大學入學考試中心
八十三學年度學科能力測驗
第一部分：選擇題
壹、單一選擇題
解答：$$6\lt \sqrt{47}\lt 7 \Rightarrow 13\lt 7+\sqrt{47}\lt 14 \Rightarrow \sqrt{13} \lt \sqrt{7+\sqrt{47}} \lt \sqrt{14} \Rightarrow 3 \lt \sqrt{7+\sqrt{47}} \lt 4\\，故選\bbox[red, 2pt]{(D)}$$
解答：$$L:{x-2\over 3}={y+1\over -1} ={z-1\over 2} \Rightarrow L的方向向量\vec u=(3,-1,2)\\(A) \times: 2x-y+z=1\Rightarrow 法向量\vec n=(2,-1,1) \Rightarrow \vec n\cdot \vec u=6+1+2\ne 0 \\(B)\bigcirc: x+y-z=2 \Rightarrow 法向量\vec n=(1,1,-1) \Rightarrow \vec n\cdot \vec u= 3-1-2=0 \\(C)\times: 3x-y+2z=1 \Rightarrow 法向量\vec n=(3,-1,2 ) \Rightarrow \vec n\cdot \vec u= 9+1+4 \ne 0 \\(D)\times: 3x+2y+z= 2 \Rightarrow 法向量\vec n=(3,2,1) \Rightarrow \vec n\cdot \vec u= 9-2+2=9 \ne 0\\ (E)\times: x-3y+z=1 \Rightarrow 法向量\vec n=( 1,-3,1) \Rightarrow \vec n\cdot \vec u=3+3+2 \ne 0\\，故選\bbox[red, 2pt]{(B)}$$
解答：$$兩次都反面的機率={1\over 4} \Rightarrow 至少一正面的機率=1-{1\over 4} ={3\over 4}，故選\bbox[red, 2pt]{(D)}$$
解答：$$\cases{A,B,C取一點、C,D,E取兩點:C^3_1C^3_2= 9\\ A,B,C取兩點、C,D,E取一點:C^3_2C^3_1= 9} \Rightarrow 共有9+9=18個三角形，故選\bbox[red, 2pt]{(D)}$$
解答：$$令\cases{X:甲生成績\\ Y:乙生成績\\ Z:丙生成績} ,則\cases{X=Y+10\\ Z=0.8X} \Rightarrow \cases{S_甲=S_乙\\ S_丙=0.8S_甲} \Rightarrow S_甲=S_乙\gt S_丙，故選\bbox[red, 2pt]{(E)}$$
解答：$$x={\sqrt[3]{88.3} \over \sqrt{2.56}} \Rightarrow \log x={1\over 3}\log 88.3-{1\over 2}\log 2.56= {1\over 3}(\log 8.83+1)-{1\over 2}\log 2.56 \\ ={1\over 3}(0.9460+1)-{1\over 2}\times 0.4082 = 0.44457 \Rightarrow x=2.78，故選\bbox[red, 2pt]{(B)}\\ 註:試題附表可查得:\log 8.83=0.9460,\log 2.56=0.4082, \log 2.78=0.444$$
解答：$$\cases{\cos \angle AOB= {\overline{OA}^2 +\overline{OB}^2-\overline{AB}^2 \over 2\cdot \overline{OA}\cdot \overline{OB}}= {2-1.6^2\over 2} = -{7\over 25} \\[1ex] \cos \angle POQ = {\overline{OP}^2 +\overline{OQ}^2-\overline{PQ}^2 \over 2\cdot \overline{OP}\cdot \overline{OQ}} ={2-1.2^2\over 2} ={7\over 25}} \Rightarrow \angle AOB+\angle POQ=180^\circ\\ \angle AOP = {1\over 2}(\angle AOP-\angle POQ) \Rightarrow \cos (2\angle AOP)= \cos(\angle AOP-\angle POQ) =-({7\over 25})^2+ ({24\over 25})^2 ={527\over 625} \\ \Rightarrow \cos \angle AOP =\sqrt{({527\over 625}+1)\div 2} ={24\over 25} \Rightarrow \angle AOP \approx 16.2(查表)，故選\bbox[red, 2pt]{(D)}$$
解答：$$(A)\times: 圖形凹向上\Rightarrow a\gt 0 \\(B)\times: 極小值出現在x=-{b\over 2a}\lt 0 \Rightarrow b\gt 0\\ (C)\bigcirc: f(0)= c\lt 0\\ (D)\times: f(x)=0有相異二實根\Rightarrow b^2-4ac \gt 0\\ (E)\bigcirc: f(-1)=a-b+c \lt 0\\，故選\bbox[red, 2pt]{(CE)}$$
解答：$$(B)-(D)皆有無限個，非一個，故選\bbox[red, 2pt]{(AE)}$$
解答：$$(C)\times: \cot 0=\infty，無此圖形\\(D) \times:5x^2+4x-6y-3=0 圖形為凹向上拋物線，無此圖形\\(E) \times: (x+2)^2-(y+2)^2=5為一雙曲線，無此圖形\\，故選\bbox[red, 2pt]{(CDE)}$$
解答：$$4^x = 2^{3x+2} \Rightarrow \log_2 4^x= \log_2 2^{3x+2} \Rightarrow 2x=3x+2 \Rightarrow x=-2 \Rightarrow y=4^x={1\over 16}\\ \Rightarrow 交點為\bbox[red, 2pt]{(-2,{1\over 16})}$$
解答：$$10+{10\over 3}+ {10\over 3}+{10\over 9} +{10\over 9}+{10\over 27}+\cdots =10+2({10\over 3} +{10\over 9}+\cdots )\\ =10+2\cdot {10/3\over 1-1/3 } = \bbox[red, 2pt]{20}$$
解答：
$$令\cases{L_1= \overleftrightarrow{BD} \\L_2= \overleftrightarrow{OA}} \Rightarrow \cases{L_1斜率m_1=-1/2\\ L_2斜率m_2=-2} \Rightarrow \cases{L_1: y=-{1\over 2}(x-4)+2\\ L_2: y=-2x} \\ \Rightarrow D=L_1\cap L_2 \Rightarrow -{1\over 2}(x-4)+2=-2x \Rightarrow x=-{8\over 3} \Rightarrow y={16\over 3} \Rightarrow D=\bbox[red,2pt]{(-{8\over 3},{16\over 3})}$$
解答：$$P在x+2y=3上\Rightarrow P(3-2a,a),又\overline{PA} =\overline{PB} \Rightarrow (2-2a)^2+(a-2)^2 =(-2a)^2 +(a-4)^2\\ \Rightarrow a=-2 \Rightarrow P\bbox[red, 2pt]{(7,-2)}$$
解答：$$圓\Gamma: x^2+y^2+ 2x=3 \Rightarrow (x+1)^2 +y^2 =4 \Rightarrow \cases{圓心O(-1,0)\\ 半徑r=2} \\ \Rightarrow d(O,L) =r \Rightarrow \left|{-m+3\over \sqrt{m^2+1}} \right| =2 \Rightarrow m^2-6m+9 = 4m^2+4 \Rightarrow m= \bbox[red, 2pt]{-1\pm {2\over 3}\sqrt 6}$$
解答：$$球:x^2+y^2 +z^2+2x -4y-11=0 \Rightarrow (x+1)^2 +(y-2)^2+z^2 =16 \\ \Rightarrow \cases{球心O(-1,2,0)\\ 球半徑R=4} \Rightarrow d(E,O)= \left|{-1+6+0-1\over \sqrt{11}} \right| ={4\over \sqrt{11}} \Rightarrow 相交圓半徑 r=\sqrt{R^2-d(E,O)^2} \\ \Rightarrow r^2 =16-{16\over 11} ={160\over 11} \Rightarrow 相交圓面積=\bbox[red, 2pt]{{160\over 11}\pi}$$
解答：$$L:\cases{x-y+z=1\\ x+y-z=1} \Rightarrow \cases{x=1\\ y=z=t, t\in \mathbb{R}} ，令\cases{P\in L \Rightarrow P(1,t,t)\\ Q(1,2,3)} \\\Rightarrow \overline{PQ}= \sqrt{f(t)}=\sqrt{(t-2)^2+(t-3)^2}=\sqrt{2t^2-10t+13} \Rightarrow f'(t)=0 \Rightarrow t={10\over 4}={5\over 2}\\ \Rightarrow P\bbox[red, 2pt]{(1,{5\over 2}, {5\over 2})}$$
解答：$$假設三邊長為a,b,c，則a+b+c=20且a+b\gt c\\ \begin{array}{} a & b & c\\\hline 9 & 9 & 2\\ & 8 & 3\\ & 7 & 4\\ & 6 & 5\\\hdashline 8 & 8 & 4\\ & 7 & 5\\ & 6 & 6\\\hdashline 7 & 7 & 6\\\hline\end{array} \Rightarrow 共有\bbox[red, 2pt]{8}種三角形$$
解答：$$\sin\theta+\cos \theta = \sqrt{1-\cos^2\theta} +\cos \theta ={1\over 5} \Rightarrow 1-\cos^2\theta =({1\over 5}-\cos\theta)^2 \\\Rightarrow 50\cos^2\theta-10\cos \theta-24 =0 \Rightarrow (5\cos\theta -4)(10\cos\theta +6)=0\\ \Rightarrow \cos\theta =\bbox[red,2pt]{4\over 5} (-{6\over 10}不合,違反\theta 在第三象限角)$$
解答：$$令\cases{f(x)= x^2+px+6\\ g(x)=x^3+px+6} \Rightarrow 公因式m(x) \mid (g(x)-f(x)) \Rightarrow m(x)\mid (x^3-x^2)\\ \Rightarrow m(x)=x,x^2,x-1，但f(0)=g(0)=6\ne 0，因此m(x)\ne x，即m(x)=x-1 \\ \Rightarrow f(1)=0 \Rightarrow p=\bbox[red, 2pt]{-7}$$
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