114年中壢家商教甄-數學詳解

 桃園市立中壢家事商業高級中等學校 114 學年度教師甄選

一、填充題(每題 6 分，共 60 分)

解答:$$5(a_1+\cdots +a_k) =(k+4)a_k \Rightarrow 5(a_1+\cdots+a_{k+1})=(k+5)a_{k+1} \\ \Rightarrow 5a_{k+1}=(k+5)a_{k+1}-(k+4)a_k \Rightarrow ka_{k+1}=(k+4)a_k \Rightarrow {a_{k+1} \over a_k} ={k+4\over k} \\ \Rightarrow {a_{21} \over a_{20}}\times {a_{20} \over a_{19}} \times\cdots \times{a_2\over a_1} ={24\over 20} \times{23\over 19} \times{22\over 18} \times \cdots \times{5\over 1} \\ \Rightarrow {a_{21} \over a_1=1} ={24\times 23\times 22 \times 21\over 4\times 3\times 2\times 1}\Rightarrow a_{21} = \bbox[red, 2pt]{10626 }$$

解答:$$假設 \cases{\log_2 x=a\\ \log_6x =b} \Rightarrow \log_4 x={1\over 2}a \Rightarrow a\cdot {1\over 2}a\cdot b= a\cdot {1\over 2}a+ ab+{1\over 2}ab \\ \Rightarrow {1\over 2}a^2b={1\over 2}a^2+{3\over 2}ab \Rightarrow ab=a+3b \Rightarrow \log_2 x\cdot \log_6 x=\log_2 x+3\log_6 x \\ \Rightarrow {(\log x)^2 \over (\log 2)(\log 6)} ={\log x\over \log 2}+{3 \log x\over \log 6} ={ \log x(\log 6+3\log 2)\over (\log 2)(\log 6)} \\ \Rightarrow (\log x)^2=\log x (\log 48) \Rightarrow \log x(\log x-\log 48)=0\Rightarrow \cases{\log x=0\\ \log x=\log 48} \Rightarrow x= \bbox[red, 2pt]{1或 48}$$

解答:$$10!\times 9!\times \cdots \times 1! =10\times 9^2\times 8^3\times 7^4\times 6^5\times 5^6\times 4^7 \times 3^8\times 2^9 =2^{38} \times 3^{17} \times 5^7 \times 7^4 \\\Rightarrow 正因數中完全平方數有:\overbrace{2^0,2^2, 2^4, \dots, 2^{38}}^{20個}, \overbrace{3^0,3^2, 3^4, \dots, 3^{16}}^{9個}, \\ \overbrace{5^0,5^2, 5^4, \dots, 5^{6}}^{4個},\overbrace{7^0,7^2, 7^4 }^{3個} \Rightarrow 共有20\times 9\times 4\times 3= \bbox[red, 2pt]{2160}個$$

解答:$$f(x)=(x-1)^2p(x)+3x+2 \Rightarrow f'(x)=2(x-1)p(x)+ (x-1)^2p'(x)+3 \Rightarrow \cases{f(1)=5\\ f'(1)=3} \\ \Rightarrow \lim_{x\to 1}{x^3f(1)-f(x^2) \over x-1} = \lim_{x\to 1}{3x^2f(1)-2xf'(x^2) \over 1} = 3f(1)-2f'(1)=3\cdot 5-2\cdot 3= \bbox[red, 2pt]9$$

解答:

$$假設\overleftrightarrow{AG}交\overline{BC}於D，由於G為重心，因此\cases{\overline{GD}=\overline{AG}/2=2\\ \overline{BD}= \overline{DC}= 10/2 =5} \\ 中線定理: \overline{GB}^2+ \overline{GC}^2 =2( \overline{GD}^2+ \overline{GD}^2) =2(4+25) =58\\ \Rightarrow \cos \angle BGC={\overline{GB}^2+ \overline{GC}^2-\overline{BC}^2 \over 2\overline{BG} \cdot \overline{GC}} \Rightarrow -{\sqrt 2\over 2} = {58-100 \over 2\overline{BG} \cdot \overline{GC}} \Rightarrow \overline{BG} \cdot \overline{GC}=21\sqrt 2 \\ \Rightarrow S_{\triangle GBC}={1\over 2}\overline{BG} \cdot \overline{GC} \sin \angle 135^\circ ={1\over 2}\cdot 21\sqrt 2\cdot {\sqrt 2\over 2} ={21\over 2} \Rightarrow S_{\triangle ABC}=3 S_{\triangle GBC} = \bbox[red, 2pt]{63\over 2}$$

解答:$$\sum_{k=1}^n C^{2k}_2 =\sum_{k=1}^n (2k^2-k) = 2\cdot {n(n+1)(2n+1) \over 6}-{n(n+1) \over 2} ={n(n+1)\over 6} (4n+2-3) \\= \bbox[red, 2pt]{n(n+1)(4n-1) \over 6}$$

解答:$$(1+x)^k \approx 1+kx \\ \lim_{n\to \infty} (\sqrt[3]{n^3+2n^2+3}-\sqrt{n^2+5}) = \lim_{n\to \infty} (n\sqrt[3]{1+2/n+3/n^3}- n\sqrt{1+5/n^2}) \\= \lim_{n\to \infty} (n(1+2/3n +1/n^3)- n(1+5/2n^2))  = \lim_{n\to \infty} ({2\over 3}+{1\over n^2}-{5\over 2n}) = \bbox[red, 2pt]{2\over 3}$$

解答:$$取x=2^n \Rightarrow 2^{2^n} =(2^n)^{32}=2^{32n} \Rightarrow 2^n= 32n =2^5\cdot n \Rightarrow n=2^3=8 \Rightarrow x=2^8=\bbox[red, 2pt]{256}$$

解答:$$2^k 除以100的餘數為2,4,8,16,32,64, 28,56,12,24,48,96,92,84,68,36,72,44,88,76,52,4,... \\ \Rightarrow 2^m \equiv 2^{20+m} \pmod {100} \Rightarrow 2^{2025} =2^{101\cdot 20+5}\equiv 2^5 \equiv 32 \pmod{100} \\ \Rightarrow 十位數字為\bbox[red, 2pt]3$$

解答:$$n!=23n^3-64n^2+41n =n(n-1)(23n-41) \\ 取f(n)=n(n-1)(23n-41) \Rightarrow \cases{f(2)=10 \gt 2!\\ f(3) =6\times 28\gt 3!\\ \cdots\\ f(6) =2910\gt 6!=720\\ f(7)=5040=7!\\ f(8)\lt 8!\\ \cdots} \Rightarrow n=\bbox[red, 2pt]7$$

二、計算證明題(每題 10 分，共 40 分)

解答:$$S= \sum_{n=1}^\infty {n^2\over 4^n} ={1\over 4}+{2^2\over 4^2} +{3^2\over 4^3}+{4^2\over 4^4}+ \cdots \Rightarrow {1\over 4}S ={1\over 4^2}+{2^2\over 4^3} +{3^2\over 4^4}+{4^2\over 4^5}+ \cdots \\ \Rightarrow S-{1\over 4}S={3\over 4}S={1\over 4}+{3\over 4^2} +{5\over 4^3}+{7\over 4^4}+ \cdots \Rightarrow {3\over 16}S=  {1\over 4^2}+{3\over 4^3} +{5\over 4^4}+{7\over 4^5}+ \cdots \\ \Rightarrow {3\over 4}S-{3\over 16}S={1\over 4}+2\left( {1\over 4^2} +{1\over 4^3} +{1\over 4^4} +\cdots\right) \Rightarrow {9\over 16}S={1\over 4}+2\cdot {1/4^2\over 1-1/4} ={5\over 12} \\ \Rightarrow S={5\over 12}\cdot {16\over 9} = \bbox[red, 2pt]{20\over 27}$$

解答:$$柯西不等式:(a^2 +b^2+c^2 +d^2)(1^2+2^2+3^2+1^2) \ge (a+2b+3c+d)^2 \\ \Rightarrow (16-e^2)\cdot 15 \ge (8-e)^2 \Rightarrow e^2-e-11\le 0 \Rightarrow {1-3\sqrt 5\over 2} \le e\le {1+3\sqrt 5\over 2} \\ \Rightarrow e的最大值\bbox[red, 2pt]{1+3\sqrt 5\over 2}$$

解答:$$\textbf{(1) }假設\cases{\overline{BC}=a =|\overrightarrow{BC}|\\ \overline{AC} =b =|\overrightarrow{AC}|\\ \overline{AB} =c =|\overrightarrow{AB}|} \Rightarrow \cos A={b^2+c^2- a^2\over 2bc} = {b^2+c^2- a^2\over 2|\overrightarrow{AC}||\overrightarrow{AB}|} \\ \Rightarrow |\overrightarrow{AC}|| \overrightarrow{AB}| \cos A=-\overrightarrow{CA} \cdot \overrightarrow{AB}={b^2+c^2-a^2\over 2 } \\ 同理,|\overrightarrow{AB}||\overrightarrow{BC}| \cos B=-\overrightarrow{AB} \cdot \overrightarrow{BC} ={a^2 +c^2-b^2\over 2},|\overrightarrow{AC}| |\overrightarrow{BC}| \cos C=-\overrightarrow{BC} \cdot \overrightarrow{CA}={a^2+ b^2-c^2\over 2}  \\ 因此{a^2+b^2-c^2\over 2} ={b^2+c^2-a^2\over 2}  ={a^2+ c^2-b^2\over 2} \Rightarrow \cases{a^2 =c^2\\ a^2=b^2} \Rightarrow a=b=c \\\Rightarrow \triangle ABC為正\triangle \Rightarrow \angle A=60^\circ \Rightarrow \cos A= \bbox[red, 2pt]{1\over 2} \\\textbf{(2) } \cases{3\overrightarrow{CA} \cdot \overrightarrow{AB}=-{3\over 2}(b^2+c^2-a^2) \\ \overrightarrow{BC} \cdot \overrightarrow{CA}=-{1\over 2}(a^2+b^2-c^2) \\ 2\overrightarrow{AB} \cdot \overrightarrow{BC}=-(a^2+ c^2-b^2) } \\\quad \Rightarrow {3\over 2}(b^2+c^2-a^2)={1\over 2}(a^2+b^2-c^2) =(a^2+c^2-b^2) =3t \Rightarrow \cases{b^2+c^2-a^2 =2t\\ a^2+b^2-c^2=6t\\ a^2+c^2-b^2 =3t} \\ \Rightarrow a^2+b^2+c^2=11t \Rightarrow \cases{a^2=9t/2\\ b^2=4t\\ c^2=5t/2} \Rightarrow \cos A={b^2+c^2-a^2\over 2bc}={2t\over 2\sqrt {10} t}= \bbox[red, 2pt]{\sqrt{10} \over 10}$$

解答:$$\textbf{(1) }x^2+y^2=37 \Rightarrow \cases{圓心O(0,0)\\ 圓半徑r=\sqrt{37}} \Rightarrow \overleftrightarrow{OP}斜率m={2} \Rightarrow 過P弦斜率為-{1\over m} \\\quad \Rightarrow 弦直線方程式: y=-{1\over 2}(x-1)+2 \Rightarrow \bbox[red, 2pt]{x+2y=5}  \\\textbf{(2) }假設弦\overline{AB}=3a且弦中點為Q \Rightarrow \cases{\overline{PQ}=a/2 \\ \overline{OP}=\sqrt 5\\ \overleftrightarrow{AB}: y=m(x-1)+2} \Rightarrow \overline{OQ}^2=\overline{OA}^2-\overline{AQ}^2 =\overline{OP}^2-\overline{PQ}^2 \\ \quad \Rightarrow 37-{9\over 4}a^2=5-{1\over 4}a^2 \Rightarrow a=4 \Rightarrow \overline{OQ}= d(O, \overleftrightarrow{AB})=1 \Rightarrow  {|2-m|\over \sqrt{m^2+1}}=1 \Rightarrow m={3\over 4}, \infty \\ \quad \Rightarrow \overleftrightarrow{AB}:\bbox[red, 2pt]{3x-4y+5=0, y=2 } $$

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解題僅供參考，其他教甄試題及詳解