113學年度四技二專統測--數學(A)詳解

 113 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:$$3x-2\lt x+3 \Rightarrow 2x\lt 5 \Rightarrow x\lt 2.5 \Rightarrow x=2, 故選\bbox[red, 2pt]{(D)}$$

解答:$${1\over \tan B}={3\over 4} \Rightarrow \tan B={\sin B\over \cos B}={4\over 3} \Rightarrow \sin B={4\over 3}\cos B \\\Rightarrow \sin^2 B+\cos^2 B= {16\over 9}\cos^2 B+ \cos^2 B ={25\over 9}\cos^2 B=1 \Rightarrow \cos^2B={3\over 5}, 故選\bbox[red, 2pt]{(C)}$$

解答:$$200\times (1.1)^2= 200\times 1.21 = 242, 故選\bbox[red, 2pt]{(A)}$$

解答:$$f(1)=-2+7=5, 故選\bbox[red, 2pt]{(D)}$$

解答:$$P^6_3={6!\over 3!} =120, 故選\bbox[red, 2pt]{(C)}$$

解答:$$\log 3^{10}=10\log 3= 10\times 0.4771 =4.771 \Rightarrow 首數=4, 故選\bbox[red, 2pt]{(C)}$$

解答:$$L通過第一、二、三象限 \Rightarrow \cases{a\lt 0\\ b \gt 0} \Rightarrow (a,b)在第二象限, 故選\bbox[red, 2pt]{(B)}$$

解答:$$若-11\lt a\lt 1,則\cases{|a+11|=a+11\\ |a-1|=1-a} \Rightarrow a+11=1-a \Rightarrow a=-5 \Rightarrow |a-1|=6=k\\, 故選\bbox[red, 2pt]{(B)}$$

解答:$$(A)\times: \cases{x=-4\\ y=3} \Rightarrow x-y+1=-6\ne 0\\ (B)\bigcirc:  \cases{x=-4\\ y=-3} \Rightarrow \cases{x-y+1=0 \\x^2+y^2=25} \\(C)\times: \cases{x=2\\ y=3} \Rightarrow x^2+y^2=13 \ne 25\\ (D)\times: \cases{x=-3\\ y=-2} \Rightarrow x^2+y^2=13 \ne 25\\, 故選\bbox[red, 2pt]{(B)}$$

解答:$$\tan 45^\circ =1 \Rightarrow 水面上高度=水平距離=20, 故選\bbox[red, 2pt]{(A)}$$

解答:$$8^1\cdot 4^4\cdot 2^3\cdot \left( {1\over 2}\right)^a\cdot \left( {1\over 4}\right)^4 =2^3\cdot 2^8\cdot 2^3 \cdot 2^{-a}\cdot 2^{-8}=2^{6-a} =1=2^0\\ \Rightarrow 6-a=0 \Rightarrow a=6, 故選\bbox[red, 2pt]{(A)}$$

解答:$$\log_3 4 +\log_3 3+ \log_3 2+ \log_3 1+ \log_3 {1\over 2} +\log_3{1\over 4}= \log_3 (4\cdot 3\cdot 2\cdot 1 \cdot {1\over 2} \cdot {1\over 4})\\= \log_3 3=1, 故選\bbox[red, 2pt]{(D)}$$

解答:$$\cases{圓心P(3,-4)\\ 原點O(0,0)} \Rightarrow O在圓上:(x-3)^2+(y+4)^2=25 \Rightarrow \overleftrightarrow{OP}: y=-{4\over 3}x \\ \Rightarrow 欲求之直線與\overleftrightarrow{OP}垂直 \Rightarrow 該直線為 y={3\over 4}x \Rightarrow 斜率={3\over 4}, 故選\bbox[red, 2pt]{(B)}$$

解答:$$1380>900+30n \Rightarrow n\lt {480\over 30}=16, 故選\bbox[red, 2pt]{(D)}$$

解答:$$\sum_{k=1}^{10} a_k =\sum_{k=1}^{10} (3k+2) =3\sum_{k=1}^{10} k+2\sum_{k=1}^{10} 1 =3\cdot 55+2\cdot 10=185, 故選\bbox[red, 2pt]{(B)}$$

解答:$$f(x)=g(x) \Rightarrow \cases{a=3\\ b=-2\\ c=-1} \Rightarrow f(x)=g(x)=3x^2+2x-1 \Rightarrow 2f(-1)-3g(-1) \\=-g(-1)=-(3-2-1)=0, 故選\bbox[red, 2pt]{(C)}$$

解答:$$\cases{L_1:3x+2y=6\\ L_2: 3x-2y=6\\ L_3: x=0} \Rightarrow \cases{A=L_1\cap L_2 =(2,0)\\ B= L_1\cap L_3=(0,3)\\ C= L_2\cap L_3=(0,-3)} \Rightarrow \triangle ABC={1\over 2}\cdot 6\cdot 2=6, 故選\bbox[red, 2pt]{(B)}$$

解答:$$先將500個產均分成10組, 再從10組中隨機抽1組, 故選\bbox[red, 2pt]{(A)}$$

解答:$$ \begin{array}{|c|c|c|}\hline 每戶人數& 累積次數&戶數\\\hline 1& 3 & 3\\ 2 & 9 & 9-3=6\\ 3& 17 & 17-9=8\\ 4& 19& 19-17=2\\ 5 & 20 & 20-19=1 \\\hline \end{array} \\ \Rightarrow 每戶3人的有8戶最多 \Rightarrow 每戶人口數的眾數為3, 故選\bbox[red, 2pt]{(A)}$$

解答:$$茜茜與珊珊兩人可坐在AB, CD,或DE三種選擇,兩人可左右互換,因此有3\times 2=6 種\\ 又其他三人任排有3!=6種排法, 因此總共有6\times 6=36種分配, 故選\bbox[red, 2pt]{(C)}$$

解答:$$\cases{m_1\gt 0\\ m_3\lt 0} \Rightarrow m_1m_3\lt 0, 故選\bbox[red, 2pt]{(D}$$

解答:$$\cases{A攝取量=15x+5y\ge 35\\ B攝取量= 5x+15y\ge 25}, 故選\bbox[red, 2pt]{(D)}$$

解答:$$算術平均數=(100+a+ 200+ 300 +400)\div 5=200+{a\over 5} \\ 又100\le a\le 200 \Rightarrow 20\le {a\over 5}\le 40 \Rightarrow 220\le 200+{a\over 5}\le 240, 故選\bbox[red, 2pt]{(C)}$$

解答:$$175一定要被選出,小於175要選2位,高於175也要選2 位\\ 因此共有C^3_2\times c^3_2=9種組合, 故選\bbox[red, 2pt]{(C)}$$

解答:$$\cases{P(A)=0.6\\ P(B)=0.7\\ P(A\cap B)=0.5} \Rightarrow P(A\cup B)=P(A) +P(B)-P(A\cap B)= 0.8\\ \Rightarrow 完全中斷=1-P(A\cup B)=1-0.8=0.2, 故選\bbox[red, 2pt]{(A)}$$

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解題僅供參考,統測歷年試題及詳解