105 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(C)
單選題，共 20 題，每題 5 分
：$$\sin(-870^\circ)= -\sin 870^\circ =-\sin(360^\circ \times 2+150^\circ)= -\sin 150^\circ =-{1\over 2}，故選\bbox[red,2pt]{(B)}$$
解答：$$\left( 3x^2-4x+5\right)\left( 2x^3+ax+b \right) \Rightarrow \cases{x^3項係數=3a+10=7 \Rightarrow a=-1\\ x^2項係數=3b-4a=1 \Rightarrow b=-1} \Rightarrow a+b=-2\\，故選\bbox[red,2pt]{(A)}$$
解答：$$\begin{vmatrix} 1 & 2 & 0\\ 2 & 2 & 1\\ 2 & a & 3\end{vmatrix} =6+4-12-a=-2-a=1 \Rightarrow a=-3，故選\bbox[red,2pt]{(A)}$$
解答：$$\cases{3x+10y=-1 \cdots(1) \\ 5x+26y=1\cdots(2)} \xrightarrow {5\times(1)-3\times(2) }-28y=-8 \Rightarrow y=2/7代回(1) \Rightarrow 3x+20/7=-1\\ \Rightarrow x=-9/7 \Rightarrow x+y=a+b =-1，故選\bbox[red,2pt]{(B)}$$
解答：$$b_k = b_1r^{k-1} =4096\cdot (-{1\over 2})^{k-1}=-{1\over 8} \Rightarrow 2^{12}\cdot 2^{1-k} \cdot (-1)^{k-1}=-2^{-3} \Rightarrow 2^{13-k}\cdot (-1)^{k-1} =-2^{-3}\\ \Rightarrow \cases{13-k=-3\\ k-1為奇數} \Rightarrow k=16，故選\bbox[red,2pt]{(D)}$$
解答：$${6!\over 2! 3!} ={720\over 12} =60，故選\bbox[red,2pt]{(A)}$$
解答：$$y=3x^2+2x+1 = 3(x^2+{2\over 3}x+{1\over 9})+{2\over 3} =3(x+{1\over 3})^2+{2\over 3} \Rightarrow 頂點(-{1\over 3},{2\over 3})在第二象限\\，故選\bbox[red,2pt]{(B)}$$
解答：$${2\over 3}-{1\over 2}+{3\over 8}-{9\over 32}+\cdots \Rightarrow \cases{首項a_1=2/3\\ 公比r=(-1/2)\div (2/3) = -3/4} \\ \Rightarrow S={a_1\over 1-r} ={2/3\over 1+3/4} = {8\over 21}，故選\bbox[red,2pt]{(B)}$$
解答：$$ax+by+2=0通過(4,2)且斜率為2/3 \Rightarrow \cases{4a+2b+2=0 \cdots(1)\\ -a/b=2/3 \cdots(2)}，由(2)可得 2b=-3a代入(1) \\ \Rightarrow 4a-3a+2=0 \Rightarrow a=-2 \Rightarrow 2b=6 \Rightarrow b=3 \Rightarrow a+b=-2+3=1，故選\bbox[red,2pt]{(C)}$$
解答：$$\cases{\theta在第三象限\\ \sin\theta=-3/5} \Rightarrow \cos\theta =-4/5 \Rightarrow \sin(2\theta) = 2\sin \theta\cos \theta = 2\cdot (-{3\over 5})\cdot (-{4\over 5})= {24\over 25}，故選\bbox[red,2pt]{(D)}$$
解答：$$\cases{\vec a=(x,y) \\ \vec b=(-1,2) \\ \vec c=(3,-4)}，由\cases{\vec a\cdot \vec b=0\\ \vec a\cdot \vec c=-2} \Rightarrow \cases{-x+2y=0\\ 3x-4y=-2} \Rightarrow \cases{x=-2\\ y=-1} \Rightarrow x+y=-3，故選\bbox[red,2pt]{(B)}$$
解答：$$f(x)=ax^3+bx^2 +x-2 = (x^2-x-2)g(x) = (x-2)(x+1)g(x)\\ \Rightarrow \cases{f(2)=8a+4b=0\\ f(-1)=-a+b-3=0} \Rightarrow \cases{a=-1\\ b=2} \Rightarrow a+b=1，故選\bbox[red,2pt]{(C)}$$
解答：$${1+2i\over a+bi} =3+4i \Rightarrow 1+2i=(a+bi)(3+4i) = 3a-4b+(4a+3b)i \Rightarrow \cases{3a-4b=1\\ 4a+3b=2} \\ \Rightarrow \cases{a=11/25\\ b=2/25} \Rightarrow a+b=13/25，故選\bbox[red,2pt]{(D)}$$
解答：$$x^2+3x-6\lt 2x \Rightarrow x^2+x-6\lt 0 \Rightarrow (x+3)(x-2)\lt 0 \Rightarrow -3\lt x\lt 2，故選\bbox[red,2pt]{(C)}$$
解答：$$(A)\times: \cases{\log_2 8=3\\ \log_3 27=3} \Rightarrow \log_2 8\not \lt \log_3 27 \\(B)\times: 理由同(A)\\ (C)\times:\cases{\log_2 (1/3)=-\log_2 3 \Rightarrow -2\lt \log_2(1/3)\lt -1\\ \log_3(1/2) =-\log_3 2 \Rightarrow -1\lt \log_3(1/2 ) \lt 0} \Rightarrow \log_3(1/2)\gt \log_2(1/3)\\(D)\bigcirc: 理由同(C)\\，故選\bbox[red,2pt]{(D)}$$
解答：$$\begin{array}{} 牌型 & 小計\\\hline (1,6-7)& 2\\ (2,5-7) & 3\\ (3,4-7) & 4\\ (4,5-7) & 3\\ (5,6-7)& 2\\ (6,7)&1\\\hline \end{array} \Rightarrow 共有2+3+4+3+2+1 =15 \Rightarrow 機率為{15\over C^7_2} ={15\over 21}={5\over 7}，故選\bbox[red,2pt]{(A)}$$
解答：$$(2x-4)^3-8 =(2x-4)^3-2^3 = (2x-4-2)((2x-4)^2+2(2x-4)+2^2) = (2x-6)(4x^2-4x+12)\\ \Rightarrow \lim_{x\to 3}{(2x-4)^3-8 \over x-3} =\lim_{x\to 3}{2(x-3)(4x^2-12x+12) \over x-3} =\lim_{x\to 3} 2(4x^2-12x+12) =2\cdot(36-36+12)\\=2\times 12= 24，故選\bbox[red,2pt]{(D)}$$
解答：$$\cases{\int_0^1 3x^2+ ax+b\;dx=1 \\[1ex]\int_0^2 3x^2+ ax+b\;dx=4} \Rightarrow \cases{\left.\left[ x^3+ax^2/2+bx \right]\right|_0^1 =1\\[1ex] \left.\left[ x^3+ax^2/2+bx \right]\right|_0^2 =4} \Rightarrow \cases{1+a/2+ b=1\\ 8+2a+2b=4} \\ \Rightarrow \cases{a+2b=0\\ a+b=-2} \Rightarrow \cases{a=-4 \\b=2} \Rightarrow a+b=-2，故選\bbox[red,2pt]{(A)}$$
解答：$$9^{x^2-2x+3} =27^x \Rightarrow 3^{2x^2-4x+6} =3^{3x} \Rightarrow 2x^2-4x+6=3x \Rightarrow 2x^2-7x+6=0\\ \Rightarrow 兩根之和= 7/2，故選\bbox[red,2pt]{(C)}$$
解答：$$x+y=k \Rightarrow y=k-x代入圓C\Rightarrow x^2+(k-x)^2 = k \Rightarrow 2x^2-2kx+k^2-k=0\\ 圖形相切代表判別式=0 \Rightarrow 4k^2-8(k^2-k)=0 \Rightarrow k^2-2k^2+2k=0 \Rightarrow k^2-2k=0\\ \Rightarrow k(k-2)=0 \Rightarrow k=2，故選\bbox[red,2pt]{(C)}$$
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