網頁

2019年12月24日 星期二

105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 105 學年度聯合轉學考招生考試
升高二數學科試題
一、單選題


$$\begin{cases}x = { \sqrt 2+1\over \sqrt 2-1} ={ (\sqrt 2+1)^2\over (\sqrt 2-1)(\sqrt 2+1)} =3+2\sqrt 2\\ y = { \sqrt 2-1\over \sqrt 2+1} ={ (\sqrt 2-1)^2\over (\sqrt 2-1)(\sqrt 2+1)} =3 -2\sqrt 2\end{cases}  \Rightarrow x+y=6, 故選:\bbox[red,2pt]{(C)}$$


:$$\begin{cases}a_3+a_5=8 \\ a_6+a_{10} = 32\end{cases}  \Rightarrow \begin{cases} (a_1+2d) +(a_1+4d)=8 \\ (a_1+5d)+ (a_1+9d) = 32\end{cases}  \Rightarrow  \begin{cases} 2a_1+6d=8 \\ 2a_1+14d = 32\end{cases} \Rightarrow 8d= 24 \\ \Rightarrow d=3, 故選\bbox[red,2pt]{(D)}$$



$$6^x\times 8^y\times 9^z= (2^x\times 3^x)\times (2^{3y}) \times (3^{2z})=2^{x+3y}\times 3^{x+2z  } =2^8\times 3^7 \Rightarrow  \begin{cases}x+3y=8 \\ x+2z=7\end{cases} \\ \Rightarrow  \begin{cases}x+3y=8 \\ x+2z=7\end{cases}  x,y,z 皆為正整數\Rightarrow \begin{cases}x=5 \\ y=1\\ z=1\end{cases} \Rightarrow x+y+z=7, 故選\bbox[red,2pt]{(D)}$$


:$$2x^4+3x^3-2x^2-x-2= (x-1)(x+2)(2x^2+x+1), 故選\bbox[red,2pt]{(B)}$$



:$$|a+1|+|a-2|=5 \Rightarrow  \begin{cases}2a-1=5 & a \ge 2\\a+1+2-a=5 & -1\le a \le 2 \\ -2a+1=5 & a\le -1\end{cases} \Rightarrow  a=3,-2\\
同理b=3,-2; 由於a>b \Rightarrow  \begin{cases}a  = 3\\b =-2\end{cases}  \Rightarrow a-b=5, 故選\bbox[red,2pt]{(D)}$$




$$  \begin{cases}-x^2+8x-7>0\\x-2\ne 0\end{cases}  \Rightarrow   \begin{cases}(x-7)(x-1)<0\\ x\ne 2\end{cases} \Rightarrow   \begin{cases}1<x<7\\ x\ne 2\end{cases} \Rightarrow x=3,4,5,6\\其中x=3時,原式變為\log_1{8}不存在 ,剩下4,5,6,共3個, 故選\bbox[red,2pt]{(B)}$$





$$xy^4=10000 \Rightarrow \log(xy^4)= \log(10000) \Rightarrow \log x+4\log y=4在(\log x,\log y)坐標上為一直線,\\且經過(0,1)及(4,0), 故選\bbox[red,2pt]{(A)}$$


:$$(A)\bigcirc:  \begin{cases}a^3為有理數\\ a^8為有理數 \end{cases}  \Rightarrow {a^8\over a^3}=a^5為有理數 \Rightarrow {a^5\over a^3}=a^2為有理數 \Rightarrow {a^3\over a^2}=a為有理數 \\ (B)\times: \begin{cases} a= \sqrt 2 \\ b=-\sqrt 2\end{cases} \Rightarrow \begin{cases} a+b= 0 \\ ab=-2\end{cases} \Rightarrow a,b 皆不是有理數\\(C)\times: 理由同(B)\\(D)\times: \begin{cases} a= 2 \\ b=\sqrt 2\end{cases} \Rightarrow  \begin{cases} a+b= 2+\sqrt 2 \\ b=\sqrt 2\end{cases}\Rightarrow a+b與ab都是無理數,但a不是無理數\\(E)\times: \begin{cases} a= c=\sqrt 2 \\ b=d =- \sqrt 2\end{cases} \Rightarrow a+b,b+c,c+d,d+a皆為有理數,但a,b,c,d皆不是有理數\\,故選\bbox[red,2pt]{(A)} $$


:$$14位同學轉一類、10位同學轉二類,這14位及10位同學是固定的,並非從24位中挑14位出來;\\因此只要考慮14位分兩班及10位分兩班的方法,共有C^{14}_7\times C^{10}_5種,故選\bbox[red,2pt]{(C)} $$


:$$f(-1+i)=0 \Rightarrow x=-1+i \Rightarrow (x+1)^2=i^2 \Rightarrow x^2+2x+2=0 \Rightarrow x^2+2x+2為f(x)的因式\\ 利用長除法可得f(x)=(x^2+2x+2)(x+(a-2)) =x^3+ax^2+(2a-2)x+(2a-4) \\ \Rightarrow  \begin{cases}2a-2=8 \\2a-4=b \end{cases}  \Rightarrow a=5 \Rightarrow (x+(a-2))=(x+3)為f(x)因式 \Rightarrow f(-3)=0,故選\bbox[red,2pt]{(E)}$$



$${患癌症且被檢驗出\over 檢驗發現有癌症} ={1\%\times 0.9 \over 1\%\times 0.9+99\%\times 0.01} ={0.009 \over 0.009+0.0099} = {10\over 21},故選\bbox[red,2pt]{(C)}$$



$$\begin{array}{c|c|cc|ccc|cccc|ccccc}

x&2&3&3&4&4&4& 5&5&5&5& 6& 6& 6& 6&6\\\hline

y& 1& 1& 2& 3 & 1& 2& 4& 1& 2& 3& 5& 1& 4& 2& 3\\\hline

z& 1& 2& 1& 1& 3& 2&1 & 4& 3& 2& 1& 5& 2&4& 3\\\hline

有3& & V & V& V & V& & & &V & V& & & & &V

\end{array}\\

合乎x-y=z共有15種,其中至少1個3的有7種,機率為{7\over 15},故選\bbox[red,2pt]{(B)}$$


:$$將最大及最小的極端值去掉,標準差會減小,但平均值無法確定,故選\bbox[red,2pt]{(D)}$$




$$  任取相異3數,由小至大分別是(x, y,z)\\
x=1有C^8_2種;x=2有C^7_2種;\dots ;x=8有C^2_2種;\\
因此x的平均值為{1\times C^9_2+2\times C^8_2+ \cdots+8\times C^2_2 \over C^9_2+ C^8_2+ \cdots+C^2_2} ={330\over 120}={11 \over 4},故選\bbox[red,2pt]{(A)}$$

二、多重選擇題



$$\begin{cases} a>0\Rightarrow \begin{cases} 直線斜率>0\\ g(x)凹向上 \end{cases}\Rightarrow 圖(C)\\ a<0 \Rightarrow \begin{cases} 直線斜率<0\\ g(x)凹向下 \end{cases}\Rightarrow 圖(B) \end{cases},故選\bbox[red,2pt]{(BC)}$$


:$$\begin{cases}a_n = C^4_n/2^4 \\ b_n= C^8_n/2^8\end{cases} \\(A)\times: a_2=C^4_2/2^4=6/16=3/8\\(B)\times: b_4=C^8_4/2^8= 70/256 =35/128\ne a_2\\(C)\times: b_2=C^8_2/2^8=28/256=7/64 \Rightarrow 2b_2=7/32\ne a_2\\(D)\bigcirc: b_2=C^8_2/2^8= C^8_6/2^8 =b_6\\(E)\bigcirc: \begin{cases}a_3= C^4_3/2^4 =4/16=8/32 \\b_3= C^8_3/2^8 = 56/256 =7/32\end{cases} \Rightarrow a_3>b_3,故選\bbox[red,2pt]{(DE)}$$




:$$(A)\bigcirc: S_{10}=S_{15} \Rightarrow a_1+\cdots+a_{10}=a_1+\cdots+a_{10}+ a_{11}+\cdots +a_{15} \Rightarrow a_{11}+\cdots +a_{15}=0\\  \Rightarrow (a_{11}+a_{15})\times 5\div 2=0 \Rightarrow a_{11}+a_{15}=0 \Rightarrow 2a_1+24d=0 \Rightarrow d=-a_1/12<0\\ (B)\bigcirc: a_{11}+\cdots +a_{15}=0 \Rightarrow   \begin{cases} a_{11}+a_{15}=0\\a_{12}+a_{14}=0\\ a_{13}=0\end{cases}  \Rightarrow \begin{cases} a_{12}+a_{14}=0\\ a_{12}>a_{14}\end{cases} \Rightarrow a_{12}>0\\(C) \bigcirc:d<0 \Rightarrow a_{14}> a_{15} \Rightarrow a_{12}+a_{15} <a_{12}+a_{14}=0 \Rightarrow a_{12}+a_{15}<0 \\\Rightarrow (a_1+11d)+(a_1+14d)<0  \Rightarrow (a_1+d)+ (a_1+24d)<0 \Rightarrow a_2+a_{25}<0\\(D)\bigcirc: S_{14}-S_{11}= a_{12}+a_{13}+a_{14} =0\\(E)\times:由(A)知2a_1+24d=0 \Rightarrow a_1+a_{25}=0 \Rightarrow S_{25}=0 \Rightarrow S_{26}=S_{25}+a_{26}=a_{26}<0\\ ,故選\bbox[red,2pt]{(ABCD)}$$



$$(1+\sqrt 2)^4=C^4_0+C^4_1\sqrt 2+ C^4_2(\sqrt 2)^2+ C^4_3(\sqrt 2)^3+C^4_4 (\sqrt 2)^4 \Rightarrow  \begin{cases}a_4  =C^4_0+2C^4_2+4C^4_4\\ b_4 =C^4_1+2C^4_3 \end{cases}\\(A)\times:a_4  =C^4_0+2C^4_2+4C^4_4\\(B)\bigcirc: b_4 =C^4_1+2C^4_3\\(C)\times: a_4+b_4= 17+12=29\ne 2^4\\(D)\bigcirc: a_4^2-2b_4^2=17^2-2\times 12^2=289-288=1\\(E)\bigcirc: (1-\sqrt 2)^4= C^4_0-C^4_1\sqrt 2+ C^4_2(\sqrt 2)^2- C^4_3(\sqrt 2)^3+C^4_4 (\sqrt 2)^4 \\ =(C^4_0+2C^4_2+4C^4_4)-(C^4_1+2C^4_3)\sqrt 2= a_4-b_4\sqrt 2\\,故選\bbox[red,2pt]{(BDE)} $$


:$$ (A)\bigcirc:  \begin{cases}\sigma(X+5)=\sigma(X) \\\sigma(Y+10)=\sigma(Y)\\ Cov(X+5,Y+10)=Cov(X,y) \end{cases}\Rightarrow 相關係數\rho= {Cov(X,Y)\over \sigma_X\sigma_Y} = {Cov(X+5,Y+10)\over \sigma_{X+5}\sigma_{Y+10}}\\(B)\times:  \begin{cases}\sigma(2X)=2\sigma(X) \\\sigma(-3Y)=3\sigma(Y)\\ Cov(2X,-3Y)=-6Cov(X,y) \end{cases}\Rightarrow 相關係數\rho= {Cov(2X,-3Y)\over \sigma_{2X}\sigma_{-3Y}} = {-6Cov(X,Y) \over 6\sigma_X\sigma_Y}\\ =-{Cov(X,Y)\over \sigma_X\sigma_Y}  \\(C)\bigcirc: \begin{cases}相關係數\rho= {Cov(X,Y)\over \sigma_X\sigma_Y}=0.634\\ 迴歸直線斜率\beta_1={Cov(X,Y) \over \sigma_X^2} =0.635 \end{cases} \Rightarrow {\beta_1\over \rho} ={\sigma_Y \over \sigma_X}={0.635 \over 0.634}>1 \Rightarrow \sigma_Y> \sigma_X \\(D)\times: \sigma_X \ne \sigma_Y  \Rightarrow 迴歸直線斜率不同\\(E)\times: 迴歸直線只是推估,並非百分之百正確\\,故選\bbox[red,2pt]{(AC)}$$




解題僅供參考

沒有留言:

張貼留言