104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104 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

解：
$$ \overline{AB}:\overline{BC} =2:1 \Rightarrow \cases{x={ 2\cdot 4-5\over 3} =1\\ y={2\cdot (-2)+10\over 3}=2 } \Rightarrow x+y=1+2=3，故選：\bbox[red,2pt]{(D)}$$

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解：$$x^2+6x+5 = -x-7 \Rightarrow x^2+7x+12=0 \Rightarrow (x+4)(x+3)=0\\ \Rightarrow \cases{ x=-3 \Rightarrow y=3-7=-4 \\ x=-4 \Rightarrow y=4-7=-3} \Rightarrow \cases{A(-3,-4) \\ B(-4,-3)} \Rightarrow \overline{AB}=\sqrt{(-1)^2+1^2} =\sqrt 2， 故選\bbox[red,2pt]{(A)}$$

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解：
$$\cases{A(6,2)\\ B(4,-2)} \Rightarrow \cases{\vec n=\overrightarrow{AB}=(-2,-4) \\ \overline{AB}} 中點C=({6+4\over 2},{2-2\over 2}) =(5,0) \\\Rightarrow 過C且法向量為\vec n的直線方程式: -2(x-5) -4(y-0)=0 \Rightarrow x+2y-5=0， 故選\bbox[red,2pt]{(C)}$$

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解：$$f(-1)=8  \Rightarrow a+8+b=8 \Rightarrow a+b=0， 故選\bbox[red,2pt]{(B)}$$

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$$\color{blue}{\textbf{解}}：\\
\cases{\angle C=90^\circ \\ \sin A=3/5} \Rightarrow \cases{\cos A=4/5 \\ \tan A= 3/4} \Rightarrow \cos A-\tan A= {4\over 5}- {3\over 4} = {1\over 20}， 故選\bbox[red,2pt]{(B)}$$

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$$\color{blue}{\textbf{解}}：\\
 (1+\sin 45^\circ -\sin 30^\circ)(1-\cos 60^\circ-\cos 45^\circ)+\tan 45^\circ = (1+{\sqrt 2\over 2} -{1\over 2}) (1-{1\over 2} -{\sqrt 2\over 2})+1\\ = ({1\over 2} +{\sqrt 2\over 2})({1\over 2} -{\sqrt 2\over 2}) +1 = {1\over 4}-{2\over 4}+1 ={3\over 4}， 故選\bbox[red,2pt]{(C)}$$

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$$\color{blue}{\textbf{解}}：\\
\cases{\tan \theta= {\sin \theta \over \cos \theta} =-{3\over 4}<0 \\ \cos \theta <0} \Rightarrow \sin \theta >0 \Rightarrow \cases{\sin \theta =3/5 \\ \cos \theta=-4/5} \Rightarrow 5\sin \theta-4\sec \theta \\= 5\times {3\over 5}-4\times{5\over -4} =3+5=8， 故選\bbox[red,2pt]{(A)}$$

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$$\color{blue}{\textbf{解}}：\\
\tan \theta 的週期為\pi \Rightarrow f(x)= 4\tan \left(2x+ \cfrac \pi 2\right)+3 的週期2x=\pi  \Rightarrow x=\pi/2 ，故選\bbox[red,2pt]{(A)} $$

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$$\color{blue}{\textbf{解}}：\\
\vec a= \vec b \Rightarrow \cases{x+y=7 \\ 3x-y=-3 } \Rightarrow \cases{x=1\\ y=6} \Rightarrow 2x-y= 2-6=-4，故選\bbox[red,2pt]{(A)} $$

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解：
$$\overrightarrow{AC} =\overrightarrow{AB} +\overrightarrow{BC} =(-1,-2)+(3,6) =(2,4) \Rightarrow |\overrightarrow{AC}| = \sqrt{2^2+4^2} =\sqrt{20} =2\sqrt 5  ，故選\bbox[red,2pt]{(B)}$$

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$$\color{blue}{\textbf{解}}：\\
\vec a \cdot \vec b = |\vec a||\vec b|\cos 120^\circ = 7\times 4\times (-{1\over 2}) =-14，故選\bbox[red,2pt]{(A)}$$

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$$\color{blue}{\textbf{解}}：\\
\vec a \bot \vec b \Rightarrow \vec a\cdot \vec b=0 \Rightarrow (4,-5)\cdot (k+3,4)=0 \Rightarrow 4k+12-20 = 0 \Rightarrow k=2，故選\bbox[red,2pt]{(B)}$$

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$$\color{blue}{\textbf{解}}：\\
\cfrac{|2-2+5|}{\sqrt{ 1^2+2^2}} = \cfrac{5}{\sqrt 5} =\sqrt 5，故選\bbox[red, 2pt]{(D)}$$

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$$\color{blue}{\textbf{解}}：\\
\cases{f(x)=x^3 +2x^2-x+1 \\ g(x)=-x^2+6x-2} \Rightarrow f(x)-2g(x)=x^3 +2x^2-x+1+2x^2-12x+4\\ =x^3+4x^2 -13x+5，故選\bbox[red, 2pt]{(C)}$$

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解：
$$令f(x)=3x^3 +2x^2-5x+7 =a(x+2)^3 +b(x+2)^2 +c(x+2)+d\\ \Rightarrow \cases{f(-2)=-24+8 +10+7 =d \\ f(-3) =-81+18+15+7 =-a+b-c+d} \Rightarrow \cases{d=1 \\ a-b+c-d= 41} \\\Rightarrow a-b+c =41+d=42 \Rightarrow a-b+c+d =42+1=43，故選\bbox[red,2pt]{(D)}$$

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解：

$$f(x)=x^3 +ax^2+bx+4 \Rightarrow \cases{f(-1)=-1+a-b+4=0 \Rightarrow a-b=-3 \\ f(-2) =-8+4a-2b+4 =6 \Rightarrow 2a-b=5 } \Rightarrow \cases{a=8 \\ b=11}\\，故選\bbox[red,2pt]{(D)}$$

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$$\color{blue}{\textbf{解}}：\\\cfrac{2x}{(x+1)(x-1)(x+2)} = \cfrac{A}{x+1} +\cfrac{B}{x+2} +\cfrac{C}{x-1}\\ \Rightarrow A(x+2)(x-1) +B(x+1)(x-1) +C(x+1)(x+2) \Rightarrow x=1 \Rightarrow 2=6C \Rightarrow  C=1/3\\，故選\bbox[red, 2pt]{(B)}$$

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$$\color{blue}{\textbf{解}}：\\
(\sqrt 6-1)^2 = 7-2\sqrt 6 = 7-\sqrt{24} \Rightarrow \sqrt{7-\sqrt{24}} = \sqrt{(\sqrt 6-1)^2} =\sqrt 6-1，故選\bbox[red, 2pt]{(D)} $$

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解：$$ \cases{\alpha+\beta =3/2 \\ \alpha\beta = -2} \Rightarrow  (\alpha +\beta)\times \alpha\beta = {3\over 2}\times (-2)  =-3 ，故選\bbox[red,2pt]{(A)}$$

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$$\color{blue}{\textbf{解}}：\\
\begin{vmatrix}a & -b \\ c & d \end{vmatrix}= 10 \Rightarrow ad+bc=10 \Rightarrow  \begin{vmatrix}6a & 3b \\ -2c & d \end{vmatrix} =6ad+6bc =6(ad+bc) =6\times 10=60，故選\bbox[red, 2pt]{(C)}$$

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$$\color{blue}{\textbf{解}}：\\
\begin{vmatrix}5 & -1 & 2\\ 4& 2 & -1\\ 1& 4& 2 \end{vmatrix}=20+32+1 -4+8+20 =77，故選\bbox[red, 2pt]{(C)}$$

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$$\color{blue}{\textbf{解}}：\\
所圍區域如上圖，其頂點為\cases{A(2,0) \\ B(4,0) \\ C(0,6) \\ D(0,2)}，並令f(x,y)=2x+y-1 \Rightarrow \cases{f(A)=3 \\ f(B) =7 \\ f(C) =5 \\ f(D)=1}\\ \Rightarrow f(D)=7最大，故選\bbox[red, 2pt]{(B)}$$

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$$\color{blue}{\textbf{解}}：\\
3x+4< x^2  \Rightarrow x^2-3x-4 >0 \Rightarrow (x-4)(x+1)>0 \Rightarrow x>4或x<-1，故選\bbox[red, 2pt]{(A)}$$

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$$\color{blue}{\textbf{解}}：\\
\left(x^2 +{4\over y^2} \right)\left( {9\over x^2}+ y^2\right) = \left(x^2 +({2\over y})^2 \right)\left( ({3\over x})^2 + y^2\right) \ge \left( x \cdot {3\over x} +{2\over y} \cdot y\right)^2 = (3+2)^2=25\\ \Rightarrow \left(x^2 +{4\over y^2} \right)\left( {9\over x^2}+ y^2\right) \ge 25，故選\bbox[red, 2pt]{(C)}$$

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$$\color{blue}{\textbf{解}}：\\
{2x+2x+y \over 3} \ge \sqrt[3] {2x\cdot 2x \cdot y} \Rightarrow {4x+y \over 3} \ge \sqrt[3]{4x^2y} \Rightarrow \left({6\over 3}\right)^3 \ge 4x^2y \Rightarrow 2\ge x^2y，故選\bbox[red, 2pt]{(D)}$$

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解題僅供參考