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:2100=(210)10=102410≡410mod10≡(42)5mod10≡65mod10≡36⋅36⋅6mod10≡6⋅6⋅6mod10≡6mod10,故選(4)解答:{(1)21/3(2)(2−3)−2=26(3)2−1/4(4)(12)1/2=2−1/2(5)8−1/3=2−1⇒2−1最小,故選(5)
解答:
{A(1/2,1,1)B(1,1,1/2)C(1/2,0,1)⇒{→AB=(1/2,0,−1/2)→AC=(0,−1,0)⇒{→AB⋅→AC=0→n=→AB×→AC=(−1/2,0,−1/2)⇒{∠B為直角過A,B,C之平面E:2x+2z=3⇒E與正立方體交於另一點D(1,0,1/2)⇒→CD=(1/2,0,−1/2)⇒→CD⋅→AC=0⇒∠C為直角,又¯AB≠¯AC⇒ABCD為長方形,故選(4)
貳、多重選擇題
解答:{直角△AHC⇒¯AH=bsinC直角△AHB⇒¯AH=csinB⇒¯AH=bsinC=csinB,故選(34)解答:(1)×:0.3¯43=0.¯34=3499為有理數(2)◯:3499>3399=13(3)◯:0.3434⋯>0.343(4)◯:0.3434⋯<0.35(5)◯:{a=0.¯34⇒100a=34.¯34b=0.3¯43⇒100b=34.¯34⇒a=b,故選(2345)
解答:f(x)=x3+x2−2x−1⇒{f(−2)=−1<0f(−1)=1>0f(0)=−1<0f(1)=−1<0f(2)=7>0f(3)=29>0(1)◯:f(−2)f(−1)<0(2)◯:f(−1)f(0)<0(3)×:f(0)f(1)>0(4)◯:f(1)f(2)<0(5)×:f(2)f(3)>0,故選(124)
解答:√(x−1)2+(y−2)2+√(x+1)2+(y+2)2=6⇒{F1(1,2)F2(−1,−2)2a=6(1)◯:中心坐標=(F1+F2)÷2=(0,0)(2)◯:焦點{F1(1,2)F2(−1,−2)(3)◯:2c=¯F1F2=√4+16=2√5⇒c=√5⇒b=√a2−c2=√9−5=2⇒短軸2b=4(4)×:↔F1F2:2x=y為對稱軸⇒另一對稱軸x=−2y(5)◯:橢圓對稱於長軸,故選(1235)
解答:(1)×:差一個負號(2)◯:det
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)}
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
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
========================= END ==========================
解答僅供參考,其他歷屆試題及詳解
沒有留言:
張貼留言