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2020年1月12日 星期日

106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106 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


$$ \begin{vmatrix}\sin \theta & -\cos \theta  \\\cos \theta  & \sin \theta  \end{vmatrix} = \sin^2 \theta +\cos^2 \theta=1, 故選:\bbox[red,2pt]{(D)}$$


:$$\begin{cases}P在y軸上 \Rightarrow P(0,a) \\ \overline{PA}= \overline{PB} \Rightarrow 1^2+(a-2)^2 = 3^2+(a-4)^2\end{cases}  \Rightarrow a^2-4a+5= a^2-8a+25 \\ \Rightarrow 4a=20 \Rightarrow a=5 \Rightarrow P(0,5), 故選\bbox[red,2pt]{(B)}$$



$$垂直3x-4y+1=0的直線L:4x+3y=k \Rightarrow 只有(C)符合要求, 故選\bbox[red,2pt]{(C)}$$


:$$直線L的斜率為-2 \Rightarrow L: y=-2x+b;又截距和為15 \Rightarrow b+b/2=15 \Rightarrow b=10\\  \Rightarrow L: y=-2x+10 \Rightarrow 2x+y-10=0, 故選\bbox[red,2pt]{(D)}$$


$$\sin 180^\circ -\tan (-135^\circ) +\cos(-540^\circ)= 0+\tan(135^\circ)+ \cos(540^\circ) = -\tan(45^\circ) +\cos(180^\circ) \\ =-1-1=-2, 故選\bbox[red,2pt]{(D)}$$

$$(\sin 45^\circ+\cos 60^\circ) (\cos 45^\circ -\sin 30^\circ) = \left( {\sqrt 2\over 2} + {1\over 2}\right)\left( {\sqrt 2\over 2} - {1\over 2}\right) =\left( {\sqrt 2\over 2} \right)^2-\left(  {1\over 2}\right)^2\\ = {1\over 2}-{1\over 4} = {1\over 4} , 故選\bbox[red,2pt]{(B)}$$


$$\begin{cases}90^\circ < \theta < 180^\circ \\ \sin \theta=1/2 \end{cases}  \Rightarrow \theta =150^\circ \Rightarrow \tan (180^\circ -\theta)= \tan 30^\circ =1/\sqrt 3, 故選\bbox[red,2pt]{(A)}$$


:$$ \begin{vmatrix}1 & 0 & x \\ -1 & x & 1 \\2 & 3 & 2 \end{vmatrix} =-13 \Rightarrow 2x-3x-2x^2 -3=-13 \Rightarrow 2x^2+x-10=0 \Rightarrow (2x+5)(x-2)=0\\  \Rightarrow x=2, -5/2,故選\bbox[red,2pt]{(C)} $$



$$f(x)=2x^2 -8x+12 =2(x^2-4x+4)+4 = 2(x-2)^2+4 \Rightarrow  \begin{cases}最小值f(2)=4 =m\\最大值 f(0)=12 =M\end{cases} \\ \Rightarrow M-m=12-4=8,故選\bbox[red,2pt]{(A)} $$


:$$\left( (\sqrt 2x)^2 +(\sqrt 3y)^2\right)\left( (\sqrt 2)^2 +(\sqrt 3)^2\right) \ge (2x+3y)^2  \Rightarrow 25 \ge (2x+3y)^2 \\ \Rightarrow 5\ge 2x+3y \ge -5 \Rightarrow  \begin{cases} M=5  \\ m=-5 \end{cases}  \Rightarrow M-m=10,故選\bbox[red,2pt]{(B)}$$



$$(\sin \theta+\cos \theta)^2 =1 +2\sin\theta \cos \theta = 1+2\cdot {3\over 8}= {7\over 4} \Rightarrow \sin \theta+\cos \theta=-{\sqrt 7\over 2}\\ 由於\theta位於第3象限,\sin \theta+\cos \theta為負值,故選\bbox[red,2pt]{(D)}$$


$$\begin{cases}\vec a \bot \vec b\\ \vec a //\vec c\end{cases}  \Rightarrow \begin{cases}\vec a \cdot \vec b=0 \\ \vec c =k\vec a\end{cases}  \Rightarrow \begin{cases} 4x+2y=0\\ {4\over 2}= {x+3 \over y+4}\end{cases}  \Rightarrow  \begin{cases} 2x+y=0\\ x-2y=5\end{cases}  \Rightarrow \begin{cases} x=1\\ y=-2\end{cases}  \Rightarrow x+y=-1,故選\bbox[red,2pt]{(B)}$$


:$$f(x,y)= nx+2y-7 \Rightarrow f(2,m)=2n+2m-7=g(m,n) \Rightarrow  \begin{cases}g(-1,-3) =-15 \\g(1,1) =-3 \\ g(1,3)=1 \ge 0\\ g(-1,-4) =-17\end{cases} ,故選\bbox[red,2pt]{(C)}$$




$$由長除法可知: \begin{cases} 8+(a-5)=0 \\ b-2(a-5)=0 \end{cases} \Rightarrow \begin{cases} a= -3 \\ b= -16 \end{cases}  \Rightarrow a+3b=-3-48 =-51,故選\bbox[red,2pt]{(B)}$$



$$f(x)=3x^{101}+4x^{100}+5x^{50} -1  \Rightarrow f(-1)=-3+4+5-1=5,故選\bbox[red,2pt]{(A)}$$



$$f(x)=\begin{cases}(x-1)p(x)+3 \\ (x+3)q(x)-5 \\ (x-1)(x+3)r(x)+ax+b \end{cases}  \Rightarrow  \begin{cases}f(1)=3=a+b \\  f(-3)=-5 =-3a+b  \end{cases}  \Rightarrow \begin{cases}a=2 \\ b=1  \end{cases} \\\Rightarrow 2a+b=4+1=5,故選\bbox[red,2pt]{(C)}$$


:$$\begin{cases}\vec a  = (1,3) \\ \vec b =(-6,2) \end{cases}  \Rightarrow \vec a\cdot \vec b=  -6+6=0  \Rightarrow \vec a\bot\vec b,故選\bbox[red,2pt]{(B)}$$



$$\overrightarrow{BA} \cdot \overrightarrow{BC}=|\overrightarrow{BA}| |\overrightarrow{BC}| \cos \angle B =2\times 2\times \cos 60^\circ =4\times {1\over 2}=2\\ \Rightarrow \overrightarrow{AB} \cdot \overrightarrow{BC}= -\overrightarrow{BA} \cdot \overrightarrow{BC}= -2,故選\bbox[red,2pt]{(A)} $$


:$$ 1根為a,另1根為4a \Rightarrow (x-a)(x-4a)=0 \Rightarrow x^2-5ax+4a^2=0\\ \Rightarrow  \begin{cases} -5a= -5\\k=4a^2\end{cases} \Rightarrow  \begin{cases} a=1\\k=4 \end{cases} ,故選\bbox[red,2pt]{(A)}$$


:$$1\le x\le 2 \Rightarrow (x-2)(x-1)\le 0 \Rightarrow x^2-3x+2\le 0\Rightarrow 2x^2-6x+4 \le 0\\ \Rightarrow \begin{cases} a=2\\ b=-6\end{cases} \Rightarrow a+b=-4,故選\bbox[red, 2pt]{(B)}$$



:$$|2x+1| \le 3 \Rightarrow -3\le 2x+1 \le 3 \Rightarrow -2\le x\le 1 \Rightarrow x=-2,-1,0,1,共4個整數解,故選\bbox[red, 2pt]{(D)}$$




:$$\begin{cases} x\ge 0, y\ge 0\\ 2x-y\le 10 \\ x+y\le 8 \end{cases},交點為\begin{cases} A(0,8) \\ B(6,2) \\ C(0,-10) \end{cases}\\ 令f(x,y) =2y-x \Rightarrow \begin{cases} f(A)=16 \\ f(B) =10 \\ f(C)=-20 \end{cases} \Rightarrow 最大值為16,故選\bbox[red, 2pt]{(A)}$$



:$${\overline{AB} \over \overline{BD}}= {1\over 2} \Rightarrow B\left({-3\times 2+1\times 4 \over 3},{ 2\times 1+1\times 2 \over 3} \right) =\left( -{2 \over 3}, {4\over 3}\right),故選\bbox[red, 2pt]{(C)}$$



:$$A至\overline{BC}的距離= \left|{ 3+8+4 \over \sqrt{3^2+4^2}} \right| ={15 \over 5}=3,故選\bbox[red, 2pt]{(B)}$$



:$${x+x+y \over 3}\ge \sqrt[3]{x^2y} \Rightarrow {2x+y \over 3} \ge \sqrt[3]{8}=2 \Rightarrow 2x+y \ge 6,故選\bbox[red, 2pt]{(D)}$$


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