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：$$4xy=1 \Rightarrow y={1\over 4x}代入x^2+xy+y^2=1 \Rightarrow x^2+{1\over 4}+{1\over 16x^2}=1 \Rightarrow 16x^4-12x^2+1=0\\ \Rightarrow x^2={3\pm \sqrt 5\over 8} \Rightarrow x=\pm \sqrt{3\pm \sqrt 5\over 8} \Rightarrow 有4交點，故選\bbox[red,2pt]{(4)}$$

解答：$$10莫耳米的粒子數為10\times 6\times 10^{23}=6\times 10^{24}相當於{6\times 10^{24}\over 100}=6\times 10^{22}立方公分=6\times 10^{16}立方公尺\\地球表面積=4\pi\times 6400^2=1.6384\times 10^{8}\pi平方公里 =1.6384\times 10^{8}\pi\times 10^6=1.6384\times 10^{14}\pi平方公尺\\ 因此{6\times 10^{16}\over 1.6384\times 10^{14}\pi} \approx 116公尺，故選\bbox[red,2pt]{(3)}$$

二、多選題

解答：$${1\over 143}={a\over 11}+{b\over 13}={13a+11b\over 143}\Rightarrow 13a+11b=1\\ 又\gcd(11,13)=1，因此(1)-(4)皆正確，故選\bbox[red,2pt]{(1234)}$$

解答：$$(1)\bigcirc: 0\lt \alpha,\beta \lt \pi \Rightarrow \cases{\sin \alpha \gt 0\\ \sin \beta \gt 0} \\(2)\bigcirc: {\alpha-\beta \over 2} \lt {\alpha+\beta \over 2} \lt {\pi \over 2} \Rightarrow \cases{\cos{\alpha+\beta\over 2} \gt 0 \\\cos{\alpha-\beta\over 2} \gt 0} \Rightarrow \cos \alpha+\cos \beta = 2\cos{\alpha+\beta \over 2} \cos{\alpha-\beta \over 2} \gt 0\\(3)\times: \tan  120^\circ +\tan  30^\circ= -\sqrt 3+{\sqrt 3\over 3} \lt 0\\(4)\bigcirc: \cot \alpha+\cot \beta={\cos \alpha\over \sin \alpha} +{\cos \beta \over \sin \beta} ={\sin(\alpha+\beta)\over \sin \alpha\sin \beta} \gt 0\\ 故選\bbox[red, 2pt]{(124)}$$

解答：$$\cases{C_1:x^2+y^2=1 \Rightarrow 圓心O_1(0,0), 半徑r_1=1\\ C_2:(x-4)^2+y^2=9 \Rightarrow 圓心O_2(4,0),半徑r_2=3}\\(1) \times: L_1:-{\sqrt 2\over 2}x+ {\sqrt 2\over 2}y=1 \Rightarrow \cases{d(O_1,L_1)=1=r_1 \\ d(O_2,L_1)=2\sqrt 2\lt r_2} \Rightarrow \cases{L_1與C_1相切\\ L_1與C_2不相切} \\(2) \times: L_2:-{\sqrt 3\over 2}x+ {1\over 2}y=1 \Rightarrow \cases{d(O_1,L_1)=1=r_1 \\ d(O_2,L_1)=2\sqrt 3+1\gt r_2} \Rightarrow \cases{L_1與C_1相切\\ L_1與C_2不相切} \\(3) \bigcirc: L_3:-{1\over 2}x- {\sqrt 3\over 2}y=1 \Rightarrow \cases{d(O_1,L_1)=1=r_1 \\ d(O_2,L_1)=3= r_2} \Rightarrow \cases{L_1與C_1相切\\ L_1與C_2相切} \\(4) \bigcirc: L_3:x=1 \Rightarrow \cases{d(O_1,L_1)=1=r_1 \\ d(O_2,L_1)=3= r_2} \Rightarrow \cases{L_1與C_1相切\\ L_1與C_2相切}\\故選\bbox[red, 2pt]{(34)}$$

解答：$$(1)\times: 不一定，也可能8次都正面\\(2)\times: 後4次與前4次無關，且出現正面與反面機率相等\\(3)\bigcirc: C^8_4\times {1\over 2^8} ={70\over 256} \gt {1\over 4} \\(4)\times: \cases{正反交錯:正反正反正反正反 \Rightarrow 機率=1/256 \\正面集中前4次:正正正正反反反反\Rightarrow 機率=1/256 \\正面集中後4次:反反反反正正正正\Rightarrow 機率=1/256} \\\qquad \Rightarrow {1\over 256} \not \gt {1\over 256} +{1\over 256}\\故選\bbox[red,2pt]{(3)}$$

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)$$\Gamma對稱軸為y軸，即x=0 \Rightarrow 圓心坐標O(0,a) \Rightarrow \text{dist}(O,L)=\overline{OP} \Rightarrow {a\over \sqrt 5} = \sqrt{1+(a-2)^2} \\ \Rightarrow 4a^2-20a+25=0 \Rightarrow a=5/2 \Rightarrow 圓心坐標\bbox[red,2pt]{(0,{5\over 2})}$$

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

解題僅供參考，其他指考試題及詳解