94年大學學測-數學詳解

大學入學考試中心九十四學年度學科能力測驗試題

第 一 部 分 ： 選 擇 題 

壹 、 單 選 題 

解答：$$43659= 3^4\times 7^2\times 11 \Rightarrow 共三個質因數，故選\bbox[red,2pt]{(3)}$$

解答：$$11^3+12^3 +\cdots +20^3 = (1^2+2^3 +\cdots +20^3)-(1^2+2^3 +\cdots+10^3)\\ = \left( {20\cdot 21\over 2}\right)^2- \left( {10\cdot 11\over 2}\right)^2 =210^2-55^2 = 41075，故選\bbox[red,2pt]{(1)}$$

解答：$$\cases{R={1\over C^{42}_6} \\ r={1\over C^{39}_5}} \Rightarrow {r\over R}={C^{42}_6\over C^{39}_5} ={42!\over 6!36!} \cdot {5!34!\over 39!} = {82\over 9} \approx 9，故選\bbox[red,2pt]{(4)}$$

解答：$$\cases{\log_7 a=11\\ \log_7 b=13} \Rightarrow \cases{a=7^{11}\\ b=7^{13}} \Rightarrow \log_7(a+b) =\log_7 (7^{11}+7^{13}) =\log_7 7^{11}(1+7^2) \\ =\log_7 7^{11}+ \log_7 50=11 +\log_7 50 \approx 11+\log_7 49=11+2=13，故選\bbox[red,2pt]{(2)}$$

解答：$$假設\cases{原始成績X\\ 週整後成績Y} \Rightarrow Y=10\sqrt{X} \Rightarrow \cases{E(Y)=E(10\sqrt X) =65\\ \sigma(Y)= \sigma(10 \sqrt X)=15} \Rightarrow \cases{E(\sqrt X)=6.5\\ \sigma(\sqrt X)=1.5} \\ \Rightarrow Var(\sqrt X) = E((\sqrt X)^2)-(E(\sqrt X))^2 \Rightarrow 1.5^2 = E(X)-6.5^2 \Rightarrow E(X)=1.5^2+6.5^2 =44.5\\，故選\bbox[red,2pt]{(5)}$$

貳、多選題

解答：$$\alpha \overrightarrow{OA}+ \beta \overrightarrow{OB} 落在陰影區\Rightarrow \alpha+\beta \ge 1\\ (1)\bigcirc: 1+2 \ge 1\\ (2) \bigcirc: {3\over 4}+{1\over 3} ={13\over 12}\ge 1\\(3) \times: {3\over 4}-{1\over 3}\lt 1\\(4) \times: {3\over 4}+{1 \over 5} ={19\over 20}\lt 1 \\(5)\times: {3\over 4}-{1\over 5}\lt 1\\，故選\bbox[red,2pt]{(12)}$$

解答：

$$(1)\times: \cases{a^2=2^2-b^2\\ (5-a)^2 =4^2-b^2} \Rightarrow 5-a\gt a \Rightarrow \cases{m_{CD}= (5-a)/b\\ m_{AB}= a/b} \Rightarrow m_{CD}\gt m_{AB}\\ (2) \bigcirc: \cases{m_{AD}=-a/b\\ m_{BC}=-(5-a)/b} \Rightarrow m_{BC}最小\\ (3)\bigcirc: \cases{m_{BC}=-(5-a)/b\\ m_{CD}=(5-a)/b} \Rightarrow m_{BC}=-m_{CD}\\(4) \times:\cases{\overline{AB}^2+ \overline{BC}^2 =2^2+4^2=20\\ \overline{AC} ^2 =5^2=25} \Rightarrow  \overline{AB}^2+ \overline{BC}^2\ne \overline{AC}^2 \Rightarrow \angle ABC\ne 90^\circ \\(5) \bigcirc: m_{CD}+ m_{DA}={5-a\over b}-{a\over b} ={5-2a\over b}\gt 0(5-a\gt a \Rightarrow 5-2a\gt 0)\\，故選\bbox[red,2pt]{(235)}$$

解答：$$\cases{A(-1,2,0)\\ B(3,0,2)} \Rightarrow L=\overleftrightarrow{AB}:{x+1\over 4}={y-2\over -2}={z\over 2} \\(1) \times: (2,2,2)\not \in L\\(2)\bigcirc: (1,1,1) \in L \\(3)\times: (4,-2,2)\not \in L\\(4)\times: (-2,4,0)\not \in L\\ (5)\times: (-5,-4,-2)\not \in L\\，故選\bbox[red,2pt]{(2)}$$

解答：

$$假設直角三角形，兩股長分別為a及b，斜邊長為c，如上圖;\\(1)\bigcirc: \theta \lt 45^\circ \Rightarrow a\lt b \Rightarrow {b\over c}\gt {a\over b}，即\cos\theta \gt \sin \theta \\(2)\times: c\gt b \Rightarrow {a\over c} \lt {a\over b} \Rightarrow \sin \theta \lt \tan \theta \\(3) \times: 若\cases{a=3\\ b=4\\c=5} \Rightarrow \cases{\tan \theta =3/4\\ \cos\theta =4/5} \Rightarrow \cos\theta \not \lt \tan\theta\\(4)\times: 若\theta=30^\circ \Rightarrow \cases{\cos 2\theta =1/2\\ \sin 2\theta =\sqrt 3/2} \Rightarrow \sin 2\theta \not \lt \cos 2\theta \\(5) \bigcirc: \tan \theta ={2\tan \theta/2\over 1-\tan^2 \theta/2} \Rightarrow {1\over 2}\tan \theta-\tan \theta /2= {\tan^3 \theta/2\over 1-\tan^2 \theta/2} \gt 0\\，故選\bbox[red,2pt]{(15)}$$

解答：

$${x^2\over 9}-{y^2\over 16}=1 \Rightarrow \cases{a=3\\ b=4} \Rightarrow c=5 \Rightarrow 兩焦點\cases{F_1(-5,0)\\ F_2(5,0)} \\ \cases{\triangle AF_1F_2: \overline{F_1F_2}= \overline{AF_2}=2c=10 \Rightarrow \overline{AF_1}-\overline{AF_2}= 2a \Rightarrow \overline{AF_1}=6+10=16\\ \triangle BF_1F_2: \overline{BF_1} =\overline{F_1F_2}=2c=10 \Rightarrow \overline{BF_1}-\overline{BF_2}=2a \Rightarrow \overline{BF_2}=10-6=4} \\ \Rightarrow \cases{\triangle AF_1F_2周長=10+10+16 = 36\\ \triangle BF_1F_2周長= 10+ 10+ 4=24}\\，故選\bbox[red,2pt]{(25)}$$

解答：$$(1)\times: P\in \overline{AB} \Rightarrow \overline{PA}+\overline{PB}= \overline{AB}=10\ne 14\\ (2)\bigcirc: 若\cases{\overline{PA}=2\\ \overline{PB}=12} \Rightarrow \overline{PA}+\overline{PB}=14，此時P在直線AB上，但不在線段AB上\\(3) \bigcirc: 若\cases{\overline{PA}=8\\ \overline{PB}=6} \Rightarrow 6^2+8^2=10^2，此時P在球面上\\ (4)\bigcirc: 以A、B為橢圓焦點，長軸長2a=14，短軸長2b\lt 20；\\\qquad 此時取P為短軸上的頂點，P在球內且符合\overline{PA}+\overline{PB}=14\\(5)\bigcirc: 理由同(4)，取短軸長2b=20、長軸長2a\gt 20，取P為長軸上的頂點，\\\qquad P在球外且符合\overline{PA}+\overline{PB}=14\\，故選\bbox[red,2pt]{(2345)}$$

第 二 部 分 ： 填 充 題

解答：$$利用長除法:x^5+x^4+x^3 +px^2 +2x+q =(x^2 +x+1)(x^3 -x+(p+1))\\ \Rightarrow \cases{4=p+1\\ q=2(p+1)} \Rightarrow p=\bbox[red,2pt]{3},q=\bbox[red, 2pt]{8}$$

解答：$$P(x,y) \Rightarrow \cases{\triangle PDA:\triangle PBC=1:2 = 1-y:y \Rightarrow y=2/3\\ \triangle PAB:\triangle PCD=2:3 = x:1-x \Rightarrow x=2/5} \Rightarrow P(\bbox[red,2pt]{2\over 5},\bbox[red,2pt]{2\over 3})$$

解答：$$假設向右跳a次、向左跳b次，則\cases{a+b=6\\ a-b=4} \Rightarrow \cases{a=5\\ b=1};\\ 5個a,1個b的排列數=\bbox[red,2pt]{6}$$

解答：$$z=1-i= \sqrt 2({1\over \sqrt 2}-{1\over \sqrt 2}i) = \sqrt 2(\cos 45^\circ-i\sin 45^\circ) \Rightarrow z^{10}= 2^5(\cos 450^\circ-i\sin 450^\circ)= -2^5 i\\ 1+z+ z^2+\cdots +z^9= a+bi  \Rightarrow (1-z)(1+z+ z^2+\cdots +z^9)=1-z^{10}= (a+bi)(1-z) \\ \Rightarrow a+bi  ={1-z^{10}\over 1-z} ={1+2^5 i\over i} =2^5-i \Rightarrow a=\bbox[red,2pt]{32},b=\bbox[red, 2pt]{-1}$$

解答：$$\cases{A(a,0)\\ B(0,b) \\P(2,1)} \Rightarrow \cases{\overrightarrow{PA}= (a-2,-1)\\ \overrightarrow{PB} =(-2,b-1)} \Rightarrow \overrightarrow{PA} \cdot \overrightarrow{PB}=-2a-b+5=0 \Rightarrow 2a+b=5 \\ 算幾不等式: {2a+b\over 2} \ge  \sqrt{2ab} \Rightarrow {5\over 2} \ge \sqrt{2ab} \Rightarrow ab\le {25\over 8 } \Rightarrow \triangle OAB面積={1\over 2}ab \le \bbox[red,2pt]{25\over 16}$$

解答：$$假設\overline{AB}=\overline{AD}=a，由於\overline{AD}為\angle BAC的角平分線，因此\cfrac{\overline{AB}}{\overline{AC}} =\cfrac{\overline{BD}}{\overline{DC}} =\cfrac{3}{6} \Rightarrow \overline{AC}= 2\overline{AB}= 2a\\ \cases{\cos \angle BAD = {a^2+a^2-3^2 \over 2a^2} \\ \cos \angle DAC= {a^2+(2a)^2-6^2\over 4a^2}} \Rightarrow {a^2+a^2-3^2 \over 2a^2}={a^2+(2a)^2-6^2\over 4a^2} \Rightarrow a^2=18 \\ \Rightarrow \cos \angle BAD ={18+18-9\over 2\cdot 18} =\bbox[red,2pt]{3\over 4}$$

解答：$$P在\Gamma:y^2=4x上 \Rightarrow P(t^2/4,t); 並假設Q(a,b)，再由{\overline{PF}\over \overline{QF}}={3\over 2} \Rightarrow F=(2P +3Q)\div 5\\ \Rightarrow Q=(5F-2P)\div 3 = ({5\over 3}-{t^2\over 6},-{2\over 3}t) 代入\Gamma \Rightarrow {4\over 9}t^2 = 4({5\over 3}-{t^2\over 6}) \Rightarrow t^2=6\\ \Rightarrow P的x坐標=t^2/4 = \bbox[red,2pt]{3\over 2}$$

解答：$$x\cdot 3^x = 3^{18} \Rightarrow \log_3(x\cdot 3^x)=\log_3 3^{18} \Rightarrow \log_3 x+x=18\\ 由於\cases{18\lt \log_3 16+ 16 \lt 19 \\ 17\lt \log_3 15+15\lt 18} \Rightarrow 15\lt x\lt 16 \Rightarrow k=\bbox[red,2pt]{15}$$

解答：$$假設\cases{D(0,0,0)\\ C(1,0,0)\\ A(0,1,0)\\ B(1,1,0)\\ E(0,1,1)\\ P(a,b,c)}  \Rightarrow \cases{\overrightarrow{AP}= (a,b-1,c)\\ \overrightarrow{AB}= (1,0,0)\\ \overrightarrow{AD}=(0,-1,0)\\ \overrightarrow{AE}= (0,0,1)} \Rightarrow {3\over 4}\overrightarrow{AB}+ {1\over 2}\overrightarrow{AD}+ {2\over 3}\overrightarrow{AE}=({3\over 4},-{1\over 2},{2\over 3})=(a,b-1,c)\\ \Rightarrow \cases{a=3/4 \\b=1/2 \\c=2/3} \Rightarrow P(3/4,1/2,2/3)至\overline{AB}(y軸)距離=\sqrt{ (1/2)^2+( 2/3)^2} = \bbox[red,2pt]{5\over 6}$$

========================= END ==========================

解答僅供參考，其他歷屆試題及詳解