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2020年2月12日 星期三

101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 101 學年度聯合轉學考招生考試
升高三數學科試題
一、單選題


$$x^2+y^2 -6x+4y+m=0 \Rightarrow (x-3)^2+(y+2)^2 =13-m \\\Rightarrow  \begin{cases}圓心(3,-2)在直線y=nx+7 \\半徑\sqrt{13-m} =2\end{cases} \Rightarrow  \begin{cases}-2=3n+7 \\ 13-m =4\end{cases} \Rightarrow  \begin{cases} n=-3 \\ m =9 \end{cases} , 故選:\bbox[red,2pt]{(B)}$$

2. 設空間中四點: A (0 , 2 , 2 ), B (1, 2 , 4 ), C (3, 4 , 1), D (1, 5, 2 )。
試求由\(\overrightarrow{AB},\overrightarrow{AC}, \overrightarrow{AD}\)所圍成平行六面體體積為何?
(A) 11 (B) 13 (C) 15 (D) 17 (E) 19

:$$\begin{cases} A(0,2,2) \\ B(1,2,4) \\ C(3,4,1) \\ D(1,5,2) \end{cases}  \Rightarrow \begin{cases} \overrightarrow{AB}= (1,0,2) \\ \overrightarrow{AC}= (3,2,-1) \\ \overrightarrow{AD}= (1,3,0) \end{cases}  \Rightarrow  體積=\begin{vmatrix}1 & 0 & 2 \\ 3 & 2 & -1 \\ 1 & 3 & 0\end{vmatrix} = 17, 故選\bbox[red,2pt]{(D)}$$



$$E_1//E_2 \Rightarrow 需找法向量為相互平行的平面: \\\begin{cases} x-y-z=1 \\ y-z-x=2 \\ z-x+y=3  \end{cases}  \Rightarrow \begin{cases} x-y-z=1 \cdots(1) \\ -x+y-z=2\cdots(2) \\ -x+y+z=3\cdots (3)  \end{cases} \Rightarrow (1)//(3), 故選\bbox[red,2pt]{(D)}$$



:$$\begin{cases} 3x+ay-2z=10\cdots (1) \\ 2x+y+z=2 \cdots(2) \\ x+2y-z=b\cdots (3)  \end{cases}  \xrightarrow{2\times(2)+(1), (2)+(3)} \begin{cases} 7x+(2+a)y =14 \cdots (3) \\ 3x+3y =b+2 \cdots(4)    \end{cases} \\ \xrightarrow{無限多組解} {7\over 3} ={2+a\over 3} = {14 \over b+2} \Rightarrow \begin{cases} a=5 \\ b=4 \end{cases} \Rightarrow a+b=9, 故選\bbox[red,2pt]{(E)}$$



$$\begin{cases} \begin{cases} \alpha 為第2象限角 \\ \cos \alpha=-3/5  \end{cases}  \Rightarrow \sin \alpha =4/5 \\ \begin{cases} \beta 為第3象限角 \\ \sin \beta =-5/13  \end{cases}  \Rightarrow \cos \beta = -12/13\end{cases} \Rightarrow \cos(\alpha+\beta ) = \cos \alpha \cos \beta-\sin \alpha \sin \beta \\=(-{3\over 5})\cdot (-{12 \over 13})- {4\over 5} \cdot (-{5\over 13}) = {36+20 \over 65} ={56 \over 65}, 故選\bbox[red,2pt]{(D)}$$


$$\begin{cases} \overline{AF_1} =3 \\  \overline{AF_2}=15 \end{cases} \Rightarrow 2a=\overline{AF_1} +\overline{AF_2}=18  \Rightarrow a=9  \Rightarrow c= \overline{OF_1} =9-\overline{AF_1} =6 \\ \Rightarrow a^2=b^2+c^2 \Rightarrow 81=b^2+36 \Rightarrow b=3\sqrt 5  \Rightarrow {x^2 \over 9^2}+ {y^2 \over (3\sqrt 5)^2}=1 \Rightarrow {x^2 \over 81}+ {y^2 \over 45}=1, 故選\bbox[red,2pt]{(A)}$$


$$令 \begin{cases}A(a,b) \\ \overline{AB}=m \\ \overline{BC}=n \\ f(x,y)=3x+y\end{cases}  \Rightarrow \begin{cases}B(a+{m \over \sqrt 2},b+{m \over \sqrt 2}) \\ C( a+{m \over \sqrt 2} -{n \over \sqrt 2},b+{m \over \sqrt 2}+{n \over \sqrt 2})\\ D(a-{n \over \sqrt 2},b+{n \over \sqrt 2})\end{cases}  \Rightarrow \begin{cases} f(A)=3a+b \\ f(B)=3a+b+{4m \over \sqrt 2} \\ f(C)= 3a+b+{4m-2n \over \sqrt 2} \\ f(D)= 3a+b-{2n\over \sqrt 2}\end{cases}\\ \Rightarrow f(B) 最大, 故選\bbox[red,2pt]{(B)}$$


:$$x^2+y^2-6x-8y-11=0 \Rightarrow (x-3)^2 +(y-4)^2=6^2 \Rightarrow  \begin{cases} x = 6\cos \theta+3\\ y=6\sin \theta +4\end{cases} \\ \Rightarrow \overline{OP}^2= (6\cos \theta+3)^2 +(6\sin \theta +4)^2 = 36\cos^2 \theta +36 \cos \theta+9 +36\sin^2 \theta+48 \sin \theta+16  \\ =36\cos \theta + 48 \sin \theta+61 = 60({36\over 60}\cos \theta+{48 \over 60} \sin \theta)+61 = 60 \sin(\alpha+\theta)+61\\ \Rightarrow \overline{OP}^2 \le 60+61=121 \Rightarrow \overline{OP} \le 11,故選\bbox[red,2pt]{(B)} $$

9. 平行四邊形ABCD,\(\overline{AB}=3, \overline{BC}=4\),則\( \overrightarrow{AC} \cdot \overrightarrow{BD}\)之值為何?

(A) -25 (B) 25  (C) -12  (D) 5  (E) 7

$$\overrightarrow{AC} \cdot \overrightarrow{BD} = (\overrightarrow{AB}+ \overrightarrow{BC}) \cdot (\overrightarrow{BC} -\overrightarrow{AB}) = \overrightarrow{AB} \cdot \overrightarrow{BC} -|\overrightarrow{AB}|^2 +| \overrightarrow{BC}|^2 -\overrightarrow{BC} \cdot \overrightarrow{AB}\\ =| \overrightarrow{BC}|^2 -|\overrightarrow{AB}|^2 =4^2-3^2 =7,故選\bbox[red,2pt]{(E)} $$




$$令O(0,0,0)如上圖,則 \begin{cases} A(1,0,3/4) \\ D(0,0,1/2) \\ B(1,1,1/2) \\ C(0,1,k)\end{cases} \Rightarrow (\overrightarrow{AB} \times \overrightarrow{AD}) // (\overrightarrow{CD} \times \overrightarrow{CB})  \\\Rightarrow (0,1,-1/4) \times (-1,0,-1/4) // (0,-1,1/2-k)\times (1,0,1/2-k)  \\\Rightarrow (-1/4,1/4,1) // (k-1/2,1/2-k,1) \Rightarrow k=1/4 \Rightarrow C(0,1,1/4) \\ \Rightarrow \cos \angle ADC = {\overrightarrow{DA} \cdot \overrightarrow{DC} \over |\overrightarrow{DA}||\overrightarrow{DC}|} = {(1,0,1/4) \cdot (0,1,-1/4) \over |(1,0,1/4) || (0,1,-1/4)|} = {-1/16 \over 17/16}= -{1\over 17},故選\bbox[red,2pt]{(B)}$$

11. 如右圖所示,兩直線\(\overleftrightarrow{OA}\)與\(\overleftrightarrow{OB}\)相交於O點,則向量\(\overrightarrow{OQ}=-{3\over 2}\overrightarrow{OA} +{2\over 3} \overrightarrow{OB}\)其終點會落在那一個區域內?

$$ -{3\over 2}\overrightarrow{OA} (紅箭頭) +{2\over 3} \overrightarrow{OB} 落在III區,故選\bbox[red,2pt]{(C)}$$


$$A= \begin{bmatrix}{1\over 2} & -{\sqrt 3\over 2} \\ {\sqrt 3\over 2} & {1\over 2} \end{bmatrix}  = \begin{bmatrix} \cos 60^\circ & -\sin 60^\circ \\ \sin 60^\circ & \cos 60^\circ \end{bmatrix} ,相當於逆時鐘旋轉60^\circ,故選\bbox[red,2pt]{(B)}$$

13. 某人長年流浪於甲、乙、丙三地。根據觀察﹕今年若此人停留在甲地,則明年留在甲地的機率為80% ,轉向乙地與丙地的機率分別為10% 與10% ﹔若今年留在乙地,則明年留在乙地的機率為60%,轉向甲地與丙地的機率分別為10% 與30%﹔若今年留在丙地,則明年留在丙地的機率為70% ,轉向甲地與乙地的機率分別為 20% 與10% 。試問當狀態穩定時,此人留在甲地之機率為多少?
(A) 45% (B) 35% (C) 30% (D) 25% (E) 20%

:$$ \begin{array}{c|ccc} & 甲& 乙& 丙\\\hline 甲& 0.8 & 0.1 & 0.1 \\\hline 乙& 0.1 & 0.6 & 0.3 \\\hline 丙& 0.2 & 0.1 & 0.7 \end{array} \Rightarrow 轉移矩陣A=\begin{bmatrix}0.8 & 0.1 & 0.2 \\ 0.1 & 0.6 & 0.1 \\ 0.1 & 0.3 & 0.7 \end{bmatrix} \\ \Rightarrow AX=X \Rightarrow \begin{bmatrix}0.8 & 0.1 & 0.2 \\ 0.1 & 0.6 & 0.1 \\ 0.1 & 0.3 & 0.7 \end{bmatrix} \begin{bmatrix}a \\ b \\ c \end{bmatrix} = \begin{bmatrix}a \\ b \\ c \end{bmatrix} \Rightarrow  \begin{cases}-0.2a+0.1b+0.2c =0 \\ 0.1a-0.4b+0.1c = 0 \\ 0.1a+0.3b-0.3c =0 \\ a+b+c=1 \end{cases} \\ \Rightarrow \begin{cases}a =9/20 =0.45\\ b = 4/20 \\ c=7/20 \end{cases}  \Rightarrow k=-2,故選\bbox[red,2pt]{(A)}$$



$$t+25 > t+16 \Rightarrow 該圖形為左右形且-25<t< -16,因此選項(D),(E)皆不正確\\  c^2=(t+25)-(t+16) =9 \Rightarrow c=3 \Rightarrow 焦點坐標(\pm 3,0),故選\bbox[red,2pt]{(B)}$$

二、多重選擇題


$$(A) \bigcirc: \begin{cases} x=2\cos 60^\circ = 1 \\ y=2\sin 60^\circ =\sqrt 3\end{cases} \\ (B) \times: \begin{cases} x=4\cos 150^\circ = -2\sqrt 3 \\ y=4\sin 150^\circ =2 \end{cases} \\ (C) \bigcirc: \overline{AB}= \sqrt{(1+2\sqrt 3)^2 + (2-\sqrt 3)^2} =\sqrt{20} =2\sqrt 5 \\ (D) \times: \angle AOB = 150^\circ -60^\circ =90 ^\circ \\(E) \times: \triangle AOB = {1\over 2} \overline{OA}\times \overline{OB} = {1\over 2} \times 2\times 4 =4\\,故選\bbox[red,2pt]{(AC)}$$



$$(A) \bigcirc: \begin{cases} x=1+2t \\ y=4-3t \end{cases} \Rightarrow {x-1\over 2}= {y-4\over -3} \Rightarrow 方向向量為(2,-3)\\ (B) \bigcirc: t=3 \Rightarrow \begin{cases} x=1+2t =7\\ y=4-3t =-5\end{cases} \Rightarrow L通過(7,-5) \\ (C) \times: {x-1\over 2}= {y-4\over -3} \Rightarrow -3x+3=2y-8 \Rightarrow 斜率為 -{3\over 2} \\ (D) \bigcirc: -3x+3=2y-8 \Rightarrow 3x+2y-11=0 \\(E) \bigcirc: \begin{cases} x=3+2t \\ y=1-3t \end{cases} \Rightarrow {x-3\over 2}= {y-1\over -3} \Rightarrow -3x+9=2y-2 \Rightarrow 3x+2y-11=0\\,故選\bbox[red,2pt]{(ABDE)}$$




$$(A) \times: \cos \angle ABO = {\overrightarrow{BA} \cdot \overrightarrow{BO} \over |\overrightarrow{BA}|| \overrightarrow{BO}|} = {(-2,0,4) \cdot (-2,-2,0) \over |(-2,0,4) || (-2,-2,0)|} = {4\over 4\sqrt{10}} = {1\over \sqrt{10}} \\ (B) \times: \triangle OAB 面積= {1\over 2}\sqrt{|\overrightarrow{BA}|^2| \overrightarrow{BO}|^2 -(\overrightarrow{BA} \cdot \overrightarrow{BO})^2} = {1\over 2} \sqrt{20\times 8-4^2} =7 \\ (C) \bigcirc: \overrightarrow{BA} \times \overrightarrow{BO} =(-2,0,4) \times (-2,-2,0) = (8,-8,4) \Rightarrow E過原點且法向量為(8,-8,4):  \\   \qquad 8x-8y+4z=0 \Rightarrow 2x-2y+z=0 \\ (D) \bigcirc: \begin{cases} \overline{AO} = \sqrt{2^2+4^2} =\sqrt{20} \\ \overline{AB} =\sqrt{2^2+4^2 }=\sqrt{20} \end{cases} \Rightarrow \overline{AO} =\overline{AB} \Rightarrow \triangle OAB為等腰 \\(E) \bigcirc: 令P為\overline{OB}的中點,即P(1,1,0); \overline{PA}直線方程式為{x \over 1} ={y-2 \over -1} = {z-4 \over -4} \\\qquad \Rightarrow \begin{cases} x=t\\ y=-t+2 \\ z=-4t+4\end{cases},t\in R;\\,故選\bbox[red,2pt]{(CDE)}$$




$$(A) \bigcirc: \begin{cases} x+2y-z=1\\ 2x+y-z=0 \end{cases} \Rightarrow \begin{cases} x=y-1\\ z=3y-2 \end{cases} \Rightarrow L:(t-1,t,3t-2),t\in R \Rightarrow 方向向量為(1,1,3)\\ (B) \bigcirc: t=3 \Rightarrow (t-1,t,3t-2) =(2,3,7) \Rightarrow (2,3,7)在L上\\(C) \times: 直線{ x+1 \over 1 }= {y \over 2} = {z+2 \over -1}的方向向量為(1,2,-1)與(1,1,3)不平行\\ (D) \bigcirc: 2x+y-z=0 與2x+y-z=1 平行 \Rightarrow L與 2x+y-z=1平行\\ (E) \times: 將(t-1,t,3t-2)代入可得3(t-1)+3t-2(3t+2) = -7\ne 4 \Rightarrow L不在該平面上\\,故選\bbox[red,2pt]{(ABD)} $$ 註: 公布的答案是(AB)


:$$(A) \times: x+2y=3為一平面\\ (B) \times: A(3,0,0)及B(3,0,1)均在x+2y=3上, \overrightarrow{AB}=(0,0,1)與(4,-2,0)並不平行\\(C) \bigcirc: (1,2,0)為平面的法向量\\ (D) \bigcirc: (3,0,4)符合x+2y=3 \\(E)\times: (1,2,-3) 與(1,2,0)並不平行\\,故選\bbox[red,2pt]{(CD)}$$




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