臺北市立南湖高級中學 115 學年度第 1 次正式教師甄選
壹、 複選題(4 題，每題 5 分，錯一個選項扣 2 分，每題扣完為止)
解答:$$\cases{\sin x-2\lt 0\\ \cos x+3\gt 0} \Rightarrow f(x)=2-\sin x+\cos x+3=5-(\sin x-\cos x)=5-\sqrt 2\sin (x-{\pi\over 4}) \\(A)\times: f(x)=5-\sqrt 2(\sin(x-\alpha) 週期為2\pi\\ (B)\times: f(\pi)=5-(\sin \pi-\cos \pi)=5-(0+1)=4 \not \gt 4 \\(C)\times: \cases{f(\pi/3+x)=5-\sqrt 2\sin(x+\pi/12) \\f(\pi/3-x)=5-\sqrt 2\sin(-x+\pi/12} \Rightarrow f(\pi/3+x)\ne f(\pi/3-x) \\(D)\bigcirc: f(x)=5-\sqrt 2\sin(x-\pi/4)\Rightarrow f(x)最大值=5+\sqrt 2 \\ (E)\bigcirc: f(x)最小值=5-\sqrt 2\gt 3 \Rightarrow 與y=3沒有交點\\故選\bbox[red, 2pt]{(DE)}$$
解答:$$(A)\times: f(t)={確診且顯示有抗體\over 確診且顯示有抗體+無確診且顯示有抗體} ={0.4t\over 0.4t+0.6\times 0.06} \\(B)\bigcirc: f(t)={0.4t\over 0.4t+0.036} =1-{0.036\over 0.4t+0.036} 遞增,t\ge 0 \\(C)\bigcirc: f(t)\ge 0.9 \Rightarrow 1-{0.036\over 0.4t+0.036}\ge 0.9 \Rightarrow 0.1\ge {0.036\over 0.4t+0.036} \Rightarrow 0.04t+0.0036\ge 0.036 \\\qquad \Rightarrow t\ge {0.0324\over 0.04}=0.81 \\(D)\times:t=0.81 \Rightarrow \cases{P(確診顯示有抗體)=0.4(1-t)=0.076 \\P(未確診顯示有抗體)=0.6\times 0.06=0.036} \Rightarrow 合計:0.112\gt 0.1 \\(E)\times: \cases{P(確診顯示有抗體)= 0.4t=0.324\\ P(未確診顯示無抗體)=0.6\times 0.94=0.564} \Rightarrow 合計:0.888 \Rightarrow {0.888\over 0.112} \lt 8 \\ 故選\bbox[red, 2pt]{(BC)}$$
解答:$$f(y)=(1+y)^{21} =C^{21}_0y^{21}+ C^{21}_1 y^{20}+ \cdots+C^{21}_{20}y +C^{21}_{21}y^0 \\ (A)\times: x=f(\sqrt 5-1)=(\sqrt 5)^{21}為無理數\\ (B)\times: x=10^a= (\sqrt 5)^{21}\Rightarrow a=\log (\sqrt 5)^{21} ={21\over 2}\log 5 \ne {21\over 2}\\ (C)\bigcirc: 由(B)可知:x=10^{10.5\log 5}\gt 10^{10\log 5} \\(D) \bigcirc:\log x=10.5\log 5 \approx10.5\cdot (1-0.301)=7.3395 \Rightarrow 八位數 \\(E)\times: x=10^{7.3395} =10^{0.3395}\times 10^7,由於\cases{\log2=0.301\\ \log 2.1=\log3+\log 7-1=0.3221 } \\ \Rightarrow \log 2.1\lt 0.3395 \Rightarrow x最高兩位數字至少是21\\ 故選\bbox[red, 2pt]{(CD)}$$
貳、 填充題(10 題，每題 5 分， 每題須完全答對才給分，答錯不倒扣。各題答案若非整數，請以最簡分數或最簡分式作答)
解答:$$假設\cases{紅球r顆\\ 白球w顆} \Rightarrow \cases{r+w=16\\ r\ge 2\\ w\ge 2} \Rightarrow 2\le r\le 14\\抽出 2 球皆為紅球的機率大於{1\over 4} \Rightarrow {{r\choose 2} \over {16\choose 2}} \gt {1\over 4} \Rightarrow {r(r-1)/2\over 120} \gt {1\over 4} \Rightarrow r(r-1)\gt 60 \\ \Rightarrow r\ge 9 \Rightarrow 9\le r\le 14\\ 抽3球同時包含紅球與白球:\cases{2紅1白: {r\choose 2} \cdot {16-r\choose 1}\\ 1紅2白:{r\choose 1} \cdot {16-r\choose 2}} \Rightarrow 合計:7r(16-r) \\\Rightarrow 機率={7r(16-r) \over {16\choose 3}} ={r(16-r)\over 80}={-(r-8)^2+64\over 80}\Rightarrow r=8有最大值,但9\le r\le 14 \\ 取r=9 \Rightarrow 機率={9(16-9)\over 80} = \bbox[red, 2pt]{63\over 80}$$
解答:$$假設矩陣S= \begin{bmatrix}2& 1/2\\1 & 5/2 \end{bmatrix} \Rightarrow \begin{bmatrix}1& 1 \end{bmatrix} S= \begin{bmatrix}3 & 3 \end{bmatrix} =3 \begin{bmatrix}1& 1 \end{bmatrix} \Rightarrow \begin{bmatrix}1& 1 \end{bmatrix} S^5= 3\begin{bmatrix}1& 1 \end{bmatrix} S^4 =3^2\begin{bmatrix}1& 1 \end{bmatrix} S^3\\ = \cdots=3^5 \begin{bmatrix}1& 1 \end{bmatrix}; 又矩陣T= \begin{bmatrix}1& 2& 3\\ 3& 2& 1 \end{bmatrix} \Rightarrow T \begin{bmatrix}1\\1\\1 \end{bmatrix} = \begin{bmatrix}6\\6 \end{bmatrix} =6 \begin{bmatrix}1\\ 1 \end{bmatrix}\\矩陣A的元素總和= \begin{bmatrix}1& 1 \end{bmatrix} A \begin{bmatrix}1\\1\\1 \end{bmatrix}= \begin{bmatrix}1& 1 \end{bmatrix} S^5T \begin{bmatrix}1\\1\\1 \end{bmatrix} =3^5 \begin{bmatrix}1& 1 \end{bmatrix} \cdot 6 \begin{bmatrix}1\\1 \end{bmatrix} =3^5\cdot 6\cdot 2= \bbox[red, 2pt]{2916}$$
解答:$$假設L: {x\over a}+{y\over b}=1 \Rightarrow \cases{A(a,0) \\B(0,b)},其中\;a,b\gt 0,又L過(3,2) \Rightarrow {3\over a}+{2\over b}=1 \\ 柯西不等式 (3a+8b) \left( {3\over a}+{2\over b} \right) \ge \left( \sqrt 9+\sqrt {16} \right)^2 \Rightarrow 3a+8b\ge 49\\ \Rightarrow \cases{\overrightarrow{AB}=(-a,b) \\ 8x+3y+12=0的方向向量\vec v=(3,-8)} \Rightarrow \overline{CD}=\overrightarrow{AB}在\vec v上的正射影長 \\ \Rightarrow \overline{CD}={|\overrightarrow{AB}\cdot \vec v|\over |\vec v|} ={|3a+8b|\over \sqrt{73}} \ge { {49} \over \sqrt{73}} =\bbox[red, 2pt]{49\sqrt{73}\over 73}$$
解答:$$假設正四面體的稜長\ell,且\cases{\vec a=\overrightarrow{OA} \\\vec b=\overrightarrow{OB} \\\vec c=\overrightarrow{OC} } \Rightarrow |\vec a|=|\vec b|=|\vec c|=\ell\\ 又正四面體兩向量夾角為60^\circ \Rightarrow \vec a\cdot \vec b=\vec b\cdot \vec c= \vec c\cdot \vec a=\ell\cdot \ell \cos 60^\circ ={1\over 2}\ell^2\\ \cases{\overline{DA}:\overline{DO}=1:3 \Rightarrow \overrightarrow{OD}={3\over 4}\overrightarrow{OA} ={3\over 4}\vec a \Rightarrow |\overrightarrow{OD}|^2=9\ell^2/16\\ E=\overline{BC}中點\Rightarrow \overrightarrow{OE} ={1\over 2}(\overrightarrow{OB}+ \overrightarrow{OC})={1\over 2}(\vec b+\vec c) \Rightarrow |\overrightarrow{OE}|^2 =12\ell^2/16}\\ \Rightarrow \overrightarrow{OD}\cdot \overrightarrow{OE}= {3\over 4}\vec a\cdot {1\over 2}(\vec b+\vec c) ={3\over 8}(\vec a\cdot \vec b+\vec a\cdot \vec c) =6\ell^2/16 \\ 題意:P為\overline{DE} 上的一點\Rightarrow x+y=1 \Rightarrow y=1-x \Rightarrow \overrightarrow{OP}= x\overrightarrow{OD}+(1-x) \overrightarrow{OE} \\ |\overrightarrow{OP}|有最小值\Rightarrow \overrightarrow{OP} \cdot \overrightarrow{DE}=0 \Rightarrow \overrightarrow{OP}\cdot (\overrightarrow{OE}-\overrightarrow{OD}) =0 \Rightarrow \left( x\overrightarrow{OD}+(1-x)\overrightarrow{OE} \right)\cdot (\overrightarrow{OE}-\overrightarrow{OD}) =0 \\ \Rightarrow -3x+12(1-x)-6(1-x)=0 \Rightarrow x=\bbox[red, 2pt]{2\over 3}$$
解答:
$$f(x)=\sqrt{x^4-x^2-10x+26}-\sqrt{x^4-5x^2+9 } = \sqrt{(x^2-1)^2+(x-5)^2}-\sqrt{(x^2-3)+(x-0)^2} \\ \qquad =\overline{PA}-\overline{PB}\le \overline{AB} =\sqrt{(3-1)^2+(0-5)^2} =\bbox[red, 2pt]{\sqrt{29}},其中\cases{P(x^2,x) \\ A(1,5)\\B(3,0)}$$
解答:$$\cases{A(5,1,12) \\B(4,2,3)\\P(0,y,0)} \Rightarrow \overline{PA}+\overline{PB} =\sqrt{25+(y-1)^2+144}+\sqrt{16+(y-2)^2+9} \\=\sqrt{(y-1)^2+13^2}+\sqrt{(y-2)^2+5^2} =\overline{QC}+\overline{QD},其中\cases{Q(y,0) \\C(1,13) \\D(2,5)}\\ 取D'=D與x軸的對稱點 \Rightarrow D'(2,-5) \Rightarrow \overline{QD}=\overline{QD'} \Rightarrow \overline{QC}+\overline{QD'}要最小 \\\Rightarrow C,Q,D'在一直線 \Rightarrow Q=\overline{CD'}與x軸的交點 \Rightarrow y={31\over 18} \Rightarrow P= \bbox[red, 2pt]{ \left( 0,{31\over 18},0 \right)}$$
解答:$$\cases{g(x)=x^4-2x^3+ 3x^2-7x+7\\ f(x)=x^3-x^2+2x-3} \Rightarrow g(x)= (x-1)f(x)-2x+4 \\ \Rightarrow \cases{g(\alpha)=(\alpha-1)f(\alpha)-2\alpha+4=-2\alpha+4=2(2-\alpha) \\g(\beta)=2(2-\beta) \\g(\gamma)=2(2-\gamma)} \\ \Rightarrow S={1\over g(\alpha)}+ {1\over g(\beta)}+ {1\over g(\gamma)} ={1\over 2(2-\alpha)} +{1\over 2(2-\beta)} +{1\over 2(2-\gamma)} \\\Rightarrow 2S={(2-\beta)(2-\gamma) +(2-\alpha)(2-\gamma) +(2\alpha)(2-\beta) \over (2-\alpha) (2-\beta) (2-\gamma)} ={ 12-4(\alpha+\beta+ \gamma) +(\alpha\beta+ \beta\gamma+ \gamma\alpha)\over (2-\alpha) (2-\beta) (2-\gamma)}\\ \alpha,\beta,\gamma 為f(x)=0的三根 \Rightarrow \cases{\alpha+ \beta+\gamma=1\\ \alpha\beta+ \beta\gamma+ \gamma\alpha=2\\ \alpha\beta\gamma=3 \\f(x)=(x-\alpha) (x-\beta)(x-\gamma)} \\\Rightarrow 2S={12-4\cdot 1+2\over f(2)} ={10\over 5} \Rightarrow S= \bbox[red, 2pt]1$$
解答:$$x-1\lt \lfloor x \rfloor \le x \Rightarrow \cases{ \displaystyle {1\over 2}a-1 \lt \left \lfloor {1\over 2}a \right \rfloor \le {1\over 2}a \\[1ex] \displaystyle {2\over 3}a-1 \lt \left \lfloor {2\over 3}a\right \rfloor \le {2\over 3}a} \\\Rightarrow \left( {1\over 2}a-1 \right)+ \left( {2\over 3}a-1 \right) \lt \left \lfloor {1\over 2}a \right \rfloor+ \left \lfloor {2\over 3}a\right \rfloor \le {1\over 2}a+{2\over 3}a \Rightarrow {7\over 6}a-2\lt a\le {7\over 6}a \\ \Rightarrow \cases{7a/6-2\lt a \Rightarrow a\lt 12\\ a\le 7a/6 \Rightarrow a\ge 0}\Rightarrow a=0,1,2,\dots,10,11 代回原式\\ \Rightarrow \cases{a=11 \Rightarrow \left \lfloor {11\over 2} \right \rfloor+ \left \lfloor {22\over 3} \right \rfloor=12 \ne 11\\ a=10 \Rightarrow \left \lfloor {10\over 2} \right \rfloor+ \left \lfloor {20\over 3} \right \rfloor= 11 \ne 10\\ \cdots\\a=7 \Rightarrow \left \lfloor {7\over 2} \right \rfloor+ \left \lfloor {14\over 3} \right \rfloor= 7=a} \Rightarrow a的最大值\bbox[red, 2pt]7$$
解答:$$面積A=\int_1^4 \sqrt{3-\sqrt x}\,dx =\int_1^2 \sqrt{3-u}\cdot 2u\,du = 2\int_1^2 u\sqrt{3-u}\,du \;(取u=\sqrt x) \\=2\int_2^1 (3-v)\sqrt v(-dv) =2\int_1^2(3-v)\sqrt v\,dv\; (取v=3-u) \\=2\int_1^2 \left( 3\sqrt v-v^{3/2} \right)\,dv = 2 \left. \left[ 2v^{3/2} -{2\over 5}v^{5/2} \right] \right|_1^2 =2 \left( {12\sqrt 2\over 5}-{8\over 5} \right) = \bbox[red, 2pt]{24\sqrt 2-16\over 5}$$
解答:$$AA^T= \begin{bmatrix}64& \alpha& 24\\ \alpha& 9& \beta\\24& \beta& 36 \end{bmatrix} \Rightarrow \cases{|\vec a|^2=64\\ |\vec b|^2=9\\ |\vec c|^2=36\\ \vec a\cdot \vec c=24\\ \vec a\cdot \vec b=\alpha\\ \vec b\cdot \vec c = \beta} \Rightarrow \vec a\cdot \vec c=|\vec a||\vec c| \cos \theta \Rightarrow \cos \theta={24\over 8\times 6}={1\over 2} \Rightarrow \theta=60^\circ \\ 取\cases{\vec a=(8,0,0)\\ \vec c=(6\cos 60^\circ, 06\sin 60^\circ)=(3,0,3\sqrt 3) \\\vec b=(x,y,z)} \Rightarrow \vec a\cdot (\vec b\times \vec c)= \det \begin{bmatrix}8&0&0\\ x& y& z\\ 3&0& 3\sqrt 3 \end{bmatrix} =24\sqrt 3y=48\sqrt 3 \\ \Rightarrow y=2 \Rightarrow x^2+y^2+z^2=9 \Rightarrow x^2+z^2=5 \Rightarrow \alpha+\beta =(\vec a\cdot \vec b)+(\vec b\cdot \vec c) =8x+3x+3\sqrt 3 z \\=11x+3\sqrt 3z \; \left( 柯西不等式:(11x+ 3\sqrt 3z)^2 \le (11^2+(3\sqrt 3)^2) \cdot (x^2+z^2)=740 \right) \\ \Rightarrow \alpha+\beta \le \sqrt{740}= \bbox[red, 2pt]{2\sqrt{185}}$$
叁、 計算證明題(3 題，每題 10 分)
解答:$$\textbf{(1) }\cases{圓心O(2,4) \\A(6,7)} \Rightarrow 圓半徑r= \overline{AB}=5 \Rightarrow 通過A的直線L:y=m(x-2)+4 \\ \qquad d(O,L) = {|2m-4-6m+7|\over \sqrt{m^2+1}} =\sqrt 5 \Rightarrow (-4m+3)^2=5(m^2+1) \Rightarrow 11m^2-24m+4=0\\ \qquad \Rightarrow (11m-2)(m-2)=0 \Rightarrow 圖形顯\overline{AB}的斜率比\overline{AC}大\Rightarrow \cases{\overline{AB}斜率= \bbox[red, 2pt]{2}\\ \overline{AC}斜率=2/11} \\ \textbf{(2) }d(O,\overline{AB}) =d(O,\overline{AC}) \Rightarrow \overline{AB}=\overline{AC} \Rightarrow \overline{AR}是\overline{BC}的中垂線 \Rightarrow \overline{AP}= \sqrt{ \overline{OA}^2- \overline{OP}^2} =2\sqrt 5\\ \qquad \Rightarrow \overline{AB}=4\sqrt 5 \Rightarrow \cos \angle OAP={2\sqrt 5\over 5} \Rightarrow \overline{AR}= \overline{AB} \cos \angle OAP=8 \\\qquad\Rightarrow \overline{OR}=\overline{AR}-\overline{OA}=8-5=3 \Rightarrow {\overline{AO} \over \overline{OR}} = \bbox[red, 2pt]{5\over 3} \\ \textbf{(3) } \overline{BR}=\overline{AB} \sin \angle OAP=4 \Rightarrow \overline{BC}=8 \Rightarrow \cos C={\overline{CR} \over \overline{AC}} ={1\over \sqrt 5} \\\qquad 經過某段時間後，乙從B向C移動了距離s (0\le s\le 8), 此時乙距離C的長度為8-s\\ \qquad \Rightarrow 甲從C向A移動了\sqrt 5 s \Rightarrow \overline{甲乙}^2= (\sqrt 5s)^2+(8-s)^2-2(\sqrt 5s)(8-s) \cos C\\\qquad =8s^2-32s+64 =8(s-2)^2+32 \Rightarrow \overline{甲乙}最短距離為\sqrt{32}= \bbox[red, 2pt]{4\sqrt 2}$$
解答:$$\textbf{(1) }\triangle ABC面積={1\over 2}bc\sin A= {1\over 2}ac\sin B= {1\over 2}ab\sin C \\\Rightarrow {2\over abc}\cdot {1\over 2}bc\sin A={2\over abc}\cdot {1\over 2}ac\sin B= {2\over abc}\cdot{1\over 2}ab\sin C \Rightarrow {\sin A\over a}={\sin B\over b}={\sin C\over c} \\ \Rightarrow {a\over \sin A}= {b\over \sin B}= {c \over \sin C},\bbox[red, 2pt]{故得證} \\\textbf{(2) } {a\over \sin A}= {b\over \sin B}= {c \over \sin C}=k \\\Rightarrow \cases{a=k\sin A=k\sin(B+C) =k(\sin B\cos C+\sin C\cos B)=b\cos C+c\cos B\\ b=k\sin B=k\sin(A+C)=k(\sin A\cos C+\sin C\cos A)=a\cos C+c\cos A\\ c=k\sin C=k\sin(A+B)=k(\sin A\cos B+\sin B\cos A)=a\cos B+ b\cos A} \\ \Rightarrow \cases{a^2=ab\cos C+ ac\cos B\\ b^2=ab\cos C+bc\cos A\\ c^2=ac\cos B+bc\cos A} \Rightarrow b^2+c^2-a^2=2bc\cos A \\ \Rightarrow a^2=b^2+c^2-2bc\cos A, \bbox[red, 2pt]{故得證} \\\textbf{(3) } \cases{\sin(\alpha+ \beta)= \sin \alpha\cos \beta+ \sin \beta \cos \alpha\\ \sin(\alpha-\beta)=\sin \alpha\cos \beta-\sin \beta \cos \alpha} \Rightarrow 兩式相加:\sin(\alpha+\beta)+ \sin(\alpha-\beta)=2\sin \alpha \cos \beta \\ 取 \cases{x=\alpha+\beta\\ y= \alpha-\beta} \Rightarrow \cases{\alpha=(x+y)/2\\ \beta=(x-y)/2} 代入上式 \Rightarrow \sin(x)+\sin (y)= 2\sin{x+y\over 2} \cos{x-y\over 2} \\ \bbox[red, 2pt]{故得證}$$
解答:$$$$
解題僅供參考，其他教甄試題及詳解
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