臺北市立南港高級中學 115 學年度正式教師甄試
1~13 題每題 7 分, 第 14 題每小題 3 分, 滿分 100 分
解答:$$\Gamma:y=f(x)={x^3\over 3}-ax \Rightarrow f'(x)=x^2-a \Rightarrow f'(0)=-a \Rightarrow 過原點的L:y=-ax \\ M也過原點且M\bot L \Rightarrow M:y={1\over a}x \Rightarrow 求\Gamma 與 M的交點 \Rightarrow {x^3\over 3}-ax={1\over a}x \\ \Rightarrow x \left( {x^3\over 3}-(a+{1\over a}) \right)=0 \Rightarrow x=0, \pm \sqrt{3(a+{1\over a})} \\ 由於\Gamma,M都對稱原點(奇函數) \Rightarrow 所圍面積S=2 \int_0^{\sqrt{3(a+1/a)}} \left[{1\over a}x- \left( {x^3\over 3}-ax \right) \right]\,dx \\=2 \left. \left[ \left( a+{1\over a} \right){x^2\over 2}-{x^4 \over 12} \right] \right|_0^{\sqrt{3(a+1/a)}} ={3\over 2} \left( a+{1\over a} \right)^2 \ge {3\over 2} \left( 2\sqrt{a\cdot {1\over a}} \right)^2 = \bbox[red, 2pt]6$$
解答:$$M= I-{\vec n \vec n^T\over \vec n^T \vec n}, \vec n為x+y+z=0的法向量 \begin{bmatrix}1\\1\\1 \end{bmatrix} \\ \Rightarrow M= \begin{bmatrix}1&0& 0\\0& 1& 0\\ 0& 0& 1 \end{bmatrix}-{\begin{bmatrix}1\\1\\1 \end{bmatrix} \begin{bmatrix}1& 1& 1 \end{bmatrix} \over \begin{bmatrix}1& 1& 1 \end{bmatrix} \begin{bmatrix}1\\1\\1 \end{bmatrix}} =\begin{bmatrix}1&0& 0\\0& 1& 0\\ 0& 0& 1 \end{bmatrix}-{ \begin{bmatrix}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix} \over 3} = \bbox[red, 2pt]{{1\over 3} \begin{bmatrix}2&-1& -1\\-1& 2& -1\\ -1&-1& 2 \end{bmatrix}}$$
解答:$$\overline{BD}=a \Rightarrow 餘弦定理:\cases{\cos A=(5^2+1^2-a^2)/(2\cdot 5\cdot 1) \\ \cos C=(7^2+5^2-a^2)/(2\cdot 7\cdot 5)} \Rightarrow \cases{a^2=26-10\cos A\\ a^2=74-70\cos C} \\A+C=180^\circ \Rightarrow \cos A=-\cos C \Rightarrow 26-10\cos A=74+70\cos A \Rightarrow \cos A=-{3\over 5} \\ \Rightarrow \overrightarrow{AB} \cdot \overrightarrow{AD} =\overline{AB}\cdot \overline{AD} \cdot \cos A=-3 \\\overline{AC}=b ,同理可得\cases{b^2=50-14\cos B\\ b^2=50+50\cos B} \Rightarrow \cos B=0 \Rightarrow \angle B=\angle D=90^\circ \Rightarrow \overline{AC}^2= 1^2+7^2=50 \\ 又\cases{|\overrightarrow{BC}|^2 = |\overrightarrow{AC}- \overrightarrow{AB}|^2 = (\overrightarrow{AC}-\overrightarrow{AB}) \cdot (\overrightarrow{AC}-\overrightarrow{AB}) \\ |\overrightarrow{CD}|^2 = |\overrightarrow{AC}-\overrightarrow{AD}|^2 = (\overrightarrow{AC}- \overrightarrow{AD}) \cdot (\overrightarrow{AC}-\overrightarrow{AD})} \Rightarrow \cases{7^2 =50+1^2-2 \overrightarrow{AC} \cdot \overrightarrow{AB} \\ 5^2= 50+5^2-2\overrightarrow{AC}\cdot \overrightarrow{AD}} \\ \Rightarrow \cases{\overrightarrow{AC} \cdot \overrightarrow{AB} =1 =(\alpha \overrightarrow{AB}+ \beta \overrightarrow{AD}) \cdot \overrightarrow{AB}\\ \overrightarrow{AC}\cdot \overrightarrow{AD}= 25 =(\alpha \overrightarrow{AB}+ \beta \overrightarrow{AD}) \cdot \overrightarrow{AD}} \Rightarrow \cases{\alpha-3b =1\\ -3\alpha+25\beta=25} \Rightarrow (\alpha,\beta) = \bbox[red, 2pt]{ \left( {25\over 4},{7\over 4} \right)}$$
解答:$$$$
解答:
$$此題相當於求兩圖形\cases{\Gamma: y=f(x)= |\log_3 x| \\ L: y=ax+b} 的相異三交點 \\ 假設三交點的x坐標為t,3t,9t \Rightarrow t\lt 1\lt 3t\lt 9t \Rightarrow \cases{-\log_3 t=at+b \\ \log_3(3t)=a(3t)+b\\ \log_3(9t)=a(9t)+b} \Rightarrow at={1\over 6}\\ \Rightarrow b={1\over 2}+\log_3 t \Rightarrow -\log_3 t={1\over 6}+ \left( {1\over 2}+\log_3 t \right) \Rightarrow -2\log_3 t={2\over 3} \Rightarrow t=3^{-1/3} \\ \Rightarrow 三解:\bbox[red, 2pt]{3^{-1/3},3^{2/3},3^{5/3}}$$
解答:$$\angle B=\pi-{\pi\over 8}-{\pi\over 4}={5\over 8} \pi \Rightarrow {8\over \sin(\pi/4)} = {\overline{AC} \over \sin(5\pi /8)} \Rightarrow \overline{AC}= 8\sqrt 2 \sin{5\pi\over 8} \\ \Rightarrow \triangle ABC ={1\over 2} \cdot \overline{AB}\cdot \overline{AC} \sin \angle A ={1\over 2}\cdot 8\cdot 8\sqrt 2 \sin{5\pi\over 8} \sin {\pi\over 8} =32\sqrt 2\cdot (-{1\over 2}) \left( \cos{3\pi\over 4}-\cos {4\pi\over 8} \right) \\=-16\sqrt 2 \left( -{1\over \sqrt 2}-0 \right) = \bbox[red, 2pt]{16}$$
解答:$$\cases{\vec a=\overrightarrow{AB} \\\vec b=\overrightarrow{AD} \\\vec c=\overrightarrow{AE} } \Rightarrow |\vec a|= |\vec b|= |\vec c|=3 \Rightarrow \cases{ \vec a\cdot \vec b=|\vec a||\vec b|\cos 60^\circ =3\times 3\times {1\over 2} ={9\over 2} \\ \vec a\cdot \vec c = |\vec a||\vec c| \cos 60^\circ= 9/2\\ \vec b\cdot \vec c=9/2} \\ \overrightarrow{AP} = \overrightarrow{AE}+ \overrightarrow{EH}+ \overrightarrow{HP}= \vec c+\vec b+{1\over 3} \vec a \Rightarrow |\overrightarrow{AP}|^2 = \left|{1\over 3}\vec a+ \vec b+ \vec c \right|^2 \\={1\over 9}|\vec a|^2+ |\vec b|^2 +|\vec c|^2+ 2 \left( {1\over 3}\vec a\cdot \vec b +{1\over 3}\vec a\cdot \vec c+ \vec b\cdot \vec c \right) =1+9+9+3+3+9=34 \Rightarrow \overline{AP}= \bbox[red, 2pt]{\sqrt{34}}$$
解答:$$\cases{\sin 15^\circ =\sin(45^\circ-30^\circ)=(\sqrt 6-\sqrt 2)/4\\ \cos 15^\circ=\cos(45^\circ-30^\circ)= (\sqrt 6+\sqrt 2)/4} \Rightarrow \cases{\csc 15^\circ=\sqrt 6+\sqrt 2\\ \cot 15^\circ =2+\sqrt 3} \\ \cot (\theta/2) ={\cos(\theta/2) \over \sin(\theta/2)} ={ 2\cos^2(\theta/2) \over 2\sin(\theta/2) \cos (\theta/2)} = {\cos \theta+1\over \sin \theta} =\cot \theta+\csc\theta\\ \theta={\pi \over 12}=15^\circ 代入上式 \Rightarrow \cot {\pi\over 24}=\cot 15^\circ + \csc 15^\circ = \bbox[red, 2pt]{2+ \sqrt 2+ \sqrt 3+ \sqrt 6}$$
解答:$$P(t,t^2/4) \in \Gamma:x^2=4y \Rightarrow \overline{OP} =\sqrt{t^2+(t^2/4)^2} =t\sqrt{1+t^2/16} \\ Q在x軸上且\overline{OQ} =\overline{OP} \Rightarrow Q= \left( t\sqrt{1+t^2/16},0 \right) \Rightarrow \overline{PQ}斜率={t\over 4 \left( 1-\sqrt{1+t^2/16} \right)} \\ \Rightarrow 過Q的\overline{PQ}方程式: y= {t\over 4 \left( 1-\sqrt{1+t^2/16} \right)}(x- t\sqrt{1+t^2/16}) \\ \Rightarrow y(0)= {t\over 4 \left( 1-\sqrt{1+t^2/16} \right)}( - t\sqrt{1+t^2/16}) ={t^2 \sqrt{1+t^2/16} \over 4(\sqrt{1+t^2/16}-1)} \\ \Rightarrow R \left( 0,{t^2 \sqrt{1+t^2/16} \over 4(\sqrt{1+t^2/16}-1)} \right) \Rightarrow \lim_{t\to 0^+} {t^2 \sqrt{1+t^2/16} \over 4(\sqrt{1+t^2/16}-1)} =\lim_{u\to 1} {16(u^2-1)u\over 4(u-1)} \;(u=\sqrt{1+t^2/16}) \\= \lim_{u \to 1} 4(u+1)u=8 \Rightarrow R= \bbox[red, 2pt]{(0,8)}$$
解答:$$1-{1\over 1+t}={t\over 1+t} \Rightarrow f \left( {t\over 1+t} \right)+f \left( {1+t\over t} \right) \log|1+t| =f \left( 1+t\over t \right)\log|t|+ 2026 \\ \Rightarrow f \left( {t\over 1+t} \right)=f \left( 1+t\over t \right) (\log|t|-\log|1+t|)+2026 =f \left( 1+t\over t \right) \log \left|{t\over 1+t} \right| +2026 \\ \Rightarrow f(x)=f({1\over x}) \log x+2026 \Rightarrow f({1\over x}) =f(x)\log{1\over x}+2026 =-f(x)\log x+2026 \\ \Rightarrow f(x) = [-f(x)\log x+2026] \log x+2026 \Rightarrow f(x)={2026(\log x+1) \over 1+(\log x)^2} \\ \Rightarrow f(1000)={2026(3+1) \over 1+3^2} = {8104\over 10} = \bbox[red, 2pt]{810.4}$$
解答:$$假設n次丟擲中,出現k次正面 \Rightarrow 獎金=1+2+3+ \cdots+k={k(k+1)\over 2} ={n(n-1) \over 2} \\ \Rightarrow k=n-1 \Rightarrow n次丟擲中,出現n-1次正面,也就是只有反面一次 \\ \Rightarrow 機率= C^n_{n-1} p^{n-1}(1-p) =np^{n-1}(1-p) = \bbox[red, 2pt]{n(p^{n-1}-p^n)}$$
解答:$$假設\cases{圓心O(0,0)\\ 圓半徑r \\ \overline{AB}在x軸上\\ L:x=k \\P(x_1,y_1)} \Rightarrow\cases{A(-r,0) \\B(r,0) \\ C(k,0) \\ x_1^2+y_1^2=r^2} \Rightarrow D(k,y_1) \Rightarrow \ell_1=\overline{PD}=k-x_1 \Rightarrow x_1=k-\ell_1 \\ \Rightarrow \ell_1^2=\overline{BP}^2=(x_1-r)^2+y_1^2=x_1^2+y_1^2-2rx_1+r^2 =r^2-2rx_1+r^2= 2r(r-x_1) =2r(r-(k-\ell_1)) \\ \Rightarrow \ell_1^2-2r\ell_1+2r(k-r)=0\\ 同理,Q(x_2,y_2) \Rightarrow \ell_2=\overline{BQ}=\overline{QE} \Rightarrow \ell_2^2 -2r\ell_2+2r(k-r)=0\\ 也就是說\ell_1,\ell_2 是x^2-2rx+2r(k-r)=0的兩根\Rightarrow 兩根之和:\ell_1+\ell_2 =2r \\ \Rightarrow \overline{BP}+ \overline{BQ}=2r=\overline{AB} \Rightarrow \overline{BP}+ \overline{BQ}= \overline{AB}, \bbox[red, 2pt]{故得證}$$
解答:$$\textbf{(1) } \cases{M \begin{bmatrix}1\\ 0 \end{bmatrix} = \begin{bmatrix}1\\ \sqrt 2 \end{bmatrix} \\[1ex] M \begin{bmatrix}0\\1 \end{bmatrix} = \begin{bmatrix}-1\\ \sqrt 2 \end{bmatrix}} \Rightarrow M= \bbox[red, 2pt]{\begin{bmatrix}1& -1\\\sqrt 2& \sqrt 2 \end{bmatrix}} \\\textbf{(2) }假設\cases{\vec a= \overrightarrow{OA} \\\vec b= \overrightarrow{OB} \\\vec c= \overrightarrow{OC} \\ \triangle ABC的重心G} 及\cases{\vec a'= \overrightarrow{OA'} =M(\vec a)\\\vec b'= \overrightarrow{OB'} = M(\vec b)\\\vec c'= \overrightarrow{OC'} =M(\vec c)}\Rightarrow \vec g=\overrightarrow{OG} ={1\over 3}(\vec a+\vec b+\vec c) \\\Rightarrow M(\vec g) ={1\over 3}[M(\vec a) +M(\vec b)+ M(\vec c)] ={1\over 3}[\vec a' + \vec b'+ \vec c'] = \triangle A'B'C'的重心向量 \\\Rightarrow 變換M將\triangle ABC的重心映射至\triangle A'B'C'的重心, \bbox[red, 2pt]{故得證} \\\textbf{(3) }\det(M)=\sqrt 2+\sqrt 2=2\sqrt 2 \Rightarrow \triangle A'B'C'= 2\sqrt 2\cdot \triangle ABC=6 \sqrt 2\\ 已知\cases{A'(1,\sqrt 2) \\ B'(-1,\sqrt 2)} \Rightarrow \overline{A'B'}=2 \Rightarrow \triangle A'B'C'= {1\over 2}\cdot 2\cdot h=6\sqrt 2 \Rightarrow h= 6\sqrt 2 \\ 由於\overline{A'B'}為一水平線，因此d(C',\overline{A'B'})= \bbox[red, 2pt]{6\sqrt 2}$$
解題僅供參考，其他教甄試題及詳解
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