大學入學考試中心
九十學年度學科能力測驗
第一部分：選擇題
壹、單一選擇題
解答：$$\cases{c=\left({1\over 4}\right)^{1/4} =\left({1\over 2}\right)^{1/2}=a\\ \cases{(\sqrt 2)^6=8\\ (\sqrt[3]3)^6=9} \Rightarrow \sqrt[3] 3\gt \sqrt 2 \Rightarrow a\gt b} \Rightarrow a=c\gt b，故選\bbox[red, 2pt]{(3)}$$
解答：
$$假設C、D、E為焦點，畫出相對應的準線L_C、L_D、L_E\\，並比較哪一個最接近？\cases{\overline{C'C}=\overline{C'P} ?\\ \overline{D'D}=\overline{D'Q} ?\\ \overline{E'E}=\overline{E'R} ?}\\，故選\bbox[red, 2pt]{(3)}$$
解答：$$R_{WY}={Cov(W,Y)\over \sigma(W)\sigma(Y)} ={Cov(168-7X,Y)\over \sigma(168-7X)\sigma(Y)} ={-7Cov( X,Y)\over 7\sigma(X) \sigma(Y)} =-R_{XY}，故選\bbox[red, 2pt]{(5)}$$
貳 、 多 重 選 擇 題
解答：$$x為第二象限角\Rightarrow \cos x,\tan x 皆為負數，故選\bbox[red, 2pt]{(345)}$$
解答：$$(1)\bigcirc: 圖形通過(0,-1)且與x軸相切\Rightarrow 圖形向凹向下 \Rightarrow a\lt 0\\ (2)\times: 無法判定\\(3)\bigcirc: f(0)=-1 \Rightarrow c=-1\\ (4)\times: 與x軸相切\Rightarrow f(x)=0只有一解\Rightarrow b^2-4ac=0 \Rightarrow b^2+4ac = 8ac \gt 0\\ (5) \bigcirc: f(x)極大值為0\Rightarrow f(1)=a+b+c \le 0\\，故選\bbox[red, 2pt]{(135)}$$
解答：$$(1)\bigcirc: (a,b)= (bq+r, b)= (r,b)=(b,r )\\ (2)\times: 8=3\cdot 2+2 \equiv a=b\cdot q+r \Rightarrow \cases{(a,b)=(8,3)=1 \\ (q,r)= (2,2)=2} \Rightarrow (a,b)\ne (q,r)\\(3) 同上例\cases{(a,q)= (8,2)=2 \\ (b,r)=(3,2)=1} \Rightarrow (a,q) \ne (b,r)\\ (4)\bigcirc: (a,q) =(bq+r,q)=(r,q) =(q,r)\\(5)\times: 同(2)例\cases{(a,r)= (8,2)=2\\ (b,q)=(3, 2)=1} \Rightarrow (a,r)\ne (b,q)\\，故選\bbox[red, 2pt]{(14)}$$
解答：$$(2) 16a+6b =28 \Rightarrow a=1,b=2\\ (3)16a+6b=82 \Rightarrow a=4,b=3 \\(5) 16a+6b=284 \Rightarrow a=14,b=10，故選\bbox[red, 2pt]{(235)}$$
解答：$$(1)\bigcirc: \cases{\overrightarrow{AB} =(-4,3)\\ \overrightarrow{OC}=(-4,3)\\ \overrightarrow{OA} = (150,200)\\ \overrightarrow{CB} =(150,200) } \Rightarrow \overrightarrow{AB} = \overrightarrow{OC} 且\overrightarrow{OA} =\overrightarrow{CB} \Rightarrow OABC為平行四邊形 \\ (2)\bigcirc: \overrightarrow{AB} \cdot \overrightarrow{OA}= -600 +600=0 \Rightarrow \overline{AB} \bot \overline{OA} \Rightarrow OABC為矩形 \\(3) \times: \cases{\overline{AB}= 5\\ \overline{OA} = \sqrt{150^2+200^2}} \Rightarrow \overline{AB} \ne \overline{OA} \Rightarrow 對角線不互垂 \\(4)\times: \overline{AB}^2+\overline{OA}^2 = 5^2+ 150^2+200^2 =62525 \lt 251^2\\ (5)\bigcirc: \overline{AB}\times \overline{AO}= 5\times 250 =1250\\，故選\bbox[red, 2pt]{(125)}$$
解答：$${x^2\over 25}-{y^2\over 4}=1 \Rightarrow 漸近線2x = \pm 5y \Rightarrow 過原點且斜率介於兩漸近線之間均與雙曲線不相交\\，故選\bbox[red, 2pt]{(124)}$$
解答：$$\cases{z^6=1 \\z\ne 1} \Rightarrow z_k= \cos {k\pi\over 3}+i\sin {k\pi \over 3},k=1-5\\(1)\bigcirc: |z_k|=1,k=1-5\\ (2)\times: z=z_1=\cos {\pi\over 3} +i\sin {\pi\over 3} \Rightarrow z^2=-{1\over 2}+i{\sqrt 3\over 2}\ne 1\\ (3) \bigcirc: z_k^3 = \cos k\pi +i\sin k\pi =\pm 1 \\(4)\bigcirc: |z_k^4| = |\cos {4k\pi\over 3}+i\sin {4k\pi \over 3}|=1\\(5)\bigcirc: z^6-1=0 \Rightarrow 1-z^6=(1-z)(1+z +z^2+z^3+z^4+z^5)=0\\\qquad \quad \Rightarrow 1+z +z^2+z^3+z^4+z^5=0 \\，故選\bbox[red, 2pt]{(1345)}$$
第二部分：填充題
解答：$$B4\cases{長邊:36.4\\ 短邊:a} \Rightarrow B5\cases{長邊:a\\ 短邊:36.4/a} \Rightarrow {36.4\over a}={a\over 36.4/2} \Rightarrow a^2=36.4^2/ 2 \Rightarrow a=36.4 /\sqrt 2 \\ =18.2\times 1.414 \approx \bbox[red, 2pt]{25.7}$$
解答：$${600\times 36\%+ 400\times 46\%\over 600+400}=\bbox[red, 2pt]{40}\%$$
解答：$$取兩數(a,b)，滿足ab=k^3 \Rightarrow (a,b,k)=(1,8,2),(2,4,2),(3,9,3)，有三種選擇\\，機率為{3\over C^9_2}=\bbox[red,2pt]{1\over 12}$$
解答：$$f(x)=\cases{(x^2-5x+4)p(x)+x+2= (x-4)(x-1)p(x)+ x+2\\ (x^2-5x+6)q(x)+3x+4=(x-2)(x-3)q(x)+ 3x+4\\ (x^2-4x+3)r(x)+ ax+b =(x-1)(x-3)r(x)+ ax+b} \\ \Rightarrow \cases{f(1) = 0\cdot p(1)+3=0\cdot r(1)+a+b\\ f(3)=0\cdot q(3)+13=0\cdot r(3)+3a+b} \Rightarrow \cases{a+b= 3\\ 3a+b= 13} \Rightarrow \cases{a=5\\ b=-2} \Rightarrow 餘式=\bbox[red, 2pt]{5}x-\bbox[red,2pt]{2}$$
解答：
$$假設O為圓心，則\cases{\angle OBC=\angle BAC=90^\circ\\ \angle COB=60^\circ\\ \angle OCB=30^\circ} \Rightarrow 圓半徑\overline{OB}= \overline{BC}\div \sqrt 3=150\sqrt 3\\ \Rightarrow \stackrel{\Large{\frown}}{AB} =150\sqrt 3\cdot {2\pi\over 3} \approx \bbox[red, 2pt]{544}$$
解答：
$$\overline{PQ}^2= \overline{PB}^2-\overline{BQ}^2=15 \Rightarrow \overline{PO}^2 = \overline{PQ}^2-\overline{OQ}^2=15-1=14 \Rightarrow \overline{PO} = \sqrt{ \bbox[red,2pt]{14}}$$
$$假設P在x軸上，且\overline{CP}\bot \overline{AB}，並令\overline{AP}=a，則\overline{BP}=6-a;\\ \cases{\angle \tan BAC= h/a= 8/9 \\ \tan \angle ABC= h/(6-a)= 8/3} \Rightarrow \cases{a=9/2\\ h=4} \Rightarrow C(-5/2,4)\\ \Rightarrow \overline{CD} =\sqrt{(-5/2-5/2)^2+ 4-(-8)^2} =\sqrt{5^2+12^2} = \bbox[red, 2pt]{13}$$
解答：$$正四面體稜長為a \Rightarrow 體積為{\sqrt 2\over 12}a^3=12\\ 小四面體稜長為{1\over 2}a \Rightarrow 體積為{\sqrt 2\over 12}({a\over 2})^3= {12\over 8 }\Rightarrow 四個小四面體體積=4\times {12\over 8}\\ \Rightarrow 正八面體體積=12-4\times {12\over 8}= \bbox[red, 2pt]{6}$$
解答：$${不良品檢驗為良品 \over 良品檢驗為良品+不良品檢驗為良品} ={5\%\times 0.16\over 95\%\times 0.8 +5\%\times 0.16}={1\over 96} \approx 0.\bbox[red,2pt]{01}$$
解答：$$甲乙在同一隊:剩下7人找一人與甲乙合組一隊，有7種組法，剩下6人組2隊有C^6_3/2=10種組法；\\\qquad \qquad 因此甲乙同一隊有7\times 10=70種組法;\\9人任組三隊，有C^9_3C^6_3\div 3! =280；因此甲乙不同隊有280-70=\bbox[red, 2pt]{210}種組法。$$
========================= END ==========================
解答：$$R_{WY}={Cov(W,Y)\over \sigma(W)\sigma(Y)} ={Cov(168-7X,Y)\over \sigma(168-7X)\sigma(Y)} ={-7Cov( X,Y)\over 7\sigma(X) \sigma(Y)} =-R_{XY}，故選\bbox[red, 2pt]{(5)}$$
解答：
沒有留言:
張貼留言