臺中市立臺中第一高級中等學校115學年度第1次教師甄選
一、填充題甲:(每格 5 分, 共 20 分)
解答:$$a+b+c+ab+ca=376 \Rightarrow a+b(a+1)+c(a+1)=376 \Rightarrow (a+1)+b(a+1)+c(a+1)=377 \\ \Rightarrow (a+1)(b+c+1)=377 =13\times 29 \Rightarrow \cases{a+1=13\\ b+c+1=29} \Rightarrow \cases{a=12\\ b+c=28} \Rightarrow \cases{a=12\\b=13\\c=15} \\ \Rightarrow (a,b,c)= \bbox[red, 2pt]{(12,13,15)}$$
解答:$$假設有四張桌子，每張桌子有四個座位，共有16個座位\\甲先入座，剩下15個座位，乙要與甲同桌(組)，只有3個空位，因此甲、乙同桌的機率={3\over 15}={1\over 5}\\ 現在甲乙同桌，剩下14個空位，丙要入座要兩種情形:\\ 甲乙丙同桌:機率為{2\over 14}={1\over 7}，為了滿足丙丁不同桌，丁只能入座其他桌的12個空位，機率為{12\over 13} \\\qquad 因此機率為{1\over 7}\times {12\over 13}={12\over 91}\\ 丙與(甲乙)不同桌:機率為{12\over 14}={6\over 7},丙丁不同桌,機率為{10\over 13},因此機率為{6\over 7}\times {10\over 13}={60\over 91} \\ 兩種情況的機率和就是甲乙同桌的情況下,丙丁不同桌的機率={12\over 91}+{60\over 91}={72\over 91} \\ \Rightarrow 甲乙同桌且丙丁不同桌的機率={1\over 5}\times {72\over 91}= \bbox[red, 2pt]{72\over 455}$$
解答:$$[\sqrt n] =k \Rightarrow k\le \sqrt n\lt k+1 \Rightarrow k^2\le n\lt (k+1)^2 \Rightarrow (k+1)^2-k^2=2k+1 \\ \Rightarrow 有2k+1個n值滿足[\sqrt n]=k \\ 由於\cases{44^2=1936\\ 45^2=2025} \Rightarrow \cases{[\sqrt{2025}]=45\\ [\sqrt{2026}]=45} \Rightarrow \sum_{n=1}^{2026} {1\over 2^{[\sqrt n]}} =\sum_{k=1}^{44} \left( \sum_{n=k^2}^{(k+1)^2-1} {1\over 2^k} \right) +\sum_{n=2025}^{2026}{1\over 2^{45}} \\= \sum_{k=1}^{44} {2k+1\over 2^k}+{2\over 2^{45}} = \left( 5-{93\over 2^{44}} \right)+{1\over 2^{44}}= 5-{92\over 2^{44}} =5-{23\over 2^{42}} \Rightarrow \cases{a=5\\ b=23} \Rightarrow a+b=\bbox[red, 2pt]{28}$$
解答:$$\ell=k-n \Rightarrow \lim_{n\to \infty} \sum_{k=n+1}^{2n} {\sqrt{3n^2-k^2+2nk} \over n^2} =\lim_{n\to \infty} \sum_{\ell=1}^{n} {\sqrt{3n^2-(\ell+n)^2+2n(\ell+n)} \over n^2} \\=\lim_{n\to \infty} \sum_{\ell=1}^{n} {\sqrt{4n^2-\ell^2} \over n^2} =\lim_{n\to \infty} \sum_{\ell=1}^{n} {\sqrt{4-(\ell/n)^2} \over n}=I =\int_0^1 \sqrt{4-x^2}\,dx \\ x=2\sin \theta \Rightarrow dx=2\cos \theta \Rightarrow I=\int_0^{\pi/6} 4\cos^2\,d\theta =\int_0^{\pi/6} (2\cos 2\theta+2)\,d\theta = \left. \left[ \sin 2\theta+2\theta \right] \right|_0^{\pi/6} \\= \bbox[red, 2pt]{{\sqrt 3\over 2}+{\pi\over 3}}$$
二、填充題乙:(每格 6 分, 共 60 分)
解答:
$$f(x)=2^x \Rightarrow g(x)=f^{-1}(x)=\log_2 x \Rightarrow h(x)=g(-x)= \log_2(-x) \\ \Rightarrow -x=\log_2(-(-y)) \Rightarrow y=k(x)=2^{-x} \\\cases{ B=(y=-x+11) \cap (y=\log_2 x) \Rightarrow B(8,3)\\ C=(y=x+11)\cap (y= \log_2(-x)) \Rightarrow C=(-8,3) \\ D=(y=x+11)\cap (y=2^{-x} ) \Rightarrow D(-3,8)} \Rightarrow 等腰梯形ABCD \Rightarrow \cases{上底\overline{AD}=6\\ 下底\overline{BC}=16 \\高=5} \\ \Rightarrow 面積={1\over 2}(6+16)\times 5= \bbox[red, 2pt]{55}$$
解答:$$\cases{a_{n+1} =\alpha a_n+ \beta b_n\\ b_{n+1}=\gamma a_n+\delta b_n} \Rightarrow \begin{bmatrix}a_{n+1} \\b_{n+1} \end{bmatrix} =\begin{bmatrix}\alpha& \beta\\ \gamma & \delta\end{bmatrix} \begin{bmatrix}a_n\\ b_n \end{bmatrix} \Rightarrow A=\begin{bmatrix}\alpha& \beta\\ \gamma & \delta\end{bmatrix} =\begin{bmatrix}\alpha& \beta\\ 1-\alpha & 1-\beta\end{bmatrix} \\ A不可逆\Rightarrow \det(A)=0 \Rightarrow \alpha(1-\beta)-\beta(1-\alpha)=\alpha-\beta=0 \Rightarrow \alpha=\beta \Rightarrow A= \begin{bmatrix}\alpha& \alpha\\ 1-\alpha& 1-\alpha \end{bmatrix}\\\begin{bmatrix}a_5\\ b_5 \end{bmatrix} = \begin{bmatrix}2/3\\ 4/3 \end{bmatrix} \Rightarrow a_5+b_5=2 \Rightarrow a_n+b_n=2 \Rightarrow \begin{bmatrix}2/3\\ 4/3 \end{bmatrix} = \begin{bmatrix}\alpha& \alpha\\ 1-\alpha& 1-\alpha \end{bmatrix} \begin{bmatrix}a_4\\ 2-a_4 \end{bmatrix} = \begin{bmatrix}2\alpha\\ 2-2\alpha \end{bmatrix}\\\Rightarrow \cases{2\alpha=2/3\\ 2-2\alpha=4/3} \Rightarrow \alpha={1\over 3} \Rightarrow A= \bbox[red, 2pt]{\begin{bmatrix}1/3& 1/3\\ 2/3& 2/3 \end{bmatrix}}$$
解答:
$$假設C(a,b) \Rightarrow \triangle ABC面積={1\over 2} \begin{Vmatrix}1&5& 1\\7& 2& 1\\ a& b& 1 \end{Vmatrix} ={1\over 2} |3a+6b-33| =15 \Rightarrow 3a+6b-33=\pm 30\\ \Rightarrow \cases{a+2b=21\\ a+2b= 1} \Rightarrow C\in L_1:x+2y=21 或C\in L_2:x+2y=1 \\圓:x^2+y^2+14x+12y+80=0 \Rightarrow (x+7)^2+(y+6)^2=5 \Rightarrow 圓心O(-7,-6) \\ \Rightarrow d(O,L_2)\lt d(O,L_1) \Rightarrow C\in L_2 \Rightarrow C是O在L_2的垂足 \\\Rightarrow 過O且斜率為2的直線L_3:y=2(x+7)-6 \Rightarrow C=L_2\cap L_3=\bbox[red, 2pt]{(-3, 2)}$$
解答:$$A(-5,a,b) \in L_1:x+4={y-11\over 7}={z-7\over 2} \Rightarrow -1={a-11\over 7}={b-7\over 2} \Rightarrow \cases{a=4\\ b=5} \Rightarrow A(-5,4,5) \\ B(c,d,0 )\in L_2 \Rightarrow \cases{c-0=7\\d=3} \Rightarrow B(7,3,0) \\ 假設\cases{P(x,y,z)\\ L_1方向向量 \vec u_1=(1,7,2) \\ L_2方向向量\vec u_2=(1,0,1) } \Rightarrow \cases{\overrightarrow{PA} \cdot \vec u_1 =0\\ \overrightarrow{PB} \cdot \vec u_2 =0} \Rightarrow \cases{x+7y+2z=33\cdots(1)\\ x+z=7 \cdots(2)} \\又\overline{PA}=\overline{PB} \Rightarrow P在A,B的中垂面E上 \Rightarrow \cases{E的法向量\vec n=\overrightarrow{AB} =(12,-1,-5) \\M=\overline{AB}中點=(1,7/2,3/2)} \\ \Rightarrow E:12x-y-5z=-4 \cdots(3) \\ 由(1),(2),(3) 可得P= \bbox[red, 2pt]{(2,3,5)}$$
解答:
$$假設\Gamma_1: {x^2\over a^2}+{y^2\over b^2}=1 \Rightarrow \cases{F_1(-c,0) \\F_2(c,0)\\ A(0,b)\\ O(0,0)}, 又\Gamma_2滿足\cases{頂點A(0,b) \\焦點O(0,0)} \Rightarrow 開口向下且焦距=b \\ \Rightarrow \Gamma_2: x^2=-4b(y-b) 通過F_2(c,0) \Rightarrow c^2=4b^2\Rightarrow c=2b \Rightarrow a^2=b^2+(2b)^2 \Rightarrow a=\sqrt 5b \\ \Rightarrow {x^2\over 5b^2}+{y^2\over b^2}=1\Rightarrow x^2+5y^2=5b^2 \\ 將\Gamma_2:x^2=-4by+4b^2代入\Gamma_1 \Rightarrow (-4by+4b^2)+5y^2=5b^2 \Rightarrow (5y+b)(y-b)=0 \\ \Rightarrow y=-{b\over 5} \Rightarrow x^2=-4b(-{b\over 5})+4b^2 \Rightarrow x=\pm \sqrt{24\over 5}b \Rightarrow P(\sqrt{24\over 5}b,-{b\over 5}) \\ \Rightarrow {\overline{OP} \over \overline{OA}} = {\sqrt{{24\over 5}b^2+{b^2\over 25}} \over b} ={{11\over 5}b\over b} = \bbox[red, 2pt]{11\over 5}$$
解答:$$2a_n-a_{n-1}=4n+2 \Rightarrow a_n={1\over 2}a_{n-1}+2n+1 \\假設a_n-(pn+q) ={1\over 2}(a_{n-1}-(p(n-1)+q)) \Rightarrow a_n={1\over 2}a_{n-1}+{1\over 2}pn+({1\over 2}p+{1\over 2}q) \\ \Rightarrow \cases{{1\over 2}p=2\\ {1\over 2}p+{1\over 2}q=1} \Rightarrow \cases{p=4\\ q=-2} \Rightarrow a_n-(4n-2) ={1\over 2}(a_{n-1}-(4(n-1)-2)) \\ 取b_n=a_n-(4n-2) \Rightarrow \cases{b_n={1\over 2}b_{n-1} \\ b_1=a_1-2 =1/2}\Rightarrow b_n= {1\over 2^{n-1}}b_1 ={1\over 2^n} \Rightarrow a_n-(4n-2) ={1\over 2^n} \\ \Rightarrow \bbox[red, 2pt]{a_n={1\over 2^n}+(4n-2)}$$
解答:$$f(x)除以(x-2)^2的餘式為12x-20 \Rightarrow \cases{f(2)=24-20=4\\f'(2)=12} \\f(x)除以(x-4)^2的餘式為12x-36 \Rightarrow \cases{f(4)=48-36=12\\f'(4)=12} \\ f為三次式\Rightarrow f'為二次式，又f'(2)=f'(4)=12 \Rightarrow y=f'(x)頂點的x坐標={2+4\over 2}=3 \\\Rightarrow f''(3)=0 \Rightarrow h=3 \Rightarrow f(3) ={f(2)+f(4)\over 2}=8 \\ g(x)=xf(x)= A(x-2)(x-3)(x-4)+ px^2+qx+r \Rightarrow \cases{g(2)=2\cdot f(2)=8=4p+2q+r\\ g(3)=3\cdot f(3)=24= 9p+3q+r\\ g(4)=4\cdot f(4)=48=16p+4q+r} \\ \Rightarrow \cases{p=4\\q=-4\\ r=0} \Rightarrow 餘式:\bbox[red, 2pt]{4x^2-4x}$$
解答:$$假設\cases{O(0,0,0) \\A(13,0,0)\\ B(0,13,0) \\地平面E:x+ay+bz=0} \Rightarrow \cases{d(A,E)={13\over \sqrt{1+a^2+b^2}}=3\\ d(B,E)={13a\over \sqrt{1+a^2+b^2}} =4} \Rightarrow a={4\over 3} \Rightarrow b=4 \\ \Rightarrow E=x+{4\over 3}y+4z=0 \Rightarrow d((13,13,13),E)= {13+{52\over 3}+52\over \sqrt{1+{16\over 9}+16}} =\bbox[red, 2pt]{19}$$
解答:
$$\overline{AD} 為\angle A 的角平分線\Rightarrow \angle CAD=\angle BAD=\theta \Rightarrow {\overline{CD} \over \overline{BD}} ={\overline{AC} \over \overline{AB}} \Rightarrow \cases{\overline{CD} = 2\sqrt{11}/3\\ \overline{BD}= 4\sqrt{11} /3}\\ 又\cases{\overline{AB}=8 \\\overline{AE} =3\overline{BE}} \Rightarrow \cases{\overline{AE}=6\\ \overline{BE}=2} \Rightarrow \cos 2\theta ={8^2+4^2 -(2\sqrt{11})^2 \over 2\cdot 4\cdot 8} ={9 \over 16} ={6^2+ 4^2-\overline{CE}^2 \over 2\cdot 4\cdot 6} \\ \Rightarrow \overline{CE}= 5 \Rightarrow {\overline{CG} \over \overline{GE}}={\overline{AC} \over \overline{AE}} ={4\over 6} \Rightarrow \overline{CG}=2 \Rightarrow {\overline{CG} \over \sin \theta}=2R \Rightarrow R= {1\over \sin \theta} \\ \Rightarrow 外接圓面積=R^2\pi ={\pi\over \sin^2\theta} ={2\pi\over 1-\cos 2\theta}={2\pi\over 1-9/16} = \bbox[red, 2pt]{32\pi\over 7}$$
解答:$$\cases{f=\sqrt{3xy}+5z \\g=x^2+4y^2+4z^2 -12} \Rightarrow \cases{f_x= \lambda g_x\\f_y= \lambda g_y\\ g=0} \Rightarrow \cases{{3y\over 2\sqrt{3xy}}= \lambda(2x) \\ {3x\over 2\sqrt{3xy}} =\lambda(8y) \\5=\lambda(8z)} \Rightarrow \cases{x=2y\\ z={5\over 3}\sqrt 6y} \\ \Rightarrow 4y^3+4y^2+{600\over 9}y^2=12 \Rightarrow y={3\over 28}\sqrt{14} \Rightarrow \cases{x=3\sqrt{14}/14\\ z=5\sqrt{21}/14} \\ \Rightarrow \sqrt{3xy}+5z= {3\sqrt{21}\over 14}+{25\sqrt{21}\over 14} = \bbox[red, 2pt]{2\sqrt{21}}$$
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臺中市立臺中第一高級中等學校115學年度第1次教師甄選數學科標準答案
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