國立鳳新高級中學 114 學年度第 1 次教師甄選
解答:$$\cases{A(1,0,1) \in L_1\\ B(2,1,0) \in L_1} \Rightarrow \cases{A在E的投影點A'(2/3,1/3,1/2)\\ B在E的投影點B'(2,1,0)} \Rightarrow M_1=\overleftrightarrow{A'B'}: {x-2\over 4}={y-1\over 2}=-z \\ \cases{P(-1,2,1) \in L_2\\ Q(1,1,3) \in L_2} \Rightarrow \cases{P在E的投影點P'(-2/3,5/3,5/3)\\ Q在E的投影點Q'(1/6,11/6,4/3)} \\\Rightarrow M_2=\overleftrightarrow{P'Q'}:{x-1/6\over 5}={y-11/6}={z-4/3\over -2} \Rightarrow M_1\cap M_2= \bbox[red, 2pt]{(6,3,-1)}$$
解答:$$\cases{7個相異數字任排:7!種\\ 出現「12」:6! \Rightarrow 出現「12」,「23」,...「67」共{6\choose 1}6! 種\\出現「12」及「23」:5! 種\Rightarrow 出現「12」,「23」,...,「67」其中任兩個共{6\choose 2}5! 種\\出現「12」,「23」,...,「67」其中任三個共{6\choose 3}4! 種\\出現「12」,「23」,...,「67」其中任四個共{6\choose 4}3! 種\\出現「12」,「23」,...,「67」其中任五個共{6\choose 5}2! \\「12」,「23」,...,「67」六個皆出現,只有1種} \\ \Rightarrow 共有7!-{6\choose 1}6!+{6\choose 2}5!-{6\choose 3}4!+{6\choose 4}3!-{6\choose 5}2!+1 \\ =5040- 4320+1800-480+90-12+1 =\bbox[red, 2pt]{2119}$$
解答:
$$z=a+bi, a,b\in \mathbb R \Rightarrow \cases{z/2=(a/2)+(b/2)i\\ 2/z=2/(a+bi) =2a/(a^2+ b^2)- 2bi/ (a^2+b^2)} \\ \Rightarrow \cases{-1\le a/2\le 1\\ -1\le b/2\le 1 \\ -1\le 2a/(a^2+b^2)\le 1\\ -1\le -2b/(a^2+b^2) \le 1} \Rightarrow \cases{-2\le a\le 2\\ -2\le b\le 2\\ (a-1)^2+b^2 \ge 1 \\(a+1)^2+b^2 \ge 1 \\ a^2+(b+1)^2 \ge 1\\ a^2+(b-1)^2 \ge 1} \Rightarrow 所圍區域為正方形內且四個圓外 \\也就是上圖斜線區域四倍的面積= 4\left(\int_0^1 \left[2-(\sqrt{1-x^2}+1) \right]\,dx +\int_1^2 \left[2-\sqrt{1-(x-1)^2} \right]\,dx \right) \\=4 \left((1-{\pi\over 4}) +(2-{\pi\over 4})\right) = \bbox[red, 2pt]{12-2\pi}$$
解答:$$\cases{\Gamma_1:y=f(x)=x^3-x\\ \Gamma_2: y=g(x)=x^2-a^2+a} \Rightarrow \cases{f'(x)=3x^2-1\\ g'(x)=2x}\\ 假設公切線L= \overleftrightarrow{AB}, 其中\cases{A\in \Gamma_1\\ B\in \Gamma_2} \Rightarrow \cases{A(s,s^3-s) \\B(t,t^2-a^2+a)} \Rightarrow 斜率m_L= {t^2-a^2+a-s^3+s\over t-s} \\ \Rightarrow m_L=f'(s)= g'(t) \Rightarrow m_L=3s^2-1=2t \Rightarrow t={3s^2-1\over 2} \\m_L= {t^2-a^2+a-s^3+s\over t-s} =2t \Rightarrow s^3-s+a^2-a=-t^2+2st=-\left( {3s^2-1\over 2}\right)^2+2s\cdot {3s^2-1\over 2} \\ \Rightarrow -9s^4+8s^3+6s^2-1=4a^2-4a \\ 欲求h(s)=-9s^4+8s^3+6s^2-1的極值 \Rightarrow f'(s)=-36s^3+24s^2+12s =-12s(s-1)(3s+1) \\ \Rightarrow f''(s)=-108s^2+48s+12 \Rightarrow \cases{f''(0)\gt 0\\ f''(1)\lt 0\\ f''(-1/3)\lt 0} \Rightarrow \cases{f(1)=4\\ f(-1/3)=-20/27} \Rightarrow f(s)最大值為4 \\ \Rightarrow 4a^2-4a\le 4\Rightarrow a^2-a+1 =(a-{1-\sqrt 5\over 2})(a-{1+\sqrt 5\over 2})\le 0 \Rightarrow \bbox[red, 2pt]{{1-\sqrt 5\over 2}\le a\le {1+\sqrt 5\over 2}}$$
解答:
$$假設P至\overline{BC}, \overline{AC}, \overline{AB}的垂足分別為D,E,F \Rightarrow \cases{\overline{PD} =k \\\overline{PE} =3k \\\overline{PF} =2k } \\ \Rightarrow 正\triangle ABC面積={\sqrt 3 \over 4} ={1\over 2}(k+3k+2k) \Rightarrow k={\sqrt 3\over 12} \Rightarrow \cases{\overline{PD} = \sqrt 3/12 \\ \overline{PE} =\sqrt 3/4 \\\overline{PF} = \sqrt 3/6 } \\過P作平行三邊的直線,即\cases{\overline{MN} \parallel \overline{BC} \\ \overline{ST} \parallel \overline{AB} \\ \overline{UV} \parallel \overline{AC}} \Rightarrow \cases{\triangle PSV為正\triangle 且\overline{PD}為高\\ \triangle PTN為正\triangle 且\overline{PE}為高 \\ \triangle PMU為正\triangle 且\overline{PF}為高}\\ \Rightarrow \cases{\triangle PSV的邊長為\overline{PD}\cdot 2/\sqrt 3 =1/6\\ \triangle PTN的邊長為\overline{PE}\cdot 2/\sqrt 3= 1/2\\ \triangle PMU的邊長為\overline{PF} \cdot 2/\sqrt 3=1/3} 且\cases{ \overline{NC} =\overline{MB} =\overline{PV}= 1/6\\ \overline{AT}= \overline{BS} =\overline{PM} =1/3\\ \overline{AU}= \overline{CV} =\overline{PN} =1/2} \\ \Rightarrow \cases{\triangle APT:\cos \angle ATP= \cos 120^\circ=-1/2= \displaystyle {(1/3)^2+(1/2)^2-\overline{AP}^2\over 2\cdot (1/2)\cdot (1/3)} \\ \triangle PBS: \cos \angle PSB=\cos 120^\circ=-1/2= \displaystyle {(1/6)^2+(1/3)^2-\overline{PB}^2 \over 2\cdot (1/6)\cdot (1/3)} \\ \triangle PVC: \cos \angle PVC= \cos 120^\circ=-1/2= \displaystyle {(1/6)^2 +(1/2)^2- \overline{PC}^2 \over 2\cdot (1/6)\cdot (1/2)}} \Rightarrow \cases{\overline{AP} =\sqrt{19}/6 \\ \overline{PB} =\sqrt 7/6 \\ \overline{PC} = \sqrt{13} /6} \\ \Rightarrow \overline{PA} +\overline{PB} +\overline{PC} ={1\over 6}(\sqrt{19} +\sqrt 7+ \sqrt {13}) \lt {1\over 6}(5+3+4) =2 \Rightarrow \bbox[red, 2pt]{否}$$
解答:
$$假設B沿著\overline{CP}折起來後的點為B'(欲求\overline{AB'}最小值)，且B'在\overline{CP}的垂足為Q\\ 若\angle PCB=\theta \Rightarrow \cases{\angle B'CP=\theta\\ \angle ACP=90^\circ-\theta\\\overline{CB'} =\overline{CB}=3 } \Rightarrow \cases{\overline{B'Q}= 3\sin \theta \\ \overline{CQ}=3\cos \theta} \\ \triangle ACQ: \cos \angle ACQ = \cos(90^\circ-\theta) =\sin \theta={4^2+(3\cos \theta)^2-\overline{AQ}^2 \over 24\cos \theta} \\ \Rightarrow \overline{AQ}^2 = 16+ 9\cos^2 \theta- 24\sin \theta\cos \theta \\ 直角\triangle AQB': \overline{AB'}^2= \overline{B'Q}^2+\overline{AQ}^2 =9\sin^2\theta+16+9\cos^2 \theta-24\sin \theta\cos \theta =25- 24\sin \theta\cos \theta \\=25-12\sin 2\theta \ge 13 \Rightarrow \overline{AB'}\ge \bbox[red, 2pt]{\sqrt{13}}$$
解答:$$假設\cases{P=\begin{bmatrix}x\\ y \end{bmatrix} \\[1ex]A=\begin{bmatrix}a & b \\c & d\end{bmatrix}} \Rightarrow Q=AP = \begin{bmatrix}ax+by\\ cx+dy \end{bmatrix} \Rightarrow \overline{OP}^2=\overline{OQ}^2 \Rightarrow x^2+y^2 =(ax+by)^2+ (cx+dy)^2 \\ =(a^2+ c^2)x^2+(b^2+d^2)y^2 +(2ab+2cd)xy \Rightarrow \cases{a^2+c^2=1 \cdots(1)\\ b^2+d^2=1 \cdots(2)\\ ab+cd=0 \cdots(3)} \\(3) \Rightarrow c=-{ab\over d} 代入(1) \Rightarrow a^2+{a^2b^2\over d^2} ={a^2d^2+a^2b^2\over d^2} ={a^2(b^2+d^2)\over d^2}={a^2\over d^2}=1 \\ \Rightarrow \cases{a=d \\ a=-d } 代入(3) \Rightarrow \cases{b=-c\\ b=c} \Rightarrow \cases{A=\begin{bmatrix}a & -c \\c & a\end{bmatrix} 為旋轉矩陣\\[1ex] A=\begin{bmatrix}a & b \\b & -a\end{bmatrix} 為鏡射矩陣}\;\bbox[red, 2pt]{故得證}$$
解答:$${(1-x)+x/2+x/2\over 3} \ge \sqrt[3]{x^2(1-x)\over 4} \Rightarrow \left( {(1-x)+x/2+x/2\over 3}\right)^3 \ge {x^2(1-x)\over 4} \\ \Rightarrow 1-x \le {4\over x^2} \left( {(1-x)+x/2+x/2\over 3}\right)^3 \Rightarrow {2\over 1-x}\ge {x^2\over 2} \cdot {1\over \left( {(1-x)+x/2+x/2\over 3}\right)^3} ={27x^2\over 2} \cdots(1)\\{y+(10-y)\over 2} \ge \sqrt{y(10-y)} \Rightarrow (10-y)y \le \left( {y+(10-y)\over 2}\right)^2\\ \Rightarrow {75\over 10-y} ={75\over 10}\left( {y\over 10-y}+1\right) ={15\over 2}\cdot {y\over 10-y}+{{15\over 2}} ={15\over 2}\cdot {y^2\over (10-y)y}+{{15\over 2}}\\ \ge {15\over 2}\cdot {y^2\over ({y+(10-y)\over 2})^2} +{15\over 2} ={15\over 2}\cdot {y^2\over 25}+{15\over 2} \cdots(2) \\ (1)+(2) \Rightarrow {2\over 1-x}+{75\over 10-y} \ge {27x^2\over 2} +{15\over 2}\cdot {y^2\over 25}+{15\over 2}={27\over 2} \left(x^2 +{y^2\over 45} \right)+{15\over 2} =\bbox[red, 2pt]{21}$$
解答:$$\textbf{(1) } 取f(x)=x+1-{\pi\over 4}-\tan x \Rightarrow f'(x)=1-\sec^2 x =\tan^2 x \ge 0 \Rightarrow f(x) 遞增在0\le x\le {\pi\over 4}\\ \quad 且f(0)=1-{\pi\over 4}\gt 0, 因此f(x)\ge 0 \Rightarrow x+1-{\pi\over 4}\ge \tan x, \bbox[red, 2pt]{故得證} \\ \textbf{(2) } t=\tan x \Rightarrow \arctan t=x \Rightarrow dx={1\over 1+t^2}dt \Rightarrow I_n=\int_0^1{t^n\over 1+t^2}\,dt \le \int_0^1 t^n\,dt ={1\over n+1} \\\quad \Rightarrow \lim_{n\to \infty } I_n= \bbox[red, 2pt]0 \\ \textbf{(3) } I_n=\int_0^1{t^n\over 1+t^2}\,dt \Rightarrow I_n+I_{n+2} =\int_0^1{t^n +t^{n+2}\over 1+t^2}\,dt =\int_0^1{t^n(t^2+1)\over 1+t^2}\,dt = \int_0^1 t^n\,dt = \bbox[red, 2pt]{1\over n+1} \\\textbf{(4) }\sum_{n=1}^\infty {(-1)^{n+1} \over 2n} ={1\over 2}-{1\over 4}+{1\over 6}-{1\over 8}+ \cdots =(I_1+I_3)-(I_3+I_5)+(I_5+I_7)-(I_7+I_9)+ \cdots \\\qquad =I_1=\int_0^1 {t\over 1+t^2}\,dt =\left. \left[{1\over 2}\ln(1+t^2) \right] \right|_0^1 ={1\over 2}\ln 2 = \bbox[red, 2pt]{\ln \sqrt 2}$$
解答:$$\textbf{(1) }假設以\alpha,\beta,\gamma為三根的三次多項式f(x)= (x-\alpha) (x-\beta)(x-\gamma) =x^3+tx^2 +vx+1\\ 假設以\alpha^{1/3},\beta^{1/3}, \gamma^{1/3}為三根的三次多項式g(y)= (y-\alpha^{1/3})(y-\beta^{1/3}) (y-\gamma^{1/3})=y^3+ay^2+by+1 \\g(y)=0 \Rightarrow y^3+1=-y(ay+b) \Rightarrow (y^3+1)^3 =-y^3(ay+b)^3 =-y^3(a^3y^3+b^3+3aby(ay+b)) \\=-y^3(a^3y^3+b^3-3ab(y^3+1)) \\ 由於g(y)=0的根的三次方為f(x)=0的根,因此將y^3=x代入上式\\\Rightarrow (x+1)^3=-x(a^3x+b^3-3ab(x+1)) \Rightarrow x^3+(a^3-3ab+3)x^2+(b^3-3ab+3)x+1=0 \\ \Rightarrow \cases{a^3-3ab+3=t\\ b^3-3ab+3=v} \Rightarrow \cases{a^3=t+3ab-3 \cdots(1)\\ b^3=v+3ab-3} \Rightarrow a^3b^3=(t+ 3ab-3)(v+3ab-3) \\ 取z=ab \Rightarrow z^3=(t+3z-3)(v+3z-3) =tv+3z(t+v)-3(t+v)+9z^2-18z+9 \\=t(-3-t)+3z(-3)-3(-3)+9z^2-18z+9=9z^2-27z+(-t^2-3t+18) \\ \Rightarrow (z-3)^3=-t^2-3t-9 \Rightarrow z=3-\sqrt[3]{t^2+3t+9} \cdots(2) \\ 由(1) \Rightarrow -a =-\sqrt[3]{t+3z-3} =-\sqrt[3]{t+6-3\sqrt[3]{t^2+3t+9}} =\sqrt[3]{(-t-6)+3\sqrt[3]{t^2+3t+9}} \\ \Rightarrow \alpha^{1/3} +\beta^{1/3} +\gamma^{1/3} = \sqrt[3]{(-t-6)+3\sqrt[3]{t^2+3t+9}}\; \bbox[red, 2pt]{QED} \\\textbf{(2) } \cos {2\pi\over 9} \cos {4\pi\over 9} \cos {8\pi\over 9} = {1\over 2} \left(\cos {6\pi\over 9}+\cos{2\pi\over 9} \right)\cos {8\pi\over 9} = {1\over 2} \left(-{1\over 2}+\cos{2\pi\over 9} \right)\cos {8\pi\over 9} \\ \quad = {1\over 2} \left(-{1\over 2} \cos{8\pi\over 9}+ \cos{2\pi\over 9}\cos {8\pi\over 9} \right) = {1\over 2} \left(-{1\over 2} \cos{8\pi\over 9}+ {1\over 2}(\cos{10\pi\over 9} +\cos {6\pi\over 9}) \right) \\\quad = {1\over 2} \left({1\over 2} \cos{\pi\over 9}- {1\over 2}\cos{\pi\over 9} -{1\over 4} \right) =-{1\over 8} \Rightarrow 8\cos {2\pi\over 9} \cos {4\pi\over 9} \cos {8\pi\over 9}=-1\\ \Rightarrow 取\cases{\alpha=2\cos(2\pi/9) \\\beta=2\cos(4\pi/9) \\\gamma=2\cos (8\pi/9)} \Rightarrow t=-(\alpha+\beta+ \gamma) =-2\left(\cos{2\pi\over 9}+ \cos{4\pi\over 9}+ \cos{8\pi\over 9} \right) \\\quad =-2\left(2\cos{3\pi\over 9}\cos{\pi\over 9}+ \cos{8\pi\over 9} \right) =-2\left(\cos{\pi\over 9}+ \cos{8\pi\over 9} \right) =-2\cdot 2\cos{\pi\over 2}\cos{7\pi\over 18}=0 \\ \Rightarrow \sqrt[3]{\cos{2\pi\over 9}} +\sqrt[3]{\cos{4\pi\over 9}} +\sqrt[3]{\cos{8\pi\over 9}} = \sqrt[3]{\alpha\over 2} +\sqrt[3]{\beta \over 2} +\sqrt[3]{\gamma\over 2} = {1\over \sqrt[3]2}\left( \alpha^{1/3} +\beta^{1/3} +\gamma^{1/3} \right) \\ ={1\over \sqrt[3]2} \sqrt[3]{-6+3\sqrt[3]9} =\sqrt[3]{{3\over 2}(\sqrt[3]9-2)}\; \bbox[red, 2pt]{QED}$$
解題僅供參考，其他教甄試題及詳解
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