110年台北市聯合轉學考-升高二(技高)數學詳解

臺北市立技術型高中暨進修部110學年度聯合招考轉學生

招生考試-升高二數學科試題

一、 單選題：共 25 題，每題 4 分，共 100 分，答錯不倒扣。

解答：$$|-x+5| \le 2 \Rightarrow -2\le -x+5\le 2 \Rightarrow -2\le x-5\le 2 \Rightarrow 3\le x\le 7，故選\bbox[red,2pt]{(B)}$$

解答：$$\cases{A(-1,-3)\\ B(6,4)} \Rightarrow \overline{AB}=\sqrt{7^2+7^2} =7\sqrt 2\\ 又4\overline{AC} =3\overline{BC} \Rightarrow \overline{BC}={4\over 7}\overline{AB} ={4\over 7}\times 7\sqrt 2=4\sqrt 2，故選\bbox[red,2pt]{(C)}$$

解答：$$當x=0時，y=40，只有(A)與(B)符合；又斜率為負值，故選\bbox[red,2pt]{(A)}$$

解答：$$y=2x^2-80x+1000= 2(x^2-40x+400)+200 = 2(x-20)^2+200\\ \Rightarrow x=20時，y有最小值200，故選\bbox[red,2pt]{(C)}$$

解答：$$f(x)=kx^2+2x+k \gt 0 \Rightarrow \cases{k\gt 0 (圖形凹向上)\\判別式4-4k^2 \lt 0 \Rightarrow k\gt 1或k\lt -1} \\ \Rightarrow k\gt 1，故選\bbox[red,2pt]{(A)}$$

解答：$$\cases{A(2,1)\\ B(1,3) \\C(4,k)} \Rightarrow \cases{\overleftrightarrow{AB}斜率m_1=(3-1)/(1-2)=-2 \\ \overleftrightarrow{AC}斜率m_2= (k-1)/(4-2)=(k-1)/2} \\ 又\overleftrightarrow{AB} \bot \overleftrightarrow{AC} \Rightarrow m_1m_2 = -1 \Rightarrow -2\times {k-1\over 2}= -1 \Rightarrow k=2，故選\bbox[red,2pt]{(B)}$$

解答：$$L: y=-{5\over 3}x+{10\over 3} \Rightarrow y=0時，{5\over 3}x={10\over 3} \Rightarrow x=2，故選\bbox[red,2pt]{(D)}$$

解答：$$\cases{P(-1,a)\\ L:3x+4y=3} \Rightarrow d(P,L)={|-3+4a-3|\over 5}=2 \Rightarrow |-6+4a|=10 \\ \Rightarrow \cases{-6+4a=10\\ -6+4a=-10} \Rightarrow \cases{a=4\\ a=-1}；由於P在第二象限，即a\gt 0，因此a=4，故選\bbox[red,2pt]{(D)}$$

解答：$$利用長除法:x^3+3x^2+ax+b =(x^2+x+1)(x+2)+(a-3)x+(b-2)\\ \Rightarrow \cases{a=3\\b=2} \Rightarrow a+b=5，故選\bbox[red,2pt]{(C)}$$

解答：$$f(x)= x^3-2x^2+x+6 = A(x+2)^3 +B(x+2)^2 +C(x+2)+D\\ \Rightarrow \cases{f(0)=6=8A+4B+2C+D\\ f(1)=6=27A+9B+3C+D\\ f(-1)=2=A+B+C+D\\ f(-2)=-12=D} \Rightarrow \cases{8A+4B+2C=18\\ 27A+9B+3C =18\\ A+B+C=14} \Rightarrow \cases{A=1\\ B=-8 \\ C=21\\ D=-12}，故選\bbox[red,2pt]{(D)}$$

解答：$$f(x)=(x+1)^{200}+2x+1 =g(x)(x+2)+k\Rightarrow f(-2)= (-1)^{200}-4+1=-2 =k \\ \Rightarrow 餘式為k=-2，故選\bbox[red,2pt]{(A)}$$

解答：$${2\over x+1}-{3\over x-2} ={2(x-2)-3(x+1) \over (x+1)(x-2)} ={-x-7\over (x+1)(x-2)}，故選\bbox[red,2pt]{(B)}$$

解答：$$由圖形可知:最大值為1；且x=0時，y=0，故選\bbox[red,2pt]{(B)}$$

解答：$$\cases{\tan \theta=3\\ \cos \theta\lt 0} \Rightarrow \cases{\sin=-3/\sqrt{10} \\\cos \theta =-1/\sqrt{10}} \Rightarrow 3\sin \theta+\cos \theta =-{10\over \sqrt{10}}= -\sqrt{10}，故選\bbox[red,2pt]{(D)}$$

解答：$$\cases{a=\sqrt 2\\ b=2 \\ c=\sqrt 3-1} \Rightarrow b\gt a\gt c \Rightarrow 最大內角為 \angle B \Rightarrow \cos \angle B={a^2+c^2-b^2 \over 2ac} ={2-2\sqrt 3\over 2(\sqrt 6-\sqrt 2)} \\=-{\sqrt 2\over 2}  \Rightarrow \angle B=135^\circ，故選\bbox[red,2pt]{(C)}$$

解答：$$\cos \angle BCA= {\overline{AC}^2 +\overline{BC}^2-\overline{AB}^2\over 2\times \overline{AC}\times \overline{BC}} \Rightarrow \cos 120^\circ ={3^2+5^2-\overline{AB}^2 \over 2\times 3\times 5} \Rightarrow -{1\over 2}={34-\overline{AB}^2 \over 30} \Rightarrow \overline{AB}=7\\ \triangle ABC面積={1\over 2}\overline{AC}\times \overline{BC}\sin \angle BCA={1\over 2}\overline{AB}\times \overline{CD} \Rightarrow 15\sin 120^\circ=7\overline{CD} \\ \Rightarrow 15\times {\sqrt 3\over 2}=7\overline{CD} \Rightarrow \overline{CD}={15\over 14}\sqrt 3，故選\bbox[red,2pt]{(B)}$$

解答：$$\sin{8\pi \over 3} +\cos\left(-{\pi\over 6}\right)+ \tan{13\pi\over 4} =\sin {2\over 3}\pi +\cos {\pi\over 6}+ \tan{5\over 4}\pi  ={\sqrt 3\over 2}+{\sqrt 3\over 2} +1=\sqrt 3+1\\，故選\bbox[red,2pt]{(D)}$$

解答：$$\cases{\vec a=(3,4)\\ \vec a\parallel \vec b} \Rightarrow \vec b=k\vec a=(3k,4k)；又\vec a\cdot \vec b=-50 \Rightarrow 9k+16k=-50 \Rightarrow k=-2\\ \Rightarrow \vec b=(-6,-8) \Rightarrow |2\vec a+3\vec b|=|(6,8)+(-18,-24)| =|(-12,-16)| \\ =\sqrt{12^2+16^2} =\sqrt{400} =20，故選\bbox[red,2pt]{(A)}$$

解答：$$\vec a\cdot \vec b=|\vec a||\vec b|\cos 60^\circ =1\cdot 1\cdot {1\over 2}={1\over 2} \Rightarrow \vec a\cdot (-\vec a+2\vec b)=-|\vec a|^2+2\vec a\cdot \vec b=-1+1=0 =\cos{\pi\over 2}\\ \Rightarrow \vec a與 (-\vec a+2\vec b)的夾角為{\pi\over 2} =90^\circ，故選\bbox[red,2pt]{(C)}$$

解答：$$ \cases{\overrightarrow{AB} =(4,8) \\\overrightarrow{AD} =(1,4)} \Rightarrow \cases{\overrightarrow{AC} = \overrightarrow{AB}+\overrightarrow{BC}= \overrightarrow{AB}+\overrightarrow{AD} = (5,12) \\ \overrightarrow{BD}= \overrightarrow{BA}+ \overrightarrow{AD}= (-4,-8)+(1,4)=(-3,-4) } \\ \Rightarrow |\overrightarrow{AC}|+ |\overrightarrow{BD}| =\sqrt{5^2+12^2} +\sqrt{3^2+4^2} =13+5=18，故選\bbox[red,2pt]{(B)}$$

解答：$$圓C:(x-3)^2 +(y-2)^2=1 \Rightarrow \cases{圓心O(3,2)\\ 半徑r=1} \Rightarrow d(O,L)={9+8+8\over 5}=5 \gt r\\ \Rightarrow d(P,L)= d(O,L)-r=5-1=4，故選\bbox[red,2pt]{(D)}$$

解答：$$圓 :2x^2+2y^2-4x+6y+1=0 \Rightarrow 2(x-1)^2+2(y+3/2)^2=11/2 \\\Rightarrow (x-1)^2+(y+3/2)^2 = ({\sqrt{11}\over 2})^2 \Rightarrow \cases{圓心O(1,-3/2)\\ 半徑r=\sqrt{11}/2} \Rightarrow \overline{OP}= \sqrt{2^2+({11\over 2})^2} ={\sqrt{137}\over 2} \\ \Rightarrow 切線長=\sqrt{\overline{OP}^2-r^2} =\sqrt{{137\over 4}-{11\over 4}} ={\sqrt{126}\over 2} ={3\sqrt{14}\over 2} ={a\sqrt{14} \over 2} \Rightarrow a=3，故選\bbox[red,2pt]{(C)}$$

解答：$$圓心需在直線L:y=x-1上\\(A)\times: 圓心(1,-2)不在L上\\(B)\times: 圓心(2,3)不在L上\\(C)\times: 圓心(1,2)不在L上\\ (D)\bigcirc: 圓心(2,1)在L上\\，故選\bbox[red,2pt]{(D)}$$

解答：$$\cases{a_1=1\\ a_4=a_1+3d=10 \Rightarrow d=3} \Rightarrow S_{10}= (a_1+a_{10})\times 10\div 2= 5(a_1+a_1+9d)\\=5(2+27)= 145，故選\bbox[red,2pt]{(A)}$$

解答：$$S_n=1{1\over 1}+2{1\over 2} +3{1\over 4}+\cdots +\left(n+{1\over 2^{n-1}}\right) =(1+2+\cdots +n)+ \left({1\over 1} +{1\over 2} +\cdots +{1\over 2^{n-1}} \right)\\ ={n(n+1)\over 2} +{1-1/2^n\over 1/2} \Rightarrow S_{10}=55+ 2(1-{1\over 1024})=57-{1\over 512} =56{511\over 512}，故選\bbox[red,2pt]{(A)}$$

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解題僅供參考，其他轉學考試題及詳解