高雄市 115 學年度市立高級中等學校聯合教師甄選
一、計算證明題 ( 1 至 6 題每題 5 分， 7 至 16 題每題 7 分，共 100 分 )
解答:$$由\overrightarrow{OA}, \overrightarrow{OB}, \overrightarrow{OC}所張出的平行六面體體積V_1= \begin{Vmatrix} -11& 1& 2\\1& 1& 5\\7& -2& -1 \end{Vmatrix} =81 \\ 滿足\cases{0\le x\le 1\\ 0\le y\le x\\ y\le z\le 1} 的體積放大比例s=\int_0^1 \int_0^x \int_y^1 dz\,dydx ={1\over 3} \Rightarrow 欲求之體積=s\cdot V_1= \bbox[red, 2pt]{27}$$
解答:$$由圖形可知f(x)的\cases{最大值M=4\\ 最小值m=1} \Rightarrow \cases{k=(4+1)/2=5/2\\ |a|=(4-1)/2=3/2} \\ 半週期{T\over 2}={11\pi\over 18}-{5\pi\over 18} \Rightarrow T={2\pi\over 3} ={2\pi\over b} \Rightarrow b=3 \Rightarrow f(x)=a\cos(3x+c)+{5\over 2} \\ \Rightarrow f \left( {5\pi\over 18} \right)=1 \Rightarrow a\cos \left( {5\pi\over 6}+c \right)+{5\over 2}=1 \Rightarrow a\cos \left( {5\pi\over 6}+c \right) =-{3\over 2} \\ \Rightarrow \cases{a=3/2 \Rightarrow \cos(5\pi/6+c)=-1 \Rightarrow c=\pi/6+ 2n\pi \not \in (\pi, 2\pi], n\in Z\\ a=-3/2 \Rightarrow c=7\pi/6+2n\pi \in (\pi, 2\pi], n=0} \Rightarrow \cases{a=-3/2\\c=7\pi/6} \\ \Rightarrow (a,b,c,k)= \bbox[red, 2pt]{\left( -{3\over 2},3,{7\pi\over 6},{5\over 2} \right)}$$
解答:$$假設g(x)={(x-1)(x-4)(x-5) \over x-3} \Rightarrow f(x)=(x-2)g(x) \Rightarrow f'(x)=g(x)+(x-2)g'(x) \\ \Rightarrow f'(2)=g(2)={1\cdot (-2)\cdot (-3)\over (-1)}= \bbox[red, 2pt]{-6}$$
解答:$$10個數字任選3個有C^{10}_3=120種選法，依所選三數字進行分類計算:\\\textbf{Case I }不含偶數:從\{1,3, 5, 7, 9\}任選3數有C^5_3=10種，但3,9不能同時被選取，\\\qquad 需扣除(1,3,9), ( 5,3,9), (7,3,9)三種，即10-3=7種\\ \textbf{Case II }含一個偶數:\cases{偶數是\{2,4,8\}之一，另外奇數選2個但不能是(3,9), \\\qquad 此類有C^3_1(C^5_2-1)=27種\\ 偶數是6, 另外兩數從\{1,5,7\}選2個,此類有C^3_2=3種\\ 偶數是10,另外兩數從\{1,3,7,9\}選2個,但不包含(3,9) \\ \qquad 此類有C^{4}_2-1=5種} \\ 以上合計:7+27+3+5=42\Rightarrow 機率={42\over 120} = \bbox[red, 2pt]{7\over 20}$$
解答:$$依題意可假設\cases{g(x) =(x+k)f(x)+ x+2\\ (f(x))^2=(x+m)g(x)+2x-1} \Rightarrow (f(x))^2=(x+m) [(x+k)f(x)+ x+2]+2x-1 \\ \Rightarrow f(x)[f(x)-(x+m)(x+k)]=(x+m)(x+2)+2x-1=x^2+(m+4)x+2m-1\\ 由於f(x)是二次式，而右式也是二次式，因此\cases{f(x)-(x+m)(x+k)=1\\ f(x)= x^2+(m+4)x+2m-1} \\ \Rightarrow f(x)=(x+m)(x+k)+1=x^2+(m+4)x+2m-1 \Rightarrow \cases{k=4\\ m=-1} \\ \Rightarrow \bbox[red, 2pt]{f(x)=x^2+3x-3} \Rightarrow g(x)=(x+4)f(x)+x+2 \Rightarrow \bbox[red, 2pt]{g(x)=x^3+7x^2+10x-10}$$
解答:$$8x^2-36x+34={k\over x} \Rightarrow 8x^3-36x^2+34x-k=0的三根為a-d,a,a+d(等差數列) \\ \Rightarrow 三根之和=3a={36\over 8} \Rightarrow a={3\over 2} 為其中一根 \Rightarrow 8\left( {3\over 2} \right)^3-36\left( {3\over 2} \right)^2+ 34\left( {3\over 2} \right)=k \\ \Rightarrow k=27-81+51= \bbox[red, 2pt]{-3 }$$
解答:$$假設\cases{y=x的斜角為\alpha\\ y=2x的斜角為\beta\\ y=3x的斜角為\gamma} \Rightarrow \cases{\tan \alpha=1\\ \tan \beta=2\\ \tan \gamma=3} \\假設P的極坐標角為\theta \Rightarrow \cases{P與y=x對稱後角度變為2\alpha-\theta ,再與y=2x對稱後角度變為2\beta-(2\alpha-\theta) \\ P與y=4x對稱後角度變為2\gamma-\theta} \\ \Rightarrow 2\beta-(2\alpha-\theta)=2\gamma-\theta+2n\pi \Rightarrow \theta=\gamma-\beta+\alpha +n\pi \Rightarrow \tan \theta=\tan (\gamma-\beta+\alpha)=m \\ 由於\tan (\gamma-\beta)={\tan \gamma-\tan \beta\over 1 +\tan \gamma \tan \beta} ={3-2\over 1+3\times 2}={1\over 7} \Rightarrow \tan(\gamma-\beta+\alpha) ={\tan(\gamma-\beta)+ \tan \alpha\over 1-\tan(\gamma-\beta)\tan \alpha} \\={1/7+1\over 1-(1/7)\times 1}={4\over 3} \Rightarrow m= \bbox[red, 2pt]{4\over 3}$$
$$x^2+y^2= 6x+ 6\sqrt 3|y| \Rightarrow (x-3)^2+(|y|-3\sqrt 3)^2=36 \\\Rightarrow \cases{y\ge 0 \Rightarrow 圓\Gamma_1:(x-3)^2+(y-3\sqrt 3)^2=36 \Rightarrow 圓心O_1(3,3\sqrt 3),半徑6\\ y\le 0 \Rightarrow 圓\Gamma_2:(x-3)^2+(y+3\sqrt 3)^2=36 \Rightarrow 圓心O_2(3,-3\sqrt 3),半徑6} \\ \Rightarrow 兩圓交於\cases{P(0,0) \\Q(6,0)} \Rightarrow \cases{\overline{O_1P} =\overline{O_1Q} =\overline{PQ}=6 \\ \overline{O_2P} =\overline{O_2Q} =\overline{PQ}=6}\Rightarrow \angle PO_1Q= \angle PO_2Q={\pi\over 3} \\ \Rightarrow \stackrel{\Large \frown}{PQ} =6\cdot (2\pi-{\pi\over 3})=10 \pi \Rightarrow 周長=2\times 10\pi=\bbox[red, 2pt]{20\pi}$$
解答:$$偵察機P在直線L_1上，L_1方向向量\overrightarrow{AB} =(4,4,-2) \Rightarrow P(1+2t,2t,2-t) \\ 攔截機Q在直線L_2上，L_2方向向量\overrightarrow{CD}=(1,-2,-3) \Rightarrow Q(1+t,11-2t,13-3t) \\ 兩機在t時刻的距離d=|P_1(t)-P_2(t)| =|(t,4t-11,2t-11)| \\ \Rightarrow f(t)=d^2=t^2+(4t-11)^2+(2t-11)^2=21t^2-132t+242 \\ \Rightarrow 極小值發生在t={132\over 42} ={22\over 7} \Rightarrow f(22/7)={242\over 7} \Rightarrow d的最小值為\sqrt{242\over 7}= \bbox[red, 2pt]{11\sqrt{14}\over 7}$$
解答:$$L_1: {2-x\over 2}={1+y\over -1}={1-z\over 2} \Rightarrow L_1方向向量\vec u_1=(2,1,2)且經過P_1(2,-1,1)\\ L_2:{1-x\over 2} ={1-y\over -2}={2+z\over 1} \Rightarrow L_2方向向量\vec u_2=(-2,2,1)且經過P_2(1,1, -2)\\ 假設平面E的的法向量\vec n=(a,b,c) \Rightarrow \vec n\bot \vec u_1 \Rightarrow \vec n\cdot \vec u_1=0 \Rightarrow 2a+b+2c=0\\ L_2與E的夾角\theta 滿足\cos \theta={2\sqrt 5\over 5} \Rightarrow \sin \theta={1\over \sqrt 5} ={|\vec u_2\cdot \vec n|\over |\vec u_2||\vec n|} ={|-2a+2b+c| \over 3\cdot\sqrt{a^2+b^2+c^2}} \\ 將b=-2a-2c代入上式\Rightarrow {(2a+c)^2 \over 5a^2+8ac+5c^2}={1\over 5} \Rightarrow 15a^2+12ac=0 \Rightarrow 3a(5a+4c)=0 \\ \textbf{Case I }a=0 \Rightarrow b=-2c,取\vec n=(0,-2,1) \Rightarrow E:0(x-2)-2(y+1)+(z-1)=0 \Rightarrow 2y-z+3=0\\ \textbf{Case II }5a+4c=0 ,取\cases{a=4\\ c=-5} \Rightarrow b=2,取 \vec n(4,2,-5) \Rightarrow E:4(x-2)+2(y+1)-5(z-1)=0 \\ \qquad \Rightarrow 4x+2y-5z=1\\ 平面E方程式:\bbox[red, 2pt]{2y-z+3=9, 4x+2y-5z=1}$$
主辦單位未公告答案，解題僅供參考，其他教甄試題及詳解
解答:$${2z-(9+3\sqrt 3i) \over z}= \sqrt 3(\cos 150^\circ+ i\sin 150^\circ) \Rightarrow 2-{9+3\sqrt 3i\over z}=\sqrt 3 \left( -{\sqrt 3\over 2}+{1\over 2} \right)i \\ \Rightarrow {9+3\sqrt 3i\over z} ={7\over 2}-{\sqrt 3\over 2}i \Rightarrow Arg \left( {z-(3+\sqrt 3i)\over z} \right) = Arg \left( 1-{3+\sqrt 3i\over z} \right) \\= Arg \left( 1-{7\over 6}+{\sqrt 3\over 6}i \right) =Arg \left( -{1\over 6} +{\sqrt 3\over 6}i\right) =Arg \left( {1\over 3}(\cos {2\pi\over 3}+i\sin {2\pi\over 3}) \right) =\bbox[red, 2pt]{2\pi\over 3}$$
解答:$$\lim_{n\to \infty} \sum_{k=1}^n{1\over 3n+6k} =\lim_{n\to \infty} \sum_{k=1}^n{1\over n(3+6(k/n))} = \int_0^1 {1\over 3+6x}\,dx = \left. \left[{1\over 6} \ln(3+6x) \right] \right|_0^1 \\={1\over 6}(\ln 9-\ln 3) = \bbox[red, 2pt]{{1\over 6}\ln 3}$$
解答:$${2\over \sqrt{n+2}+\sqrt n} \lt {1\over \sqrt n} \lt {2\over \sqrt{n}+\sqrt{n-2}} \Rightarrow \sqrt{n+2}-\sqrt n\lt {1\over \sqrt n} \lt \sqrt n-\sqrt {n-2} \\ \Rightarrow \sum_{k=57}^{1012} (\sqrt{2k+3}-\sqrt{2k+1}) \lt \sum_{k=57}^{1012} {1\over \sqrt{2k+1}} \lt \sum_{k=57}^{1012} (\sqrt{2k+1}-\sqrt{2k-1}) \\ \Rightarrow \sqrt{2027}-\sqrt{115} \lt \sum_{k=57}^{1012} {1\over \sqrt{2k+1}} \lt\sqrt{2025}-\sqrt{113}\Rightarrow 34.3 \lt \sum_{k=57}^{1012} {1\over \sqrt{2k+1}} \lt 34.37 \\ \Rightarrow \left\lfloor \sum_{k=57}^{1012} {1\over \sqrt{2k+1}}\right \rfloor=\bbox[red, 2pt]{34}$$
解答:$$\cot \theta-\cot 2\theta={\cos \theta\over \sin \theta}-{\cos 2\theta\over \sin 2\theta} ={2\cos^2\theta-\cos 2\theta\over \sin 2\theta} ={2\cos^2\theta-(2\cos^2\theta -1) \over \sin 2\theta}={1\over \sin 2\theta} \\ \Rightarrow {1\over \sin 2\theta} = \cot \theta-\cot 2\theta \Rightarrow \cases{{1\over \sin24^\circ} =\cot12^\circ-\cot 24^\circ\\ {1\over \sin48^\circ} =\cot 24^\circ-\cot 48^\circ \\{1\over \sin96^\circ} =\cot 48^\circ-\cot 96^\circ}\\ \Rightarrow {1\over \sin24^\circ}+{1\over \sin 48^\circ}+{1\over \sin96^\circ}= \cot 12^\circ-\cot96^\circ ={\cos 12^\circ \over \sin12^\circ}-{\cos 96^\circ\over \sin 96^\circ} \\={\cos 12^\circ \sin 96^\circ-\sin12^\circ \cos96^\circ \over \sin 12^\circ \sin 96^\circ} ={\sin(96^\circ-12^\circ) \over \sin12^\circ \sin 96^\circ} = {\sin 84^\circ \over \sin12^\circ \sin 96^\circ} = {\sin 84^\circ \over \sin12^\circ \sin (180^\circ-84^\circ)} \\={1\over \sin 12^\circ} ={1\over \sin(180^\circ-168^\circ)} ={1\over \sin 168^\circ}, \bbox[red, 2pt]{故得證}$$
====================== END ==========================主辦單位未公告答案，解題僅供參考，其他教甄試題及詳解
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