2020年1月3日 星期五

107學年度高雄區公立高職聯合轉學考-升高二數學科詳解


高雄區107 學年度公立高職聯合招考轉學生
升高二數學科試題詳解
單選題


$$\begin{cases} A(-4,3)\\ B(8,-1) \\ C(x,y) \\ G(2,-1)\end{cases} \Rightarrow G=(A+B+C)/3 \Rightarrow \begin{cases} 2=(-4+8+x)/3\\ -1= (3-1+y)/3\end{cases} \Rightarrow \begin{cases} x=2\\ y=-5\end{cases} , 故選:\bbox[red,2pt]{(D)}$$


:$$\begin{cases} A(1,-3)\\ B(4,1) \\ C(4,6) \\ D(-2,2a)\end{cases} \Rightarrow \begin{cases} \overline{AB}斜率:m_1 ={1-(-3) \over 4-1}=4/3\\ \overline{CD}斜率: m_2 ={6-2a\over 4-(-2)} =(6-2a)/6\end{cases} \Rightarrow\overline{AB}// \overline{CD} \Rightarrow m_1=m_2 \\\Rightarrow{4\over 3} ={6-2a \over 6} \Rightarrow a=-1, 故選\bbox[red,2pt]{(A)}$$




$$f(x)=-2x^2+4x-3 = -2(x^2-2x+1)-1 =-2(x-1)^2-1 \Rightarrow \begin{cases}最大值f(1)=-1=M\\ 最小值f(-2)= -19=m\end{cases} \\ \Rightarrow M+m =-1-19=-20, 故選\bbox[red,2pt]{(C)}$$


:$$L: y=ax+b, 由\begin{cases}斜率為-{1\over 2} \Rightarrow a= -{1\over 2}\\ x截距為{3\over 4} \Rightarrow 0={3\over 4}a+b \end{cases} \Rightarrow 0={3\over 4}\times {-1\over 2}+b \Rightarrow b= {3 \over 8} \\\Rightarrow L: y=-{1\over 2}x+ {3\over 8} \Rightarrow 4x+8y-3=0 , 故選\bbox[red,2pt]{(A)}$$



$$r\times {2\pi \over 3}=2 \Rightarrow 半徑r={3\over \pi} \Rightarrow 面積=({3\over \pi})^2\pi \times {2\pi \over 3}\div 2\pi = {3\over \pi}, 故選\bbox[red,2pt]{(D)}$$




$$\tan A=2.4 = {12\over 5} \Rightarrow \begin{cases} \overline{BC}=12\\ \overline{AC}=5\end{cases} \Rightarrow \overline{AB}=13 \Rightarrow \csc B= {\overline{AB} \over \overline{AC}}= {13 \over 5}, 故選\bbox[red,2pt]{(A)}$$




$$\sin 60^\circ \times \tan 330^\circ + \cos (-120^\circ)\times \cot 225^\circ = \sin 60^\circ \times \tan (-30^\circ) + \cos (120^\circ)\times \cot 45^\circ \\= {\sqrt 3\over 2}\times {-1\over \sqrt 3}+ {-1 \over 2}\times 1 =-{1\over 2}- {1\over 2}= -1 , 故選\bbox[red,2pt]{(B)}$$


:$$0< \theta <{\pi \over 4} \Rightarrow \cos \theta > \sin \theta \Rightarrow (\sin \theta-\cos \theta)^2= 1-2\sin \theta \cos \theta=1-((\sin\theta +\cos \theta)^2-1)\\ = 2- ({\sqrt 7 \over 2})^2 = {1\over 4} \Rightarrow \sin \theta-\cos \theta = -{1\over 2},故選\bbox[red,2pt]{(C)} $$



$$\begin{cases}a=\tan 250^\circ = \tan 70^\circ \\ b=\cos 380^\circ = \cos 20^\circ = \sin 70^\circ\\ c=\sin 770^\circ =\sin 50^\circ\end{cases} \Rightarrow a>b>c,故選\bbox[red,2pt]{(D)} $$


:$$\begin{cases}\vec a=(3,x+1) \\ \vec b= (-2,5)\\ \vec c=(y-1,3) \end{cases} \Rightarrow \vec c=\vec a+\vec b \Rightarrow (y-1,3) =(1,x+6) \Rightarrow \begin{cases} x=-3 \\ y=2\end{cases}\Rightarrow x+y=-1,故選\bbox[red,2pt]{(B)}$$



$$ \begin{cases}\vec u =(-5,-12) \\ \vec v= k\vec u\\ |\vec v|=3 \end{cases} \Rightarrow \vec v=(-5k,-12k) \Rightarrow 25k^2+144k^2=9 \Rightarrow 169k^2=9 \Rightarrow 13k= \pm 3\\ \Rightarrow k= {3\over 13}(k<0不合, \vec u,\vec v需同向) \Rightarrow \vec v= ({-15 \over 13},{-36 \over 13}),故選\bbox[red,2pt]{(A)}$$


$$\vec u\cdot \vec v=|\vec u||\vec v|\cos \theta \Rightarrow -10\sqrt 2= 5\times 4\times \cos \theta \Rightarrow \cos \theta =-{\sqrt 2\over 2} \Rightarrow \theta ={3\pi \over 4},故選\bbox[red,2pt]{(C)}$$


:$$\vec u\cdot \vec v=|\vec u||\vec v|\cos 120^\circ =1\times 1\times {-1\over 2}= {-1\over 2} \Rightarrow |\vec u+\vec v|^2 =(\vec u+\vec v) \cdot (\vec u+\vec v) =|\vec u|^2+2\vec u \cdot \vec v+|\vec v|^2\\ =2+2\vec u \cdot \vec v =2+2 \times {-1\over 2} =1 \Rightarrow |\vec u+\vec v|=1,故選\bbox[red,2pt]{(B)}$$



$$f(x)=2x^6-12x^5-101x^4-102x^3 +35x^2-25x+50\\ \Rightarrow f(11)= 2\times 11^6-12\times 11^5-101\times 11^4-102\times 11^3 +35\times 11^2-25\times 11+50 \\ =(22-12)\times 11^5-101\times 11^4-102\times 11^3 +35\times 11^2-25\times 11+50\\ =10 \times 11^5-101\times 11^4-102\times 11^3 +35\times 11^2-25\times 11+50 \\=(110-101)\times 11^4-102\times 11^3 +35\times 11^2-25\times 11+50\\ = 9\times 11^4-102\times 11^3 +35\times 11^2-25\times 11+50 =(99-102)\times 11^3 +35\times 11^2-25\times 11+50 \\= -3\times 11^3 +35\times 11^2-25\times 11+50 =(-33+35)\times 11^2-25\times 11+50 \\=2 \times 11^2-25\times 11+50 =(22-25)\times 11+50 =-3\times 11+50 =-33+50=17,故選\bbox[red,2pt]{(B)}$$



$$f(x)=x^4+mx^2 +nx+6 = p(x)(x^2-2x-3) = p(x)(x-3)(x+1) \\ \Rightarrow \begin{cases} f(-1)=0 \\ f(3)=0\end{cases} \Rightarrow \begin{cases} 1+m-n+6=0 \\ 81+9m+3n+6=0\end{cases} \Rightarrow \begin{cases} m-n=-7 \\ 3m +n = -29\end{cases} \Rightarrow \begin{cases} m=-9 \\ n = -2\end{cases} \Rightarrow m+n=-11\\,故選\bbox[red,2pt]{(D)}$$



$$由長除法(如上圖)可得\begin{cases}a=3\\ b=-4 \\c=2 \\d =-2 \\e=-3 \end{cases} \Rightarrow a+b +c +d +e =3-4+2-2-3=-4,故選\bbox[red,2pt]{(C)}$$




:$$f(x)=p(x)(x^2-1)+ax+b,由\begin{cases} f(-1)=7\\ f(1)=3\end{cases} \Rightarrow \begin{cases} -a+b=7\\ a+b=3\end{cases} \Rightarrow  \begin{cases} a=-2\\ b=5 \end{cases} \\\Rightarrow 餘式為-2x+5 \Rightarrow a=0,故選\bbox[red,2pt]{(A)}$$




$$\begin{cases} (a+1)x +ay=5\\ 12x +(a+6)y =15\end{cases} 相依 \Rightarrow {a+1\over 12}= {a\over a+6} ={5\over 15} \Rightarrow a=3,故選\bbox[red,2pt]{(C)} $$


:$$ \begin{vmatrix}x & x-1 & x-2 \\x-2 & x & x-1 \\ x-1 & x-2 &x \end{vmatrix} \xrightarrow{-r_1+r_2, -r_1+r_3}\begin{vmatrix}x & x-1 & x-2 \\-2 & 1 & 1 \\ -1 & -1 &2 \end{vmatrix} \\=2x +2(x-2)-(x-1)+(x-2)+4(x-1)+x =8x-8=0 \Rightarrow x=1,故選\bbox[red,2pt]{(C)}$$




:$$ \begin{vmatrix}4 & -12 & 8 \\0 & 3 & 14 \\ 0 & 0 & 2 \end{vmatrix} + \begin{vmatrix}-2 & 0 & 0 \\9 & 4 & 0 \\ 11 & 13 & 5 \end{vmatrix}=k \Rightarrow 24-40=k \Rightarrow k=-16,故選\bbox[red, 2pt]{(A)}$$




:$$|2x+y-1|+(x+2y+4)^2=0 \Rightarrow \begin{cases} 2x+y-1=0 \\ x+2y+4=0\end{cases} \Rightarrow \begin{cases} x=2 \\ y=-3 \end{cases} \Rightarrow x-y=5,故選\bbox[red, 2pt]{(B)}$$




:$${10-x \over x-3}\ge 2 \Rightarrow {10-x \over x-3}- 2\ge 0 \Rightarrow {10-x -2x+6\over x-3}\ge 0 \Rightarrow  {16-3x \over x-3}\ge 0\\ (16-3x)(x-3) \ge 0 \Rightarrow (3x-16)(x-3) \le 0 \Rightarrow 3\le x\le 16/3 \Rightarrow x=4,5(x\ne 3, 分母x-3不能為0)\\ \Rightarrow 有2個整數解,故選\bbox[red, 2pt]{(B)}$$




:$$有解 \Rightarrow 判別式\ge 0 \Rightarrow (k-3)^2-4(3-k) \geq 0 \Rightarrow k^2-6k+9-12+4k \ge 0 \\ \Rightarrow k^2-2k-3\ge 0 \Rightarrow (k-3)(k+1) \ge 0 \Rightarrow k\ge 3或k\le -1,故選\bbox[red, 2pt]{(D)}$$




:$$\begin{cases} x -2= 0\\ y+3=0 \\ x-2y+4=0 \end{cases}  ,三直線交點 \Rightarrow \begin{cases} A(2,-3) \\ B(2,3) \\ C(-10,-3) \end{cases} \Rightarrow \triangle ABC面積={1\over 2} \times \overline{AB}\times \overline{AC} \\ ={1\over 2}\times 6 \times 12 =36,故選\bbox[red, 2pt]{(C)}$$




:$$2<|3x-5|<13 \Rightarrow  \begin{cases}|3x-5|<13  \Rightarrow -13<3x-5 <13 \Rightarrow  -8/3< x< 6\\ 2<|3x-5| \Rightarrow 3x-5>2 或3x-5<-2 \Rightarrow x>7/3或x< 1\end{cases} ;\\取交集 \Rightarrow -8/3<x<1 或7/3<x<6 \Rightarrow x=-2,-1,0,3,4,5 \Rightarrow 6個整數解\\,故選\bbox[red, 2pt]{(D)}$$


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