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:xy+x+y=39⇒xy=39−(x+y)⇒x2y+xy2=xy(x+y)=(39−(x+y))(x+y)=308⇒(x+y)2−39(x+y)+308=0⇒(x+y−11)(x+y−28)=0⇒{x+y=11⇒xy=28⇒x2+y2=(x+y)2−2xy=112−56=65x+y=28⇒xy=11⇒x2+y2=(x+y)2−2xy=282−22=762⇒最大值762
解答:{圓C1:a2+b2=49圓C2:c2+d2−16c−12d=−96⇒(c−8)2+(d−6)2=4⇒{圓C1:圓心O1(0,0),半徑r1=7圓C2:圓心O2(8,6),半徑r2=2√16c+12d−2ac−2bd−47=√(a−c)2+(b−d)2−(c−8)2−(d−6)2−a2−b2+53=√(a−c)2+(b−d)2−4−49+53=√(a−c)2+(b−d)2=¯PQ,其中{P(a,b)∈圓C1Q(c,d)∈圓C2⇒¯PQ最小值=¯O1O2−r1−r2=10−7−2=1
解答:(a,b,c,d)排列數(1,5,6,6)4!2!=12(2,4,6,6)4!2!=12(2,5,5,6)4!2!=12(3,3,6,6)4!2!2!=6(3,4,5,6)4!=24(3,5,5,5)4!3!=4(4,4,4,6)4!3!=4(4,4,5,5)4!2!2!=6⇒合計=12×3+(6+4)×2+24=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'(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}
============== END =====================
解題僅供參考,其他教甄試題及詳解
沒有留言:
張貼留言