114年彰化高中教甄-數學詳解

 國立彰化高級中學 114 學年度第一次教師甄選初試

一、填充題 (每題 5 分，共 70 分：不用詳述計算過程，寫答即可，全對始計分)

解答:$$x\gt 0時，f(x)\gt 0 \Rightarrow \cases{圖形凹向上 \Rightarrow 3-k \ge 0 \Rightarrow 3\ge k\\ f(0)\ge 0 \Rightarrow k+4 \ge 0 \Rightarrow k\ge -4} \Rightarrow \bbox[red, 2pt]{-4\le k\le 3}$$

解答:$$\cases{(x^2+y^2)^3 = x^6+y^6 +3x^4y^2 +3x^2y^4 \\ (x^3-3xy^2)^2 +(y^3-3x^2y)^2= x^6-6x^4y+9x^2y^4 +y^6-6x^2y^4+9x^4y^2 =x^6+y^6+3x^2y^4+3x^4y^2} \\ \Rightarrow (x^2+y^2)^3 =(x^3-3xy^2)^2 +(y^3-3x^2y)^2 =5^2+1^2=26 \Rightarrow x^2+y^2 = \bbox[red, 2pt]{\sqrt[3]{26}}$$

解答:$$\cases{E_1與稜邊交點為(1/2,0,1), (1,0,1/2), (1/2,1,0), (0,1,1/2), (1,1/2,0), (0,1/2,1) \Rightarrow a=6\\ E_2與稜邊交點為 (1,0,2/3), (2/3,0,1),( 1,1,1/3), (1/3,1,1) \Rightarrow b=4 \\E_3與稜邊交點為(2/3,0,1), (0,2/3,1), (1,0,3/4),(0,1,3/4), (1,1,0) \Rightarrow c=5} \\ \Rightarrow (a,b,c)= \bbox[red, 2pt]{(6,4,5)}$$

解答:

$$假設A(5,2,-2)在L的投影點為A' \Rightarrow A'(2t+1, t+2,-2t+3) \Rightarrow \overrightarrow{AA'} =(2t-4,t,-2t+5) \\ \Rightarrow \overrightarrow{AA'} 與L的方向向量(2,1,-2) 垂直\Rightarrow 2(2t-4)+t-2(-2t+5)=0 \Rightarrow t=2 \Rightarrow A'(5,4,-1)\\ 同理，B(1,-3,5)在L上的投影點B'=(-1,1,5) \\ \triangle PAA' \sim \triangle PBB' \Rightarrow {\overline{PA'} \over \overline{PB'}} ={\overline{AA'} \over \overline{BB'}} ={\sqrt 5\over 2\sqrt 5} ={1\over 2}\Rightarrow P={2A'+B'\over 3} ={1\over 3}(9,9,3) =\bbox[red, 2pt]{(3,3,1)}$$

解答:$$t=\log x \Rightarrow t^2-[t]-6=0 \Rightarrow [t]= t^2-6 \Rightarrow t-1\lt t^2-6\le t \\ 右式:t^2-6\le 6 \Rightarrow t^2-t-6=(t-3)(t+2) \le 0 \Rightarrow -2\le t\le 3 \cdots(1) \\左式:t-1\lt t^2-6 \Rightarrow t^2-t-5\gt 0 \Rightarrow t\gt {1+\sqrt{21} \over 2} 或t\lt {1-\sqrt{21} \over 2} \cdots(2) \\(1) \cap (2) \Rightarrow \cases{-2\le t\lt \displaystyle {1-\sqrt{21}\over 2} (\approx -1.8) \Rightarrow 3.2\lt t^2\le 4 \Rightarrow -2.8 \lt [t]\le -2 \Rightarrow [t]=-2\\ \displaystyle {1+\sqrt{21}\over 2}(\approx 2.8) \lt t\le 3 \Rightarrow 7.8\lt t^2\le 9 \Rightarrow 1.8\lt [t]\le 3  \Rightarrow [t]=2,3}  \\ \Rightarrow \cases{[t]=-2 \Rightarrow -2=t^2-6 \Rightarrow t=-2 \Rightarrow x=10^{-2}\\ [t]=2 \Rightarrow 2=t^2-6 \Rightarrow t=2\sqrt 2 \Rightarrow x=10^{2\sqrt 2}\\ [t]=3 \Rightarrow 3=t^2-6 \Rightarrow t=3 \Rightarrow x=10^3} \Rightarrow x= \bbox[red, 2pt]{{1\over 100},10^{2\sqrt 2}, 1000}$$

解答:$$f(x)=x^3+(-a^2+2a+2)x-2a^2-2a =(x-a)(x^2+ax+2a+2) \\ \Rightarrow x^2+ax+2a+2=0有二虛根\Rightarrow a^2-4(2a+2)=a^2-8a-8\lt 0 \\ \Rightarrow 4-2\sqrt 6\lt a\lt 4+2\sqrt 6 又a\gt 0, 因此 \bbox[red, 2pt]{0\lt a\lt 4+2\sqrt 6}$$

解答:

$$假設\cases{\angle BAC=\theta \\ A(0,0)}\Rightarrow \cases{B(7\cos \theta, 7\sin \theta)\\ C(3,0) \\ P(4\cos \theta, 4\sin \theta) \\ Q(2,0)} \Rightarrow M={B+C\over 2} =  \left( {3+7\cos \theta\over 2}, {7\sin \theta\over 2} \right) \\ \Rightarrow \cases{\overrightarrow{MP} = \left( \displaystyle {\cos \theta-3\over 2}, {\sin \theta\over 2} \right) \\ \overrightarrow{MQ} = \left( \displaystyle {1-7\cos \theta\over 2}, -{7\sin \theta\over 2} \right)} \Rightarrow \overrightarrow{MP} \cdot \overrightarrow{MQ}=0 \Rightarrow {1\over 2}(11 \cos \theta-5)=0 \Rightarrow \cos \theta= \bbox[red, 2pt]{5\over 11}$$

解答:$$(2^6-2^4)\times 2=(64-16)\times 2= \bbox[red, 2pt]{96}\\ \href{https://zhidao.baidu.com/question/1836376399825521020.html}{相關說明}$$

解答:$$\cases{x=1 \Rightarrow f(1)+f(0)=1 \Rightarrow f(1)=1\\ x=1/2 \Rightarrow f(1/2)+f(1/2)=1 \Rightarrow f(1/2)=1/2} \Rightarrow f({x=1\over 4}) ={1\over 2}f(1) \Rightarrow f({1\over 4})={1\over 2} \\ \Rightarrow \cases{f({1\over 2})=f({1\over 4}) ={1\over 2} \\ f為遞增} \Rightarrow f(x)={1\over 2}, \forall x\in \left[{1\over 4}, {1\over 2} \right] \\\Rightarrow f({1\over 2025}) ={1\over 2}f({4\over 2025})= \cdots ={1\over 2^5}f({4^5\over 2025}) ={1\over 32} f({1024\over 2025}) = {1\over 32} \left( 1-f(1-{1024\over 2025}) \right) \\={1\over 32} \left( 1-f({1001\over 2025}) \right) ={1\over 32} \left( 1-{1\over 2} \right) = \bbox[red, 2pt]{1\over 64}$$

解答:$$假設L與拋物線y=f(x)=x^2-2x-5交點為\cases{P(\alpha, f(\alpha)\\ Q(\beta, f(\beta)}, \\過A(-1,0)的直線L: y=m(x+1)代入拋物線 \Rightarrow x^2-2x-5=m(x+1)  \\ \Rightarrow x^2-(m+2)x-5-m=0 \Rightarrow \cases{\alpha+\beta=m+2\\ \alpha \beta=-5-m} \\ 所圍面積最小\Rightarrow \overline{PQ}中點為A \Rightarrow \alpha+\beta=-2=m+2 \Rightarrow m=-4 \Rightarrow L: \bbox[red, 2pt]{y=-4(x+1)}$$

解答:

$$假設\cases{B(0,0) \\ \overleftrightarrow{BE}為x軸 \\ F=\overline{BE} \cap \overline{AD}} \Rightarrow \overline{BF}為\overline{AD}的中垂線 \Rightarrow \cases{A(a,4) \\F(a,0) \\D(a,-4) \\E(8,0)} \\ 又D={B+C\over 2} \Rightarrow C=2D-B=(2a,-8) \\ \overline{BE}為\angle B的角平分線 \Rightarrow {\overline{AE} \over \overline{EC}} = {\overline{BA} \over \overline{BC}} ={1\over 2} \Rightarrow E={2A+C\over 3} \Rightarrow 8={4a\over 3} \Rightarrow a=6 \\ \Rightarrow \cases{A(6,4) \\ C(12,-8)} \Rightarrow \cases{\overline{AB} = 2\sqrt{13} \\ \overline{BC}=2\overline{AB}=4\sqrt{13} \\ \overline{AC}= 6\sqrt 5} \Rightarrow 周長= \bbox[red, 2pt]{6\sqrt{13}+6\sqrt 5}$$

解答:$$\lim_{n\to \infty} \sum_{k=1}^n {\sqrt{4n^2-3k^2}\over 2n^2} =\lim_{n\to \infty} \sum_{k=1}^n {\sqrt{4-3(k/n)^2}\over 2n } = \int_0^1 {\sqrt{4-3x^2}\over 2}\,dx \\ x={2\over \sqrt 3} \sin u \Rightarrow \cases{dx =(2/\sqrt 3) \cos u\,du\\ u= \sin^{-1}(\sqrt 3 x/2)} \Rightarrow I=\int_0^1 {\sqrt{4-3x^2}\over 2}\,dx = \int_0^{\pi/3} {1\over \sqrt 3} \cos u\cdot \sqrt{4-4\sin^2 u}\,du \\={2\over \sqrt 3} \int_0^{\pi/3} \cos^2 u\,du = {1\over \sqrt 3} \int_0^{\pi/3} (\cos 2u+1)\,du ={1\over \sqrt 3} \left. \left[ {1\over 2}\sin 2u+u \right] \right|_0^{\pi/3}= {1\over \sqrt 3} \left( {\sqrt 3\over 4}+{\pi\over 3} \right) \\= \bbox[red, 2pt]{9+4\sqrt 3 \pi\over 36}$$

解答:$$S= \sum_{n=1}^{2025} \left[{2025+2^n\over 2^{n+1}}  \right] = \sum_{n= 1}^{2025} \left[{2025\over 2^{n+1}} +{1\over 2} \right] \\\Rightarrow \cases{n=1 \Rightarrow S_1=506\\ n=2 \Rightarrow S_2 =253\\ n=3 \Rightarrow S_3= 127\\ n=4 \Rightarrow S_4 =63\\ n=5 \Rightarrow S_5=32\\ n=6 \Rightarrow S_6=16\\ n=7 \Rightarrow S_7=8\\ n=8 \Rightarrow S_8=4\\ n=9 \Rightarrow S_9=2 \\ n=10 \Rightarrow S_{10}=1 \\n\ge 11 \Rightarrow S_n=0} \Rightarrow \sum_{n=1}^{10} \left[{2025\over 2^{n+1}} +{1\over 2} \right]= \bbox[red, 2pt]{1012}$$

解答:$$假設\cases{A(0,3,0) \\B(0,0,4) \\C(1,0,0)} \Rightarrow \cases{d(C,A) =|1-0|+|0-3|+|0-0|=4\\ d(C,B)=|1-0|+ |0-0|+|0-4| =5} \Rightarrow s=4+5=9 \\ P(x,y,z)在橢圓上 \Rightarrow d(P,A)+d(P,B)=s \Rightarrow 2|x|+(|y|+ |y-3|) +(|z|+|z-4|) =9 \\ \Rightarrow \cases{f(y)=|y|+|y-3| \\ g(z)=|z|+ |z-4|} \Rightarrow \cases{\cases{f(y)=3, 0\le y\le 3\\ f(y)=3+2d_y, d_y=y到區間[0,3]的距離} \\ \cases{g(z)=4, 0\le z\le 4\\ g(z)=4+2d_z, d_z=z到區間[0,4]的距離}} \\ \Rightarrow 2|x|+3+2d_y+4+2d_z=9 \Rightarrow |x|+d_y+d_z=1 \\ \textbf{Case I }\cases{y\in [0,3]\\ z\in [0,4]} \Rightarrow |x|\le 1 \Rightarrow \cases{x範圍長度=2\\ y範圍長度=3\\ z範圍長度=4 } \Rightarrow 體積V_1=2\times 3\times 4=24\\ \textbf{Case II }\cases{z\in[0,4] \\ y\ge 3}  \Rightarrow |x|+d_y\le 1 \Rightarrow V_2= (\text{底面積} \times \text{柱高}) \times 2 = (1 \times 4) \times 2 = 8 \\ \textbf{Case III }\cases{y\in[0,3] \\z\ge 4}  \Rightarrow |x|+d_z\le 1 \Rightarrow V_3=(1\times 3)\times 2=6\\ \textbf{Case IV }\cases{y\ge 3\\ z\ge 4} \Rightarrow |x|+d_y+d_z \le 1 \Rightarrow V_4= 4\times \int_{-1}^1 {1\over 2}(1-|x|)^2\,dx =4\times {1\over 3}={4\over 3} \\ \Rightarrow V= V_1+ V_2+V_3 +V_4 =24+8+6+{4\over 3} =\bbox[red, 2pt]{118\over 3}\\ 絕對橢球經由電腦繪圖如下:$$

二、計算證明題 (配分如各小題，共30分，要有計算或證明過程)

解答:$$f(x)=x+3+\sqrt{5-x^2} \Rightarrow f'(x)=1-{x\over \sqrt{5-x^2}} =0 \Rightarrow {x\over \sqrt{5-x^2}} =1\cdots(1) \\ \Rightarrow x^2=5-x^2 \Rightarrow x=\sqrt{5\over 2} \left( x=-\sqrt{5\over 2} 不符式(1) \right) \Rightarrow \cases{f(\sqrt{5/2}) =3+\sqrt{10} \\邊界點\cases{f(\sqrt 5) =3+\sqrt 5\\ f(-\sqrt 5)=3-\sqrt 5}} \\ \Rightarrow \bbox[red, 2pt]{\cases{最大值:3+\sqrt{10} \\最小值:3-\sqrt 5}}$$

解答:$$\textbf{(1) } {x^2 \over y+z}+{y^2\over z+x}+{z^2\over x+y}=0 \Rightarrow  x+y+z=\left( x+{x^2\over y+z}  \right)+ \left( y+{y^2\over z+x} \right) + \left( z+{z^2\over x+y} \right) \\\qquad =x \left( {x+y+z\over y+z} \right) +y \left( {x+y+z\over z+x} \right) +z \left( {x+y+z\over x+y} \right) =(x+y+z) \left( {x\over y+z} + {y\over z+x}+{z\over x+y} \right) \\\qquad \Rightarrow {x\over y+z} + {y\over z+x}+{z\over x+y}= \bbox[red, 2pt]1 \\\textbf{(2) }   {x\over y+z} + {y\over z+x}+{z\over x+y}=  1 \Rightarrow {x(z+x)(x+y) + y(y+z)(x+y) +z(y+z)(z+x) \over (x+y) (y+z)(z+x)}=1 \\ \quad \Rightarrow (x+y) (y+z)(z+x) =(x+y+z)(xy+ yz+zx)-xyz \\ \quad 用力展開\Rightarrow  x^3+y^3+z^3 +3xyz=2xyz \Rightarrow x^3+y^3+z^3=-xyz \\ \quad \Rightarrow {x^2\over yz}+{y^2\over xz} +{z^2\over xy} =-{xyz\over xyz} = \bbox[red, 2pt]{-1}$$

解答:

$$\textbf{(1) }\cases{\Gamma_1 圖形凹向上，最低頂點A(0,-k)\\ \Gamma_2圖形凹向左，最右頂點B(k,30)} \Rightarrow 有相異四交點代表B必需在\Gamma_1的外側(右側)\\\quad  \Gamma_1 對稱y軸，右側圖形的方程式為x=\sqrt{y+k} \Rightarrow k\ge \sqrt{30+k} \Rightarrow k^2-k-30\ge 0 \\ \quad \Rightarrow (k-6)(k+5)\ge 0 \Rightarrow k\ge 6 \Rightarrow k的最小值為6\quad \bbox[red, 2pt]{QED} \\\textbf{(2) }k=6 \Rightarrow \cases{\Gamma_1: y=x^2-6 \Rightarrow x^2-y-6=0\\ \Gamma_2:x=-2(y-30)^2+6 \Rightarrow 2y^2+x-120y+1794=0}  \\\quad \Rightarrow 2(x^2-y-6) +(2y^2+x-120y+1794) =0 \Rightarrow x^2+y^2+{1\over 2}x-61y+891=0 \\ \Rightarrow \left( x+{1\over 4} \right)^2+  \left( y-{61\over 2} \right)^2 ={629\over 16} \Rightarrow r= \bbox[red, 2pt]{\sqrt{629}\over 4}$$

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