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2021年6月20日 星期日

93年大學指考(補考)-數學甲詳解

九十三學年度指定科目考試

敏督利颱風受災地區考生補救考試-數學甲

一、單選題

解答4xy=1y=14xx2+xy+y2=1x2+14+116x2=116x412x2+1=0x2=3±58x=±3±584(4)
解答1010×6×1023=6×10246×1024100=6×1022=6×1016=4π×64002=1.6384×108π=1.6384×108π×106=1.6384×1014π6×10161.6384×1014π116(3)

二、多選題

解答1143=a11+b13=13a+11b14313a+11b=1gcd(11,13)=1(1)(4)(1234)
解答(1):0<α,β<π{sinα>0sinβ>0(2):αβ2<α+β2<π2{cosα+β2>0cosαβ2>0cosα+cosβ=2cosα+β2cosαβ2>0(3)×:tan120+tan30=3+33<0(4):cotα+cotβ=cosαsinα+cosβsinβ=sin(α+β)sinαsinβ>0(124)
解答{C1:x2+y2=1O1(0,0),r1=1C2:(x4)2+y2=9O2(4,0),r2=3(1)×:L1:22x+22y=1{d(O1,L1)=1=r1d(O2,L1)=22<r2{L1C1L1C2(2)×:L2:32x+12y=1{d(O1,L1)=1=r1d(O2,L1)=23+1>r2{L1C1L1C2(3):L3:12x32y=1{d(O1,L1)=1=r1d(O2,L1)=3=r2{L1C1L1C2(4):L3:x=1{d(O1,L1)=1=r1d(O2,L1)=3=r2{L1C1L1C2(34)
解答(1)×:8(2)×:44(3):C84×128=70256>14(4)×:{:=1/2564:=1/2564:=1/2561256

解答令實驗結果為\langle(x,y)\rangle,其中x的單位為英噸,y的單位為英吋;又x'=x/1.06,y'=2.54y\\(1)\bigcirc: m=r\cdot {\sigma(y) \over \sigma(x)} \Rightarrow 相關係數與斜率同號 \Rightarrow r\cdot m\gt 0\\(2)\times:由實驗結果得知:x越大,則y越小,即r\lt 0\\(3) \bigcirc:  {1\over 1.06} \times 2.54 \gt 0 \Rightarrow r=R\\ (4)\times: \cases{m=r\cdot {\sigma(y)\over \sigma(x)} \\ M= R\cdot {\sigma(y')\over \sigma(x')}= r\cdot {2.54\sigma(y)\over \sigma(x)/1.06}} \Rightarrow m\ne M\\故選\bbox[red,2pt]{(13)}
解答(1)\bigcirc: k=1 \Rightarrow \cases{-z=-2\\ y-4z=-9\\ -2x+y+2z=6} \Rightarrow \cases{x=-3/2\\ y=-1\\ z=2}恰有一解 \\(2)\bigcirc: k=\sqrt{10} \Rightarrow \cases{{1\over 2}y-z=-2\\ x+y-4z=-9\\ -2x+y+2z=6} \Rightarrow \cases{ y-2z=-4 \cdots(1)\\ 2x+2y-8z=-18 \cdots(2)\\ -2x+y+2z=6 \cdots(3)},\\\qquad {(2)+(3)\over 3}\Rightarrow y-2z=-4 \equiv (1)\Rightarrow 有無限多解 \\(3)\times: k=10 \Rightarrow \cases{ y-z=-2 \cdots(1)\\ 2x+y-4z=-9 \cdots(2)\\ -2x+y+2z=6 \cdots(3)} ,{(2)+(3)\over 2}\Rightarrow y-z=-{3\over 2}與(1)矛盾\Rightarrow 無解 \\(4)\times: k=100 \Rightarrow \cases{ 2y-z=-2  \\ 4x+y-4z=-9  \\ -2x+y+2z=6  } \Rightarrow \cases{x=3/2\\ y=1\\ z=4}恰有一解\\故選\bbox[red, 2pt]{(12)}
解答A=\begin{bmatrix}a & b\\ c& d \end{bmatrix} 為轉移矩陣 \Rightarrow \cases{a+c=1\\ b+d=1}\\ A^2=\begin{bmatrix}a^2+bc & ab+bd\\ ac+dc & bc+d^2 \end{bmatrix}  = \begin{bmatrix}5/9 & 4/9\\ 4/9& 5/9 \end{bmatrix} \Rightarrow \cases{a^2+ bc=5/9 \cdots(1) \\ ac+dc=4/9 \cdots(2)\\ ab+bd=4/9\cdots(3)\\ bc+d^2=5/9 \cdots(4)}\\(1)\bigcirc: a={1\over 3}代入a+c=1 \Rightarrow c={2\over 3},將\cases{a=1/3\\ c=2/3} 代入(2) \Rightarrow a+d=2/3 \Rightarrow d=1/3 \\(2) \times: a={2\over 3} \Rightarrow c={1\over 3},將\cases{a=2/3\\ c=1/3} 代入(2) \Rightarrow a+d={4\over 3} \Rightarrow d={2\over 3}\ne {1\over 2} \\(3)\times:式(1)-式(4) \Rightarrow a^2-d^2=0 \Rightarrow a=d (a\ne -d,\because a,d \gt 0) \Rightarrow b=c(\because a+c=b+d) \\ \qquad \Rightarrow ac-bd =0\\(4)\bigcirc: 理由同(3)\\ 故選\bbox[red,2pt]{(14)}

三、選填題

解答\cases{P(\sqrt{n+5},\sqrt{n-1}) \\ Q(\sqrt{n-1},\sqrt{n+5})\\ L:y-x=0} \Rightarrow d_n=d(P,L) = {\sqrt{n+5}-\sqrt{n-1}\over \sqrt 2} \\={(\sqrt{n+5}-\sqrt{n-1})(\sqrt{n+5}+\sqrt{n-1}) \over \sqrt 2(\sqrt{n+5}+\sqrt{n-1})}  ={3\sqrt 2\over \sqrt{n+5}+\sqrt{n-1}}\\ \Rightarrow {3\sqrt 2\over 2\sqrt{n+5}}\lt  d_n\lt {3\sqrt 2\over 2\sqrt{n-1}} \Rightarrow \bbox[red,2pt]{\lim_{n\to \infty}d_n=0} \left(\because \cases{\lim_{n\to \infty} {3\sqrt 2\over 2\sqrt{n+5}}=0 \\ \lim_{n\to \infty} {3\sqrt 2\over 2\sqrt{n-1}}=0} \right)\\ 又\cases{\overline{PQ}=\sqrt{(\sqrt{n+5}-\sqrt{n-1})^2 +(\sqrt{n-1}-\sqrt{n+5})^2} =\sqrt 2(\sqrt{n+5}- \sqrt{n-1})\\ R=(P+Q)/2=\left({\sqrt{n+5}+\sqrt{n-1}\over 2}, {\sqrt{n+5}+\sqrt{n-1}\over 2} \right)} \\ \Rightarrow a_n=\triangle OPQ ={1\over 2}\cdot \overline{PQ}\cdot \overline{OR} ={1\over 2}\cdot \sqrt 2(\sqrt{n+5}-\sqrt{n-1}) \cdot {\sqrt{n+5}+\sqrt{n-1}\over \sqrt 2}  =3 \\ \Rightarrow \bbox[red,2pt]{\lim_{n\to \infty} a_n=3}

第貳部分:非選擇題

解答
令\overline{AD}=x \Rightarrow \cases{\cos \angle ADB ={x^2+d^2-c^2\over 2dx} \\ \cos \angle ADC = {x^2+e^2-b^2\over 2ex}} ,由於\angle ADB+\angle ADC=\pi \Rightarrow \cos\angle ADB =-\cos \angle ADC\\ \Rightarrow {x^2+d^2-c^2\over 2dx}={b^2-x^2-e^2 \over 2ex}={c^2-d^2-x^2  \over 2ex} \Rightarrow x^2e+e(d^2-c^2) =d(c^2-d^2)-x^2d \\  \Rightarrow x^2(d+e)= (d+e)(c^2-d^2) \Rightarrow x^2=c^2-d^2 \Rightarrow \angle ADB=90^\circ\Rightarrow \overline{AD}\bot \overline{BC},\bbox[red,2pt]{故得證}
解答

(1)P(1,2)在拋物線\Gamma:y=x^2+1上,而y'=2x \Rightarrow y'(1)=2 \Rightarrow 切線斜率為2 \\\Rightarrow 切線L方程式: y=2(x-1)+2 \Rightarrow \bbox[red,2pt]{y=2x}(2)\Gamma對稱軸為y軸,即x=0 \Rightarrow 圓心坐標O(0,a) \Rightarrow \text{dist}(O,L)=\overline{OP} \Rightarrow {a\over \sqrt 5} = \sqrt{1+(a-2)^2} \\ \Rightarrow 4a^2-20a+25=0 \Rightarrow a=5/2 \Rightarrow 圓心坐標\bbox[red,2pt]{(0,{5\over 2})}
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解題僅供參考,其他指考試題及詳解


1 則留言:

  1. 第二題地球表面積似乎數值出來不對

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