臺中市立文華高級中等學校 115 學年度第 1 次教師甄選
一、 填充題： (共 80 分)
Ⅰ .填充一(每格 4 分， 共 32 分， 每格全對才給分。 )
解答:$$25,26,28三個數子的差距為1與2 \Rightarrow 任取三數字的差距也是1,2 (或2,1)\\ \textbf{Case I (a,a+1,a+3):}(11,12,14), (12,13,15), (13,14,16), (14,15,17), (16,17,19) \\ \textbf{Case II (a,a+2,a+3): }(11,13,14),(12,14,15), (13,15,16),(14,16,17) \\ \text{Case I 有5種、Case II 有4種，合計9種} \Rightarrow 機率={9\over C^8_3} = \bbox[red, 2pt]{9\over 56}$$
解答:$$G為\triangle ABC重心\Rightarrow \overrightarrow{PG} ={1\over 3}(\overrightarrow{PA} +\overrightarrow{PB}+ \overrightarrow{PC}) \Rightarrow 3 \overrightarrow{PG} = \overrightarrow{PA} +\overrightarrow{PB}+ \overrightarrow{PC} \\ \Rightarrow \overrightarrow{PA} +2 \overrightarrow{PB}+ ( \overrightarrow{PA} +\overrightarrow{PB}+ \overrightarrow{PC}) =k\overrightarrow{AB} \Rightarrow 2\overrightarrow{PA} + 3 \overrightarrow{PB}+ \overrightarrow{PC} =k \overrightarrow{AB} \\ \Rightarrow 2(-\overrightarrow{AP}) +3( \overrightarrow{AB}- \overrightarrow{AP})+ (\overrightarrow{AC} -\overrightarrow{AP}) =k \overrightarrow{AB} \Rightarrow \overrightarrow{AP}={3-k\over 6}\overrightarrow{AB} +{1 \over 6}\overrightarrow{AC} \\ \Rightarrow \cases{3-k \gt 0 \\(3-k)/6+1/6 \lt 1} \Rightarrow \bbox[red, 2pt]{-2\lt k\lt 3}$$
解答:$$L:{x-5\over 4}={y-5\over 3 } ={z-2\over 1} \Rightarrow P(4t+5,3t+5,t+2) \in L\\ 將P代入E \Rightarrow 2(4t+5)+(3t+5)+3(t+2)=7 \Rightarrow t=-1 \Rightarrow B(1,2,1) \\ 通過A且與E垂直的直線(2k+5, k+5, 3k+2) 代入E求A在E上的垂足H(3,4,-1) \\ \Rightarrow \cases{\overline{AH}=\sqrt{14} \\ \overline{BH}=\sqrt{12}} \Rightarrow 面積= {1\over 2}\overline{AB}\cdot \overline{AC} \sin \theta \; (\theta =\angle BAC) \\ 又 \overrightarrow{AB} \cdot \overrightarrow{AC} = (\overrightarrow{AH}+ \overrightarrow{HB}) \cdot (\overrightarrow{AH} +\overrightarrow{HC}) =\overline{AH}^2+ 0+0 +\overrightarrow{HB}\cdot \overrightarrow{HC} =14+ \overline{HB} \cdot \overline{HC}\cos \phi \\ C在以H為圓心,\overline{HB}為半徑的圓上\Rightarrow \overline{HC} =\overline{HB}=\sqrt{12} \Rightarrow \overrightarrow{AB} \cdot \overrightarrow{AC} = 14+12\cos \phi \\ 最小值發生在\cos \phi=-1 \Rightarrow H={1\over 2}(B+C) \Rightarrow C=2H-B= \bbox[red, 2pt]{(5,6,-3)}$$
解答:$$假設\cases{1元硬幣有x個\\ 5元硬幣有y個\\ 10元硬幣有z個} \Rightarrow \cases{x\ge y\ge z\ge 8\\ x+3y-z=60} \Rightarrow 欲求f(x,y,z)=x+5y+10z的最大值 \\ \Rightarrow f(x,y,z)=(60-3y+z)+5y+10z =2y+11z+60 \Rightarrow z越大越好\\ x=60-3y+z \ge y \Rightarrow y\le 15 +{z\over 4} \Rightarrow z\le y\le 15 +{z\over 4} \Rightarrow z-{z\over 4}\le 15 \Rightarrow z\le 20 \\ z=20\Rightarrow 20\le y\le 15+{20\over 4} \Rightarrow 20\le y\le 20 \Rightarrow y=20 \Rightarrow x=60-60+20=20 \\ \Rightarrow f(20,20,20) =20+100+200 = \bbox[red, 2pt]{320}$$
解答:$$(-8+4s+2t, 3+s-3t,27+5s+t) =(-8,3,27)+s(4,1,5)+ t(2,-3,1) \\ 代表某個平面E通過P(-8,3,27)，且由兩個向量\cases{\vec a=(4,1,5) \\\vec b=(2,-3,1)} 所張成的 \\ \Rightarrow E的法向量\vec n平行於(\vec a\times \vec b) =(16,6,-14) \Rightarrow 取\vec n=(8,3,-7) \\ \Rightarrow E: 8(x+8)+3(y-3)-7(z-27)=0 \Rightarrow 8x+3y-7z+244=0 \\ \Rightarrow ||(-8+4s+2t, 3+s-3t,27+5s+t)||的最小值=d(O,E) ={244\over \sqrt{8^2+3^2+7^2}} = \bbox[red, 2pt]{2\sqrt{122}}$$
解答:
$$\triangle ACQ: \cos \angle Q={4^2+5^2- \overline{AC}^2 \over 2\cdot 4\cdot 5} \Rightarrow {1\over 2}={41-\overline{AC}^2 \over 40} \Rightarrow \overline{AC} = \sqrt{21} \\ \Rightarrow \cos \angle ACQ ={28+25-16\over 2\cdot 2\sqrt 7\cdot 5} = {\sqrt{21} \over 7}\\ 直角 \triangle ABC: \overline{BC} = \sqrt{\overline{AB}^2-\overline{AC}^2} =\sqrt{49-21} =2\sqrt 7 \\ 作\overline{BD} \bot \overline{PC} \Rightarrow \cases{\angle DBC+\angle BCD =90^\circ \\ \angle ACQ+\angle BCD=90^\circ} \Rightarrow \angle DBC=\angle ACQ=\theta \\ \Rightarrow \cos \theta={\sqrt{21} \over 7} ={\overline{BD} \over \overline{BC}=2\sqrt 7} \Rightarrow \overline{BD} =2\sqrt 3 \Rightarrow \overline{CD} =\sqrt{28-12}=4 \\\Rightarrow \tan \angle P= \sqrt 3={2\sqrt 3\over \overline{PD}} \Rightarrow \overline{PD}=2 \Rightarrow \overline{PC} =\overline{PD}+\overline{CD}=2+4= \bbox[red, 2pt]6$$
解答:$$(g(x))^2=f(x)Q(x)+a\cdot g(x) \\ \deg(g)=2 \Rightarrow \cases{\deg(g^2) =4\\\deg(a\cdot g) =2} \Rightarrow \deg(f)\ge 3,又f(x)=0無解 \Rightarrow \deg(f)=4 \Rightarrow Q(x)=c(常數) \\ \Rightarrow g(x)(g(x)-a) =c\cdot f(x) \Rightarrow \cases{g(x)=0無實數解\\ g(x)-a=0 無實數解} \\ 由於\cases{g(x)首項係數為正的二次式\\ y=g(x)頂點y坐標為9} \Rightarrow y=g(x)\ge 9且圖形凹向上\Rightarrow g(x)=0無解 \\ \Rightarrow g(x)-a=0也無實數解的條件為 9-a\gt 0 \Rightarrow 0\lt a\lt 9, a\in \mathbb N \\ \Rightarrow a=1,2,\dots,8 \Rightarrow 1+2+\cdots +8=\bbox[red, 2pt]{36}$$
解答:$$\cases{A(1,-2,-1) \\B(3,1,0)} \Rightarrow \cases{C=\overline{AB}中點=(2,-1/2,-1/2) \\\overline{AB} =\sqrt{14}} \\ 在\triangle PAB中，依中線定理:d=\overline{PA}^2+\overline{PB}^2= 2\overline{PC}^2+ {1\over 2} \overline{AB}^2 = 2\overline{PC}^2+ 7 \\ d要最小值代表\overline{PC}要最小\Rightarrow P為C在E上的投影點\\平面E:2x-y-2z-1=0的法向量\vec n=(2,-1,-2) \\\Rightarrow 過C且方向向量為\vec n的直線\{(2+2t,-1/2-t, -1/2-2t) \mid t\in \mathbb R\}代入E \Rightarrow t=-{1\over 2}\\ \Rightarrow P(x_0,y_0,z_0) =(1,0,{1\over 2}) \Rightarrow \overline{PC}^2 =1+{1\over 4}+1={9\over 4} \\ \Rightarrow M=2\overline{PC}^2+7={23\over 2} \Rightarrow (M,x_0,y_0,z_0) = \bbox[red, 2pt]{\left( {23\over 2},1,0,{1\over 2} \right)}$$
Ⅱ .填充二(每格 6 分， 共 48 分， 每格全對才給分)
解答:$$取\cases{u=2x+y\\ v=x-3y} \Rightarrow \left| {\partial(u,v) \over \partial(x,y)}\right| = \begin{Vmatrix} 2& 1\\1& -3 \end{Vmatrix} =7 \\ |u|+|v|\le 8 面積=4\times \left( {1\over 2}\cdot 8\cdot 8 \right) =128 \Rightarrow |2x+y|+ |x-3y|\le 8面積= \bbox[red, 2pt]{128\over 7}$$
解答:$$\textbf{Case I }\sin \theta\gt \cos \theta : \lim_{n\to \infty} {4\sin^n\theta-5\cos^n\theta\over 3\sin^n \theta+7\cos^n\theta} =\lim_{n\to \infty} {4 -5(\cos \theta /\sin \theta)^n \over 3 +7(\cos\theta/\sin \theta)^n} = {4\over 3} \\\textbf{Case II }\sin \theta\lt \cos \theta : \lim_{n\to \infty} {4\sin^n\theta-5\cos^n\theta\over 3\sin^n \theta+7\cos^n\theta} =\lim_{n\to \infty} {4 (\sin \theta/\cos\theta)^n -5 \over 3(\sin \theta/\cos\theta)^n +7 } = -{5\over 7} \\ \textbf{Case III }\sin \theta= \cos \theta : \lim_{n\to \infty} {4\sin^n\theta-5\cos^n\theta\over 3\sin^n \theta+7\cos^n\theta} =\lim_{n\to \infty} {4\sin^n\theta-5\sin^n\theta\over 3\sin^n \theta+7\sin^n\theta} =\lim_{n\to \infty} {-\sin^n\theta\over 10\sin^n \theta } =-{1\over 10} \\ \Rightarrow 所有可能的值為\bbox[red, 2pt]{{4\over 3},-{5\over 7},-{1\over 10}}$$
解答:
$$\Gamma \xrightarrow{繞原點旋轉\theta } \Gamma_1 \Rightarrow \Gamma與\Gamma_1對稱於兩者的對稱軸的角平分線上\\ \Rightarrow 對稱軸的斜率 m=\tan {\theta\over 2} = \sqrt{1-\cos \theta\over 1+\cos \theta} = \sqrt{1-3/5\over 1+3/5} = {1\over 2} \Rightarrow L_1:y={1\over 2}x \\ \Rightarrow 另一條對稱軸L_2\bot L_1 \Rightarrow L_2:y=-2x \Rightarrow \cases{y=x/2代入\Gamma \Rightarrow A(12/5,6/5)\\ y=-2x代入\Gamma \Rightarrow B(-3/\sqrt{10},6/\sqrt{10})} \\ \Rightarrow \cases{\overline{OA} = 6/\sqrt 5 \\ \overline{OB}= 3/\sqrt 2} \Rightarrow 菱形ABCD面積= 4\times {1\over 2}\cdot \overline{OA}\cdot \overline{OB} = \bbox[red, 2pt]{{18\over 5}\sqrt{10}}$$
解答:$$\int_0^1 {x^3+2x^2+3x+4\over (x^2+1) (x^2+2x+3)} \,dx = \int_0^1 \left( {1\over x^2+1} + {x+1\over x^2+2x+3} \right) \,dx \\= \left. \left[ \tan^{-1}x +{1\over 2} \ln(x^2+2x+3) \right] \right|_0^1 ={\pi\over 4}+{1\over 2}\ln 6-{1\over 2}\ln 3= \bbox[red, 2pt]{{\pi\over 4}+{1\over 2}\ln 2}$$
解答:$$假設a+b=k \in \mathbb Z \Rightarrow b=k-a \Rightarrow {k\over a^2+a(k-a)+(k-a)^2} ={3\over 32} \\ \Rightarrow 3a^2-3ka+(3k^2-32k) =0 \Rightarrow \Delta=(-3k)^2-12(3k^2-32)\ge 0 \Rightarrow 9k^2-128k\le 0 \\ \Rightarrow k(9k-128)\le 0 \Rightarrow 0\le k\le {128\over 9} \approx 14.2 \Rightarrow 最大整數k= \bbox[red, 2pt]{14}$$
解答:
$$取\cases{A(-a,h) \\B(-b,0)\\C(b,0) \\D(a,h)} \Rightarrow \cases{M=(A+B)/2= (-(a+b)/2,h/2) \\N=(C+D)/2 =((a+b)/2,h/2)} \Rightarrow \cases{\overrightarrow{MN} =(a+b,0) \\ \overrightarrow{AC}=(a+b,-h)} \\\Rightarrow \overrightarrow{AC} \cdot \overrightarrow{MN} =(a+b)^2=13 \Rightarrow \overline{MN}=a+b=\sqrt{13} \Rightarrow \overline{AC}^2= 7^2 =(a+b)^2+h^2 \\ \Rightarrow 49=13+h^2 \Rightarrow h=6 \Rightarrow 等腰梯形面積=\overline{MN} \times h= \bbox[red, 2pt]{6\sqrt{13}}$$
解答:$$取\cases{y=x^2\\ t=\log_3 k} \Rightarrow y^2+2\sqrt 3ty+2-t^2=0 \Rightarrow \cases{判別式\Delta \gt 0\\ 兩根之和\gt 0\\ 兩根之積\gt 0} \Rightarrow \cases{t^2\gt 1/2 \Rightarrow t\lt -\sqrt 2/2或t\gt \sqrt 2/2\\ -2\sqrt 3 t\gt 0 \Rightarrow t\lt 0\\ 2-t^2\gt 0 \Rightarrow -\sqrt 2\lt t\lt \sqrt 2} \\ \Rightarrow 三條件取交集: -\sqrt 2\lt t\lt -{\sqrt 2\over 2} \Rightarrow -\sqrt 2 \lt \log_3 k\lt -{\sqrt 2\over 2} \Rightarrow \bbox[red, 2pt]{3^{-\sqrt 2} \lt k\lt 3^{-\sqrt 2/2}}$$
解答:$$k=C^n_2 ={n(n-1)\over 2} \gt 2026 \Rightarrow n(n-1)=4052 \Rightarrow \cases{n=64: 64\times 63=4032 \lt 4052\\ n=65:65\times 64=4160\gt 4052} \\ \Rightarrow n=\bbox[red, 2pt]{65}$$
二、 計算證明題： (每題 10 分， 共 20 分， 須詳列計算過程或說明理由。 )
解答:$$想辧法把\; a_n= 2a_{n-1}+n 改寫成 a_n+ \alpha n+ \beta =2(a_{n-1}+ \alpha(n-1)+ \beta)\\ \Rightarrow a_n=2a_{n-1}+\alpha n+(\beta-2\alpha) \Rightarrow \cases{\alpha =1\\ \beta =2} \Rightarrow a_n+n +2=2(a_{n-1}+ (n-1)+2) \\ 取b_n=a_n+n+2 \Rightarrow b_n=2b_{n-1} =b_1\cdot 2^{n-1} =4\cdot 2^{n-1}\; (b_1=a_1+1+2=4) \\ \Rightarrow b_n=2^{n+1} \Rightarrow a_n+n+2=2^{n+1} \Rightarrow \bbox[red, 2pt]{a_n=2^{n+1}-n-2}$$
解答:$$\cases{|\alpha|=1\\ |\beta|=1\\ |\gamma|=1} \Rightarrow \cases{\bar \alpha=1/\alpha\\ \bar \beta =1/\beta\\ \bar \gamma= 1/\gamma} \Rightarrow z={(\alpha+\beta) (\beta+\gamma) (\gamma+\alpha) \over \alpha \beta \gamma} \Rightarrow \bar z= {(\bar\alpha+ \bar\beta) (\bar\beta+ \bar\gamma) ( \bar\gamma+ \bar\alpha) \over \bar \alpha \bar\beta \bar\gamma} \\={(1/\alpha+ 1/\beta)( 1/\beta +1/\gamma) (1/\gamma+1/\alpha) \over 1/\alpha\beta \gamma} ={(\alpha+\beta) (\beta+\gamma) (\gamma+\alpha) /\alpha^2 \beta^2 \gamma^2\over 1/\alpha \beta \gamma}\\ ={(\alpha+\beta) (\beta+\gamma) (\gamma+\alpha) \over \alpha \beta \gamma} =z \Rightarrow \bar z=z \Rightarrow z\in \mathbb R\; \bbox[red, 2pt]{故得證}$$
====================== END ==========================
解題僅供參考，其他教甄試題及詳解
沒有留言:
張貼留言