新竹市立香山高級中學 115 學年度教師甄選
科目： 國中數學科
單選題(共 25 題， 每題 4 分， 計 100 分)
解答:$$41^{48} =(1 + 40)^{48} = \binom{48}{0} \cdot 1^{48} + \binom{48}{1} \cdot 1^{47} \cdot 40^1 + \binom{48}{2} \cdot 1^{46} \cdot 40^2 + \dots \\ 只需考慮前兩項:\binom{48}{0}+\binom{48}{1} \cdot 40= 1+ 1920= 1921 \equiv {21} \text{ (mod 100) }，故選\bbox[red, 2pt]{(B)}$$
解答:$$a\log_{200} 2+b \log_{200} 5=c \Rightarrow \log_{200} (2^a\cdot 5^b)=c \Rightarrow 2^a\cdot 5^b=200^c= \left( 5^2\cdot 2^3 \right)^c =2^{3c} \cdot 5^{2c} \\ \Rightarrow \cases{a=3c\\ b=2c} , 由於a,b,c互質 \Rightarrow \cases{a=3\\ b=2\\ c=1} \Rightarrow a+b+c= 6，故選\bbox[red, 2pt]{(A)}$$
解答:$$假設\cases{ a = \sqrt[3]{x-1} \\b = \sqrt[3]{x+1} } \Rightarrow f(x) = \frac{1}{\sqrt[3]{x^2 - 2x + 1} + \sqrt[3]{x^2 - 1} + \sqrt[3]{x^2 + 2x + 1}}= {1\over a^2+ab+ b^2} \\= {a-b \over (a-b)(a^2+ab+b^2)} ={a-b\over a^3-b^3} ={1\over 2} (\sqrt[3]{x+1} -\sqrt[3]{x-1}) \Rightarrow \sum_{x=1}^{999} f(x) \\= \frac{1}{2} \left[ (\sqrt[3]{2} - 0) + (\sqrt[3]{3} - \sqrt[3]{1}) + (\sqrt[3]{4} - \sqrt[3]{2}) + \cdots + (\sqrt[3]{999} - \sqrt[3]{997}) + (\sqrt[3]{1000} - \sqrt[3]{998}) \right]\\= {1\over 2} \left[ -\sqrt[3]1+ \sqrt[3]{999}+ \sqrt[3]{1000} \right] ={1\over 2} \left( -1+3\sqrt[3]{37}+10 \right)= {3\over 2} \left( 3+\sqrt[3]{37} \right)，故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{假設第差數列\{a_n\} 的公差為d \Rightarrow a_n=a_1+(n-1)d \\假設第比數列\{b_n\} 的公比為r \Rightarrow b_n=b_1r^{n-1} } \\ \Rightarrow \cases{a_1b_1= 20 \cdots(1)\\ a_2 b_2 = (a_1 + d) b_1 r = a_1 b_1 r + d b_1 r = 20r + d b_1 r = 19 \cdots(2) \\ a_3 b_3 = (a_1 + 2d) b_1 r^2 = a_1 b_1 r^2 + 2d b_1 r^2 = 20r^2 + 2d b_1 r^2 = 14 \cdots(3)} \\ 式(2) \Rightarrow db_1r=19-20r代入(3) \Rightarrow 20r^2 + 2(d b_1 r)r = 14 \Rightarrow 10r^2-19r+7=0 \\ \Rightarrow(2r-1)(5r-7)=0 \Rightarrow r=1/2,7/5\\ 欲求 a_4 b_4 = (a_1 + 3d) b_1 r^3 = 20r^3 + 3(d b_1 r) r^2 = 20r^3 + 3(19 - 20r)r^2 =57r^2 -40r^3 \\ 若r={1\over 2} \Rightarrow a_4 b_4 = 57 \left(\frac{1}{2}\right)^2 - 40 \left(\frac{1}{2}\right)^3 = 57 \times \frac{1}{4} - 40 \times \frac{1}{8} = \frac{57}{4} - 5 = \frac{37}{4} \\ 若r={7\over 5} \Rightarrow a_4 b_4 = 57 \left(\frac{7}{5}\right)^2 - 40 \left(\frac{7}{5}\right)^3 = 57 \times \frac{49}{25} - 40 \times \frac{343}{125}={49\over 25} \\ 因此a_4b_4 最大可能值為{37\over 4}，故選\bbox[red, 2pt]{(A)}$$
解答:$$S=1+ {2\over 2026}+{3\over 2026^2}+{4\over 2026^3}+ {5\over 2026^4}+\cdots \\\Rightarrow {1\over 2026}S ={1 \over 2026}+ {2\over 2026^2}+{3\over 2026^3}+{4\over 2026^4}+ {5\over 2026^5}+\cdots \\ \Rightarrow S-{1\over 2026}S=1+{1\over 2026}+{1\over 2026^2}+{1\over 2026^3}+{1\over 2026^4}+ \cdots\\ \Rightarrow {2025\over 2026}S={1\over 1-(1/2026)} ={2026\over 2025} \Rightarrow S= \left( {2026\over 2025} \right)^2，故選\bbox[red, 2pt]{(C)}$$
解答:$$取S(n)= f(1)+f(2)+ \cdots+f(n)=n^2f(n) \Rightarrow S(n-1)=(n-1)^2f(n-1) \\ \Rightarrow f(n)=S(n)-S(n-1) =n^2f(n)-(n-1)^2f(n-1) \Rightarrow (n^2-1)f(n)=(n-1)^2f(n-1) \\ \Rightarrow (n+1)f(n)=(n-1)f(n-1) \Rightarrow f(n)={n-1\over n+1}f(n-1) \\ \Rightarrow f(n)={n-1\over n+1} \cdot {n-2\over n} \cdot {n-3\over n-1} \cdots {3\over 5}\cdot {2\over 4} \cdot {1\over 3} \cdot f(1) ={2\cdot 1\over (n+1)n}f(1) \\ \Rightarrow f(100)={2\over 101\cdot 100}\cdot 100={2\over 101}，故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{x= a/c= \sin A\\ y=b/c = \cos A}\Rightarrow (\sin A+\cos B)^2=1+2 \sin A\cos B \Rightarrow 13xy=15(x+y)-15\\ \Rightarrow 13\sin A\cos A=15(\sin A+\cos A)-15 \\ \Rightarrow \sin A+\cos A={13\over 15}(\sin A\cos A)+1={13\over 30} \left( (\sin A+\cos A)^2-1 \right)+1 \\ \Rightarrow a={13\over 30}(a^2-1)+1 \Rightarrow 13a^2-30a+17=0 \Rightarrow (13a-17)(a-1)=0 \\ \Rightarrow a= \sin A+\cos A=x+y={17\over 13}，故選\bbox[red, 2pt]{(D)}$$
解答:$${a+6\over b}+{13\over ab}={4-b\over a} \Rightarrow {a^2+6a+13\over ab}={4b-b^2\over ab} \\ \Rightarrow a^2+6a+13=4b-b^2 \Rightarrow (a+3)^2+(b-2)^2=0 \Rightarrow \cases{a=-3\\ b=2} \Rightarrow a+b=-1，故選\bbox[red, 2pt]{(A)}$$
解答:$$假設放入的紅球數x \Rightarrow 放入的白球數x+80 \\ 原來\cases{紅球數:19k\\ 白球數:13k} \Rightarrow 放入紅球後\cases{紅球數:19k+x\\ 白球數:13k} \Rightarrow {19k+x\over 13k}={5\over 3 } \cdots(1)\\ \Rightarrow 再放入白球後\cases{紅球數19k+x\\ 白球數13k+x+80} \Rightarrow {19k+x\over 13k+x+80}={13\over 11} \cdots(2) \\\cases{式(1)\Rightarrow x=8k/3\\ 式(2) \Rightarrow x=20k-520} \Rightarrow {8k\over 3}=20k-520 \Rightarrow k=30 \\\Rightarrow 原來袋子有19k+13k=32k=32\cdot 30= 960個球，故選\bbox[red, 2pt]{(C)}$$
解答:$$ \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = 1 \Rightarrow (a+b+c) \cdot \left( \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \right) = 1 \cdot (a+b+c) \\ 左式: \left( \frac{a^2}{b+c} + \frac{ab}{c+a} + \frac{ac}{a+b} \right) + \left( \frac{ab}{b+c} + \frac{b^2}{c+a} + \frac{bc}{a+b} \right) + \left( \frac{ac}{b+c} + \frac{bc}{c+a} + \frac{c^2}{a+b} \right) \\= \frac{a^2+ab+ac}{b+c} + \frac{ab+b^2+bc}{c+a} + \frac{ac+bc+c^2}{a+b} \\= \left( \frac{a^2}{b+c} + a \right) + \left( \frac{b^2}{c+a} + b \right) + \left( \frac{c^2}{a+b} + c \right) = \frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} + a + b + c \\ 左式=右式\Rightarrow \frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} + a + b + c=a+b+c \\ \Rightarrow \frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} =0，故選\bbox[red, 2pt]{(A)}$$
解答:
$$假設\cases{\triangle APG= \triangle BQH=\triangle CER= \triangle DFS= \alpha\\ 四邊形APQH=BQRE= CRSF= DSPG =\beta \\四邊形PQRS=\gamma =1/421}\\ \Rightarrow \cases{正方形ABCD:4\alpha+ 4\beta+\gamma=1 \cdots(1)\\ 直角\triangle ADF= 2\alpha+ \beta=(n-1)/2n \cdots(2)} \\ 又{\triangle APG\over \triangle ASD}={\alpha\over \alpha+\beta} ={\overline{AG}^2\over \overline{AD}^2} = \left( 1-{1\over n} \right)^2= \left( {n-1\over n} \right)^2 \cdots(3)\\ 由式(1)及(2)可得 \alpha= {n-1 \over 2n}-{105\over 421}, \beta={210\over 421}-{n-1\over 2n} 代入(3) \Rightarrow n^2-n-210=0\\ \Rightarrow (n-15)(n+14)=0 \Rightarrow n=15，故選\bbox[red, 2pt]{(B)}$$
解答:$$取\cases{A(0,c) \\B(-b,0) \\C(a,-d)},a,b,c,d\gt 0 \Rightarrow \cases{\overline{AB}= \sqrt{b^2+c^2} \\ \overline{AC} = \sqrt{a^2+ (c+d)^2} \\ \overline{BC} = \sqrt{(a+b)^2+ d^2}} \Rightarrow \triangle ABC面積= {1\over 2}\begin{Vmatrix} 0&c & 1\\ -b& 0& 1\\ a&-d& 1 \end{Vmatrix} \\= {1\over 2}|bd+ac+bc| ={1\over 2}(ac+bc+bd)，故選\bbox[red, 2pt]{(A)}$$
解答:$$\cases{a^2=13\\ b^2 =29\\ c^2=34} \Rightarrow \cos C={13+29-34\over 2\sqrt{13}\cdot \sqrt{29}}={4\over \sqrt{377}} \Rightarrow \sin C ={19\over \sqrt{377}} \\ \Rightarrow \triangle ABC面積={1\over 2}ab \sin C={1\over 2}\cdot \sqrt{13}\cdot \sqrt{29}\cdot {19\over \sqrt{377}}= {19\over 2}，故選\bbox[red, 2pt]{(D)}$$
解答:$$\cases{\sqrt{64 + 32\sqrt{3}}= \sqrt{64+ 2\sqrt{48\times 16}} =\sqrt{48}+ \sqrt{16} =4\sqrt 3+4 \\ \sqrt{97 + 56\sqrt{3}} =\sqrt{97+2 \sqrt{49\times 48}} =\sqrt{49}+ \sqrt{48} =7+4\sqrt 3} \\ \Rightarrow \sqrt{64 + 32\sqrt{3}} - \sqrt{97 + 56\sqrt{3}} = (4\sqrt{3} + 4) - (7 + 4\sqrt{3})=-3，故選\bbox[red, 2pt]{(A)}$$
解答:$$\log_2 \log_4 x+\log_4 \log_2 x= \log_2 \left( {\log_2 x\over 2} \right) +{1\over 2}\log_2 \log_ 2x={1\over 2} \\ \Rightarrow \log_2 {1\over 2}+ \log_2 \log_2 x+{1\over 2}\log_2 \log_ 2x={1\over 2} \Rightarrow -1+{3\over 2}\log_2 \log_2 x= {1\over 2} \\ \Rightarrow {3\over 2}\log_2 \log_2 x={3\over 2} \Rightarrow \log_2 \log_2 x=1 \Rightarrow \log_2 x=2 \Rightarrow x=4，故選\bbox[red, 2pt]{(C)}$$
解答:$$\cases{a = 2b + c \cdots(1)\\b = 2c + d \cdots(2) \\2c = d + a + 1 \cdots(3)\\d = a - c \cdots(4)}\;將(4)代入(2)及(3) \Rightarrow \cases{b=2c+(a-c)=a+c \cdots(5)\\ 2c=(a-c)+a+1 \Rightarrow 3c=2a+1 \cdots(6)}\\ 將(5)代入(1) \Rightarrow a=2(a+c)+c=2a+3c \Rightarrow a=-3c 代入(6) \Rightarrow 3c=-6c+1 \Rightarrow c={1\over 9} \\ \Rightarrow a = -3 \left(\frac{1}{9}\right) = -\frac{1}{3} \Rightarrow b=a+c= -{2\over 9}，故選\bbox[red, 2pt]{(D)}$$
解答:$$ a_n = 3^{\frac{1}{3} \cdot 3^{\frac{n-1}{3}}} = 3^{3^{-1} \cdot 3^{\frac{n-1}{3}}} = 3^{3^{\frac{n-1}{3} - 1}} = 3^{3^{\frac{n-4}{3}}} \Rightarrow \frac{n-4}{3} 最小的非負整數為0 \Rightarrow n=4，故選\bbox[red, 2pt]{(B)}$$
解答:$$假設正方形 DGEF 的邊長為 x\Rightarrow \overline{DE} = \overline{EF} = \overline{FG} = \overline{DG} = x\\ \cases{直角\triangle ADE: \tan A = \frac{ \overline{DE}}{ \overline{AE}} = \frac{x}{\overline{AE}} \implies \overline{AE} = \frac{x}{\tan A} \\ 直角\triangle BGF: \tan B = \frac{\overline{FG}}{ \overline{FB}} = \frac{x}{\overline{FB}} \implies \overline{ FB} = \frac{x}{\tan B} } \\ 由於\angle A+\angle B=90^\circ \Rightarrow \tan B={1\over \tan A} \Rightarrow \overline{FB} =x \tan A \\ \overline{AB} = \overline{AE}+ \overline{EF}+\overline{FB}= {x\over \tan A}+x+x\cdot \tan A=x \left( 1+\tan A+{1\over \tan A} \right) \\\ge x(1+2\sqrt{\tan A\cdot {1\over \tan A}}) =x(1+2)=3x \Rightarrow \overline{AB}\ge 3\overline{EF}，故選\bbox[red, 2pt]{(D)}$$
解答:$$\textbf{Case I }a\ge 0且b\ge 0 \Rightarrow \cases{a+a+b=2a+b=10 \\ a+b-b=a=12} \Rightarrow \cases{a=12 \\ b=-14不合} \\\textbf{Case II }a\ge 0且b\le 0 \Rightarrow \cases{2a+b=10 \\a-2b=12} \Rightarrow \cases{b=-14/5\\ a=32/5} \Rightarrow a+b= 18/5\\ \textbf{Case III }a\le 0且b\ge 0 \Rightarrow \cases{b=10\\ a=12 不合} \\ \textbf{Case IV }a\le 0且b\le 0 \Rightarrow \cases{b=10 不合\\ a-2b=12}\\，故選\bbox[red, 2pt]{(B)}$$
解答:$$ \gcd(a,b) = c \Rightarrow \cases{a=c\cdot m\\ b=c\cdot n},其中\gcd(m,n)=1 \Rightarrow \text{lcm}(a,b) = d = c \cdot m \cdot n \\ \Rightarrow c+d=c+c\cdot m\cdot n =187 \Rightarrow c(1+mn)=187 =11\times 17 \\\cases{c=17 \Rightarrow 1+mn=11 \Rightarrow mn=10 \Rightarrow \{m,n\} =\{2,5\}\Rightarrow a+b=cm+cn= 119 \\ c=11 \Rightarrow 1+mn=17 \Rightarrow mn=16無法拆成互質整數的乘積} \\ \Rightarrow a+b=119，故選\bbox[red, 2pt]{(C)}$$
解答:$$\cases{x^2 + ax + 2b = 0有實數解 \Rightarrow a^2-8b \ge 0 \Rightarrow a^2 \ge 8b\\ x^2+2bx+a=0有實數解 \Rightarrow 4b^2-4a \ge 0 \Rightarrow b^2\ge a} \Rightarrow (b^2)^2 \ge a^2 \ge 8b \Rightarrow b^3 \ge 8\\ \Rightarrow b\ge 2\Rightarrow b=最小值2 \Rightarrow \cases{a^2 \ge 8b=16 \Rightarrow a\ge 4 \\ a\le b^2 =4 \Rightarrow a\le 4} \Rightarrow a=4 \Rightarrow a+b=6，故選\bbox[red, 2pt]{(B)}$$
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