109年中壢高中教甄-數學詳解

 國立中央大學附屬中壢高級中學109學年度第 1 次教師甄選數學科筆試題目卷

解：

$$\cases{直角\triangle ADO:\overline{OA}^2 =\overline{OD}^2 +\overline{AD}^2 \\直角\triangle OBD:\overline{OB}^2 =\overline{OD}^2 +\overline{BD}^2} \Rightarrow \cases{\overline{AD}=1 \\ \overline{BD}=5 };\\ 又D,E,F為切點\Rightarrow \cases{\overline{AF} =\overline{AD} =1\\ \overline{BE} =\overline{BD}=5 \\ \overline{CE}=\overline{CF}=a } \Rightarrow \triangle ABC周長=2S=2a+12\\ \triangle ABC面積=\cases{\triangle OAB+\triangle OBC +\triangle OAC = \sqrt 3(a+6)\\ \sqrt{s(s-\overline{AB}) (s-\overline{BC})(s-\overline{AC})} =\sqrt{5a(a+6)}} \\ \Rightarrow \sqrt 3(a+6)= \sqrt{5a(a+6)} \Rightarrow 3(a+6)=5a \Rightarrow a=9 \Rightarrow \triangle ABC面積= \sqrt 3(9+6) =\bbox[red,2pt]{15\sqrt 3}$$

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解：

$$\sum_{k=1}^8 \overrightarrow{PA_k} =\sum_{k=1}^8 (\overrightarrow{PO}+\overrightarrow{OA_k}) =8\overrightarrow{PO} +\sum_{k=1}^8\overrightarrow{OA_k} = 8\overrightarrow{PO} \\ \Rightarrow |\sum_{k=1}^8 \overrightarrow{PA_k}| = |8\overrightarrow{PO}| = \bbox[red,2pt]{8}$$

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解：

$$P為A逆時鐘旋轉60^\circ,即P=\begin{bmatrix} 1/2 & -\sqrt 3/2\\ \sqrt 3/2 & 1/2\end{bmatrix} \begin{bmatrix} 5\\ 1\end{bmatrix} =\begin{bmatrix} {5-\sqrt 3\over 2}\\ {1+5\sqrt 3\over 2}\end{bmatrix}\\ \overrightarrow{AC}= \overrightarrow{OP} \Rightarrow \overrightarrow{AB} = 3\overrightarrow{AC} =3({5-\sqrt 3\over 2}, {1+5\sqrt 3\over 2}) =({15-3\sqrt 3\over 2}, {3+15\sqrt 3\over 2}) \\ \Rightarrow B=({15-3\sqrt 3\over 2}, {3+15\sqrt 3\over 2})+(5,1)=\bbox[red,2pt]{({25-3\sqrt 3\over 2}, {5+15\sqrt 3\over 2})}$$

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解：

$$正立方體\Rightarrow \overleftrightarrow{AC} \bot \overleftrightarrow{HF} \Rightarrow (-2,2,1) \bot (2,1,a) \Rightarrow (-2,2,1) \cdot (2,1,a) =-4+2+a=0 \Rightarrow a=2;\\ \overleftrightarrow{AC} 與 \overleftrightarrow{HF}的距離即為立方體的邊長,因此令\vec n = (-2,2,1) \times (2,1,2) = (3,6,-6)\\ \cases{P(3,-3,-5)在\overleftrightarrow{AC}上 \\Q(0,-2,2)在\overleftrightarrow{HF}上} \Rightarrow \overrightarrow{PQ}=(-3,1,7) \\\Rightarrow \overrightarrow{PQ}在\vec n上的正射影長= {\overrightarrow{PQ} \cdot \vec n \over |\vec n|} ={|-9+6-42| \over \sqrt{9+36+36}} ={45\over 9} =5 \\ \Rightarrow 正立方體體積=5^3 =\bbox[red,2pt]{125}$$

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解：

$$假設\cases{A(a,0,0)\\ B(0,b,0)\\ C(0,0,c)} \Rightarrow E: {x\over a} +{y\over b}+ {z\over c}=1; 又E過P(3,1,2) \Rightarrow {3\over a} +{1\over b}+ {2\over c}=1\\由柯西不等式:((\sqrt{3\over a})^2 +(\sqrt{1\over b})^2+ (\sqrt{2\over c})^2) ( (\sqrt{3a})^2+ (\sqrt{9b})^2 +(\sqrt{2c})^2)  \ge (3+3+2)^2 \\ \Rightarrow ({3\over a} +{1\over b}+ {2\over c})(3a+9b+2c) \ge 8^2 \Rightarrow m = 3\overline{OA}+9\overline{OB} +2\overline{OC}=3a+9b+2c \ge 64\\ \Rightarrow m之最小值為\bbox[red,2pt]{64}$$

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解：

$$\begin{array}{} 前5字母\not A& 中5字母\not B& 後5字母\not C & 排列數\\\hline 5B & 5C & 5A & 1\\ 4B1C & 4C1A & 4A1B & 5^3 \\ 3B2C & 3C2A & 3A2B & C^5_3C^5_3C^5_3 \\ 2B3C & 2C3A & 2A3B & C^5_2C^5_2C^5_2 \\ 1B4C & 1C4A & 1A4B & 5^3\\ 5C & 5A & 5B & 1\\\hline\end{array} \\ \Rightarrow 排列數共有1+125+ 1000+1000 + 125+1 =\bbox[red,2pt]{2252}$$

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解：

$$P,Q,R為切點(見上圖)\Rightarrow \overline{NP}= \overline{NQ} =\overline{NR}=a\\ 直角\triangle O_1SO_2: \overline{O_1O_2}^2 = \overline{O_1S}^2 +\overline{SO_2}^2 \Rightarrow (9+r_2)^2 =(9-r_2)^2 +(25-9-r_2)^2 \\\Rightarrow r_2^2-68r_2+256=0\Rightarrow r_2=4(64不合) \Rightarrow a=(25-9-r_2)\div 2=6\\ N為切點 \Rightarrow \angle SO_1O_2 +\angle PNQ = 180^\circ \Rightarrow \angle MNC = \angle O_2O_1S \\\Rightarrow \overline{MN}的斜率= \tan \angle MNC = \tan \angle O_2O_1S = {2a\over 9-r_2} = \bbox[red,2pt]{12\over 5}$$

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解：

$$\begin{Vmatrix} 1 & 2 & 3\\ 1 & 0 & 3\\ 0 & -1 & 5\end{Vmatrix} = |-10|=10 \Rightarrow T=10\times (1-(-1)\times (2-0) \times (3-(-2)) = 10\times 20 =\bbox[red,2pt]{200}$$

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解：

$$\triangle ABC \cong \triangle EDC \Rightarrow \overline{AC} =\overline{EC}=a \Rightarrow \cases{\cos \alpha = {32-a^2\over 32} \\ \cos \beta ={2a^2-1\over 2a^2}} \\ \Rightarrow (1-\cos\alpha)(1-\cos \beta) = {a^2\over 32}\times {1\over 2a^2} =\bbox[red,2pt]{1\over 64}$$

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解：

$$\cases{f(x)=x^3+2x^2 -3x-1 \\ g(x)=x^4+3x^3-x^2-5x+1} \Rightarrow g(x)=(x+1)f(x)-x+2\\ 又\alpha,\beta,\gamma 為f(x)=0之三根\Rightarrow \cases{\alpha+\beta +\gamma = -2\\ \alpha\beta +\beta\gamma +\gamma\alpha=-3\\ \alpha\beta\gamma = 1}\\ 因此 {1\over g(\alpha)} +{1\over g(\beta)} +{1\over g(\gamma)} ={1\over (\alpha+1)f(\alpha)-\alpha+2} +{1\over (\beta+1)f(\beta)-\beta+2} +{1\over (\gamma+1)f(\gamma)-\gamma+2} \\= {1\over 2-\alpha}+{1\over 2-\beta} +{1\over 2-\gamma}  = {(2-\beta)(2-\gamma)+(2-\alpha)(2-\gamma) +(2-\alpha)(2-\beta) \over (2-\alpha)(2-\beta)(2-\gamma)} \\={12-4(\alpha+\beta +\gamma) +(\alpha\beta +\beta\gamma +\gamma\alpha) \over 8-4(\alpha+\beta +\gamma)+2(\alpha\beta +\beta\gamma +\gamma\alpha)-\alpha\beta \gamma}  ={12+8-3 \over 8+8-6-1} =\bbox[red,2pt]{17\over 9}$$

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解：

$$\triangle APC \Rightarrow \cases{\overline{PA} =\overline{PC}=a \\ \overline{AC}=\overline{AB}\times \sqrt 2=4\sqrt 2}\Rightarrow \cos \beta = {2a^2-(4\sqrt 2)^2\over 2a^2} =-{1\over 4} \Rightarrow a={8\over \sqrt 5} \\ \angle BPC=90^\circ \Rightarrow \overline{BC}^2 =\overline{BP}^2+\overline{PC}^2 \Rightarrow 16=\overline{BP}^2 + ({8\over \sqrt 5})^2 \Rightarrow \overline{BP}= {4\over \sqrt 5}\\ 同理,\angle CPE=90^\circ \Rightarrow \overline{EC}^2 =\overline{PE}^2(=(\overline{EC}-\overline{PB})^2)+\overline{PC}^2 \\\Rightarrow \overline{EC}^2 =(\overline{EC}-{4\over \sqrt 5})^2+{8\over \sqrt 5}^2 \Rightarrow \overline{EC}= 2\sqrt 5\\ 直角\triangle EQC: (2\sqrt 5)^2 =2^2 +\overline{EQ}^2 \Rightarrow \overline{EQ}=4 \Rightarrow \cos \alpha={\overline{OQ} \over \overline{EQ}} ={2\over 4}=\bbox[red,2pt]{1\over 2}$$

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解：

$$令\cases{P(x,3x^2) \\ Q(2\cos y,2\sin y-8)} \Rightarrow \cases{P為拋物線\Gamma_1:y=3x^2上的點\\ Q為圓\Gamma_2: x^2+(y+8)^2=4上的點\\\overline{PQ}^2 =(x-2\cos y)^2+(3x^2+8-2\sin y)^2}\\ 因此\overline{PQ}最短的距離=\Gamma_2圓心至\Gamma_1頂點的距離再減去半徑長=8-2=6 \Rightarrow \overline{PQ}^2 =\bbox[red,2pt]{36}$$

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解：

此題相當於求上圖兩塊著色的面積和，也相當於下圖著色部份的面積 = $\overline{AB}\times \overline{BC}\div 2= 2\sqrt 2\times 2\sqrt 2\div 2=\bbox[red,2pt]{4}$

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解：

$$兩平面\cases{3x+4y=5\\ 3x+4y=0}距離為1，兩平面\cases{4x-3y=0\\ 4x-3y=5}距離為1，兩平面\cases{z=0\\ z=2}距離為2;\\ 該長方體邊長分別為1,1,2，見上圖；任取三頂點共有C^8_3=56種取法，其中\triangle面積比1大的取法:\\(1)\triangle 邊長為1,例\overline{AB},有2種選法,如:ABG,ABH;共有8\times 2=16種取法\\(2)\triangle 邊長為2,例\overline{AE},有2種選法,如:AEG,AEC;共有4\times 2=8種取法\\(3)\triangle 邊長為\sqrt 2,例\overline{AC},有2種選法,如:ACF,ACH;共有4\times 2=8種取法\\(1)+(2)+(3) =16+8+8 = 32 \Rightarrow 機率為{32\over 56} =\bbox[red,2pt]{4\over 7}$$

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解： $$\cases{f(x)=0有虛根\\ \lim_{x\to 1}{f(x)\over x^3-1}={1\over 3}} \Rightarrow f(1)=0;由於 f(x)=(x^2-2ax+b-2a)(x-1)-2a+b-c \\ \Rightarrow f(1)=-2a+b-c=0 \Rightarrow 2a-b+c=0\cdots(1)\\ 又\lim_{x\to 1}{f(x)\over x^3-1}= \lim_{x\to 1}{ x^2-2ax+b-2a\over x^2+x+1} = {1-2a+b-2a\over 3} ={1\over 3} \Rightarrow b=4a代入(1) \Rightarrow c=2a\\ f(x)=0有虛根\Rightarrow x^2-2ax+b-2a=0 無實根 \Rightarrow 判別式 < 0 \Rightarrow 4a^2-4b+8a< 0 \\ \Rightarrow a^2-b+2a < 0 \Rightarrow a^2-4a+2a < 0 \Rightarrow a(a-2) < 0 \Rightarrow 0 < a < 2 \Rightarrow a=1 \\\Rightarrow \cases{b=4a=4\\ c=2a=2} \Rightarrow a+b+c = 1+4+2 = \bbox[red,2pt]{7}$$

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解： $$\cases{a^b=b^a \\ a=89b} \Rightarrow \cases{ b\log a=a\log b \\ \log a = \log 89 +\log b}\\ \log_{89}(ab) = {\log ab \over \log 89} ={\log a+\log b \over \log a-\log b} = {\log a+{b\over a}\log a \over \log a-{b\over a}\log a} ={1+b/a \over 1-b/a} ={a+b\over a-b} ={89b+b\over 89b-b} ={90\over 88} \\= \bbox[red,2pt]{45\over 44}$$

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解： $$S=\sum_{n=1}^\infty {n(n+1)\over 2^n} = {2\over 2}+{6\over 2^2} +{12\over 2^3} +{20\over 2^4}+\cdots+{(n-1)n\over 2^{n-1}} +{n(n+1)\over 2^n} +\cdots \\ \Rightarrow {1\over 2}S={2\over 2^2}+{6\over 2^3} +{12\over 2^4} +{20\over 2^5}+\cdots+{(n-1)n\over 2^{n}} +{n(n+1)\over 2^{n+1}} +\cdots \\ \Rightarrow S-{1\over 2}S ={2\over 2}+{4\over 2^2} +{6\over 2^3} +{8\over 2^4 }+\cdots +{2n\over 2^n}+\cdots \\ \Rightarrow {S\over 2}= 2\left( {1\over 2}+{2\over 2^2} +{3\over 2^3} +{4\over 2^4 }+\cdots +{n\over 2^n}+\cdots\right)=2\times {1\over 1-1/2}=4\\ \Rightarrow S=\bbox[red,2pt]{8}$$

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解： $$單位圓上兩點z_1,z_2,且\overline{z_1z_2}={\sqrt 6-\sqrt 2\over 2},則\cos \angle z_1Oz_2={1+1-({\sqrt 6-\sqrt 2\over 2})^2 \over 2} ={\sqrt 3\over 2} \\\Rightarrow \angle z_1Oz_2=30^\circ \Rightarrow |z_1-z_2| < {\sqrt 6-\sqrt 2\over 2} \Rightarrow \angle z_1Oz_2 < 30^\circ;\\現在正2020邊形z_1z_2...z_{2020},每一內角為{360^\circ \over 2020} \Rightarrow {360^\circ \over 2020}\times n < 30^\circ \Rightarrow n\le 168\\ \Rightarrow z_i的左邊有168個z_j(i\ne j),右邊也有168個z_j(i\ne j),使得|z_i-z_j|< {\sqrt 6-\sqrt 2\over 2}\\也就是說對任一z_i而言,在其它2019個z_j(i\ne j)有168\times 2=336個符合條件\\,其機率為{336\over 2019} =\bbox[red,2pt]{112\over 673}$$

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解：

$$S_n={3\over 2!}- {4\over 3!}+ {5\over 4!}-{6\over 5!}+ \cdots +(-1)^n\cdot {n+2 \over (n+1)!} \\ =({2\over 2!}+{1\over 2!})- ( {3\over 3!}+ {1\over 3!})+ ({4\over 4!} +{1\over 4!})-({5\over 5!}+{1\over 5!}) \cdots +(-1)^n\cdot ({n+1 \over (n+1)!} +{1 \over (n+1)!}) \\ =(1+{1\over 2!})- ( {1\over 2!}+ {1\over 3!})+ ({1\over 3!} +{1\over 4!})-({1\over 4!}+{1\over 5!}) \cdots +(-1)^n\cdot ({1 \over n!} +{1 \over (n+1)!}) \\ =1+(-1)^n\cdot {1\over (n+1)!} \\ \Rightarrow \lim_{n\to \infty} S_n = \bbox[red,2pt]{1}$$

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解：

$$三圖形\cases{\Gamma_1: {x^2\over 4}+y^2=1 \\ \Gamma_2: y+1=({\sqrt 3\over 2}+1)x\\ \Gamma_3:y+1=-({\sqrt 3\over 2}+1)x}的交點\cases{A(-1,{\sqrt 3\over 2})\\ B(1,{\sqrt 3\over 2})\\ C(0,1)}，見上圖；\\著色面積=2\int_0^1 \sqrt{1-{x^2\over 4}}-(({\sqrt 3\over 2}+1)x -1)\;dx \\ =2\left. \left[ {1\over 4}\sqrt{4-x^2}\cdot x+\sin^{-1}{x\over 2}-(({\sqrt 3\over 4}+{1\over 2})x^2-x)\right] \right|_0^1 =2({1\over 4}\sqrt 3+{\pi \over 6}-{2+\sqrt 3\over 4}+1) \\ =2({\pi \over 6}+{1\over 2}) = \bbox[red,2pt]{{\pi \over 3}+1}$$

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