114年台北市國中教甄聯招-資優數學詳解

 臺北市 114 學年度市立國民中學正式教師聯合甄選

特殊教育-資優數學科題本

貳、專業科目
選擇題（共 40 題，每題 1.75 分，共 70 分）

解答:$$\cases{AB= 10A+B\\ BA=10B+A} \Rightarrow 10A+B+10B+A=k^2 \Rightarrow 11(A+B)=k^2 \\ \Rightarrow (A,B)=(2,9), (3,8),(4,7),(5,6),( 6,5), (7,4),(8,3), (9,2)共8個，故選\bbox[red, 2pt]{(C)}$$

解答:$$52張牌中有四張3及四張8，最差的情形下，\\抽49張(剩3張8, 或剩3張3)一定會抽到一張3及一張8，故選\bbox[red, 2pt]{(D)}$$

解答:$$假設x^2-Ax+B=0的兩根分別為\alpha, \beta (\alpha\ge \beta)\Rightarrow x^2+Px+19=0的兩根分別為\alpha+1, \beta+1 \\ \Rightarrow (\alpha+1)(\beta+1)=19 為質數\Rightarrow \cases{\alpha+1=19\\ \beta+1=1} \Rightarrow \cases{\alpha =18\\ \beta=0} \Rightarrow \cases{A=\alpha+ \beta=18 \\B= \alpha\beta=0} \\ \Rightarrow A+B=18，故選\bbox[red, 2pt]{(A)}$$

解答:$$圓心(1,-1) \xrightarrow{左移兩單位} (-1,-1) \xrightarrow{向上三單位}(-1,2) \\ \Rightarrow 圓方程式:(x+1)^2+(y-2)^2=25，故選\bbox[red, 2pt]{(A)}$$

解答:

$$2 (四分之一圓-直角三角形) =2 \left( {1\over 4}\cdot 1^2\pi-{1\over 2}\cdot 1\cdot 1 \right) ={1\over 2}\pi-1，故選\bbox[red, 2pt]{(C)}$$

解答:$$\cases{k_1 \equiv {1 \mod 7} \\k_2 \equiv {5 \mod 7} \\k_3 \equiv {6 \mod 7} \\ k_4 \equiv {5 \mod 7} \\k_5 \equiv {2 \mod 7} \\ k_6 \equiv {0 \mod 7} \\k_7 \equiv {1 \mod 7} \\ \cdots} \Rightarrow 循環數6 \Rightarrow \cases{2013 \equiv { 3\mod 6}\\ 2014 \equiv {4 \mod 6}\\ 2015 \equiv {5 \mod 6} \\ 2016 \equiv {0 \mod 6}} \Rightarrow k_{2016} 可被7整除，故選\bbox[red, 2pt]{(D)}$$

解答:$$[1,8,9,7,6,3,9,2,1,3,4,7],[1,8,9,\dots \Rightarrow 循環數12 \Rightarrow \cases{前12項的和=60 \\前7項的和=43 \\前6項的和=34}\\ S_n=1000=60\times 16+40 \Rightarrow n=12\times 16+7=199，故選\bbox[red, 2pt]{(B)}$$

解答:$$f(x)=3x^4-16x^3+6x^2 +72x+100 \Rightarrow f'(x)=12x^3-48x^2+12x+72 =12(x+1)(x-2)(x-3) \\ \Rightarrow \begin{cases} f'(x) \gt 0& x\gt 3 ,-1\lt x\lt 2 \\ f'(x)\lt 0& x\lt -1, 2\lt x\lt 3\end{cases}  \Rightarrow f(x)遞增區域:-1\lt x\lt 2, x\gt 3，故選\bbox[red, 2pt]{(C)}$$

解答:$$f(x)= 2x^3-12x^2+30x+3 \Rightarrow f'(x) =6x^2-24x+30=6(x-2)^2+6 \gt 0\\ \Rightarrow \cases{f(x)為嚴格遞增函數\\ f(0)=3 \gt 0\\ f(-1)=-41\lt 0} \Rightarrow f(x)=0只有一實根, 實根在區間(-1,0)，故選\bbox[red, 2pt]{(B)}$$

解答:$$k(x)=\cos x^3 \Rightarrow k(-x)=\cos(-x)^3 =\cos(-x^3)=\cos x^3 \Rightarrow k(-x)=k(x) \Rightarrow k(x)為偶數函\\，故選\bbox[red, 2pt]{(D)}$$

解答:$$f(x)={1\over (x-1)^2} \Rightarrow x\ne 1, x\in \mathbb R \\ f_4(x)= \sqrt{x-1} \Rightarrow x-1\ge 0 \Rightarrow x\ge 1 \Rightarrow f_4(x)定義域與f(x)定義域不同，故選\bbox[red, 2pt]{(D)}$$

解答:$$\int_{-88}^{88} x^{11}\,dx + \int_{-10}^{10}2|x|\,dx + \int_{-1}^1 3x^2\,dx = 0+4 \int_0^{10}x\,dx +6\int_0^1 x^2\,dx \\=4 \left. \left[ {1\over 2}x^2 \right] \right|_0^{10}+ 6 \left. \left[ {1\over 3}x^3 \right] \right|_0^1=4\cdot 50+6\cdot {1\over 3} =202，故選\bbox[red, 2pt]{(B)}$$

解答:$$(A)\times: \cases{\begin{bmatrix}1& 0\\0& 0 \end{bmatrix} \begin{bmatrix}0& 1\\ 0& 0 \end{bmatrix} =\begin{bmatrix}0&1 \\ 0& 0 \end{bmatrix} \\ \begin{bmatrix}0& 1\\ 0& 0 \end{bmatrix} \begin{bmatrix}1& 0\\ 0& 0 \end{bmatrix} = \begin{bmatrix}0& 0\\0& 0 \end{bmatrix}} \Rightarrow AB\ne BA \\(B) \times: \begin{bmatrix}0& 1\\ 0& 0 \end{bmatrix} \begin{bmatrix}1& 0\\ 0& 0 \end{bmatrix} = \begin{bmatrix}0& 0\\0& 0 \end{bmatrix}, 但\begin{bmatrix}0& 1\\ 0& 0 \end{bmatrix}\ne 0, \begin{bmatrix}1& 0\\ 0& 0 \end{bmatrix} \ne 0\\ (D) \times: A0=0 \ne I \Rightarrow 零矩陣不是任何矩陣的反矩陣\\，故選\bbox[red, 2pt]{(C)}$$

解答:$$K的數字總和=45\times 19+1= 856 =95\times 9+1，故選\bbox[red, 2pt]{(B)}$$

解答:$$\sqrt x+ \sqrt y=\sqrt{336} =4 \sqrt{21} \Rightarrow \cases{x=21m^2\\ y=21 n^2} \Rightarrow m+n=4, m,n\in \mathbb Z\\ \cases{m=0\\ n=4} \Rightarrow (m,n)=(0,4),(1,3),(2,2), (3,2), (4,0) ,共5組解，故選\bbox[red, 2pt]{(D)}$$

解答:$$18x+5y=48 \Rightarrow \cases{x=1 \Rightarrow y=6\\ x=2 \Rightarrow y=12/5 \not \in \mathbb N \\ x=3 \Rightarrow y\lt 0} \Rightarrow 只有一組解，故選\bbox[red, 2pt]{(B)}$$

解答:$$ \left( x^2+{1\over x} \right)^{12} = \sum_{n=0}^{12} {12\choose n} x^{2n} \cdot {1\over x^{12-n}} \Rightarrow 2n=12-n\Rightarrow n=4時, 即為常數項, \\係數={12\choose 4} =495，故選\bbox[red, 2pt]{(C)}$$

解答:$$f(x)=a(x-4)^2-11 \Rightarrow \cases{圖形凹向上\Rightarrow a\gt 0\\ 兩根之和=-b/a \gt 0 \Rightarrow b\lt 0\\ 兩根之積=c/a\lt 0 \Rightarrow c\lt 0} \Rightarrow 只有a\gt 0，故選\bbox[red, 2pt]{(A)}$$

解答:

$$假設\cases{正方形邊長為a \\ d(A,\overline{DG})=h} \Rightarrow \cases{\triangle BDE=3=a\cdot \overline{BE}/2 \Rightarrow \overline{BE}=6/a\\ \triangle CFG=2 =a\cdot \overline{CF}/2 \Rightarrow \overline{CF}=4/a\\ \triangle ADG=4 =a\cdot h/2 \Rightarrow h=8/a} \\ \Rightarrow \triangle ABC={1\over 2}\cdot \overline{BC}\cdot d(A,\overline{BC}) =\triangle ADG+\triangle BDE+\triangle CFG + \square DEFG \\ \Rightarrow {1\over 2} \left( a+{10\over a} \right)\cdot \left( a+{8\over a} \right) =4+3+2+a^2 \Rightarrow a^2+18+{80\over a^2}=18+2a^2\\ \Rightarrow a^4=80 \Rightarrow \square DEFG=a^2=\sqrt{80}=4\sqrt 5，故選\bbox[red, 2pt]{(A)}$$

解答:

$$假設\cases{\overline{PB}= a \\  \overline{QC}=b} \Rightarrow \cases{\overline{AP} =8-a\\ \overline{AQ}=7-b} \Rightarrow \cases{\triangle APQ周長=15-(a+b)+ \overline{PQ}\\ 梯形PBCQ周長=9+a+b+ \overline{PQ}} \\ \Rightarrow 2(a+b)=6 \Rightarrow a+b=3 \cdots(1)\\ 又\overline{PQ} \parallel \overline{BC}  \Rightarrow {\overline{AP} \over \overline{AB}} ={\overline{AQ} \over \overline{AC}} \Rightarrow {8-a\over 8} ={7-b\over 7} \Rightarrow 7a=8b \Rightarrow a={8b\over 7} 代入(1) \Rightarrow b={7\over 5} \Rightarrow a={8\over 5} \\ \Rightarrow  {\overline{AP} \over \overline{AB}} ={\overline{PQ} \over \overline{BC}} \Rightarrow {8-8/5\over 8} ={\overline{PQ} \over 9} \Rightarrow \overline{PQ} ={36\over 5}，故選\bbox[red, 2pt]{(D)}$$

解答:

$$假設\cases{\overline{GA}= a \\  \overline{GB}=b \\ \overline{GC}= c \\  \overline{GD}=d \\ \overline{GE}= e \\  \overline{GF}=f } \Rightarrow \cases{ \triangle ABG\sim \triangle EDG\\ \triangle BCG \sim \triangle FEG\\ \triangle AFG \sim \triangle CDG} \Rightarrow \cases{a/e= 1/7 \cdots(1)\\  c/e=3/9 \cdots(2)\\ a/c=   \overline{AF}/5 \dots(3)} \\ 式(1) \Rightarrow e=7a代入(2) \Rightarrow c/7a=3/9 \Rightarrow a/c=3/7代入(3) \Rightarrow 3/7=\overline{AF}/5 \Rightarrow \overline{AF}=15/7\\，故選\bbox[red, 2pt]{(C)}$$

解答:$$事件A:前二次和等於第三 \Rightarrow A =\{(1,1,2), (1,2,3), (1,3,4), (1,4,5),(1,5,6),\\\qquad (2,1,3), (2,2,4),(2,3,5), (2,4,6), (3,1,4), (3,2,5), (3,3,6), (4,1,5),(4,2,6), (5,1,6) \} \\事件B=事件A中符合三次點數至少有一次為2 \Rightarrow B=\{(1,1,2), (1,2,3), \\\qquad (2,1,3),(2,2,4),(2,3,5), (2,4,6),(3,2,5), (4,2,6)\} \\ \Rightarrow {n(B)\over n(A)} ={8\over 15}，故選\bbox[red, 2pt]{(C)}$$

解答:$$(A)\times: a=0 \Rightarrow \cases{x-2=1 \Rightarrow x=3\\ x-2=-1 \Rightarrow x=1} \Rightarrow 只有2個解\\ (B)\bigcirc: a=1 \Rightarrow \cases{|x-2|-1=1 \Rightarrow \cases{x-2=2 \Rightarrow x=4\\ x-2=-2 \Rightarrow x=0}\\ |x-2|-1=-1 \Rightarrow x=2} \Rightarrow 有3個解\\ (C)\times: a=2 \Rightarrow \cases{|x-2|-1=2 \Rightarrow \cases{x-2=3 \Rightarrow x=5\\ x-2=-3 \Rightarrow x=-1}\\ |x-2|-1=-2  \Rightarrow 無解} \Rightarrow 只有2個解 \\(D)\times: a=3 \Rightarrow \cases{|x-2|-1=3 \Rightarrow x=6,-2\\ |x-2|-1=-3 \Rightarrow 無解}\Rightarrow 只有2個解 \\，故選\bbox[red, 2pt]{(B)}$$

解答:$$\cases{X=x-1\\ Y=y-1 \\ Z=z-2} \Rightarrow X+Y+Z=14的非負整數解個數=H^3_{14} =C^{16}_{14} =120，故選\bbox[red, 2pt]{(A)}$$

解答:$$\cases{P(取A袋黑球, B袋白球)= (2/6) \cdot (3/8)=1/8\\ P(取A袋白球,B袋黑球) =(4/6) \cdot (5/8) =5/12} \Rightarrow {1\over 8}+{5\over 12} ={13\over 24}，故選\bbox[red, 2pt]{(D)}$$

解答:

$$\triangle PQC為正\triangle \Rightarrow 假設\overline{DP} =\overline{BQ} =a \Rightarrow \cases{\overline{PA} =\overline{AQ} =1/\sqrt 2\\ \overline{CD} =\sqrt{\overline{CP}^2-\overline{DP}^2} = \sqrt{1-a^2}} \\ \Rightarrow \overline{AD} =\overline{CD} \Rightarrow a+{1\over \sqrt 2} =\sqrt{1-a^2} \Rightarrow \left( a+ {1\over \sqrt 2} \right)^2=1-a^2\Rightarrow a={\sqrt 6-\sqrt 2\over 4} \\ \Rightarrow \square ABCD面積=1-a^2 =1- \left( {\sqrt 6-\sqrt 2\over 4} \right)^2= {2+\sqrt 3\over 4}，故選\bbox[red, 2pt]{(C)}$$

解答:$$5人入座有4!=24種方法, 剩下5個間隔插入兩人有P^5_2=20種插入方法\\ 一共24\times 20=480種入座方式，故選\bbox[red, 2pt]{(C)}$$

解答:$$x^2-3xy +y^2=1 \Rightarrow 2x-3y-3xy'+2yy'=0 \Rightarrow y'={3y-2x\over 2y-3x} \Rightarrow y'(3,1)={3-6\over 2-9} ={3\over 7} \\ \Rightarrow 切線: y={3\over 7}(x-3)+1 ={3\over 7}x-{2\over 7}，故選\bbox[red, 2pt]{(D)}$$

解答:$$f(x)={x\over x^2-1} \Rightarrow f(-x)=-{x\over x^2-1} =-f(x) \Rightarrow f(x)為奇函數\Rightarrow 不對稱y軸，故選\bbox[red, 2pt]{(D)}$$

解答:$$假設\cases{\overline{AP}=p\\ \overline{AQ}=q} \Rightarrow {\triangle APQ\over\triangle ABC} ={pq\over 23\cdot 21} ={1\over 2} \Rightarrow pq={483\over 2} \\ 內心I\in \overline{PQ} \Rightarrow \overrightarrow{AI} ={\overline{AC} \over \triangle ABC周長}\cdot \overrightarrow{AB}+ {\overline{AB} \over \triangle ABC周長}\cdot \overrightarrow{AC} ={21\over 66}\cdot \overrightarrow{AB} +{23\over 66}\cdot \overrightarrow{AC}\\ ={21\over 66}\cdot {23\over p}\cdot \overrightarrow{AP} +{23\over 66}\cdot {21\over q} \cdot \overrightarrow{AQ} \Rightarrow {21\over 66}\cdot {23\over p}+{23\over 66}\cdot {21\over q}=1 \Rightarrow {483\over 66} \left( {1\over p}+{1\over q} \right)=1 \\\Rightarrow {1\over p}+{1\over q}={66\over 483} \Rightarrow{p+q\over pq} ={p+q\over 483/2} ={66\over 483} \Rightarrow p+q=33，故選\bbox[red, 2pt]{(D)}$$

解答:$$\lim_{x\to 0} {\tan 3x\over 2x} =\lim_{x\to 0} {(\tan 3x)'\over (2x)'} =\lim_{x\to 0} {3\sec^2 3x\over 2} ={3\over 2}，故選\bbox[red, 2pt]{(A)}$$

解答:$$\textbf{Case I }n=2k為偶數\Rightarrow n^5=32k^5為4的倍數\Rightarrow n除以4的餘數=0\\ \textbf{Case II }n=2k+1為奇數 \Rightarrow n^2=4(k^2+k)+1 \Rightarrow n^2 \equiv1 {\mod 4} \Rightarrow n^5 \equiv n {\mod 4} \\ \Rightarrow \cases{n=1 \Rightarrow n^5 \equiv {1\mod 4} \\n=2 \Rightarrow n^5 \equiv {0\mod 4} \\n=3 \Rightarrow n^5 \equiv {3\mod 4} \\n=4 \Rightarrow n^5 \equiv {0\mod 4} \\n=5 \Rightarrow n^5 \equiv {1\mod 4} \\n=6 \Rightarrow n^5 \equiv {0\mod 4} \\ \cdots} \Rightarrow 循環數4且 (1^5+2^5+3^5+4^5) \equiv 0 {\mod 4} \\ \Rightarrow 1^5+2^5+ \cdots +100^5 \equiv 0\times 25 {\mod 4 =0}，故選\bbox[red, 2pt]{(A)}$$

解答:

$$假設\cases{A(0,0) \\\overline{AB}=4} \Rightarrow \cases{B(2, 2\sqrt 3) \\C(4,0) } \Rightarrow \cases{P=(A+C)/2=(2,0) \\Q=(B+C)/2=(3,\sqrt3 )} \Rightarrow O=(2P+B)/3=(2,{2\over 3}\sqrt 3) \\ \Rightarrow \cases{圓半徑\overline{OP} =2\sqrt 3/3 \\ \overline{PQ}=2} \Rightarrow \cases{\triangle PQR={\sqrt 3\over 4}\cdot 2^2\\ 圓O=4\pi/3\\ \triangle ABC={\sqrt 3\over 4}\cdot 4^2} \Rightarrow \triangle PQR:圓O:\triangle ABC =\sqrt 3:4\pi/3:4\sqrt 3 \\ =1:{4\over 3\sqrt 3}\pi:4，故選\bbox[red, 2pt]{(D)}$$

解答:$$\cases{L_1=\{(s+2,-3s,-s+4), s\in \mathbb R\} \\L_2=\{(2t+1,-t-2,3t), t\in \mathbb R\}} \Rightarrow \cases{s+2=2t+1\\ -3s=-t-2\\ -s+4=3t} \Rightarrow s=t=1\\ \Rightarrow 交點(3,-3,3)，故選\bbox[red, 2pt]{(D)}$$

解答:$$f(x)=a(x-1)(x+2)+b(x+2)+ c(x-1)^2=1 \Rightarrow \cases{f(1)=3b=1 \\ f(-2)=9c=1 \\ f(0)=-2a+2b+c=1}\\ \Rightarrow   \cases{a=-1/9\\ b=1/3\\ c=1/9} \Rightarrow a+b+c={1\over 3}，故選\bbox[red, 2pt]{(C)}$$

解答:$$\cases{\alpha+\beta+ \gamma=-a\\ \alpha\beta+ \beta\gamma+ \gamma\alpha =b\\ \alpha \beta\gamma=-c} \Rightarrow \cases{(1/\alpha)+(1/\beta)+ (1/\gamma)= ( \alpha\beta+ \beta\gamma+ \gamma\alpha)/(\alpha \beta\gamma) =-b/c\\ (1/\alpha)\cdot (1/\beta)\cdot  (1/\gamma)= 1/(\alpha\beta \gamma) =-1/c} \\ \Rightarrow  {1\over \alpha},  {1\over \beta},  {1\over \gamma}為三根的方程式:x^3+{b\over c}x^2 + \square x+{1\over c}=0 \Rightarrow cx^3+bx^2+c\square x+1=0，故選\bbox[red, 2pt]{(A)}$$

解答:$$假設三邊長為a,b,c, 需滿足\cases{a+b+c=30\\ 1\le a\lt b\lt c\\ a+b\gt c} \Rightarrow \cases{a+b=30-c\gt c \Rightarrow c\lt 15 \\ 3c\gt a+b+c=30 \Rightarrow c\gt 10} \\ \Rightarrow \cases{c=14 \Rightarrow (a,b) =(3,13),(4,12),(5,11), (6,10),(7,9) \\ c=13 \Rightarrow (a,b)=(5,12),(6,11),(7,10),(8,9)\\c=12 \Rightarrow (a,b)=(7,11), (8,10) \\c=11 \Rightarrow (a,b)=(9,10)} \Rightarrow 共12組，故選\bbox[red, 2pt]{(A)}$$

解答:

$$M=\overline{AD}中點\Rightarrow \overline{AM} =\overline{MD}=a,\;又 假設\overline{CN}=c \Rightarrow \overline{BN}=2a-c \\\cases{\overline{MP} 為對折軸\Rightarrow \overline{MO} =\overline{MD} =a\\ \overline{NP} 為對折軸\Rightarrow \overline{NO} =\overline{NC} =c } \Rightarrow \overline{SN} 為對折軸\Rightarrow \overline{BN} =\overline{MN}  \Rightarrow 2a-c=a+c \Rightarrow a=2c$$

$$MQNP為矩形 \Rightarrow 假設\overline{MQ} = \overline{NP}=b \Rightarrow \overline{BQ}=b \Rightarrow \overline{AB}= \sqrt{\overline{MB}^2-\overline{AM}^2} =2\sqrt{b^2-c^2} \\又\cases{\overline{MP} =\overline{QN} = \sqrt{\overline{MN}^2-\overline{QM}^2} = \sqrt{9c^2-b^2} \Rightarrow \overline{DP}= \sqrt{5c^2- b^2}\\ \overline{PC} = \sqrt{\overline{NP}^2-\overline{NC}^2} =\sqrt{b^2- c^2}} \\ \Rightarrow \overline{AB} =\overline{DP}+ \overline{PC} \Rightarrow 2\sqrt{b^2-c^2} =\sqrt{5c^2-b^2}+ \sqrt{b^2-c^2} \Rightarrow b^2-c^2=5c^2-b^2 \Rightarrow b^2=3c^2 \\ \Rightarrow b=\sqrt 3c \Rightarrow \cases{\overline{AB} =2\sqrt{2c^2} =2\sqrt 2c \\ \overline{MP} =\sqrt 6c} \Rightarrow \cases{矩形MPNQ面積= \sqrt 6c\times \sqrt 3c\\ 矩形ABCD面積=4c\times 2\sqrt 2c} \\ \Rightarrow {MPNQ\over ABCD} ={\sqrt{18} \over 8\sqrt 2}={3\over 8}，故選\bbox[red, 2pt]{(C)}$$

解答:$$(x^2+25x+52)^2=9(x^2+25x+80) \Rightarrow x^4+ \cdots +(52^2-9\cdot 80)=0 \\\Rightarrow  四根之積=常數項=1984，故選\bbox[red, 2pt]{(D)}$$

解答:

$$過H作垂線分別交\overleftrightarrow{AD} 及\overleftrightarrow{BG}於P及Q \\ EFGH為正方形\Rightarrow ABQP正方形\Rightarrow \cases{A(0,8) \\B(0,0) \\C(11,0) \\D(6,8) \\Q(8,0)\\ P(8,8)}  \\ \triangle DPH \sim \triangle CQH (AAA)  \Rightarrow {\overline{DH} \over \overline{CH}} = {\overline{DP} \over \overline{CQ}} ={2\over 3}，故選\bbox[red, 2pt]{(B)}$$

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