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2021年5月24日 星期一

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x=abc{a=b=c={M(x):a,b,cm(x):a,b,cH(x)=M(x)m(x)x使x=H(x)a,b,cH(x)xx=692H(693)=963369=594x=594H(594)=954459=495x=495H(495)=954459=495495:Youtube
解答f(x)=3xf(t+16)f(t16)=23t+163t16=2(3t+16)3=(2+3t16)3t+16=8+(t16)+63t16(2+3t16)63(t16)2+123t1624=03t16=51t=(51)3+16=8516+16=85
解答x22x+1262+x212x+3922=(x1)2+1162+(x6)2+322=¯PA+¯PB,{P(x,0)A(1,1162)B(6,322)PxABx¯PA+¯PB=¯AB¯AB=(61)2+(1162+322)2=25+(32+21)2=25+4=29
解答


195499C95=9126=114


解答Lm(1,2)L:y=m(x1)+2xy=1x(m(x1)+2)=1f(x)=mx2+(2m)x1=0α,β(α>β){P(α,f(α))Q(β,f(β)){α+β=(m2)/mαβ=1/m(αβ)2=(α+β)24αβ=m2+4m2αβ=m2+4m2¯PQ=(αβ)×m2+1=m4+5m2+4m2=5+m2+4m25+4=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



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