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2022年3月29日 星期二

111年身障升大學-數學B詳解

111 學年度身心障礙學生升學大專校院甄試

甄試類(群)組別:大學組-數學 B

單選題,共 20 題,每題 

解答:$$A(3,-2)相對於y=x的對稱點B(-2,3),B相對於x軸的對稱點C(-2,-3)\\,C相對於y軸的對稱點D(2,-3),因此\overline{AD} =\sqrt{(3-2)^2+(-2-(-3))^2} =\sqrt 2,故選\bbox[red, 2pt]{(A)}$$
解答:$$1-9已抽到到8,剩下8個數取2個,有C^8_2=28種取法;\\可以構成\triangle 三邊長的取法:(8,9,2-7)、(8,7,2-6)、(8,6,3-5)、(8,5,4),\\因此共有6+5+3+1= 15種取法,機率p=15/28 \Rightarrow {1\over 2}\lt p\le 1,故選\bbox[red, 2pt]{(D)}$$
解答:$$y=x^2-3x-2 =(x-{3\over 2})^2 -{17\over 4} \Rightarrow 頂點為({3\over 2},-{17\over 4}) \Rightarrow f(x)=-3(x-{3\over 2})^2-{17\over 4}\\ \Rightarrow f(1)= -3\cdot {1\over 4}-{17\over 4}= -5,故選\bbox[red, 2pt]{(C)}$$
解答:$$b_{n-1} =2^{a_{n-1}-1} \Rightarrow \color{blue}{2^{a_{n-1}}= 2b_{n-1}} \Rightarrow b_n= 2^{a_n-1} =2^{3a_{n-1}+1} =2\cdot (\color{blue}{2^{a_{n-1}}})^3 =2\cdot (2b_{n-1})^3 \\=2\cdot 8b_{n-1}^3 =16b_{n-1}^3,故選\bbox[red, 2pt]{(B)}$$
解答:$$沿著向量(3,4)移動{5\over 2}單位長,相當於\cases{向右移動 {5\over 2}\times {3\over 5} ={3\over 2}單位長\\ 向上移動 {5\over 2}\times {4\over 5} =2單位長};\\也就是n天後,暴風圈變為(x+10-{3\over 2}n)^2 +(y+13-2n)^2 = (5+n)^2\\ 因此n天後暴風圈中心為P(-10+{3\over 2}n, -13+2n),與原點O(0,0)(城市)的距離為\overline{OP}\\ (A)n=1 \Rightarrow \overline{OP}^2 =({3\over 2}-10)^2 +11^2 = {773\over 4} \gt (5+1)^2\\ (B)n=2  \Rightarrow \overline{OP}^2 =(3-10)^2 +9^2 = 130 \gt (5+2)^2 \\(C)n=3 \Rightarrow \overline{OP}^2 =({9\over 2}-10)^2 +7^2 = {317\over 4} \gt (5+3)^2 \\(D)n=4 \Rightarrow \overline{OP}^2 =(6-10)^2 +5^2 = 41 \lt (5+4)^2 \\ ,故選\bbox[red, 2pt]{(D)}$$
解答:$$假設第二個區域的圓心角為\alpha,則r^2\pi \times {\theta\over 2\pi} =(2r)^2\pi \times {\alpha\over 2\pi } \Rightarrow \alpha ={\theta \over 4},故選\bbox[red, 2pt]{(A)}$$
解答:$$每人先分1顆,因此題目變成甲+乙+丙=4;\\因此(甲,乙,丙)= \cases{(0,1,3) \Rightarrow 排列數=6\\ (0,2,2)\Rightarrow 排列數=3\\ (1,1,2) \Rightarrow 排列數=3} \Rightarrow 共有6+3+3=12種分法,故選\bbox[red, 2pt]{(B)}$$
解答:$$假設\overline{AC}=a,令=(\overline{AB} +\overline{BC}+ \overline{CA})\div 2= 9+a/2 ={a+18\over 2}\\\Rightarrow \triangle ABC 面積= \sqrt{s (s-\overline{AB}) (s-\overline{BC}) (s-\overline{CA})}=32 \\\Rightarrow \sqrt{{a+18\over 2} \cdot {a-2\over 2}\cdot {a+2\over 2}\cdot {18-a\over 2}} ={1\over 4}\sqrt{(18^2-a^2)(a^2-2^2)} =32 \Rightarrow (324-a^2)(a^2-4)=16384 \\ \Rightarrow a^4-328a^2+17680=0 \Rightarrow (a^2-68)(a^2-260)=0 \Rightarrow \cases{a=\sqrt{68} \lt 10=\overline{AB},不合\\ a=\sqrt{260}},故選\bbox[red, 2pt]{(C)}$$
解答:$${ 有此基因且試劑顯示有帶\over 有此基因且試劑顯示有帶+ 無此基因且試劑顯示有帶}={10\% \times 95\% \over 10\% \times 95\% +90\%\times 2.5\%} \\= {38\over 47} =0.809 \approx 80\%,故選\bbox[red, 2pt]{(A)}$$
解答:$$\cases{\vec u=(-1,1)\\[1ex] \cos\theta ={\vec u\cdot \vec v\over |\vec u||\vec v|} } \Rightarrow \cases{(A) \vec v=(-1,2) \Rightarrow \cos\theta ={3\over \sqrt{10}} \Rightarrow \cos^2\theta =9/10\\(B)\vec v=(-2,3) \Rightarrow \cos\theta ={5\over \sqrt{26}} \Rightarrow \cos^2\theta =25/26 (最接近1)\\(C) \vec v=(-3,5) \Rightarrow \cos\theta ={8\over \sqrt{68}} \Rightarrow \cos^2\theta =64/68 \\(D)\vec v=(-5,8) \Rightarrow \cos\theta ={13\over \sqrt{178}} \Rightarrow \cos^2\theta =169/178 }\\ \Rightarrow {5\over \sqrt{26}}最大,故選\bbox[red, 2pt]{(B)}$$
解答:$$r = \sqrt[3]{40\times 60\times 70} =10\sqrt[3]{4\times 6\times 7} =10\sqrt[3] {168}\\ 由於125(5^3)\lt 168 \lt 216(6^3) \Rightarrow 10\cdot \sqrt[3]{125}\lt r\lt 10\cdot \sqrt[3]{216} \Rightarrow 50\lt r\lt 60,故選\bbox[red, 2pt]{(B)}$$
解答:$$長度倍率r={1.2公尺\over 5公分} ={120\over 5} =24 \Rightarrow 面積倍率=r^2 =24^2 \\\Rightarrow 牆上面積=0.14\times 0.1\times 24^2 = 8.064,故選\bbox[red, 2pt]{(D)}$$
解答:$$假設此正弦波為M\sin(kx),由於\cases{振幅=2 \Rightarrow M=2\\ 週期=0.2 \Rightarrow k = 2\pi/0.2 = 10\pi} \Rightarrow 2\sin (10\pi x),故選\bbox[red, 2pt]{(C)}$$
解答:$$把圓壓扁變為橢圓,故選\bbox[red, 2pt]{(D)}$$
解答:$$立方體的頂點為三個稜邊的交點,因此L與12個稜邊相交,剩下12-6=6個不相交,故選\bbox[red, 2pt]{(C)}$$
解答:$$|x-3| \le 8 \Rightarrow -8\le x-3\le 8 \Rightarrow -5\le x\le 11,令A=\{-5\le x\le 11, x\in \mathbb{Z}\} \\(A) a=-8 \Rightarrow |x+8| \gt 10 \Rightarrow B=\{x \gt 2,x\lt -18\} \Rightarrow A\cap B=\{3\le x\le 11 \},9個整數\\(B) a=-9 \Rightarrow |x+9| \gt 10 \Rightarrow B=\{x \gt 1,x\lt -19\} \Rightarrow A\cap B=\{2\le x\le 11 \},10個整數 \\(C) a=-10 \Rightarrow |x+ 10| \gt 10 \Rightarrow B=\{x \gt 0,x\lt -20\} \Rightarrow A\cap B=\{1\le x\le 11 \},11個整數 \\(D) a=-11 \Rightarrow |x+ 11| \gt 10 \Rightarrow B=\{x \gt -1,x\lt -21\} \Rightarrow A\cap B=\{0\le x\le 11 \},12個整數\\,故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{a^{1/2}b^{1/3} =100\\[1ex] \cfrac{a^{1/3}}{b^{1/2}} =10} \Rightarrow \cases{\log (a^{1/2}b^{1/3}) = \log(100) \\[1ex] \log(\cfrac{a^{1/3}}{b^{1/2}}) = \log(10)} \Rightarrow \cases{{1\over 2}\log a+ {1\over 3}\log b= 2\\[1ex] {1\over 3}\log a-{1\over 2}\log b=1} \\ \Rightarrow \cases{\log a=48/13\\ \log b=6/13},故選\bbox[red, 2pt]{(D)}$$
解答:$$以五位同學為例,國英數社四科得分分別為\cases{甲:2,2,2,2\\ 乙:1,1,1,100\\ 丙:1,1,100,1 \\ 丁:1,100,1,1 \\ 戊:100,1,1,1 };\\則各科中位數皆是1,甲每一科得分都高於中位數,但總分卻是最後一名,故選\bbox[red, 2pt]{(D)}$$
解答:$$\cases{A=(-1,-4) \\ \overline{AB}斜率=1\\ \overline{AC}斜率=-1}\Rightarrow \cases{B(-1+a,-4+a)\\ C(-1-b,-4+b)},又\overline{BC}斜率=-3 \Rightarrow {b-a\over -b-a}=-3 \Rightarrow b=-2a\\ 因此{\overline{AB} \over \overline{AC}} ={\sqrt{a^2+a^2} \over \sqrt{b^2+b^2}} = {\sqrt 2|a|\over \sqrt 2|b|}  =\left|{a\over b} \right| ={1\over 2},故選\bbox[red, 2pt]{(B)}$$
解答:$$x\begin{bmatrix} 3\\ 1\end{bmatrix} +y \begin{bmatrix} -5\\ -1\end{bmatrix} = \begin{bmatrix} 2\\ 1 \end{bmatrix} \Rightarrow \cases{3x-5y=2\\ x-y=1} \Rightarrow \cases{x=3/2\\ y=1/2} \\\Rightarrow x\begin{bmatrix} 2\\ -1\end{bmatrix} +y \begin{bmatrix} 4\\ 3\end{bmatrix} ={3\over 2}\begin{bmatrix} 2\\ -1\end{bmatrix} +{1\over 2} \begin{bmatrix} 4\\ 3\end{bmatrix} =\begin{bmatrix} 5\\ 0\end{bmatrix},故選\bbox[red, 2pt]{(A)}$$

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