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2022年2月7日 星期一

105年身障生升大學-數學乙詳解

105 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-數學乙

單選題,共 20 題,每題 5 分

解答:$$\cases{x=|a-1|\\ y=2(b-1)^2\\ z= 3\sqrt{c-1}} \Rightarrow \cases{x=0,1,2,3,\dots\\ y=0,2,4,6,\dots\\ c=0,3,6,9,\dots(只需考慮c為整數)}  \Rightarrow x+y+z=2 \\ \Rightarrow (x,y,z)=\cases{(2,0,0)\Rightarrow (a,b,c)=(-1,1,1),(3,1,1)\\(0,2,0)\Rightarrow (a,b,c)=(1,0,1),(1,2,1)} \Rightarrow 共四組解,故選\bbox[red,2pt]{(C)}$$
解答:$$f(x)= 5x^3+ax^2-3x+b = (x^2-x-2)p(x)+x+4 =(x-2)(x+1)p(x)+x+4\\ \Rightarrow \cases{f(2)=40+4a-6+b=6\\ f(-1)=-5+a+3+b=3} \Rightarrow \cases{4a+b=-28\\ a+b=5} \Rightarrow \cases{a=-11\\ b=16} \Rightarrow a+b=5,故選\bbox[red,2pt]{(D)}$$
解答:$$x^3-ax^2+a^2x-a^3= x^2(x-a)+a^2(x-a)= (x^2+a^2)(x-a)=0\\ \Rightarrow \cases{x=a為一實數解\\ x^2=-a^2 為複數解(非實數)},故選\bbox[red,2pt]{(B)}$$
解答:$$-{2\over 3}\lt x\lt 4 \Rightarrow (x+{2\over 3})(x-4)\lt 0 \Rightarrow x^2-{10\over 3}x-{8\over 3}\lt 0 \\ \Rightarrow {3\over 2}x^2-5x-4\lt 0\Rightarrow \cases{(A)\times:a=3/2 \gt 0\\ (B)\times:b=-4\lt 0 \\(C)\times: a+b=-5/2\\(D)\bigcirc: ab=-6},故選\bbox[red,2pt]{(D)}$$
解答:$$a=10^{-0.8} \Rightarrow \log a=-0.8 \\ b=\log 8=3\log 2 =3\times 0.301=0.903 \Rightarrow \log b\approx \log {900\over 1000} \\\; =2\log 3+2-3= 2\cdot 0.4771-1=-0.0458\\ c=\sqrt 3/2 \Rightarrow \log c={1\over 2}\log 3-\log 2={1\over 2}0.4771-0.301=-0.06245\\ 因此b\gt c\gt a,故選\bbox[red,2pt]{(B)}$$
解答:$$100^x-6\cdot 10^x+5=0 \Rightarrow (10^x)^2-6\cdot 10^x+5=0 \Rightarrow (10^x-5)(10^x-1)=0\\ \Rightarrow \cases{10^x=5 \Rightarrow x=\log 5\\ 10^x=1 \Rightarrow x=0},故選\bbox[red,2pt]{(B)}$$
解答:$$\cases{固定利率:100\times (1+2\%)^3= 106.1208\\ 機動利率:100(1+1.8\%)(1+2\%)(1+2.2\%) =106.120392};或\\ (1+1.8\%)(1+2.2\%)= (1+2\%-0.2\%)(1+2\%+0.2\%) =(1+2\%)^2-(0.2\%)^2\\ \Rightarrow (1+2\%)^2 \gt (1+1.8\%)(1+2.2\%) \Rightarrow 固定利率\gt 機率利率,故選\bbox[red,2pt]{(A)}$$
解答:$$\cases{A(1,1)\\ P(x,y)} \Rightarrow \overrightarrow{AP}= (x-1,y-1)= 2\vec u-\vec v=2(2,-1)-(0,3)=(4,-5)\\ \Rightarrow \cases{x-1=4\\ y-1=-5} \Rightarrow \cases{x=5\\ y=-4},故選\bbox[red,2pt]{(B)}$$
解答:$$本題\bbox[red,2pt]{(送分)}$$
解答:$$\cases{SSO的排列數為3\\ 4458的排列數為6 (最後一碼有2種選擇,剩下44X排列數為3,合計2\times 3=6)} \\ \Rightarrow 共有3\times 6=18種車牌號碼,故選\bbox[red,2pt]{(A)}$$
解答:$$奇數有3個、偶數也有3個\\\Rightarrow \cases{P(A)=C^3_1/6=1/2\\ B=(奇,奇)或(偶,偶) \Rightarrow P(B)= (C^3_1C^3_1+ C^3_1C^3_1)/6^2 =18/36=1/2 \\C=2奇1偶或3偶 \Rightarrow P(C)=(3\cdot C^3_1C^3_1C^3_1+C^3_1C^3_1C^3_1)/6^3 =108/216=1/2}\\ \Rightarrow 三個機率值都是0.5,故選\bbox[red,2pt]{(D)}$$
解答:$$\cases{A=\begin{bmatrix} 2 & -4\\ 1 & -2 \end{bmatrix} \\[1ex] B=\begin{bmatrix} -2 & -4\\ 1 & 2 \end{bmatrix} } \Rightarrow \cases{A+B= \begin{bmatrix} 0 & -8\\ 2 & 0 \end{bmatrix} \\[1ex] AB= \begin{bmatrix} -8 & -16\\ -4 & -8 \end{bmatrix} \\[1ex]BA= \begin{bmatrix} -8 & 16\\ 4 & -8 \end{bmatrix}}  \Rightarrow A^2+2AB+ B^2 =(A+B)^2-BA+AB \\=\begin{bmatrix} 0 & -8\\ 2 & 0 \end{bmatrix}\begin{bmatrix} 0 & -8\\ 2 & 0 \end{bmatrix}-\begin{bmatrix} -8 & 16\\ 4 & -8 \end{bmatrix} + \begin{bmatrix} -8 & -16\\ -4 & -8 \end{bmatrix} = \begin{bmatrix} -16 & 0\\ 0 & -16 \end{bmatrix}+ \begin{bmatrix} 0 & -32\\ -8 & 0 \end{bmatrix}\\ = \begin{bmatrix} -16 & -32\\ -8 & -16 \end{bmatrix},故選\bbox[red,2pt]{(C)}$$
解答:$${0-(-2)\over 2-(-2)} ={2\over 4}={1\over 2},故選\bbox[red,2pt]{(C)}$$
解答:$$\cases{原始分數x\\ 新分數y} \Rightarrow y={1\over a}x+b \Rightarrow \cases{\mu_Y={1\over a}\mu_X+b\\ \sigma_Y = \lvert {1\over a}\rvert \sigma_X} \Rightarrow \cases{68={1\over a} \cdot 40+b \\ 12= \lvert {1\over a}\rvert\cdot 10} \\ \Rightarrow \cases{1/a=1.2  \Rightarrow b=20 \Rightarrow b/a=24\\1/a=-1.2 \Rightarrow b=116 \Rightarrow b/a= -139.2},故選\bbox[red,2pt]{(D)}$$
解答:$$\begin{bmatrix} 下學期吃自助餐\\ 下學期吃麵\end{bmatrix}=\begin{bmatrix} 80\% & 30\%\\ 20\% & 70\%\end{bmatrix}\begin{bmatrix} 這學期吃自助餐\\ 這學期吃麵 \end{bmatrix}\\ \Rightarrow 轉移矩陣= \begin{bmatrix} 80\% & 30\%\\ 20\% & 70\%\end{bmatrix},故選\bbox[red,2pt]{(C)}$$
解答:$$車子被偷\cases{65\%慣竊所為\cases{25\% 兩天內找回\\ 15\%超過兩天找回\\ 60\%找不回}\\[1ex] 35\%非慣竊所為 \cases{36\%兩天內找回\\ 56\%超過兩天找回\\ 8\%找不回}} \\\Rightarrow 被偷一直找不回的機率=65\%\times 60\%+ 35\%\times 8\% = 41.8\%,故選\bbox[red,2pt]{(C)}$$
解答:$$假設同時籌辦園遊會及成果展的女生有a人,男生有5-a人;\\因此22 = 8+6+10-(5-a) \Rightarrow a=3,故選\bbox[red,2pt]{(C)}$$
解答:$$令\cases{R:取到的紅球數\\ W:取到的白球數\\ B:取到的黑球數} \Rightarrow  \begin{array}{} X=(R,W,B) & P(X)\\\hline (3,0,0),(0,3,0),(0,0,3) & 1/C^9_3\\ \hdashline (2,1,0),(2,0,1),(1,0,2) &  C^3_2C^3_1C^3_1/C^9_3\\ (1,2,0),(0,1,2),(0,2,1) &\\\hdashline (1,1,1) & C^3_1C^3_1C^3_1/C^9_3\\\hline\end{array}\\ \Rightarrow 期望值={1\over C^9_3}(60+15 -30) + {C^3_2C^3_1C^3_1\over C^9_3}(25+ 10-10+ 0-15+ 0) +{C^3_1C^3_1C^3_1 \over C^9_3}\cdot 5\\ ={1\over 84}\cdot 45+ {27\over 84}\cdot 10+{27\over 84}\cdot 5 = {450\over 84} \approx 5.36,故選\bbox[red,2pt]{(A)}$$
解答:$$\begin{array}{c|rrrrr}X&7 & 11 & 12 & 13 & 17\\Y& 14 & 23 & 25 & 26 & 32\\\hdashline X^2 &49 & 121 & 144 & 169 & 289\\Y^2 &196 & 529 & 625 & 676 & 1024\\XY & 98 & 253 & 300 & 338 & 544\end{array} \Rightarrow \cases{\sum X=60\\ \sum Y=120\\ \sum X^2=772\\ \sum Y^2=3050\\ \sum XY=1533} \\ \Rightarrow r={\sum XY-(\sum X)(\sum Y)/n\over \sqrt{\sum X^2-(\sum X)^2/n}\cdot \sqrt{\sum Y^2-(\sum Y)^2/n}} ={1533-60\cdot 120/5\over \sqrt{772-60^2/5}\cdot \sqrt{ 3050-120^2/5}} \\={93\over \sqrt{52}\cdot \sqrt{170}}\approx 0.989,故選\bbox[red,2pt]{(D)}\\\bbox[blue,2pt]{註}:由於(x,y) 幾乎滿足直線y=2x,也就是r=1$$
解答:$$,故選\bbox[red,2pt]{(C)}$$

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