3^{n-2}+3^{n-3}+\cdots + 1) =3^{n-1}-{ 3^{n-1}-1\over 2} ={1\over 2}(3^{n-1}+1)，故選\bbox[red, 2pt]{(C)}$$

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解： $$\cases{F_n是偶數 \Rightarrow n是3的倍數\\ F_n是3的倍數數\Rightarrow n是4的倍數} \Rightarrow n是12的倍數\\ \Rightarrow n=12,24, 36,48,60,72,84，共七個，故選\bbox[red, 2pt]{(B)}$$

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解：

$$\cases{\overline{RU}=a \\\overline{RU} ={1\over 4}\overline{RS}} \Rightarrow \overline{SU}=3a \Rightarrow \overline {AU} = {1\over 2}\overline{RS} -\overline{RU}=a;\\令\overline{BT}=b，由於A、B、F皆為圓切點\Rightarrow \cases{\overline{UA}= \overline{UF}=a\\ \overline{BT}= \overline{TF}=b} \Rightarrow \overline{TR}=2a-b\\ 直角\triangle TUR \Rightarrow (a+b)^2 = a^2 + (2a-b)^2 \Rightarrow b={2\over 3}a \Rightarrow {\overline{RT} \over \overline{RQ}} ={2a-b \over 4a} = {2a-2a/3 \over 4a}={1\over 3}\\，故選\bbox[red, 2pt]{(A)}$$

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解：

$$\cases{\cos \angle ADC = {60^2+52^2-\overline{AC}^2 \over 2\times 60\times 52} ={  6304-\overline{AC}^2 \over 6240}\\ \cos \angle ABC = {25^2+ 39^2-\overline{AC}^2 \over 2\times 25\times 39} ={2146- \overline{AC}^2\over 1950}}; \\由於\cos \angle ADC = -\cos \angle ABC \Rightarrow {  6304-\overline{AC}^2 \over 6240}=-{2146- \overline{AC}^2\over 1950} \Rightarrow \overline{AC}^2=3136 \Rightarrow \overline{AC} =56  \\ \Rightarrow \cos \theta= {  6304-\overline{AC}^2 \over 6240}={  6304-3136 \over 6240} ={33\over 65} \Rightarrow \sin \theta ={56\over 65}\\正弦定理: {\overline{AC} \over \sin \theta} =2r = {56\over 56/65} =65 \Rightarrow 周長=2r\pi = 65\pi，故選\bbox[red, 2pt]{(D)}$$

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解：

$$作直線\overline{AF}交\overline{BC}於G點，見上圖；又\cases{\overline{AD}:\overline{DB} =1:3 \Rightarrow \cases{\overline{AD}=m,\overline{DB}=3m \\ \triangle ADF=a, \triangle BDF=3a} \\\overline{AE}:\overline{ED} =2:1 \Rightarrow \cases{\overline{AE}=2n,\overline{ED}=n \\ \triangle AEF=2b, \triangle CEF=b}};\\ \triangle BCF ={1\over 2}\triangle ABC \Rightarrow \overline{AF}:\overline{FG}=1:1 \Rightarrow \cases{\triangle BFG=\triangle AFB=4a \\ \triangle CFG =\triangle ACF=3b};\\ {\triangle ADE \over \triangle ABC} ={\overline{AD}\times \overline{AE} \over \overline{AB}\times \overline{AC}} \Rightarrow {a+2b \over 8a+6b}={m\times 2n \over 4m\times 3n}={1\over 6} \Rightarrow a=3b \\\Rightarrow {\triangle BDF \over \triangle ABC} ={3a \over 8a+6b}={9b \over 30b}={3\over 10}，故選\bbox[red, 2pt]{(B)}$$

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