113年身障生升四技二專-數學(A)詳解

113 學年度身心障礙學生升學大專校院甄試

甄試類(群)組別：四技二專組-數學(A)

單選題，共 20 題，每題 5 分

解答:$$當x＝3時有最小值n \Rightarrow f(x)=(x-3)^2+n= x^2-6x+9+n =x^2+mx-6\\ \Rightarrow \cases{m=-6\\ 9+n=-6 \Rightarrow n=-15} \Rightarrow m-n=-6+15= 9,故選\bbox[red, 2pt]{(B)}$$

解答:$$A(2,5)至直線x+4y=8的距離即為高 \Rightarrow {2+20-8\over \sqrt{1+4^2}} ={14\over \sqrt{17}} ={k\over \sqrt{17}} \Rightarrow k=14,故選\bbox[red, 2pt]{(C)}$$

解答:$$\cases{f(x)除以x-2得餘式14 \\f(x)除以x+1得餘式5 } \Rightarrow \cases{f(2)=14\\ f(-1)=5} \Rightarrow \cases{4+2b+c=14\\ 1-b+c=5}\\ \Rightarrow \cases{b=2\\c=6} \Rightarrow c-b= 6-2=4,故選\bbox[red, 2pt]{(A)}$$

解答:$$|2x+3|\lt 5 \Rightarrow -5\lt 2x+3\lt 5 \Rightarrow -4\lt x\lt 1 \cdots (1)\\ 1\lt |2x+3| \Rightarrow \cases{2x+3\gt 1\\ 2x+3\lt -1} \Rightarrow \cases{x\gt -1\\ x\lt -2} \cdots(2)\\ (1)\cap (2) \Rightarrow \cases{-1\lt x\lt 1\\ -4\lt x\lt -2} \Rightarrow |b-d|=|1-(-2)|=3,故選\bbox[red, 2pt]{(C)}$$

解答:$$2^{2x}-9\times 2^x+8=0 \Rightarrow (2^x-8) (2^x-1)=0 \Rightarrow \cases{2^x=8\\ 2^x=1} \Rightarrow \cases{x=3\\ x=0} \Rightarrow a+b=3+0=3\\ ,故選\bbox[red, 2pt]{(D)}$$

解答:$$\cases{a=\log_{0.3} 0.5 ={\log_{10} 0.5 \over \log_{10} 0.3 } ={\log_{10} {5\over 10} \over \log_{10}{3\over 10}} ={\log_{10} 5-1\over \log_{10} 3-1} ={1-\log_{10} 5 \over 1-\log_{10}3} \\ b=\log_{0.3} 0.05  ={\log_{10} 0.05 \over \log_{10} 0.3 } ={\log_{10} {5\over 100} \over \log_{10}{3\over 10}} ={\log_{10} 5-2\over \log_{10} 3-1}  ={2-\log_{10} 5 \over 1-\log_{10}3} \\ c=\log_{0.3} 0.8  ={\log_{10} 0.8 \over \log_{10} 0.3 } ={\log_{10} {8 \over 10} \over \log_{10}{3\over 10}} ={\log_{10} 8-1\over \log_{10} 3-1}  ={1-\log_{10} 8 \over 1-\log_{10}3} } \\ \Rightarrow 2-\log_{10}5 \gt 1-\log_{10} 5 \gt 1-\log_{10}8  \Rightarrow b \gt a\gt c,故選\bbox[red, 2pt]{(B)}$$

解答:$$x^2+2x-15=(x-3) (x+5) \lt 0 \Rightarrow -5\lt x \lt 3 \\ \Rightarrow x=-4,-3,-2,-1,0, 1 ,2, 共7個整數解,故選\bbox[red, 2pt]{(A)}$$

解答:$$假設f(x,y)=x-2y-4 \Rightarrow \cases{f(A)=-14\lt 0\\ f(B)=-4\lt 0\\ f(C)=1\gt 0\\ f(D)=-1\lt 0} \Rightarrow \cases{A,B,D在左側\\ C在右側}\\,故選\bbox[red, 2pt]{(C)}$$

解答:$$斜率3, 通過(1,2) \Rightarrow 直線方程式: y-2=3(x-1) \Rightarrow 3x-y-1=0\\ 令f(x,y)=3x-y-1 \Rightarrow \cases{f(4,11)=0\\ f(7,20)=0\\ f(11,32)=0\\ f(15,43)= 1 \ne 0},故選\bbox[red, 2pt]{(D)}$$

解答:

$$\bbox[cyan,2pt]{送分}$$

解答:$$\cases{a_1=11\\ a_{21}=a_1+20d=71} \Rightarrow a_{11}=a_1+10d={1\over 2}(a_1+ a_{21})=41,故選\bbox[red,2pt]{(B)}$$

解答:$$\sin{\pi\over 3}+ \cfrac{\tan{4\pi\over 3}}{ \cos{2\pi\over 3}} ={\sqrt 3\over 2}+ \cfrac{\sqrt 3}{-{1\over 2}}=-{3\over 2}\sqrt 3,故選\bbox[red,2pt]{(C)}$$

解答:$$1-20的質數: 2,3,5, 7, 11, 13,17, 19,共8個, 機率為{8\over 20}={2\over 5} \Rightarrow 期望值=100\times {2\over 5}=40\\ ,故選\bbox[red,2pt]{(B)}$$

解答:$$\begin{array}{c}643\\ 634 \\ 463\\ 436\\ 346\\ 364 \\\hline 2886\end{array} ,故選\bbox[red,2pt]{(C)}$$

解答:$$每兩個頂點連成一直線,需扣除邊線,即C^9_2-9=36-9=27,故選\bbox[red,2pt]{(A)}$$

解答:$${\pi\over 2} \lt \theta\lt \pi \Rightarrow \cos \theta \lt 0\\ 5\cos^2 \theta-2\cos\theta-3=(5\cos \theta+3)(\cos \theta-1)=0 \Rightarrow \cos\theta= -{3\over 5},故選\bbox[red,2pt]{(A)}$$

解答:$$\cases{a_1=1\\ a_2=r\\ a_3=r^2\\ a_4=r^3\\ a_5=r^4\\ a_6=r^5=243} \Rightarrow r=3 \Rightarrow \cases{a_1=1\\ a_2=3 \\ a_3=9\\ a_4= 27\\ a_5=81\\ a_6 =243} \\ (A)\times: r=3\ne -3\\ (B)\times: 總和={3^6-1\over 3-1}=365 \ne 182\\(C) \times: a_4\gt 0 \ne -27\\ (D) \bigcirc: 3+9+27+81=120\\,故選\bbox[red,2pt]{(D)}$$

解答:$$只有4出現2次,其他數字只有出現1次,故選\bbox[red,2pt]{(D)}$$

解答:

$$x^2+y^2-4x+6y-12=0 \Rightarrow (x-2)^2+(y+3)^2=5^2 \Rightarrow \cases{圓心O(2,-3)\\ 圓半徑r=5}\\ (A) \times: 圓心O至直線距離=d(O,L)={6+12+2\over \sqrt{3^2+4^2}} =4\ne 5 \\(B)\times: d(O,L)=4\lt 5 \Rightarrow 相交兩點\\ (C)\bigcirc: 5^2=4^2+3^2 \Rightarrow 所圍三角形一半面積={1\over 2}\times 3\times 4=6 \Rightarrow 面積=12 \\(D)\times： d(O,L)=4\ne 3\\ 故選\bbox[red, 2pt]{(C)}$$

解答:$$(A\cup B)'=A'\cap B'=\{2,3\}, 故選\bbox[red, 2pt]{(B)}$$

=================== END =======================

解題僅供參考，其他歷年試題及詳解