國立竹北高中114 學年度第1 次教師甄選數學科試題卷
解答：$$f(x)=\int_{-1}^1 (x-t)f(t)\,dt +1 =x\int_{-1}^1 f(t)\,dt -\int_{-1}^1 tf(t)+1\Rightarrow f(x)=ax+b, a,b\in \mathbb R \\ \Rightarrow \cases{a=\int_{-1}^1 f(t)\,dt \\b=-\int_{-1}^1 tf(t)\,dt+1} \Rightarrow \cases{a=\int_{-1}^1 (at+b)\,dt =2b\\ b=-\int_{-1}^1t(at+b)\,dt+1 =-{2\over 3}a+1} \\ \Rightarrow b=-{2\over 3}\cdot (2b)+1 \Rightarrow {7\over 3}b=1 \Rightarrow b={3\over 7} \Rightarrow a=2b={6\over 7}\Rightarrow \bbox[red, 2pt]{f(x)={6\over 7}x+{3\over 7}}$$
解答：$$依題意\cases{球心O(t,2t,t-1) \\ 球半徑R=2} \Rightarrow \cases{d(O,xy平面)=|t-1|\\ d(O, yz平面)=|t|\\ d(O, xz平面)=|2t| } \Rightarrow 截圓面積\cases{A_{xy}=\pi [2^2-(t-1)^2]\\ A_{yz}=\pi[2^2-t^2] \\ A_{xz} =\pi[2^2-4t^2]} \\ \Rightarrow 面積和f(t)= \pi \left( 12-(t-1)^2-t^2-4t^2 \right) \Rightarrow f'(t)=0 \Rightarrow t={1\over 6} \Rightarrow \cases{f(1/6) =67\pi/6\\ O=(1/6,1/3,-5/6)} \\ \Rightarrow (M,a,b,c) = \bbox[red, 2pt]{\left( {67\pi\over 6},{1\over 6},{1\over 3},-{5\over 6} \right)}$$
解答：$$\cases{z_1=x+iy\\ z_2=x-iy} \Rightarrow |z_1-z_2|=|2yi|=4\sqrt 3 \Rightarrow y=\pm 2\sqrt 3 \Rightarrow y^2=12\\ {z_1\over z_2^2} ={x+iy\over (x^2-y^2)+ i(-2xy)} \in \mathbb R \Rightarrow {x\over x^2-y^2} ={y\over -2xy} \Rightarrow {x\over x^2-12} =-{1\over 2x} \Rightarrow x^2=4\\ \Rightarrow |z_1|= \sqrt{x^2+y^2} =\sqrt{4+12} =\bbox[red, 2pt] 4$$
解答：
$$G=正\triangle OAB的重心= 正\triangle C_1C_2C_3的重心 \Rightarrow \cases{\overline{GC_2}=6\times {\sqrt 3\over 2}\times {2\over 3} =2\sqrt 3 \\ \overline{GP} =2\times {\sqrt 3\over 2} \times {1\over 3}={\sqrt 3\over 3}} \\ \Rightarrow \overline{PC_3}=2\sqrt 3-{\sqrt 3\over 3}={5\over 3}\sqrt 3 \Rightarrow 三角錐的高h=\sqrt{\overline{PC_3}^2-\overline{PG}^2} = 2\sqrt 2 \\ \Rightarrow 三角錐體積={1\over 3}\cdot \triangle OAB面積\cdot h={1\over 3} \cdot {\sqrt 3\over 4}\cdot 2^2 \cdot 2\sqrt 2 = \bbox[red, 2pt]{{2\over 3}\sqrt 6}$$
解答：
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解答：$$b_n={2^n\over a_n} \Rightarrow a_n={2^n\over b_n} \Rightarrow a_{n+1}={2^{n+1}\over b_{n+1} } \Rightarrow {2^{n+1}\over b_{n+1} } ={2^{n+1} \cdot {2^n\over b_n} \over (n+{1\over 2}){2^n\over b_n} +2^n} \\ \Rightarrow {2\over b_{n+1}} \Rightarrow {1\over b_{n+1} } ={1\over (n+1/2)+b_n} \Rightarrow b_{n+1}=b_n+ n+{1\over 2} \Rightarrow b_{n+1}-b_n=n+{1\over 2} \\ \Rightarrow b_n=b_1+ \sum_{k=1}^{n-1} (b_{k+1}-b_k) =1+ \sum_{k=1}^{n-1} \left( k+{1\over 2} \right) =1+{n(n-1)\over 2}+{1\over 2}(n-1) =\bbox[red, 2pt]{n^2+1\over 2}$$
解答：
$$D=\overline{BC}中點 \Rightarrow \angle BAD=\angle CAD=\theta \Rightarrow\cases{ \overline{AD}=\displaystyle {1\over \tan \theta} = {\cos \theta\over \sin \theta} \\ \overline{AB}=\displaystyle {1\over \sin \theta}} \\\Rightarrow \cases{\triangle ABC面積= {1\over 2} \overline{BC}\cdot \overline{AD}= \displaystyle {\cos \theta \over \sin \theta} \\ \triangle ABC周長=2\overline{AB}+ \overline{BC} = \displaystyle {2\over \sin \theta}+2} \Rightarrow r={\triangle ABC面積\over \triangle ABC周長/2} = {\cos \theta\over 1+\sin \theta} \\ 正弦定理: {\overline{BC} \over \sin \angle A} =2R \Rightarrow R={1\over \sin 2\theta} \Rightarrow {r\over R}=f(\theta) ={\cos \theta \sin 2\theta\over 1+\sin \theta} ={2\sin \theta \cos^2 \theta\over 1+\sin \theta}\\ ={2\sin \theta(1+\sin\theta)(1-\sin \theta) \over 1+\sin \theta} =2\sin \theta(1-\sin \theta) =2\sin \theta-2\sin^2\theta \\ \Rightarrow f'(\theta)=2\cos \theta-4\sin \theta \cos \theta =0 \Rightarrow 2\cos \theta(1- 2\sin \theta) =0 \Rightarrow \sin \theta={1\over 2} \Rightarrow \cos \theta ={\sqrt{3} \over 2}\\ \Rightarrow {r\over R}={2\cdot {1\over 2}\cdot {3\over 4} \over 1+{1\over 2}} = \bbox[red, 2pt]{1\over 2}$$
解答：$$$$
解答：
$$\cases{D(0,0,0) \\ A(1,0,0)\\ B(1,1,0)\\ C(0,2,0)\\ P(0,0,\sqrt 2) \\ E(0,2 ,\sqrt 2)} \Rightarrow F={A+P\over 2} =({1\over 2},0,{\sqrt 2\over 2}) \Rightarrow \cases{E=\triangle PBC: x+ y+\sqrt 2z=2 \\ L=\overleftrightarrow{EF}: -x={y-2\over 4}={z-\sqrt 2\over \sqrt 2}} \\ \Rightarrow Q(-t,4t+2, \sqrt 2t+\sqrt 2) \Rightarrow \cases{d(Q,E) =\overline{QQ'} ={|5t+2|\over 2} \\ \overline{BQ} =\sqrt{19t^2+14t+4}} \Rightarrow \sin {\pi\over 6}={d(Q,E) \over \overline{BQ}}\\ \Rightarrow {|5t+2|/2 \over \sqrt{19t^2+14t+4}} ={1\over 2} \Rightarrow t^2+t=0 \Rightarrow t=-1 \Rightarrow Q(1,-2, 0) \\ \Rightarrow \overline{QF} =\sqrt{{1\over 4}+4+ {1\over 2}} = \bbox[red, 2pt]{\sqrt{19}\over 2}$$
解答：$$g(x)=f(x+2)-2是奇函數 \Rightarrow g(-x)=-g(x) \Rightarrow f(-x+2)-2,=-f(x+2)+2\\ \Rightarrow f(2+x)+ f(2-x)=4 \cdots(1) \\h(x)=f(2x+1)是偶函數\Rightarrow h(-x)=h(x) \Rightarrow f(-2x+1)=f(2x+1) \Rightarrow f(1-x)=f(1+x) \\ \Rightarrow f(x+2)= f(-x) \xrightarrow{代入(1)} f(-x)+f(2-x)=4 \Rightarrow f(x+2)=4-f(x)\\ \Rightarrow f(x+4)=f((x+2)+2)=4-f(x+2) =4-(4-f(x))=f(x) \Rightarrow f(x+4)=f(x) \Rightarrow 週期為4\\\cases{f(1)=0 \\ f(x)+f(x+2)=4 \Rightarrow f(1)+f(3)=4 \Rightarrow f(3)=4\\f(2-x)+f(2+x)=4 \Rightarrow f(2)+f(2)=4 \Rightarrow f(2)=2 \\ f(x)+f(x+2)=4 \Rightarrow f(2)+f(4)=4 \Rightarrow f(4)=2} \Rightarrow f(1)+ f(2)+f(3) +f(4)=8\\2025=4\times 506+1 \Rightarrow f(1)+f(2)+...+f(2025)=506\times 8+f(1)= \bbox[red, 2pt]{4048}$$
解答：$$(y-x) \left( y-{18\over 25x} \right) \ge 0 \Rightarrow \cases{條件一:y\ge x且y \ge {18\over 25x} \\條件二:y\le x且y \le {18\over 25x}} ,由於要找2x-y的最小值,因此以條件一為主\\ y={18\over 25x} \Rightarrow xy={18\over 25} \Rightarrow 2x-y的最小值發生在兩圖形\cases{(x-1)^2+ (y-1)^2=1\\ xy=18/25} 的交點 \\ \Rightarrow \cases{x=\cos \theta+1\\ y=\sin \theta+1} \Rightarrow xy=\cos \theta \sin \theta+ \cos \theta+\sin \theta+1={18\over 25} \\ 取t=\sin \theta+ \cos \theta \Rightarrow \sin \theta\cos \theta={t^2-1\over 2} \Rightarrow {t^2-1\over 2}+t+1={18\over 25} \Rightarrow 25t^2+50t-11=0 \\ \Rightarrow (5t-1)(5t+11)=0 \Rightarrow t=1/5 \Rightarrow \cos \theta+\sin \theta=1/5 \Rightarrow x+y=1/5+2={11/5 } \\ \Rightarrow \cases{x +y=11/5\\ xy=19/25} \Rightarrow 25x^2-55x+18=0 \Rightarrow \cases{x=2/5 \Rightarrow y=9/5\\ x=9/5 \Rightarrow y=2/5} \\ \Rightarrow 2x-y的最小值={4\over 5}-{{9\over 5}}=\bbox[red, 2pt]{-1}$$
解答：
$$z=x+iy \Rightarrow |z-3i|= {z-\bar z\over 2i} \Rightarrow \sqrt{x^2+(y-3)^2}={2yi\over 2i}=y \Rightarrow \overline{PF}=d(P,L:x=0) \\ \Rightarrow \Gamma 為一拋物線,其中\cases{焦點F(0,3)\\ 準線L:x=0} \Rightarrow |z+2-4i|+|z-3i|= \overline{PA}+\overline{PF},其中A(-2,4) \\ \Rightarrow \overline{PA}+\overline{PF}= \overline{PA}+d(P,L) 最小值=d(A,L)= \bbox[red, 2pt]4$$
解答：$$2x^4-6x^2+8x+16=2(x^4-3x^2+4x+8)=2[(x^4-4x^2+4)+(x^2+4x+4)] \\=2[(x^2-2)^2+(x+2)^2] \Rightarrow 取\cases{A=x^2-2\\ B=x+2} \Rightarrow f(x)=A-B+4+ \sqrt 2\cdot \sqrt{A^2+B^2} \\取\cases{\vec u=(1,-1)\\ \vec v=(A,B)} \Rightarrow \cases{|\vec u|=\sqrt 2 \\ |\vec v|= \sqrt{A^2+B^2} \\ \vec u\cdot \vec b=A-B} \Rightarrow f(x)=4+ \vec u\cdot \vec v+|\vec u||\vec v|= 4+ |\vec u||\vec v|\cos \theta+|\vec u||\vec v| \\ \Rightarrow f(x)=4+ |\vec u||\vec v|(\cos \theta+1) 最小值發生在\cos \theta=-1 \Rightarrow \vec u與\vec v反向 \Rightarrow \vec v=-k\vec u , k\ge 0 \\ \Rightarrow (A,B)=-k(1,-1)=(-k,k) \Rightarrow \cases{A=-B \Rightarrow x^2-2=-(x+2) \Rightarrow x=0,-1\\ B\ge 0 \Rightarrow \cases{x=0 \Rightarrow B=2\ge 0\\ x=-1 \Rightarrow B=1 \ge 0}} \\ \Rightarrow x=0,x=-1均符合條件\Rightarrow \cases{f(0)=4\\ f(-1)=4} \Rightarrow (a,m)= \bbox[red, 2pt]{(0,4) 或(-1,4)}$$
解答：$$g(x)={\sin \pi x\over x^2}+ {\sin \pi(1-x)\over (1-x)^2} ={\sin \pi x\over x^2}+ {\sin (\pi-\pi x)\over (1-x)^2} ={\sin \pi x\over x^2}+ {\sin \pi x\over (1-x)^2} \\\qquad =\sin \pi x \left( {1\over x^2}+{1\over (1-x)^2} \right) \Rightarrow \cases{y=g(x)圖形對稱x={1\over 2} \\ \lim_{x\to 0} g(x) =\lim_{x\to 1}g(x)=\infty} \Rightarrow g(1/2)=\bbox[red, 2pt]8為最小值$$
解答：
$$假設直線\overleftrightarrow{O_1O_2} 斜率為m \Rightarrow \cases{O_1(a, ma) , 半徑r_1=ma\\O_2(b,mb), 半徑r_2=mb} ,其中b\gt a\gt 0, m\gt 0 \\ \Rightarrow \cases{r_1r_2=2=m^2ab\\ \overline{O_1P}=r_1 \Rightarrow (a-2)^2+(ma-2)^2=m^2a^2\\ \overline{O_2P}=r_2 \Rightarrow (b-a)^2+ (mb-2)^2 =m^2b^2} \Rightarrow {a^2-4a+8\over 4a}=m= {b^2-4b+8\over 4b} \\ \Rightarrow a^2b-ab^2+8b-8a=0 \Rightarrow (ab-8)(a-b)=0 \Rightarrow ab=8 \Rightarrow 8m^2=2 \Rightarrow m={1\over 2} \\ \Rightarrow \tan \angle O_1OA =\tan \theta={1\over 2} \Rightarrow \tan \angle BOA=\tan 2\theta={1\over 1-1/4} ={4\over 3} \Rightarrow L: \bbox[red, 2pt]{y={4\over 3}x}$$
第二部份、計算證明題(共20分)
解答：
$$\Gamma: {x^2\over 9}+{y^2\over 5}=1 \Rightarrow \cases{a=3\\ b=\sqrt 5} \Rightarrow c=2 \Rightarrow F_1(-2,0) \\ B(x_0,y_0 ) \Rightarrow m_{\overline{QB}} \cdot m_{\overline{F_1B}} =-1 \Rightarrow {y_0-0\over x_0-(-9/2)} \cdot {y_0-0\over x_0-(-2)}=-1 \Rightarrow y_0^2=-(x_0^2+{13\over 2}x_0+9) \\ B\in \Gamma \Rightarrow {x_0^2\over 9}+{-(x_0^2+{13\over 2}x_0+9)\over 5}=1 \Rightarrow 8x_0^2+117x_0+252=0 \\ \Rightarrow (x_0+12)(8x_0+21)=0 \Rightarrow \cases{x_0=-21/8\\ x_0=-12 \not \in [-3,3]} \Rightarrow x_0=-{21\over 8} \Rightarrow y_0^2={75\over 64} \Rightarrow y_0= -{5\sqrt{3} \over 8} \\ \Rightarrow m={y_0-0\over x_0-(-2)} ={-5\sqrt 3/8\over -5/8} = \bbox[red, 2pt]{\sqrt 3}$$
解答：$$A= \begin{bmatrix}7&18\\ -3&-8 \end{bmatrix} \Rightarrow \det(A-\lambda I)=\lambda^2+\lambda-2=0 \Rightarrow (\lambda-1)(\lambda+2)=0 \\ \lambda_1=1 \Rightarrow (A-\lambda _1 I)v=0 \Rightarrow x+3y=0 ,取v_1= \begin{bmatrix}3\\ -1 \end{bmatrix} \\ \lambda_2=-2 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow x+2y=0, 取v_2= \begin{bmatrix}2\\ -1 \end{bmatrix} \\ P_1= \begin{bmatrix}1\\ 0 \end{bmatrix} =v_1-v_2 \Rightarrow P_n=A^{n-1}(v_1-v_2)=1^{n-1}v_1-(-2)^{n-1} v_2= \begin{bmatrix} 3-2\cdot (-2)^{n-1} \\ -1+(-2)^{n-1} \end{bmatrix} \\ \Rightarrow \bbox[red, 2pt]{\cases{x_n =3-2\cdot (-2)^{n-1} \\y_n=-1+(-2)^{n-1}}} \Rightarrow \bbox[red, 2pt]{x_n+2y_n=1} \Rightarrow \bbox[red, 2pt]{L:x+2y=1}$$
解題僅供參考，其他教甄試題及詳解
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