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=1⇒y=14x代入x2+xy+y2=1⇒x2+14+116x2=1⇒16x4−12x2+1=0⇒x2=3±√58⇒x=±√3±√58⇒有4交點,故選(4)解答:10莫耳米的粒子數為10×6×1023=6×1024相當於6×1024100=6×1022立方公分=6×1016立方公尺地球表面積=4π×64002=1.6384×108π平方公里=1.6384×108π×106=1.6384×1014π平方公尺因此6×10161.6384×1014π≈116公尺,故選(3)
二、多選題
解答:1143=a11+b13=13a+11b143⇒13a+11b=1又gcd(11,13)=1,因此(1)−(4)皆正確,故選(1234)解答:(1)◯:0<α,β<π⇒{sinα>0sinβ>0(2)◯:α−β2<α+β2<π2⇒{cosα+β2>0cosα−β2>0⇒cosα+cosβ=2cosα+β2cosα−β2>0(3)×:tan120∘+tan30∘=−√3+√33<0(4)◯:cotα+cotβ=cosαsinα+cosβsinβ=sin(α+β)sinαsinβ>0故選(124)
解答:{C1:x2+y2=1⇒圓心O1(0,0),半徑r1=1C2:(x−4)2+y2=9⇒圓心O2(4,0),半徑r2=3(1)×:L1:−√22x+√22y=1⇒{d(O1,L1)=1=r1d(O2,L1)=2√2<r2⇒{L1與C1相切L1與C2不相切(2)×:L2:−√32x+12y=1⇒{d(O1,L1)=1=r1d(O2,L1)=2√3+1>r2⇒{L1與C1相切L1與C2不相切(3)◯:L3:−12x−√32y=1⇒{d(O1,L1)=1=r1d(O2,L1)=3=r2⇒{L1與C1相切L1與C2相切(4)◯:L3:x=1⇒{d(O1,L1)=1=r1d(O2,L1)=3=r2⇒{L1與C1相切L1與C2相切故選(34)
解答:(1)×:不一定,也可能8次都正面(2)×:後4次與前4次無關,且出現正面與反面機率相等(3)◯:C84×128=70256>14(4)×:{正反交錯:正反正反正反正反⇒機率=1/256正面集中前4次:正正正正反反反反⇒機率=1/256正面集中後4次:反反反反正正正正⇒機率=1/256⇒1256≯
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}
(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
第二題地球表面積似乎數值出來不對
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