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2020年2月5日 星期三

104學年度臺北市聯合轉學考-高中升高二-數學科詳解


臺北市高級中等學校 104 學年度聯合轉學考招生考試
升高二數學科試題
一、單選題


$$\begin{cases} a=\sqrt {11}+\sqrt 7 \\ b=4+\sqrt 2 \\ c=2\sqrt 3+\sqrt 6\end{cases} \Rightarrow \begin{cases} a^2= 18+  2\sqrt {77}>18+2\sqrt{72} =18+12\sqrt 2 \\ b^2= 18+8\sqrt 2 \\ c^2 =18 +12\sqrt {2}\end{cases} \\\Rightarrow a^2>c^2 >b^2 \Rightarrow a>c>b, 故選:\bbox[red,2pt]{(E)}$$


:$${11+3i \over a+bi} =2+i \Rightarrow (a+bi)(2+i)= 11+3i \Rightarrow 2a-b+(a+2b)i=11+3i \\ \Rightarrow \begin{cases} 2a-b =11 \\ a+2b=3\end{cases} \Rightarrow \begin{cases} a=5 \\ b=-1\end{cases}, 故選\bbox[red,2pt]{(D)}$$



$$99999 = 9\times 41 \times 271 \Rightarrow {1\over 41} = {9\times 271 \over 99999} = {2439 \over 99999} = 0.\overline{02439} \\ 2014 = 5\times 402+4 \Rightarrow 第2014位數字=第4位數字=3, 故選\bbox[red,2pt]{(C)}$$


:$$1^2-2^2+3^2-4^2+5^2-6^2+ \cdots+49^2-50^2 \\ = 1^2+(3^2-2^2)+(5^2-4^2)+ \cdots+(49^2-48^2)-50^2 \\=1+5+9+\cdots + 97-50^2 = {(97+1)\times 25 \over 2}-50^2 =1225-2500 =-1275 , 故選\bbox[red,2pt]{(C)}$$



$$\begin{cases}\tan \theta< 0\\\sin \theta< 0\end{cases}  \Rightarrow \begin{cases}\cos \theta> 0\\\sin \theta< 0\end{cases}  \Rightarrow \tan 2\theta = {\sin 2\theta \over \cos 2\theta} ={2\sin \theta \cos \theta \over 2\cos^2\theta-1} \Rightarrow {分子<0 \over 分母無法判定}, 故選\bbox[red,2pt]{(E)}$$




$$x(x-\log_3 2)(x- \log_3 12)(x-\log_3 54) \le 0 \Rightarrow 0\le x\le \log_3 2 或 \log_3 12 \le x\le \log_3 54\\ \Rightarrow x的區間長度為\log_3 2+ (\log_3 54-\log_3 12) = \log_3 2+\log_3{54 \over 12} =\log_3 (2\times{54 \over 12}) \\ =\log_3 9 =2, 故選\bbox[red,2pt]{(A)}$$





$$\log \left({12 \over 7} \right)^{100} =100(\log 12-\log 7) =100(2\log 2+\log 3-\log 7) =100(2\times 0.301+0.4771-0.8451) \\ =23.4 \Rightarrow K=23+1=24; 又\log 2< 0.4<\log 3 \Rightarrow S=2 \Rightarrow (K,S)=(24,2), 故選\bbox[red,2pt]{(D)}$$


:$$x_1+x_2+x_3+x_4+x_5 =8 有H^5_8=C^{12}_8 = 495組非負整數解,故選\bbox[red,2pt]{(B)} $$



$$50 \times C^{49}_6 = 699190800,故選\bbox[red,2pt]{(C)} $$


:$$(甲,乙,丙) = (0,0,5),(0,1,4),(0,2,3),(1,1,3),(1,2,2)\\\Rightarrow 3C^5_0C^5_0C^5_5 + 3!\times C^5_0C^5_1C^4_4 + 3!\times C^5_0C^5_2C^3_3 + 3\times C^5_1C^4_1C^3_3 + 3\times C^5_1C^4_2C^2_2 =243\\
(甲=2,乙,丙) =(2,0,3),(2,1,2),(2,2,1),(2,3,0) \Rightarrow C^5_2C^3_0C^3_3 +C^5_2C^3_1C^2_2 +C^5_2C^3_2C^1_1 +C^5_2C^3_3C^0_0\\ =C^5_2(C^3_0C^3_3 +C^3_1C^2_2 +C^3_2C^1_1 +C^3_3C^0_0) =10(1+3+3+1)=80\\ 因此所求機率為 {80 \over 243 },故選\bbox[red,2pt]{(B)}$$



$$P(A\cup B)=P(A) +P(B)-P(A\cap B) =P(A) +P(B)-P(A)\times P(B) \\\Rightarrow 0.7=0.4+P(B) -0.4P(B) \Rightarrow 0.3=0.6P(B) \Rightarrow P(B)=0.5,故選\bbox[red,2pt]{(C)}$$


$$(2+i)^6 =a+bi= C^6_02^0i^6 +C^6_12^1i^5 +C^6_22^2i^4 +C^6_32^3i^3 +C^6_42^4i^2 +C^6_52^5i^1 +C^6_62^6i^0 \\ \Rightarrow b= C^6_12^1-C^6_32^3+C^6_52^5,故選\bbox[red,2pt]{(B)}$$


:$$ {合唱團高三男生人數 \over 合唱團男生人數} ={0.3\times {2\over 5} \over 0.4\times {1\over 3}+0.3\times {1\over 5}+0.3\times {2\over 5}} ={{18\over 150} \over {20+9+18 \over 150}} = {18\over 47},故選\bbox[red,2pt]{(B)}$$





$$\begin{cases} 第1組 \begin{cases} \overline{X}=40 \\ \sigma_X=5 \end{cases} \Rightarrow 5^2= EX^2-40^2 \Rightarrow EX^2 = 1625\\ 第2組 \begin{cases} \overline{Y}=70 \\ \sigma_Y=10 \end{cases} \Rightarrow 10^2 =EY^2 - 70^2 \Rightarrow EY^2 =5000\end{cases} \\ \Rightarrow 第1組與第2組合併\Rightarrow 平均值\overline{Z}= {n_X\cdot \overline{X} +n_Y \cdot \overline{Y} \over n_X+n_Y} ={20\times 40+ 10\times 70 \over 20+10} = 50\\ \Rightarrow 標準差\sigma_Z= \sqrt{EZ^2-(EZ)^2}  = \sqrt{{1\over n_X+n_Y}(n_XEX^2 +n_YEY^2)-50^2} \\=\sqrt{{1\over 30}( 20\times 1625+10 \times 5000)- 2500 } = \sqrt{250}\\  \Rightarrow \sqrt {225}< \sigma_Z< \sqrt{256} \Rightarrow 15<\sigma_Z < 16,故選\bbox[red,2pt]{(D)}$$

二、多重選擇題



$$(A)\times: \sum_{k=1}^{20}5 = 5+5+\cdots+5 = 5\times 20=100\\(B)\bigcirc: \sum_{k=1}^{20}k = 21\times 20 \div 2=210\\(C) \bigcirc: \sum_{k=3}^{22}(k-2)^2 = 1^2+2^2+\cdots +20^2 =\sum_{k=1}^{20}k^2\\(D) \bigcirc: \sum_{k=1}^{20}k^3 = \left( {21\times 20\over 2}\right)^2 =210^2 =44100\\(E)\times: 1^4+2^4+3^4+\cdots 20^4 \ne \left( 1+2+4+\cdots + 20\right)^4\\,故選\bbox[red,2pt]{(BCD)}$$


:$$(A)\bigcirc: b^2-4ac<0 \Rightarrow f(x)與x軸無交點;又f(-1)<0,因此圖形為凹向下\Rightarrow a<0\\(B)\times: \begin{cases} f(-1)=-4 \\ f(3)=-1\end{cases} \Rightarrow \begin{cases} a-b+c=-4\cdots(1) \\ 9a+3b+c=-1 \cdots(2)\end{cases} ,(2)-(1)\Rightarrow 8a+4b=3,由於a<0\\ \qquad,因此b>0\\(C) \bigcirc: 圖形為凹向下且與x軸無交點\Rightarrow f(x)<0,\text{for all }x \Rightarrow f(0)=c<0\\(D) \bigcirc: 由(B)知8a+4b=3且b>0 \Rightarrow a+b/2>0 \Rightarrow a+b/2+b/2 =a+b>0 \\\qquad \Rightarrow f(1)-f(0)=a+b >0\\(E)\times: 圖形為凹向下,頂點坐標為\left(-{b\over 2a},f(-{b\over 2a})\right) \Rightarrow  f(x) 為\begin{cases}遞減 & x > -{b\over 2a} \\ 遞增 & x < -{b\over 2a}\end{cases} \\ \Rightarrow   \begin{cases}f(4)>f(5) & 4> -{b\over 2a} \\f(4)<f(5) & 5<-{b\over 2a} \end{cases} ;現在由(B)知8a+4b=3 \Rightarrow  b={3\over 4}-2a 代回(1) \\\qquad \Rightarrow c=-{13\over 4}-3a <0  \Rightarrow a>-{13 \over 12} \Rightarrow {3\over 8a}<-{9 \over 26} \\ \qquad 因此 4- (-{b\over 2a}) =4+ {3\over 8a}-1 =3+{3\over 8a} < 3-{9 \over 26}= {69\over 26} >0\\ \qquad \Rightarrow  4> -{b\over 2a} \Rightarrow f(4)>f(5)\\,故選\bbox[red,2pt]{(ACD)}$$






$$(A)\times: 等差數列為遞增或遞減數列,不可能同時成立\\(B) \bigcirc: -1,1,-1 為公比r=-1的等比數列,符合b_2>b_1且b_2>b_3\\(C)\times: -10,-1,8為公差d=9的等差數列,a_1+a_2=-11<0,但a_2+a_3=7>0\\ (D)\bigcirc: \begin{cases}b_1b_2 = b_1^2r\\ b_2b_3 = b_1^2r^3 \end{cases},因此b_1b_2<0 \Rightarrow r<0 \Rightarrow r^3<0\Rightarrow b_2b_3<0\\(E)\times: b_1=4, 公比r={3\over 2} \Rightarrow b_1,b_2,b_3 = 4,6,9,但4不能整除6\\,故選\bbox[red,2pt]{(BD)}$$


$$(A)\bigcirc: a^0=1 \Rightarrow 圖形通過(0,1)\\(B)\times: 0.5^2=0.25<1 \\(C) \bigcirc: 交於(k,a^k) \\(D) \times: y=({1\over 2})^x, 其圖形上任兩點連線的斜率小於0\\(E)\bigcirc: {a^{100}+a^{200} \over 2} > \sqrt {a^{100}\times a^{200}} =\sqrt{a^{300}} =a^{150}\\,故選\bbox[red,2pt]{(ACE)} $$


:$$ 迴歸直線必經(\bar x, \bar y)=(65,70),並由題意知迴歸直線經過(5,46);\\因此可求得該直線方程式為y=0.4x+44\\(A) \times: 斜率為0.4\\(B)\bigcirc:迴歸直線必經(\bar x, \bar y)=(65,70)\\(C) \times: y=0.4x+44 \Rightarrow \sigma(Y)=0.4\sigma(X) \Rightarrow \sigma(Y)<\sigma(X)\\(D) \bigcirc: 理由同(C)\\(E)\times: 不一定,迴歸直線只是推估\\,故選\bbox[red,2pt]{(BD)}$$




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