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2022年8月11日 星期四

86年大學學測-數學詳解

大學入學考試中心
八十六學年度學科能力測驗

第一部分:選擇題  

壹、單一選擇題 

解答:$$\tan \theta = {m_1-m_2\over 1+m_1m_2} ={\sqrt 3-1/\sqrt 3\over 1+1} ={1\over \sqrt 3} \Rightarrow \theta =30^\circ ,故選\bbox[red, 2pt]{(1)}$$
解答:$$\overrightarrow{PQ} \bot (a,b,c) \Rightarrow \overrightarrow{PQ}\cdot (a,b, c) =0,故選\bbox[red, 2pt]{(4)}$$
解答:$$-2\lt x\lt 4 \Rightarrow (x+2)(x-4)\lt 0 \Rightarrow f(x)=-(x+2)(x-4)\\ \Rightarrow f(2x)=-(2x+2)(2x-4)\lt 0 \Rightarrow (x+1)(x-2)\gt 0 \Rightarrow x\gt 2或x\lt -1,故選\bbox[red, 2pt]{(2)}$$
解答:$$假設\cases{首項a\\ 公比r} \Rightarrow \cases{{a\over 1-r}={8\over 9} \cdots(1) \\ ar^3={3\over 32} \cdots(2)}; {(2)\over (1)} \Rightarrow r^3(1-r)={3^3\over 4^4} \Rightarrow r={3\over 4}  ,故選\bbox[red, 2pt]{(3)}$$
解答

$$假設\cases{正六邊形邊長\overline{AB}=3\\ 正12邊形邊長=3-2a},見上圖;則\cos A= {a^2+a^2-(3-2a)^2\over 2\cdot a\cdot a} \\\Rightarrow -{1\over 2}= {-2a^2+12a-9\over 2a^2} \Rightarrow a^2-12a+9=0 \Rightarrow a=6-3\sqrt 3 \\\Rightarrow 3-2a =6\sqrt 3-9,故選\bbox[red, 2pt]{(5)}$$
解答:$$\vec a=(1,0,0),(0,1,0),(0,0,1),(-1,0,0),(0,-1,0),(0,0,-1),共6個,故選\bbox[red, 2pt]{(2)}$$
解答:$$將正立方體的中心視為空間原點,則與xy-平面、yz-平面及xz-平面\\平行的四中心點連線皆為正方形,因此共有3個,故選\bbox[red, 2pt]{(1)}$$
解答:$$40人全都沒有連續丟5局的機率:({31\over 32})^{40} \Rightarrow 至少有一人的機率p=1-({31\over 32})^{40},利用試題附表:\\ \log({31\over 32})^{40} =40(\log 31-\log 32)=40(\log 3.1+1-(\log 3.2+1)) =40(0.4914+1-(0.5051+1))\\ =-40\times 0.0137 = -0.548 =-1+ 0.452 =\log{2.83\over 10} =\log 0.283 \Rightarrow ({31\over 32})^{40}=0.283\\ \Rightarrow p=1-0.283=0.717,故選\bbox[red, 2pt]{(4)}$$


貳、多重選擇題

解答:$$f(x) =\sum_{n=1}^3(x-n)^2 +\sum_{n=8}^{10}(x-n)^2 \\=(x-1)^2+(x-2)^2 +(x-3)^2+(x-8)^2 +(x-9)^2+ (x-10)^2 \\ f'(x)=0 \Rightarrow 2(x-1) +2(x-2)+ 2(x-3)+ 2(x-8) +2(x-9)+ 2(x-10)=0\\ \Rightarrow 6x-33=0 \Rightarrow x=5.5\\(1)\times: a=5.5 \not \in \mathbb{Z}\\ (2)\bigcirc: 5.5\lt 5.9\\ (3)\bigcirc: 5.5\gt 5.1 \\(4)\times: |a-4|=|5.5-4|=1.5 \not \lt 5\\ (5) \times: |a-6|=|5.5-6|=0.5 \not \lt 0.5\\,故選\bbox[red, 2pt]{(23)}$$
解答:$$令\cases{直線L:3x+y-19=0\\ F(-1,2)\\ P(x,y)},則\left|{3x+y-19\over \sqrt{10}} \right| =\sqrt{(x+1)^2 +(y-2)^2}相當於 \overline{PF}=d(P,L)\\(1)\bigcirc: \cases{F為焦點\\ L為準線},該方程式符合拋物線的定義\\(2) \times: 焦點為(-1,2)\ne (1,-2)\\ (3) \times: L為準線非漸近線\\ (4)\bigcirc: 對稱軸與L垂直且通過F,即為x-3y+7=0\\(5)\times: Q=(3,1) \Rightarrow \cases{\overline{QF}= \sqrt{17}\\ d(Q,L)={9/\sqrt{10}}} \Rightarrow \overline{QF}\ne d(Q,L) \Rightarrow Q\not \in \Gamma\\,故選\bbox[red, 2pt]{(14)}$$

解答:$$(1)\times: 標準跑道有直線區域,橢圓沒有\\(2)\times 將橢圓往外擴張固定長度並非橢圓 \\(3)\bigcirc: 軌跡(x,y)為橢圓 \Rightarrow 軌跡(2x,2y)也是橢圓\\ (4)\bigcirc: 直圓錐與斜平面的截面為橢圓\\(5) \bigcirc: 圓柱體與斜平面的截面為橢圓\\,故選\bbox[red, 2pt]{(345)}$$
解答:$$(1)\bigcirc: r_1=r_2=0\\ (2)\bigcirc: r_3\gt 0=r_2\\(3) \times: r_3=r_4,X,Y互換\\(4)\times: r(x,y)=r(x,2y) \Rightarrow r_3=r_5\\(5)\bigcirc: r_3=r_4=r_5\\,故選\bbox[red, 2pt]{(125)}$$

第二部分:填充題

解答:$$利用長除法:f(x)=(x+7)(x^4-x^3+3x^2+4x+2)+6 \Rightarrow f(-7)=\bbox[red, 2pt]6$$
解答:$$\cases{E_1:2x-y+2z=6 \\ E_2: 3x-4z=2} \Rightarrow \cases{\vec n_1=(2,-1,2)\\ \vec n_2= (3,0,-4)} \Rightarrow \cos \theta = {\vec n_1\cdot \vec n_2\over |\vec n_1||\vec n_2|} =-{2\over 15} \Rightarrow \theta 為鈍角\\ \Rightarrow \cos(\pi -\theta)={2\over 15} \Rightarrow \theta \approx \bbox[red, 2pt]{82}(\text{查試題附表})$$
解答:$$\overrightarrow{AB} \cdot \overrightarrow{AD}= |\overrightarrow{AB}||\overrightarrow{AD}| \cos \angle A =2\cos 120^\circ =-1 \\ \Rightarrow |\overrightarrow{AC}|^2 =\overrightarrow{AC} \cdot \overrightarrow{AC} =(3\overrightarrow{AB} + 2\overrightarrow{AD}) \cdot (3\overrightarrow{AB} + 2 \overrightarrow{AD}) =9 |\overrightarrow{AB}|^2 + 12 \overrightarrow{AB} \cdot \overrightarrow{AD} +4 |\overrightarrow{AD}|^2 \\=25-12=13 \Rightarrow |\overrightarrow{AC}| = \sqrt{\bbox[red, 2pt]{13}}$$
解答:$$三直線\cases{y=0\\ 3x-2y+3=0\\ x+y-4=0} 相交於\cases{A(1,3)\\ B(-1,0)\\ C(4,0)} \Rightarrow \cases{\overline{AB}= \sqrt{13} \\ \overline{AC}= 3\sqrt 2\\ \overline{BC}=5} \Rightarrow \cos \angle A={1 \over \sqrt {16}} \\ \Rightarrow \sin \angle A={5\over \sqrt{26}} \Rightarrow 直徑={\overline{BC}\over \sin \angle A} ={5\over 5/\sqrt{26}} = \sqrt{\bbox[red, 2pt]{26}}$$
解答:$$\angle A+ \angle C=180^\circ \Rightarrow \cos \angle A=-\cos \angle C \Rightarrow {1+16-\overline{BD}^2\over 2\cdot 1\cdot 4} =- {2^2+3^2-\overline{BD}^2 \over 2\cdot 2\cdot 3} \\ \Rightarrow 51-3\overline{BD}^2 = -26+2\overline{BD}^2 \Rightarrow \overline{BD}^2 ={77\over 5} \Rightarrow \overline{BD}= \bbox[red, 2pt]{\sqrt{77\over 5}}$$
解答:$$3^{100}= a\times 10^m \Rightarrow \log 3^{100}= \log(a\times 10^m) \Rightarrow 100\log 3 =100\times 0.4771 = 47.71=m+\log a\\ \Rightarrow \log a=0.71 ; 由於\cases{\log 5=1-\log 2= 1-0.301=0.699\\  \log 6=0.301+0.4771= 0.7781},因此\log 5\lt \log a\lt \log 6\\ \Rightarrow a的整數部份為\bbox[red, 2pt]{5}$$
解答:$$\cases{P(甲\to 甲\to 甲)={9\over 10}\times {9\over 10}\\ P(甲\to 乙\to 甲)={1\over 10}\times {1\over 5}} ,兩路線機率之和={81\over 100}+{1\over 50}={\bbox[red, 2pt]{83} \over 100}$$
解答:$$\cases{P(三粒骰子都是n)=1/216\\ P(三粒骰子中恰有二粒都是n)=15/216\\ P(三粒骰子只有一粒是n)=75/216\\ P(三粒骰子都不是n)=125/216 } \Rightarrow 期望值=3\times {1\over 216}+ 2\times {15\over 216}+{75\over 216}-{125\over 216}\\ ={\bbox[red, 2pt]{-17}\over 216}$$ 

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