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2020年3月25日 星期三

109年身心障礙學生大學甄試-數學乙-詳解


109學年度身心障礙學生升學大專校院甄試試題
甄試類(群)組別:大學組
考試科目(編號):數學乙
單選題,共 20 題,每題 5 分

:$$\cases{\sqrt {10} > 3.1 \\ 2>\sqrt 3 > 1.7}\\(A) |-2-\sqrt 3|>3.7 \\(B) \cases{|\sqrt{10}-1|<3 \\ |-1-\sqrt 3|< 3} \\(C) |0-\sqrt{10}|> 3.1 \\ (D) |1+\sqrt{10}|> 3\\,故選\bbox[red,2pt]{(B)} $$



$$令f(x)=x^3+x-5 \Rightarrow \cases{f(-1)=-7 \\ f(0)=-5 \\ f(1)=-3 \\ f(2)=5 \\ f(3)=25} \Rightarrow f(1)\cdot f(2)<0,故選\bbox[red,2pt]{(C)}$$


:$$(A) |x-3| \le 2 \Rightarrow -2 \le x-3\le 2 \Rightarrow 1\le x\le 5 \\(B) |x+3| \le 2 \Rightarrow -2 \le x+3\le 2 \Rightarrow -5\le x\le -1\\ (C)|x-3| \ge 2 \Rightarrow \cases{x-3\ge 2\\ x-3\le -2} \Rightarrow \cases{x\ge 5\\ x\le 1}\\ (D)|x+3| \ge 2 \Rightarrow \cases{x+3\ge 2\\ x+3\le -2} \Rightarrow \cases{x\ge -1\\ x\le -5}\\,故選\bbox[red,2pt]{(A)}$$


解:
$$P(A\cap B') = P(A)-P(A\cap B) = P(A)-P(A)P(B) ={1\over 3} -{1\over 3}\cdot {1\over 4} ={1\over 3}-{1\over 12} ={1\over 4},故選\bbox[red,2pt]{(B)}$$



:$$加分前後,大家的分數與平均數的距離都沒改變,所以只有標準差不變;\\其它的平均數、中位數及最低分都增加10分,故選\bbox[red,2pt]{(D)}$$




$$A=\left[ \matrix{1 & 0 & 1 \\0 & 1& 1}\right]\left[ \matrix{1 \\ 2 \\ 3}\right] =\left[ \matrix{4 \\ 5}\right],只有(A)與A都是2\times 1,故選\bbox[red,2pt]{(A)}$$




:$$f(x)=ax^2 +bx+4 經過(1,2) \Rightarrow f(1)=a+b+4=2 \Rightarrow a+b=-2\\ (A)g_1(3)=25a+5b+5 =20a+5(a+b)+5 =20a-5 \\(B)g_2(3) =25a+5b+3= 20a-7 \\(C)g_3(3) = a+b+5=3 \Rightarrow g_3經過(3,3) \\(D)g_4(3)= a+b+3=1 \Rightarrow g_4經過(3,1) ,故選\bbox[red,2pt]{(D)}$$



$$f(x)=3^x, \cases{f(a)=100 \\f(b)=5} \Rightarrow \cases{3^a=100 \\ 3^b=5} \Rightarrow f(a-2b)= 3^{a-2b} = \cfrac{3^a}{(3^b)^2} = \cfrac{100}{5^2}=4,故選\bbox[red,2pt]{(A)}$$


:$$\cases{A=\{-2,-1,0,1,2,3,4,5,6,7\}\\ B=\{-5,-4,-3,-2,-1,0,1,2,3\}} \Rightarrow A\cap B=\{-2,-1,0,1,2,3\},有6個元素,故選\bbox[red,2pt]{(B)} $$


:$$令\cases{f(x,y)=2x+3y-2 \\ g(x,y)=2x+3y-6} \Rightarrow \cases{f(A)=7-2>0 \\ f(A)=7-6>0}\\(A)P=(0,0) \Rightarrow \cases{f(P)=-2<0 \\ f(P)=-6<0} \Rightarrow \cases{f(A)f(P)<0 \\ g(A)g(P)<0} \\(B)P=(1,2) \Rightarrow f(P)=6>0 \Rightarrow f(A)f(P) \nless 0 \\(C)P=(0,2) \Rightarrow f(P)=4>0 \Rightarrow f(A)f(P) \nless 0\\ (D)P=(2,0) \Rightarrow f(P)=2>0 \Rightarrow f(A)f(P) \nless 0\\,故選\bbox[red,2pt]{(A)}$$



$$M=1+ 5\log \left({32 \over 27} \right) = 1+5(5\log 2-3\log 3) = 1+5(5\times 0.301-3\times 0.4771) = 1.3685\\,故選\bbox[red,2pt]{(B)}$$


:$$(A) \vec u \cdot (12,4) = -12+12=0\\ (B)\vec u\cdot (12,6)=-12+18>0 \\(C) \vec u\cdot (1,1)=-1+3>0\\ (D) \vec u\cdot (15,4) = -15+12<0\\,故選\bbox[red,2pt]{(D)}$$



$$\{(1,2,1),(1,3,2),(1,4,3),(1,5,4),(1,6,5),\\(2,1,1),(2,3,1),(2,4,2),(2,5,3),(2,6,4),\\\cdots\\(6,1,5), (6,2,4),(6,3,3),(6,4,2),(6,5,1)\}\\ 共有5\times 6=30種,機率為{30\over 6\times 6\times 6}= {5\over 36},故選\bbox[red,2pt]{(D)}$$


:$$(A) y=3為一水平線,相關係數為0\\ (B) 接近y=x \Rightarrow 相關係數接近1 \\(C) 符合y=2x-1 \Rightarrow 相關係數=1\\ (D) y=x^2 \Rightarrow 相關係數遠離1\\,故選\bbox[red,2pt]{(C)}$$


:$$X=1 \Rightarrow S=\{(正,反,反),(反,正,反),(反,反,正)\} \Rightarrow P(S)=3\times {1\over 4}\times ({3\over 4})^2 ={27 \over 64}\\ X=2 \Rightarrow S=\{(正,正,反),(正,反,正),(反,正,正)\} \Rightarrow P(S)=3\times ({1\over 4})^2\times {3\over 4} ={9 \over 64}\\ X=3 \Rightarrow S=\{(正,正,正)\} \Rightarrow P(S)= ({1\over 4})^3 ={1 \over 64}\\因此期望值為1\times {27\over 64} +2 \times {9\over 64} +3\times {1\over 64} = {48\over 64} ={3\over 4},故選\bbox[red,2pt]{(B)}$$



:$$\triangle OPQ 面積={1\over 2}\left|\matrix{6 & 8 \\2 & 4}\right| =\left|\matrix{3 & 4 \\2 & 4}\right|,故選\bbox[red,2pt]{(D)}$$


:$$令\cases{x=第1天的景點數 \\y=第2天的景點數 \\z=第3天的景點數 } \Rightarrow \cases{x+y+z=7 \\ x,y,z\ge 2}\\ \Rightarrow \begin{array}{c|ccc}(x,y,z) & (2,2,3) & (2,3,2) & (3,2,2) \\\hline 規劃數 & C^7_2C^5_2C^3_3 & C^7_2C^5_3C^2_2 & C^7_3C^4_2C^2_2\end{array} \\\Rightarrow 總規劃數=C^7_2C^5_2C^3_3 + C^7_2C^5_3C^2_2 + C^7_3C^4_2C^2_2 = 210+210+210=630,故選\bbox[red,2pt]{(C)}$$





$$A(a,0) \Rightarrow \overline{OA}=a \Rightarrow \overline{OF}=\sqrt 3a \Rightarrow F(0,\sqrt 3)\\ 令f(x,y)=2x+y \Rightarrow \cases{f(A)=2a \\ f(F)=\sqrt 3 a}\Rightarrow f(F)<f(A),故選\bbox[red,2pt]{(D)}$$



:$$組合方式(x,y,z,w),即\cases{50元硬幣x個,0\le x\le 2\\ 10元硬幣y個,0\le y\le 8\\ 5元硬幣z個,0\le z\le 6\\ 1元硬幣w個,0\le z\le 10 } \Rightarrow 50x+10y+5z+w=128 \\ \Rightarrow \begin{array}{}x & y & z& w \\\hline 2 & 2 & 1& 3\\  & & 0 & 8 \\ & 1 & 3& 3 \\ & & 2 & 8 \\ & 0 & 5 & 3 \\ & & 4 & 8\\\hdashline 1 & 7 & 1& 3\\ & & 0 & 8 \\  & 6 & 3 & 3 \\ & & 2 & 8 \\ & 5 & 5 & 3\\ & & 4 & 8 \\ & 4 & 6 & 8 \\\hline\end{array} \Rightarrow 共有13組,故選\bbox[red,2pt]{(B)}$$




$$\cases{{\overline{BE} \over \overline{EC}} =2 \Rightarrow \overrightarrow{AE}= {2\over 3}\overrightarrow{AC} +{1\over 3} \overrightarrow{AB} \\ {\overline{AD} \over \overline{DB}} =1 \Rightarrow \overrightarrow{CD}= {1\over 2} \overrightarrow{CA} +{1\over 2} \overrightarrow{CB} =  {1\over 2} \overrightarrow{CA} +{1\over 2}(\overrightarrow{CA}+ \overrightarrow{AB}) =-\overrightarrow{AC} +{1\over 2}\overrightarrow{AB}} \\ \Rightarrow \overrightarrow{AE} \cdot \overrightarrow{CD} =\left( {2\over 3}\overrightarrow{AC} +{1\over 3} \overrightarrow{AB}\right) \cdot \left(-\overrightarrow{AC} +{1\over 2}\overrightarrow{AB} \right) = -{2\over 3}|\overrightarrow{AC}|^2 +{1\over 6} |\overrightarrow{AB}|^2 =0 \\\Rightarrow {2\over 3} \overline{AC}^2 ={1\over 6} \overline{AB}^2 \Rightarrow {\overline{AB}^2 \over \overline{AC}^2} =4  \Rightarrow {\overline{AB} \over \overline{AC}} =2  ,故選\bbox[red,2pt]{(C)}$$


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