113年武陵高中教甄-數學詳解

桃園市立武陵高級中等學校 113 學年度第一學期第 1 次正式教師甄選(未完)

一、 填充題 (每小題 5 分； 共 75 分)

解答: $$\lim_{n\to \infty} \left({1\over \sqrt{n^2+3n}}+ {1\over \sqrt{n^2+6n}}+ {1\over \sqrt{n^2+9n}} +\cdots +{1\over \sqrt{n^2+3n^2}} \right) \\ =\lim_{n\to \infty} \left({1\over n \sqrt{1+3/n}}+ {1\over n\sqrt{1+6/n}}+ {1\over n\sqrt{n^2+9/n}} +\cdots +{1\over n\sqrt{n^2+3n/n}} \right) \\=\lim_{n\to \infty} \sum_{k=1}^n {1\over n\sqrt{1+{3k\over n}}} =\int_0^1 {1\over \sqrt{1+3x}} \,dx =\left. \left[ {2\over 3}\sqrt{3x+1} \right] \right|_0^1 ={4\over 3}-{2\over 3} =\bbox[red, 2pt]{2\over 3}$$

解答: $$y=f(x)=x^3+ax^2+1 \Rightarrow f'(x)=3x^2+2ax\\ 假設切點P(\alpha,f(\alpha)) \Rightarrow \overleftrightarrow{OP}: y=f'(\alpha)(x-\alpha)+f(\alpha) 通過原點 \Rightarrow f'(\alpha)(-\alpha)+f(\alpha)=0 \\ \Rightarrow (3\alpha^2+2a\alpha)(-\alpha)+\alpha^3+a\alpha^2+1= -2\alpha^3-a\alpha^2+1=0 有三相異實根 \\令g(x)=2x^3+ax^2-1 \Rightarrow g'(x)=6x^2+2ax=2x(3x+a)=0 \Rightarrow x=0,-a/3 \\ \Rightarrow g(0)g(-a/3) \lt 0 \Rightarrow {a^3\over 26}-1 \gt 0 \Rightarrow \bbox[red, 2pt]{a\gt 3}$$

解答:

解答: $$\cases{O(0,0)\\ A(1,0)\\ B(-{1\over 2},{\sqrt 3\over 2})} \Rightarrow x\overrightarrow{OA} +y\overrightarrow{OB} =(x-{1\over 2}y,{\sqrt 3\over 2}y) \Rightarrow C(x-{1\over 2}y,{\sqrt 3\over 2}y) \\C在圓上\Rightarrow (x-{1\over 2}y)^2+{3\over 4}y^2=1 \Rightarrow x^2-xy+y^2=1 \Rightarrow (x-y)^2=1-xy\\ C在\overline{AB}的外側\Rightarrow x+y\ge 1, x,y\ge 0 \Rightarrow (x-y)^2最大值＝1\Rightarrow \bbox[red, 2pt]{-1\le x-y\le 1}$$

解答: $$x^2+ax+b =0有實根 \Rightarrow a^2-4b\ge 0, 又\cases{\alpha+\beta =-a\\ \alpha \beta=b} \Rightarrow \alpha^2+\beta^2 = (\alpha+\beta)^2-2\alpha\beta=a^2-2b \lt 11\\ 同時符合\cases{ a^2-4b \ge 0\\ a^2-2b\lt 11} \Rightarrow (a,b) =(2,1), (3,1),(3,2),(4,3),(4,4),共五組\Rightarrow 機率=\bbox[red, 2pt]{5\over 36}$$

解答: $$\cases{f(x+3)\ge f(x)+3 \cdots(1)\\ f(x+1) \le f(x)+1 \cdots(2)}\\(1) \Rightarrow f(x)\le f(x+3)-3 \le f(x+2)+1-3\le f(x+1)+1+1-3\\ \Rightarrow f(x)\le f(x+1)-1 \cdots(3)\\ (2) \Rightarrow f(x+1)-1\le f(x) \cdots(4) \\ (3) \text{ and }(4) \Rightarrow f(x)=f(x+1)-1 \Rightarrow f(x)=f(x-1)+1 \\ \Rightarrow f(2024)=f(2023)+1= f(2022)+2 = \cdots =f(1)+2023 = 2+2023= \bbox[red, 2pt]{2025}$$

解答: $$邊長10\Rightarrow 圓半徑=5\sqrt 2\\ 假設底面中心位於原點,則邊長=\sqrt 2x, 又圓方程式x^2+y^2=r^2 \Rightarrow x^2=r^2-y^2 \\ \Rightarrow 正方形面積=(\sqrt 2 x)^2=2x^2= 2(r^2-y^2) \Rightarrow 容積= \int_0^{5\sqrt 2} 2(50-y^2)\,dy =\left. \left[ 100y-{2\over 3} y^3 \right]  \right|_0^{5\sqrt 2} \\=500\sqrt 2-{500\over 3}\sqrt 2 = \bbox[red, 2pt]{{1000\over 3}\sqrt 2}$$

解答: $$\sqrt{n+{1\over 2}} -\sqrt{n-{1\over 2}} =\cfrac{1}{\sqrt{n+{1\over 2}} +\sqrt{n-{1\over 2}}} \approx {1\over 2\sqrt n} \\ 因此取d_n={1\over \sqrt n}-2\left(\sqrt{n+{1\over 2}} -\sqrt{n-{1\over 2}}\right)\\則a=\sum_{k=1}^{99} {1\over \sqrt k} = 2\sum_{k=1}^{99}\left( \sqrt{k+{1\over 2}}-\sqrt{k-{1\over 2}} \right) +\sum_{k=1}^{99}d_k =2\left( \sqrt{99+{1\over 2}}-\sqrt{1\over 2}\right)+\sum_{k=1}^{99}d_k \\= (\sqrt{398}-\sqrt 2)+\sum_{k=1}^{99}d_k \approx (20-1.414)+\sum_{k=1}^{99}d_k = 18.586+\sum_{k=1}^{99}d_k\\ \Rightarrow a的整數部份=\bbox[red, 2pt]{18}\\ 註:\href{https://math.stackexchange.com/questions/2168520/find-the-value-of-left-frac1-sqrt-2-frac1-sqrt-3-frac1-s}{參考資料}$$

解答: $$\textbf{(1)}\;P(3\sqrt 3,1) \in \Gamma \Rightarrow {27\over a^2}+{1\over b^2}=1 \Rightarrow \text{Let } \cases{f(a,b)=a+b\\ g(a,b)={27\over a^2}+{1\over b^2}-1},\text{ then by Lagrange multiplier, we have} \\ \cases{f_a=\lambda g_a\\ f_b=\lambda g_b\\ g=0} \Rightarrow \cases{1= \lambda\left( -{54\over a^3}\right) \\1 = \lambda\left( -{2\over b^3}\right)} \Rightarrow 54b^3= 2a^3 \Rightarrow a=3b \Rightarrow g(3b,b)=0 \Rightarrow {3\over b^2}+{1\over b^2}=1 \Rightarrow b^2=4\\ \Rightarrow b=2\;( b\gt 0) \Rightarrow a=3b=6 \Rightarrow a+b= 6+2=\bbox[red, 2pt]8 \\ \textbf{(2)}; \Gamma: \bbox[red, 2pt]{{x^2\over 36}+{y^2\over 4}=1}$$

解答: $$\sum_{k=1}^{113} \left(2\cdot {k^3\over 113^3}-3\cdot {k^2\over 113^2}+4\cdot {k\over 113} -2 \right) \\={2\over 113^3}\sum_{k=1}^{113} k^3- {3\over 113^2} \sum_{k=1}^{113}k^2 +{4\over 113}\sum_{k=1}^{113}k-2\times113 \\={2\over 113^3}\left({113\cdot 114\over 2}\right)^2- {3\over 113^2} \cdot{113\cdot 114 \cdot 227\over 6} +{4\over 113}\cdot {113\cdot 114\over 2}-226 \\={114^2\over 113\cdot 2} -{114\cdot 227\over 113\cdot 2}+{228-226} ={114\cdot (-113) \over 113\cdot 2}+2 =-57+2=\bbox[red, 2pt]{-55}$$

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解答:

$$\cases{圓心O(0,0)\\ P(2,4)} \Rightarrow 直線L=\overleftrightarrow{OP}: y=2x \Rightarrow L為直角\angle MPN的角平分線 \Rightarrow \angle MPQ=45^\circ\\ \Rightarrow \angle OPM=135^\circ \Rightarrow \cos \angle OPM ={\overline{OP}^2+ \overline{PM}^2-\overline{OM}^2 \over 2 \cdot \overline{OP}\cdot \overline{PM}} \Rightarrow \cos {\pi\over 4} ={20+ \overline{PM}^2-36\over 2\cdot 2\sqrt 5\cdot \overline{PM}} \\ \Rightarrow -{\sqrt 2\over 2} ={\overline{PM}^2-16 \over 4\sqrt 5\cdot \overline{PM}} \Rightarrow \overline{PM} =\sqrt{26}-\sqrt{10} \Rightarrow |\overrightarrow{PM} +\overrightarrow{PN}|的最小值=\sqrt 2\overline{PM} =\sqrt{52}-\sqrt{20} \\= \bbox[red, 2pt]{2\sqrt{13}-2\sqrt 5}$$

解答: $$假設\cases{L_1:y=2\\ L_2:y=0\\ C(a,b)} \Rightarrow \cases{M(a-2,0)\\ N(a+2,0)\\ A(x,2)\\ B(x+4,2)} \Rightarrow \overline{AC}^2=\overline{MC}^2 \Rightarrow (x-a)^2+(b-2)^2=4+b^2\\ \Rightarrow x=a-2\sqrt b \Rightarrow B(a-2\sqrt b+4,2) \Rightarrow d+\overline{BC} =f(b)= b+\sqrt{(4-2\sqrt b)^2+(b-2)^2} \\ \Rightarrow f'(b)=0 \Rightarrow 4b^3-25b^2+40b-16=0 \Rightarrow (b-4)(4b^2-9b+4)=0 \\ \Rightarrow b=4,b={9\pm \sqrt{17}\over 8} \Rightarrow b={9+\sqrt{17}\over 8} 有極小值 \\b= {9+\sqrt{17}\over 8} \Rightarrow \sqrt b= \sqrt{18+ 2\sqrt{17} \over 16} ={\sqrt{17}+1\over 4} \\ \Rightarrow f(b={9+\sqrt{17}\over 8})= b+\sqrt{b^2-16\sqrt b+20} ={9+\sqrt{17} \over 8}+\sqrt{1122-238\sqrt{17}\over 64} \\={9+\sqrt{17} \over 8} +{1\over 8}(7\sqrt{17}-17) =\bbox[red, 2pt]{\sqrt{17}-1}$$

另解:

$$假設\cases{L_1:y=2\\ L_2:y=0\\ C(x,y)} \Rightarrow \cases{M(x-2,0)\\ N(x+2,0)\\ A(0,2)\\ B(4,2)} \Rightarrow \overline{AC}^2=\overline{MC}^2 \Rightarrow x^2+(y-2)^2=4+y^2 \\ \Rightarrow x^2=4y為一拋物線,焦點F(0,1), 準線y=-1\\ 欲求\overline{BC}+\overline{CD}之最小值,也就是求(\overline{BC}+\overline{CE}-1)之最小值, 而\overline{CE}=\overline{CF},\\因此最小值出現在C=\overline{BF}與拋物線的焦點上,此時(\overline{BC}+\overline{CE}-1)=\overline{BF}-1= \bbox[red, 2pt]{\sqrt{17}-1}$$

解答: $$\cases{z_1=a+bi\\ z_2=c+di} \Rightarrow \cases{z_1+2\bar z_2= (a+2c)+(b-2d)=-i\\ z_1\times z_2=(ac-bd)+(ad+bc)i=-3+i} \Rightarrow \cases{a+2c=0 \cdots(1)\\ b-2d=1 \cdots(2)\\ ac-bd=-3 \cdots(3) \\ ad+bc=1 \cdots(4)} \\ \cases{(1)\\(2)} \Rightarrow \cases{c=-a/2\\ d=(b+1)/2}代入 (4) \Rightarrow a\cdot {b+1\over 2}-{ab\over 2}=1 \Rightarrow a=2 \Rightarrow c=-1\\ 將\cases{a=2\\ c=-1} 代入 (3)  \Rightarrow -2-bd=-3 \Rightarrow bd=1 \cdots(4) \\將(2)代入(4) \Rightarrow (2d-1)d=1 \Rightarrow (2d+1)(d-1)=0 \Rightarrow \cases{d=1\\ d=-1/2} \Rightarrow \cases{b=1\\ b=-2} \\ \Rightarrow |z_1|=\sqrt{a^2+b^2 } =\cases{\sqrt{2^2+1^2} =\sqrt 5\\ \sqrt{2^2+(-2)^2} =2\sqrt 2} \Rightarrow |z_1|=\bbox[red, 2pt]{\sqrt 5或2\sqrt 2}$$

二、計算證明題： (每題 15 分；共 45 分)

解答:

解答: $$a,b \gt 0 \Rightarrow (a+b)(a-b)^2 \ge 0 \Rightarrow (a^2-b^2)(a-b)\ge 0 \Rightarrow a^3-a^2b-b^2a+b^3\ge 0\\ \Rightarrow a^3+b^3\ge a^2b+ab^2 \Rightarrow \cases{a^3+b^3\ge a^2b+ ab^2 \cdots(1)\\ b^3+c^3\ge b^2c+ bc^2 \cdots(2)\\ c^3+a^3 \ge c^2a+ ca^2 \cdots(3)} \\ (1)+(2)+(3) \Rightarrow 2(a^3+b^3+c^3) \ge a^2b+ab^2+ b^2c+ bc^2+ c^2a +ca^2\\ \Rightarrow 2(a^3+b^3+c^3)+a^3+b^3+c^3 \ge a^2b+ab^2+ b^2c+ bc^2+ c^2a +ca^2+a^3+b^3+c^3 \\ \Rightarrow 3(a^3+b^3+c^3)\ge (a+b+c)(a^2+b^2+c^2) \\ \Rightarrow {a^3+b^3+c^3\over a^2+b^2 +c^2} \ge {1\over 3}(a+b+c) \Rightarrow \cases{{a^3+b^3+c^3\over a^2+b^2 +c^2} \ge {1\over 3}(a+b+c) \cdots(4) \\ {b^3+c^3+d^3\over b^2+c^2 +d^2} \ge {1\over 3}(b+c+d) \cdots(5) \\{c^3+d^3+a^3\over c^2+d^2 +a^2} \ge {1\over 3}(c+d+a) \cdots(6) \\{d^3+a^3+b^3\over d^2+a^2 +b^2} \ge {1\over 3}(d+a+b) \cdots(7) } \\ (4)+(5)+(6)+(7) \Rightarrow {a^3+b^3+c^3\over a^2+b^2 +c^2}+{b^3+c^3+d^3\over b^2+c^2 +d^2}+ {c^3+d^3+a^3\over c^2+d^2 +a^2}{d^3+a^3+b^3\over d^2+a^2 +b^2} \ge a+b+c+d. \bbox[red, 2pt]{QED}$$

解答: $$令S(n)=\sum_{k=1}^na_k, 已知S(n+1)\le ra_n=r(S(n)-S(n-1)) \\ \Rightarrow S(n)\ge {S(n+1)\over r}+S(n-1) \ge 2\sqrt{{1\over r}S(n+1)S(n-1)} \\ \Rightarrow S^2(n)\ge {4\over r}S(n+1)S(n-1) \Rightarrow {S(n+1)\over S(n)} \le {r\over 4}{S(n)\over S(n-1)} \\ \Rightarrow {S(n+1)\over S(n)} \le \left( {r\over 4}\right)^{n-1} {S(2)\over S(1)}\\  因此當{r\over 4}\lt 1時,\lim_{n\to \infty}{S(n+1)\over S(n)}=0不合,因此{r\over 4}\ge 1 \Rightarrow r\ge 4 \Rightarrow 最小實數r=\bbox[red, 2pt]4$$