114年嘉義高中教甄2-數學詳解

 國立嘉義高級中學 114 學年度第一學期第2次教師甄選

一、 填充題：共 15 題，每題 6 分，合計 90 分

解答:$$  4\sin \left( {\pi\over 6} +x\right)-4\cos x=1  \Rightarrow 4 \left({1\over 2}\cos x+{\sqrt 3\over 2}\sin x-\cos x \right) =4 \left( {\sqrt 3\over 2}\sin x-{1\over 2}\cos x \right) \\=4 \sin \left( x-{\pi\over 6}\right)=1\Rightarrow \sin \left( x-{\pi\over 6}\right) ={1\over 4} \\ 假設\theta=x-{\pi\over 6} \Rightarrow 0\le x=\theta+ {\pi\over 6} \le 4\pi \Rightarrow -{\pi\over 6}\le \theta\le {23\over 6}\pi \\  \textbf{Case I } \theta=2k\pi+ \sin^{-1} \left({1\over 4} \right) \Rightarrow \cases{k=0 \Rightarrow \theta_1= \alpha =\sin^{-1}(1/4) \\ k=1 \Rightarrow \theta_2=2\pi+\alpha  } \\ \textbf{Case II }\theta=(2k+1)\pi-\sin^{-1} \left({1\over 4} \right) \Rightarrow \cases{k=0 \Rightarrow \theta_3= \pi-\alpha\\ k=1 \Rightarrow \theta_4=3\pi-\alpha} \\ \Rightarrow 所有的根\cases{\theta_1= \alpha \Rightarrow x_1=\alpha+\pi/6\\ \theta_2=2\pi+\alpha \Rightarrow x_2=(2\pi+\alpha)+\pi/6\\ \theta_3=\pi-\alpha \Rightarrow x_3=(\pi-\alpha)+\pi/6\\ \theta_4=3\pi-\alpha \Rightarrow x_4 = (3\pi-\alpha)+\pi/6} \Rightarrow  \sum_{i=1}^4 \theta_i =6\pi+{2\over 3}\pi = \bbox[red, 2pt]{{20\over 3}\pi}$$

解答:

$$P(2,2)對稱x軸的P'(2,-2) \Rightarrow 過P'的直線L:y=m(x-2)-2 \\ 若L為切線\Rightarrow d(圓心,L) =r \Rightarrow {|4m-6| \over \sqrt{m^2+1}} =2 \Rightarrow m= {6\pm2\over 3}\sqrt 3 \\\Rightarrow \cases{L_1: y= {6+2\sqrt 3\over 3}(x-2)-2\\ L_2: y={6-2\sqrt 3\over 3} (x-2)-2} \Rightarrow \cases{A= L_1 \cap (y=0) =(7-\sqrt 3)/2\\ B=L_1 \cap (y=0) =(7+\sqrt 3)/ 2} \Rightarrow \overline{AB}= \bbox[red, 2pt]{\sqrt 3}$$

解答:$$[4x]=\begin{cases}4& 1\le x\lt 5/4\\ 5& 5/4\le x\lt 3/2\\ 6&3/2 \le x\lt 7/4\\ 7&7/4 \le x\lt 2 \end{cases} \Rightarrow [x]+[4x]=6 \Rightarrow \bbox[red, 2pt]{{5\over 4}\le x\lt {3\over 2}}$$

解答:$$|x|+3|y|=6 所圍面積為24, 又{\partial(a',b') \over \partial (a,b)} =\begin{Vmatrix} 5& 1\\ 2&-1\end{Vmatrix} =7 \Rightarrow \Gamma'面積=24\times 7= \bbox[red, 2pt]{168}$$

解答:$$假設\cases{ 5 隻牛 (C)\\4 隻豬 (P) \\ 7 隻馬 (H)}，配對必須是(C,P),(C,H),(P,H)三種可能\\ 因此假設\cases{x=(C,P)數量\\ y=(C,H)數量\\ z=(P,H) 數量} \Rightarrow \cases{x+y=5 (5個C) \\ x+z=4(4個P) \\ y+z=7(7個H) \\ x+y+z=8} \Rightarrow \cases{x=1\\ y=4\\ z=3} \\ \cases{4個P要挑1個跟C配\\ 7個H要挑4個跟C配}\Rightarrow {4 \choose 1} {7\choose 4}\cdot 5! \cdot 3!= \bbox[red, 2pt]{100800}$$

解答:$$直線L_1:{x-9\over 2}={y+10\over 1} ={z-11\over -1}  \Rightarrow 方向向量 \vec u=(2,1,-1) \\直線L_2:{x-9\over 2}={y+10\over 2} ={z-11\over -1}  \Rightarrow 方向向量 \vec v=(2,2,-1) \\ 直線L_3: \cases{x=1\\ y+z=3} \Rightarrow 方向向量\vec w=(0,1,-1) \\ \Rightarrow 平面E法向量:(\vec u\times \vec v)\times \vec w=(2,-1,-1) 又E通過(1,1,2) \Rightarrow E: 2(x-1)-(y-1)-(z-2)=0 \\ \Rightarrow \bbox[red, 2pt]{2x-y-z=-1}$$

解答:$$假設\cases{x=\sqrt{\log a} \\y=\sqrt{\log b}} \Rightarrow \cases{\log a=x^2\\ \log b=y^2} \Rightarrow x-y+{1\over 2}x^2+{1\over 2}y^2\le 0 \Rightarrow (x^2+2x)+(y^2-2y)\le 0 \\ \Rightarrow (x+1)^2+ (y-1)^2 \le 2 \Rightarrow (x+1)^2 \le 2-(y-1)^2 \le 2 \Rightarrow -\sqrt 2-1\le x\le \sqrt 2-1 \\ \Rightarrow 0\le x\le \sqrt 2-1  \Rightarrow 0\le \sqrt{\log a}\le \sqrt 2-1 \Rightarrow 0\le \log a\le 3-2\sqrt 2 \approx 0.172 \\ \Rightarrow a=1 \Rightarrow x=0 \Rightarrow (0+1)^2 +(y-1)^2\le 2 \Rightarrow (y-1)^2 \le 1 \Rightarrow 0\le y\le 2 \Rightarrow 0\le \sqrt{\log b}\le 2 \\ \Rightarrow 0\le \log b\le 4 \Rightarrow 1\le b\le 10^4 \Rightarrow (a,b)=(1,1-10^4)共\bbox[red, 2pt]{10000}組$$

解答:$$\int_0^1 f(x)f'(x)\,dx = \left. \left[ {1\over 2}(f(x))^2\right] \right|_0^1 ={1\over 2} \left( f(1)^2-f(0)^2\right) =1 \Rightarrow f(1)^2-f(0)^2=2 \cdots(1)\\ \int_0^1 (f(x))^3f'(x)\,dx = \left. \left[ {1\over 4}(f(x))^4\right] \right|_0^1  ={1\over 4} \left( f(1)^4-f(0)^4\right)=2 \Rightarrow f(1)^4-f(0)^4=8 \cdots(2)\\ (1) \Rightarrow f(0)^2=f(1)^2-2代入(2) \Rightarrow f(1)^2=3 \Rightarrow f(0)^2=1\\\Rightarrow  \int_0^1 (f(x))^5f'(x)\,dx = \left. \left[ {1\over 6}(f(x))^6\right] \right|_0^1 ={1\over 6} \left( f(1)^6-f(0)^6\right) ={1\over 6}(3^3-1^3) = \bbox[red, 2pt]{13\over 3}$$

解答:$$\cases{A=2I+B \\ AB=O } \Rightarrow A^2= AA= A(2I+B)=2A+AB=2A \Rightarrow A^2=2A \\ \Rightarrow (A+I)^2=A^2 +2A+I=4A+I \Rightarrow (A+I)^4 =(4A+I)^2= 16A^2+8A+I=40A+I \\ \Rightarrow (A+I)^8 =(40A+I)^2= 1600A^2+80A+I=3280A+I \Rightarrow k= \bbox[red, 2pt]{3280}$$

解答:$$橢圓:{x^2\over m}+y^2 =1 \Rightarrow \cases{a=\sqrt m\\ b=1} \Rightarrow c=\sqrt{m-1} \Rightarrow \overline{PF_1}+ \overline{PF_2}=2\sqrt m \cdots(1) \\ 雙曲線:{x^2\over n}-{y^2\over 3}=1 \Rightarrow \cases{a= \sqrt n\\ b=\sqrt 3} \Rightarrow c=\sqrt{n+3} \Rightarrow \overline{PF_1}-\overline{PF_2}=2\sqrt n \cdots(2)\\ \Rightarrow \sqrt{m-1} =\sqrt{n+3} \Rightarrow m-1=n+3 \Rightarrow m-n=4\cdots(3) \\ 由(1)及(2)可得\cases{\overline{PF_1} =\sqrt m+\sqrt n\\ \overline{PF_2} =\sqrt m-\sqrt n} \Rightarrow \cos \angle F_1PF_2 ={(\sqrt m+\sqrt n)^2+(\sqrt m-\sqrt n)^2-4c^2\over 2(\sqrt m+\sqrt n)(\sqrt m -\sqrt n)} \\={2m+2n-4c^2\over 2(m-n)} ={m+n-2(m-1) \over m-n} ={n-m+2\over m-n} ={-4+2\over 4}=-{1\over 2} \\ \Rightarrow \tan \angle F_1PF_2= \bbox[red, 2pt]{-\sqrt 3}$$

解答:$$\textbf{Case I }a\ge b\ge c \Rightarrow \cases{a+2b= 2025\cdot 3\\ b+2c=2026\cdot 3\\ a+2c=2027\cdot 3} \Rightarrow (a,b,c)=(2027,2024,2027) 不合 \\ \textbf{Case II }a\ge c\ge b \Rightarrow \cases{a+2b=2025\cdot 3\\ c+2b=2026\cdot3\\ a+2c=2027\cdot 3} \Rightarrow a-c \lt 0不合\\ \textbf{Case III }b\ge a\ge c \Rightarrow \cases{b+2a = 2025\cdot 3\\ b+2c=2026\cdot 3 \\ a+2c=2027\cdot 3} \Rightarrow a-b=3不合 \\ \textbf{Case IV }b\ge c\ge a \Rightarrow \cases{b+2a = 2025\cdot 3\\ b+2c=2026\cdot 3 \\ c+2a=2027\cdot 3}  \Rightarrow b-c\lt 0不合 \\ \textbf{Case V }c\ge a\ge b  \Rightarrow \cases{a+2b = 2025\cdot 3\\ c+2b=2026\cdot 3 \\ c+2a=2027\cdot 3}  \Rightarrow (a,b,c)= \bbox[red, 2pt]{(2026,{4049\over 2}, 2029)} \\ \textbf{Case VI }c\ge b\ge a\Rightarrow \cases{b+2a = 2025\cdot 3\\ c+2b=2026\cdot 3 \\ c+2a=2027\cdot 3}  \Rightarrow b-a\lt 0不合\\ $$

解答:$$A=\{a,a+1,a+3,a+4\} \Rightarrow A \subseteq S且A符合要求\\ \Rightarrow 任意連續9個數中,必有一組像A的數列， 因此共有\lfloor {2025\over 9} \rfloor =225組,合計225\times 4= \bbox[red, 2pt]{900}個數字$$

解答:$$假設O至正方形頂點距離為x \Rightarrow 正方形面積=(\sqrt 2x)^2=2x^2 \\ \Rightarrow 容積=\int_0^{5\sqrt 2} 2x^2\,dx = \left. \left[ {2\over 3}x^3 \right] \right|_0^{5\sqrt 2} = \bbox[red, 2pt]{1000\sqrt 2\over 3}$$

解答:$${1\over x}+{8\over y}=1 \Rightarrow 8x+y=xy \Rightarrow 令\cases{f(x,y)=(x-1)^2+(y-8)^2\\ g(x,y)= 8x+y-xy} \Rightarrow \cases{f_x= \lambda g_x\\ f_y= \lambda g_y\\ g=0} \\ \Rightarrow \cases{2(x-1) =\lambda(8-y)\\ 2(y-8)=\lambda(1-x) \\ 8x+y=xy} \Rightarrow {x-1\over y-8}={8-y\over 1-x} \Rightarrow (x-1)^2=(y-8)^2 \Rightarrow x=y-7 \\ \Rightarrow 8(y-7)+y=(y-7)y \Rightarrow y^2-16y+56=0 \Rightarrow y=8\pm 2\sqrt 2 \Rightarrow x=1\pm 2\sqrt 2 \\ \Rightarrow f(1\pm 2\sqrt 2, 8\pm 2\sqrt 2) =(2\sqrt 2)^2 +(2\sqrt 2)^2= \bbox[red, 2pt]{16}$$

解答:

$$假設\cases{大圓半徑R_1\\ 小圓半徑R_2 \\ \triangle ABC外接圓半徑R_3\\  \angle ABC=\beta\\ \angle ACB=\gamma},並\cases{在\stackrel{ \Large\frown}{AB}取一點P\\ 在\stackrel{\Large\frown}{AC}取一點Q} \Rightarrow \cases{ \angle APB= \angle ABC =\beta\\ \angle AQC= \angle ACB=\gamma} \\ \cases{\triangle ABP: \overline{AB}/\sin \beta= 2R_1 \Rightarrow \overline{AB}= 2R_1\sin \beta\\ \triangle ACQ: \overline{AC}/ \sin \gamma =2R_2 \Rightarrow \overline{AC} = 2R_2\sin \gamma\\ \triangle ABC: \cases{ \overline{AC}/ \sin \beta=2R_3\\ \overline{AB} /\sin \gamma =2R_3} } \Rightarrow 4R_3^2 ={\overline{AC} \over \sin \beta}\cdot {\overline{AB} \over \sin \gamma} ={2R_2\sin \gamma\over \sin \beta} \cdot {2R_1\sin \beta\over \sin \gamma} \\ \Rightarrow 4R_3^2=4R_1R_2 \Rightarrow R_3^2=R_1R_2 \Rightarrow \triangle ABC外接圓面積= R_3^2\pi=R_1R_2\pi =\sqrt{m\over \pi} \cdot \sqrt{n\over \pi} \cdot \pi = \bbox[red, 2pt]{\sqrt{mn}}$$

二、計算題(10 分，需有計算過程，否則不予計分)

解答:$$\bbox[cyan,2pt]{學校提供}$$

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解題僅供參考，其他教甄試題及詳解