113學年度四技二專統測--數學(B)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(B)

 

解答: $$\cases{\vec a=(x,1) \\ \vec b=(2,y) }\Rightarrow \vec a\cdot \vec b=2x+y=10,故選\bbox[red, 2pt]{(A)}$$

解答: $$|x-2| \lt |2-3| =|3-2|,故選\bbox[red, 2pt]{(A)}$$

解答: $$\cases{\sin(2\theta-360^\circ) = \sin(2\theta) \\ \theta=50^\circ\to100^\circ \Rightarrow 2\theta=100^\circ \to 200^\circ} \Rightarrow \sin(2\theta)從第二象限到第三象限\\ \Rightarrow 從正遞減到負,故選\bbox[red, 2pt]{(A)}$$

解答: $$\cases{L_1\parallel L_2\\ L_2\bot L_3} \Rightarrow L_1\bot L_3 \Rightarrow L_3:x-y+k=0\\ 又L_3通過(113,114) \Rightarrow 113-114+k=0  \Rightarrow k=1 \Rightarrow L_3:x-y+1=0 \\ 當y=0 \Rightarrow x=-1 \Rightarrow  L_3的x截距=－1,故選\bbox[red, 2pt]{(B)}$$

解答: $$(A)\times:f(x)= 2024(x-1)(x-2)-2(x-2)+1 \Rightarrow f(1)=2+1=3\ne 2\\ (B) \times: f(x)= 2024(x-1)(x-2)-3(x-2)+2 \Rightarrow f(1)=3+2=5\ne 2\\(C)\times: f(x)= 2024(x-1)(x-2)+(x-1)+2 \Rightarrow f(2)=1+2=3\ne 1\\ (D)\bigcirc: f(x)= 2024(x-1)(x-2)-(x-2)+1 \Rightarrow \cases{f(1)=1+1= 2\\ f(2)=1}\\ ,故選\bbox[red, 2pt]{(D)}$$

解答: $$正弦定理: {5\over \sin 30^\circ} =2R \Rightarrow 外接圓半徑R=5 \Rightarrow 圓形噴水池面積=25\pi ,故選\bbox[red, 2pt]{(C)}$$

解答: $$40000\times (1.02)^n=80000 \Rightarrow (1.02)^n=2 \Rightarrow \log (1.02)^n =n\log 1.02=\log 2 \\ \Rightarrow n={\log 2\over \log 1.02} ={0.301\over 0.0086} =35 \\ 註:試題有附:\log 2=0.301, \log 1.02=0.0086,故選\bbox[red, 2pt]{(D)}$$

解答: $$假設\cases{A:僅擁有汽車\\ B:僅擁有機車\\C:兩者都擁用} \Rightarrow \cases{A+B+C=100\\ B=A+10\\ C=20} \Rightarrow A+(A+10)+20)=100\\ \Rightarrow A=35,故選\bbox[red, 2pt]{(C)}$$

解答: $$\cases{\alpha+\beta =-3 \\ \alpha \beta=c} \Rightarrow {1\over \alpha} +{1\over \beta}={\alpha+ \beta\over \alpha\beta} ={-3\over c} =1 \Rightarrow c=-3,故選\bbox[red, 2pt]{(D)}$$

解答: $$令f(x,y)=2x-3y-6, 則f(C)=f(1,3)= -13 \lt 0\\ 又\cases{A,C同側 \Rightarrow f(A)f(C) \gt 0\\ B,C異側 \Rightarrow f(B)f(C) \lt 0 } \Rightarrow  \cases{(2a_1-3a_2-6)(-13) \gt 0 \\ (2b_1-3b_2-6)(-13) \lt 0}  \Rightarrow  \cases{(2a_1-3a_2-6)(-13) \gt 0 \\ (2b_1-3b_2-6)(-13) \lt 0} \\ \Rightarrow \cases{(2a_1-3a_2-6)  \lt 0 \\ (2b_1-3b_2-6)  \gt 0}  \Rightarrow (負,正)在第二象限,故選\bbox[red, 2pt]{(B)}$$

解答: $$(x-10)打七折再加上x=x+(x-10)\times 0.7 ,故選\bbox[red, 2pt]{(D)}$$

解答: $$20x+12y \le 70 \Rightarrow  \begin{array}{c}x & y & 20x+12y \\\hline  1 & 1 & 32\\ & 2 & 44\\ & 3& 56\\ & 4 &68 \\\hdashline 2& 1& 52\\ & 2& 64\\\hline \end{array} \Rightarrow 共六種,故選\bbox[red, 2pt]{(B)}$$

解答: $$a_6=88 \Rightarrow a_1+5d=88\\ a_1+a_2+\cdots +a_{11}=a_1+(a_1+d)+\cdots +(a_1+10d)= 11a_1+55d =11a_6=968, 故選\bbox[red, 2pt]{(D)}$$

解答: $$\stackrel{\huge \frown}{AB} =200\pi\times{240\over 360} ={400\pi\over 3}公尺 \Rightarrow 一趟={800\pi\over 3}公尺\\ \Rightarrow {800\pi \over 3}x \gt 8000 \Rightarrow x \gt {30\over \pi} \approx 9.5 \Rightarrow x=10,故選\bbox[red, 2pt]{(C)}$$

解答: $$C^3_1 \times C^5_2=3\times 10=30, 故選\bbox[red, 2pt]{(B)}$$

解答: $$邊長=L_1與L_2的距離={|-11-(-6)|\over \sqrt{3^2+4^2}} ={5\over 5} =1,故選\bbox[red, 2pt]{(A)}$$

解答: $$\cases{y=f(x)=ax^2+1為上方的圖形 \\y=g(x) =bx^2為中間的圖形} \Rightarrow 切線斜率\cases{f'(x)=2ax\\ g'(x)=2bx} \\\Rightarrow 依圖形可知: 對所有x\gt 0,我們有f'(x) \gt g'(x) \Rightarrow a\gt b\\ 又y=cx^2-1圖形為凹向下 \Rightarrow c\lt 0, 因此a\gt b\gt c,故選\bbox[red, 2pt]{(B)}$$

解答: $$餘弦定理: \cos \angle A={\overline{AC}^2 +\overline{AB}^2-\overline{BC}^2 \over 2\cdot \overline{AC}\cdot \overline{AB}}\Rightarrow {1\over 2} ={300^2+800^2-x^2\over 2\cdot 300\cdot 800} \Rightarrow x= \overline{BC}=700 \\ \Rightarrow \overline{AB} +\overline{BC}-\overline{AC} =800+700-300=1200,故選\bbox[red, 2pt]{(C)}$$

解答: $$圓心至兩直線\cases{L_1:5x+12y=16\\ L_2: 5x+12y=-12}的距離都必須大於等於半徑\\(A) \bigcirc:㘣心(-2,1)\cases{至L_1距離=14/13 \gt 1 \\至L_2距離=14/13\gt 1} \\(B) \times: 㘣心(1,1) 至L_1距離=11/13 \lt 1\\(C)  \times: 㘣心(0,0) 至L_2距離=12/13 \lt 1\\ (D)\times: 㘣心(3,-2) 至L_2距離= 3/13 \lt 1\\,故選\bbox[red, 2pt]{(A)}$$

解答: $$\overrightarrow{AB} +\overrightarrow {BC}的方向約為右上往左下 \\(A)\times: 方向為由右向左\\ (B)\times: 方向為往右上\\(C) \bigcirc: 方向為往左下\\ (D)\times: 方向向左\\故選\bbox[red, 2pt]{(C)}$$

解答: $$\cases{E(X)=47500 \\ \sigma(X)=3000} \Rightarrow \cases{x=E(X+5000)= 52500\\ y=\sigma(X+5000)=\sigma(X)=3000} \Rightarrow x\gt 47500;y=3000,故選\bbox[red, 2pt]{(D)}$$

解答: $$假設\cases{小慈投出點數為x \\小巴投出點數為y } \Rightarrow x+y=9,10 \Rightarrow \begin{array}{}x & y\\\hline 3 & 6\\\hdashline 4& 5\\ & 6\\\hdashline 5 & 4\\\hdashline & 5\\\hdashline 6& 3\\6& 4\\\hline\end{array}\\ \Rightarrow 共七種情況,其中小慈勝出的情形(x\gt y): (x,y)=(5,4),(6,3),(6,4)有三種,機率為{3\over 7}\\,故選\bbox[red, 2pt]{(C)}$$

解答: $$此題相當於A,B,C,D,E,F六個字母任排,但A,B不相鄰的排列數\\ =全部-AB相鄰=6!-5!\times 2=720-240=480,故選\bbox[red, 2pt]{(B)}$$

解答: $$假設\cases{A口味生產a個 \\B口味生產b個 } \Rightarrow 在條件\cases{100a+120b\le 3000\\ a+b\le 28\\a\ge 0\\ b\ge 0 }下求f(a,b)=80a+90b的最大值\\ 兩直線\cases{100x+120y=3000\\ x+y=28} 交點為(18,10) \Rightarrow f(18,10)=2340,故選\bbox[red, 2pt]{(A)}$$

解答: $$(A)\bigcirc: 50\times \log 100= 50\times 2=100\\ (B) \bigcirc: 50\times \log 2=50\times 0.301\approx 15.05\lt 60\\(C) \times: \lfloor 50\times \log 2\rfloor =15 \ne 16 \\(D) \bigcirc: 50\times \log 16 =50\times 4\log 2=200\times 0.301=60.2 \Rightarrow \lfloor 60.2 \rfloor=60,故選\bbox[red, 2pt]{(C)}$$

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解題僅供參考,統測歷年試題及詳解