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2019年12月29日 星期日

108學年度高雄區公立高職聯合轉學考-升高二數學科詳解


高雄區公立高職 108 學年度聯合招考轉學生
升高二數學科試題詳解
單選題


$$|x-3|\le 5 \Rightarrow -5\le x-3 \le 5 \Rightarrow -2\le x \le 8 \Rightarrow  \begin{cases}最大值 M= 8\\ 最小值m=-2\end{cases}  \Rightarrow M-m=10\\, 故選:\bbox[red,2pt]{(A)}$$


:$$\begin{cases}x+y=7\cdots(1) \\ y+z=8\cdots(2) \\z+x=9 \cdots(3)\end{cases} \xrightarrow{(1)+(2)+(3)}2(x+y+z)=24 \Rightarrow x+y+z=12\cdots (4)\\ \Rightarrow \begin{cases}(4)-(1) \Rightarrow z=5 \\ (4)-(2) \Rightarrow x=4 \\ (4)-(3) \Rightarrow y=3\end{cases} \Rightarrow x+y+2z =4+3+10=17, 故選\bbox[red,2pt]{(D)}$$




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


:$$\text{dist}(P,L)= \left|  \frac{5+2\times 12-3}{ \sqrt{5^2+12^2} } \right|= \frac{26}{13} =2, 故選\bbox[red,2pt]{(C)}$$



$${2x+3y \over 2} \geq  \sqrt{2x\times 3y}  \Rightarrow  \frac{6}{2}  \geq  \sqrt{6xy}  \Rightarrow 9\ge 6xy \Rightarrow  \frac{9}{6} \ge xy \Rightarrow \frac{3}{2} \ge xy, 故選\bbox[red,2pt]{(A)}$$




$$\theta =30^\circ  \Rightarrow  \begin{cases}\sin \theta=1/2 \\ \cos \theta=\sqrt 3/2\end{cases}   \Rightarrow  \begin{vmatrix}\cos^2 \theta & \sin^2 \theta \\ \sin^2 \theta & \cos^2 \theta \end{vmatrix} =\cos^4 \theta-\sin^4 \theta =9/16-1/16= 8/16=1/2\\, 故選\bbox[red,2pt]{(C)}$$




$$\overline{AP} :\overline{BP}=1:3 \Rightarrow P={3\over 4}A+ {1\over 4}B ={3\over 4}(-2,4) + {1\over 4}(10,-20) = (4/4,-8/4)= (1,-2)\\, 故選\bbox[red,2pt]{(B)}$$


:$$\begin{cases}A(2,3)\\ B(-4,3) \\C(6,-1) \end{cases}  \Rightarrow \overline{BC}的中點D=(B+C)/2 = (2/2,2/2) =(1,1); \\ 假設經A、D的直線方程式為y=ax+b,將A、D代入 \Rightarrow \begin{cases}3=2a+b \\1 =a+b \end{cases} \Rightarrow \begin{cases}a=2 \\b=-1\end{cases} \\ \Rightarrow y=2x-1,故選\bbox[red,2pt]{(D)} $$



$$2x-1=5 \Rightarrow x=3 \Rightarrow 3x-2=9-2=7,故選\bbox[red,2pt]{(A)} $$


:$$\sin 30^\circ + \sin 150^\circ -\sin 210^\circ -\sin 330^\circ =\sin 30^\circ + \sin 30^\circ + \sin 30^\circ +\sin 30^\circ =  4\times \frac{1}{2} =2\\,故選\bbox[red,2pt]{(C)}$$



$$ -1\le \cos x \le 1 \Rightarrow -3+1\le 3\cos x+1 \le 3+1 \Rightarrow -2\le 3\cos x+1 \le 4 \Rightarrow 最大值為4\\,故選\bbox[red,2pt]{(D)}$$


$$\sin (180^\circ +\theta) =-\sin \theta = \frac{1}{3},故選\bbox[red,2pt]{(B)}$$


:$$ \begin{vmatrix}1 & 4 &8\\2& 5 &7\\3 &6 &9 \end{vmatrix} =45+96+84 - 120-72-42 =-9,故選\bbox[red,2pt]{(A)}$$



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



$$\begin{cases}-3 \le x \le 5\\-7 \le y \le 4\end{cases}  \Rightarrow x\cdot y= \begin{cases} M=21 & x = -3,y=-7 \\ m=-35 & x=5,y=-7\end{cases}  \Rightarrow M+m=-14,故選\bbox[red,2pt]{(C)}$$


:$$\tan \theta+ {1\over \tan \theta} ={\sin \theta \over \cos \theta}+ {\cos \theta \over \sin \theta} = {\sin^2 \theta +\cos^2 \theta \over \sin\theta \cos \theta} ={1 \over 1/3}=3,故選\bbox[red,2pt]{(D)}$$




:$$令f(x) =x^4+3x^3 -2x^2+ax+4,由於x+1是f(x)的因式 \Rightarrow f(-1)=0 \Rightarrow 1-3-2-a+4 =0 \\ \Rightarrow a=0,故選\bbox[red,2pt]{(A)}$$




$$利用長除法,如上\Rightarrow \begin{cases} a+5=6 \\ b=6\end{cases} \Rightarrow \begin{cases} a=1 \\ b=6\end{cases},故選\bbox[red,2pt]{(C)} $$


:$$x=1代入多項式\Rightarrow 4+3+2-1+29 =37,故選\bbox[red,2pt]{(D)}$$




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




:$$(4x^3 +2x+5)(2x^2+6x-1)中x^3係數=4\times (-1)+2\times 2 =0,故選\bbox[red, 2pt]{(A)}$$




:$$\begin{cases}\vec a=(1,2)\\ \vec b =(-4,2)\end{cases}  \Rightarrow \vec a+\vec b=(-3,4) \Rightarrow |\vec a+\vec b|=  \sqrt{(-3)^2+4^2} =5,故選\bbox[red, 2pt]{(D)}$$




:$$\begin{cases}\vec a=(-2,k)\\ \vec b =(-3,5)\end{cases}  \Rightarrow \vec a\cdot\vec b=0 \Rightarrow 6+5k=0 \Rightarrow k=-{6\over 5},故選\bbox[red, 2pt]{(B)}$$




:$$\begin{cases}\vec a=(5,-7)\\ \vec b =(-3,-5)\end{cases}  \Rightarrow \vec a\cdot\vec b= |\vec a||\vec b| \cos \theta \Rightarrow -15+35=\sqrt{5^2+(-7)^2}\times \sqrt{(-3)^2+(-5)^2}\times \cos \theta \\ \Rightarrow 20=\sqrt{74} \times \sqrt{34}\times \cos \theta  \Rightarrow \cos \theta ={10\over \sqrt {17\times 37}} \Rightarrow \sin \theta ={ 23\over \sqrt {17\times 37}} \\ \Rightarrow \triangle 面積={1\over 2}|\vec a||\vec b|\sin \theta ={1\over 2}\sqrt{74} \times \sqrt{34}\times { 23\over \sqrt {17\times 37}}=23,故選\bbox[red, 2pt]{(C)}$$




:$$\begin{cases}2x+ay=1 \\ ax+18y=-3\end{cases}  \Rightarrow  \frac{2}{a} = \frac{a}{18} = \frac{1}{-3}  \Rightarrow a=-6,故選\bbox[red, 2pt]{(B)}$$


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