110年台中女中教甄-數學詳解

臺中市立臺中女子高級中等學校 110 學年度第一次教師甄選

一、 填充題：每格 4 分，全對才給分

解答：$$ 設x=abc為三位數，其中\cases{a=千位數\\ b=佰位數\\ c=個位數}；並令\cases{M(x):將a,b,c順序改變得到最大的三位數 \\ m(x):將a,b,c順序改變得到最小的三位數\\ H(x)=M(x)-m(x)}\\ 依題意，就是要找x，使得x=H(x)；\\任取a,b,c，多作幾次H(x)運算就能找到所需的x；例x=692 \Rightarrow  H(693)=963-369= 594，\\ 再令x=594 \Rightarrow H(594)=954-459=495；再令x=495 \Rightarrow H(495)=954-459=495\\ 因此\bbox[red, 2pt]{495}即為所求。參考資料:\href{https://www.youtube.com/watch?v=qX9RIU-FrYk}{Youtube 影片}$$

解答：$$f(x)= \sqrt[3]{x} \Rightarrow f(t+16)-f(t-16)=2 \Rightarrow \sqrt[3]{t+16}-\sqrt[3]{t-16}=2\\ \Rightarrow (\sqrt[3]{t+16})^3 = (2+\sqrt[3]{t-16})^3  \Rightarrow t+16= 8+(t-16) + 6\sqrt[3]{t-16}(2+\sqrt[3]{t-16})\\ \Rightarrow 6\sqrt[3]{(t-16)^2} +12\sqrt[3]{t-16} -24=0\\ \Rightarrow \sqrt[3]{t-16}= \sqrt 5-1 \Rightarrow t=(\sqrt 5-1)^3+16= 8\sqrt 5-16+16=\bbox[red, 2pt]{8\sqrt 5}$$

解答：$$\sqrt{x^2-2x+12-6\sqrt 2} +\sqrt{x^2-12x+39-2\sqrt 2} \\=\sqrt{(x-1)^2 +11-6\sqrt 2} +\sqrt{(x-6)^2 +3-2\sqrt 2} =\overline{PA} +\overline{PB},其中\cases{P(x,0)\\ A(1, \sqrt{11-6\sqrt 2}) \\ B(6,-\sqrt{3-2\sqrt 2})}\\ P在x軸上，A、B在x軸的異側，因此\overline{PA} +\overline{PB}的最小值=\overline{AB}\\ 而\overline{AB}=\sqrt{(6-1)^2 +(\sqrt{11-6\sqrt 2} +\sqrt{3-2\sqrt 2})^2} =\sqrt{25+(3-\sqrt 2+\sqrt 2-1)^2} \\ =\sqrt{25+4} =\bbox[red, 2pt]{\sqrt{29}}$$

解答：

$$1-9有5個奇數4偶數，而三個數字和為奇數，只能是三奇或一奇二偶；\\因此排法就是三行中有一行全是奇數，三列中有一列全是奇數；\\上圖藍色代表奇數的位置，有9種擺放奇數的可能，；\\因此機率為{9\over C^9_5} ={9\over 126}=\bbox[red,2pt]{1\over 14}$$

解答：$$假設直線L斜率m，且過(1,2)，則L:y=m(x-1)+2 代入xy=1 \Rightarrow x(m(x-1)+2)=1\\ \Rightarrow f(x)=mx^2+(2-m)x-1=0 之二根為\alpha,\beta (假設\alpha \gt \beta)\\ \Rightarrow  兩交點\cases{P(\alpha,f(\alpha))\\ Q(\beta,f(\beta))} 且\cases{\alpha+\beta  = (m-2)/m\\ \alpha \beta=-1/m } \Rightarrow (\alpha-\beta)^2 =(\alpha+\beta)^2 -4\alpha\beta = {m^2+4 \over m^2}\\ \Rightarrow \alpha-\beta =  \sqrt{m^2+4 \over m^2}\Rightarrow \overline{PQ} = (\alpha-\beta) \times   \sqrt{m^2+1} =\sqrt{m^4+5m^2+4 \over m^2} =\sqrt{5+m^2+{ 4 \over m^2}} \\ \ge \sqrt{5 +4} =3 \left(\because m^2+{4\over m^2} \ge 2\sqrt{m^2 \cdot {4\over m^2}} =4\right)  \Rightarrow \overline{PQ}的最小值為\bbox[red, 2pt]{3}$$

解答：$$\cases{z=2e^{i\alpha} \\ \omega =e^{i\beta}} \Rightarrow \cases{z/\omega = 2e^{i(\alpha-\beta) }  \\\omega/z=1/2\cdot e^{i(\beta-\alpha} \\z\omega =2e^{i(\alpha+\beta)}} \Rightarrow \lvert z^2-4\omega^2 +7z\omega \rvert =\lvert (z\omega)(z/\omega-4\omega/z +7) \rvert \\= \lvert z\omega\rvert \lvert z/\omega-4\omega/z +7\rvert =2\lvert 2e^{i(\alpha-\beta)} -2e^{i(\beta-\alpha)} +7 \rvert =2\lvert 7+ 4\sin(\alpha-\beta) i\rvert \\ \Rightarrow \cases{最大值M=2\cdot \sqrt{7^2+4^2} = 2\sqrt{65} \\最小值m=2\cdot \sqrt{7^2+0^2} =14} \Rightarrow (M,m)= \bbox[red, 2pt]{(2\sqrt{65},14)}$$

解答：

$$底部正\triangle ABC頂點坐標\cases{A(-{27\over 2},0,0) \\ B({27\over 2},0,0) \\C(0,{27\over 2}\sqrt 3,0)} \Rightarrow 重心G(0,{9\over 2}\sqrt 3,0) \\\Rightarrow 頂面\triangle DEF重心G'(0,{9\over 2}\sqrt 3,\sqrt{141}) \Rightarrow F=G'+(0,-5\sqrt 3,0) =(0,-{\sqrt 3\over 2},\sqrt{141})\\ \Rightarrow \overline{AF} = \sqrt{({27\over 2})^2 +({\sqrt 3\over 2})^2+ 141}  = \sqrt{183+141} =\sqrt{324} = \bbox[red, 2pt]{18}$$

解答：$$g(x)=\int_1^x f(t)\;dt = {1\over 12}x^3-2x^2+px +q \Rightarrow g'(x)=f(x)= {1\over 4}x^2-4x+p \Rightarrow f'(x)={1\over 2}x-4\\ 令過原點的切線為y=mx代入f(x) \Rightarrow mx={1\over 4}x^2-4x+p \Rightarrow {1\over 4}x^2-(4+m)x+p=0 \\\Rightarrow 判別式: (m+4)^2-p=0 \Rightarrow m^2+8m+16-p=0 \Rightarrow 兩根之積=16-p=-1 \\ \Rightarrow p=17\\ 又g(1)= \int_1^1f(t)\;dt = 0={1\over 12}-2+p+q = -{23\over 12}+17+q \Rightarrow q=-{181\over 12} \\ \Rightarrow (p,q) = \bbox[red, 2pt]{(17,-{181\over 12} )}$$

解答：$$孟氏定理:{\overline{AE} \over \overline{EB}} \times{\overline{BC} \over \overline{CD}} \times{\overline{DO} \over \overline{OA}} =1 \Rightarrow {1\over 2}\times {2\over 1}\times{\overline{DO} \over \overline{OA}} =1 \Rightarrow \overline{DO} =\overline{OA}\\ 依題意: \overrightarrow{AB}\cdot \overrightarrow{AC} =6\overrightarrow{AO} \cdot \overrightarrow{EC} =3\overrightarrow{AD} \cdot (-\overrightarrow{CE}) ={3\over 2}(\overrightarrow{AB} +\overrightarrow{AC}) \cdot -({1\over 3}\overrightarrow{CB} +{2\over 3}\overrightarrow{CA}) \\ ={3\over 2}(\overrightarrow{AB} +\overrightarrow{AC}) \cdot ({1\over 3}\overrightarrow{BC} + {2\over 3}\overrightarrow{AC})  = {3\over 2}(\overrightarrow{AB} +\overrightarrow{AC}) \cdot ({1\over 3}(-\overrightarrow{AB} +\overrightarrow{AC}) + {2\over 3}\overrightarrow{AC}) \\={3\over 2}(\overrightarrow{AB} +\overrightarrow{AC}) \cdot (-{1\over 3} \overrightarrow{AB} +\overrightarrow{AC} ) =-{1\over 2}\overline{AB}^2+ \overrightarrow{AB}\cdot \overrightarrow{AC} +{3\over 2}\overline{AC}^2 \\ \Rightarrow {1\over 2}\overline{AB}^2= {3\over 2}\overline{AC}^2 \Rightarrow {\overline{AB} \over \overline{AC}} =\bbox[red, 2pt]{\sqrt 3}$$    

解答：

$$\cases{旋轉矩陣 R=\begin{bmatrix} \cos \alpha & -\sin \alpha\\ \sin \alpha & \cos \alpha\end{bmatrix} \\\tan \alpha=3/4}  \Rightarrow R=\begin{bmatrix} 4/5 & -3/5\\ 3/5& 4/5 \end{bmatrix} \\ \cases{對稱直線L'矩陣S'=\begin{bmatrix} \cos 2\beta & \sin 2\beta\\ \sin 2\beta & -\cos 2\beta \end{bmatrix} \\L':y=5x \Rightarrow \tan \beta=5 \Rightarrow \tan 2\beta ={5+5\over 1-5^2}=-{5\over 12}} \Rightarrow S'=\begin{bmatrix} -12/13 & 5/13\\ 5/13 & 12/13 \end{bmatrix}\\ \cases{對稱直線L矩陣S= \begin{bmatrix} \cos 2\gamma & \sin 2\gamma\\ \sin 2\gamma & -\cos 2\gamma \end{bmatrix} \\ L:y=mx \Rightarrow \tan \gamma=m \Rightarrow \tan 2\gamma=2m/(1-m^2)}\\ 由題意知: SR=S' \Rightarrow S=S'R^{-1} = \begin{bmatrix} -12/13 & 5/13\\ 5/13 & 12/13 \end{bmatrix}\begin{bmatrix} 4/5 & 3/5\\ -3/5& 4/5 \end{bmatrix}\\ =\begin{bmatrix} -63/65  & -16/65\\ -16/65 & 63/65 \end{bmatrix} =\begin{bmatrix} \cos 2\gamma & \sin 2\gamma\\ \sin 2\gamma & -\cos 2\gamma \end{bmatrix} \Rightarrow \cases{\cos 2\gamma = -63/65\\ \sin 2\gamma = -16/65}\\ \Rightarrow \tan 2\gamma = 16/63 = 2m/(1-m^2) \Rightarrow 8m^2+63m-8=0 \Rightarrow (8m-1)(m+8) =0\\ \Rightarrow m=\bbox[red, 2pt]{-8}(1/8不合,P與Q需在L的異側)$$

解答：

$$BPQC面積=2\triangle APQ \Rightarrow {\triangle APQ \over \triangle ABC} ={\triangle APQ \over \triangle APQ +2\triangle APQ} ={1\over 3} = {ab\over 2\cdot 3} \Rightarrow ab=2\\ BPQC周長=2(\triangle APQ 周長) \Rightarrow \overline{PQ}+9-(a+b) = 2(a+b+\overline{PQ}) \\ \Rightarrow \overline{PQ}= 9-3(a+b) \cdots(1)\\ 又\cases{\cos \angle A={2^2+3^2 -4^2 \over 2\cdot 2\cdot 3} =-{1\over 4} \\ \cos \angle A= {a^2+b^2 -\overline{PQ}^2  \over 2ab}= {a^2+b^2 -\overline{PQ}^2  \over 4}} \Rightarrow \overline{PQ}^2= a^2+b^2+1=(a+b)^2-2ab+1= (a+b)^2-3\\ \Rightarrow \overline{PQ}=\sqrt{(a+b)^2-3} \cdots(2)\\ 由(1)及(2)\Rightarrow 9-3x=\sqrt{x^2-3} \Rightarrow 4x^2-27x+42=0,其中x=a+b \\ \Rightarrow x=a+b=({27-\sqrt{57}})/8 代回(2) \Rightarrow \overline{PQ} =\sqrt{594-54\sqrt{57} \over 64} =\bbox[red, 2pt]{3\sqrt{57}-9\over 8} \\ 但\cases{a+b=({27-\sqrt{57}})/8\approx 2.43\\a+b \ge 2\sqrt{ab}=2\sqrt 2 \approx 2.8}兩者矛盾$$

解答：

$$兩直線\cases{L_1:3x+2y=7 \\ L_2:x-2y=13} 交點G(5,-4)即為\triangle ABC之重心，\\又\cases{B在L_1上\\ C在L_2上} \quad \Rightarrow \cases{B(s,(7-3s)/2)\\ C(2t+13,t)},s,t\in \mathbb{R}\\ \Rightarrow G=(A+B+C)/3 \Rightarrow \cases{5=(-1+s+2t+13)/3\\ -4=(2+(7-3s)/2+ t)/3} \Rightarrow \cases{s=19/2\\ t=-13/4} \\ \Rightarrow C(-13/2+13,-13/4)= \bbox[red, 2pt]{\left({13\over 2},-{13\over 4} \right)}$$

解答：

$$假設\cases{\overline{MN}\bot \overline{AI}\\ \angle M'IM=\theta} \Rightarrow \cases{\triangle M'IM= \overline{MI}\cdot \overline{M'I}\sin \theta\div 2\\ \triangle N'IN= \overline{NI}\cdot \overline{N'I}\sin \theta\div 2} \\ \Rightarrow \triangle NIN' \gt \triangle MIM' \Rightarrow \triangle AMN \lt \triangle AM'N' (\because \overline{IM}=\overline{IN})\\ \Rightarrow 當\overline{AI}\bot \overline{MN}時，\triangle AMN面積最小$$

$$\cases{\overline{AB}=6 \\ \overline{AC} =9\\ \overline{BC}=12} \Rightarrow  \cos A={6^2+ 9^2-12^2 \over 2\cdot 6\cdot 9}=-{1\over 4} \Rightarrow \sin A={\sqrt{15}\over 4} \Rightarrow \tan {A\over 2} ={\sin A\over 1+\cos A}={\sqrt{15} \over 3}\\ I為內心 \Rightarrow \overrightarrow{AI} = {9\over 6+9+12} \overrightarrow{AB} +{6\over 6+9+12} \overrightarrow{AC} ={1\over 3}\overrightarrow{AB} +{2\over 9} \overrightarrow{AC} \\ \Rightarrow |\overrightarrow{AI} |^2 = ({1\over 3}\overrightarrow{AB} +{2\over 9} \overrightarrow{AC}) \cdot ({1\over 3}\overrightarrow{AB} +{2\over 9} \overrightarrow{AC}) = {1\over 9}\overline{AB}^2 +{4\over 81}\overline{AC}^2 + {4\over 27}\overrightarrow{AB} \cdot \overrightarrow{AC} \\ ={1\over 9}\cdot 36+{4\over 81} \cdot 81 + {4\over 27}|\overrightarrow{AB}| \cdot |\overrightarrow{AC}| \cos A =4+4+{4\over 27}\cdot 6\cdot 9\cdot (-{1\over 4})= 6  \\ 因此\triangle AMN = \overline{AI}\times \overline{IN}= \overline{AI}\times \overline{AI} \tan {A\over 2}= 6\cdot {\sqrt{15}\over 3} =\bbox[red, 2pt]{2\sqrt{15}}$$

解答：

$$令\cases{D(0,0)\\ B(-1,0)\\ C(1,0)\\ I(0,r)\\ \angle IBD=\theta} \Rightarrow \cases{\tan \theta =r \\ H(0,2r)} \Rightarrow \tan 2\theta ={2r \over 1-r^2} \Rightarrow A(0,{2r\over 1-r^2}) \Rightarrow \cases{\overrightarrow{BH}=(1,2r) \\ \overline{AC}=(1,{2r\over r^2-1})}\\ 又H為垂心 \Rightarrow \overrightarrow{BH} \cdot \overrightarrow {AC}=0 \Rightarrow   1+{4r^2 \over r^2-1} ={5r^2-1\over r^2-1} =0 \Rightarrow r^2={1\over   5} \\ \Rightarrow {\overline{AD} \over \overline{HD}} ={2r/(1-r^2) \over 2r} ={1\over 1-r^2} ={1\over 1-1/5} = \bbox[red, 2pt]{5\over 4}$$

解答：$$假設監測站O在原點，則依題意颱風中心P在y=-7x的直線上，且距離300公里，即\\ \cases{P(a,-7a)\\ \overline{OP}=300}  \Rightarrow P(30\sqrt 2,-210\sqrt 2)；\\又中心移動速度20公里/時，方向西偏北45^\circ \Rightarrow t小時後，P(30\sqrt 2-10 \sqrt 2t,-210\sqrt 2+10 \sqrt 2t)；\\ 每小時颱風圈半徑增加10公里，因此t小時後，半徑變為60+10t；\\ 若\overline{OP} \le r(颱風半徑)，則監測站受到颱風侵襲，即\\ (30\sqrt 2-10\sqrt 2t)^2 +(10\sqrt t-210\sqrt 2)^2 \le (60+10t)^2 \\ \Rightarrow t^2-36t+288\le 0 \Rightarrow (t-24)(t-12)\le 0 \Rightarrow 12\le t\le 24 \Rightarrow 持續24-12=\bbox[red, 2pt]{12}小時$$

解答：$$6x+2y-3x\lfloor {x^2+y^2 \over x^2}\rfloor=0 \Rightarrow 6x+2y = 3x\lfloor 1+ {y^2\over x^2}\rfloor =3x\left(1+ \lfloor {y^2 \over x^2}\rfloor \right) \\ \Rightarrow 3x+2y = 3x\lfloor {y^2 \over x^2}\rfloor \Rightarrow 1+{2\over 3}t= \lfloor t^2 \rfloor , \text{where }t=y/x \Rightarrow \lfloor t^2 \rfloor \ge 0 \\ 可得: \begin{array}{} \lfloor t^2 \rfloor & t^2 & t & 1+{2\over 3}t & t=\\\hline 0 & [0,1) & [0,1) &[1,{5\over 3}) & \lfloor t^2 \rfloor \lt 1+{2\over 3}t \\ 1 & [1,2)& [1,\sqrt 2) & [{5\over 3},{3+2\sqrt 2\over 3}) & \lfloor t^2 \rfloor \lt 1+{2\over 3}t \\ 2 & [2,3) & [\sqrt 2, \sqrt 3)&[{3+2\sqrt 2\over 3},{3+2\sqrt 3\over 3}) &t=3/2\\ 3 & [3,4) & [\sqrt 3,2) & [{3+2\sqrt 3\over 3},{7\over 3}) & \lfloor t^2 \rfloor \gt 1+{2\over 3}t\\ \dots & \dots &\dots &\dots &  \lfloor t^2 \rfloor \gt 1+{2\over 3}t \\\hline\end{array}\\ t={3\over 2}={y\over x}\Rightarrow (x,y)=(2k,3k),1\le k\le 16 \Rightarrow 共\bbox[red, 2pt]{16}組$$

解答：$$3m+4n要最大，即奇數與偶數的數值要小，因此先計算最小的奇數和與偶數和；\\\cases{m個正偶數和=2+4+\cdots + 2m= m^2+m\\ n個正奇數和=1+3+\cdots + 2n-1=n^2} \quad \Rightarrow m^2+m+n^2 \le 2025  \\ \Rightarrow (m+{1\over 2})^2 +n^2 \le 2025{1\over 4} \cdots(1)\\ 柯西不等式 \left( (m+{1\over 2})^2 +n^2 \right)(3^2+4^2) \ge (3(m+{1\over 2})+4n)^2 \Rightarrow 2025{1\over 4}\times 25 \ge (3m+4n+{3\over 2})^2\\ \Rightarrow 225.X \ge  3m+4n+{3\over 2} \Rightarrow 3m+4n\lt 224 \Rightarrow 3m+4n= \bbox[red, 2pt]{223}$$

解答：$$執白棋(先下)贏後，比賽次數的期望值E_1={2\over 5}\times 1(再贏) +{3\over 5}(E_1+1)(輸了再贏) \Rightarrow E_1={5\over 2}\\ 執黑棋(後下)贏後，比賽次數的期望值E_2={3\over 5}\times 1(再贏) +{2\over 5}(E_2+1)(輸了再贏) \Rightarrow E_2={5\over 3}\\ 因此整個比賽的次數期望值={3\over 5}(E_1+1)+ {2\over 5}(E_2+1) = \bbox[red, 2pt]{19\over 6}$$

二、計算證明題：請寫出詳細計算與證明過程，否則不予給分， 共 24 分。

解答：$${a\over b+c} +{b\over c+a} +{c\over a+b} =\left({a+b+c\over b+c}-1 \right) +\left({a+b+c\over c+a}-1 \right) +\left({a+b+c\over a+b}-1 \right) \\ =(a+b+c)\left( {1\over b+c} +{1\over c+a} +{1\over a+b}\right)-3 \\ ={1\over 2}((b+c) +(c+a) +(a+b))\left( {1\over b+c} +{1\over c+a} +{1\over a+b}\right)-3 \\ \ge {1\over 2}(1+1+1)^2-3 ={9\over 2}-3={3\over 2}\\ \Rightarrow {a\over b+c} +{b\over c+a} +{c\over a+b} \ge {3\over 2}，\bbox[red, 2pt]{故得證}$$

解答：

$$y=f(x)=x^3 \Rightarrow f'(x)=3x^2，並假設\cases{P(a,a^3)\\ Q(b,b^3)}，其中a\gt 0\\ 因此\overleftrightarrow{PQ}斜率={a^3-b^3 \over a-b} =f'(b) \Rightarrow a^2+ab+b^2 =3b^2 \Rightarrow a^2+ab-2b^2=0 \Rightarrow (a-b)(a+2b)=0\\ \Rightarrow a=-2b (P\ne Q \Rightarrow a\ne b) \Rightarrow \cases{\overleftrightarrow{PA} 斜率m_1=f'(a)=3a^2 = 12b^2\\ \overleftrightarrow{PB} 斜率m_2= f'(b)=3b^2} \\ \Rightarrow \tan \angle APB = {m_1- m_2 \over 1+m_1m_2} ={9b^2 \over 1+ 36b^4} = {9\over 1/b^2+ 36b^2}\\ 由於{1\over b^2} +36b^2 \ge 2\sqrt{{1\over b^2} \cdot 36b^2} =12 \Rightarrow \tan \theta的最大值={9\over 12} =\bbox[red, 2pt]{3\over 4}$$

解答：

(1)$$\cases{a_1=4\\ a_2=5\\ a_n={a_{n-1}^2-1\over a_{n-2}},n\ge 3} \Rightarrow a_3={5^2-1\over 4}=6 \\ 又\cases{a_n^2 =a_{n+1} a_{n-1}+1 \\ a_{n-1}^2 = a_na_{n-2}+1} \Rightarrow a_n^2-a_{n-1}^2 =a_{n+1}a_{n-1} -a_na_{n-2} \Rightarrow a_n^2+ a_na_{n-2} = a_{n-1}^2+ a_{n-1}a_{n+1} \\ \Rightarrow a_n(a_n+a_{n-2}) = a_{n-1}(a_{n-1}+a_{n+1}) \Rightarrow {a_{n+1}+a_{n-1} \over a_{n }} = {a_{n}+a_{n-2} \over a_{n-1}} = \cdots ={a_3+a_1\over a_2} ={6+4\over 5}=2\\ \Rightarrow {a_{n+1}+ a_{n-1}\over a_n} =2 \Rightarrow a_{n+1}+a_{n-1} =2a_n \Rightarrow a_{n+1}-a_n=a_n-a_{n-1} \\ \Rightarrow a_n-a_{n-1} = a_{n-1}-a_{n-2} =\cdots = a_2-a_1=1\Rightarrow \langle a_n \rangle 為等差數列，公差=1\\ \Rightarrow a_n =a_1+(n-1) d= 4+n-1 =n+3 \Rightarrow \bbox[red,2pt]{a_n= n+3,n\in \mathbb{N}}$$(2)$$\cases{{1\over \sqrt n} ={2\over \sqrt n+\sqrt n} \le {2\over \sqrt n+\sqrt{n-1}} =2(\sqrt n-\sqrt{n-1}) \\ {1\over \sqrt n} ={2\over \sqrt n+\sqrt n} \ge {2\over \sqrt{n+1} +\sqrt n} =2(\sqrt{n+1}- \sqrt n)} \\ \Rightarrow 2(\sqrt{n+1}- \sqrt n)\le {1\over \sqrt n} \le 2(\sqrt n-\sqrt{n-1}) \\因此{1\over \sqrt{a_n}} ={1\over \sqrt{n+3}} \Rightarrow 2(\sqrt{n+4}- \sqrt {n+3})\le {1\over \sqrt{a_n}}  \le 2(\sqrt {n+3}-\sqrt{n+2}) \\ \Rightarrow 2\sum_{n=1}^{2021} (\sqrt{n+4}- \sqrt {n+3})\le \sum_{n=1}^{2021} {1\over \sqrt{a_n}}  \le 2\sum_{n=1}^{2021}  (\sqrt {n+3}-\sqrt{n+2}) \\ \Rightarrow  2 (\sqrt{2025}- \sqrt 4)\le \sum_{n=1}^{2021} {1\over \sqrt{a_n}}  \le 2(\sqrt {2024}-\sqrt{3}) \\ \Rightarrow 2(45-2)\le \sum_{n=1}^{2021} {1\over \sqrt{a_n}}  \Rightarrow 86 \le \sum_{n=1}^{2021} {1\over \sqrt{a_n}} \\\Rightarrow \sum_{n=1}^{2021} {1\over \sqrt{a_n}}的整數部分為\bbox[red, 2pt]{86}$$

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解題僅供參考，其它教甄試題及詳解