104 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(S)
單選題，共 20 題，每題 5 分
解答：$$\cases{A(a,8)\\ B(1,4)\\ C(-2,b)}，B為\overline{AC}中點\Rightarrow B=(A+C)\div 2 \Rightarrow \cases{1=(a-2)\div 2\\ 4=(8+b)\div 2} \Rightarrow \cases{a=4\\ b=0} \\ \Rightarrow \cases{A(4,8)\\ C(-2,0)} \Rightarrow \overline{AC} =\sqrt{6^2+8^2} =10，故選\bbox[red,2pt]{(D)}$$
解答：$$L與4x+3y=5平行 \Rightarrow L:4x+3y=k；又L通過(2,3) \Rightarrow 8+9=k \Rightarrow k=17 \\\Rightarrow L:4x+3y=17 \\(A) \times:(-2,0) \Rightarrow -8+0\ne 17\\ (B)\times:(1,-1)\Rightarrow 4-3\ne 17\\ (C)\bigcirc:(5,-1)\Rightarrow 20-3=17 \\(D)\times: (3,3)\Rightarrow 12+9\ne 17\\，故選\bbox[red,2pt]{(C)}$$
解答：$$\sin\theta-\cos\theta={1\over 4} \Rightarrow (\sin\theta-\cos\theta)^2 =1-2\sin\theta \cos\theta={1\over 16} \Rightarrow \sin\theta \cos\theta= {15\over 32}\\ \Rightarrow (\sin\theta+\cos\theta)^2 =1+2 \sin\theta\cos\theta =1+{15\over 16} ={31\over 16} \Rightarrow \sin\theta +\cos\theta ={\sqrt{31}\over 4}(負值違反\theta為銳角)\\(A)\times: \sin^2\theta-\cos^2\theta =(\sin\theta+\cos\theta) (\sin\theta -\cos\theta)= {\sqrt{31}\over 4}\cdot {1\over 4}={\sqrt{31}\over 16} \\(B)\bigcirc: \tan\theta+\cot\theta = {\sin\theta\over \cos\theta} +{\cos\theta\over \sin \theta} ={1\over \sin\theta \cos\theta} ={32\over 15} \\(C)\times: \sin\theta+\cos\theta = {\sqrt{31}\over 4} \ne {31\over 16} \\(D)\times: \sin\theta \cos\theta={15\over 32} \ne {15\over 16}\\，故選\bbox[red,2pt]{(B)}$$
解答：$$\cos A= {b^2+c^2-a^2\over 2bc} ={25-13\over 24} ={1\over 2} \Rightarrow \sin A={\sqrt 3\over 2}，故選\bbox[red,2pt]{(D)}$$
解答：$$\cos{\pi\over 3} ={\vec a\cdot \vec b\over |\vec a||\vec b|} ={\vec a\cdot \vec b\over 20} \Rightarrow \vec a\cdot \vec b=20\times {1\over 2}= 10\\ \Rightarrow \left|2\vec a-\vec b\right|^2 =(2\vec a-\vec b)\cdot (2\vec a-\vec b) =4|\vec a|^2+ |\vec b|^2-4\vec a\cdot \vec b\\ =64+25-40 = 49 \Rightarrow \left|2\vec a-\vec b\right|=\sqrt{49}=7，故選\bbox[red,2pt]{(B)}$$
解答：$$\cases{\vec a=(-4,1)\\ \vec b=(-1,7)\\ \vec c=(x,y)} \Rightarrow \cases{-2(\vec a-\vec c)=-2(-4-x,1-y)\\ \vec b-\vec c=(-1-x,7-y)} \Rightarrow \cases{8+2x= -1-x\\ -2+2y=7-y} \Rightarrow \cases{x=-3\\ y=3} \\ \Rightarrow x+y= -3+3=0，故選\bbox[red,2pt]{(C)}$$
解答：$$f(x)=2x^3-x^2-ax+b \Rightarrow \cases{f(0)=-3\\ f(-1)=3} \Rightarrow \cases{b=-3\\ -3+a+b=3} \Rightarrow \cases{a=9\\ b=-3} \Rightarrow a+b=6\\，故選\bbox[red,2pt]{(D)}$$
解答：$$\alpha,\beta 為3x^2-2x-4=0的兩根\Rightarrow \cases{\alpha+\beta = 2/3\\ \alpha\beta=-4/3} \\(A)\bigcirc: {\beta\over \alpha}+ {\alpha\over \beta} ={\alpha^2+\beta^2\over \alpha\beta} ={(\alpha+\beta)^2-2\alpha\beta\over \alpha\beta} ={4/9+8/3\over -4/3} =-{7\over 3} \\(B)\times: \alpha^2+\beta^2\ge 0 \ne -{20\over 9}\\ (C)\times: \alpha+\beta = 2/3 \ne -2/3\\(D)\times: {1\over \alpha}+{1\over \beta} ={\alpha+\beta \over \alpha\beta} ={2/3\over -4/3} =-{1\over 2}\ne {1\over 2}\\，故選\bbox[red,2pt]{(A)}$$
解答：$$\cases{a=\log_2 0.1\\ b=2\log_4 0.01 =\log_2 0.01\\ c=3\log_8 0.001 =\log_2 0.001} \Rightarrow a\gt b\gt c，故選\bbox[red,2pt]{(A)}$$
解答：$${9^{7/2}\cdot 3^{-2}\over 27^{5/3}} ={3^7\cdot 3^{-2}\over 3^5} = {3^5\over 3^5} =1，故選\bbox[red,2pt]{(A)}$$
解答：$$半徑r =\left|{12-6+1 \over \sqrt{4^2+3^2}} \right| ={7\over 5}，故選\bbox[red,2pt]{(B)}$$
解答：$$x^2+y^2+dx+ey +f=0 通過(1,0)及(0,2) \Rightarrow \cases{1+d+f=0 \\ 4+2e+f=0} \Rightarrow \cases{f=-1-d\\ f=-2e-4}\\由於f\ge 5，因此\cases{-1-d\ge 5\\ -2e-4\ge 5} \Rightarrow \cases{d\le -6 \\ e\le -9/2} \Rightarrow |d-e| \ge 6-{9\over 2}={3\over 2}\\ \Rightarrow (d-e)^2 \ge ({3\over 2})^2 ={9\over 4}=2.25，故選\bbox[red,2pt]{(C)}$$
解答：$$a_{n+1}-a_n=-1 \Rightarrow a_{n+1}=a_n-1 \Rightarrow \cases{\langle a_n\rangle為等差數列\\ 公差d=-1\\ 首頁a_1=10} \Rightarrow a_n=10-(n-1) \lt 0 \Rightarrow n\ge 12\\，故選\bbox[red,2pt]{(B)}$$
解答：$$1+{1\over 2}+{1\over 4}+\cdots +{1\over 32} ={1-(1/2)^6\over 1-1/2} =2-{1\over 32} ={63\over 32} \Rightarrow x+y= 63+32=95，故選\bbox[red,2pt]{(C)}$$
解答：$$從低排到高只有一種排法，但3人等高，這3人相鄰任排有3!=6種排法，故選\bbox[red,2pt]{(A)}$$
解答：$$C^{10}_2C^{13}_2 = 3510，故選\bbox[red,2pt]{(C)}$$
解答：$$\begin{array}{}(a,b)&小計\\\hline (1,5-6) & 2\\ (2,4-5)& 2\\ (3,3-4)& 2\\ (4,2-3) & 2\\ (5,1-2) & 2\\ (6,1) & 1\\\hline\end{array} \Rightarrow 共有2+2+2+2+2+1 =11種情形，機率為{11 \over 36}，故選\bbox[red,2pt]{(D)}$$
解答：$$(10+12+14+\cdots +28)\div 10=190\div 10=19，故選\bbox[red,2pt]{(B)}$$
解答：$$算術平均數\bar x=13 =(15+12 +x+10+18)\div 5 \Rightarrow x=10\\ \Rightarrow \sum(x_i-\bar x)^2 =2^2+(-1)^2 +3^2+(-3)^2 +5^2= 48 \Rightarrow \sigma =\sqrt{{1\over 5}\cdot 48} =\sqrt{9.6} ={4\over 5}\sqrt{15}\\，故選\bbox[red,2pt]{(D)}$$
解答：$$(A)\bigcirc: (3,-2)\Rightarrow 3x-4y=9+8=17\ge 15\\ (B)\times: (7,3)\Rightarrow 3x-4y=21-12=9 \not \ge 15\\(C)\times: (\sqrt {17},1) \Rightarrow 3x-4y=3\sqrt{17}-4\not \ge 15\\ (D)\times: (-5,-\sqrt{28}) \Rightarrow 4x-3y=-20+6\sqrt 7\not \ge 15\\，故選\bbox[red,2pt]{(A)}$$
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