朱式幸福: 115年桃園陽明高中教甄-數學詳解

桃園市立陽明高中數學科教甄試題

一、填充題(每格 5 分，共 80 分)

解答:$$2^{x^2+4y+2}+ 2^{4y^2+2z+2} +2^{z^2+2x+2} \ge 3\sqrt[3]{2^{x^2+2x +4y^2+4y+z^2+ 2z+6}} =3 \sqrt[3]{2^{(x+1)^2+4(y +1/2)^2+(z+1)^2+3}}\\ 當\cases{x=-1\\y=-1/2\\ z=-1} 時，最小值3\sqrt[3]{2^{3}} =\bbox[red, 2pt]6$$

解答:$$2x+y+k=0 \Rightarrow y=-2x-k代入y=x^2+kx+1 \Rightarrow x^2+(k+2)x+(k+1)=0 \\ \cases{有兩相異實根\Rightarrow \Delta =(k+2)^2-4(k+1)\gt 0 \Rightarrow k^2\gt 0 \Rightarrow k\ne 0 \\ 兩根之積\lt 0 \Rightarrow k+1\lt 0\Rightarrow k\lt -1} \Rightarrow \bbox[red, 2pt]{k\lt -1}$$

解答:$$假設新的戰鬥力z=x+y+xy \Rightarrow z+1=(x+1)(y+1) \\ \Rightarrow k+1=(3+1)(7+1)(15+1)(31+1)(63+1)(127+1) (255+1) =2^{35} \\ \Rightarrow \log_2(k+1)= \bbox[red, 2pt]{35}$$

解答:$$假設\cases{事件A:xy為偶數\\ 事件B:x+y+z為奇數} \Rightarrow P(A)=1-P(x=奇)\times P(y=奇) =1-{2\over 3}\times {3\over 4}={1\over 2}\\ \Rightarrow A\cap B:xy為偶數且x+y+z為奇數有以下情況:\\ \Rightarrow \cases{x=偶且y=偶 \Rightarrow z=奇 \Rightarrow 機率={1\over 3} \times {1\over 4}\times {3\over 5}={3\over 60} \\ x=偶且y=奇 \Rightarrow z=偶 \Rightarrow 機率={1\over 3}\times {3\over 4} \times {2\over 5}={6\over 60} \\ x=奇且y=偶\Rightarrow z=偶\Rightarrow 機率= {2\over 3} \times {1\over 4}\times {2\over 5} ={4\over 60}} \Rightarrow P(A\cap B)={3+6+4\over 60} ={13\over 60} \\ \Rightarrow P(B\mid A)= {P(A\cap B) \over P(A)}={13/60\over 1/2}= \bbox[red, 2pt]{13\over 30}$$

解答:$$假設\cases{A(0,10) \\ B(0,0) \\C(10,0) \\D(10,10)} \Rightarrow \cases{\overline{AD}落在y=10上\\ \overline{BC} 落在y=0上\\ \overline{CD} 落在x=10上}\\ A\xrightarrow{翻摺後} A'在\overline{CD}上 \Rightarrow A'(10,h), 0\le h\le 10 \Rightarrow \overline{PQ}是\overline{AA'}的中垂線 \\\Rightarrow \overleftrightarrow{PQ}: y-{10+h\over 2}={10\over 10-h}(x-5) \\ 又P在\overline{AD}上,將y=10代入\overleftrightarrow{PQ} \Rightarrow P的x坐標x_P=5+{(10-h)^2\over 20} \\Q在\overline{BC}上, 將y=0 代入\overleftrightarrow{PQ} \Rightarrow Q的x坐標x_Q= {h^2\over 20} \\ 四邊形 ABQP 是一個直角梯形,面積={1\over 2}(\overline{AP}+ \overline{BQ})\times \overline{AB} ={1\over 2} \left( 5+{(10-h)^2\over 20}+ {h^2\over 20} \right)\times 10 \\={1\over 2}(h^2-10h+100)={1\over 2}((h-5)^2+ 75) \Rightarrow h=5時,面積有最小值\\ 此時, \overline{AP}=5+{(10-5)^2\over 20} = \bbox[red, 2pt]{25\over 4}$$

解答:$$(A)\times: 最大的平均數=(2+6+6 +6+6+6+6+6)/8=5.5 \gt 4 \\(B)\times: 2,3,3,3,5,6,6,6符合要求，但平均數={34\over 8}=4.25 \gt 4\\ (C)\bigcirc: 2至少出現3次，總和最大的情況:2,2,2,4,5,5,6,6 \Rightarrow 平均數=4\\ \qquad 最大極限就是4 ,因此算術平均數必定\le 4 \\(D)\bigcirc: 變異數最大的情況:2,2,2,2,6,6,6,6 \Rightarrow 變異數4 \Rightarrow 最大標準差2\\ (E)\bigcirc: 假設8位學生吃的粽子數分別為x_1,x_2, \dots, x_8 \Rightarrow \cases{2\le x_i\le 6\\ \sum x_i=23} \\\qquad y_i=x_i-2 \Rightarrow 0\le y_i\le 4 \Rightarrow \sum y_i=7\Rightarrow 有學生吃6顆,即y_i=4\\ \qquad 8人中選1人給他4，有C^8_1= 8種; 剩下3顆分給7個人,有H^7_3=84種\Rightarrow 總共8\times 84=672\\ 故選\bbox[red, 2pt]{(CDE)}$$

解答:$$假設\cases{A(-1,2,1) \\B(-4,1,3) \\ C(2,0,-3)} \Rightarrow \cases{\overrightarrow{AB }= (-3,-1,2) \\ \overrightarrow{AC}=(3,-2,-4)} \Rightarrow \vec n=\overrightarrow{AB} \times \overrightarrow{AC}=(8,-6,9) \Rightarrow 底面積 =|\vec n|= \sqrt{181}\\ 底面方程式:8(x+1)-6(y-2)+9(z-1)=0 \Rightarrow E:8x-6y+9z+11=0\\ 另一面之一頂點P在yz平面上\Rightarrow P(0,y,z)，且與原點距離\sqrt{13} \Rightarrow y^2+z^2=13 \\ \Rightarrow 高h=d(P,E)={|-6y+9z+11| \over \sqrt{181}} \Rightarrow 體積V=\sqrt{181} \cdot h= |-6y+9z+11|\\ 柯西不等式:(y^2+z^2)((-6)^2+9^2) \ge (-6y+9z)^2 \Rightarrow (-6y+9z)^2\le 1521=39^2 \\ \Rightarrow -39\le -6y+9z\le 39 \Rightarrow \cases{-6y+9z=39 \Rightarrow V=50\\ -6y+9z=-39 \Rightarrow V=28} \Rightarrow 最大體積為\bbox[red, 2pt]{50}$$

解答:$$取複數z_n= a_n+b_ni \Rightarrow z_{n+1} =a_{n+1}+ b_{n+1}i= (\sqrt 3a_n-b_n)+(a_n+ \sqrt 3b_{n})i= (\sqrt 3+i)(a_n+b_ni) \\ \Rightarrow z_{n+1}=(\sqrt 3+i)z_n =2(\cos{\pi\over 6}+i\sin{\pi\over 6})z_n \\ \Rightarrow z_{115}=z_4\times (\sqrt 3+i)^{111} =z_4 \times 2^{111} \left( \cos{111\pi\over 6} +i\sin {111\pi\over 6} \right) =z_4 \times 2^{111} \left( \cos{\pi\over 2} +i\sin { \pi\over 2} \right) \\ \Rightarrow z_{115}=1+5i=z_4\times 2^{111}i=(a_4+ b_4i)\times 2^{111} i\Rightarrow =-2^{111}b_4+2^{111}a_4i \Rightarrow \cases{b_4=-1/2^{111} \\a_4=5/2^{111}} \\ \Rightarrow a_4+b_4={5-1\over 2^{111}}= \bbox[red, 2pt]{1\over 2^{109}}$$

解答:$$為了讓數字容易處理，一開始先借給豬兄弟3顆虛擬蘋果，總數變為x+3\\ 新數量={3\over 4}(原數量-1) \Rightarrow 新數量 +3 ={3\over 4}(原數量-1) +3 ={3\over 4}(原數量+3) \\ \Rightarrow x_3+3=(x+3)\times \left( {3\over 4} \right)^3={27\over 64}(x+3) 為一整數且x_3是4的倍數,即\cases{x_3=4k\\ x+3=64m} \\\Rightarrow 4k+3= 27m \Rightarrow 3\equiv 27m {\mod 4} \equiv 3m \mod 4 \Rightarrow m=1,5,9,\dots \\ \cases{m=1 \Rightarrow x+3=64m \Rightarrow x=61不合，小於三百多顆 \\m=5 \Rightarrow x+3=320 \Rightarrow x= \bbox[red, 2pt]{317}}$$

解答:$$A = \begin{bmatrix}4/5& -3/5\\ 3/5& 4/5 \end{bmatrix} = \begin{bmatrix}\cos \theta&-\sin \theta\\ \sin \theta& \cos \theta \end{bmatrix} 為一旋轉矩陣 \Rightarrow \overline{OP_1} =\overline{OP2} =\overline{OP_3} =r \\ S=\triangle P_1P_2 P_3= \triangle OP_1P_2 +\triangle OP_2P_3-\triangle OP_1P_3 ={1\over 2}r^2\sin \theta +{1\over 2}r^2\sin \theta -{1\over 2}r^2\sin 2\theta \\=r^2\sin \theta -r^2\sin \theta \cos \theta=r^2 \left( {3\over 5}-{3\over 5} \cdot {4\over 5}\right) ={3\over 25}r^2 \\ P_1(x,y)在y={1\over 2}x^2-5上 \Rightarrow x^2=2y+10 \Rightarrow r^2=x^2+y^2= 2y+10+y^2=(y+1)^2+9 \\ \Rightarrow r^2的最小值9\Rightarrow S的最小值={3\over 25}\cdot 9= \bbox[red, 2pt]{27\over 25}$$

解答:$$x^2+x-1=0 \Rightarrow x^2=1-x \Rightarrow x^4=1-2x+x^2=1-2x+1-x=2-3x \\ \Rightarrow x^8= 13-21x \Rightarrow x^{16}=610-987x \Rightarrow x^{17}=610x-987x^2 =-987+1597x \\ \Rightarrow ax^{17}+bx^{16}+1= (1597a-987b)x+610b-987a+1=0\\ \Rightarrow \cases{1597a=987b\\ 610b-987a+1=0} \Rightarrow \left( {1597\over 987} \times 610-987 \right)a+1=0 \Rightarrow a= \bbox[red, 2pt]{-987}$$

解答:$$S_n=2a_n-1 \Rightarrow S_1=a_1=2a_1-1\Rightarrow a_1=1 \Rightarrow a_n=S_n-S_{n-1 }=(2a_n-1)-(2a_{n-1}-1) \\ \Rightarrow a_n=2a_{n-1} \Rightarrow \langle a_n \rangle為一等比數列，首項a_1=1,公比r=2 \Rightarrow a_n=2^{n-1} \\ b_{n+1} =a_n+b_n \Rightarrow b_{n+1}-b_n= a_n=2^{n-1} \Rightarrow \cases{b_n-b_{n-1}= 2^{n-2} \\b_{n-1}-b_{n-2}= 2^{n-3} \\ \cdots \\b_2-b_{1}= 2^{0} } \\ \Rightarrow b_n-b_1=2^0+2^1+2^2+ \cdots+2^{n-2} =2^{n-1}-1 \Rightarrow b_n=2^{n-1}-1+3=2^{n-1}+2 \\ \Rightarrow N=\sum_{k=1}^{30}b_k =\sum_{k=1}^{30} \left( 2^{k-1}+2 \right) =2^{30}+59 \\由於2^{10}= 1024 \equiv 24 {\mod 100} \Rightarrow 2^{30} \equiv 24^3 {\mod 100} \equiv 24 {\mod 100} \\ \Rightarrow N \equiv {\mod 100} \equiv (24+59) = \bbox[red, 2pt]{83} \equiv {\mod 100}$$

解答:

$${\sqrt{x^4+x^2+2x+1} + \sqrt{x^4-2x^3+5x^2-4x+1} \over x} \\=\sqrt{x^2+1+2/x+1/x^2} + \sqrt{x^2-2x+5-4/x+1/x^2}\\ = \sqrt{x^2+ \left( 1+{1\over x} \right)^2} + \sqrt{(x-1)^2+ \left( 2-{1\over x} \right)^2}\\= \sqrt{(x-0)^2+ \left( {1\over x}-(-1) \right)^2} + \sqrt{(x-1)^2+ \left( {1\over x} -2\right)^2} =\overline{PA}+ \overline{PB},其中\cases{P(x,1/x) \in xy=1\\ A(0,-1)\\B(1,2)} \\ \Rightarrow 最小值=\overline{AB}=\sqrt{1+9}= \bbox[red, 2pt]{\sqrt{10}}$$

解答:

$$假設\cases{\overline{PA}=x\\ \overline{PB}=y} \Rightarrow \angle APB=180^\circ-\angle ACB=120^\circ \Rightarrow \overline{AB}^2=x^2+y^2-2xy\cos 120^\circ \\ \Rightarrow x^2+y^2+xy=25 \Rightarrow y^2+xy+(x^2-25)=0 \Rightarrow y={1\over 2}(-x+\sqrt{100-3x^2})\\ 托勒密定理: \overline{PC} \cdot \overline{AB} =\overline{PA}\cdot \overline{BC}+ \overline{PB}\cdot \overline{AC} \Rightarrow \overline{PC}=x+y \\ S=\triangle APB+\triangle APC= {1\over 2}xy \sin 120^\circ +{1\over 2}x(x+y)\sin 60^\circ ={\sqrt 3\over 4}(x^2+2xy) \\={\sqrt 3\over 4}(x^2+2x\cdot {1\over 2}(-x+\sqrt{100-3x^2})) ={\sqrt 3\over 4}x\sqrt{100-3x^2} \\ \Rightarrow x^2={50\over 3}時，S有最大值\bbox[red, 2pt]{25\over 2}$$

解答:$$\cases{A\in L_1:\displaystyle {x\over 2}={y+1\over 3}=z+3 \\ B\in L_2: \displaystyle {x+1\over 4}={y-4\over -2}={z+2\over -1} \\P(1,2,-1)} \Rightarrow \cases{A=(2t,3t-1,t-3) \\B=(4s-1,-2s+4,-s-2)} \\ \Rightarrow \cases{ \overrightarrow{PA} =(2t-1,3t-3,t-2) \\ \overrightarrow{PB} =(4s-2,-2s+2,-s-1)} \Rightarrow \overrightarrow{PA} \parallel \overrightarrow{PB} \Rightarrow s=t=1 \Rightarrow B= \bbox[red, 2pt]{(3,2,-3)}$$

解答:

$$假設點O滿足\angle AOB= \angle BOC= \angle COA=120^\circ, 並令\cases{x =\overline{OA} \\ y=\overline{OB} \\ z=\overline{OC}} \\ \Rightarrow \cases{\overline{AB}^2= x^2+y^2-2xy\cos 120^\circ =x^2+y^2+xy=2 \\ \overline{BC}^2=y^2+z^2-2yz\cos 120^\circ=y^2+z^2+yz=3\\ \overline{CA}^2=x^2+z^2-2xz\cos 120^\circ=x^2+z^2+xz=5} \Rightarrow \cases{\overline{AB}=\sqrt 2\\ \overline{BC}=\sqrt 3\\ \overline{CA}=\sqrt 5} \\ \Rightarrow \overline{CA}^2= \overline{AB}^2+ \overline{BC}^2 \Rightarrow \angle ABC=90^\circ \Rightarrow \triangle ABC面積=\cases{\overline{AB}\cdot \overline{BC}/2 \\ \triangle OAB+ \triangle OBC+ \triangle OCA} \\ \Rightarrow {1\over 2}\cdot \sqrt 2\cdot \sqrt 3= {1\over 2}(xy+ yz+zx)\sin 120^\circ={\sqrt 3\over 4}(xy+yz+ zx) \Rightarrow xy+yz+ zx=2\sqrt 2 \\ 原式\cases{x^2+y^2 +xy=2\\ y^2+z^2 +yz=3\\ z^2+x^2+xz=5} \Rightarrow 三式相加\Rightarrow 2(x^2+y^2+z^2)+xy+yz+zx=10 \\ \Rightarrow x^2+y^2+z^2=(10-2\sqrt 2)/2=5-\sqrt 2 \Rightarrow (x+y+z)^2=5-\sqrt 2+2\cdot 2\sqrt 2= 5+3\sqrt 2 \\ \Rightarrow x+y+z= \bbox[red, 2pt]{\sqrt{5+3\sqrt 2}}$$