網頁

2019年12月15日 星期日

106學年度臺北市聯合轉學考-高中升高三-數學科詳解


臺北市高級中等學校 106 學年度聯合轉學考招生考試
升高三數學科試題
一、單選題


$${\sin 3\theta \over \sin \theta}- {\cos 3\theta \over \cos \theta} = {\cos\theta \sin 3\theta- \sin\theta \cos 3\theta \over \sin \theta \cos \theta} ={\sin (3\theta-\theta) \over {1\over 2}\cdot 2\sin \theta \cos \theta } = {\sin 2\theta \over {1\over 2}\sin 2\theta }=2, 故選:\bbox[red,2pt]{(C)}$$


:$$\begin{cases}a=\cos 80^\circ \\ b=\cos 140^\circ =-\cos 40^\circ \\ c=\cos 230^\circ = -\cos 50^\circ \\d=\cos 290^\circ = \cos 70^\circ \\ e=\cos 340^\circ =\cos 20^\circ\end{cases} \Rightarrow e>d>a>c>b \Rightarrow 中位數為a, 故選\bbox[red,2pt]{(A)}$$



$$\begin{cases}x=2+2t \\ y=4+3t \\ z=-3+4t \end{cases} 代入E:3x-2y+z=3 \Rightarrow 3(2+2t) -2(4+3t)+ (-3+4t)=3 \\ \Rightarrow 4t=8 \Rightarrow t=2 \Rightarrow 交點=(2+2\times 2, 4+3 \times 2,-3+4\times 2) =(6,10,5), 故選\bbox[red,2pt]{(A)}$$


:$$\left( {x\over 9}+{ y\over 4}\right) \left( {x\over 9}-{ y\over 4}\right)=k\\
(A)\times:k=0\Rightarrow \left( {x\over 9}+{ y\over 4}\right) \left( {x\over 9}-{ y\over 4}\right)=0 \Rightarrow {x\over 9}=\pm {y\over 4}為兩直線\\
(B)\times:k=1\Rightarrow \left( {x\over 9}+{ y\over 4}\right) \left( {x\over 9}-{ y\over 4}\right)=1 \Rightarrow {x^2\over 9^2}-{y^2\over 4^2}=1為雙曲線\\
(C)\times:k=0為兩不平行的直線,說明如(A)\\
(D)\times:k=0為兩不平行的直線,說明如(A)\\
(E)\bigcirc:{x^2\over 9^2}-{y^2\over 4^2}=k為雙曲線, 故選\bbox[red,2pt]{(E)}$$



:$$令中心點為O,此題相當於找\begin{cases} \overline{CP}=\overline{OA}? \\ \overline{CQ}=\overline{OA}? \\ \overline{CR} =\overline{OA}? \\ \overline{CS}=\overline{OA}? \\ \overline{CT}=\overline{OA}?\end{cases} , 故選\bbox[red,2pt]{(B)}$$




$${x^2 \over 90-k}+ {y^2 \over k-30}=1  \Rightarrow \begin{cases}90-k>0 \\ k-30>0 \\ 90-k\ne k-30\end{cases}   \Rightarrow \begin{cases}90>k>30 \\ k\ne 60\end{cases}  \Rightarrow k=31,32,\dots,89 \\  \Rightarrow 共有89-31+1-1(k=60)= 58, 故選\bbox[red,2pt]{(C)}$$




將A、B及C(直升機)投影至地面的相對位置如上圖,只要能找出\(\overline{AC}\)及\(\overline {BC}\)投影至地面的距離就可求出\(\overline{AB}\);
A在C西方俯角60度,如上圖,因此\(\overline{AC}\)在地面的投影為\(\overline{AC_1} = h/\sqrt 3\);
另一方面,B在C的俯角30度,如上圖,因此\(\overline{BC}\)在地面的投影為\(\overline{CB'} = \sqrt 3 h\);
最後再由餘弦定理(代入最上圖): $$\cos \angle ACB = {\overline{AC}^2 +\overline{BC}^2- \overline{AB}^2 \over 2\times \overline{AB}\times \overline{AC}}\Rightarrow -{1\over 2} ={h^2/3+3h^2 - \overline{AB}^2 \over 2h^2} \Rightarrow \overline{AB}^2={13 \over 3}h^2 \\\Rightarrow \overline{AB}= \sqrt {13\over 3}h, 故選\bbox[red,2pt]{(E)}$$


:$$\tan^2 \angle AEB  ={25 \over 24}\Rightarrow 假設 \begin{cases} \overline{AB} =5k\\ \overline{EB}=\sqrt{24}\end{cases} \Rightarrow   \overline{CE}^2 = \overline{EB}^2 +\overline{BC}^2  =24k^2+25k^2 =49k^2 \\ \Rightarrow  \overline{CE}=7k \Rightarrow \tan \angle CED ={\overline {DC} \over \overline{EC}} ={5k\over 7k} \Rightarrow \tan^2 \angle CED={25 \over 49},故選\bbox[red,2pt]{(B)} $$


:$$假設P(x, {x^2\over 4})  \Rightarrow \overrightarrow {AB}\cdot \overrightarrow{AP} = (6,3) \cdot (x+2, {x^2\over 4}-1) =  6x+12+ {3\over 4}x^2-3 =0 \\\Rightarrow x^2+8x+12 =0 \Rightarrow (x+6)(x+2)=0 \Rightarrow x=-2或-6,故選\bbox[red,2pt]{(B)} $$


:$$\begin{cases} P(6,s,3) \\ Q(0,16,t)\\ R(-3,19,6)\end{cases} \Rightarrow \begin{cases} \overrightarrow{P,R} =(-9,19-s, 3) \\ \overrightarrow{QR} = (-3,3,6-t) \end{cases} \Rightarrow {-9 \over -3}= {19-s \over 3}= {3 \over 6-t} \Rightarrow \begin{cases} 19-s=9 \\ 6-t=1\end{cases} \\\Rightarrow \begin{cases} s=10 \\ t=5\end{cases}\Rightarrow s+t=15,故選\bbox[red,2pt]{(A)}$$




$$考慮原點不在所圍區域,例原點在3x+y=6的異側,因此3x+6\ge 6,故選\bbox[red,2pt]{(D)}$$


:$$x^2+y^2-2x-2ky+k^2-31=0 \Rightarrow (x-1)^2+(y-k)^2=32 \Rightarrow  \begin{cases}圓心O(1,k)  \\半徑r=\sqrt{32}=4\sqrt 2\end{cases} \\\text{dist}(O,L)=r \Rightarrow \left|{ 1-k-6\over \sqrt{1^2+1^2}} \right| =4\sqrt 2  \Rightarrow |-5-k|=8 \Rightarrow  k =\begin{cases}-13\\3\end{cases} ,故選\bbox[red,2pt]{(D)}$$


:$$ (A+B)^2=A^2+2AB+B^2 \Rightarrow AB=BA \Rightarrow  \begin{bmatrix}
 2 & 4 \\
 2 & 6
 \end{bmatrix}\begin{bmatrix}
 k & 4 \\
 2 & 10
 \end{bmatrix}=\begin{bmatrix}
 k & 4 \\
 2 & 10
 \end{bmatrix} \begin{bmatrix}
 2 & 4 \\
 2 & 6
 \end{bmatrix} \\  \Rightarrow  \begin{bmatrix}
 2k+8 & 48 \\
 2k+12 & 68
 \end{bmatrix} =\begin{bmatrix}
 2k+8 & 4k+24 \\
 24 & 68
 \end{bmatrix} \Rightarrow 2k+12=24 \Rightarrow k=6,故選\bbox[red,2pt]{(C)}$$




$$  A=\begin{bmatrix}
 a & b \\
 c & 1/3
 \end{bmatrix}為轉移矩陣 \Rightarrow  \begin{cases}a +c=1\\ b +1/3=1\end{cases}   \Rightarrow \begin{cases} c = 1-a\\ b=2/3\end{cases}  \\\Rightarrow  a+b+c=(a+c)+b=1+2/3=5/3,故選\bbox[red,2pt]{(E)}$$

二、多重選擇題



$$\begin{cases} A(1,0)\\ B(-2,5) \end{cases}代入y=ax+b \Rightarrow \begin{cases} 0=a+b\\ 5=-2a+b \end{cases}\Rightarrow \begin{cases} a=-5/3\\ b=5/3 \end{cases} \Rightarrow y=-{5\over 3}x+{5\over 3}\\  \Rightarrow y=-{5\over 3}(x-1) \Rightarrow y-5=-{5\over 3}(x+2) \Rightarrow 5x+3y=5,故選\bbox[red,2pt]{(BCE)}$$




只要符合任兩邊皆平行的條件,如上圖之紅點,即(1,0), (3,0), (3,2),故選\(\bbox[red,2pt]{(CD)}\)




:$$只有選項(A)及(B)未限制z值的範圍,故選\bbox[red,2pt]{(AB)}$$


:$$(A)\bigcirc: \overrightarrow{OP}=(6,3,3)為L的法向量 \Rightarrow (6,3,3)\cdot (-2,1,3)=-12 +3+9 =0 \\ \qquad\Rightarrow(6,3,3) \bot (-2,1,3) \Rightarrow (-2,1,3)為L的方向向量\\(B)\bigcirc: Q=(2,-2,2) \Rightarrow \overrightarrow{OQ} \cdot \overrightarrow{QP} = (2,-2, 2) \cdot (4,5,1)=8-10+2=0  \Rightarrow \overrightarrow{OQ} \bot \overrightarrow{QP}\\ \qquad \Rightarrow Q為原點在E上的投影點\\(C)\times: (2,-2,2) 才是投影點\\(D)\times: \overline{OQ}= \sqrt{2^2+(-2)^2+2^2}= \sqrt{12} =2\sqrt 3\\(E)\bigcirc: 理由同(D)\\,故選\bbox[red,2pt]{(ABE)} $$


:$$  \left[\begin{array}{rrr|r}
 1 & -3 & 2 & 1 \\
 -2 & 5 & -3 &-3 \\
3 &-9 & 8 &1
 \end{array}\right] \xrightarrow{-3r_1+r_3,2r_1+r_2}  \left[\begin{array}{rrr|r}
 1 & -3 & 2 & 1 \\
 0 & -1 & 1 &-1 \\
0 & 0 & 2 & -2
 \end{array}\right]\\ \xrightarrow{-r2}  \left[\begin{array}{rrr|r}
 1 & -3 & 2 & 1 \\
 0 & 1 & -1 &1 \\
0 & 0 & 2 & -2
 \end{array}\right]=
 \left[\begin{array}{rrr|r}
 1 & \alpha & 2 & 1 \\
 0 & 1 & \beta & 1 \\
0 &0  & 2 & \gamma
 \end{array}\right]   \Rightarrow   \begin{cases} \alpha =-3\\ \beta =-1 \\ \gamma =-2\end{cases} \\
由\left[\begin{array}{rrr|r}
 1 & -3 & 2 & 1 \\
 0 & 1 & -1 &1 \\
0 & 0 & 2 & -2
 \end{array}\right] \Rightarrow \begin{cases} x-3y+2z=1\\ y-z=1\\ 2z=-2\end{cases} \Rightarrow \begin{cases} x=3\\ y=0\\ z=-1\end{cases}
\\
(A)\bigcirc:\alpha=-3\\(B)\times: \beta=-2\ne 1\\(C)\times: \gamma=-2 \ne -1\\(D)\times: 只有一組解\\(E)\bigcirc:共有一組解\\,故選\bbox[red,2pt]{(AE)}$$




解題僅供參考

沒有留言:

張貼留言