113年中科實中教甄-數學詳解

 國立中科實驗高級中學 113 學年度第 1 次教師甄選

雙語部-數學科專業知能試題

一、 填充題： 共 16 題，每題 5 分

解答:$$沒有好方法，只能一個一個算: 正\triangle:100, 等腰\triangle: 7400, 其它\triangle: 79625, 合計:\bbox[red, 2pt]{87125}$$

解答:$$\cases{-\sqrt 2\le \cos \theta-\sin \theta \le \sqrt 2 \\ -\sqrt 2\le \cos \theta+\sin \theta \le \sqrt 2} \Rightarrow 所圍區域面積相當於以O為圓心,半徑\sqrt 2的圓面積= \bbox[red, 2pt]{2\pi}$$

解答:$${\sin A+\sqrt 3\cos A\over \cos A-\sqrt 3 \sin A} ={2({1\over 2}\sin A+{\sqrt 3\over 2}\cos A ) \over 2( {1\over 2}\cos A-{\sqrt 3\over 2} \sin A)} = {\cos 60^\circ \sin A+ \sin 60^\circ \cos A\over \cos 60^\circ \cos A- \sin 60^\circ \sin A} \\ ={\sin(A+60^\circ) \over \cos(A+60^\circ)} = \tan(A+60^\circ) =\tan{7\pi\over 12} =\tan 105^\circ \Rightarrow A=45^\circ \Rightarrow C=180^\circ-45^\circ -B=135^\circ -B\\ \Rightarrow \sin 2B+ 2\cos C= \sin 2B+ 2\cos (135^\circ -B)= 2\sin B\cos B+2 (-{1\over \sqrt 2}\cos B+ {1\over \sqrt 2}\sin B) \\=2\sin B\cos B+\sqrt 2(\sin B-\cos B) ,取x=\sin B-\cos B, 則2\sin B\cos B=1-x^2 \\ \Rightarrow  \sin 2B+ 2\cos C=f(x) =1-x^2+\sqrt 2x \Rightarrow f'(x)=-2x+\sqrt 2=0 \Rightarrow x={1\over \sqrt 2} \\\Rightarrow f({1\over \sqrt 2}) =1-{1\over 2}+1= \bbox[red, 2pt]{3\over 2}$$

解答:$$1\times 2+2\times 3+\cdots + n(n+1)= \sum_{k=1}^n k(k+1)= \sum_{k=1}^n (k^2+k) ={n(n+1)(2n+1)\over 6}+{n(n+1)\over 2} \\= {n(n+1)(n+2)\over 3} \Rightarrow \sum_{n=1}^{25} \left( {1\over 1\times 2+2\times 3+\cdots + n(n+1)} \right) =\sum_{n=1}^{25} \left( {3\over n(n+1)(n+2)} \right) \\ ={3\over 2}\sum_{n=1}^{25} \left( {1\over n(n+1)} -{1\over (n+1)(n+2)} \right) ={3\over 2}\left( {1\over 2}-{1\over 26\cdot 27}\right) ={3\over 2}\cdot {175\over 351}= \bbox[red, 2pt]{175 \over 234}$$

解答:$$u=\sqrt{x^2-6x+10} \Rightarrow f(x)={4u^2+81\over u} =4u+{81\over u} \ge 2\sqrt{4u\cdot {81\over u}} =\bbox[red, 2pt]{36}$$

解答:

$$\cases{L_1: 2x=4y-4=2z-3\\ L_2: {3\over 2}x+3=6-y= 12-3z} \Rightarrow \cases{L_1: \displaystyle {x\over {1\over 2}}={x-1\over {1\over 4}}= {z-{3\over 2}\over {1\over 2}} \\ L_2:\displaystyle  {x+2\over {2\over 3}}= {y-6\over -1}= {z-4\over {-1\over 3}}} \Rightarrow \cases{L_1方向向量 \vec u_1=(1/2,1/4,1/2) \\ L_2方向向量\vec u_2= (2/3,-1,-1/3)} \\ \Rightarrow 平面E的法向量 \vec n=\vec u_1\times \vec u_2 =({5\over 12},{1\over 2},-{2\over 3}) \Rightarrow E: {5\over 12}x+{1\over 2}(y-1)-{2\over 3}(z-{3\over 2}) =0 \\ \Rightarrow E:5x+6y-8z+6=0\\ 又P=L_1 \cap L_2=(1,{3\over 2},{5\over 2}) \Rightarrow \cases{d(A,E) = \overline{AH} =\sqrt 5\\ \overline{AP} =\sqrt{21}} \Rightarrow \overline{PH} =\sqrt{\overline{AP}^2- \overline{AH}^2} =4\\ 假設L_1與L_2的夾角為\theta \Rightarrow |\cos \theta|={\vec u_1\cdot \vec u_2 \over |\vec u_1||\vec u_2|} = {1\over 3\sqrt{14}} \Rightarrow \sin \theta= {5\sqrt 5\over 3\sqrt{14}} \\ 由於\angle PMH=\angle PNH =90^\circ \Rightarrow PMHN共圓 \Rightarrow \overline{PH}=4為直徑 \Rightarrow {\overline{MN} \over \sin \theta}=4 \\ \Rightarrow \overline{MN} =\bbox[red, 2pt]{10\sqrt{70} \over 21}$$

解答:$$這也沒有好方法,只能一個一個算,共\bbox[red, 2pt]{1404}個$$

解答:

$$\pi\int_{-5}^3 \left({x+5\over 2} \right)^2 \,dx + \pi \int_3^5 (25-x^2)\,dx =\pi \left. \left[ {2\over 3}\left({x+5\over 2} \right)^3\right] \right|_{-5}^3+ \pi \left. \left[ 25x-{1\over 3}x^3 \right] \right|_3^5 \\={128\over 3}\pi +{52\over 3}\pi =\bbox[red, 2pt]{60\pi}$$

解答:$$z=-{2\over 1+\sqrt 3i}= -{2(1-\sqrt 3i)\over 4}=-{1\over 2}+{\sqrt 3\over 2}i =\cos({2\over 3}\pi)+ i\sin({2\over 3}\pi) =e^{2\pi i/3 } \\ \Rightarrow z^3=1  \Rightarrow S={1\over 5}+z+z^2+ \cdots+ z^{2025}={1\over 5}+{z-z^{2026}\over 1-z} ={1\over 5}+{z-z\over 1-z} =\bbox[red, 2pt]{1\over 5}$$

解答:$$(骰子1點數,骰子2點數)=(2,1), (3,1-2),(4,1-3),(5,1-4),(6,1-5) \\ \Rightarrow 機率p={1\over 36}(1+2+3+4+5) ={5\over 12} \\\Rightarrow X\sim Bin(p={5\over 12},n=24) \Rightarrow Var(X) =np(1-p) =24\cdot {5\over 12} \cdot {7\over 12} = \bbox[red, 2pt]{35\over 6}$$

解答:$$\lfloor{10000\over 3}  \rfloor +\lfloor{10000\over 5}  \rfloor +\lfloor{10000\over 7}  \rfloor- \lfloor{10000\over 3\times 5}  \rfloor -\lfloor{10000\over 3\times 7}  \rfloor -\lfloor{10000\over 5\times 7}  \rfloor +\lfloor{10000\over 3\times 5\times 7}  \rfloor \\ =3333+2000+1428-666-476-286+95 = \bbox[red, 2pt]{5429}$$

解答:$$\cases{球:x^2+y^2+z^2=9 \\點P(1,1,1)}\Rightarrow \cases{球心O(0,0,0)\\ 球半徑r=3} \Rightarrow L=\overleftrightarrow{OP} :x=y=z \\ \Rightarrow 球與直線L交點:\cases{A(\sqrt 3,\sqrt 3,\sqrt 3)\\ B(-\sqrt 3,-\sqrt 3,-\sqrt 3)} \Rightarrow 離P最遠的點B \bbox[red, 2pt]{(-\sqrt 3,-\sqrt 3,-\sqrt 3)}$$

解答:$$令\cases{x'=x-4\\ y'=y-5\\ z'=z-3} \Rightarrow x+y+z=24 \Rightarrow x'+y'+z'=12 \Rightarrow 有H^3_{12}=\bbox[red, 2pt]{91}組解$$

解答:

$$如上圖，共\bbox[red, 2pt]{11}個$$

解答:$$\cases{\tan \alpha+\tan \beta=3/5\\ \tan \alpha \tan \beta=-1/5} \Rightarrow {\sin(\alpha+ \beta) \over \cos(\alpha-\beta)} ={\sin \alpha \cos \beta+ \sin \beta \cos \alpha\over \cos \alpha \cos \beta +\sin \alpha \sin \beta} \\={(\sin \alpha \cos \beta+ \sin \beta \cos \alpha)/ \cos \alpha \cos \beta\over (\cos \alpha \cos \beta +\sin \alpha \sin \beta )/ \cos \alpha \cos \beta} ={\tan \alpha+ \tan \beta \over 1+\tan \alpha \tan \beta} ={3/5\over 1-1/5} = \bbox[red, 2pt]{3\over 4}$$

解答:$$代公式:2+2C^8_2 =\bbox[red, 2pt]{58}\\ \href{https://blog.csdn.net/weixin_45750972/article/details/124160923}{參考資料}$$

二、計算題： 共 2 題，每題 10 分

解答:$$\textbf{(1)}\;A-I_2=\begin{bmatrix}{4\over 3} & 1 \\-{2\over 3} & -1 \end{bmatrix} -\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix} =\begin{bmatrix}{1\over 3} & 1 \\-{2\over 3} & -2 \end{bmatrix} \\ \Rightarrow A(A-I_2)= \begin{bmatrix} \frac{-2}{9} & \frac{-2}{3} \\\frac{4}{9} & \frac{4}{3} \end{bmatrix} =k(A-I_2)= k\begin{bmatrix}{1\over 3} & 1 \\-{2\over 3} & -2 \end{bmatrix} \Rightarrow \bbox[red, 2pt]{k=-{2\over 3}} \\\textbf{(2)}\;由(1)可知A的特徴值為-{2\over 3},1 ,\\因此令f(x)=x^n=ax+b \Rightarrow \cases{f(-2/3)=(-2/3)^n=-2a/3+b\\ f(1)=1=a+b} \Rightarrow \cases{a=3/5(1-(-2/3)^n)\\ b=2/5+3/5(-2/3)^n} \\ \Rightarrow A^n=aA+bI \Rightarrow A^\infty ={3\over 5}A+{2\over 5}I =\begin{bmatrix}{4\over 5} & {3\over 5} \\-{2\over 5} & -{3\over 5} \end{bmatrix} +\begin{bmatrix}{2\over 5} & 0 \\0 & {2\over 5} \end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix}{6\over 5} & {3\over 5} \\-{2\over 5} & -{1\over 5} \end{bmatrix}}$$

解答:$$取\cases{u=x+y\\ v=x-y} \Rightarrow \cases{v^2=\sqrt 2 u \\u=4\sqrt 2} \Rightarrow 所圍面積=2\int_0^{2\sqrt 2} (4\sqrt 2-{v^2\over \sqrt 2})\,dv={64\over 3}\\ 又{\partial(u,v)\over \partial(x,y)} =\begin{Vmatrix} 1 & 1\\ 1& -1 \end{Vmatrix} =2 \Rightarrow 欲求之面積={1\over 2}\cdot {64\over 3}= \bbox[red, 2pt]{32\over 3}$$

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解題僅供參考,教甄歷年試題及詳解