112年竹北高中教甄－數學詳解

國立竹北高中 112 學年度 第 1 次教師甄選

一、填充題：(每格 6 分，共 60 分)

解答： $$\cases{(1,2-6,7)有5種\\ (1,2-7,8)有6種 \\ (1,2-8,9)有7種\\ (1,2-8,10)有8種 \\ (2,3-7,8)有5種 \\ (2,3-8,9)有6種  \\ (2,3-9,10)有7種 \\ (3,4-8,9)有5種\\ (3,4-9,10)有6種\\ (4,5-9,10)有5種} \Rightarrow 共有4\cdot 5+3\cdot 6+2\cdot 7+8=60 \\ \Rightarrow 機率＝{60\over C^{10}_3}={60\over 120} =\bbox[red,2pt]{1\over 2}$$

解答： $$z=x+yi \Rightarrow |z-i|=|z| \Rightarrow \sqrt{x^2+(y-1)^2} =\sqrt{x^2+y^2} \Rightarrow (y-1)^2=y^2 \Rightarrow y={1\over 2} \\ 又|z-1|+|z|=\sqrt 2為一橢圓，其中\cases{焦點F_1(1,0)\\ 焦點F_2  (0,0) \\ 2a=\sqrt 2} \Rightarrow \cases{a=\sqrt 2/2\\ c=1/2} \Rightarrow b={1\over 2}\\ \Rightarrow 直線y={1\over 2}與橢圓相切，切點為(0，{1\over 2}),因此解只有\bbox[red,2pt]1個$$

解答： $$\sum_{m=1}^9 \sum_{k=1}^m k = {1\over 2}\sum_{m=1}^9(m^2+m) = {1\over 2}\left({1\over 6\cdot 9\cdot 10\cdot 19}+45 \right) ={1\over 2}\cdot 330 =\bbox[red,2pt]{165}$$

解答：

$$令\angle AOP=\theta \Rightarrow \angle AOC=90^\circ-\theta,又\triangle OAB\cong \triangle OCB (RHS) \Rightarrow \angle AOB=45^\circ-{1\over 2}\theta \\ \Rightarrow \overline{AB}=\overline{AB}\tan \angle AOB = \tan\left( 45^\circ-{1\over 2}\theta\right)={1-\tan(\theta/2)\over 1+\tan(\theta/2) }\\ 重疊面積＝2\triangle OAB= \overline{OA}\times \overline{AB} =\overline{AB}={1-\tan(\theta/2)\over 1+\tan(\theta/2) }={2\over 3} \Rightarrow \tan(\theta/2)={1\over 5} \\ \Rightarrow \tan \theta={2\tan(\theta/2) \over 1-\tan^2(\theta/2)} = \bbox[red, 2pt]{5\over 12}$$

解答： $$\cases{\log_4 x+\log_8(yz)=2 \\ \log_4 y+\log_8(xz)=4\\ \log_4 z+\log_8 (xy)=5} \Rightarrow \cases{{1\over 2}\log_2 x+{1\over 3}\log_2(yz)=2 \\ {1\over 2}\log_2 y+ {1\over 3}\log_2(xz) =4\\ {1\over 2}\log_2 z+ {1\over 3}\log_2 (xy)=5} \Rightarrow \cases{ \log_2 x^{1/2}(yz)^{1/3} =2 \\ \log_2 y^{1/2}(xz)^{1/3} =4 \\ \log_2 z^{1/2}(xy)^{1/3} =5} \\ \Rightarrow \cases{x^{1/2}(yz)^{1/3}  =4 \\y^{1/2}(xz)^{1/3} =16 \\ z^{1/2}(xy)^{1/3} =32},三式相乘\Rightarrow (xyz)^{7/6}=2^{11} \Rightarrow xyz=(2^{11})^{6/7} =2^{66/7}\\ \Rightarrow k =\bbox[red, 2pt]{66\over 7}$$

解答： $$a+b=k \Rightarrow a^2+b^2=k^2-2ab\\ \left(a+{1\over a} \right)\left(b+{1\over b} \right) =ab+{b\over a}+ {a\over b}+{1\over ab} =ab+{1\over ab}+{a^2+b^2\over ab} =ab+{1\over ab}+{k^2-2ab\over ab} \\=ab+{k^2+1\over ab}-2 \ge 2\sqrt{ab\cdot {k^2+1\over ab}}-2 = \bbox[red, 2pt]{2\sqrt{k^2+1}-2}$$

解答： $$f(x)=x^{100}=(x^3+x^2+x)p(x)+r(x) \Rightarrow f(0)=0=r(0) \Rightarrow r(x)沒有常數項\\ \Rightarrow r(x)=ax^2+bx \Rightarrow g(x)=x^{99} =(x^2+x+1)p(x)+ax+b\\ x^3-1=0的三根為1,\omega,\omega^2 \Rightarrow \cases{g(\omega)=1=a\omega+b \\ g(\omega^2)=1=a\omega^2+b} \\ \Rightarrow \cases{a=0\\ b=1} \Rightarrow 餘式r(x)=\bbox[red, 2pt]x$$

解答： $$取\cases{A(12\cos \theta,12\sin\theta) \\B(12\cos \phi,12\sin \phi) \\P(8,0)} \Rightarrow \cases{\overrightarrow{PA} =(12\cos\theta-8,12\sin\theta) \\ \overrightarrow{PB} =(12\cos \phi-8),12\sin \phi)}\\ \angle APB=90^\circ \Rightarrow \overrightarrow{PA}\cdot \overrightarrow{PB} =0 \Rightarrow (12\cos\theta-8)(12\cos \phi-8) +144\sin \theta \sin \phi=0 \\ \Rightarrow 9\cos(\phi-\theta)-6(\cos \theta+\cos \phi)+4=0 \Rightarrow \cos(\phi-\theta)={1\over 9}(6(\cos\theta+\cos \phi)-4)\\ 而M=\overline{AB}中點＝(6(\cos\theta+\cos \phi),6(\sin\theta+\sin \phi)) \equiv (x,y)\\ \Rightarrow x^2+y^2= 36(2+2\cos(\phi-\theta)) =36(2+2\cdot {1\over 9}(6(\cos\theta+\cos \phi)-4))\\ =72+48(\cos\theta+\cos \phi)-32=40+48(\cos\theta+\cos \phi)=40+8x \\ \Rightarrow \bbox[red,2pt]{x^2+y^2-8x-40=0}$$

解答： $$取E＝x-y平面\Rightarrow \cases{A(-5,0,6)\\ B(5,0,6)\\ P(\cos\theta,\sin\theta)\\ Q(\cos \phi,\sin \phi)} \Rightarrow \cases{\overrightarrow{AB}=(10,0,0) \\\overrightarrow{AP} =(\cos\theta+5, \sin \theta,-6) \\\overrightarrow{AQ} =(\cos\phi +5, \sin \phi,-6)} \\ \Rightarrow 四面體體積＝{1\over 6}\begin{Vmatrix} 10 & 0 & 0\\ \cos\theta+5 & \sin \theta, & -6\\ \cos \phi+5 & \sin \phi& -6\end{Vmatrix} =10|\sin \phi-\sin \theta| 最大值=\bbox[red, 2pt]{20}$$

解答：

$$旋轉矩陣T=\begin{bmatrix}\cos 60^\circ & -\sin 60^\circ\\ \sin 60^\circ & \cos 60^\circ \end{bmatrix} =\begin{bmatrix}1/2 & -\sqrt 3/2\\ \sqrt 3/2 & 1/2 \end{bmatrix} \\ \Rightarrow \begin{bmatrix}X\\ Y\end{bmatrix} =\begin{bmatrix}1/2 & -\sqrt 3/2\\ \sqrt 3/2 & 1/2 \end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix} =\begin{bmatrix}{1\over 2}(x-\sqrt 3y)\\ { 1\over 2}(\sqrt 3x+y)\end{bmatrix} \Rightarrow \cases{x-\sqrt 3y=2X\\ \sqrt 3x+y= 2Y} \\ 因此3||x-\sqrt 3y|-8|+ 2||\sqrt 3x+y|-18|=72 \Rightarrow 3||2X|-8|+2||2Y|-18|=72\\ \Rightarrow |6|X|-24|+|4|Y|-36|=72\\ \Rightarrow 各頂點坐標為\cases{(\pm16, \pm 9)\\ (\pm 4,\pm 27)\\ (0,\pm 21)\\ (\pm 10,0)} \Rightarrow 所圍面積＝4(\triangle OMN-\triangle AMC-\triangle BDN)\\ =4\cdot {1\over 2}(22\cdot 33-12\cdot 9-12\cdot 3)=\bbox[red,2pt]{1140}$$

二、計算證明題：(每題 10 分，共 40 分)

解答：

$$y=f(x)={x^2+4\over 2x} \Rightarrow 漸近線\cases{x=0\\ y={1\over 2}x},又f'(x)= {(x+2)(x-2)\over 2x^2} =0 \Rightarrow x=\pm 2\\ \Rightarrow \cases{f(2)=2為極小值\\ f(-2)=(-2)為極大值},圖形如上；$$

解答：

解答： $$令\cases{\overrightarrow{OA}=\vec a \\\overrightarrow{OA}=\vec a \\\overrightarrow{OB}=\vec b \\\overrightarrow{OC}=\vec c \\\overrightarrow{OD}=\vec d },又\cases{\vec c=0.6\vec a+0.4\vec b \\ 3\overline{AD}= 8\overline{BD}} \Rightarrow \overline{AC}: \overline{CB}: \overline{BD}=2:3: 3\\ \vec c\cdot \vec d=0 \Rightarrow \vec c\cdot \vec d= (0.6\vec a+0.4\vec b)(\vec b+\overrightarrow{BD})= (0.6\vec a+0.4\vec b)(\vec b+ 0.6\overrightarrow{AB})\\= (0.6\vec a+0.4\vec b)(\vec b+ 0.6(-\vec a+\vec b))= (0.6\vec a+0.4\vec b)(-0.6\vec a+1.6\vec b)) =-0.36|\vec a|^2+ 0.64|\vec b|^2=0\\ \Rightarrow {|\vec b|^2 \over |\vec a|^2}={0.36\over 0.64} \Rightarrow {|\vec b| \over|\vec a|}={\overline{OB} \over \overline{OA}}={6\over 8}= \bbox[red,2pt]{3\over 4}$$

 ============ END ============

解題僅供參考,其他教甄試題及詳解