114年新竹高中教甄2-數學詳解

國立新竹中學114學年度第二次教師甄選

一、填充題Ⅰ (每題 5 分，共 25 分)

解答:$$假設\alpha為x^2-x+2=0的根 \Rightarrow \alpha^2-\alpha+2=0 \Rightarrow \alpha^2=\alpha-2 \\ \Rightarrow \alpha^3=\alpha^2-2\alpha=\alpha-2-2\alpha=-\alpha-2\\ f(x)=x^{26}=(x^2-x+2)P(x)+181x+8098 \Rightarrow f(\alpha)=\alpha^{26} =181\alpha+8098\\ g(x)=x^{23} =(x^2-x+2)Q(x)+ ax+b \Rightarrow g(\alpha) = \alpha^{23}=a\alpha+b\\ \Rightarrow \alpha^{26} =\alpha^3(a\alpha+b)= (-\alpha-2)(a\alpha+b) =-a\alpha^2-(2a+b)\alpha-2b =-a(\alpha-2)-(2a+b)\alpha-2b \\=-(3a+b)\alpha+2a-2b=181\alpha+8098 \Rightarrow \cases{3a+b=-181\\ 2a-2b=8098} \Rightarrow \cases{a=967\\ b=-3082} \\\Rightarrow 餘式為\bbox[red, 2pt]{967x-3082}$$

解答:$$f(x,y)=2x^2-4xy+4y^2-4x-4y+5 \Rightarrow \cases{f_x=4x-4y-4\\ f_y=-4x+8y-4} \Rightarrow \cases{f_{xx}=4\\ f_{yy}=8\\ f_{xy}=-4} \\ \Rightarrow D(x,y)= f_{xx}f_{yy}-f_{xy}^2 =32-16=16\gt 0 \\ \cases{f_x=0\\ f_y=0} \Rightarrow \cases{x=3\\ y=2} \Rightarrow f(3,2)=-5為極小值 \Rightarrow (m,n,k)=\bbox[red, 2pt]{(3,2,-5)}\\ \bbox[red, 2pt]{另解}: f(x,y)=(x-2y+1)^2+(x-3)^2-5 \Rightarrow 當\cases{x-2y+1=0\\ x-3=0}時,f(x,y)有最小值-5\\,此時\cases{x=3\\ y=2}$$

解答:$$\log \log a+\log \log b={1\over 2} \Rightarrow \log (\log a)(\log b)={1\over 2} \Rightarrow (\log a)(\log b) = \sqrt{10}\\ \Rightarrow  L=(a^{\log b})\cdot (b^{\log a}) \Rightarrow \log L=\log a^{\log b}+\log b^{\log a} =2(\log a)(\log b) =2\sqrt{10} \approx 6.3 \\ \Rightarrow L整數部份為\bbox[red, 2pt]7位數$$

解答:$$(A-2I)^{-1}=pA+qI \Rightarrow (pA+qI)(A-2I) =I \\ \Rightarrow pA^2-2pA+qA-2qI=I \Rightarrow 114pA-2pA+qA-2qI=112pA+qA -2qI=I\\ \Rightarrow \cases{112p+q=0\\ -2q=1} \Rightarrow (p,q)= \bbox[red, 2pt]{({1\over 224},-{1\over 2})}$$

解答:

$$假設\cases{\vec a= \overrightarrow{OA} \\ \vec b=\overrightarrow{OB}  \\ \vec c= \overrightarrow{OC}}, 其中O為原點，由於\cases{|\vec a|=|\vec b|=1 \\ \vec a\cdot \vec b=1/2} \Rightarrow A,B均在單位圓上且\angle AOB=60^\circ\\ |\vec c|的最小值發生在|\vec a-\vec c|=|\vec b-\vec c| \Rightarrow C在\overline{AB}的中垂線上,且靠近原點O \\ 令\cases{A(1,0) \\B(1/2,\sqrt 3/2)} \Rightarrow P=\overline{AB}中點 \Rightarrow \overline{OP}為正\triangle OAB的高 \Rightarrow \overline{OP}=\sqrt 3/2\\ \triangle ACP三角為30^\circ-60^\circ-90^\circ \Rightarrow \overline{CP}={1\over 2\sqrt 3} \Rightarrow |\vec c| =\overline{OC} =\overline{OP}-\overline{PC} ={\sqrt 3\over 2}-{1\over 2\sqrt 3} = \bbox[red, 2pt]{\sqrt 3\over 3}$$

二、填充題Ⅱ (每題 6 分，共 48 分)

解答:

$$六個頂點代表六個人，兩頂點有連線代表兩人互為好友，依圖形可分為兩類:\\\cases{\textbf{Case I }一個環狀:共有\displaystyle {5!\over 2}=60種\\ \textbf{Case II }兩個環狀: 共有\displaystyle {C^6_3\over 2}= 10種}\Rightarrow  因此共有60+10= \bbox[red, 2pt]{70}種組合方式$$

解答:

$$\cases{P(5,13) \\Q(-16,10)} \Rightarrow 切線L_1=\overleftrightarrow{PQ}: x-7y+86=0 \Rightarrow 切點A\in L_1 \Rightarrow A(7t-86,t) \\ \Rightarrow \overline{AK}^2 =f(t) =(7t-86)^2+(t+2)^2 \Rightarrow f'(t)=0 \Rightarrow t=12 \Rightarrow A(-2,12) \\ \Rightarrow 切點A對稱於L_1的對稱點B即為另一切點\\ \cases{P(5,13) \\K(0,-2)} \Rightarrow 直線L_2=\overleftrightarrow{AK}: y=3x-2 \Rightarrow L_3=\overleftrightarrow{AB}: y=-{1\over 3}(x+2)+12 \\ \Rightarrow \overline{AB}中點C=L_2\cap L_3=(4,10) ={A+B\over 2}\Rightarrow B=(8,20)-(-2,12)= \bbox[red, 2pt]{(10,8)}$$

解答:

$$\cos A={4^2+6^2-5^2 \over 2\cdot 4\cdot 6}={9\over 16} \Rightarrow \sin A={5\sqrt 7\over 16} \Rightarrow {5\over 5\sqrt 7/16}=2R \Rightarrow R={8\over \sqrt 7} \\ \Rightarrow P到平面ABC的距離=P到\triangle ABC外心的距離=h \Rightarrow h^2+R^2=8^2 \Rightarrow h^2=64-{64\over 7}={384\over 7} \\ \Rightarrow h=\sqrt{384\over 7}= \bbox[red, 2pt]{8\sqrt{42}\over 7}$$

解答:$$\overline{AC}=x \Rightarrow \cases{\cos D={2-x^2\over 2} \\ \cos B={5-x^2\over 4}} \Rightarrow \cases{\sin^2 D=1-\cos^2 D=\displaystyle {4x^2-x^4\over 4} \\ \sin^2B =1-\cos^2B =\displaystyle {-9+10x^2-x^4\over 16}} \\ \Rightarrow \cases{T^2={1\over 4}\sin^2D =\displaystyle {4x^2-x^4\over 16}\\ S^2=\sin^2B =\displaystyle {-9+10x^2-x^4\over 16}} \Rightarrow S^2+T^2 =f(x)={-2x^4+14x^2-9\over 16} \\ \Rightarrow f'(x)={-8x^3+28x\over 16} =0 \Rightarrow 4x(-2x^2+7) =0 \Rightarrow x^2={7\over 2}\\ \Rightarrow f(x^2=7/2) ={-{49\over 2}+40\over 16} = \bbox[red, 2pt]{31\over 32}$$

解答:$$\Gamma:{x^2\over a^2}-{y^2\over b^2}=1且a=2b \Rightarrow c=\sqrt{a^2+b^2} ={\sqrt 5\over 2}a, \\ 又\cases{F_1(-c,0) \\ F_2(c,0)} \Rightarrow 過F_2之直線L:y=m(x-c) \Rightarrow 在L上的點可表示為(c+t,mt) 代入\Gamma \\ \Rightarrow (1-4m^2)t^2+2ct+{a^2\over 4}=0, 假設兩根為t_1,t_2\\ 由於\overline{AF_2}: \overline{BF_2}=3:1 \Rightarrow \cases{t_1=3s\\ t_2=-s}, s\gt 0 \Rightarrow \cases{兩根之和=t_1+t_2=2s=2c/(4m^2-1) \\ 兩根之積=-3s^2=a^2/4(1-4m^2)} \\ \Rightarrow c^2=-{a^2 \over 12}(1-4m^2)\\ \triangle F_1AB面積={1\over 2}\cdot \overline{AB}\cdot d(F_1,L) ={1\over 2} \cdot (2c)\cdot |t_1-t_2| =4c|m|s = {a^2\over 6} \sqrt{1+{12c^2\over a^2}} \\={a^2\over 6} \cdot \sqrt{1+15}= {2\over 3}a^2=96 \Rightarrow a=12 \Rightarrow 貫軸長2a=\bbox[red, 2pt]{24}$$

解答:$$  \cases{A \begin{bmatrix} 217\\ 175 \end{bmatrix} =\begin{bmatrix} 83\\ 125 \end{bmatrix} \\[1ex]A \begin{bmatrix} 188\\ 410 \end{bmatrix} =\begin{bmatrix} 112\\ -110 \end{bmatrix} } \Rightarrow  A \left(\begin{bmatrix} 217\\ 175 \end{bmatrix}+  \begin{bmatrix} 188\\ 410 \end{bmatrix} \right) =\begin{bmatrix} 83\\ 125 \end{bmatrix}+\begin{bmatrix} 112\\ -110 \end{bmatrix} \\ \Rightarrow A\begin{bmatrix} 405\\ 585 \end{bmatrix} =\begin{bmatrix} 195\\ 15 \end{bmatrix} \Rightarrow A(3P)= \begin{bmatrix} 195\\ 15 \end{bmatrix} \Rightarrow AP= \begin{bmatrix} 195/3\\ 15/3 \end{bmatrix} \Rightarrow P'=\begin{bmatrix} 65\\ 5 \end{bmatrix} \\ 對稱軸L:y=2x \Rightarrow \tan \theta=2\Rightarrow \cases{\sin 2\theta =4/5\\ \cos 2\theta=-3/5} \Rightarrow 鏡射矩陣B=\begin{bmatrix} -3/5& 4/5\\ 4/5& 3/5 \end{bmatrix} \Rightarrow BP'= \begin{bmatrix} -35\\ 55 \end{bmatrix} \\ \Rightarrow 鏡射後的坐標\bbox[red, 2pt]{(-35,55)}$$

解答:$$z^8-1=\prod_{k=0}^7(z-z_k) \Rightarrow \prod_{k=1}^8 \overline{PA_k} =\prod_{k=0}^7|z-z_k| =|z^8-1| =|e^{i8\theta}-1|=|\cos 8\theta+i\sin 8\theta-1| \\=\sqrt{(\cos 8\theta-1)^2+\sin^28\theta} =\sqrt{2-2\cos 8\theta} \Rightarrow 最大值為\sqrt{2+2}= \bbox[red, 2pt]2$$

解答:$$f(x)= \int_a^x g(t)\,dt +x^3+6x^2-3x+4 \Rightarrow f'(x)=g(x)+3x^2+12x-3  \\g(x)=\int_a^x f(t)\,dt -{1\over 2}x^4-x^3+4x^2-11x+m \Rightarrow g'(x)=f(x)-2x^3-3x^2+8x-11 \\ \Rightarrow g''(x)= f'(x)- 6x^2-6x+8=(g(x)+3x^2+12x-3)-6x^2-6x+ 8  \\ \Rightarrow g''-g(x)=-3x^2+6x+5 \Rightarrow 解微分方程可得 g(x)=3x^2-6x+1 \Rightarrow g'(x)=6x-6\\ \Rightarrow f(x)=g'(x)+2x^3+3x^2-8x+11 \Rightarrow f(x)= 2x^3+3x^2-2x+5 \\ \Rightarrow 2x^3+3x^2-2x+5=\int_a^x (3t^2-6t+1)\,dt+x^3+6x^2-3x+4 \Rightarrow a^3-3a^2+a+1=0\\ \Rightarrow (a-1)(a^2-2a-1)=0 \Rightarrow a=1 (a是整數) \\ \Rightarrow 3x^2-6x+1=\int_{a=1}^x 2t^3+3t^2-2t+5\,dt-{1\over 2}x^4-x^3+4x^2-11x+m \\ \Rightarrow m-{11\over 2}=1 \Rightarrow m={13\over 2} \Rightarrow (a,m)= \bbox[red, 2pt]{(1, {13\over 2})}$$

三、計算證明題(每題 9 分，共 27 分，請詳列計算與證明過程。) 

解答:$$\textbf{(1) }\cases{a_n=b_{n-1} \\ b_n=2a_{n-1}+b_{n-1}} \Rightarrow T= \bbox[red, 2pt]{\begin{bmatrix} 0& 1\\ 2& 1\end{bmatrix}}\\ \textbf{(2) }\cases{a_n =b_{n-1} \\ b_n=2a_{n-1}+b_{n-1}} \Rightarrow a_{n+1}=a_n+2a_{n-1} \Rightarrow \alpha^2-\alpha-2 =0 \Rightarrow (\alpha-2)(\alpha+1)=0 \\\quad \Rightarrow \alpha= 2,-1 \Rightarrow a_n=c_12^n+c_2(-1)^n \Rightarrow \cases{a_2=0\\ a_3=b_2= 6} \Rightarrow \cases{0=4c_1 +c_2\\ 6=8c_1-c_2} \Rightarrow \cases{c_1=1/2\\ c_2=-2} \\ \Rightarrow a_n={1\over 2}\cdot 2^n-2(-1)^n \Rightarrow \bbox[red, 2pt]{a_n=2^{n-1}+2\cdot (-1)^{n-1}, n\ge 2}$$

解答:$$\textbf{(1) }3顆紅球、n顆黑球任排有C^{n+3}_3 ={(n+3)(n+2)(n+1)\over 6}排法\\在第k次取到第2顆紅球:第k次為紅球，前(k-1)次為1紅球、k-2顆黑球，有(k-1)種排法；\\後(n+3-k)顆球有1紅球、(n+2-k)黑球，有(n+3-k)排法；因此\\ P(X=k)={(k-1)(n+3-k) \over (n+3)(n+2 )(n+1)/6} = \bbox[red, 2pt]{6(k-1)(n-k+3) \over (n+3)(n+2)(n+1)} \\\textbf{(2) } E(X)= \sum_{k=1}^{n+3}kP(X=k)= {6\over (n+3)(n+2) (n+1)} \sum_{k =1}^{n+3} \left[k(k-1)(n-k+3) \right]\\= {6\over (n+3)(n+2) (n+1)} \sum_{k=1}^{n+3} \left[-k^3+(n+4)k^2-(n+3)k \right] \\= {6\over (n+3)(n+2) (n+1)} \left[(n+3)(n +4) \left({1\over 6}(n+4)(2n+7)-{1\over 4}(n+3)(n+6) \right) \right] \\={6(n+4) \over (n+1)(n+2)} \left[ {1\over 6}(2n^2+15n+28)-{1\over 4}(n^2+9n+18)\right] = \bbox[red, 2pt]{n+4\over 2}$$

解答:$$f(x)為2n(偶數)次多項式\Rightarrow f(-x)為2n(偶數)次多項式\Rightarrow f(x)與f(-x)有相同的x^{2n}係數 \\因 取g(x)=f(x)-f(-x)  \Rightarrow deg(g)\lt 2n\\ 但g(x)=0有2n個相異實根\pm x_1,\pm x_2, \dots, \pm x_n  \Rightarrow g(x)=0 \Rightarrow f(x)=f(-x) \quad\bbox[red, 2pt]{故得證}$$

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