網頁

2020年7月21日 星期二

109年臺北市聯合轉學考-技術型高中(升高二)數學詳解


臺北市高級中等學校109學年度
聯合轉學考招生考試技術型高中數學科試題

單選題: 共 25 題,總分 100 分,每題 4 分

解:
$$f(x)=3x^2+ax+b \Rightarrow 頂點X坐標: -{a\over 6}=2 \Rightarrow a=-12 ;\\又 f(2)=-1 \Rightarrow 12+2a+b=-1 \Rightarrow 12-24+b=-1 \Rightarrow b=11 \Rightarrow a+b=-12+11=-1\\,故選\bbox[red, 2pt]{(B)}$$


解:
$$\cases{L_1:2x=3y-5\\ L_2:kx=4y+1} \Rightarrow \cases{m_{L_1}=2/3 \\ m_{L_2}= k/4} \Rightarrow m_{L_1} \times m_{L_2} = -1 = k/6 \Rightarrow k=-6,故選\bbox[red, 2pt]{( B)}$$


解:
$$圖形過一、二、三象限 \Rightarrow \cases{f(0)> 0\\ 斜率大於0} \Rightarrow \cases{-b>0 \Rightarrow b<0 \\a>0} \Rightarrow \cases{a-b >0 \\ ab <0} \\\Rightarrow (a-b,ab)在第四象限,故選\bbox[red, 2pt]{(D)}$$


解:

$$\cases{\text{dist}(A,L)={6-3+5\over \sqrt{10}}={8\over \sqrt{10}} \\\text{dist}(B,L)={|-12-1+5|\over \sqrt{10}}={8\over \sqrt{10}}} \Rightarrow {\overline{AP} \over \overline{BP}} ={\overline{AC} \over \overline{BD}}={1\over 1}=1 ,故選\bbox[red, 2pt]{( C)}$$


解:
$$x^2+y^2+4x-2y+1=0 \Rightarrow (x^2+4x+4)+(y^2-2y+1)=4 \Rightarrow (x+2)^2+(y-1)^2=2^2\\ \Rightarrow 半徑=2,故選\bbox[red, 2pt]{( A)}$$


解:
$$f(x)=3x^3-4x^2-5x+2 \Rightarrow \cases{f(-1)=-3-4+5+2=0 \\ f(2)=24-16-10+2=0 \\ 三根之積=-2} \Rightarrow 另一根為正值(-3/2不是根)\\,故選\bbox[red, 2pt]{(C )}$$


解:
$$\cases{a_7=a_1r^6=5 \\ a_{10}=a_1r^9=40} \Rightarrow {a_1r^9 \over a_1r^6}={40 \over 5} \Rightarrow r^3=8 \Rightarrow r=2,故選\bbox[red, 2pt]{(B )}$$


解:
$$f(x)=x^2-x+a \Rightarrow f(2)=5 \Rightarrow 4-2+a=5 \Rightarrow a=3 \Rightarrow f(x)=x^2-x+3\\ \Rightarrow f(x+1) = (x+1)^2-(x+1)+3 \Rightarrow f(x+1)除以x+1的餘式為3,故選\bbox[red, 2pt]{(B )}$$


解:
$$-{3\over 2} < x <1 \Rightarrow (x-1)(x+{3\over 2}) < 0 \Rightarrow x^2+{1\over 2}x-{3\over 2}< 0 \Rightarrow 2x^2+x-3 < 0 \\ \Rightarrow \cases{a=2\\ b=1} \Rightarrow a+b=2+1=3,故選\bbox[red, 2pt]{(C )}$$


解:
$$2\le |x+3| \le 5 \Rightarrow |x+3| = 2,3,4,5 \Rightarrow x+3=\pm 2,\pm 3,\pm 4,\pm 5 \Rightarrow x=\cases{-1,0,1,2\\ -5,-6,-7,-8}\\ \Rightarrow 共有8個解,故選\bbox[red, 2pt]{(C )}$$


解:



$$利用長除法可得餘式為28=f(9),故選\bbox[red, 2pt]{(D )}$$


解:
$$若a>b>0,圖形為左上右下,斜率為負值,故選\bbox[red, 2pt]{(A)}$$


解:
$$斜率為正值的直線為\overline{DE}及\overline{BC},其中\overline{BC}較為陡峭,故選\bbox[red, 2pt]{(A)}$$


解:
$$\cos 2x=\sin x \Rightarrow 1-2\sin^2 x=\sin x \Rightarrow 2\sin^2x + \sin x -1=0 \Rightarrow \sin x=-1,{1\over 2}\\ 由於 -\pi \le x \le 2\pi \Rightarrow  \cases{\sin x=-1 \Rightarrow x=-\pi/2,3\pi/2\\ x=1/2 \Rightarrow x=\pi/6,5\pi/6},共有4個解,故選\bbox[red, 2pt]{(C )}$$


解:
$$\cases{\angle A=\pi/12=15^\circ \\ \angle B=30^\circ} \Rightarrow \angle C=180^\circ-15^\circ -30^\circ= 135^\circ,\\ 再利用正弦定理 {c\over \sin \angle C} =2R,R是外接圓半徑 \Rightarrow {10\sqrt 2\over \sin 135^\circ =\sqrt 2/2} =20=2R \Rightarrow R=10 \\\Rightarrow 圓面積=R^2\pi =100\pi,故選\bbox[red, 2pt]{( B)}$$


解:
$$餘弦定理: \cos \angle A={b^2+c^2 -a^2 \over 2bc} \Rightarrow \cos 120^\circ ={3^2+5^2-a^2 \over 2\times 3\times 5} \Rightarrow -{1\over 2}= {34-a^2 \over 30 } \\\Rightarrow 34-a^2=-15  \Rightarrow a^2=49 \Rightarrow a=7(長度無負值),故選\bbox[red, 2pt]{(C )}$$


解:

$$假設A為原點,則\cases{C(3,\sqrt 3)\\ E(0,2\sqrt 3} \Rightarrow \cases{\overrightarrow{AC} =(3,\sqrt 3) \\ \overrightarrow{AE} =(0,2\sqrt 3)} \Rightarrow \overrightarrow{AC} \cdot \overrightarrow{AE} =0+6=6,故選\bbox[red, 2pt]{(A )}$$


解:
$$餘弦定理: \cos \angle B= {\overrightarrow{BA}\cdot \overrightarrow{BC} \over |\overrightarrow{BA}||\overrightarrow{BC}|} = {-\overrightarrow{AB}\cdot \overrightarrow{BC} \over \overline{AB}\times |\overrightarrow{BC}|} ={-15\sqrt 3\over 6\times 5} =-{\sqrt 3\over 2} \Rightarrow \angle B=150^\circ ,故選\bbox[red, 2pt]{(D )}$$


解:
$$|\vec a-2\vec b|^2 = (\vec a-2\vec b)\cdot (\vec a-2\vec b) =|\vec a|^2-4\vec a\cdot \vec b+4|\vec b|^2 =(2\sqrt 3)^2+0+4 =16 \\\Rightarrow |\vec a-2\vec b|=\sqrt {16}=4,故選\bbox[red, 2pt]{(A )}$$


解:
$$f(x)=2x^3-3x^2+4x-5 = a(x-1)^3 + b(x-1)^2 +c(x-1)+d\\ \Rightarrow f(2)=16-12+8-5=7=a+b+c+d,故選\bbox[red, 2pt]{(A )}$$


解:
$$\overline{PQ}的最大值=\overline{PO}+半徑=\sqrt{4^2+3^2}+3=5+3=8,故選\bbox[red, 2pt]{(D )}$$


解:
$$令P為\overline{AB}的中點,則\triangle OAP為直角\triangle,因此\overline{OA}^2 = \overline{OP}^2 + \overline{AP}^2 \Rightarrow 5^2=\overline{OP}^2 + (8/2)^2 \\\Rightarrow \overline{OP}=3 ,故選\bbox[red, 2pt]{(B )}$$


解:
$$被3除餘1的數:1,4,7,...,3n+1,...3\times 36+1,n為非負整數\\\Rightarrow \sum_{n=0}^{36} 3n+1 =3\times 36\times 37\div 2+37 =1998 +37=2035,故選\bbox[red, 2pt]{(D)}$$



解:
$$\cases{a=(1+9)\div 2=5 \\ 4=\sqrt{2b} \Rightarrow b=8} \Rightarrow a+b=5+8=13,故選\bbox[red, 2pt]{( D)}$$


解:
$${x-2\over x-3} +{x-5\over x-6} ={x-3\over x-4} +{x-4\over x-5} \Rightarrow {x-2\over x-3} -{x-3\over x-4} = {x-4\over x-5}- {x-5\over x-6} \\\Rightarrow {-1\over (x-3)(x-4)} ={-1 \over (x-5)(x-6)}  \Rightarrow (x-3)(x-4) =(x-5)(x-6) \\\Rightarrow -7x+12=-11x+30 \Rightarrow 4x=18 \Rightarrow x=9/2,故選\bbox[red, 2pt]{(D)}$$


-- END   (僅供參考)  --



沒有留言:

張貼留言