113年嘉科實中雙語部教甄-數學詳解

 國立嘉科實驗高級中學113學年度雙語部教師甄選

第一部份 填充題 50%

解答: $$|z-2|=\sqrt 3 \Rightarrow (x-2)^2+y^2=3\\ 令k={y\over x} \Rightarrow y=kx 代入圓\Rightarrow (x-2)^2+k^2x^2=3 \Rightarrow k=\sqrt{3-(x-2)^2 \over x^2} \\ 取f(x) ={3-(x-2)^2 \over x^2} \Rightarrow f'(x)={-2(x-2) \over x^2}+{-6+2(x-2)^2\over x^3} \Rightarrow x={1\over 2} \\\Rightarrow k=\sqrt{f(1/2)} = \bbox[red, 2pt] {\sqrt 3}$$

解答: $$\cases{\log_9 x+\log_9 y+ \log_3 z=2\\ \log_{16} x+ \log_4 y+ \log_{16}z= 1\\ \log_5 x+ \log_{25} y+ \log_{25}z=0} \Rightarrow \cases{\log_3 z\sqrt{xy} =2\\ \log_4 y\sqrt{xz}=1 \\ \log_5 x\sqrt{yz} = 0} \Rightarrow \cases{z\sqrt {xy}=9\\ y\sqrt{xz}=4\\ x\sqrt{yz} =1} \\ 三式相乘\Rightarrow (xyz)^2=36 \Rightarrow xyz=6 \Rightarrow \cases{z \sqrt{6\over z} =9\\ y\sqrt{6\over y}=4 \\ x\sqrt{6\over x} =1} \Rightarrow \cases{z=27/2\\ y=8/3\\ x=1/6} \Rightarrow (x,y,z)= \bbox[red, 2pt]{({1\over 6},{8\over 3}, {27\over 2})}$$

解答: $$f(x)={2x\over ax+b} \Rightarrow \cases{f(1) ={2\over a+b}=1\\ f({1\over 2}) ={1\over a/2+b}={2\over 3}} \Rightarrow \cases{a+b=2\\ a+2b=3} \Rightarrow \cases{a=1\\ b=1} \Rightarrow f(x)={2x \over x+1} \\ \Rightarrow x_n ={2x_{n-1}\over x_{n-1}+1} \Rightarrow {1\over x_n} ={x_{n-1}+1\over 2x_{n-1}} ={1\over 2}\cdot {1\over x_{n-1}} +{1\over 2}\\ 取y_n={1\over x_n},則y_n={1\over 2}y_{n-1}+{1\over 2} ={1\over 2}({1\over 2}y_{n-2}+{1\over 2})+{1\over 2} ={1\over 2^2}y_{n-2}+{1\over 2^2}+{1\over 2} \\ =\cdots = {1\over 2^{n-1}}y_1+{1\over 2^{n-1}} +{1\over 2^{n-2}} +\cdots +{1\over 2}= {2 \over 2^{n-1}} +{1\over 2^{n-1}} +{1\over 2^{n-2}} +\cdots +{1\over 2} \\={2\over 2^{n-1}}+1-{1\over 2^{n-1}}  =1 + {1 \over 2^{n-1}} \Rightarrow x_n= \bbox[red, 2pt]{1\over 1+({1\over 2})^{n-1}}$$

解答: $$假設O為原點,則S_{\triangle ABC} ={1\over 2}\cdot \overline{OB} \cdot \overline{AC} =2\overline{AC}=4 \Rightarrow \overline{AC} =2 \Rightarrow A(2,0) \\ \Rightarrow 2,4為y=f(x)的兩根 \Rightarrow f(x)=k(x-2)(x-4) \Rightarrow f(0)=8k=-4 \Rightarrow k=-{1\over 2 } \\ \Rightarrow y=f(x)=-{1\over 2}(x-2)(x-4) \Rightarrow f(3)={1\over 2} \Rightarrow D(3,{1\over 2}) \Rightarrow S_{\triangle DBC} ={1\over 2} \begin{Vmatrix} 3& {1\over 2} & 1\\ 0& -4 & 1\\ 4& 0 & 1 \end{Vmatrix} =\bbox[red, 2pt]3$$

解答: $$甲先擲硬幣獲勝的情形:甲正+甲反乙反甲正+ 甲反乙反甲反乙反甲正+\cdots \\\Rightarrow 機率={1\over 2}+ ({1\over 2})^2{1\over 2} + ({1\over 2})^3{1\over 2} +\cdots = {1\over 2}(1+ {1\over 2^2} +{1\over 2^4} + \cdots) ={1\over 2}\cdot {1\over 1-{1\over 2^2}} ={2\over 3}\\第n局甲勝的機率P(n) \Rightarrow P(n) = (1-P(n-1))\cdot {2\over 3} +P(n-1)\cdot {1\over 3} =-{1\over 3}P(n-1)+{2\over 3} \\= (-{1\over 3})^{n-1}P(1)+{2 \over 3}\left( (-{1\over 3})^{n-2} +(-{1\over 3})^{n-3}+ \cdots +1\right)= (-{1\over 3})^{n-1}P(1)+{2 \over 3}\cdot {3\over 4} \left(1-(-{1\over 3})^{n-1} \right)\\ =(-{1\over 3})^{n-1}\cdot {2\over 3}+ {1\over 2}-{1 \over 2}  (-{1\over 3})^{n-1} ={1\over 2}+{1\over 6} (-{1\over 3})^{n-1} ={1\over 2}-{1\over 2} (-{1\over 3})^{n} = \bbox[red, 2pt]{{1\over 2}\left[1-\left(-{1\over 3} \right)^n \right]}$$

解答: $${x^2\over 25} +{y^2\over 9}=1 \Rightarrow \cases{a=5\\ b=3} \Rightarrow c=4 \Rightarrow \cases{左焦點F(-4,0) \\右焦點A(4,0)} \\ \triangle MBF: \overline{MB} \lt \overline{MF}+\overline{FB} \Rightarrow   \overline{MA} +\overline{MB} \lt \overline{MA} +\overline{MF} + \overline{FB} =2a+\overline{FB} =10+2\sqrt{10} \\ \Rightarrow \overline{MA} +\overline{MB}的最大值為 \bbox[red, 2pt]{10+2\sqrt{10}}$$

解答: $$\textbf{Case I }各項參加人數為3,1,1,1,1: 共有C^7_3\cdot 5!種組合，需扣除甲乙皆同在三人組，\\\qquad 即C^5_1 (五人挑一人與甲乙同組)\times 5!，因此實際為(C^7_3-C^5_1)5!= 3600\\ \textbf{Cases II }各項參加人數為2,2,1,1,1:共有{C^7_2C^5_2\over 2!}\times 5!，需扣除甲乙兩人同在二人組，\\\qquad 即C^5_2\times 5!，因此實際為 \left({C^7_2C^5_2\over 2!}-C^5_2\right)\times 5! =11400 \\ \text{Case I + Case II} =3600+11400= \bbox[red, 2pt]{15000}$$

解答: $${5\sin(10k^\circ)-2 \over \sin^2(10k^\circ)} \ge 2 \Rightarrow 2\sin^2(10k^\circ)-5 \sin(10k^\circ)+2\le 0 \Rightarrow (2\sin(10k^\circ)-1)( \sin(10k^\circ)-2) \le 0 \\ \Rightarrow {1\over 2} \le \sin(10k^\circ) \le 2 \Rightarrow k=3,4,5,6,7,8,9,10,11,12,13,14,15，共\bbox[red, 2pt]{13}可能值$$

解答: $$甲供應商哈密瓜直徑大於137的機率=P(X\gt 137) =P(Z\gt{137-133\over 5}) =P(Z\gt 0.8) =0.2119\\ \\{甲供應商 \over 甲供應商+乙供應商} = {0.7\times 0.2119\over 0.7\times 0.2119+ 0.3\times 0.8413} = {7\times 2119\over 7\times 2119+ 3\times8413} = \bbox[red, 2pt]{14833\over 40072}$$

解答:

$$假設\cases{\overline{DP} =\overline{PQ} =\overline{QB}=a\\ \angle DPR=\angle QBC= \theta\\ \angle RPQ=\alpha} \Rightarrow \cases{\overline{CD} =\overline{DR} +\overline{MQ}+ \overline{QN} =2a\sin \theta+ a\sin \alpha\\ \overline{BC}= \overline{BN}+\overline{BN}-\overline{PM} =2a\cos \theta-a\cos \alpha} \\ 正方形邊長相等: \overline{CD} =\overline{BC} \Rightarrow 2\sin \theta+\sin \alpha=2\cos \theta-\cos \alpha \Rightarrow 2(\sin \theta-\cos \theta) =-(\sin \alpha+ \cos \alpha) \\ \Rightarrow 4(\sin \theta-\cos \theta)^2=(\sin \alpha+\cos \alpha)^2 \Rightarrow 4-4 \sin 2\theta=1+\sin 2\alpha \Rightarrow \sin 2\theta={3-\sin 2\alpha \over 4} \\ \Rightarrow {1\over 2}\le \sin 2\theta \le 1 \Rightarrow 30^\circ \le 2\theta \le 90^\circ \Rightarrow 15^\circ \le \theta \le 45^\circ \Rightarrow \theta 最小值\bbox[red, 2pt]{15^\circ}$$

第二部份 計算證明題 30%

解答: $$\textbf{(1)}\; P到F(1,0)的距離與P到y軸的距離差為1 相當於P到F(1,0)的距離與P到直線x=-1的距離相等\\ \quad 依拋物線定義，\cases{準線x=-1 \\焦點F(1,0)} \Rightarrow \bbox[red, 2pt]{y^2=4x}\\\textbf{(2) }極值出現在\overline{AB}=\overline{CD}, 可假設\cases{A({a^2\over 4},a) \\B({b^2\over 4},-b) \\C({b^2\over 4},b)\\ D({a^2\over 4},-a)} \Rightarrow \cases{\overrightarrow{AD} =(0,-2a) \\\overrightarrow{CB} =(0,-2b)} \Rightarrow \overrightarrow{AD} \cdot \overrightarrow{CB}=4ab \\ 假設P,Q分別為A,B在x軸的垂足，則\triangle APF \sim \triangle BQF(AAA) \Rightarrow {\overline{AP} \over \overline{BQ}} ={\overline{AF}=d(L,A)\over \overline{BF}= d(L,B)} \\ \Rightarrow {a\over b} ={a^2/4+1\over b^2/4+1} ={a^2+4\over b^2+4} \Rightarrow ab^2+4a=a^2b+4b \Rightarrow ab(a-b)-4(a-b)=0 \\ \Rightarrow (ab-4)(a-b)=0 \Rightarrow ab=4 (若a=b,則\overline{CD}=x軸,不合)\Rightarrow \overrightarrow{AD} \cdot \overrightarrow{CB}= \bbox[red, 2pt]{16}$$

解答:

解答: $$令s={a+b+c\over 2} \Rightarrow   \left(1 -{a\over s} \right) + \left(1 -{b\over s} \right) +\left(1 -{c\over s} \right) \ge 3\sqrt[3]{\left(1 -{a\over s} \right) \left(1 -{b\over s} \right) \left(1 -{c\over s} \right)} \\ \Rightarrow 3-{a+b+c\over s} =3-{2s\over s} =1 \ge 3\sqrt[3]{\left(1 -{a\over s} \right) \left(1 -{b\over s} \right) \left(1 -{c\over s} \right)} \\ \Rightarrow \left(1 -{a\over s} \right) \left(1 -{b\over s} \right) \left(1 -{c\over s} \right) \le {1\over 27} \Rightarrow \sqrt{\left(1 -{a\over s} \right) \left(1 -{b\over s} \right) \left(1 -{c\over s} \right)} \le {1\over 3\sqrt 3} ={\sqrt 3\over 9}\\ \Rightarrow 三角形面積S_\triangle =\sqrt{s(s-a) (s-b)(s-c)} =s^2 \sqrt{\left(1 -{a\over s} \right) \left(1 -{b\over s} \right) \left(1 -{c\over s} \right)}\le {\sqrt 3\over 9}s^2 \\ \qquad ={\sqrt 3\over 9}\left({a+b+c\over 2} \right)^2 ={\sqrt 3\over 4} \left({a+b+c\over 3} \right)^2 \\ \Rightarrow S_\triangle \le {\sqrt 3\over 4} \left({a+b+c\over 3} \right)^2. \bbox[red, 2pt]{故得證}$$

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解題僅供參考，教甄歷年試題及詳解