113年復興高中第2次教甄-數學詳解

臺北市立復興高級中學 113 學年度第二次專任教師甄選

一、填充題： (80%。共 10 題，每題 8 分)

解答: $$取g(x)=xf(x)-1,則1,2,\dots,50皆為g(x)=0的根\\ 因此g(x)=a(x-1)(x-2)\cdots(x-50) \Rightarrow g(0)=a\cdot 50!=0\cdot f(0)-1\\ \Rightarrow a=-{1\over 50!} \Rightarrow f(x)={g(x)+1\over x} \Rightarrow f(-2)={g(-2)+1\over -2 } \\={1\over -2}\left( -{1\over 50!}(-3)\cdot(-4)\cdots (-52)+1\right) =-{1\over 2}\left( -{52\cdot 51\over 2}+1\right) =\bbox[red, 2pt]{1325\over 2}$$

解答: $$xy+x+y=39 \Rightarrow xy=39-(x+y) \Rightarrow x^2y+ xy^2=xy(x+y) =(39-(x+y))(x+y)=308 \\ \Rightarrow (x+y)^2-39(x+y) +308=0 \Rightarrow (x+y-11)(x+y-28)=0 \\ \Rightarrow \cases{x+y=11 \Rightarrow xy=28 \Rightarrow x^2+y^2=(x+y)^2-2xy=11^2-56=65\\ x+y=28 \Rightarrow xy=11 \Rightarrow x^2+y^2=(x+y)^2-2xy=28^2-22=762} \\ \Rightarrow 最大值\bbox[red, 2pt]{762}$$

解答: $$\cases{圓C_1:a^2+b^2=49\\ 圓C_2:c^2+d^2-16c-12d=-96 \Rightarrow (c-8)^2+(d-6)^2=4}\\ \Rightarrow \cases{圓C_1: 圓心O_1(0,0), 半徑r_1=7\\ 圓C_2:圓心O_2(8,6),半徑r_2=2 }\\ \sqrt{16c+12d-2ac-2bd-47}\\= \sqrt{(a-c)^2+(b-d)^2-(c-8)^2-(d-6)^2-a^2-b^2+53} \\=\sqrt{(a-c)^2+(b-d)^2-4-49+53} =\sqrt{(a-c)^2+(b-d)^2} \\=\overline{PQ},其中\cases{P(a,b) \in 圓C_1\\ Q(c,d) \in 圓C_2} \Rightarrow \overline{PQ}最小值=\overline{O_1O_2}-r_1-r_2=10-7-2=\bbox[red, 2pt]1$$

解答: $$\begin{array}{c}(a,b,c,d)&排列數\\\hline ( 1 , 5 , 6 , 6 ) & {4!\over 2!}=12\\( 2 , 4 , 6 , 6 ) & {4!\over 2!}=12\\( 2 , 5 , 5 , 6 ) & {4!\over 2!}=12\\( 3 , 3 , 6 , 6 ) &{4!\over 2!2!}=6\\ ( 3 , 4 , 5 , 6 )&4!=24\\( 3 , 5 , 5 , 5 ) & {4!\over 3!}=4\\( 4 , 4 , 4 , 6 ) &{4!\over 3!}=4\\( 4 , 4 , 5 , 5 ) & {4!\over 2!2!} = 6\end{array} \Rightarrow 合計=12\times 3+(6+4)\times 2+24= \bbox[red, 2pt]{80}$$

解答: $${5\over 5+4}+{3\over 3+4}-{5+3\over 5+3+4}= \bbox[red, 2pt]{20\over 63}\\ \href{https://math.ntnu.edu.tw/~horng/letter/hpm17010.pdf}{公式來源}$$

解答: $$令\cases{\overline{AB}=a\\ \overline{AC}=b \\ \overline{AD}=3為角平分線} \Rightarrow \triangle ABC面積=\triangle ABD+\triangle ACD \\ \Rightarrow ab\sin A=3a\sin {A\over 2}+3b\sin {A\over 2} \Rightarrow 15\sin A=3(a+b)\sin{A\over 2} \\ \Rightarrow 30\cos{A\over 2}\sin {A\over 2}=3(a+b)\sin{A\over 2} \Rightarrow a+b=10\cos {A\over 2} \ge 2\sqrt{ab} =2\sqrt{15} \\ \Rightarrow 極值出現在\cos{A\over 2}={\sqrt{15}\over 5}\Rightarrow \cos A={1\over 5} \Rightarrow \sin A={\sqrt{24}\over 5} \\ \Rightarrow \triangle ABC面積={1\over 2}\cdot 15\cdot {\sqrt{24}\over 5} =\bbox[red, 2pt]{3\sqrt 6}$$

解答: $$f(x)=\left|\log{x-k\over 3} \right| \Rightarrow \begin{cases} f(x)遞減& x-k\le 3 \\ f(x)遞增& x-k\ge 3\end{cases} \Rightarrow f(1)=0 \Rightarrow 1-k=3 \Rightarrow k= \bbox[red, 2pt]{-2}$$

解答:

$$令G=\overline{EF}中點，則ADEG與BCFG均為稜長為1的正四面體\\ 因此體積=三角錐ABCDG+兩個正四面體={1\over 3}\cdot 1\cdot {\sqrt 2\over 2}+ 2\times {\sqrt 2\over 12} = \bbox[red, 2pt]{\sqrt 2\over 3}$$

解答: $$f'(x)=x^2+ax+b=0的兩根\alpha,\beta分別在區間[-1,1)及(1,3]內 \\ 且滿足\cases{\alpha +\beta =-a\\ \alpha\beta=b} \Rightarrow a^2-4b=(\alpha+\beta)^2-4\alpha\beta = (\alpha-\beta)^2 \\ 最大值發生在\cases{\alpha=-1\\ \beta=3} ,此時(\alpha-\beta)^2= \bbox[red, 2pt]{16}$$

解答: $$f(t)=\int_0^t (\sin 2x-\sqrt 3\cos 2x)\,dx \Rightarrow f'(t)=\sin 2t-\sqrt 3\cos 2t =2({1\over 2}\sin 2t-{\sqrt 3\over 2}\cos 2t) \\=2\sin(2t-{\pi\over 3})  \Rightarrow f''(t)= 4\cos(2t-{\pi\over 3}) \\ f'(t)=0 \Rightarrow \cases{2t-\pi/3=0\\ 2t-\pi/3=\pi} \Rightarrow \cases{t=\pi/6 \Rightarrow f''(\pi/6)= 4\gt 0\\ t=2\pi/3 \Rightarrow f''(2\pi/3)=-4\lt 0} \Rightarrow t= \bbox[red, 2pt]{2\pi\over 3}$$

二、計算證明題： (20%。共 2 題，每題 10 分)

解答: $$假設(I_m-AB)^{-1}=C \Rightarrow \color{blue}{C-ABC}=I_m\\ \Rightarrow (I_n-BA)(BCA)= BCA-BABCA =B(\color{blue}{C-ABC})A = BA\\ 因此 (I_n-BA)(BCA+I_n)= (I_n-BA)(BCA)+ (I_n-BA) =BA+(I_n-BA) =I_n\\ \Rightarrow BCA+I_n =B(I_m-AB)^{-1}A+I_n=(I_n-BA)^{-1}. \bbox[red, 2pt]{QED} \\ \href{https://math.stackexchange.com/questions/237779/i-m-ab-is-invertible-if-and-only-if-i-n-ba-is-invertible}{參考資料}$$

解答: $$\;假設\cases{移1格有a次\\移2格有b次 } \Rightarrow a+2b=6 \Rightarrow \begin{array}{} a& b& 移動次數 &排列數\\\hline 0& 3& 3&1\\ 2& 2& 4&{4!\over 2!2!}=6 \\ 4& 1& 5&5\\ 6& 0 & 6& 1\\\hline \end{array} \\ \textbf{(1)}  共有1+6+5+1= \bbox[red, 2pt]{13}種移動方法 \\ \textbf{(2)}\; 移動期望值{1\over 13}(3\cdot 1+4\cdot 6+ 5\cdot 5+6\cdot 1)=\bbox[red, 2pt]{58\over 13}$$

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