115年花蓮女中教甄-數學詳解

 國立花蓮女子高級中學 115 學年度第 1 次正式教師甄選

一、填充題（每題 5 分）

解答:$$擲骰子三次，所有的可能情形總數為 6^3 = 216 種\\ \textbf{Case I }最大點數為1:代表三次擲出的點數都必須是1 \Rightarrow 只有1種情況\\ \textbf{Case II }最大點數為3: 代表三次擲出的點數都必須小於等於 3，但不能全部都小於等於 2\\ \qquad \Rightarrow 共有3^3-2^3=19種\\ \textbf{Case III }最大點數為5: 代表三次擲出的點數都必須小於等於 5，但不能全部都小於等於 4\\ \qquad \Rightarrow 共有5^3-4^3= 61種 \\ 三種\text{ Case }的合計=1+19+ 61=81 \Rightarrow 機率={81\over 216} = \bbox[red, 2pt]{3\over 8}$$

解答:$$假設\cases{\vec u= \begin{bmatrix}1\\-1 \end{bmatrix} \\ \vec v= \begin{bmatrix}1\\1 \end{bmatrix}} \Rightarrow \cases{A\vec u=\vec v\\ A^2 \vec u=-\vec u \Rightarrow A\vec v=-\vec u} \Rightarrow A [\vec u \; \vec v] = [\vec v\; -\vec u] \\ \Rightarrow A= [\vec v\; -\vec u] [\vec u\; \vec v]^{-1}  \Rightarrow A^{-1} = [\vec u\; \vec v] [\vec v \; -\vec u]^{-1} = \begin{bmatrix}1& 1\\-1& 1 \end{bmatrix} \begin{bmatrix}1& -1\\1& 1 \end{bmatrix}^{-1} = \begin{bmatrix}1& 1\\-1& 1 \end{bmatrix} \begin{bmatrix}1/2& 1/2\\-1/2& 1/2 \end{bmatrix} \\= \bbox[red, 2pt]{\begin{bmatrix}0& 1\\-1& 0 \end{bmatrix}}$$

解答:$$假設取出紅球的機率為 p \Rightarrow X\sim B(n,p) \Rightarrow \cases{E(X)=np= 6\\ Var(X)=np(1-p)= 2^2=4} \Rightarrow p={1\over 3} \\\Rightarrow n= \bbox[red, 2pt]{18}$$

解答:$$無論勝負情形，一場比賽雙方得分加總都是1 \Rightarrow 總得分=總比賽場數 \\ 假設高二學生數為a \Rightarrow 參賽總人數a+2 \Rightarrow 總比賽場數=C^{a+2}_2={(a+2)(a+1)\over 2}\\ 假設每位高二學生得分皆為k \Rightarrow 高二總得分ak \Rightarrow {(a+2)(a+1)\over 2} =ak+8 \\ \Rightarrow k={a\over 2}+{3\over 2}-{7\over a} \Rightarrow 2k= \left( a+3-{14\over a} \right) \in \mathbb N \Rightarrow a=1,2,7 (a\lt 10 \Rightarrow a\ne 14)\\ \Rightarrow \cases{a=1 \Rightarrow 總得分=C^3_2=3矛盾, 高一已得8分\\ a=2 \Rightarrow 總得分=C^4_2= 6矛盾, 高一已得8分 \\a=7 \Rightarrow 總得分=C^9_2 = 36} \Rightarrow a=\bbox[red, 2pt]7$$

解答:$$|\overrightarrow{BC}|^2=5^2 =|-\overrightarrow{AB}+ \overrightarrow{AC}|^2 = (\overrightarrow{AC} -\overrightarrow{AB}) \cdot  (\overrightarrow{AC} -\overrightarrow{AB})  =36-2\overrightarrow{AB} \cdot  \overrightarrow{AC}+16 \\ \Rightarrow \overrightarrow{AB} \cdot  \overrightarrow{AC} ={52-25\over 2}={27\over 2} \\P 在 \overline{AB} 的中垂線上 \Rightarrow \overrightarrow{AP} \cdot \overrightarrow{AB} ={1\over 2} ={1\over 2}|\overrightarrow{AB}|^2= 8 \Rightarrow (x\overrightarrow{AB} +y \overrightarrow{AC}) \cdot \overrightarrow{AB}=8 \\ \Rightarrow 16x+{27\over 2}y=8 \Rightarrow 32x+27y=16 \cdots(1) \\ \overrightarrow{BP} \bot \overrightarrow{AC} \Rightarrow \overrightarrow{BP} \cdot \overrightarrow{AC}=0 \Rightarrow (\overrightarrow{AP}-\overrightarrow{AB}) \cdot \overrightarrow{AC} = 0 \Rightarrow \overrightarrow{AP} \cdot \overrightarrow{AC} = \overrightarrow{AB} \cdot \overrightarrow{AC} ={27\over 2} \\ \Rightarrow (x\overrightarrow{AB} +y \overrightarrow{AC}) \cdot \overrightarrow{AC}={27\over 2} \Rightarrow {27\over 2}x+36y={27\over 2} \Rightarrow 3x+8y=3 \cdots(2) \\ 由(1)及(2)可得(x,y) = \bbox[red, 2pt]{ \left( {47\over 175}, {48\over 175} \right)}$$

解答:$$\overrightarrow{PQ} \cdot \overrightarrow{AB} = \overrightarrow{PQ} \cdot (\overrightarrow{AP}+ \overrightarrow{PQ} +\overrightarrow{QB}) =\overrightarrow{PQ} \cdot \overrightarrow{AP}+ |\overrightarrow{PQ}|^2 + \overrightarrow{PQ} \cdot \overrightarrow{QB} =|\overrightarrow{PQ}|^2 \\ \cases{L_1方向向量\vec u_1=(1,1,3)且通過P_1(1,-1,1) \\L_2方向向量\vec u_2=(1,2,2) 且通過P_2(0,2,3)} \Rightarrow \cases{\overrightarrow{P_1P_2} =(-1,3,2) \\ \vec n=\vec u_1\times \vec u_2=(-4,1,1)} \\ \Rightarrow |\overrightarrow{PQ}| ={|\overrightarrow{P_1P_2} \cdot \vec n| \over |\vec n|} ={3\over \sqrt 2} \Rightarrow \overrightarrow{PQ} \cdot \overrightarrow{AB} =|\overrightarrow{PQ}|^2= \bbox[red, 2pt]{9\over 2}$$

解答:$$f(x) = x^3 + x^2 - 4x + k  = (x^2 + ax + 2)(x + c) = x^3 + (a+c)x^2 + (ac+2)x + 2c \\ \Rightarrow \cases{a+c=1\\ ac+2=-4\\ 2c=k} \Rightarrow c=1-a \Rightarrow a(1-a)+2=-4 \Rightarrow a^2-a-6=0 \Rightarrow (a-3)(a+2)=0 \\ \Rightarrow a=3,-2, 但g(x)=0有虛根\Rightarrow \Delta=a^2-8 \lt 0 \Rightarrow -2\sqrt 2\lt a\lt 2\sqrt 2 \Rightarrow a=-2 \\ \Rightarrow c=3 \Rightarrow f(x)=(x^2-2x+2)(x+3) =0 \Rightarrow x= \bbox[red, 2pt]{-3, 1\pm i}$$

解答:$$\lim_{n\to \infty} {n^2+(n+1)^2+ (n+2)^2+\cdots +(3n-1)^2\over n^3} =\lim_{n\to \infty} {1\over n} \sum_{k=0}^{2n-1} {(n+k)^2\over n^2} =\lim_{n\to \infty} {1\over n} \sum_{k=0}^{2n-1} \left( 1+{k\over n} \right)^2 \\= \int_0^2 (1+x)^2\,dx = \left. \left[ {1\over 3}(1+x)^3 \right] \right|_0^2 = \bbox[red, 2pt]{26\over 3}$$

解答:$$小於等於 30 的質數有：2, 3, 5, 7, 11, 13, 17, 19, 23, 29，總共有 10 個質數\\ 為了使 \gcd(a,b)=1，將這 10 個質因數分配給 (a, b) 的方法數共有 2^{10} = 1024 種組合 \\ q={b\over a} \lt 1 \Rightarrow b\lt a \Rightarrow 滿足條件的組合數為 \frac{1024}{2} = \bbox[red, 2pt]{512}$$

解答:$$f(x)= x^{2026} =Q(x)(x-1)^2(x^2+1) +r(x) \Rightarrow 取r(x)= ax^3+bx^2+ cx+d \\ \Rightarrow f(i)=i^{2026}=-1=-ai-b+ci+d=(d-b)+(c-a)i \Rightarrow \cases{d-b=-1\\ c=a} \\ \Rightarrow f(1)=1=a+b+c+d \Rightarrow a+(d+1)+a+d=1 \Rightarrow a=-d \Rightarrow \cases{b=d+1\\ c=-d} \\ f'(x)=2026x^{2025}=Q'(x)(x-1)^2(x^2+1) +2Q(x)(x-1)(x^2+1)+ 2Q(x)x(x-1)+r'(x) \\ \Rightarrow f'(1)=2026 =r'(1)=3a+2b+c=       3(-d)+2(d+1)+(-d) \Rightarrow d= \bbox[red, 2pt]{-1012}$$

解答:$$取\cases{A=\log y\\ u=2^x+2^{-x}} \Rightarrow A^2+2uA+ 2(u^2-2)=0 \Rightarrow 判別式\Delta=4u^2-8(u^2-2)\ge 0 \\ \Rightarrow u^2\le 4, 由於u=2^x+2^{-x}\ge 2\sqrt{2^x\cdot 2^{-x}}=2 \Rightarrow u\ge 2\\ 同時滿足\cases{u^2\le 4\\ u\ge 2} \Rightarrow u=2 \Rightarrow 2^x+2^{-x}=2 \Rightarrow x=0\\ 同時u=2 \Rightarrow A^2+4A+4=0 \Rightarrow (A+2)^2 =0 \Rightarrow \log y=A=-2 \Rightarrow y={1\over 100} \\ \Rightarrow x+y=0+{1\over 100} = \bbox[red, 2pt]{1\over 100}$$

解答:$$\cases{A(9/2, 2, -1/2) \\ B(2,1,0)\\ C(1,0,-1)} \Rightarrow \cases{\overrightarrow{AB} =(-5/2,-1,1/2) \\ \overrightarrow{AC} =(-7/2,-2,-1/2) \\ \overrightarrow{BC}= (-1, -1,-1)} \Rightarrow \vec n = \overrightarrow{BC}\times \overrightarrow{AB} =(-{3\over 2},3,-{3\over 2}) \\ \Rightarrow 平面方程式:-{3\over 2}(x-2)+3(y-1)-{3\over 2}(z-0)=0 \Rightarrow x-2y+z=0 \cdots(1)\\H(x,y,z) \Rightarrow \cases{\overrightarrow{AH} \cdot \overrightarrow{BC} =0 \\ \overrightarrow{BH} \cdot \overrightarrow{AC} =0} \Rightarrow \cases{x+y+z=6 \cdots(2)\\ 7x+4y+z=18 \cdots(3)}, 由(1),(2), (3)可得 (x,y,z)= \bbox[red, 2pt]{(1,2,3)}$$

解答:

$$兩曲線\cases{\Gamma_1: x^2-4x+y+5=0 \Rightarrow y=-x^2+4x-5\\ \Gamma_2: y^2-4y+x+5=0 \Rightarrow x=-y^2+4y-5} \Rightarrow 兩曲線對稱於\;y=x \\ \Rightarrow 兩圖形上的最短距離發生在兩曲線上切線斜率皆為 1 的對稱點之間 \\ \Rightarrow  \Gamma_1: 2x-4+y'=0 \Rightarrow y'=4-2x =1 \Rightarrow x={3\over 2} \Rightarrow y=-{5\over 4} \Rightarrow 斜率為1的切點A_1({3\over 2},-{5\over 4}) \\ \Rightarrow A_1對稱於y=x的對稱點A_2(-{5\over 4},{3\over 2}) \in \Gamma_2 \Rightarrow \overline{A_1A_2} = \bbox[red, 2pt]{11\sqrt 2\over 4}$$


解答:$$|z-(3-i)|-|z-(-1+3i)|=\pm 2\sqrt 2 \Rightarrow \overline{PF_1}-\overline{PF_2}=\pm 2\sqrt 2, 其中\cases{P(z) \\ F_1(3,-1) \\F_2(-1,3)} \\ \Rightarrow 圖形為一雙曲線\Rightarrow 中心點O={1\over 2}(F_1+F_2)=(1,1) \Rightarrow 2c= \overline{F_1F_2}=4\sqrt 2 \Rightarrow c=2\sqrt 2 \\ \Rightarrow 2a=2\sqrt 2 \Rightarrow a=\sqrt 2 \Rightarrow b=\sqrt{c^2-a^2}=\sqrt 6 \\貫軸\overleftrightarrow{F_1F_2} 斜率m_0={-1-3\over 3-(-1)}=-1 \Rightarrow 與x軸夾角-45^\circ \\ 假設漸近線與貫軸夾角\theta \Rightarrow \tan \theta=\pm {b\over a}=\pm \sqrt 3 \Rightarrow \theta=\pm 60^\circ \\\Rightarrow 漸近線與x軸夾角為\cases{60^\circ-45^\circ =15^\circ \Rightarrow 斜率=\tan 15^\circ=2-\sqrt 3\\ -60^\circ-45^\circ= -105^\circ \Rightarrow 斜率=\tan(-105^\circ)=2+\sqrt 3} \Rightarrow \bbox[red, 2pt]{2\pm \sqrt 3}$$

解答:$$f'(x)的最小值為-12 \Rightarrow f'(x)=k(x-x_0)^2-12, k\gt 0 \Rightarrow f''(x)=2k(x-x_0) \\ \cases{f''(5)=-2 \Rightarrow 2k(5-x_0)=-2 \\f'(5)=0 \Rightarrow k(5-x_0)^2-12=0} \Rightarrow \cases{k=1/12\\ x_0=17} \Rightarrow f'(x)={1\over 12}(x-17)^2-12\\ f'(x)=0 \Rightarrow x=5(極大值點),29(極小值點) \Rightarrow f(29)=2 \\ \Rightarrow f(29)-f(5) = \int_5^{29} \left[ {1\over 12}(x-17)^2-12 \right]\,dx = -192 \Rightarrow f(5)=2+192 =194 \\ \Rightarrow f(17)={f(5)+f(29)\over 2}=98 \Rightarrow 反曲點(17,f(17)) =\bbox[red, 2pt]{(17,98)}$$

解答:$$假設\cases{事件A:抽到偶數\\ 事件B:抽到3的倍數} \Rightarrow \cases{N=3 \Rightarrow P(A\cap B)=0 \ne P(A)\times P(B)={1\over 3}\times {1\over 3} \\N=4 \Rightarrow P(A\cap B)=0 \ne P(A)\times P(B)={2\over 4}\times {1\over 4} \\ N=5 \Rightarrow P(A\cap B)=0 \ne P(A)\times P(B)={2 \over 5}\times {1\over 5} \\ N=6 \Rightarrow P(A\cap B)={1\over 6} = P(A)\times P(B)={3\over 6}\times {2\over 6} \\ N=7 \Rightarrow P(A\cap B)={1\over 7} \ne P(A)\times P(B)={3\over 7}\times {2\over 7} \\ N=8 \Rightarrow P(A\cap B)={1\over 8} = P(A)\times P(B)={4\over 8}\times {2 \over 8} } \\ \Rightarrow 每 6 個一組，一組裡面有 2 個符合要求 \Rightarrow 2026=6\times 337+4 \Rightarrow 337\times 2= \bbox[red, 2pt]{674}$$

二、計算題（每題 6 分）

解答:$${x+y\over x^2-xy+y^2} ={3\over 13} \Rightarrow 13(x+y) =3(x^2-xy+y^2) \Rightarrow x+y是3的倍數 \Rightarrow x+y=3k\\ \Rightarrow 13\cdot 3k= 3(x^2-xy+y^2) =3((x+y)^2-3xy) \Rightarrow 13k=9k^2-3xy \\ \Rightarrow xy={k(9k-13) \over 3} \Rightarrow k也是3的倍數\Rightarrow k=3m \Rightarrow \cases{x+y=9m\\ xy=m(27m-13)}\\ 對於以x,y為兩根之方程式:t^2-(x+y)t+xy=0,其判別式\Delta=(x+y)^2-4xy\ge 0 \\ \Rightarrow 81m^2-4m(27m-13)\ge 0 \Rightarrow -27m^2+52m\ge 0 \Rightarrow 0\le m\le {52\over 27} \Rightarrow m=0,1\\ m=0\Rightarrow \cases{x+y=0\\ xy=0} \Rightarrow 分母x^2-xy+y^2=0, 不合\\ m=1 \Rightarrow \cases{x+y=9\\ xy=14} \Rightarrow (x,y) = \bbox[red, 2pt]{(2,7)或(7,2)}$$

解答:$$f(x)=x^3-x \Rightarrow f'(x)=3x^2-1 \\ 若P在圖形y=f(x)上\Rightarrow P(t,t^3-t) \Rightarrow 過P之切線方程式: y-(t^3-t)=(3t^2-1)(x-t) \\ \Rightarrow y= (3t^2-1)(x-t)+(t^3-t) 代入y=f(x) \Rightarrow (3t^2-1)(x-t)+(t^3-t)=x^3-x \\ \Rightarrow x^3-(3t^2)x+(2t^3)=0,其三根為t,t及另一根假設為x' \Rightarrow 三根之和:2t+x'=0 \\ \Rightarrow x'=-2t \Rightarrow x_1=-2a \Rightarrow x_2=-2x_1=(-2)^2 a \Rightarrow \bbox[red, 2pt]{x_n=a(-2)^n, n\ge 1}$$

解答:$$x^2-(1+4i)x-(9-7i)=0 \Rightarrow x={1+4i\pm \sqrt{21-20i}\over 2} ={1+4i\pm (5-2i)\over 2} =3+i, -2+3i\\ \Rightarrow \cases{x_1=-2+3i\\ x_2= 3+i} \Rightarrow {x_1\over x_2}={-2+3i\over 3+i} ={-3+11i\over 10} \\ 假設z=r(\cos \theta+i \sin \theta) \Rightarrow {x_1\over x_2}\cdot z= \left( {-3\over 10}+{11\over 10}i \right)r(\cos \theta+i\sin \theta)\\ \Rightarrow 虛部為r \left( {-3\over 10} \sin \theta+{11\over 10}\cos \theta \right) =0 \Rightarrow {-3\over 10} \sin \theta+{11\over 10}\cos \theta=0 \Rightarrow \tan \theta={11\over 3} \\ \Rightarrow Arg(z)= \bbox[red, 2pt]{\tan^{-1}{11\over 3} 或\tan^{-1}{11\over 3}-\pi}$$

解答:$$(1+x)^{11} =11Q(x)+x^{11}+1 \Rightarrow (1+x)^{11} \equiv 1+x^{11} \text{ (mod 11)} \\ \Rightarrow (1+x)^{115}=((1+x)^{11})^{10} \cdot (1+x)^5 \equiv (1+x^{11})^{10}\cdot (1+x)^5 \text{ (mod 11)} \\ (1+x^{11})^{10}\cdot (1+x)^5展開式中x^{25}的係數為C^{10}_2\times C^5_3 \Rightarrow C^{115}_{25} \equiv C^{10}_2\times C^5_3 \text{ (mod 11)} = \bbox[red, 2pt]{10}$$

解答:$$\cases{\angle A為共用角\\ \angle APB= \angle AQC=90^\circ} \Rightarrow \triangle ABP \sim \triangle ACQ (AA) \Rightarrow {\overline{AP} \over \overline{AB}}= {\overline{AQ} \over \overline{AC}} \Rightarrow \triangle APQ \sim \triangle ABC \\ \Rightarrow {\overline{PQ} \over \overline{BC}}={\overline{AP} \over \overline{AB}} =\cos A \Rightarrow \overline{PQ}= \overline{BC} \cos A =8\cdot {5^2+7^2-8^2\over 2\cdot 5\cdot 7}=8\times {1\over 7}= \bbox[red, 2pt]{8\over 7}$$

解答:$$假設\vec c=(x,y) \Rightarrow {\vec c\cdot \vec a\over |\vec a|^2} \vec a= \left( {3x+4y\over 3^2+4^2} \right) (3,4) = \left( -{6\over 25},-{8\over 25} \right) \Rightarrow L:3x+4y+2=0 \\ \Rightarrow |\vec b-\vec c|的最小值=d(B, L) ={|6-4+2|\over 5} = \bbox[red, 2pt]{4\over 5}$$

解答:$$將圓心I平移至原點(0,0) \Rightarrow \cases{A(-1,0) \\B(1,0) \\ D(\cos 120^\circ,\sin 120^\circ) = \left( -{1\over 2},{\sqrt 3\over 2} \right) \\E(\cos \theta, \sin \theta), 0\le \theta\le 180^\circ} \\ \Rightarrow \cases{\overline{EA} = \sqrt{(\cos\theta+1)^2+\sin^2\theta} = 2\cos (\theta/2) \\ \overline{EB} =\sqrt{(\cos\theta -1)^2 +\sin^2\theta} =2\sin (\theta/2) \\ \overline{ED}= \sqrt{(\cos\theta+1/2)^2+(\sin \theta-\sqrt 3/2)^2} = \pm 2\cos(30^\circ+\theta/2)} \\ 若30^\circ\le 30^\circ +{\theta\over 2}\le 90^\circ  \Rightarrow 0^\circ \le \theta\le 120^\circ \Rightarrow \overline{ED}= \sqrt 3 \cos {\theta\over 2}-\sin{\theta \over 2} \\ \qquad \Rightarrow S= \overline{EA}+ \overline{EB} +\overline{ED} =(2+\sqrt 3)\cos{\theta\over 2}+\sin{ \theta\over 2} \Rightarrow 最大值=\sqrt{(2+\sqrt 3)^2+1^2} =\sqrt 6+\sqrt 2 \\ 若90^\circ\lt  30^\circ+{\theta\over 2} \le 120^\circ  \Rightarrow 120^\circ \lt \theta\le 180^\circ \Rightarrow \overline{ED}= -\sqrt 3\cos {\theta\over 2}+ \sin {\theta\over 2} \\ \qquad \Rightarrow S=(2-\sqrt 3) \cos {\theta\over 2}+3 \sin {\theta\over 2} \Rightarrow 最大值=\sqrt{16-4\sqrt 3} \lt \sqrt 6+\sqrt 2 \\ 因此S的最大值為\bbox[red, 2pt]{\sqrt 6+\sqrt 2}$$

解答:

$$設該直線方程式為 y = g(x) = mx + k。因為這條直線與四次函數 f(x) = x^4 - 2x^2 + x \\恰好有兩個切點，我們假設這兩個切點的 x 坐標分別為 a 與 b（其中 a \neq b）\\ \Rightarrow  f(x) - g(x) = 0，在 x=a 與 x=b 處都有重根 \\\Rightarrow f(x) - g(x)=x^4-2x^2+x-(mx+k) =(x-a)^2(x-b)^2 \\=x^4-2(a+b)x^3+ ((a+b)^2+2ab)x^2-2ab(a+b)x+a^2b^2 \Rightarrow \cases{-2(a+b)=0\\(a+b)^2+2ab=-2 \\-2ab(a+b)=1-m \\ a^2b^2=-k} \\ \Rightarrow \cases{(a,b)=(1,-1)或(-1,1) \\m=1\\ k=-1} \Rightarrow \bbox[red, 2pt]{g(x)=x-1} \\ 封閉區域面積A=\int_{-1}^1 (f(x)-g(x))\,dx = \int_{-1}^1 (x^4-2x^2+1)\,dx =2 \int_0^1 (x^4-2x^2+1)\,dx \\= 2 \left. \left[ {x^5\over 5}-{2x^3\over 3} +x\right] \right|_0^1 = \bbox[red, 2pt]{16\over 15}$$

====================== END ==========================
解題僅供參考，其他教甄試題及詳解