108學年度臺北市聯合轉學考-高職升高二-數學科詳解

臺北市高級中等學校 108 學年度聯合轉學考招生考試
高職數學科試題

 單選題：共25 題，每題 4 分，共 100 分，答錯不倒扣。

解：$$P\left({a\over b},b-a\right)在第四象限 \Rightarrow \begin{cases} a/b>0 \\ b-a<0 \Rightarrow a>b\end{cases} \Rightarrow  \begin{cases}b<a<0 & a<0\\a>b>0 & a > 0\end{cases} ， 故選：\bbox[red,2pt]{(B)}$$

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解：$$由(a-b)(a^2+ab+b^2)=a^3-b^3可知 \left(\sqrt[3]{5}-2\right) \left(\sqrt[3]{25}+2\sqrt[3]{5}+4\right) =(\sqrt[3]{5})^3-2^3=5-8=-3 \\， 故選\bbox[red,2pt]{(A)}$$

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解：
$$(x^3-2x^2+3x-1)(x^2-x+1)=x^5+(-1-2)x^4+(1+2+3)x^3+(-2-3)x^2+(3+1)x-1\\ =x^5-3x^4+6x^3-5x^2+4x-1=x^5+ax^4+bx^3+cx^2+dx-1\\ \Rightarrow \begin{cases} a=-3\\b=6 \end{cases} \Rightarrow a+2b=-3+12=9， 故選\bbox[red,2pt]{(C)}$$

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解：$${3x+4 \over 2x-3}+ {2-x\over 3-2x} ={3x+4 \over 2x-3}+ {x-2\over 2x-3} = {4x+2\over 2x-3}， 故選\bbox[red,2pt]{(B)}$$

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解：$$\sin 320^\circ=\sin (360^\circ-40^\circ)= \sin (-40^\circ) =-\sin 40^\circ = -\cos 50^\circ= \cos (180^\circ+50^\circ) =\cos 230^\circ\\， 故選\bbox[red,2pt]{(A)}$$

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解：
$$\cos A=-{\sqrt 2\over 2} \Rightarrow \angle A=225^\circ \Rightarrow \triangle ABC={1\over 2}\overline{AB}\times \overline{AC}\times \sin A ={1\over 2}\times 4\times 12\times {\sqrt 2\over 2}= 12\sqrt 2\\，故選\bbox[red,2pt]{(C)}$$

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解：$$\begin{cases} \theta為第二象限角 \\ \sin \theta=3/4\end{cases}\Rightarrow \cos \theta = -{\sqrt 7 \over 4} \Rightarrow {3\csc \theta-1 \over 3+\sqrt 7 \cos \theta} ={3\times {4\over 3}-1 \over 3+\sqrt 7(-{\sqrt 7\over 4})}= {3\over 3-{7\over 4}} = {3\over {5\over 4}}=12/5\\，故選\bbox[red,2pt]{(B)}$$

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解：$$\angle A: \angle B:\angle C=1:4:1 \Rightarrow \begin{cases} \angle A=k \\ \angle B=4k \\ \angle C=k\end{cases},k為常數 \Rightarrow k+4k+k=180^\circ \Rightarrow k=30^\circ \Rightarrow \begin{cases} \angle A=30^\circ \\ \angle B=120^\circ \\ \angle C=30^\circ \end{cases}\\ 由{a\over \sin A}= {b\over \sin B}= {c\over \sin C} =2R \Rightarrow {a\over 1/2}= {b\over \sqrt 3/2}= {c\over 1/2} =2R \Rightarrow \begin{cases} a=R \\ b= \sqrt 3R \\ c= R \end{cases} \\ \Rightarrow {3a+b-c \over a+b+c} ={3R+\sqrt 3R-R \over R+\sqrt 3R+R} = {2+\sqrt 3 \over 2+\sqrt 3}=1，故選\bbox[red,2pt]{(D)} $$

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解：$$由三點坐標可知\angle C=90^\circ \Rightarrow \begin{cases} G=\left( {24-12+24 \over 3},{ -12-24-24\over 3}\right) =(12,-20)\\ P=\left({24-12\over 2} ,{-12-24 \over 2}\right) =(6,-18)\end{cases} \Rightarrow \\\overline{PG}= \sqrt{6^2+2^2} =2\sqrt{10}，故選\bbox[red,2pt]{(A)} $$

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解：

$$\theta=40^\circ \Rightarrow c>b>a \Rightarrow \begin{cases} \csc 40^\circ =c/a \\ \sec 40^\circ =c/b \\ \tan 40^\circ=a/b \end{cases} \Rightarrow \csc 40^\circ> \sec 40^\circ >\tan 40^\circ，故選\bbox[red,2pt]{(A)}$$

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解：

$$\begin{cases}L_1:12x-5y=m，且過(99,-37) \Rightarrow m=12\times 99+5\times 37\\ L_2:12x-5y=n，且過(34,-37) \Rightarrow n=12\times 34+5\times 37\end{cases} \Rightarrow \text{dist}(L_1,L_2)={ m-n\over \sqrt{12^2+5^2}} \\ ={12(99-34) \over 13} = {12\times 65 \over 13} =60，故選\bbox[red,2pt]{(C)}$$

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解：

$$\overrightarrow {OA} \cdot \overrightarrow{OB}= |
\overrightarrow {OA}| \cdot |\overrightarrow{OB}| \cos\theta \Rightarrow 3\sin 105^\circ=3\sqrt{\sin^2 105^\circ+\cos^275^\circ}\cos\theta \\ \Rightarrow \cos \theta =\cfrac{ \sin 105^\circ }{\sqrt{\sin^2 105^\circ+\cos^275^\circ}}=\cfrac{ \sin 105^\circ }{\sqrt{1-\cos^2 105^\circ+\cos^2105^\circ}}= \sin 105^\circ=\sin (90^\circ+15^\circ) \\=\cos 15^\circ \Rightarrow \theta=15^\circ，故選\bbox[red,2pt]{(A)}$$

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解：

假設F為原點(0,0,0)，各頂點坐標如上圖；
$$(A)\overrightarrow{AB}\cdot \overrightarrow{AE} =(0,0,-2)\cdot (0,-2,0)=0\\
(B)\overrightarrow{BD}\cdot \overrightarrow{AC} =(2,0,2)\cdot (2,0,-2)=0\\
(C)\overrightarrow{AD}\cdot \overrightarrow{DE} =(2,0,0)\cdot (-2,-2,0)= -4\\
(D)\overrightarrow{DH}\cdot \overrightarrow{DG} =(0,-2,0)\cdot (0,-2,-2)=4\\，故選\bbox[red,2pt]{(D)}$$

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解：

$$\triangle ABD:\triangle ADC=2:1 \Rightarrow \overline{BD}: \overline{DC}=2:1 \Rightarrow D=\left ({2\times 3+(-3)\over 3},{2\times 2+(-1) \over 3}\right) =(1,1)\\ \Rightarrow 過A(-1,3)及D(1,1)的直線方程式為x+y=2，故選\bbox[red,2pt]{(B)}$$

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解：

$$6\cos^2 x-5\sin x-6=0 \Rightarrow 6(1-\sin^2x)-5\sin x-6=0 \Rightarrow 6\sin^2x+5\sin x=0\\  \Rightarrow \sin x(6\sin x+5)=0 \Rightarrow \sin x=-{5\over 6} \Rightarrow |\cos x|={\sqrt {11}\over 6}，故選\bbox[red,2pt]{(D)}$$

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解：$$\begin{cases}-x=2\pi \\kx=\pi\end{cases}  \Rightarrow k=-{1\over 2}，故選\bbox[red,2pt]{(C)}$$

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解：$$\left||\vec a|\cos \theta \right|= \left|\sqrt{24^2+10^2}\cdot \cos 120^\circ \right| =\left| 26\cdot (-{1\over 2})\right|=13，故選\bbox[red,2pt]{(A)}$$

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解：$$(\vec a+\vec b)\cdot (\vec a-\vec b)=|\vec a|^2-\vec a\cdot \vec b+\vec b\cdot \vec a-|\vec b|^2 = 0 \Rightarrow (\vec a+\vec b)\bot (\vec a-\vec b) \\\Rightarrow \begin{cases} 斜邊向量為(\vec a+\vec b)- (\vec a-\vec b)=2\vec b \\ 斜邊長度為 \sqrt{4^2+3^2}=5\end{cases} \Rightarrow |2\vec b|=5 \Rightarrow |\vec b|=5/2=|\vec a|，故選\bbox[red,2pt]{(C)} $$

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解：$$x^2+8x+c=0的二根分別為\alpha、\beta \Rightarrow \begin{cases}\alpha+\beta = -8\\\alpha\beta =c \end{cases} ，又\alpha-\beta=6；\\因此\begin{cases}\alpha+\beta = -8\\\alpha-\beta =6 \end{cases} \Rightarrow \begin{cases}\alpha = -1\\ \beta =-7 \end{cases} \Rightarrow c=\alpha\beta=7，故選\bbox[red,2pt]{(D)}$$

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解：$$f(-3)=9a^2+2=38 \Rightarrow a=\pm 2\\又f(x)有最大值 \Rightarrow a<0 \Rightarrow a=-2，故選\bbox[red,2pt]{(A)}$$

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解：$$令f(x)=2x^3+x^2-5x-3=a(x+1)^3+b(x+1)^2+c(x+1)+d \\  \Rightarrow f(-2)=-16+4+10-3=-a+b-c+d \Rightarrow 5=a-b+c-d，故選\bbox[red,2pt]{(D)}$$

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解：$$x^2-1是f(x)的因式 \Rightarrow  \begin{cases}f(1)=0 \\ f(-1)= 0\end{cases} \Rightarrow  \begin{cases}a+b+c+d=0\cdots (1) \\-a+b-c+d= 0\cdots(2)\end{cases} \\ \Rightarrow \begin{cases}(1)+(2) \Rightarrow b+d=0\cdots(3) \\(1)-(2) \Rightarrow a+c=0 \cdots(4)\end{cases} \xrightarrow {(4)\times 2+(3)} 2a+b= -(2c+d)\\又
f(x)除以x-2餘6 \Rightarrow f(2)=6 \Rightarrow 8a+4b+(2c+d)=6 \Rightarrow 8a+4b-(2a+b)=6\\ \Rightarrow 3(2a+b)=6 \Rightarrow 2a+b=2，故選\bbox[red,2pt]{(C)}$$

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解：$$6^x\times 8^y\times 9^z =(2\times 3)^x\times (2^3)^y\times (3^2)^z = 2^x\times 3^x \times 2^{3y}\times 3^{2z} =2^{x+3y}\times 3^{x+2z} =2^{11}\times 3^6\\  \Rightarrow  \begin{cases}x+3y = 11\\ x+2z=6\end{cases}  \xrightarrow{x,y,z皆為正整數} \begin{cases}x = 2\\ y =3 \\z=2\end{cases} \Rightarrow x+y+z=7，故選\bbox[red,2pt]{(D)}$$

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解：$$\log_{10}(100x) = \log_{10}100+\log_{10}x=2+\log_{10}x=2+2=4，故選\bbox[red,2pt]{(B)}$$

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解：
$$ \begin{vmatrix}7 & \sqrt{11}+\sqrt 7 \\\sqrt{11}-\sqrt 7 & 11 \end{vmatrix} =77-(\sqrt{11}+\sqrt 7)(\sqrt{11}-\sqrt 7) =77-(11-7)=73，故選\bbox[red,2pt]{(B)}$$

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解題僅供參考