109年大學指考數學乙詳解

109學年度指定科目考試試題

數學乙

第壹部分：選擇題
一、單選題

解：

$$A=\left[ \matrix{-1 & 0 \\ 1 & -1}\right] \Rightarrow A^2=\left[ \matrix{-1 & 0 \\ 1 & -1}\right] \left[ \matrix{-1 & 0 \\ 1 & -1}\right] =\left[ \matrix{1 & 0 \\ -2 & 1}\right] \\ \Rightarrow A^4 = \left[ \matrix{1 & 0 \\ -2 & 1}\right] \left[ \matrix{1 & 0 \\ -2 & 1}\right] =\left[ \matrix{1 & 0 \\ -4 & 1}\right] \\ \Rightarrow A^5= \left[ \matrix{1 & 0 \\ -4 & 1}\right]  \left[ \matrix{-1 & 0 \\ 1 & -1}\right] =\left[ \matrix{-1 & 0 \\ 5 & -1}\right]，故選\bbox[red, 2pt]{(5)}$$

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解：

$$\begin{array}{}  A(3) & B(3) & C(2) & 數量\\\hline 甲乙+1 & 3 & 2 & C^6_1C^5_3C^2_2 \\ 3 & 甲乙+1 & 2 & C^6_3C^3_1C^2_2 \\ 3 & 3& 甲乙& C^6_3C^3_3 \\\hline \end{array}\\ \Rightarrow 共有 C^6_1C^5_3C^2_2+ C^6_3C^3_1C^2_2+C^6_3C^3_3 = 60+60 +20=140種分法，故選\bbox[red,2pt]{(1)}$$

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解： $$大體而言，X與Y為正相關，X大則Y大，故選\bbox[red,2pt]{(4)}$$

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二、多選題

解：
$$f(x)=0沒有實根\Rightarrow \cases{凹向上,且f(x)>0 ,x\in R\\ 凹向下,且f(x)<0, x\in R} \\(1)\times: 若圖形凹向下，則f(0)<0; \\(2)\bigcirc: \cases{凹向上\Rightarrow \cases{f(1)> 0 \\ f(2) > 0} \Rightarrow f(1)f(2) >0 \\凹向下\Rightarrow \cases{f(1)< 0 \\ f(2) < 0} \Rightarrow f(1)f(2) >0} \Rightarrow f(1)f(2) >0\\ (3)\bigcirc: \cases{f(x)=0無實根\\ f(x)-1=0有實根} \Rightarrow y=f(x)圖形為凹向上 \Rightarrow f(x)-2=0有實根\\ (4) \bigcirc: f(x)-1=0有重根 \Rightarrow 圖形凹向上且y=f(x)-1與x軸相切\\\qquad \Rightarrow  y=f(x)-1/2與x軸不相交 \Rightarrow f(x)-1/2=0無實根 \\ (5) \times: 圖形無法確定凹向上或向下，因此不能確定f(x)-1/2=0是否有實根\\，故選\bbox[red,2pt]{(2,3,4)}$$

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解： $$\cases{\cases{a_1=3\\ r_1=1/3} \Rightarrow a_1=3, a_3=1, a_5=1/3, a_7=1/9,\dots,a_{2k-1}={1\over 3^{k-2}},k=1,2,...\\ \cases{a_2=2\\ r_2=1/2} \Rightarrow a_2=2,a_4=1,a_6=1/2,a_8=1/4,\dots,a_{2k}={1\over 2^{k-2}},k=1,2,...}\\(1) \times: \cases{a_4=1\\ a_5=1/3 \\ a_6=1/2 \\ a_7=1/9} \Rightarrow a_4 > a_6 > a_5 > a_7 \\(2) \bigcirc:\cases{a_{10} =a_{2\times 5} ={1\over 2^{5-2}} ={1\over 2^3} ={1\over 8} \\a_{11}= a_{2\times 6-1} ={1\over 3^{6-2}}={1\over 3^4} ={1\over 81}} \Rightarrow {a_{10} \over a_{11}} ={81\over 8} > {80\over 8}=10 \\(3)\bigcirc: 公比小於1且為正值，因此\lim_{ n\to \infty} a_n=1 \\(4) \times: 若n=2k-1 \Rightarrow {a_{n+1} \over a_n} ={ a_{2k}\over a_{2k-1}} = {1/2^{k-2} \over 1/3^{k-2}} ={3^{k-2} \over 2^{k-2}} =({3\over 2})^{k-2} \\ \qquad \lim_{n\to \infty} {a_{n+1}\over a_n} = \lim_{k\to \infty}({3\over 2})^{k-2} =\infty \ne 0 \\(5)\times: a_1+\cdots+a_{100} =(a_1+a_3+\cdots +a_{99}) +(a_2+a_4+\cdots +a_{100}) \\ \qquad =\left(3+1+\cdots +1/3^{48} \right) +\left(2+1+\cdots +1/2^{48} \right) \\ \qquad ={3-1/3^{49}\over 1-1/3} +{2-1/2^{49}\over 1-1/2} ={1\over 2}\left( 3^2-{1\over 3^{48}}\right)+2^2 -{1\over 2^{48}} ={17\over 2}-{1\over 2\cdot 3^{48}} -{1\over 2^{48}} \not \gt 9\\故選\bbox[red,2pt]{(2,3)}$$

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解：
$$(1)\times: 棋子與原點距離為2\Rightarrow 點數為3或4 \Rightarrow 機率為2/6=1/3 \\(2) \bigcirc: {1\over 6} (0+1-2+2-4+3)=0 \\(3) \times: 沒有辦法在\{0,1,-2,2,-4,3\}任挑2數(可重複挑選)，其和為-5 \\(4)\bigcirc:點數和為奇數的機率為1/2，其中(1,2),(1,4),(1,6),(2,1), (3,6),(4,1),(6,1),(6,3)共有8種，機率為8/36；\\ \qquad 因此該條件機率為{8/36 \over 1/2} =4/9 \\(5)\times: (1,1,1), (1,3,4)\xrightarrow{排列數}6,(2,2,3)\xrightarrow{排列數}3,(2,5,6)\xrightarrow{排列數}6, (4,4,5)\xrightarrow{排列數}3; \\ \qquad 棋子在原點機率為{1+6+3+6+3 \over 6^3} \ne {1\over 6^3}\\ 故選\bbox[red,2pt]{(2,4)}$$

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 三、選填題

解：
$$\cases{A(1,6)\\ B(4,5) \\ C(6,2) \\ f(x,y)=7x+2y} \Rightarrow \cases{f(A)\le k\\ f(B)\le k \\f(C)\le k} \Rightarrow \cases{19\le k\\ 38\le k \\46\le k} \Rightarrow k至少為\bbox[red,2pt]{46}$$

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解：

$$\cases{P(X=1)={1\over 1}s\\ P(X=2)={1\over 2}s\\ P(X=3)={1\over 3}s\\ P(X=4)={1\over 4}s},s為一常數 \Rightarrow \sum_{k=1}^4P(X=k)=1 \Rightarrow (1+{1\over 2} +{1\over 3} +{1\over 4})s=1\\  \Rightarrow {25\over 12}s=1 \Rightarrow  s={12\over 25} \Rightarrow P(X=3)={1\over 3}s = {1\over 3}\cdot {12\over 25} =\bbox[red, 2pt]{4\over 25}$$

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解：

$$\cases{經理薪資a \\ 秘書薪資 b\\ 業務薪資c \\s=a+b+c} \Rightarrow \cases{0.1b=0.03s\\ 0.2c=0.04s \\ 0.15a= p\cdot s} \Rightarrow \cases{b=0.3s \\ c=0.2s \\ a={20\over 3}p\cdot s } \\ \Rightarrow s=(0.3+0.2+{20\over 3}p)s \Rightarrow {20\over 3}p=0.5 \Rightarrow {3\over 40} = \bbox[red, 2pt]{7.5}\%$$

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解：

$$直線y=2x+4與x軸交於R(-2,0)，並令P(b,2b+4)，即\cases{P(b,2b+4) \\ Q(a,2a+4) \\ R(-2,0)} \\ \Rightarrow \cases{\overrightarrow{AQ} =(a,2a+4) \\ \overrightarrow{BP} =(b-1,2b+4)}；又\overrightarrow{AQ} \parallel \overrightarrow{BP} \Rightarrow \overrightarrow{AQ} =k \overrightarrow{BP} \Rightarrow {a\over b-1} ={2a+4 \over 2b+4} \Rightarrow b=(3a+2)/2 \\ \Rightarrow P({3a+2\over 2},3a+6) \Rightarrow \triangle RPB面積={1\over 2} \times \overline{RB}\times h ={1\over 2} \times 3\times (3a+6) ={9\over 2}(a+2)\\ 由於{\triangle RQA \over \triangle RPB} ={\overline{RA}^2 \over \overline{RB}^2} ={4\over 9} \Rightarrow \triangle RQA={4\over 9}\triangle RPB \Rightarrow 梯形ABPQ= (1-{4\over 9})\triangle RPB ={5\over 9}\triangle RPB \\ ={5\over 9}\times {9\over 2}(a+2) ={5\over 2}(a+2) = \bbox[red, 2pt]{ {5\over 2}a+5}$$

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第貳部份：非選擇題

解：

(1) $$\cases{A={\log P_5 -\log P_2 \over 3}= {1\over 3}(\log P_0(1+r)^5 -\log P_0(1+r)^2 ) ={1\over 3}\log {P_0(1+r)^5 \over P_0(1+r)^2} =\log (1+r)\\ B= {\log P_8 -\log P_6 \over 2} ={1\over 2}(\log P_0(1+r)^8 -\log P_0(1+r)^6) = {1\over 2} \log {P_0(1+r)^8 \over P_0(1+r)^6} =\log (1+r)} \\ \Rightarrow A=B$$ (2) $${P_{16}\over P_0}=10 \Rightarrow {P_0(1+r)^{16} \over P_0 } =10  \Rightarrow (1+r)^{16}=10 \Rightarrow (1+r)^8 =\sqrt{10} \\ \Rightarrow {P_{20} \over P_{17}} \times {P_{8} \over P_{6}} \times {P_{5} \over P_{2}} ={P_0(1+r)^{20} \over P_0(1+r)^{17} }\times {P_0(1+r)^{8} \over P_0(1+r)^{6} }\times {P_0(1+r)^{5} \over P_0(1+r)^{2} } =(1+r)^3 \times (1+r)^2 \times (1+r)^3 \\= (1+r)^8 = \bbox[red, 2pt]{\sqrt{10}}$$ (3) $$由(2)知: (1+r)^{16}=10 \Rightarrow 16\log(1+r) =1 \Rightarrow \log (1+r)={1\over 16} \\ \Rightarrow {\log P_{20}-\log P_{17} \over 3} ={1\over 3}\log {P_{20} \over P_{17}}  ={1\over 3}\log {P_{0}(1+r)^{20} \over P_{0}(1+r)^{17}} ={1\over 3}\log (1+r)^3 = \log (1+r) = \bbox[red, 2pt]{1\over 16}$$

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解：

(1) $$\overline{AB} = \text{dist}(L_1,L_2)=5 \Rightarrow \overline{AB} \bot L_1 \Rightarrow \overline{AB}斜率=\bbox[red,2pt]{-{1\over 2}}$$ (2) $$\cases{L_1: y=2x+k_1\\ A(2,-1)在L_1上} \Rightarrow -1=4+k_1 \Rightarrow k_1=-5 \Rightarrow L_1:y=2x-5 \Rightarrow 方向向量\vec u=(1,2)\\  \cases{A(2,-1) \\ B(a,b)} \Rightarrow \overrightarrow{AB} \bot \vec u \Rightarrow (2-a,-1-b)\cdot (1,2)=0 \Rightarrow 2-a-2-2b=0 \\ \Rightarrow a=-2b \Rightarrow B(-2b,b) \Rightarrow \overline{AB}=5 \Rightarrow (2+2b)^2+(b+1)^2=5^2 \Rightarrow b= -1\pm \sqrt 5 \\ 由於B在第二象限，因此 b=-1+\sqrt 5 \Rightarrow B(2-2\sqrt 5,-1+\sqrt 5) \Rightarrow \overrightarrow{AB} =\bbox[red, 2pt]{(-2\sqrt 5,\sqrt 5)}$$ (3) $$\cases{L_2: y=2x+k_2\\ B(2-2\sqrt 5,-1+\sqrt 5) 在L_2上} \Rightarrow -1+\sqrt 5=4-4\sqrt 5+k_2 \Rightarrow k_2=-5+ 5\sqrt 5 \Rightarrow L_2: y=2x-5+5\sqrt 5;\\同理 \cases{L_3: y=3x+k_3\\ A(2,-1)在L_3上} \Rightarrow -1=6+k_3 \Rightarrow k_3=-7 \Rightarrow L_3:y=3x-7 \\ 求 L_3與L_2交點: 2x-5+5\sqrt 5=3x-7 \Rightarrow x=2+5\sqrt 5 \Rightarrow y=6+15\sqrt 5-7=-1+15\sqrt 5\\ \Rightarrow C(2+5\sqrt 5,-1+15\sqrt 5) \Rightarrow  \overrightarrow{AB} \cdot \overrightarrow{AC} =(-2\sqrt 5,\sqrt 5) \cdot (5\sqrt 5,15\sqrt 5) = -50+75=\bbox[red, 2pt]{25}$$ (4) $$\cases{A(2,-1)\\ C(2+5\sqrt 5,-1+15\sqrt 5)} \Rightarrow \overrightarrow{AC} = \bbox[red, 2pt]{(5\sqrt 5,15\sqrt 5)}$$

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