112學年度四技二專統測--數學(A)詳解

112 學年度科技校院四年制與專科學校二年制

統 一 入 學 測 驗- 數學(A)

解答: $$f(x)=(9x^2+1) (2x-3)+6 \Rightarrow \cases{商式:2x-3\\ 餘式: 6},故選\bbox[red,2pt]{(D)}$$

解答: $$18.5\le {x\over 2^2} \le 24 \Rightarrow 74\le x\le 96,故選\bbox[red,2pt]{(A)}$$

解答: $$(x,y)在第二象限 \Rightarrow x\lt 0  \Rightarrow y=2x-3\lt 0 \Rightarrow y\lt 0 \Rightarrow y\le 2x-3 沒有經過第二象限\\, 故選\bbox[red,2pt]{(C)}$$

解答: $$3\times 2\times 2=12,故選\bbox[red,2pt]{(A)}$$

解答: $${ |-1/\sqrt 2-1/\sqrt 2-1|\over \sqrt{1/2+1/2}} = 1+\sqrt 2, 故選\bbox[red,2pt]{(B)}$$

解答: $$(A)\times:-5與3的距離=|-5-3|=8 \ne 2 \\ (B) \times: 0與-2的距離=2 \ne 3\\ (C)\bigcirc: \cases{1與-2的距離=|1-(-2)|=3 \\ 1與3的距離=|1-3|=2} \\(D)\times: 5與-2的距離=|5-(-2)|=7 \ne 3\\,故選\bbox[red,2pt]{(C)}$$

解答: $$a=\sin 137^\circ = \sin (180^\circ-137^\circ) =\sin 43^\circ,$故選\bbox[red,2pt]{(A)}$$

解答: $$x=0時,其值為2, 故選\bbox[red,2pt]{(D)}$$

解答: $$a_9 = S_9-S_8 = (9^2+9-4)-(8^2+8-4)=9^2-8^2+1 = 17+1=18, 故選\bbox[red,2pt]{(A)}$$

解答: $$總費用=200\times 50+5400 = 15400;\\ 令x=50+a \Rightarrow 門票收入=(200-a)\times (50+a)\ge 15400 \\ \Rightarrow (200-(x-50))(50+(x-50)) \ge 15400 \Rightarrow (250-x)(x)\ge 15400,故選\bbox[red,2pt]{(D)}$$

解答: $$\cases{醫師2人,護理師1人有C^5_2C^6_1 =60種派法\\ 醫師1人,護理師2人有C^5_1C^6_2 =75種派法\\ 醫師0人,護理師3人有 C^6_3 =20種派法 } \Rightarrow 共有60+ 75+ 20 =155種派法,故選\bbox[red,2pt]{(A)}$$

解答: $$(A) (0,0) \Rightarrow f(0)=0 \Rightarrow a+2=0 \Rightarrow a=-2 \lt 0\\ (B)\times: (1,0) \Rightarrow f(1)=0,不合,f(1)=2 \ne 0 \\(C) \times:(3,2) \Rightarrow f(3)=2 \Rightarrow 2a+2=3 \Rightarrow a=1/2 \not \lt 0 \\(D)\times: (3,4) \Rightarrow f(3)=4 \Rightarrow 4a+2=4 \Rightarrow a=1/2 \not \lt 0\\故選\bbox[red,2pt]{(A)}$$

解答: $$相切 \Rightarrow 圓心與直線的距離=半徑 \Rightarrow {|-8+3|\over \sqrt{m^2+1}}= {5\over \sqrt {m^2+1}}=\sqrt 5 \\ \Rightarrow m^2+1=5 \Rightarrow m^2=4 \Rightarrow |m|=2,故選\bbox[red,2pt]{(B)}$$

解答: $$\cases{f(x)-g(x)=2(x-1) \cdots(1)\\f(x)+g(x)= 2x(x-1) \cdots(2)} \Rightarrow \cases{f(x)=((1)+(2))\div 2=(x-1)(x+1) \\ g(x)=((2)-(1))\div 2=(x-1)^2} \\ \Rightarrow \cases{f(-1)=0\\ g(-1)=4} \Rightarrow 3f(-1)+2g(-1)=0+8=8, 故選\bbox[red,2pt]{(D)}$$

解答: $$f(x)=x^3+ 9x^2+26x+24 \Rightarrow \cases{a=f(1)=60\\ b=f(-1)=6} \Rightarrow a-b=54,故選\bbox[red,2pt]{(C)}$$

解答: $$\cases{f(x)=a^x\\ g(x)=a^{-x}} \Rightarrow \cases{f(-x)=a^{-x}=g(x)\\ g(-x)=a^x =f(x)} \Rightarrow 兩圖形對稱y軸,故選\bbox[red,2pt]{(D)}$$

解答: $$\log_9 49 \times \log_8 25\times \log_7 4\times \log_5 3 ={\log 49\over \log 9}\times {\log 25\over \log 8}\times {\log 4\over \log 7}\times {\log 3\over \log 5} \\ ={2\log 7\over 2\log 3}\times {2\log 5\over 3\log 2}\times {2\log 2\over \log 7}\times {\log 3\over \log 5} ={2\times 2\over 3} ={4\over 3},故選\bbox[red,2pt]{(C)}$$

解答: $$\cases{A=\{4,5,6\} \\B=\{ 1,2,4,5\} \\C=\{ 2,4,6\}} \Rightarrow A\cap B\cap C=\{4\},故選\bbox[red,2pt]{(B)}$$

解答: $$2\cdot {1\over 10} +4\cdot {2\over 10}+ 6\cdot {3\over 10}+ 8\cdot {4\over 10} ={60\over 10}=6, 故選\bbox[red,2pt]{(C)}$$

解答: $$\cases{臺北:35.8-20.2=15.6\\ 臺中:34.6-22.6=12\\ 臺南:33.8-23.9=9.9\\ 高雄:34.8-23.7=11.1} \Rightarrow 臺南變化最小,故選\bbox[red,2pt]{(C)}$$

解答: $$先將樣本依工作屬性分別抽樣, 各工作屬性又依人數比例隨機抽取,故選\bbox[red,2pt]{(B)}$$

解答: $$\cases{A(-2,-2) \\B(4,6)} \Rightarrow \cases{圓心O=(A+B)\div 2=(1,2) \\ 圓半徑r=\overline{AB}\div 2 =5} \Rightarrow 圓C:(x-1)^2 +(y-2)^2=5^2\\ \Rightarrow \cases{a=1\\ b=2 \\ c=5} \Rightarrow a+b+c=8, 故選\bbox[red,2pt]{(B)}$$

解答: $$\cases{b=d(P,M)=5\\ a=d(P,L_1) =2} \Rightarrow a+b=7,故選\bbox[red,2pt]{(A)}$$

解答: $$\cases{第1排座位數a_1\\ 公差d=3} \Rightarrow a_5= a_1+4d \Rightarrow 27 = a_1+12 \Rightarrow a_1=15 \\ \Rightarrow 座位總數={2a_1+(n-1)d\over 2} \times n \Rightarrow 330={30n +3n(n-1)\over 2} \\ \Rightarrow n^2+9n-220=0 \Rightarrow (n+20)(n-11)=0 \Rightarrow n=11, 故選\bbox[red,2pt]{(D)}$$

解答: $$星期二,三,四可能的考科為(國英數),(國數國),(國數英),(英國英),(英數國),(英數英),共六種\\ 而週六,週日各有2\times 2=4種考科規劃,因此共有6\times 4=24種安排,故選\bbox[red,2pt]{(B)}$$
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