88年大學學測-數學詳解

大學入學考試中心
八十八學年度學科能力測驗

第一部分：選擇題  

壹、單一選擇題 

解答：$$2^{100}=(2^{10})^{10} = 1024^{10} \equiv 4^{10} \mod 10 \equiv (4^2)^5 \mod 10 \equiv  6^5 \mod 10\\ \equiv 36\cdot 36\cdot 6 \mod 10 \equiv 6\cdot 6 \cdot 6 \mod 10 \equiv 6 \mod 10，故選\bbox[red, 2pt]{(4)}$$

解答：$$\cases{(1)2^{1/3}\\ (2)(2^{-3})^{-2}=2^6\\ (3) 2^{-1/4}\\ (4) ({1\over 2})^{1/2} =2^{-1/2}\\ (5)8^{-1/3}= 2^{-1}} \Rightarrow 2^{-1}最小，故選\bbox[red, 2pt]{(5)}$$

解答：

$$\cases{A(1/2,1,1)\\ B(1,1,1/2)\\ C(1/2,0,1)} \Rightarrow \cases{ \overrightarrow{AB} =(1/2,0,-1/2) \\ \overrightarrow{AC} =(0,-1,0) } \Rightarrow \cases{ \overrightarrow{AB} \cdot \overrightarrow{AC} =0 \\ \vec n=\overrightarrow{AB} \times \overrightarrow{AC} =(-1/2,0,-1/2)} \\ \Rightarrow \cases{ \angle B 為直角\\ 過A,B,C之平面E: 2x+2z=3} \Rightarrow E與正立方體交於另一點D(1,0,1/2) \\ \Rightarrow \overrightarrow{CD}= (1/2,0,-1/2) \Rightarrow \overrightarrow{CD} \cdot \overrightarrow{AC} =0 \Rightarrow \angle C為直角,又\overline{AB}\ne \overline{AC} \Rightarrow ABCD為長方形\\，故選 \bbox[red, 2pt]{(4)}$$

貳、多重選擇題

解答：$$\cases{直角\triangle AHC \Rightarrow \overline{AH} = b\sin C\\ 直角\triangle AHB \Rightarrow \overline{AH} = c\sin B } \Rightarrow \overline{AH}= b\sin C= c\sin B，故選\bbox[red, 2pt]{(34)}$$

解答：$$(1)\times: 0.3\overline{43} =0.\overline{34}= {34\over 99} 為有理數\\(2) \bigcirc: {34\over 99} \gt {33\over 99}={1\over 3} \\(3)\bigcirc: 0.3434\dots \gt 0.343 \\ (4) \bigcirc: 0.3434\dots \lt 0.35 \\(5)\bigcirc: \cases{a=0.\overline{34} \Rightarrow 100a= 34.\overline{34}\\b= 0.3\overline{43} \Rightarrow 100b=34.\overline{34}} \Rightarrow a=b\\，故選\bbox[red, 2pt]{(2345)}$$

解答：$$f(x)= x^3+x^2 -2x-1 \Rightarrow \cases{f(-2)=-1 \lt 0\\ f(-1)= 1 \gt 0\\ f(0)=-1\lt 0\\ f(1)=-1\lt 0\\ f(2)=7 \gt 0 \\ f(3)=29\gt 0}\\ (1) \bigcirc: f(-2)f(-1)\lt 0 \\(2)\bigcirc: f(-1)f(0) \lt 0 \\(3)\times: f(0)f(1) \gt 0 \\(4)\bigcirc: f(1)f(2) \lt 0 \\(5) \times: f(2)f(3)\gt 0\\，故選\bbox[red, 2pt]{(124)}$$

解答：$$\sqrt{(x-1)^2 +(y-2)^2} +\sqrt{(x+1)^2+ (y+2)^2}=6 \Rightarrow \cases{F_1(1,2)\\ F_2(-1,-2)\\ 2a=6} \\(1)\bigcirc: 中心坐標=(F_1+F_2)\div 2=(0,0)\\ (2) \bigcirc: 焦點\cases{F_1(1,2)\\ F_2(-1,-2) } \\ (3) \bigcirc: 2c=\overline{F_1F_2} = \sqrt{4+16}=2\sqrt 5 \Rightarrow c=\sqrt 5 \Rightarrow b=\sqrt{a^2-c^2}=\sqrt{9-5}=2 \Rightarrow 短軸2b=4\\(4) \times: \overleftrightarrow{F_1F_2}: 2x=y 為對稱軸 \Rightarrow 另一對稱軸x=-2y\\(5)\bigcirc: 橢圓對稱於長軸\\，故選\bbox[red, 2pt]{(1235)}$$

解答：$$(1)\times: 差一個負號\\(2) \bigcirc: \det(A)=\det(A^T) \\(3) \bigcirc: 列運算行列式相同\\(4) \times: 非列運算\\(5)\times: 非列運算\\，故選\bbox[red, 2pt]{(23)}$$

解答：$$假設\cases{舊數據X\\ 新數據Y }，其中Y=100X-240\\(1)\bigcirc: \bar y= (3+6+1+5+4+ 8+6+7 +5)\div =45\div 9=5 \\(2) \bigcirc: \sum(y-\bar y)^2 = 4+1 + 16+ 0+1 +9+1+4+0=36 \Rightarrow \sigma(y)=\sqrt{36\over 9}=2\\(3) \bigcirc: E(Y)= E(100X-240) = 100E(X)-240= 5 \Rightarrow E(X)=2.45 \\(4) \times: \sigma(Y) =\sigma(100X-240) = 100\sigma(X)=2 \Rightarrow \sigma(X)= 0.02 \\(5)\bigcirc: 原數據依序排列: 2.41,2.43, 2.44, 2.45,2.45, ... \Rightarrow 中位數=2.45\\，故選\bbox[red, 2pt]{(1235)}$$

解答：

$$假設正立方體邊長為1、D為空間原點，則\cases{A(0,1,0)\\ B(1,1,0)\\ C(1,0,0)\\ D(0,0,0)\\ E(0,1,1)\\ F(1,1,1)\\ G(1,0,1) \\H(0,0,1)} \\(1)\bigcirc: \cases{\overrightarrow{EA} =(0,0,-1)\\ \overrightarrow{EG}= (1,-1,0)} \Rightarrow \overrightarrow{EA} \cdot \overrightarrow{EG} =0 \\(2) \bigcirc: \cases{\overrightarrow{ED} =(0,-1,-1)\\ \overrightarrow{EF}= (1,0,0)} \Rightarrow \overrightarrow{ED} \cdot \overrightarrow{EF} =0 \\(3) \bigcirc: \cases{\overrightarrow{EF}= (1,0,0) \\\overrightarrow{EH}= (0,-1,0) \\\overrightarrow{AC}= (1,-1,0)} \Rightarrow \overrightarrow{EF} +\overrightarrow{EH} =(1,-1,0 ) =\overrightarrow{AC} \\(4)\times: \cases{ \overrightarrow{EC} =(1,-1,-1)\\ \overrightarrow{AG}= (1,-1,1)} \Rightarrow \overrightarrow{EC} \cdot \overrightarrow{AG} =1\ne 0 \\(5)\bigcirc: \overrightarrow{EF} +\overrightarrow{EA} +\overrightarrow{EH} =(1,0,0)+ (0,0,-1)+ (0,-1,0) = (1,-1,-1) =\overrightarrow{EC} \\，故選\bbox[red, 2pt]{(1235)}$$

第二部分：填充題

解答：

$$假設正\triangle 邊長為3a \Rightarrow 正六邊形邊長為a，見上圖;\\ 假設\triangle APQ面積為s \Rightarrow {\triangle APQ\over \triangle AUR}={\overline{AQ}^2\over \overline{AR}^2} ={a^2\over 4a^2}={1\over 4} \Rightarrow \triangle AUR=4s \Rightarrow 梯形PQRU面積=3s \\\Rightarrow \cases{正六邊形面積= 6s \\正\triangle ABC面積=9s=36 \Rightarrow s=4}\Rightarrow 正六邊形面積= 6\times 4=\bbox[red, 2pt]{24}$$

解答：$$年利率6\% \Rightarrow 半年利率3\% \Rightarrow 五年本利和=100(1+3\%)^{10} \approx \bbox[red, 2pt]{134}$$

解答：$$假設大王椰子為原點O(0,0) \Rightarrow \cases{第1件珠寶位置A(12,0) \\第2件珠寶位置B(4,a) \\第3件珠寶位置C(a,-8) }\\ A、B、C在一直線上\Rightarrow {12-4\over 4-a}={0-a\over a-(-8)} \Rightarrow {8\over 4-a} ={-a\over a+8} \Rightarrow a^2-12a-64=0 \\ \Rightarrow (a-16)(a+4)=0 \Rightarrow a= \bbox[red, 2pt]{16}$$

解答：$$2+\sqrt 3為x^2-(\tan \theta+\cot\theta)x+1=0的一根\Rightarrow 另一根為2-\sqrt 3\\ \Rightarrow 兩根之和=4=\tan \theta+\cot\theta ={\sin \theta\over \cos\theta}+{\cos \theta\over \sin \theta} ={2 \over \sin 2\theta} \Rightarrow \sin 2\theta ={1\over 2} \Rightarrow \theta =15^\circ \\\Rightarrow \tan \theta={\sqrt 6-\sqrt 2\over \sqrt 6+\sqrt 2} =\bbox[red, 2pt]{2-\sqrt 3}$$

解答：$$半徑r=50 \Rightarrow 周長=2\pi r=100\pi \Rightarrow 200公分相當於轉動{200 \over 100\pi  } \times 360^\circ = \bbox[red, 2pt]{229}.18^\circ$$

解答：$$\cos \angle C={1\over 2} ={3000^2 +2000^2 -\overline{AB}^2\over 2\cdot 3000\cdot 2000} \Rightarrow \overline{AB} =1000\sqrt 7\\ {\overline{AB}\over \sin \angle C} ={\overline{BC} \over \sin \angle A} \Rightarrow {1000\sqrt 7\over \sqrt 3/2} ={2000\over \sin \angle A} \Rightarrow \sin \angle A={\sqrt 3 \over \sqrt 7} ={1.732\over 2.646} =0.6546\\ 查試題附表可得\angle A\approx \bbox[red, 2pt]{41}^\circ$$

解答：$$3球任取2球有三種可能\cases{1+1=2\\ 1+5=6\\ 1+5=6} \Rightarrow 期望值{1\over 3}(2+6+6)={\bbox[red,2pt]{14}\over 3}$$

解答：$$假設\cases{白色磁磚a個\\ 咖啡磁磚b個} \Rightarrow 2a+4b=12 \Rightarrow \begin{array}{}(a,b) & 排列數\\\hline (0,3) & 1\\ (2,2) & {4!\over 2!2!}=6\\ (4,1) & 5 \\ (6,0) & 1\\\hline\end{array}\\ \Rightarrow 共有1+6+5+1 = \bbox[red, 2pt]{13}花色$$

解答：$$(a,a,b)的排列數為3，又1\le a,b\le 6且a\ne b，因此共有3\times (6\times 5)=90種可能，機率為{\bbox[red, 2pt]{90}\over 6^3}$$

解答：$$\cases{\overline{PQ}中點A=(P+Q)\div 2 = (3,3,4) \\垂直平分面之法向量\vec n= \overrightarrow{PQ}= (2,4,2)} \\\Rightarrow 過A且法向量為\vec n之平面:2(x-3)+ 4(y-3) +2(z-4)=0 \Rightarrow \bbox[red, 2pt]{x+2y+z}=13$$ 

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