114年南港高工教甄2-數學詳解

臺北市立南港高工 114 學年度第 2 次教師甄選

一、 填充題 ( 60 分，每格 5 分)

解答:$${2a+1\over 4-a}\gt 1 \Rightarrow {-3a+3\over 4-a} \lt 0 \Rightarrow 3(a-4)(a-1) \lt 0 \Rightarrow \bbox[red, 2pt]{1\lt a\lt 4}$$

解答:$$錯排D(n=5)=44 \Rightarrow {C^7_2\times 44\over 7!} = \bbox[red, 2pt]{11\over 60}\\ 錯排公式D(n) =\lfloor{n!\over e}+0.5 \rfloor$$

解答:

$$\Gamma:(y+2)^2=-4x+4 \Rightarrow \cases{頂點A(1,-2) \\ 焦點F(0,-2)\\ 對稱軸:y=-2 } \Rightarrow L=\overleftrightarrow{PQ}: y=\sqrt 3x-2 代入\Gamma\\ \Rightarrow 3x^2=-4x+4 \Rightarrow 3x^2+4x-4=0 \Rightarrow (3x-2)(x+2) \Rightarrow \cases{x=2/3 \Rightarrow y=(2\sqrt 4-6)/3\\ x=-2 \Rightarrow y=-2-2\sqrt 3} \\ \Rightarrow \cases{P(2/3, (2\sqrt 3-6)/3) \\Q(-2, -2-2\sqrt 3)} \Rightarrow \triangle APQ = {1\over 2}\begin{Vmatrix} 1& -2& 1\\ 2/3& (2\sqrt 3-6)/3 &1\\ -2&-2-2\sqrt 3 & 1\end{Vmatrix} = \bbox[red, 2pt]{4\sqrt 3\over 3}$$


解答:$$\sec \theta+ \csc \theta=1 \Rightarrow {1\over \cos \theta} +{1\over \sin \theta} ={\sin \theta +\cos \theta\over \sin \theta \cos \theta}=1 \Rightarrow \sin \theta+\cos \theta=\sin \theta \cos \theta =t \lt 1\\ \Rightarrow (\sin \theta+\cos \theta)^2=t^2= 1+2\sin \theta\cos \theta =1+2 t \Rightarrow t^2-2t-1=0 \Rightarrow t={1-\sqrt 2} \\ \Rightarrow \sec \theta \csc \theta={1\over t} = {1\over 1-\sqrt 2} =\bbox[red, 2pt]{-1-\sqrt 2}$$


解答:$$\cases{A(1,-2,2) \\B(9,-6,10) \\ D(3,1,6)} \Rightarrow \overrightarrow{AB}=(8,-4,8) \parallel \overrightarrow{CD} \Rightarrow C(2t+3,-t+1,2t+6) \\ \Rightarrow \overline{AD}^2= \overline{BC}^2 \Rightarrow 4+9+16=(2t-6)^2+(-t+7)^2+ (2t-4)^2 \Rightarrow t^2-6t+8=0 \\ \Rightarrow (t-4)(t-2)=0 \Rightarrow \cases{t=4 \Rightarrow C(11,-3,14) \Rightarrow \overrightarrow{AD} =\overrightarrow{BC} =(2,3,4)不合\\t=2 \Rightarrow C\bbox[red, 2pt]{(7,-1,10)}}$$

解答:$$\cases{f(a,b)=a+b\\ g(a,b)=a+b+{1\over a}+{9\over b}-10} \Rightarrow \cases{f_a= \lambda g_a\\ f_b= \lambda g_b \\ g=0} \Rightarrow \cases{1=\lambda(1-{1\over a^2}) \\ 1=\lambda (1-{9\over b^2}) \\ a+b+{1\over a}+{9\over b}=10} \\ \Rightarrow 1-{1\over  a^2} =1-{9\over b^2} \Rightarrow b=3a \Rightarrow a+3a+{1\over a}+{3\over a}=10 \Rightarrow 2a^2-5a+2=0 \\\Rightarrow (2a-1)(a-2)=0 \Rightarrow \cases{a=1/2 \Rightarrow b=3/2 \Rightarrow a+b=2\\ a=2\Rightarrow b=6 \Rightarrow a+b=8}  \Rightarrow (M,m)= \bbox[red, 2pt]{(8,2)}$$

解答:

$$假設\cases{C(0,0) \\D(8,0) \\A(4,4\sqrt 3)},將正四面體展開 \Rightarrow B(-4, 4\sqrt 3) \Rightarrow \cases{H=(D+2C)/4=(2,0) \\ N=(A+B)/2 =(0,4\sqrt 3)} \\ \Rightarrow \overline{NH} =\sqrt{2+48} = \bbox[red, 2pt]{2\sqrt{13}}$$

解答:$$\overrightarrow{AB} \cdot \overrightarrow{AP_1} =\overrightarrow{AB} \cdot (\overrightarrow{AB} + \overrightarrow{BP_1} )= 1+\overrightarrow{AB} \cdot \overrightarrow{BP_1} \\ \overrightarrow{AP_1} \cdot \overrightarrow{AP_2} =(\overrightarrow{AB}+ \overrightarrow{BP_1}) \cdot (\overrightarrow{AB} +\overrightarrow{BP_2}) =1+\overrightarrow{AB}\cdot \overrightarrow{BP_1}+ \overrightarrow{AB} \cdot \overrightarrow{BP_2} +{2\over n^2} \\\overrightarrow{AP_2} \cdot \overrightarrow{AP_3} =(\overrightarrow{AB}+ \overrightarrow{BP_2}) \cdot (\overrightarrow{AB} +\overrightarrow{BP_3}) =1+\overrightarrow{AB}\cdot \overrightarrow{BP_2}+ \overrightarrow{AB} \cdot \overrightarrow{BP_3} +{6\over n^2} \\ \cdots \cdots\\ \overrightarrow{AP_{n-2}} \cdot \overrightarrow{AP_{n-1}} =(\overrightarrow{AB}+ \overrightarrow{BP_{n-2}}) \cdot (\overrightarrow{AB} +\overrightarrow{BP_{n-1}}) =1+\overrightarrow{AB}\cdot \overrightarrow{BP_{n-2}}+ \overrightarrow{AB} \cdot \overrightarrow{BP_{n-1}} +{(n-2)(n-1)\over n^2}\\ \overrightarrow{AP_{n-1}} \cdot \overrightarrow{AC} =(\overrightarrow{AB}+ \overrightarrow{BP_{n-1}}) \cdot (\overrightarrow{AB} +\overrightarrow{BC}) =1+\overrightarrow{AB}\cdot \overrightarrow{BP_{n-1}}+ \overrightarrow{AB} \cdot \overrightarrow{AC} +{(n-1)n\over n^2} \\ \Rightarrow S_n=n+2(\overrightarrow{AB} \cdot  \overrightarrow{BP_1} +\overrightarrow{AB} \cdot  \overrightarrow{BP_2} +\cdots+\overrightarrow{AB} \cdot  \overrightarrow{BP_{n-1}} ) +\overrightarrow{AB} \cdot  \overrightarrow{AC} + \sum_{k=1}^{n-1} {k(k+1)\over n^2} \\\qquad = n- \left({1\over n}+{2\over n}+ \cdots+{n-1\over n} \right)-{1\over 2}+ {1\over n^2}\left({(n-1)n(2n-1)\over 6} +{n(n-1)\over 2} \right) \\\qquad =n-{n-1\over 2}-{1\over 2}+{n^2-1\over 3n} ={n\over 2}+{n^2-1\over 3n} = \bbox[red, 2pt]{ 5n^2-2\over 6n}$$



解答:$$PA=BP \Rightarrow \begin{bmatrix} s&1 \\ t& 1\end{bmatrix} \begin{bmatrix} 2& 1\\ 0& -1\end{bmatrix} =\begin{bmatrix} -1& 0\\ 0& 2\end{bmatrix} \begin{bmatrix} s& 1\\ t& 1\end{bmatrix} \Rightarrow \begin{bmatrix}2s& s-1\\ 2t& t-1 \end{bmatrix} =\begin{bmatrix}-s& -1\\ 2t& 2 \end{bmatrix} \Rightarrow \cases{s=0\\ t=3} \\ \Rightarrow P=\begin{bmatrix} 0& 1\\ 3& 1\end{bmatrix}  \Rightarrow   A-P= \begin{bmatrix}2& 0\\ -3& -2 \end{bmatrix}   \\ ACA+PCP =ACP+ PCA+I \Rightarrow AC(A-P)-PC(A-P)= (A-P)C(A-P)=I \\ \Rightarrow  \begin{bmatrix} 4a-6b&-4b\\ -6a+9b-4c+6d & 6b+4d \end{bmatrix}=I \Rightarrow \cases{4a-6b=1\\ -4b=0\\ -6a+9b-4c+6d=0\\ 6b+4d=1} \Rightarrow \cases{a=1/4\\   d=1/4} \\\Rightarrow a+d= \bbox[red, 2pt]{1\over 2}$$


解答:$$f(x)=50\log x-x  ={50\over \ln 10}\ln x-x \Rightarrow f'(x)={50\over x\ln 10}-1=0 \Rightarrow x={50\over \ln 10} =50\log e\\   e\approx 2.718 \Rightarrow \log e \approx 0.718 \cdot 0.4771+0.282\cdot 0.301 \approx 0.43 \Rightarrow 50\log e\approx 21.5\\ \Rightarrow \cases{\lfloor 50\log 21\rfloor-21 =66-21=45 \\ \lfloor 50\log 22\rfloor-22 =67-22=45} \Rightarrow k= \bbox[red, 2pt]{45}$$

解答:$$\left|z+\sqrt 3 i \right|+\left|z-\sqrt 3 i \right|=4為一橢圓，即\Gamma:x^2+{y^2\over 4}=1 \\ |z-i|=\overline{PA},其中\cases{P\in \Gamma \Rightarrow P(\cos \theta, 2\sin \theta)\\ A(0,1)} \Rightarrow \overline{PQ} =\sqrt{\cos^2\theta +(2\sin \theta-1)^2} \\=\sqrt{3\sin^2 \theta-4\sin \theta+2} =\sqrt{3\left( \sin\theta-{2\over 3}\right)^2+{2\over 3}} \Rightarrow \overline{PQ}最小值=\sqrt{2\over 3} =\bbox[red, 2pt]{\sqrt 6\over 3}$$

解答:$$

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28 :a\to e\to f\to d\to c\to b\to a \\共\bbox[red, 2pt]{28}種$$

二、 統測試題講解 (10 分)

解答:$$A,B皆在xy平面的同一側，取B(5,1,4)的對稱點B'(5,1,-4),\\ 得直線L=\overleftrightarrow{AB'}:(-1+6t,4-3t,2-6t) \Rightarrow L與xy平面的交點，即為P\\ 2-6t=0 \Rightarrow t=1/3 \Rightarrow P= \bbox[red, 2pt]{(1,3,0)}$$



解答:$$x+3y=k \Rightarrow x=k-3y代入橢圓 \Rightarrow 42y^2-(24k+12)y+4k^2-6=0 \\ \Rightarrow 判別式:(24k+ 12)^2-168(4k^2-6) =0 \Rightarrow k^2-6k-12=0 \Rightarrow k=3\pm \sqrt{21} \\ \Rightarrow x+3y最大值為 \bbox[red, 2pt]{3+\sqrt{21}}$$

三、 計算、證明題 ( 30 分) 

解答:$$0.1^x ={1\over 10^x} ={1\over \sqrt{2-\sqrt 3}} =\sqrt{2+\sqrt 3} \\ 由於2+\sqrt 3為無理數,其根號也是無理數, 因此0.1^x為無理數$$

解答:$$f(x)=1+\sin(x-{\pi\over 2}) \Rightarrow \int_0^{2\pi} \left( 1+\sin(x-{\pi\over 2})\right) \,dx = \bbox[red, 2pt]{2\pi}$$

解答:$$\cases{\lim_{x\to 0}\displaystyle {f(x)+g(x)\over x}=7 \Rightarrow f(x)+g(x)=7x \cdots(1) \\ \lim_{x\to 0}\displaystyle {2+g(x)\over xf(x)} =5 \Rightarrow 2+g(x)=5xf(x) \cdots(2)} \\ \Rightarrow \cases{f(x)=\displaystyle {7x+2\over 5x+1}\\ g(x) =\displaystyle {5x(7x+2)\over 5x+1}-2} \Rightarrow \cases{f'(x)= \displaystyle {7\over 5x+1}-{5(7x+2) \over (5x+1)^2} \\g'(x)= \displaystyle {70x+ 10\over 5x+1}-{5(35x^2+10x) \over (5x+1)^2}} \Rightarrow \cases{f(0)=2\\ f'(0)=-3\\ g(0) =-2\\ g'(0)=10} \\ \Rightarrow h'(0) =f'(0)g(0)+ f(0)g'(0)=6+20= \bbox[red, 2pt]{26}$$

解答:$$\textbf{(1) }\alpha =s+ti,s,t\in \mathbb R \Rightarrow {\alpha^2+4\over 2}=\bar \alpha \Rightarrow {s^2-t^2+4+2sti\over 2} =s-ti \Rightarrow \cases{st=-t\\ s^2-t^2+4=2s} \\ \quad \Rightarrow \cases{s=-1\\ t=\pm \sqrt 7} \Rightarrow \alpha=-1\pm \sqrt 7 \Rightarrow |\alpha|=\sqrt{8} =\bbox[red, 2pt]{2\sqrt 2} \\\textbf{(2) }假設f(x)=0的實數根為\beta  \Rightarrow 三根之和=\alpha+ \bar \alpha+\beta =-2+\beta=3 \Rightarrow \beta=5 \\ \Rightarrow y=f(x)=(x-\alpha)(x-\bar \alpha)(x-5 ) =(x^2+2x+8)(x-5) =x^3-3x^2-2x-40 \Rightarrow \cases{a=-2\\ b=-40}\\y=f(-x+3)=g(x) =(-x+3)^3-3(-x+3)^2+a(-x+3)+b \\\Rightarrow g'(x)= -3(-x+3)^2+6(-x+3) \Rightarrow g''(x)=6(-x+3)-6=0 \Rightarrow x=2 \\ \Rightarrow 對稱中心 (2,g(2)) =(2,-2+a+b) =\bbox[red, 2pt]{(2,-44)}$$

 

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