104年身心障礙學生四技二專甄試-數學(B)-詳解

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(B)

單選題，共 20 題，每題 5 分

解答：$$直線\overleftrightarrow{BC}的法向量\vec n=\overrightarrow{BA}=(2,-6) \\\Rightarrow 通過B(-1,9)且法向量為\vec n的直線: 2(x+1)-6(y-9)=0 \Rightarrow x-3y+28=0，故選\bbox[red,2pt]{(B)}$$

解答：$$\sin \theta+ \cos \theta = {\sqrt{17} \over 3} \Rightarrow (\sin \theta+\cos \theta)^2 =1+ 2\sin \theta\cos \theta ={17\over 9} \Rightarrow 2\sin \theta\cos \theta ={8\over 9} \\ \Rightarrow (\sin \theta-\cos \theta)^2 =1-2\sin\theta \cos \theta=1-{8\over 9}={1\over 9} \Rightarrow \sin \theta-\cos\theta ={1\over 3}，故選\bbox[red,2pt]{(A)}$$

解答：$$\cases{\vec b=-k\vec a,k\gt 0\\ |\vec b|=5} \Rightarrow \cases{(x,y)=(5k,-12k) \\x^2+y^2=25} \Rightarrow 25k^2+144k^2=25 \Rightarrow k^2={25\over 169} \Rightarrow k={5\over 13} \\ \Rightarrow x+y=5k-12k= -7k = -{35\over 13}，故選\bbox[red,2pt]{(A)}$$

解答：$$\cases{\vec a=(2,1)\\ \vec b=(-2,1)} \Rightarrow \cases{\vec a+2\vec b=(-2,3)\\ \vec a-\vec b=(4,0) } \Rightarrow (\vec a+2\vec b)\cdot (\vec a-\vec b)= (-2,3)\cdot (4,0)=-8，故選\bbox[red,2pt]{(C)}$$

解答：$$\log_2(x+6) +\log_2(x-1)= 3\Rightarrow \log_2(x+6)(x-1) =\log_2 8 \Rightarrow (x+6)(x-1) =8\\ \Rightarrow x^2+5x-14=0 \Rightarrow (x+7)(x-2)=0 \Rightarrow x=2 (x=-7違反x-1\gt 0)，故選\bbox[red,2pt]{(D)}$$

解答：$$a=(0.25)^{3/2}({27\over 8})^{-2/3} =(({1\over 2})^2)^{3/2}(({3\over 2})^3)^{-2/3} =({1\over 2})^3({3\over 2})^{-2 } ={1\over 8}\cdot {4\over 9} ={1\over 18}，故選\bbox[red,2pt]{(C)}$$

解答：$$\cases{a=0.\overline{21} ={21\over 99} \\ b=0.3\overline{21}= {318\over 990}} \Rightarrow a+b= {21\over 99} +{318\over 990 } ={528\over 990} ={8\over 15}，故選\bbox[red,2pt]{(D)}$$

解答：$${x-4\over (x-1)^2} ={A\over  x-1 } +{B\over (x-1)^2} \Rightarrow A(x-1)+B= x-4 \Rightarrow \cases{A=1\\ B-A=-4}  \Rightarrow B=A-4=-3\\ \Rightarrow A+B= 1-3=-2，故選\bbox[red,2pt]{(C)}$$

解答：$$f(x)=(x+1)p(x)+2 \Rightarrow f(x+1)=(x+2)p(x+1)+2 \Rightarrow 餘式為2，故選\bbox[red,2pt]{(A)}$$

解答：$$\begin{vmatrix} 8a & 8b\\ 8c & 8d\end{vmatrix} = 8\begin{vmatrix} a & b\\ 8c & 8d\end{vmatrix} = 8\times 8\begin{vmatrix} a & b\\ c & d\end{vmatrix} = 64\times 6=384，故選\bbox[red,2pt]{(D)}$$

解答：$$判別式=0 \Rightarrow (2k+1)^2 -4(k^2-3)=0 \Rightarrow 4k+13=0 \Rightarrow k=-{13\over 4}，故選\bbox[red,2pt]{(B)}$$

解答：$$ {x+y\over 2} \ge \sqrt{xy}  \Rightarrow {18\over 2} \ge \sqrt{xy} \Rightarrow xy \le 81 \Rightarrow xy的最大值為81，故選\bbox[red,2pt]{(A)}$$

解答：$$(A)\times: 取出2顆黑球的機率:C^2_2/C^4_2= 1/6 \\(B)\times: 取出2顆白球的機率:C^2_2/C^4_2= 1/6 \\(C)\bigcirc:取出1顆黑球與1顆白球的機率=C^2_1C^2_1/C^4_2 = 4/6=2/3\\ (D)\times:與(C)相同\\，故選\bbox[red,2pt]{(C)}$$

解答：$$平均數y={全班總分\over 全班人數} \Rightarrow 全班總分=50y，故選\bbox[red,2pt]{(D)}$$

解答：$$\sin A=0.6 \Rightarrow \cos^2 A= 1-0.6^2 =0.64 \Rightarrow \cos (2A)=\cos^2A- \sin^2 A=0.64-0.36=0.28\\，故選\bbox[red,2pt]{(B)}$$

解答：$$(A)\times: 總排列數=5!\\(B)\times:3!\times 3!=36 \\(C)\bigcirc: 4!\times 2=48 = 3!\times 2!\times 4\\(D)\times: 2\times 3!\times 2!=24 \ne 3!\times 2!\\，故選\bbox[red,2pt]{(C)}$$

解答：$$R^2\pi =9\pi \Rightarrow R=3 \Rightarrow {\overline{BC}\over \sin A}=2R= 6 \Rightarrow \sin A= {\overline{BC}\over 6} ={3\over 6}={1\over 2}，故選\bbox[red,2pt]{(A)}$$

解答：$$\lim_{x\to 1}{x^2-1\over \sqrt x-1} =\lim_{x\to 1}{(x+1)(\sqrt x-1)(\sqrt x+1)\over \sqrt x-1}  =\lim_{x\to 1}{(x+1) (\sqrt x+1) } =2\cdot 2=4，故選\bbox[red,2pt]{(B)}$$

解答：$$f(x)=x^3-x+1 \Rightarrow f'(x)=3x^2-1\\(A)\times: f'(1)=3-1=2\ne 1 \\(B)\times: f'(0)=-1\lt 0 \Rightarrow 非遞增\\(C)\times: f'(x)=0 \Rightarrow x=\pm {1\over \sqrt 3}\in [-2,2] \Rightarrow f(x)在該區間有極大值與極小值，非凹向下\\(D)\bigcirc: \int_0^1 f'(x)\,dx= \left.\left[ x^3-x\right] \right|_0^1 =0\\，故選\bbox[red,2pt]{(D)}$$

解答：$${x^2\over 4}+{y^2\over 9}=1 \Rightarrow \cases{a=3\\ b=2}\\(A)\times: 長軸長=2a= 6\\ (B)\times: 短軸長=2b= 4\\ (C) \bigcirc: c=\sqrt{a^2-b^2} =\sqrt 5 \Rightarrow 焦點為(0,\pm \sqrt 5)\\ (D) \times: 正焦弦長={2b^2\over a} ={8\over 3} \ne {4\over 3}\\，故選\bbox[red,2pt]{(C)}$$

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