114年內湖高中教甄2-數學詳解

臺北市立內湖高級中學114學年度第2次正式教師甄選

一、 填充題

 

解答:$$\cases{g(x)=f(f(x)) =2(2x^2-x-6)^2-(2x^2-x-6)-6\\ h(x)=f(x)\times f(x)=(2x^2-x-6)^2} \\ \Rightarrow g(c)=h(c) \Rightarrow 2(2c^2-c-6)^2-(2c^2-c-6)-6=(2c^2-c-6)^2 \\ \Rightarrow (2c^2-c-6)^2-(2c^2-c-6)-6=0 \Rightarrow (2c^2-c-6-3)(2c^2-c-6+2)=0\\ \Rightarrow (2c^2-c-9)(2c^2-c-4) =0 \Rightarrow \cases{2c^2-c=9 \Rightarrow h(c)=(9-6)^2=9\\ 2c^2-c=4 \Rightarrow h(c)=(4-6)^2=4} \Rightarrow 餘式= \bbox[red, 2pt]{9或4}$$

解答:

$$\cases{\overline{PB}=2\\ \overline{PC}=\sqrt 5 \\ \overline{BC}=3} \Rightarrow \overline{BC}^2=\overline{PB}^2+ \overline{PC}^2 \Rightarrow \angle BPC=90^\circ \Rightarrow 取\cases{P(0,0) \\B(2,0) \\C(0,\sqrt 5)} \\ \Rightarrow A=C以B為旋轉中心,旋轉60^\circ \Rightarrow \begin{bmatrix}x\\ y \end{bmatrix} = \begin{bmatrix} \cos 60^\circ &-\sin 60^\circ\\ \sin 60^\circ& \cos 60^\circ \end{bmatrix} \begin{bmatrix}0-2\\ \sqrt 5-0 \end{bmatrix} + \begin{bmatrix}2\\ 0 \end{bmatrix}\\= \begin{bmatrix}1/2& -\sqrt 3/2\\ \sqrt 3/2& 1/2 \end{bmatrix} \begin{bmatrix}-2\\ \sqrt 5 \end{bmatrix} + \begin{bmatrix}2\\ 0 \end{bmatrix} = \begin{bmatrix}1-\sqrt{15}/2\\ \sqrt 5 /2-\sqrt 3\end{bmatrix}\\ \Rightarrow \overline{PA}^2 =x^2+y^2 = \left( 1-{\sqrt {15}\over 2} \right)^2+ \left( {\sqrt 5\over 2}-\sqrt 3 \right)^2 = \bbox[red, 2pt]{9-2\sqrt{15}}$$

解答:$$\cases{3^1的個位數為3\\ 3^2的個位數為9\\ 3^3的個位數為7\\ 3^4的個位數為1\\ 3^5的個位數為3 \\ \cdots } \Rightarrow 循環數為4\quad \cases{7^1的個位數為7\\ 7^2的個位數為9\\ 7^3的個位數為3\\ 7^4的個位數為1\\ 7^5的個位數為7\\ \cdots} \Rightarrow 循環數為4\\ 2到2025共有2024個數字, 且2024為4的倍數 \Rightarrow \cases{3^a個位數字是1,3,7,9的機率皆為1/4 \\7^b個位數字是1,3,7,9的機率皆為1/4}\\ 3^a+7^b的個位數字是8的情形:(a,b)=(1,7),(7,1),(9,9) \Rightarrow 機率為3\cdot {1\over 4}\cdot {1\over 4}= \bbox[red, 2pt]{3\over 16}$$

解答:$$f(x)=(1+x)^{2n} =\sum_{k=0}^{2n}C^{2n}_k x^k \Rightarrow f(i)=(1+i)^{2n} =\sum_{k=0}^{2n}C^{2n}_k i^k \\ 左式=(1+i)^{2n} = \left( \sqrt 2(\cos {\pi\over 4} +i \sin{\pi\over 4}) \right) ^{2n}  = \left(   2(\cos {\pi\over 2} +i \sin{\pi\over 2}) \right) ^{n} =2^n(\cos {n\pi\over 2} +i \sin{n\pi\over 2}) \\  \Rightarrow 左式虛部=2^n \sin{n\pi\over 2}\\右式=C^{2n}_0+C^{2n}_1i-C^{2n}_2-C^{2n}_3i+C^{2n}_4+\cdots \Rightarrow 右式虛部=C^{2n}_1-C^{2n}_3 +C^{2n}_5- \cdots\\ 左式虛部=右式虛部 \Rightarrow 2^n \sin{n\pi\over 2} = C^{2n}_1-C^{2n}_3 +C^{2n}_5- \cdots\\ n={2025\over 2}代入上式 \Rightarrow 2^{2025/2 }\cdot {\sqrt 2\over 2} =C^{2025}_1-C^{2025}_3+C^{2025}_5 -\cdots +C^{2025}_{2025}\\ \Rightarrow  C^{2025}_1-C^{2025}_3+C^{2025}_5 -\cdots +C^{2025}_{2025} = \bbox[red, 2pt]{2^{1012}}$$

解答:$$取\cases{A(0,0,0) \\B(\sqrt 2,0,\sqrt 2) \\C(\sqrt 2,\sqrt 2,0) \\D(0,\sqrt 2,\sqrt 2)} \Rightarrow \cases{M=(C+D)/2=(\sqrt 2/2,\sqrt 2,\sqrt 2/2) \\ E_1=平面BCD:x+y+z=2\sqrt 2 \\ E_2= 平面ACD: x-y+z=0} \Rightarrow \cases{A'({4\sqrt 2\over 3},{4\sqrt 2\over 3},{4\sqrt 2\over 3})\\ B'(-{\sqrt 2\over 3},{4\sqrt 2\over 3}, -{\sqrt 2\over 3})} \\ \Rightarrow \cases{\vec u= \overrightarrow{MA'} =({5\sqrt 2\over 6},{\sqrt 2\over 3},{5\sqrt 2\over 6}) \\ \vec v= \overrightarrow{MB'} =(-{5\sqrt 2\over 6},{ \sqrt 2\over 3}, -{5\sqrt 2\over 6})} \Rightarrow \triangle AMB'={1\over 2}||\vec u\times \vec v|| ={1\over 2}||(-{10\over 9},0,{10\over 9})|| \\ ={1\over 2}\cdot {10\over 9}\sqrt 2= \bbox[red, 2pt]{{5\over 9}\sqrt 2}$$

解答:$$假設第一批測驗:\cases{n_1=100\\ \mu_1=60\\ 標準差\sigma_1}, 第二批測驗:\cases{n_2= 50\\ \mu_2=66\\ 標準差\sigma_2}\Rightarrow 全體:\cases{n_A= n_1+n_2=150\\ 平均值\mu_A\\ 標準差\sigma_A} \\小廷的T分數=50+10\times {84-60\over \sigma_1} =62 \Rightarrow \sigma_1=20 \\ 全體考生平均值\mu_A= {n_1\cdot \mu_1+ n_2\cdot \mu_2\over n_1+n_2} ={100\cdot 60+50\cdot 66\over 100+50} =62 \\ \Rightarrow 小廷在全體的T分數=50+10\times {84-62\over \sigma_A} =60 \Rightarrow \sigma_A=22 \\ \sigma_1^2={1\over n_1}\sum_{i=1}^{n_1} x_{1i}^2-\mu_1^2 \Rightarrow 20^2={1\over 100} \sum_{i=1}^{n_1} x_{1i}^2-60^2 \Rightarrow \sum_{i=1}^{n_1} x_{1i}^2 =400000\\ \sigma_A^2 ={1\over n_1+n_2} \left( \sum_{i =1}^{n_1} x_{1i}^2+ \sum_{i=1}^{n_2} x_{2i}^2 \right) -\mu_A^2 \Rightarrow 22^2={1\over 150} \left( 400000+ \sum_{i=1}^{n_2} x_{2i}^2\right) -62^2 \\ \Rightarrow \sum_{i=1}^{n_2} x_{2i}^2 =249200 \Rightarrow \sigma_2^2 =a={1\over n_2}\sum_{i=1}^{n_2} x_{2i}^2-\mu_2^2={1\over 50}\cdot 249200-66^2= \bbox[red, 2pt]{628}$$

解答:$$假設\cases{L_1= \overleftrightarrow{AC} \\ L_2 = \overleftrightarrow{BD}} \Rightarrow \cases{L_1的方向向量\vec u=(2,-2,-1) \\L_2 的方向向量\vec v=(1,2,-2)} \Rightarrow \cases{\vec u\cdot \vec v=0 \Rightarrow L_1 \bot L_2 \\ \vec n= \vec u\times \vec v=(6,3,6)} \\ \cases{P(-1,2,0) \in L_1\\ Q(1,3,5)\in L_2} \Rightarrow \overrightarrow{PQ}=(2,1,5)\Rightarrow \overline{CD} =d(L_1,L_2) = {\overrightarrow{PQ} \cdot \vec n \over |\vec n|} =5 \\ \Rightarrow \overline{AB}^2= \overline{AC}^2+ \overline{CD}^2+ \overline{BD}^2=4+25+4=33 \Rightarrow \overline{AB}= \bbox[red, 2pt]{\sqrt{33}}$$

解答:

$$假設\cases{\Gamma_1: y=f(x)= (\sqrt{13})^x\\ \Gamma_2: y=g(x)=\log_{\sqrt{13}}x \\ 直線L: y=\sqrt{13}-x} \Rightarrow \cases{\Gamma_1 \cap L =P(\alpha, (\sqrt{13})^\alpha) \\ \Gamma_2 \cap L= Q(\beta, \log_{\sqrt{13}}\beta)}\\ f(x)與g(x)互為反函數 \Rightarrow \Gamma_1與\Gamma_2對稱於y=x,又直線L與對稱軸y=x垂直 \\即P(\alpha, (\sqrt{13})^\alpha)對稱Q(\beta, \log_{\sqrt{13}}\beta) \Rightarrow (\sqrt{13})^\alpha=\beta \Rightarrow P(\alpha,\beta), 又P在L上 \Rightarrow \alpha+\beta=\sqrt{13} \\ \Rightarrow (\sqrt{13})^\alpha+ (\alpha+\beta)^2+ \log_{\sqrt{13}} \beta =\sqrt{13}-\alpha+ (\alpha+\beta)^2 + \sqrt{13}-\beta =2\sqrt{13}-(\alpha+\beta) +(\alpha+\beta)^2 \\=2\sqrt{13}-\sqrt{13}+13= \bbox[red, 2pt]{13+\sqrt{13}}$$

解答:$$z^3-2z^2-2z-3= (z-3)(z^2+z+1) =0 \Rightarrow z=3, {-1\pm \sqrt 3i\over 2} \\\Rightarrow 三角形三頂點\cases{A(3,0) \\B(-{1\over 2}, {\sqrt 3\over 2}) \\C(-{1\over 2}, -{\sqrt 3\over 2})} \Rightarrow 外心O(k,0) \Rightarrow \overline{OA}=\overline{OB} \Rightarrow 3-k= \sqrt{(k+{1\over 2})^2+ {3\over 4}} \\ \Rightarrow k^2-6k+9= k^2+k+1 \Rightarrow k={8\over 7} \Rightarrow (a,b)= \bbox[red, 2pt]{({8\over 7},0)}$$

解答:$$y={2ax+b\over x^2+1} \Rightarrow yx^2-2ax+(y-b)=0 有實數解 \Rightarrow 4a^2-4y(y-b)\ge 0 \\ \Rightarrow y^2-by-a^2\le 0 \Rightarrow y^2-by-a^2 =(y+2)(y-3)=y^2-y-6 \\ \Rightarrow \cases{b=1\\ a^2=6 \Rightarrow a=\pm \sqrt 6} \Rightarrow (a,b)= \bbox[red, 2pt]{(\pm \sqrt 6,1)}$$

解答:$$\cases{球A: x^2+y^2+z^2=1 \\ 球B: (x-r)^2 +y^2+z^2=1} \Rightarrow 平面圓C= 球A\cap 球B:x={r\over 2}\\ \Rightarrow \cases{x\lt r/2 \Rightarrow C: y=\sqrt{1-x^2} \\ x\gt r/2 \Rightarrow C: y= \sqrt{1-(x-r)^2}} \Rightarrow V(r)= 2\times \int_{-1}^{r/2} \pi (\sqrt{1-x^2})^2\,dx =2\pi \int_{-1}^{r/2}(1-x^2)\,dx \\=2\pi \left. \left[ x-{1\over 3}x^3 \right] \right|_{-1}^{r/2} =2\pi \left( {r\over 2}-{r^3\over 24}+{2\over 3} \right) = \bbox[red, 2pt]{\pi r-{\pi r^3\over 12}+{4\pi\over 3}}$$

解答:$$假設[x]=n \in \mathbb Z \Rightarrow x+{114\over x}=n+{114\over n} \Rightarrow x-n=114 \left( {1\over n}-{1\over x} \right) =114\cdot {x-n\over nx} \\ \Rightarrow 1={114\over nx} \Rightarrow x={114\over n} \Rightarrow n\lt {114\over n}\lt n+1\\ 若n\gt 0 \Rightarrow \cases{n\lt 114/n\\ 114/n\lt n+1} \Rightarrow \cases{n^2 \lt 114 \Rightarrow n=1,2,\dots, 10\\ n^2+n-114\gt 0 \Rightarrow 10^2+10-114\lt 0} \Rightarrow 無解\\ 若n\lt 0 \Rightarrow \cases{n^2 \gt 114 \Rightarrow n\le -11\\ n^2+n-114\lt 0 \Rightarrow n=-1,-2,\dots,-11} \Rightarrow n=-11 \Rightarrow x={114\over n} \\ \Rightarrow x= \bbox[red, 2pt]{-{114\over 11}}$$

解答:$$\cases{O(0,0) \\ P_i(x,y) \\A(a,b)} \Rightarrow S=\triangle OPiA ={1\over 2}|bx-ay|\\ 由於a,b,x,y皆為整數, 因此|bx-ay|為整數, 最小值為1, 因此S最小值為\bbox[red, 2pt]{1\over 2}\\ 例:A(2,1),P(1,1)或A(3,2),P(1,1) 皆可得到S={1\over 2}$$

解答:

$$\cases{\triangle ABD=\displaystyle {1\over 2} \cdot 8\cdot 6=24\\ \triangle ABC =\displaystyle {1\over 2}\cdot 6\cdot 14 \sin 60^\circ =21\sqrt 3\\ \triangle ACD =\displaystyle {1\over 2}\cdot 8\cdot 14 \sin 30^\circ=28} \Rightarrow \triangle BCD=\triangle ABC+ \triangle ACD-\triangle ABD\\=21\sqrt 3+28-24 = \bbox[red, 2pt]{4+21\sqrt 3}$$

解答:

$$假設\overline{AB} =\sqrt 2 \Rightarrow \overline{AD} =\sqrt 2\cdot \overline{AB}=2 \Rightarrow \cases{\overline{P_1D} =\overline{P_2D} =\overline{P_3D} =\sqrt 2 \\ \overline{P_1B}= \overline{P_4B} =\overline{P_3C} =\overline{P_4C} =2} \\ \Rightarrow 取\cases{A(0,0,0)\\ B(\sqrt 2,0,0) \\C(\sqrt 2, 2,0)\\ D(0,2,0) \\P(0,1,1)=P_1=P_2=P_3=P_4} \Rightarrow \cases{E_1=平面PAB: y-z=0\\E_2=平面PBC: x+\sqrt 2 z=\sqrt 2} \\\Rightarrow \cases{E_1法向量\vec u=(0,1,-1) \\E_2法向量 \vec v=(1,0,\sqrt 2)} \Rightarrow \cos \theta ={\vec u\cdot \vec v\over |\vec u||\vec v|} ={-\sqrt 2\over \sqrt 2\cdot \sqrt 3} =-{1\over \sqrt 3}\Rightarrow \sin \theta={\sqrt 2\over \sqrt 3} = \bbox[red, 2pt]{\sqrt 6\over 3}$$

解答:$$f(t)= (\vec a+t \vec b) \cdot  (\vec a+t \vec b)=|\vec a|^2+2t \vec a\cdot \vec b+|\vec b|^2 t^2 \Rightarrow f'(t)=2\vec a\cdot \vec b+2|\vec b|^2t=0 \\ \Rightarrow 2(\vec a+t\vec b)\cdot b=0 \Rightarrow (\vec a+t\vec b) \cdot b=0 \Rightarrow \vec b\cdot (\vec a+(t-2) \vec b) = \vec b\cdot (\vec a+ t\vec b-2\vec b) \\=\vec b\cdot (0-2\vec b)=-2|\vec b|^2= \bbox[red, 2pt]{-32}$$

解答:$$P(x,y,z) \Rightarrow \cases{\cases{P到圓C_1 的垂直距離=|z-\sqrt 2|\\ P到圓C_1最近的水平距離= |r-\sqrt 2|} \Rightarrow m= \sqrt{(r-\sqrt 2)^2+ (z-\sqrt 2)^2} \\\cases{P到圓C_2的垂直距離=|z+\sqrt 2| \\ P到圓C_2最遠的水平距離= r+\sqrt 2} \Rightarrow M= \sqrt{(r+\sqrt 2)^2+ (z+ \sqrt 2)^2}} \\ \Rightarrow m+M=  \sqrt{(r-\sqrt 2)^2+ (z-\sqrt 2)^2}+ \sqrt{(r+\sqrt 2)^2+ (z+ \sqrt 2)^2}=2\sqrt 6   \\ \Rightarrow (r-\sqrt 2)^2+ (z-\sqrt 2)^2 =24-4\sqrt 6\cdot \sqrt{(r+\sqrt 2)^2+ (z+ \sqrt 2)^2}+ (r+\sqrt 2)^2+ (z+ \sqrt 2)^2 \\ \Rightarrow \sqrt 6\cdot  \sqrt{(r+\sqrt 2)^2+ (z+ \sqrt 2)^2}=6+\sqrt 2 r+ \sqrt 2z \Rightarrow \bbox[red, 2pt]{r^2+z^2-rz=3}$$

解答:$$\cases{\begin{bmatrix}x_{n+1} \\y_{n+1}  \end{bmatrix} = \begin{bmatrix}a& -b\\ b& a \end{bmatrix} \begin{bmatrix}x_n\\[1ex] y_n \end{bmatrix} \\ (x_0, y_0) =(1,0) \\[1ex] P_0=P_6} \Rightarrow \begin{bmatrix}a& -b\\ b& a \end{bmatrix} 為旋轉\pm 60^\circ 矩陣\Rightarrow (a,b )= \bbox[red, 2pt]{({1\over 2}, \pm {\sqrt 3\over 2})}$$

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