114年高雄中學教甄-數學詳解

 高雄市立高雄高級中學 114 學年度正式教師甄選

第一部分： 計算證明題(每題 6 分，共計 30 分

解答:$$取到1號球的機率p={1\over 4} \Rightarrow 取到非1號球的機率=q=1-p={3\over 4}; 令X:取到1號球的次數 \\ (p+q)^{114}= \sum_{k=0}^n {114\choose k} p^kq^{n-k} =P(X=0)+P(X=1)+\cdots +P(X= 114) \\ \Rightarrow (q-p)^{114} =\sum_{k=0}^n {114\choose k} (-p)^kq^{n-k} =P(X=0)-P(X=1)+P(X=2)-\cdots+P(X=114) \\ \Rightarrow (p+q)^{114}-(q-p)^{114}=2 \left( P(X=1)+P(X=3)+ \cdots+P(X=113) \right)\\ \Rightarrow P(X=奇數)={1^{114}-(1/2)^{114} \over 2} = \bbox[red, 2pt]{{1\over 2}-{1\over 2^{115}}}$$

解答:$$\cases{\vec a\times (\vec b+\vec c) =\vec a\times \vec b+ \vec a\times \vec c=(-12,9,8) \\ \vec a\times \vec b=(-10,8,6)} \Rightarrow \vec a\times \vec c =(-12,9,8)-(-10,8,6)=(-2,1,2) \\ 又\cases{\vec a\bot (\vec a\times \vec b) \\ \vec a\bot (\vec a\times \vec c)} \Rightarrow \vec a \parallel [(\vec a\times \vec b)\times (\vec a\times \vec c)] =\vec a \parallel (10,8,6) \Rightarrow \vec a=k(5,4,3)\\ \vec b\bot (\vec a\times \vec b) \Rightarrow \vec b\cdot (\vec a\times \vec b)=0 \Rightarrow -10+16+6t=0 \Rightarrow t=-1 \Rightarrow \vec b=(1,2,-1) \\ \Rightarrow \vec a\times \vec b=(5k,4k,3k) \times (1,2,-1) = (-10k,8k,6k) =(-10,8,6) \Rightarrow k=1 \Rightarrow \vec a= \bbox[red, 2pt]{(5,4,3)}$$

解答:$$\triangle PAB周長= \overline{PA}+ \overline{PB}+ \overline{AB}=2a+2c , 令S={1\over 2}(2a+2c)=a+c \\ \Rightarrow \cases{\tan \displaystyle {\alpha\over 2}= \sqrt{(S-\overline{PA})(S-\overline{AB}) \over S(S-\overline{PB})} \\[1ex] \tan \displaystyle {\beta\over 2}= \sqrt{(S-\overline{PB})(S-\overline{AB}) \over S(S-\overline{PA})} } \Rightarrow \tan{\alpha\over 2} \cdot \tan {\beta\over 2} ={S-\overline{AB} \over S} ={a+c-2c\over a+c} =\bbox[red, 2pt]{a-c\over a+c}$$

解答:$$\sigma^2 ={n_1\sigma_1^2+ n_2\sigma_2^2\over n_1+n_2} +{n_1n_2\over (n_1+n_2)^2}(\mu_1-\mu_2)^2 = A+B,\\\qquad 其中A={n_1\sigma_1^2+ n_2\sigma_2^2\over n_1+n_2},B= {n_1n_2\over (n_1+n_2)^2}(\mu_1-\mu_2)^2;並假設\cases{\sigma_{\min} =\min\{\sigma_1, \sigma_2\}\\\sigma_{\max} =\max\{\sigma_1, \sigma_2\}} \\ \bbox[red, 2pt]{\textbf{Case I }\mu_1=\mu_2 \Rightarrow B=0 \Rightarrow \sigma^2=A \Rightarrow \sigma_{\min} \le \sigma \le \sigma_{\max}} \\ \bbox[red, 2pt]{\textbf{Case II }\mu_1\ne \mu_2 且\mu_1與\mu_2 差距小 \Rightarrow \sigma_{\min} \lt \sigma\le \sigma_{\max}} \\ \bbox[red, 2pt]{\textbf{Case III }\mu_1\ne \mu_2 且\mu_1與\mu_2 差距大 \Rightarrow \sigma \gt \sigma_{\max}}$$

解答:

$$直角\triangle ABD: \overline{AD}^2=\overline{AB}^2+\overline{BD}^2 =24^2 + 7^2 =625 \Rightarrow \overline{AD}=25\\ 假設\overleftrightarrow{BD}與\overleftrightarrow{OA}交於E點 \Rightarrow \overline{AE} 為直徑, 即\overline{BD}為\overline{AE}的中垂線\Rightarrow \overline{DE} =\overline{AD}=25 \\ 直角\triangle ABE: \overline{AE}^2= \overline{AB}^2+ \overline{BE}^2 =24^2+(25+7)^2= 1600 \Rightarrow \overline{AE}=40 \Rightarrow 圓半徑=\overline{OA} =20 \\ \Rightarrow 扇形OAC={1\over 4}圓面積={1\over 4}20^2 \pi= \bbox[red, 2pt]{100\pi}$$

第二部分：計算證明題(每題 7 分，共計 70 分

解答:$$假設A= \begin{bmatrix}{2\over 5}& {1\over 3}\\ {3\over 5} & {2\over 3}\end{bmatrix}, B= \begin{bmatrix}{3\over 5} & {4\over 5} \\ {4\over 5}& -{3\over 5}\end{bmatrix} , C= \begin{bmatrix}{24\over 25}& -{7\over 25}\\ {7\over 25}& {24\over 25} \end{bmatrix}, D = \begin{bmatrix}1& 4\\ 3& 2 \end{bmatrix} \\ \Rightarrow BC= {1\over 5} \begin{bmatrix}4& 3\\ 3& -4 \end{bmatrix} \Rightarrow (BC)^2= {1\over 25} \begin{bmatrix}25&0\\0& 25 \end{bmatrix} =I \Rightarrow (BC)^{114} =I^{57}=I \\ 將A對角化\Rightarrow A= \begin{bmatrix}5& 1\\9& -1 \end{bmatrix} \begin{bmatrix}1& 0\\ 0 &1/15 \end{bmatrix} \begin{bmatrix} 1/14& 1/14\\ 9/14& -5/14 \end{bmatrix} \\\Rightarrow A^{2025} = \begin{bmatrix}5& 1\\9& -1 \end{bmatrix} \begin{bmatrix}1^{2025}& 0\\ 0 &(1/15)^{2025} \end{bmatrix} \begin{bmatrix} 1/14& 1/14\\ 9/14& -5/14 \end{bmatrix} = \begin{bmatrix}5& 1\\9& -1 \end{bmatrix} \begin{bmatrix}1& 0\\ 0 &0 \end{bmatrix} \begin{bmatrix} 1/14& 1/14\\ 9/14& -5/14 \end{bmatrix}  \\= \begin{bmatrix}5 &0\\ 9& 0 \end{bmatrix} \begin{bmatrix} 1/14& 1/14\\ 9/14& -5/14 \end{bmatrix} = \begin{bmatrix} 5/14& 5/14\\ 9/14& 9/14 \end{bmatrix} \\ \Rightarrow A^{2025} (BC)^{114} D= \begin{bmatrix} 5/14& 5/14\\ 9/14& 9/14 \end{bmatrix} \cdot I\cdot \begin{bmatrix}1& 4\\ 3& 2 \end{bmatrix} = \begin{bmatrix}20/14& 30/14\\ 36/14& 54/14 \end{bmatrix} \Rightarrow a+b+c+d = \bbox[red, 2pt]{10}$$

解答:$$\cases{E_1: z=3\\ E_2: x-y+z=6\\ E_3:x+y-z=2} \Rightarrow \cases{E_1法向量\vec n_1=(0,0,1) \\E_2法向量\vec n_2=(1,-1,1) \\ E_e法向量\vec n_3=(1,1,-1) \\ P=E_1\cap E_2\cap E_3 =(4,1,3)}\\ \Rightarrow \cases{\vec n_1\times \vec n_2=(1,1,0) \Rightarrow 取\vec u_3=(1,1,0), |\vec u_3|=\sqrt 2 \\ \vec n_2\times \vec n_3= (0,2,2) \parallel (0,1,1) \Rightarrow 取\vec u_1=(0,1,1), |\vec u_1|=\sqrt 2 \\ \vec n_1\times \vec n_3 =(-1,1,0) \Rightarrow 取\vec u_2=(-1,1,0), |\vec u_2|=\sqrt 2} \Rightarrow \cases{L_1方向向量為\vec u_1\\ L_2方向向量為\vec u_2\\ L_3方向向量為\vec u_3} \\ \overline{PA} =\overline{PB} =\overline{PC} =\sqrt 2 \Rightarrow \cases{A=P\pm u_1 \Rightarrow A可能是A_1(4,2,4), A_2(4,0,2) \\B=P\pm u_2 \Rightarrow B可能是B_1(3,2,3), B_2(5,0,3) \\ C=P\pm u_3 \Rightarrow C可能是C_1(5,2,3), C_2(3,0,3)} \\ 四面體PABC體積V={1\over 6} \left| \overrightarrow{PA} \cdot (\overrightarrow {PB}\times \overrightarrow{PC})\right| ={1\over 6}|(0,1,1) \cdot ((-1,1,0) \times (1,1,0)) =\bbox[red, 2pt]{1\over 3} \\ \Rightarrow 過A,B,C三點的平面共有2\times 2\times 2=\bbox[red, 2pt]8種可能, 8個平面為\bbox[red, 2pt]{\cases{y=0\\ y=2\\ x-z=0\\ x-z-2=0\\ x+z-6=0\\ x+z-8=0\\ y-2z+4=0\\ y-2z+6=0}}$$

解答:$$3k\le 25 \Rightarrow k\le 8.33 \Rightarrow k=1,2,\dots,8; 又{k\over n} \approx 13\% \Rightarrow n \approx7.69k\\ 欲找 誤差e=\left| {k\over n}-13\%\right| \approx 0 \Rightarrow \cases{k=1,取n=8 \Rightarrow e=|{1\over 8}-0.13 |=0.005 \\ k=2 , 取n=15 \Rightarrow e \approx 0.0033\\ k=3,取n=23 \Rightarrow e\approx 0.000435\\ k=4, 取n=31 \Rightarrow e\approx 0.001\\ k=5,取n=38 \Rightarrow e\approx0.0015 \\ k=6 , 取n=54 \Rightarrow e\approx 0.000435\\ k=7,取n=54 \Rightarrow e=0.00037\\ k=8 n取54 \Rightarrow e\approx 0.129} \\ \Rightarrow 最接近13\%的數對(n,k)= \bbox[red, 2pt]{(54, 7)}$$

解答:

$$假設平面E垂直L_2, 也就是L_2投影到E為一個點O, 而L_1投影至E仍為一直線L_1'\\ 假設\cases{A投影至E為A' \\B投影至E為B' \\C投影至E為C' \\ O在L_1'的投影點P } 及\cases{\overline{A'B'}= \overline{B'C'}= k\\ \overline{A'P}= a\\ \overline{OP}=h} \\ \triangle OA'C'中線定理: \overline{OA'}^2+ \overline{OC'}^2 =2(\overline{OB'}^2 + \overline{A'B'}^2) \Rightarrow 1+7=2(3+k^2) \Rightarrow k=1\\ \cases{直角\triangle OPB':3=h^2+(a+1)^2 \\ 直角\triangle OPA': 1=h^2+a^2} \Rightarrow h^2=3-(a+1)^2=1-a^2 \Rightarrow a={1\over 2} \Rightarrow h=\bbox[red, 2pt]{\sqrt 3\over 2}$$

解答:$$f(x)=x^3-px+p^3 =(x-\alpha) (x-\beta)(x-\gamma) \\\Rightarrow f'(x)= 3x^2-p = (x-\alpha) (x-\beta)  + (x-\alpha)  (x-\gamma) +  (x-\beta)(x-\gamma)  \\ \Rightarrow {f'(x)\over f(x)} ={3x^2-p\over x^3-px+p^3} ={1\over x-\alpha}+{1\over x-\beta} +{1\over x-\gamma} \\\Rightarrow {f'(-p)\over f(-p)} ={3p^2-p\over -p^3+p^2+p^3}=3-{1\over p} =- \left( {1\over p+\alpha} +{1\over p +\beta}+{1\over p +\gamma} \right) \\ \Rightarrow {1\over \alpha+p} +{1\over \beta +p}+{1\over \gamma +p}={1\over p}-3 \\ 欲求{\alpha-p\over \alpha+p} +{\beta-p\over \beta +p} +{\gamma-p\over \gamma +p} =1-{2p\over \alpha+p} +1-{2p\over \beta+p}+1-{2p\over \gamma+p}\\ =3-2p \left( {1\over p+\alpha} +{1\over p+\beta}+{1\over p+\gamma} \right) =3-2p \left( {1\over p}-3 \right) =3-2+6p = \bbox[red, 2pt]{6p+1}$$

解答:$$n=\prod_{k=1}^{20}(k!)=(1! \times 2!)\times (3!\times 4!)\times \cdots\times (19!\times 20!) \\=[(1!\times 1!)\times 2] \times [(3!\times 3!)\times 4] \times \cdots \times[(19!\times 19!) \times 20] \\=(1!)^2\times 2\times (3!)^2\times 4\times \cdots\times (19!)^2\times 20 =k^2 \times(2\cdot 4\cdot 6\cdots20) \\=k^2\times 2^{10}\times 10! =(k\times 2^5)^2\times (2\cdot 3\cdot 4\cdot 5\cdot 6)\times 7 \times(8\cdot 9\cdot 10) \\=(k\times 2^5)^2\times 720^2\times 7 \Rightarrow m= \bbox[red, 2pt]7$$

解答:$$4z_1^2 +5z_2^2 +4z_3^2 =4z_1z_2 + {\color{blue} 6}z_2z_3 + 4z_3z_1 應修訂為4z_1^2 +5z_2^2 +4z_3^2 =4z_1z_2 + {\color{blue} 5}z_2z_3 + 4z_3z_1\\ 假設頂點z_3為原點 \Rightarrow 4z_1^2 +5z_2^2=4z_1z_2 \Rightarrow z_1={1\pm 2i\over 2}z_2\\ 計算三角形三邊長:\cases{c= |z_2-z_3|=|z_2|=k\\ b=|z_1-z_3| =|z_1|= \left| {1\pm 2i\over 2}\right||z_2|= {\sqrt 5\over 2}k\\ a=|z_1-z_2 |= \left|{1\pm 2i\over 2}z_2-z_2 \right| ={\sqrt 5\over 2}k} \\ \Rightarrow 2:r:s =k:{\sqrt 5\over 2}k: {\sqrt 5\over 2}k =2:\sqrt 5:\sqrt 5 \Rightarrow rs= \bbox[red, 2pt]5$$

解答:$$\cases{a_2=a_1+d= 5\\ a_6=a_1+ 5d=21} \Rightarrow \cases{a_1=1\\ d=4} \Rightarrow a_n=1+(n-1)\cdot 4=4n-3\\ 取T_n= S_{2n+1}-S_n ={1\over a_{n+1}} +{1\over a_{n+2}} + \cdots+{1\over a_{2n+1}} = \sum_{k=n+1}^{2n+1} {1\over 4k-3}\\ \Rightarrow T_{n+1}-T_n= {1\over 8n+5}+{1\over 8n+9}-{1\over 4n+1} \lt {2\over 8n+5}-{1\over 4n+1} ={1\over 4n+5/2}-{1\over 4n+1}\lt 0 \\ \Rightarrow T_n 為嚴格遞減數列 \Rightarrow T_n的最大值發生在n=1 \Rightarrow T_1=S_3-S_1={1\over a_2}+{1\over a_3} ={14\over 45}\le {k\over 15} \\ \Rightarrow \bbox[red, 2pt]{k\ge {14\over 3}}$$

解答:$$假設L: y=mx+k 代入曲線\Gamma求交點 \Rightarrow ax^3+px=mx+k \Rightarrow ax^3+(p-m)x-k=0 \\ 三交點\cases{A(x_a, f(x_a))\\ B(x_b,f(x_b)) \\C(x_c, f(x_c))} \Rightarrow ax^3+(p-m)x-k=a(x-x_a)(x-x_b)(x-x_c) \Rightarrow  x_a+x_b+ x_c =0 \cdots(1)\\ \Rightarrow m={f(x_a)-f(x_b)\over x_a-x_b} ={ax_a^3+px_a-(ax_b^3+px_b)\over x_a-x_b} =a(x_a^2+x_ax_b+x_b^2)+p \cdots(2)\\ 假設\cases{P(x_p, f(x_p))\\L_1= \overleftrightarrow{AP}: y=m_1x+k_1 } \Rightarrow ax^3+px=m_1x+k_1 \Rightarrow ax^3+(p-m_1)x-k_1=0\\ \Rightarrow 三根為x_a,x_a, x_p (x_a為切點\Rightarrow x_a是重根) \Rightarrow 2x_a+x_p=0 \Rightarrow x_p=-2x_a \Rightarrow P(-2x_a, f(-2x_a))\\ 同理, 假設\cases{Q(x_q, f(x_q)) \\R(x_r, f(x_r))} \Rightarrow \cases{x_q=-2x_b\\ x_r=-2x_c} \Rightarrow \cases{Q(-2x_b, f(-2x_b)) \\R(-2x_c, f(-2x_c))} \\ \Rightarrow \overline{PQ}斜率m_{\overline{PQ}} = {f(-2x_a)-f(-2x_b)\over -2x_a-(-2x_b)} ={-8a(x_a^3-x_b^3)-2p(x_a-x_b)\over -2(x_a-x_b)} =4a(x_a^2+x_ax_b+x_b^2)+p \\將(2)代入上式\Rightarrow m_{\overline{PQ}} =4(m-p)+p=4m-3p\Rightarrow 同理可得 m_{\overline{QR}} =4m-3p \;\bbox[red, 2pt]{QED.}\\ \Rightarrow P,Q,R三點共線,且斜率為\bbox[red, 2pt]{4m-3p}$$

解答:$$A= \begin{bmatrix}a& b\\ c& d \end{bmatrix} \Rightarrow A^TA= \begin{bmatrix}a& c\\ b& d \end{bmatrix}\begin{bmatrix}a& b\\ c& d \end{bmatrix}  = \begin{bmatrix}a^2+c^2& ab+cd\\ ab+cd & b^2+d^2 \end{bmatrix} = \begin{bmatrix}1& 0\\ 0& 1 \end{bmatrix} \\ \Rightarrow \cases{\vec u= \begin{bmatrix}a\\ c \end{bmatrix} \Rightarrow |\vec u|=1\\ \vec v= \begin{bmatrix}b\\ d \end{bmatrix} \Rightarrow |\vec v|=1 \\ \vec u\cdot \vec v=0 \Rightarrow \vec u \bot \vec v} \Rightarrow \cases{a=\cos \theta\\ c=\sin \theta} \Rightarrow b\cos\theta+d \sin \theta=0 \Rightarrow \cases{\vec v= \begin{bmatrix}-\sin \theta\\ \cos \theta \end{bmatrix} \\ \vec v= \begin{bmatrix}\sin \theta\\ -\cos \theta \end{bmatrix}} \\ \textbf{Case I }\vec v= \begin{bmatrix}-\sin \theta\\ \cos \theta \end{bmatrix} \Rightarrow \cases{b=-\sin \theta\\ d=\cos \theta} \Rightarrow A = \begin{bmatrix}\cos \theta& -\sin \theta\\ \sin \theta& \cos \theta \end{bmatrix} 為一旋轉矩陣\\ \textbf{Case II }\vec v= \begin{bmatrix}\sin \theta\\ -\cos \theta \end{bmatrix} \Rightarrow \cases{b= \sin \theta \\d= -\cos \theta} \Rightarrow A = \begin{bmatrix} \cos \theta &\sin \theta\\ \sin \theta& -\cos \theta \end{bmatrix} 為一鏡射矩陣\\\qquad (與直線y=\tan(\theta/2)\cdot x 鏡射) \\ 因此\cases{\det(A)=1 時, A為旋轉矩陣\\ \det(A)=-1時, A為鏡射矩陣} \quad \bbox[red, 2pt]{故得證}$$

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