114年台北市國中教甄聯招-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

貳、專業科目
選擇題（共 40 題，每題 1.75 分，共 70 分）

解答：$$(10A+B)+(10B+A)=11(A+B)= k^2 \Rightarrow A+B=11 \Rightarrow (A,B)=(2,9), (3,8), \dots, (9,2) \\ \Rightarrow 共8組，故選\bbox[red, 2pt]{(C)}$$


解答：$$全部52張牌，其中有4張3及4張8，因此抽出m張後剩下的牌不可以有4張3或4張8，\\因此m\ge 49，故選\bbox[red, 2pt]{(D)}$$

解答：$$假設x^2+Px+19=0的兩個解為\alpha, \beta \Rightarrow \alpha\beta=19 \Rightarrow \cases{\alpha=1\\ \beta=19 } \Rightarrow P=-(\alpha+\beta)=-20\\ 又x^2-Ax+B=0的兩個解為\alpha+1,\beta+1\Rightarrow \cases{B=(\alpha+1)(\beta+1)=2\times 20=40\\ A=-(\alpha+1+\beta+1) =-22} \\ \Rightarrow A+B=40-22=18，故選\bbox[red, 2pt]{(A)}$$

解答：$$(x-1)^2+(y+1)^2=25  \xrightarrow{向左移2單位}(x-1+2)^2+(y+1)^2=25 \Rightarrow (x+1)^2+(y+1)^2=25 \\ \xrightarrow{向上移3單位} (x+1)^2+(y+1-3)^2=25 \Rightarrow (x+1)^2+(y-2)^2=25，故選\bbox[red, 2pt]{(A)}$$


解答：

$$此題相當於求兩單位圓垂疊區域，即上圖著色面積的兩倍，也就是\\ 2\left({1\over 4}\pi- {1\over 2} \right) ={\pi\over 2}-1，故選\bbox[red, 2pt]{(C)}$$

解答：$$\cases{k_1= 1\text{ mod }7 \\k_2= 4\text{ mod }7 \\k_3= 6\text{ mod }7 \\k_4= 5\text{ mod }7 \\k_5= 2 \text{ mod }7 \\k_6= 0\text{ mod }7 \\k_7= 1\text{ mod }7 \\ \cdots} \Rightarrow 循環數為6 \Rightarrow 2016= 0 \text{ mod }6 \Rightarrow k_{2016} =0 \text{ mod }7，故選\bbox[red, 2pt]{(D)}$$

解答：$$\begin{array}{c|cccccccccccc|cc} n& 1& 2& 3& 4& 5& 6&7&8&9&10 &11&12&13 &14 &15\\\hline a_n & 1& 8& 9 & 7&6&3&9&2&1&3&4&7&1&8 &9\end{array} \\ \Rightarrow 循環數為12 \Rightarrow S_{12}=60 \Rightarrow 1000=60\times 16+40 \\ 又a_1+a_2+\cdots+a_7=43 \Rightarrow 最小的n=12\times 16+7= 199，故選\bbox[red, 2pt]{(B)}$$

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

解答：$$f(x)=2x^3-12x^2+30x+3 \Rightarrow f'(x)=6x^2-24x+30= 6(x^2-4x+5) \\ \Rightarrow f'(x)=0 無實根\Rightarrow y=f(x)圖形為左下右上, 僅有一實根，故選\bbox[red, 2pt]{(B)}$$

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

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

解答：$$f_1(x)=x^{11}為奇函數\Rightarrow \int_{-88}^{88} x^{11}\,dx =0 \\ 因此\int_{-88}^{88} x^{11}\,dx+ \int_{-10}^{10} 2|x|\,dx+ \int_{-1}^1 3x^2\,dx =\int_{-10}^{10} 2|x|\,dx+\int_{-1}^1 3x^2\,dx \\= 4 \int_{0}^{10}x\,dx +2 \int_0^1 3x^2\,dx =4\times 50 +2\times 1=202，故選\bbox[red, 2pt]{(B)}$$

解答：$$假設A= \begin{bmatrix} 1&0\\0&0 \end{bmatrix}, B=  \begin{bmatrix} 0& 1\\0&0 \end{bmatrix}\\ (A) \times: \cases{AB= \begin{bmatrix} 0&1\\0&0 \end{bmatrix} \\BA= \begin{bmatrix} 0 &0 \\0&0 \end{bmatrix} } \Rightarrow AB\ne BA \\(B)\times: BA=0, 但A\ne0且 B\ne 0\\(D)\times :反矩陣不存在\\，故選\bbox[red, 2pt]{(C)}$$

解答：$$\cases{個位數字的總和=10(1+2+ \cdots+ 9)=450 \\ 十位數字的總和=10(1+2+ \cdots+ 9)=450\\ 百位數字的總和=1} \Rightarrow 各位數字的總和=901 = 1\text{ mod }9，故選\bbox[red, 2pt]{(B)}$$

解答：$$\sqrt x+ \sqrt y =\sqrt{336} =4\sqrt{21} \\ 取\cases{\sqrt x=a\sqrt{21} \\ \sqrt y= b\sqrt{21}} \Rightarrow a+b=4 有H^2_4=C^5_4=5組非負整數解，故選\bbox[red, 2pt]{(D)}$$

解答：$$18x+5y=48 \Rightarrow \cases{x=1 \Rightarrow y=6 \\ x=2 \Rightarrow \times\\ x=3 \Rightarrow \times } \Rightarrow 只有一組正整數解，故選\bbox[red, 2pt]{(B)}$$

解答：$$\left(x^2+{1\over x} \right)^{12} =\sum_{n=0}^{12}C^{12}_n x^{2n}\cdot x^{n-12} \Rightarrow 常數項發生在n=4時, 係數=C^{12}_4=495，故選\bbox[red, 2pt]{(C)}$$

解答：$$頂點(4,-11)在x軸下方且與x軸交於兩點，表示圖形凹向上(a\gt 0)，y截距為負值(c\lt 0)\\ 又頂點在第四象限代表兩根之和為正值 \Rightarrow -{b\over a}\gt 0 \Rightarrow b\lt 0，故選\bbox[red, 2pt]{(A)}$$

解答：

$$假設正方形邊長為a \Rightarrow 正方形面積=a^2 \Rightarrow \triangle ABC面積=a^2+4+3+2=a^2+9 \\ 又\cases{\triangle BDE=3 \Rightarrow \overline{BE}= 6/a\\ \triangle CFG=2 \Rightarrow \overline{CF}=4/a} \Rightarrow \overline{BC}= {6\over a}+a+{4\over a} =a+{10\over a} \\ {\triangle ADG\over \triangle ABC} ={\overline{DG}^2 \over \overline{BC}^2} \Rightarrow {4\over a^2+9} ={a^2\over (a+{10\over a})^2} \Rightarrow a^6+5a^4-80a^2-400=0 \\ \Rightarrow (a^2+5)(a^4-80)=0 \Rightarrow a^2=\sqrt{80}=4\sqrt 5，故選\bbox[red, 2pt]{(A)}$$

解答：$$\overline{PQ}=a \Rightarrow {\overline{AP} \over \overline{AB}} = {\overline{PQ} \over \overline{BC}} = {\overline{AQ} \over \overline{AC}}  \Rightarrow {\overline{AP} \over 8} = {a \over 9} = {\overline{AQ} \over 7}  \Rightarrow \cases{\overline{AP} =8a/9\\ \overline{AQ} =7a/9} \\ \Rightarrow \cases{\triangle APQ 周長=8a/9+ 7a/9+ a= 8a/3\\ PBCQ周長=a+8-(8a/9)+9+7-7a/9 =a+24-5a/3} \\ \Rightarrow {8\over 3}a=a+ 24-{5\over 3}a \Rightarrow a={36\over 5}，故選\bbox[red, 2pt]{(D)}$$

解答：$$\triangle ABG \sim \triangle EDG (AAA) \Rightarrow {\overline{AG} \over \overline{EG}} ={\overline{BG} \over \overline{DG}} ={\overline{AB} \over \overline{DE}}={1\over 7} \Rightarrow \cases{\overline{BG}=k\\ \overline{DG}=7k\\ \overline{AG} =m \\ \overline{EG} =7m} \\ 同理\triangle BCG\sim \triangle FEG \Rightarrow {\overline{BG} \over \overline{FG}} ={\overline{CG} \over \overline{EG}} ={\overline{BC} \over \overline{EF}}={3\over 9} \Rightarrow \cases{\overline{FG}=3k\\ \overline{CG} =7m/3} \\ 又\triangle AFG \sim \triangle CDG \Rightarrow {\overline{AG} \over \overline{CG}} ={\overline{FG} \over \overline{DG}} ={\overline{AF} \over \overline{CD}} \Rightarrow {m\over 7m/3} ={3k\over 7k} ={\overline{AF}\over 5} \Rightarrow \overline{AF}={15\over 7}，故選\bbox[red, 2pt]{(C)}$$

解答：$$a+b=c \Rightarrow (a,b,c)=(1,1,2), (1,2,3),( 2,1,3),( 1,3,4),(3,1,4),(2,2,4),\\\qquad (1,4,5),(4,1,5),(2,3,5),(3,2,5),(1,5,6),(5,1,6),( 2,4,6),(4,2,6),(3,3,6) \\ \Rightarrow 共有15種情形，其中至少1個2的有8種，因此機率={8\over 15}，故選\bbox[red, 2pt]{(C)}$$

解答：$$(A)a=0 \Rightarrow ||x-2|-1|=0 \Rightarrow |x-2|=1 \Rightarrow \cases{x-2=1\\ x-2=-1} \Rightarrow  兩個整數解\\ (B) a=1 \Rightarrow ||x-2|-1|=1 \Rightarrow \cases{|x-2|-1=1 \Rightarrow |x-2|=2 \Rightarrow \cases{x-2=2 \Rightarrow x=4\\ x-2=-2  \Rightarrow x=0}\\ |x-2|-1=-1 \Rightarrow |x-2|=0 \Rightarrow x=2}  \\\qquad \Rightarrow  3個整數解\\ (C)a=2 \Rightarrow ||x-2|-1|=2 \Rightarrow \cases{|x-2|-1=2 \Rightarrow |x-2|=3 \Rightarrow \cases{x-2=3 \Rightarrow x=5\\ x-2=-3 \Rightarrow x=-1}\\ |x-2|-1=-2 \Rightarrow |x-2|=-1無解}\\\qquad  \Rightarrow  2個整數解 \\(D) a=3\Rightarrow ||x-1|-1|=3 \Rightarrow \cases{|x-1|-1=3 \Rightarrow |x-1|=4 \Rightarrow \cases{x-1=4 \Rightarrow x=5\\ x-1=-4 \Rightarrow x=-3}\\ |x-1|-1=-3 \Rightarrow |x-1|=-2 無解} \\\qquad \Rightarrow  2個整數解\\，故選\bbox[red, 2pt]{(B)}$$

解答：$$x+y+z=18 \Rightarrow (x-1)+(y-1)+(z-2)=14 \Rightarrow 有H^3_{14} =C^{16}_{14}= 120組整數解，故選\bbox[red, 2pt]{(A)}$$

解答：$$\cases{A袋取白球且B袋取黑球的機率={4\over 6}\cdot {5\over 8} ={5\over 12} \\A袋取黑球且B袋取白球的機率={2\over 6}\cdot {3\over 8} ={1\over 8} } \\ \Rightarrow 一黑一白的機率={5\over 12}+{1\over 8}={13\over 24}，故選\bbox[red, 2pt]{(D)}$$

解答：$$\triangle APQ為等腰直角三角形 \Rightarrow \angle APQ=45^\circ \Rightarrow \angle CPQ=180^\circ-60^\circ-45^\circ=75^\circ\\ \Rightarrow \overline{CD}= \overline{CP}\sin 75^\circ = \sin 75^\circ =\sin(45^\circ +30^\circ) =\sin 45^\circ \cos 30^\circ+ \sin 30^\circ \cos 45^\circ ={\sqrt 3+1\over 2\sqrt 2} \\ \Rightarrow 正方形ABCD面積=\left( {\sqrt 3+1\over 2\sqrt 2} \right)^2={4+2\sqrt 3\over 8}={2+\sqrt 3\over 4}，故選\bbox[red, 2pt]{(C)}$$

解答：$$\cases{7人環狀排列數={7!\over 7}=720 \\6人環狀排列數={6!\over 6}=120} \Rightarrow 720-2\times 120=480，故選\bbox[red, 2pt]{(C)}$$

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

解答：$$y=f(x)={x\over x^2-1} \Rightarrow f(-x)\ne f(x) \Rightarrow f(x)不是偶函數,即沒有對稱y 軸，故選\bbox[red, 2pt]{(D)}$$

解答：$$過內心等分三角形面積的直線同時也等分三角形周長, \href{https://matrix67.com/blog/archives/5313}{參考資料}\\ \triangle ABC周長=23+22+21= 66 \Rightarrow \overline{AP}+\overline{AQ} ={66\over 2} =33，故選\bbox[red, 2pt]{(D)}$$

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：$$\cases{1^5 = 1\text{ mod }4 \\2^5 = 0\text{ mod }4 \\3^5 = 3\text{ mod }4 \\4^5 = 0\text{ mod }4 \\5^5 = 1\text{ mod }4 \\6^5 = 0\text{ mod }4 \\7^5 = 3\text{ mod }4 \\8^5 = 0\text{ mod }4 \\ \cdots} \Rightarrow 循環數4且四餘數相加為4的倍數 \Rightarrow \sum_{k=1}^{100} k^5 =0 \text{ mod 4}，故選\bbox[red, 2pt]{(A)}$$

解答：

$$O是圓心, 也是\triangle ABC及\triangle PQR的重心\\ 假設\overline{CD}=1 \Rightarrow \overline{BD}=\sqrt 3 \Rightarrow \cases{\overline{OD}=\sqrt 3/3 =\overline{OP}\\ \triangle ABC=\sqrt 3} \Rightarrow \cases{ \overline{PE}=\sqrt 3/2\\ 圓面積=\pi/3} \Rightarrow \overline{ER}=1/2 \\\Rightarrow \triangle PQR=\sqrt 3/4  \Rightarrow \triangle PQR:圓O:\triangle ABC={\sqrt 3\over 4}:{\pi\over 3}:\sqrt 3=1:{4\pi\over 3\sqrt 3}:4，故選\bbox[red, 2pt]{(D)}$$

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

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

解答：$$\cases{\alpha+\beta+ \gamma= -a\\ \alpha\beta+ \beta\gamma + \gamma\alpha=b\\ \alpha\beta\gamma =-c} \Rightarrow \cases{{1\over \alpha}+{1\over \beta}+{1\over \gamma }={b\over -c} \\ {1\over \alpha\beta}+{1\over \beta\gamma}+{1\over \gamma \alpha}={a\over c} \\ {1\over \alpha \beta\gamma }=-{1\over c}}  \Rightarrow x^3+{b\over c}x^2+{ a\over c}x+{1\over c}=0 \\ \Rightarrow cx^3+bx^2+ax+1=0，故選\bbox[red, 2pt]{(A)}$$

解答：$$需符合\cases{a+b+c=30 \\ a\ne b\\ b\ne c\\ c\ne a\\ a+b\gt c\\ b+c\gt a\\ c+a\gt b},a,b,c\in \mathbb N\\符合要求的三邉長為(3,13,14),  (4,12,14), (5,11,14), (5,12,13), (6,10,14),(6,11,13), \\(7,9,14),(7,10,13), (7,11,12),  (8,9,13),(8,10,12), (9,10,11)\\ 合計共12個，故選\bbox[red, 2pt]{(A)}$$

解答：

$$假設\cases{\overline{AM}=a\\ \overline{BR}=b \\\overline{BN} =c}, \overline{RN}為折線 \Rightarrow \cases{\overline{RM}=b\\ \overline{MN}=c} ,又\cases{\overline{MP}為折線\Rightarrow \cases{\overline{MO}=a\\ \overline{OP}=\overline{PD}=d} \\ \overline{NP}為折線\Rightarrow \cases{\overline{CP}=\overline{OP}=d\\ \overline{NC} = \overline{NO}=\overline{MN}-\overline{MO}=c-a}} \\ 最後\overline{MB}為折線\Rightarrow \overline{BO}=\overline{AB}=\overline{CD}=2d, 詳如上圖；\\ \overline{AD} =\overline{BC} \Rightarrow 2a=2c-a \Rightarrow c={3\over 2}a \Rightarrow \cases{\overline{BN}=3a/2\\ \overline{NO}=\overline{NC}=a/2} \\ \angle C=90^\circ \Rightarrow \overline{BP}^2=\overline{BC}^2+\overline{CP}^2 \Rightarrow 9d^2=4a^2+d^2 \Rightarrow d={a\over \sqrt 2} \\ \Rightarrow \cases{\overline{MP}^2 =a^2+d^2 =3a^2/2\\ \overline{NP}^2=(a/2)^2+d^2 =3a^2/4} \Rightarrow \cases{ \overline{MP}= \sqrt 3a/\sqrt 2\\ \overline{NP} =\sqrt 3a/2} \Rightarrow MQNP面積={3a^2\over 2\sqrt 2} \\ \Rightarrow {MQNP\over ABCD} ={3a^2/2\sqrt 2\over 2a\cdot 2d=2\sqrt 2a^2  } ={3\over 8}，故選\bbox[red, 2pt]{(C)}$$

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

解答：

$$假設\overline{AE}=a, 由於\triangle FAE \cong \triangle GBF \Rightarrow \overline{BF}=a \Rightarrow \overline{AF}=\overline{BG}=8-a\\ 因此可假設\cases{B(0,0)\\ A(0,8)\\ C(11,0)\\ D(6,8)} \Rightarrow \cases{E(a,8) \\ F(0,a)\\ G(8-a,0)} \Rightarrow 正方形中心點O=(E+G)/2=(4,4) =(F+H)/2 \\ \Rightarrow H(8,8-a) ,並假設D,H在x軸投影點分別為D'(6,0), H'(8,0) \\ \Rightarrow {\overline{CH} \over \overline{CD}} ={\overline{CH'} \over \overline{CD'}} ={3\over 5} \Rightarrow {\overline{DH} \over \overline{CH}}={2\over 3}，故選\bbox[red, 2pt]{(B)}$$

 

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