110年彰化高中科學班甄選-數學詳解

國立彰化高級中學 110 學年度科學班甄選【數學科】試題

填充題【1~13 題，每格 6 分，14~15 題，每題8分。答案請化簡，並依序填入答案欄內】

解答：$$\color{blue}{x^5}+x-1 = \color{blue}{x^2(x^3+1)-x^2}+x-1 =x^2(x+1)(x^2-x+1)-(x^2-x+1)\\ =(x^2(x+1)-1)(x^2-x+1) = \bbox[red, 2pt]{(x^3 + x^2-1)(x^2-x+1)}$$

解答：$$\cases{首項a_1=16\\ 公差d} \Rightarrow a_{16}+a_{17} +a_{18} =a_1+15d+ a_1+16d+ a_1+17d= 3a_1+ 48d= 48+48d=0\\ \Rightarrow d=-1 \Rightarrow a_{110}= a_1+109d = 16-109= \bbox[red,2pt]{-93}$$

解答：

$$\cases{\cases{\overline{AE}=5\\ \overline{AD}=4} \Rightarrow \overline{DE}=3\\ \overline{AB}為直徑 \Rightarrow \angle C=90^\circ} \Rightarrow \triangle ADE\sim \triangle ACB (AAA) \Rightarrow \overline{AC}: \overline{AB}= \overline{AD}:\overline{AE}\\ \Rightarrow \overline{AC}:(4+11)= 4:5 \Rightarrow \overline{AC}=4\times 3=12 \Rightarrow \overline{CE} = 12-5=\bbox[red,2pt]{7}$$

解答：$$\cfrac{2021}{110} = 18+\cfrac{41}{110} = 18+\cfrac{1}{\cfrac{110}{41}} = 18+\cfrac{1}{2+\cfrac{28}{41}} = 18+\cfrac{1}{2+\cfrac{1}{\cfrac{41}{28}}} = 18+\cfrac{1}{2+\cfrac{1}{1+\cfrac{13}{28}}} \\ = 18+\cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{\cfrac{28}{13}}}}  = 18+\cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{2}{13}}}}  = 18+\cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{\cfrac{13}{2}}}}} \\  = 18+\cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{6+\cfrac{1}{2}}}}} \Rightarrow \cases{x=2 \\y=2 \\ z=6} \Rightarrow x+y+z= \bbox[red,2pt]{10}$$

解答：$$9x^2 +10xy+9y^2 =2021 = 43\times 47 \Rightarrow (3x+y)(x+3y) =43\times 47 \\ \Rightarrow \cases{3x+y= 43\\ x+3y=47} \xrightarrow {兩式相加} 4x+4y=90 \Rightarrow 2x+2y= 45，無正整數解 (兩個偶數相加\ne 奇數)\\  \Rightarrow 有\bbox[red, 2pt]{0}組正整數解$$

解答：$$符合要求的二個數:(1,4),(1,9),(2,8), (4,9)，共4組，因此機率為{4\over C^9_2} ={4\over 36} =\bbox[red, 2pt]{1\over 9}$$

解答：$$先求兩圖形\cases{L:y=-x\\ \Gamma: y=-3x^2-6x}的交點,即-3x^2-6x=-x \Rightarrow 3x^2+5x=0 \Rightarrow x=0,-5/3\\ A(a,b)以直線L為對稱軸的對稱點為C(-b,-a)，又A在\Gamma上\Rightarrow A=(a,-3a^2-6a)\\ \Rightarrow B=(3a^2+6a, -a) 也在\Gamma上 \Rightarrow -a=-3((3a^2+6a)^2 -6(3a^2+6a) \\\Rightarrow 27a^4+108a^3+126a^2+35a=0，兩圖形交點x=0,-5/3一定是其中的解\\ 即27a^4+108a^3+126a^2+35a =a(3a+5)( 9a^2+21a+7)=0 \Rightarrow a=(-7\pm \sqrt{21})/6\\ \Rightarrow   b=-(3a^2+6a) =a+{7\over 3}= (7\pm\sqrt{21})/6 \Rightarrow a-b= \bbox[red,2pt]{-7/3}\\註: 9a^2+21a+7=0 \Rightarrow 3(3a^2+7a)=-7 \Rightarrow b=-(3a^2+6a) =a+{7\over 3}$$

解答：$$20a21b 為99的倍數 \Rightarrow 20+a2+1b 為99的倍數\Rightarrow 20+a2+1b =99 \Rightarrow a2+1b=79\\ \Rightarrow \cases{a=7-1=6 \\ b=9-2=7} \Rightarrow (a,b)=\bbox[red, 2pt]{(6,7)}$$

解答：

(1)$$x^2-8x+9=0 \Rightarrow \alpha= 4+\sqrt 7 \Rightarrow \sqrt \alpha = \sqrt{4+\sqrt 7}\\ \Rightarrow \sqrt \alpha 的小數部分=\sqrt{4+\sqrt 7}-2 ={ \sqrt{8+2\sqrt 7}-2\sqrt 2\over \sqrt 2}={ \sqrt 7+1-2\sqrt 2\over \sqrt 2} = \bbox[red, 2pt]{\sqrt{14}+\sqrt 2-4\over 2}$$(2)$$\alpha^2-8\alpha+9= 0 \Rightarrow \cases{\alpha^2=8\alpha-9\\ \alpha^3=8\alpha^2-9\alpha} \Rightarrow \alpha^3-9\alpha^2+8\alpha +17 = (8\alpha^2-9\alpha)-9\alpha^2+8\alpha +17 \\ =-\alpha^2-\alpha+17 = -(8\alpha-9)-\alpha+17 = -9\alpha+26 =-9(4+\sqrt 7)+26 = \bbox[red, 2pt]{-10-9\sqrt 7}$$

解答：

$$任一捷徑將矩形分割成上下兩個連續的區塊，而且區塊個數是遞增的；\\假設區塊個數由左而右分別是a_1,a_2,\dots,a_8，其中\cases{0\le a_k\le 3,1\le k\le 8\\[1ex] a_k\le a_{k+1},1\le k\le 7\\[1ex] \sum_{k=1}^8 a_k=12 }\\ \Rightarrow (a_8,a_7,\dots,a_1)=\cases{ (3,3,3,3,0,0,0,0)\\ (3,3,3,2,1,0,0,0),(3,3,3,1,1,1,0,0)\\ (3,3,2,2,2,0,0,0),(3,3,2,2,1,1,0 ,0) ,(3,3,2,1,1,1,1,0) ,(3,3,1,1,1,1,1,1)\\ (3,2,2,2,2,1,0,0), (3,2,2,2,1,1,1,0), (3,2,2,1,1,1,1,1) \\ (2,2,2,2,2,2,0,0), (2,2,2,2,2,1,1,0) ,(2,2,2,2,1,1,1,1) }\\ \Rightarrow 共\bbox[red, 2pt]{13}種走法$$

解答：$$\cases{1位數:5\\ 2位數:40,31,22,13\\ 3位數:300,201,210, 120,102,111\\ 4位數:2000,1001,1010,1100\\ 5位數:10000 } \Rightarrow 以上合計:\bbox[red,2pt]{16266}$$

解答：$$x^2-y^2 = 2x+10y+24 \Rightarrow x^2-2x+1= y^2+10y+25 \Rightarrow (x-1)^2=(y+5)^2\\ \Rightarrow x-1=y+5 \Rightarrow x=y+6 \Rightarrow (x,y) =(17,11), (19,13),...,(89,83)\\ \Rightarrow \cases{m= 17+11=28\\ M= 89+83=172} \Rightarrow (M,m)= \bbox[red, 2pt]{(172,28)}$$

解答：

$$\cases{\triangle OAB=15/2\\ \triangle OBC= 16/2} \Rightarrow \triangle OBC\gt \triangle OAB \Rightarrow L交y軸於D(0,a)，見上圖；\\ \cases{\triangle BDC=2(4-a)\\ OABD面積= 2(a+3)+ 3/2} \Rightarrow 8-2a=2a+6+{3\over 2} \Rightarrow a={1\over 8} \Rightarrow D(0,1/8)\\ \Rightarrow L=\overleftrightarrow{BD}:y={23\over 32}x+ {1\over 8} \Rightarrow -16y+{23\over 2}x +2=0 \Rightarrow \cases{a=23/2\\ b=-16} \Rightarrow a+b = \bbox[red, 2pt]{-{9\over 2}}$$

解答：$$2+3+8+9 = 22 \Rightarrow 最小的立方數為27 \Rightarrow 此數列為2,3,8,9,a,b 循環，其中a+b=5\\ 2021=336\times 6+5 \Rightarrow 前2021項之和=\cases{336\times 27+ (2+3+8+9+a) =9094+a\\ 336\times 27+ (2+3+8+9+b) =9094+b} \\ \Rightarrow 最小值為9094+1(取a=1,b=4) = \bbox[red, 2pt]{9095}$$

解答：

$$假設\cases{\overleftrightarrow{AD}為x軸\\ \overleftrightarrow{BE}為y軸} \Rightarrow \cases{F(0,0)\\ A(-4,0)\\ D(4,0)}，並取B(0,a) \Rightarrow E(0,a-8)； 又D為\overline{BC}的中點\Rightarrow C(8,-a)；\\ 因此\overleftrightarrow{AE}: y={a-8\over 4}x+a-8 且經過C(8,-a) \Rightarrow -a= 2a-16+a-8 \Rightarrow a=6 \\ \Rightarrow \cases{B(0,6)\\ C(8,-6)} \Rightarrow \cases{\overline{AB}= \sqrt{4^2+6^2} =2\sqrt{13} \\\overline{BC} =\sqrt{8^2+ 12^2} =4\sqrt{13} \\ \overline{AC} =\sqrt{12^2 +6^2} =6\sqrt 5} \Rightarrow \triangle ABC周長= \bbox[red, 2pt]{6\sqrt{13} +6\sqrt 5}$$

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