新北市立國民中學 115 學年度教師聯合甄選
選擇題： 共 40 題，總分 100 分。 每題 2.5 分
解答:$$\cases{圖形凹向下\Rightarrow a\lt 0\\c=y截距\gt 0\\頂點的x坐標:{-b\over 2a} \lt 0 \Rightarrow ab \gt 0 \Rightarrow b\lt 0} \Rightarrow \cases{a\lt 0 \\ b\lt 0\\ c\gt 0} \Rightarrow abc\gt0 \\ 有相異二實根\Rightarrow b^2-4ac\gt 0 \Rightarrow (abc,b^2-4ac) =(+,+) 位於第一象限，故選\bbox[red, 2pt]{(D)}$$
解答:$$a_n=(n+\sqrt n)^2 \Rightarrow \cases{a_{34} \lt 1600\\ a_{35} \gt 1600 }，故選\bbox[red, 2pt]{(A)}$$
解答:$$|a|=a+3 \Rightarrow a\lt 0 \Rightarrow -a=a+3 \Rightarrow a=-{3\over 2} \Rightarrow |a+3|\times b= {3\over 2}b= -{3\over 2}-3=-{9\over 2} \\ \Rightarrow b=-3 \Rightarrow |a-b|-|a+b|={3\over 2}-{9\over 2}=-3，故選\bbox[red, 2pt]{(B)}$$
解答:$$甲、乙有相同的底面積A，因此V_1=9V_2 \\ 假設底面的周長L \Rightarrow \cases{S_1=2A+ 9L\\S_2=2A+L} \Rightarrow S_1\lt 9S_2，故選\bbox[red, 2pt]{(D)}$$
解答:$$2^2\times 3^2\times7^2\times 11-42\times 11 =2\times 3\times 7\times 11(2\times 3\times 7-1) =2\times 3\times 7\times 11\times 41 \\ \Rightarrow 質因數:2,3,7,11,41，共5個，故選\bbox[red, 2pt]{(C)}$$
解答:$$假設a_n表示第n個1出現在小數點的位數，例:a_1=2,a_2=5,a_3=9, a_4=14,\dots\\ 取\cases{b_1=a_1=2\\ b_n=a_n-a_{n-1}, n\ge 2} \Rightarrow b_n為等差數列,其公差d=1 \Rightarrow a_n=b_1+b_2+\cdots+b_n={n(n+3)\over 2} \\ \Rightarrow \cases{a_{18}=189\\ a_{19}=209} \Rightarrow 小數點第209位是1 \Rightarrow 小數點第208位至210位是0,1,0，故選\bbox[red, 2pt]{(C)}$$
解答:$$A,B,C,D,E,F任意排列，有6!=720種，若A,B相鄰有2\times 5!=240種\\ \Rightarrow A,B不相鄰有720-240=480種，故選\bbox[red, 2pt]{(B)}$$
$$假設h=d(A,\overline{BC}) =10\sin 30^\circ =5 \Rightarrow \triangle ADE= {1\over 2}\cdot \overline{AD}\cdot h={1\over 2}\cdot 18\cdot 5=45= {1\over 2}矩形AEFG面積 \\ \Rightarrow 矩形AEFG面積=90 \Rightarrow \triangle AFG={1\over 2}矩形AEFG面積 =45 \Rightarrow \triangle ADG={4\over 5} \triangle AFG \\= {4\over 5}\cdot 45=36，故選\bbox[red, 2pt]{(A)}$$
解答:
$$假設\cases{\angle AEB= \angle C =\alpha\\ \angle BAE=\angle DBE= \beta \\ \angle ABD=\theta} \;又\overline{BE}=\overline{CD}\Rightarrow \triangle ABE \cong \triangle BDC (SAS) \Rightarrow \overline{AB} =\overline{BD} \\ \Rightarrow \angle ADB= \angle BAD=\beta+33^\circ =\angle C+9^\circ= \alpha+9^\circ \Rightarrow \alpha-\beta=24^\circ \cdots(1) \\ \cases{\triangle ABE: \alpha+2\beta+ \theta=180^\circ \\ \triangle ABD: \theta+2\beta+66^\circ=180^\circ} \Rightarrow \alpha=66^\circ 代入(1) \Rightarrow \beta=42^\circ \Rightarrow 66^\circ+2\cdot 42^\circ+\theta=180^\circ\\ \Rightarrow \theta=30^\circ \Rightarrow \angle ABE=\theta+\beta=30^\circ+42^\circ=72^\circ，故選\bbox[red, 2pt]{(B)}$$
解答:$${1\over n}+{3\over n}+\cdots+{27 \over n} = {1\over n}{\sum_{k=1}^{14} (2k-1)} ={196\over n} \Rightarrow {2^2\times 7^2\over n}為整數 \\196的正因數有(2+1)(2+1)=9個，負因數也有9個，因此n值可能有9+9=18個，故選\bbox[red, 2pt]{(C)}$$
解答:$$2k^2x+k^2=(1-k)x+1 \Rightarrow (2k^2+k-1)x=1-k^2 \Rightarrow \cases{2k^2+k-1=0\\ 1-k^2\ne 0} \\ \Rightarrow \cases{(2k-1)(k+1)=0\\ (1+k)(1-k)\ne 0} \Rightarrow k={1\over 2} \Rightarrow 4k^3+k+1={1\over 2}+{1\over 2}+1=2，故選\bbox[red, 2pt]{(C)}$$
解答:$${2-x\over 2+x} =-{1\over 3} \Rightarrow x=4 \Rightarrow g(-{1\over 3})=4 \\ 又{2+x\over 2-x}= 4 \Rightarrow x={6\over 5} \Rightarrow f(4)=5\cdot {6\over 5}=6 \\ 因此f(g(-{1\over 3}))+g(-{1\over 3})=f(4)+4=6+4=10，故選\bbox[red, 2pt]{(A)}$$
解答:$$假設平行四邊形 ABCD的面積為S \Rightarrow \triangle BCE={S\over 2} \Rightarrow \triangle ABE+ \triangle CDE={S\over 2} \\ \Rightarrow {S\over 2}=14+10+12+12=48 \Rightarrow \triangle ABP+ \triangle CDP={S\over 2} \Rightarrow 14+14+12+甲=48 \\ \Rightarrow 甲=8，故選\bbox[red, 2pt]{(C)}$$
$$E是\overline{BC}中點 \Rightarrow \overline{BE} =\overline{EC}=k \Rightarrow\overline{AD}=\overline{BC}=2k \\ \overline{AD} \parallel \overline{BC} \Rightarrow \cases{\angle BAD=\angle DBE\\ \angle EAD= \angle BEA \\ \angle AFD=\angle BFE=\theta} \Rightarrow \triangle FBE \sim \triangle FDA (AAA) \Rightarrow {\overline{BF} \over \overline{FD}} ={\overline{FE} \over \overline{FA}} ={\overline{BE} \over \overline{AD}}={k\over 2k} \\ \Rightarrow \cases{\overline{FB}=\overline{FE}=6\\ \overline{FA}=\overline{FD}=12} \Rightarrow S=\triangle FBE ={1\over 2}\overline{FB} \cdot \overline{FE}\sin \theta=18\sin \theta \Rightarrow \triangle AFB=2\triangle FBE=2S\\ \Rightarrow \triangle ADF =2\triangle AFB=4S \Rightarrow \triangle ABD=4S=2S=6S \Rightarrow ABCD=12S= 216\sin \theta \\\cases{(A) 216 \sin \theta=180 \Rightarrow \sin \theta={180\over 216}= {5\over 6} \\(B) 216\sin\theta=108\sqrt 3 \Rightarrow \sin \theta={\sqrt 3\over 2} \\(C) 216\sin \theta=108 \Rightarrow \sin \theta={1\over 2} \\ (D) 216\sin \theta=90\sqrt3 \Rightarrow \sin \theta={5\sqrt{13}\over 12}} \Rightarrow \theta非唯一解，公布的答案是\bbox[cyan,2pt]{(B)}$$
解答:$${a+b\over c}={b+c\over a}+{c+a\over b}=r \Rightarrow \cases{a+b=cr\\ b+c=ar\\ c+a=br} \Rightarrow 三式相加 \Rightarrow 2(a+b+c)=r(a+b+c) \\ \Rightarrow (r-2)(a+b+c)=0 \Rightarrow \cases{r=2\\ a+b+c=0 \Rightarrow a+b=-c \Rightarrow {-c\over c}=r \Rightarrow r=-1} \\ \Rightarrow r=-1,2，共有2個r值，故選\bbox[red, 2pt]{(B)}$$
解答:$$\tan \alpha,\tan \beta 為x^2-px+q=0的兩根\Rightarrow \cases{\tan \alpha+\tan \beta=p\\ \tan \alpha \cdot \tan \beta=q} \\\cot \alpha,\cot \beta 為x^2-rx+s=0的兩根\Rightarrow \cases{\cot \alpha+\cot \beta=r\\ \cot \alpha \cdot \cot \beta= s} \Rightarrow \cases{\displaystyle {1\over \tan \alpha}+{1\over \tan \beta}={p\over q} =r\\ \displaystyle {1\over \tan \alpha}\cdot {1\over \tan \beta} ={1\over q}=s} \\ \Rightarrow rs={p\over q}\cdot {1\over q}={p\over q^2}，故選\bbox[red, 2pt]{(C)}$$
解答:$$(A)\times: \cases{2^1=2\\ 2^2=4\\a^3=8\\a^4=6\\a^5 \equiv 2 \text{ mod }10} \Rightarrow 循環數為4 \Rightarrow 2026=506\times 4+2 \Rightarrow 2^{2026} \equiv 4 \text{ mod 10} \\\qquad \Rightarrow 2^{2026}-1的個位數是3 \ne 1\\ (B)\times \cases{7^1=7\\ 7^2 \equiv 9 \text{ mod 10} \\ 7^3 \equiv 3 \text{ mod 10} \\ 7^4 \equiv 1 \text{ mod 10} \\ 7^5 \equiv 7 \text{ mod 10} } \Rightarrow 循環數為4 \Rightarrow110=27\cdot 4+2 \Rightarrow 7^{110} \equiv 9 \text{ mod 10} \\ \qquad \Rightarrow 7{110}+3的個位數字是2 \ne 6\\ (C)\times: \log_{2026}{1\over 2} \cdot \log_{2025}{1\over 3} \cdot \log_{2024}{1\over 4} \cdots \log_{3}{1\over 2025} \cdot \log_2{1\over 2026}\\\qquad =(-1)^{2025} \log_{2026}2 \cdot \log_{2025}3\cdot \log_{2024}4\cdots\log_2 2026 \\ \qquad =(-1)\cdot {\log 2\cdot \log 3\cdot \log 4\cdots \log 2025\cdot \log 2026\over \log 2026\cdot \log 2025\cdot \log 2024\cdots\log 3\cdot \log 2} =-1 \ne 1 \\(D) \bigcirc: 點數和為7:(1,6), (2,5), (3,4),(4,3), (5,2),(6,1)，共6種，機率={6\over 6^2}={1\over 6}\\，故選\bbox[red, 2pt]{(D)}$$
解答:$$3a^2+2a-4 =0 \Rightarrow a={-1\pm \sqrt{13}\over 3} \Rightarrow {2\over a}={1\pm \sqrt{13}\over 2} \\ b^4-b^2-3=0 \Rightarrow b^2 ={1+\sqrt{13}\over 2} \\ 欲求{a^2b^4+4\over a^2}=b^4+{4\over a^2} =b^2+3+{4\over (4-2a)/3} =b^2+{2\over a}+6 ={1+\sqrt{13}\over 2}+ {1\pm \sqrt{13}\over 2}+6\\=7或7+\sqrt{13}，故選\bbox[red, 2pt]{(D)}$$
解答:$$\angle A=108^\circ \Rightarrow \angle B=\angle C={180^\circ-108^\circ \over 2}=36^\circ \Rightarrow {\overline{BC} \over \sin A}={\overline{AB} \over \sin C} \Rightarrow {\overline{BC} \over \overline{AB}}={\sin 108^\circ\over \sin 36^\circ} \\={\sin(180^\circ-72^\circ) \over \sin 36}={\sin 72^\circ \over \sin 36^\circ} ={2\sin 36^\circ \cos 36^\circ \over \sin 36^\circ} =2\cos 36^\circ =2\cdot {1+\sqrt 5\over 4} ={1+\sqrt 5\over 2}，故選\bbox[red, 2pt]{(D)}$$
解答:$$6球號碼和為奇數的情況\cases{1奇5偶: C^6_1C^5_5= 6\\ 3奇3偶:C^6_3C^5_3=200\\ 5奇1偶:C^6_5C^5_1=30} \Rightarrow 合計236 \Rightarrow 機率={236\over C^{11}_6} ={118\over 231}，故選\bbox[red, 2pt]{(C)}$$
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解題僅供參考，其他教甄試題及詳解
解答:$$符合x=1代入可得2的多項式(B),(C),(D);再將x=2代入(B),(C),(D),只有(D)可得1\\，故選\bbox[red, 2pt]{(D)}$$
解答:
$$\overline{CH}^2= \left( 1+{1\over 2}+1 \right)^2+ \left( {\sqrt 3\over 2}+\sqrt 3 \right)^2={25\over 4}+{27\over 4}=13\Rightarrow \overline{CH} =\sqrt{13}，故選\bbox[red, 2pt]{(B)}$$
解答:$$所圍區域\cases{x+y\le 2\\ x\ge 0\\ y\ge 0} 頂點坐標 \cases{O(0,0)\\A(2,0) \\B(0,2)} 代入f(x,y)=x-y \Rightarrow \cases{f(0,0)=0\\ f(2,0)=2\\ f(0,2)=-2} \\ \Rightarrow -2\le a\le 2，故選\bbox[red, 2pt]{(D)}$$
解答:$$取t=2^x \Rightarrow 2^{3x+1}-17\cdot 2^{2x}+2^{x+3}=2t^3 -17t^2+ 8t=0 \Rightarrow t(2t-1)(t-8)=0 \\ \Rightarrow \cases{t=2^x= 1/2 \Rightarrow x=-1\\ t=2^x=8 \Rightarrow x=3} \Rightarrow -1\cdot 3=-3，故選\bbox[red, 2pt]{(A)}$$
解答:$${a+6\over b}+{13\over ab}={4-b\over a} \Rightarrow a(a+6)+13=b(4-b) \Rightarrow (a+3)^2+ (b-2)^2=0 \\ \Rightarrow \cases{a=-3\\ b=2} \Rightarrow b-a=5，故選\bbox[red, 2pt]{(D)}$$
解答:$${1\over \sqrt 1+\sqrt 2}+{1\over \sqrt 2+\sqrt 3}+{ 1\over \sqrt 3+\sqrt 4}+ \cdots+ {1\over \sqrt{98}+ \sqrt{99}}+ {1\over \sqrt{99}+\sqrt{100}} \\= (\sqrt 2-\sqrt 1)+(\sqrt 3-\sqrt 2)+(\sqrt 4-\sqrt 3)+ \cdots+(\sqrt{99}-\sqrt{98})+( \sqrt{100}-\sqrt{99}) \\=\sqrt{100}-\sqrt 1=10-1=9，故選\bbox[red, 2pt]{(A)}$$
解答:$$0^\circ \lt \theta\lt 45^\circ \Rightarrow \cos \theta \gt \sin \theta \gt 0 \Rightarrow \cot \theta={\cos \theta\over \sin \theta} \gt 1 \Rightarrow \cot \theta \gt \cos \theta \gt \sin \theta ，故選\bbox[red, 2pt]{(C)}$$
解答:$$\log_9 a= \log_{12}b= \log_{16}(a+b) =k \Rightarrow \cases{a=9^k\\b=12^k\\a+b=16^k} \Rightarrow 9^k+12^k=16^k \Rightarrow 1+ \left( {12\over 9} \right)^k= \left( {16\over 9} \right)^k \\ \Rightarrow 1+ \left( {4\over 3} \right)^k= \left( {4\over 3} \right)^{2k} \Rightarrow \left( {4\over 3} \right)^{2k}-\left( {4\over 3} \right)^{k}-1=0 \Rightarrow \left( {4\over 3} \right)^{k}={1+\sqrt 5\over 2} \\ \Rightarrow {b\over a}=\left( {4\over 3} \right)^{k}={1+\sqrt 5\over 2}，故選\bbox[red, 2pt]{(C)}$$
解答:$$兩邊之和大於第三邊，兩邊之差小於第三邊 \Rightarrow 24-10 \lt a\lt 24+10 \Rightarrow 14\lt a\lt 34 \\ \textbf{Case I } 最長邊是24 \Rightarrow 14\lt a\le 24 \Rightarrow 銳角三角形需滿足10^2+a^2\gt 24^2 \Rightarrow a^2\gt 476 \\\qquad \Rightarrow a=22,23,24\\ \textbf{Case II }最長邊是a \Rightarrow 24\le a \Rightarrow 10^2+24^2\gt a^2 \Rightarrow a^2\lt 676 \Rightarrow a=24,25 \\ 因此a=22,23,24,25，共4個，故選\bbox[red, 2pt]{(A)}$$
解答:$$\cases{a^3=3ab^2+11\\ b^3=3a^2b+2 } \Rightarrow (a^3-3ab^2)^2+ (b^3-3a^2b)^2 =11^2+2^2 \Rightarrow a^6+3a^4b^2+3a^2b^4+b^6=125 \\ \Rightarrow (a^2+b^2)^3=125 \Rightarrow a^2+b^2=5，故選\bbox[red, 2pt]{(B)}$$
解答:$$假設\cases{\overline{AB}=a\\ \overline{BC}=b}, 且b\gt a \Rightarrow \cases{B(0,0) \\ C(b,0) \\A(0,a)\\ F(b/2,0)\\E(b,a/2)} \Rightarrow \cases{\overrightarrow{FA}=(-b/2,a) \\ \overrightarrow{FE}=(b/2,a/2)} \Rightarrow \overline{FA} \bot \overline{FE} \Rightarrow \overrightarrow{FA} \cdot \overrightarrow{FE}=0\\ \Rightarrow -{b^2\over 4}+{a^2\over 2} \Rightarrow {b^2\over a^2} =2 \Rightarrow {b\over a} ={\overline{BC} \over \overline{AB}}=\sqrt 2，故選\bbox[red, 2pt]{(A)}$$
解答:$$由於x^4+324= x^4+4\cdot 3^4 =(x^2-6x+18)(x^2+6x+18) \\ 因此取f(x)=x^2-6x+18 \Rightarrow f(x+6)= (x+6)^2-6(x+6)+18 =x^2+6x+18 \\ \Rightarrow 原式={f(10)f(16) \cdot f(22)f(28)\cdot f(34)f(40) \cdot f(46)f(52) \over f(4)f(10)\cdot f(16)f(22) \cdot f(28)\cdot f(34) \cdot f(40)f(46)} ={f(52) \over f(4)} ={2410\over 10}=241，故選\bbox[red, 2pt]{(B)}$$
解答:
$$取\cases{A(1,0) \\B(0,1) \\C(0,0)} \Rightarrow D=(B+C)/2=(0,1/2) \Rightarrow E在\overleftrightarrow{AB}:x+y=1 \Rightarrow E=(t,1-t) \\ \Rightarrow \cases{\overrightarrow{DE}=(t,1/2-t)\\ \overrightarrow{DA}=(1,-1/2)} \Rightarrow \overrightarrow{DE} \cdot \overrightarrow{DA}=0 \Rightarrow t={1\over 6} \Rightarrow E \left( {1\over 6},{5\over 6} \right) \Rightarrow \triangle BDE={1\over 2} \cdot \overline{BD}\cdot d(E,\overline{BC}) \\={1\over 2}\cdot {1\over 2}\cdot {1\over 6}={1\over 24}，故選\bbox[red, 2pt]{(A)}$$
解答:$$\cases{ab+5=c \cdots(1)\\ bc+1=a \cdots(2)\\ ca+1=b \cdots(3)} \Rightarrow (2)-(1)=bc-ca=a-b \Rightarrow (c+1)(b-a)=0 \\ \Rightarrow \cases{c=-1 代入(2)\Rightarrow b=1-a 代入(1) \Rightarrow a(1-a)+5=-1 \Rightarrow a^2-a-6=0 \Rightarrow \cases{a=3\Rightarrow b=-2\\ a=-2 \Rightarrow b=3} \\a=b 代入(1)\Rightarrow c=a^2+5 代入(2) \Rightarrow a(a^2+5)+1=a \Rightarrow a(a^2+4)=-1無整數解} \\ \Rightarrow (a,b,c)=(3,-1,-1),(-2,3,-1)，共2組解，故選\bbox[red, 2pt]{(B)}$$
解答:$$\angle APB= \angle BPC= \angle CPA={360^\circ \over 3}=120^\circ ,並假設\overline{PB}=a\\ \Rightarrow \cases{\overline{AC}^2=8^2+6^2-2\cdot 8\cdot 6 \cos 120^\circ=148\\ \overline{BC}^2= 6^2+a^2-12a\cos 120^\circ =36+ a^2+6a \\ \overline{AB}^2 =8^2+a^2-16a\cos 120^\circ=64+ a^2+8a} \\ \angle C=90^\circ\Rightarrow \overline{AB}^2=\overline{AC}^2+ \overline{BC}^2 \Rightarrow 64+a^2+8a=148+36+a^2+6a \Rightarrow a=60，故選\bbox[red, 2pt]{(D)}$$
解答:$$\cases{a_2=a_1+2\cdot 1\\ a_3=a_2+2\cdot 2\\ a_4= a_3+2\cdot 3\\ \cdots\\ a_{99}=a_{98} +2\cdot 98\\a_{100} =a_{99}+2 \cdot 99} \Rightarrow 合計:a_{100} =a_1+2(1+2+\cdots+ 99) =2+2\cdot {100\cdot 99\over 2}=9902\\，故選\bbox[red, 2pt]{(B)}$$
解答:$$A= 69^5+ 5\times 69^4+ 10\times 69^3+ 10\times 69^2+ 5\times 69+1 =(69+1)^5= 70^5= 2^5\times 5^5\times 7^5 \\ \Rightarrow 正因數個數(5+1)(5+1)(5+1) =216，故選\bbox[red, 2pt]{(D)}$$
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解題僅供參考，其他教甄試題及詳解
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