112年高雄區高職聯合轉學考-數學詳解

高雄區112學年度公立高級職業學校聯合招考轉學生

一、單選題：共25題，每題4分，答錯不倒扣

解答：$$\cases{A(-2,0)\\ B(3,5) \\C(4,-2)} \Rightarrow \cases{\overline{AB}斜率={5-0\over 3-(-2)} =1\\ \overline{AC} 斜率 ={-2-0\over 4-(-2)} =-{1\over 3}} \Rightarrow -{1\over 3}\le m\le 1，故選 \bbox[red, 2pt]{(C)}$$

解答：$$同界角:{\pi \over 6}\pm 2k\pi  \xrightarrow{k=2} {\pi \over 6}+4\pi ={25\over 6}\pi，故選 \bbox[red, 2pt]{(D)}$$

解答：$$(3,2) \cdot (6,-9)=18-18=0 \Rightarrow (3,2) \bot(6,-9)，故選 \bbox[red, 2pt]{(C)}$$

解答：$$\cases{\cos \theta \gt 0\\ \tan \theta \lt 0} \Rightarrow \sin \theta \lt 0 \Rightarrow 第四象限，故選 \bbox[red, 2pt]{(Ｄ)}$$

解答：$$\cases{a_1+2d=-20\\ a_1+ 10d=20} \Rightarrow \cases{a=-30\\ d=5} \Rightarrow a_n=-30+(n-1)\cdot 5 \gt 100 \Rightarrow n-1\gt 26 \\ \Rightarrow n\gt 27 \Rightarrow n=28，故選 \bbox[red, 2pt]{(B)}$$

解答：$$x截距為6 \Rightarrow 直線通過(6,0) \Rightarrow y=m(x-6),且通過(2,4) \Rightarrow 4=m(2-6) \Rightarrow m=-1\\ \Rightarrow y=-(x-6) \Rightarrow x=0時，y=6\Rightarrow y截距=6，故選 \bbox[red, 2pt]{(C)}$$

解答：$$\cos(\angle B+\angle C)= \cos(180^\circ-\angle A) =-\cos \angle A =-{4^2+10^2-8^2\over 2\cdot 4\cdot 10} =-{13\over 20}，故選 \bbox[red, 2pt]{(A)}$$

解答：$$2x+4y+5=0的斜率=-{1\over 2} \Rightarrow 與其垂直的斜率=2 且通過(3,6)的直線: y=2(x-3)+6\\ \Rightarrow 2x-y=0，故選 \bbox[red, 2pt]{(D)}$$

解答：$$\cases{\vec a=(1,2) \\\vec b=(3,x)} \Rightarrow \cases{\vec a+2\vec b=(7,2+2x) \\ 2\vec a-\vec b =(-1,4-x)} \\ (\vec a+2\vec b) \parallel (2\vec a-\vec b) \Rightarrow {7\over -1}={2+2x \over 4-x} \Rightarrow 2+2x=7x-28 \Rightarrow x=6，故選 \bbox[red, 2pt]{(B)}$$

解答：$$a,b,c,d成等比 \Rightarrow \cases{b=ar\\ c=ar^2\\ d=ar^3},又\cases{a+b=32\\ c+d=288} \Rightarrow \cases{a+ar=32\\ ar^2+ ar^3=288} \\ \Rightarrow {a+ar \over ar^2+ar^3}={32\over 288} \Rightarrow {1\over r^2}={1\over 9}\Rightarrow \cases{r=3\\ r=-3(不合，a,b,c,d皆正數)}　\\ \Rightarrow a=8 \Rightarrow c=ar^2=72 \Rightarrow a+c=8+72=80 ，故選 \bbox[red, 2pt]{(C)}$$

解答：$$\cos \theta ={\vec a\cdot \vec b\over |\vec a||\vec b|} ={-2\sqrt 3\over 2\cdot 2}=-{\sqrt 3\over 2} \Rightarrow \theta=150^\circ，故選 \bbox[red, 2pt]{(D)}$$

解答：$$x^2+y^2-6x+4y+3=0 \Rightarrow (x-3)^2 +(y+2^2)=4^2 \Rightarrow 圓Ｏ(3,-2) \\ \Rightarrow 圓心至(4,1)的直線斜率=3 \Rightarrow 切線斜率=-{1\over 3}，故選 \bbox[red, 2pt]{(D)}$$

解答：$$假設\cases{首項a\\ 公差d} \Rightarrow \cases{S_{10}=5(2a+9d) =80\\ a_{10}=a+9d= 20} \Rightarrow a=-4，故選 \bbox[red, 2pt]{(B)}$$

解答：$$h=r\sin \theta = 6\cdot \sin 30^\circ= 3 \Rightarrow \cases{扇形面積= 6^2\pi\cdot {30^\circ \over 360^\circ} =3\pi \\ 三角形面積={1\over 2}rh= 9} \Rightarrow 弓形面積=3\pi-9，故選 \bbox[red, 2pt]{(C)}$$

解答：$$\cases{\sin \theta =-12/13\\ \cos \theta =-5/13} \Rightarrow \cases{\sec \theta =-13/5\\ \tan \theta = 12/5} \Rightarrow \sec \theta+\tan \theta =-{1\over 5}，故選 \bbox[red, 2pt]{(B)}$$

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

解答：$$x^3項係數=2\cdot 4+4\cdot (-1)=4，故選 \bbox[red, 2pt]{(A)}$$

解答：$$\cases{A(5,-2) \\B(1,4) \\C(0,a)} \Rightarrow \cases{\overrightarrow{CA}=(5,-2-a) \\ \overrightarrow{CB} =(1,4-a)} \Rightarrow \overrightarrow{CA} \cdot \overrightarrow{CB}=0 \Rightarrow 5-(a+2)(4-a)=0 \Rightarrow a^2-2a-3=0\\ \Rightarrow (a-3)(a+1)=0 \Rightarrow a=3或-1，故選 \bbox[red, 2pt]{(C)}$$

解答：$$25x^2-60x+36\le 0  \Rightarrow (5x-6)^2 \le 0,由於(5x-6)^2 \ge 0,因此5x-6=0 \Rightarrow x={6\over 5}，故選 \bbox[red, 2pt]{(A)}$$

解答：$$y=-x^2+6x-11= -(x-3)^2-2 \Rightarrow  y+2=-(x-3)^2 \Rightarrow 頂點(3,-2)，故選 \bbox[red, 2pt]{(A)}$$

解答：$${6x+3\over x(x+1)} ={a\over x+1}+ {b\over x} \Rightarrow ax+b(x+1) =(a+b)x+ b=6x+3 \Rightarrow \cases{a+b=6\\ b=3} \Rightarrow a=b=3\\，故選 \bbox[red, 2pt]{(B)}$$

解答：$$\overline{AB}: \overline{BC}=b+3: 11-b=3:4 \Rightarrow 3(11-b)=4(b+3) \Rightarrow b=3，故選 \bbox[red, 2pt]{(D)}$$

解答：$$x^2+y^2-2x+4y-1=0 \Rightarrow (x-1)^2+(y+2)^2=6 \Rightarrow \cases{圓心O(1,-2) \\ 㘣半徑r=\sqrt 6} \\ 又\overline{AB}中點P(2,-3) \Rightarrow \overline{OP}= \sqrt{1+1} =\sqrt 2 \Rightarrow \overline{AP}=\sqrt{r^2-\overline{OP}^2}=\sqrt{6-2} =2 \\ \Rightarrow \overline{AB}= 2\overline{AP}=4，故選 \bbox[red, 2pt]{(A)}$$

解答：$$1\lt |3x+2| \le 8 \Rightarrow \cases{|3x+2|\le 8  \Rightarrow -8\le 3x+2\le 8 \Rightarrow -10/3\le x\le 2\\ 1\lt |3x+2| \Rightarrow \cases{3x+2\gt 1 \Rightarrow x\gt -1/3 \\ 3x+2\lt -1 \Rightarrow x\lt -1} \Rightarrow x\gt -1/3 或x\lt -1} \\ \Rightarrow x=-3,-2,-1,1,2，共5個整數解，故選 \bbox[red, 2pt]{(B)}$$

解答：$$\angle A=180^\circ-30^\circ-105^\circ = 45^\circ  \Rightarrow {\overline{BC} \over \sin \angle A} ={\overline{AB} \over \sin \angle C} \Rightarrow {5 \over \sqrt 2/ 2} ={\overline{AB} \over 1/2} \\ \Rightarrow \overline{AB} ={5\sqrt 2\over 2}，故選 \bbox[red, 2pt]{(A)}$$

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解題僅供參考，其他歷年試題及詳解