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2019年12月28日 星期六

108學年度高雄區公立高中聯合轉學考-升高三數學科詳解


高雄區公立高中 108 學年度聯合招考轉學生
升高三數學科試題詳解
一、單選題


$$25\cos^2 \theta-5\sin \theta-13=0 \Rightarrow 25(1-\sin^2) \theta-5\sin \theta-13=0 \Rightarrow 25\sin^2\theta +5\sin \theta-12=0\\ \Rightarrow (5\sin \theta-3)(5\sin \theta+4)=0 \Rightarrow  \sin \theta =\begin{cases}3/5 \\-4/5 ,不合(\theta在第2象限, \sin \theta>0)\end{cases} \\ \Rightarrow \cos \theta =-4/5(\theta在第2象限, \cos \theta<0), 故選:\bbox[red,2pt]{(E)}$$


:$$\cos A={\overline{AB}^2+ \overline{AC}^2- \overline{BC}^2 \over 2\times \overline{AB} \times \overline{AC}} \Rightarrow \cos 120^\circ={1+4-\overline{BC}^2 \over 2\times 1\times 2} \Rightarrow -{1\over 2}= {5-\overline{BC}^2 \over 4}\\  \Rightarrow \overline{BC}^2=7 \Rightarrow \overline{BC} =\sqrt 7, 故選\bbox[red,2pt]{(C)}$$




$$\overline{ED}為平面,建築物A的高度\overline{AE}=150,建築物B的高度為\overline{BD},如上圖\\ 在\triangle ACD中,\overline{AC}=\overline{CD}\times \sqrt 3 = 150\sqrt 3; 在\triangle ABC中,\overline{BC}= \overline{AC}=150\sqrt 3\\ \Rightarrow \overline{BD}= \overline{BC}+\overline{CD}=150\sqrt 3+150, 故選\bbox[red,2pt]{(A)}$$


:$$圓C: x^2+y^2-4x-2y=0 \Rightarrow (x-2)^2+(y-1)^2=5 \Rightarrow 圓心O(a,b)=(2,1),半徑r_1=\sqrt 5;\\ 圓S半徑r_2= \overline{AO}= \sqrt{3^2+4^2}=5 \Rightarrow 圓S:(x-2)^2+(y-1)^2=25 \Rightarrow 面積為25\pi, 故選\bbox[red,2pt]{(D)}$$



$$\begin{cases}A = (2,-4)\\B=(5,-2) \end{cases}  \Rightarrow \begin{cases}圓心O= ((2+5)/2,(-4-2)/2) =(7/2,-3)\\半徑r={1\over 2}\overline{AB}={1\over 2}\sqrt{3^2+2^2} = {1\over 2} \sqrt{13} \end{cases} \\ \Rightarrow 圓方程式: (x-7/2)^2+(y+3)^2 = {13\over 4} \Rightarrow x^2-7x+{49\over 4} +y^2+6y+9 ={13\over 4}\\  \Rightarrow x^2+y^2-7x+6y+18=0, 故選\bbox[red,2pt]{(A)}$$




$$\begin{cases}|\vec a| = 3 \\ |\vec b|=4\\ |\vec a+\vec b|= \sqrt {13}\end{cases}  \Rightarrow (\vec a+\vec b)\cdot (\vec a+\vec b)= |\vec a|^2+2\vec a\cdot \vec b+|\vec b|^2 =| \vec a+\vec b|^2\\ \Rightarrow 3^2+2\vec a\cdot \vec b+ 4^2= (\sqrt {13})^2 \Rightarrow \vec a\cdot \vec b=-6 =|\vec a||\vec b|\cos \theta =12\cos \theta \Rightarrow \cos \theta=-{1\over 2} \Rightarrow \theta = 120^\circ\\, 故選\bbox[red,2pt]{(D)}$$




$$在\triangle ABC中,\begin{cases} \vec a= \overrightarrow {AB} \\ \vec b= \overrightarrow {AC}\\ \overline{AD}平分\angle A\end{cases},如上圖\\ {\overline{BD} \over \overline{DC}} = {\overline{AB} \over \overline{AC}} ={3 \over 5} \Rightarrow \overrightarrow {AD} ={5\over 8}\vec a+{3\over 8}\vec b ={5\over 8}(\vec a+{3\over 5}\vec b) \Rightarrow t={3\over 5}, 故選\bbox[red,2pt]{(E)}$$


:$$由L方程式可知: L與z=3平行,即xy平面平行,故選\bbox[red,2pt]{(C)} $$



$$\begin{cases}3x+2y+z=4 \cdots(1)\\ x+2y+2z=4 \cdots (2)\end{cases} \xrightarrow{(1)-(2)} 2x-z=0 \Rightarrow z=2x代回(1) \Rightarrow 3x+2y+2x=4 \Rightarrow y=-{5 \over 2}x+2\\  \Rightarrow (x,y,z)=(x,-{5 \over 2}x+2,2x) \Rightarrow {x\over 1}={y-2 \over -5/2}={z\over 2} \Rightarrow L方向向量為(1,-5/2,2),故選\bbox[red,2pt]{(D)} $$


:$$c_{23}即為C之最後一個元素=4\times 5+3\times (-3)+(-5)\times 1= 20-9-5=6,故選\bbox[red,2pt]{(B)}$$



$$ (A)\times: \begin{cases}A= \begin{bmatrix}1 & 1 \\1 & 0 \end{bmatrix}  \\ B= \begin{bmatrix}1 & 1 \\0 & 0 \end{bmatrix} \end{cases} \Rightarrow \begin{cases}AB= \begin{bmatrix}1 & 1 \\1 & 1 \end{bmatrix}  \\ BA= \begin{bmatrix}2 & 1 \\0 & 0 \end{bmatrix} \end{cases} \Rightarrow AB\ne BA\\ (C)\times: \begin{cases}A= \begin{bmatrix}1 & 0 \\0 & 0 \end{bmatrix}  \\ B= \begin{bmatrix}0 & 0 \\1 & 1 \end{bmatrix} \end{cases} \Rightarrow AB= \begin{bmatrix}0 & 0 \\0 & 0 \end{bmatrix}  ,但 A\ne 0且B\ne 0\\(D)\times: \begin{cases}A= \begin{bmatrix}1 & 0 \\0 & 0 \end{bmatrix}  \\ B= \begin{bmatrix}0 & 0 \\1 & 1 \end{bmatrix} \\C= \begin{bmatrix}0 & 0 \\1 & 0 \end{bmatrix} \end{cases}  \Rightarrow AB=AC=0,但B\ne C\\(E)\times: (A+B)^2 =A^2+AB +BA +B^2 \ne A^2+2AB +B^2 (除非AB=BA)\\,故選\bbox[red,2pt]{(B)}$$



$$\begin{cases}L_1: x-2y=0\\ L_2: x+2y=0 \end{cases}  \Rightarrow 雙曲線: (x-2y)(x+2y)=k \Rightarrow x^2-4y^2=k\\ (2\sqrt 2,2)代入上式 \Rightarrow 8-16=k \Rightarrow k=-8 \Rightarrow x^2-4y^2=-8 \Rightarrow {x^2 \over 8}-{y^2 \over 2}=-1,故選\bbox[red,2pt]{(B)}$$

二、多重選擇題

:$$(A) \bigcirc : \begin{cases}\cos 100^\circ <0\\ \sin 200^\circ <0\end{cases}  \Rightarrow 第三象限\\(B) \times :\tan 550^\circ = \tan 190^\circ >0 \Rightarrow 非第三象限\\(C) \times :\tan -2 = \tan \left(-2\times {180^\circ \over \pi}\right) \approx \tan -114^\circ >0非第三象限\\(D) \bigcirc :\begin{cases}\cos 155^\circ <0\\ \sin 350^\circ <0\end{cases}  \Rightarrow 第三象限 \\(E) \times: \tan 36 = \tan \left(36\times {180^\circ \over \pi}\right) \approx \tan 2063^\circ
 =\tan 263^\circ>0非第三象限\\,故選\bbox[red,2pt]{(AD)}$$



$$y=mx+2 代入x^2+y^2=1 \Rightarrow x^2+(mx+2)^2=1 \Rightarrow (m^2+1)x^2 +4mx+3=0 \\ \Rightarrow 判別式為0 \Rightarrow 16m^2-12(m^2+1)=0 \Rightarrow 4m^2=12 \Rightarrow m=\pm \sqrt 3,故選\bbox[red,2pt]{(CD)}$$



$$L:{1-x \over 1}= {y-2 \over 2} ={z\over 2} \Rightarrow L上的點Q可表示成(-t+1,2t+2,2t) \\\Rightarrow \overline{PQ}= \sqrt{(-t-1)^2+(2t-1)^2 +(2t+5)^2} =\sqrt{9t^2+18t +27} =\sqrt{9(t+1)^2 +18}\\  \Rightarrow t=-1 \Rightarrow \begin{cases} \overline{PQ} =\sqrt{18} =3\sqrt 2 \\ Q=(1+1,-2+2,-2)=(2,0,-2)\end{cases}\Rightarrow  \begin{cases} P至L最短距離 =3\sqrt 2 \\ Q=(2,0,-2)是離P最近的點\end{cases}\\ 令P'=(a,b,c)為P相對L的對稱點 \Rightarrow {P+P'\over 2}=Q \Rightarrow  \begin{cases}2 = (2+a)/2 \\ 0=(3+b)/2 \\ -2= (-5+c)/2 \end{cases} \Rightarrow  \begin{cases}a=2 \\ b=-3 \\ c=1 \end{cases} \\ \Rightarrow P'=(2,-3,1),故選\bbox[red,2pt]{(BCE)}$$


:$$\begin{cases}\begin{bmatrix}1 & -1 & a & 8 \\ 0 & 5 & -3 & b\\0& 1 & c &-15 \end{bmatrix} \Rightarrow \begin{cases} x-y+az=8 \\ 5y-3z=b \\ y+cz=-15\end{cases} \\ \begin{bmatrix}1 & 0 & 0 & 4 \\ 0 & 1 & 0 & -3\\0& 0 & 1 & 1 \end{bmatrix} \Rightarrow \begin{cases} x=4 \\ y =-3 \\ z=1 \end{cases}\end{cases}  \Rightarrow  \begin{cases} 4+3+a=8 \\ -15-3=b \\ -3+c=-15\end{cases} \Rightarrow  \begin{cases} a=1 \\ b=-18 \\ c=-12\end{cases}\\,故選\bbox[red,2pt]{(ABE)}$$




:$$(B)\times: A= \begin{bmatrix}1 & -1 \\1 & -1 \end{bmatrix}  \Rightarrow A^2=0,但A\ne 0\\(C) \times: (AB)^{-1}=B^{-1}A^{-1}\\(D) \times: \begin{cases}A= \begin{bmatrix}1 & 0 \\0& 0 \end{bmatrix} \\B = \begin{bmatrix}0 & 0 \\1 & 1 \end{bmatrix} \end{cases}  \Rightarrow   \begin{cases}AB = 0\\BA =  \begin{bmatrix}0 & 0 \\1 & 0 \end{bmatrix} \ne 0\end{cases} ,故選\bbox[red,2pt]{(AE)}$$



$$(B)\times: \begin{vmatrix}3a & 3b \\3c & 3d \end{vmatrix} =9ad-9bc =9(ad-bc)=9\begin{vmatrix}a & b \\c & d \end{vmatrix} ,故選\bbox[red,2pt]{(ACDE)} $$


:$$(A)\times: t不能為3\\(B)\times: t=2.5 \Rightarrow  \Gamma :{x^2 \over 0.5}+ {y^2 \over 1.5}=1 \Rightarrow 長軸在Y軸\\ (D)\times: t=4 \Rightarrow  \Gamma :-{x^2 \over 1}+ {y^2 \over 3}=1 \Rightarrow 貫軸在Y軸 \\,故選\bbox[red,2pt]{(CE)}$$




:$$2y^2-4y-3x-4=0 \Rightarrow (y-1)^2 =4\cdot{3 \over 8}(x+2) \Rightarrow  \begin{cases}頂點(-2, 1)\\ c=3/8\end{cases} \\\Rightarrow  \begin{cases}焦點(-2+3/8, 1) =(-13/8,1)\\ 準線: x= -2-3/8 =-19/8\end{cases}  \Rightarrow 正焦弦長=4c = 4\times 3/8=3/2\\ \Rightarrow 最小焦半徑=c=3/8,故選\bbox[red, 2pt]{(BCD)}$$


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