115年復興高中教甄2-數學詳解

 臺北市立復興高級中學 115 學年度第二次專任教師甄選

解答:$$f(x)= \sqrt{9^x-8\cdot3^{x+1}+x^2-10x+169} -\sqrt{9^x-16\cdot 3^x+x^2-6x+73} \\=\sqrt{(3^x-12)^2+(x-5)^2} -\sqrt{(3^x-8)^2+(x-3)^2} =\overline{PA}-\overline{PB},其中\cases{A(5,12) \\B(3,8) \\P(x,3^x)} \\ \Rightarrow \overline{PA}-\overline{PB}\le \overline{AB}= \bbox[red, 2pt]{2\sqrt 5}$$

解答:$$\cases{H_0: p\le 1\%=0.01 \\ H_1: p\gt 1\%}, X\sim B(n=20,p) \Rightarrow \cases{P(X=0)= (1-p)^{20} \approx 0.99\times 0.826=0.81744\\ P(X=1)=C^{20}_1 p(1-p)^{19} \approx 0.1652} \\我們需要找到一個最小的整數 c，累積機率 P(X \ge c) \le 5\%(顯著水準)。 \\ \Rightarrow \cases{c=1 \Rightarrow P(X\ge 1)=1-P(X=0)=0.18226  \not \le 5\%\\ c=2\Rightarrow P(X\ge 2)=1-P(X=0)-P(X=1)=0.01706\le 5\%} \\ \Rightarrow  拒絕域 \bbox[red, 2pt]{\{X\mid X\ge 2\}}$$

解答:$$\textbf{(1) }P\in L:{x-1\over 2}={y\over 1}={z-3\over -2} \Rightarrow P(2t+1,t,-2t+3)\\ \Rightarrow \cases{\overline{PA}=\sqrt{(1+2t-5)^2+(t-5)^2+(3-2t-5)^2} =3\sqrt{(t-1)^2+2^2} \\ \overline{PB} =\sqrt{(1+2t-3)^2+(t+2)^2+(3-2t-4)^2} =3\sqrt{(t-0)^2+1^2}} \\ \Rightarrow \overline{PA}+ \overline{PB} =3 \left( \overline{P'A'}+\overline{P'B'} \right), 其中\cases{A'(1,2) \\B'(0,-1) \\P'(t,0)} \\ \Rightarrow L'=\overleftrightarrow{A'B'}: y=3x-1 \Rightarrow L'與x軸交於({1\over 3},0) \Rightarrow t={1\over 3} \Rightarrow \bbox[red, 2pt]{P({5\over 3},{1\over 3},{7\over 3})} \\ \textbf{(2) }\min(\overline{PA}+ \overline{PB})= 3 \left( \overline{P'A'}+\overline{P'B'} \right) = 3\overline{A'B'}  = \bbox[red, 2pt]{3\sqrt{10}}$$

解答:$$\textbf{Case I }6個實根\Rightarrow 只有1種\\ \textbf{Case II }4個實根及1對虛根\Rightarrow {3\choose1} {6\choose 4}=45種\\ \textbf{Case III }2個實根及2對虛根\Rightarrow {3\choose 2}{6\choose 2} =45種\\ \textbf{Case IV }3對虛根\Rightarrow 只有1種 \\ 合計:1+45+45+1 =\bbox[red, 2pt]{92}$$

解答:$$x^3+ax^2+14x+b =(x^2+cx+2)(x+k) =x^3+(c+k)x^2+(ck+2)x+2k\\ \Rightarrow \cases{c+k=a\\ ck+2=14\\ 2k=b} \Rightarrow ck=12 \Rightarrow (c,k)=\cases{(1,12) \Rightarrow a=13\\(2,6) \Rightarrow a=8\\ (3,4) \Rightarrow a=7 \\ (4, 3) \Rightarrow a=7 \\(6,2) \Rightarrow a=8\\ (12,1) \Rightarrow a=13} \Rightarrow a\in \{7,8,13\} \Rightarrow 中位數\bbox[red, 2pt]8$$

解答:$$\cos A={\sqrt 3\over 2}={b^2+c^2-a^2\over 2bc} \Rightarrow a^2=b^2+c^2-\sqrt 3bc \Rightarrow k={b^2-c^2\over a^2} ={b^2-c^2\over b^2+c^2-\sqrt 3bc} \\={(b/c)^2-1\over (b/c)^2+1-\sqrt 3(b/c)} ={x^2-1\over x^2-\sqrt 3x+1} \Rightarrow x^2-1=kx^2-\sqrt 3kx+k \\ \Rightarrow (k-1)x^2-\sqrt 3kx+(k+1)=0 \Rightarrow \Delta=(-\sqrt 3k)^2-4(k-1)(k+1)\ge 0 \Rightarrow k^2\le 4\\ \Rightarrow -2\le k\le 2 \Rightarrow k的最大值2 \Rightarrow x^2-2\sqrt 3x+3=0 \Rightarrow (x-\sqrt 3)^2=0 \Rightarrow x={b\over c}=\sqrt 3 \\ \Rightarrow b=\sqrt 3c\Rightarrow a^2=3c^2+c^2-3c^2=c^2 \Rightarrow a=c \Rightarrow a:b:c= \bbox[red, 2pt]{1:\sqrt 3:1}$$

解答:$$a_n-a_{n-1}={a_{n-1} \over n}+3n+3 \Rightarrow a_n= \left( {n+1\over n} \right)a_{n-1}+3(n+1) \Rightarrow {a_n\over n+1}={a_{n-1}\over n}+3\\取b_n={a_n\over n+1} \Rightarrow b_n=b_{n-1}+3 \Rightarrow \langle b_n\rangle 為等差數列,公差為3,且b_1={a_1\over 1+1}=4 \\ \Rightarrow b_n=4+3(n-1)=3n+1 \Rightarrow a_n=(n+1)b_n=(n+1) (3n+1)  = \bbox[red, 2pt]{3n^2+4n+1}$$

解答:$$z=a+bi \Rightarrow z^{115}=\bar z^3 \Rightarrow |z^{115}|=|\bar z^3| \Rightarrow |z|^{115}=|z|^3 \Rightarrow |z|^3(|z|^{112}-1)=0\\ \Rightarrow \cases{|z|=0 \Rightarrow (a,b)=(0,0) \\|z|=1 \Rightarrow z^{115}=(1/z)^3 \Rightarrow z^{118}=1 \Rightarrow 有118種不同的(a,b)} \Rightarrow 共118+1= \bbox[red, 2pt]{119}組相異解$$

解答:$$假設\cases{O(0,0) \\A(x,y) \\B(m,n)} \Rightarrow \cases{\vec u=\overrightarrow{OB} =(m,n)\\ \vec v= \overrightarrow{OA} =(x,y)} \Rightarrow \sqrt{m^2+n^2}+ \sqrt{x^2+ y^2} = \sqrt{(m+x)^2+(n+y)^2} \\ \Rightarrow |\vec u|+|\vec v|=|\vec u+ \vec v| \Rightarrow O,A,B三點共線,假設此直線L:y=tx\\ 又B(m,n)滿足圓方程式: (m-5)^2+ (n-5)^2=10,其中\cases{圓心C(5,5)\\ 半徑r=\sqrt{10}} \Rightarrow 此圓在第一象限內\\ \Rightarrow t\gt 0 \Rightarrow d(C,L) \le r \Rightarrow {|5t-5| \over \sqrt{t^2+1}} \le \sqrt{10} \Rightarrow 3t^2-10t+3\le 0 \Rightarrow (3t-1)(t-3)\le 0\\ \Rightarrow {1\over 3}\le t\le 3 \Rightarrow {1\over 3} \le {y\over x}\le 3\\ 將正方形在第一象限的兩邊\cases{y=1\\ x=1} 代入上式\Rightarrow \cases{y=1 \Rightarrow {1\over 3}\le {1\over x} \le 3\Rightarrow {1\over 3}\le x\le 3 \Rightarrow {1\over 3}\le x\le 1\\ x=1 \Rightarrow {1\over 3}\le y\le 3} \\ \Rightarrow \bbox[red, 2pt]{{1\over 3}\le x\le 1}$$

解答:$$\cases{x=a^2+a \in \mathbb Q \Rightarrow a^2=x-a \\y=a^3+2a^2 \in \mathbb Q }\Rightarrow y=a(x-a)+2(x-a) =ax-a^2+ 2x-2a\\ =ax-(x-a)+2x-2a =a(x-1)+x \Rightarrow x-1=0 \Rightarrow x=1 \Rightarrow a^2=1-a \\ \Rightarrow a^3=a-a^2=a-(1-a)=2a-1 \Rightarrow a^6=(2a-1)^2=-8a+5 \Rightarrow a^{12} =(-8a+5)^2\\=-144a+89 \Rightarrow a^{12}+an=(n-144)a+89 \Rightarrow n= \bbox[red, 2pt]{144}$$

====================== END ==========================
解題僅供參考，其他教甄試題及詳解