114 學年度身心障礙學生升學大專校院甄試
甄試類(群)組別：四技二專組-數學(C)
單選題，共 20 題，每題 5 分
:$$ABCD為平行四邊形,因此 \overrightarrow{AB} \parallel \overrightarrow{DC} 且\overrightarrow{DA} \parallel \overrightarrow{BC}\\ \Rightarrow \cases{A(-3,-2) \\B(2,-1) \\C(3,5) \\D(a,b)} \Rightarrow \cases{(5,1) \parallel (3-a,5-b) \\ (-3-a, -2-b) \parallel (1,6)} \Rightarrow \cases{{5\over 3-a}={1\over 5-b} \\ {-3-a\over 1} ={-2-b\over 6}} \Rightarrow \cases{a-5b=-22 \\ 6a-b=-16} \\ \Rightarrow \cases{a=-2\\ b=4}，故選\bbox[red, 2pt]{(B)}$$
解答:$$假設公差k,則\cases{a=b-k\\ c=b+k\\ d=b+2k} \Rightarrow a+b+c=3b=21 \Rightarrow b=7 \Rightarrow \cases{a=7-k \\ b=7\\ c=7+k\\ d=7+2k} \\ \Rightarrow abc=7(49-k^2)=280 \Rightarrow k=3\Rightarrow d=7+6=13，故選\bbox[red, 2pt]{(A)}$$
解答:$$假設A,B,C三人的帽子為a,b,c，則都拿錯的情形只有\cases{(A,b), (B,c),(C,b)\\ (A,c), (B,a), (C,b)}兩種情形\\，故選\bbox[red, 2pt]{(B)}$$
解答:$$AB+C= \begin{bmatrix} 1& 3\\ 5& 2\end{bmatrix} \begin{bmatrix} 4&7\\ 3& -1 \end{bmatrix} +\begin{bmatrix} 1& 2\\ -4& -7\end{bmatrix} =\begin{bmatrix} 13&4 \\ 26& 33 \end{bmatrix} +\begin{bmatrix} 1& 2\\ -4& -7\end{bmatrix} = \begin{bmatrix} 14 & 6\\ 22& 26\end{bmatrix}，故選\bbox[red, 2pt]{(D)}$$
解答:$${\sqrt[3]{ab^2} \over (a^{-1}b)^2} ={a^{1/3}b^{2/3} \over a^{-2}b^2} =a^{7/3}b^{-4/3} \Rightarrow \cases{p=7/3\\ q=-4/3}，故選\bbox[red, 2pt]{(C)}$$
解答:$$x+{1\over x}=2+ \sqrt 3 \Rightarrow (x+{1\over x})^2= (2+ \sqrt 3)^2 \Rightarrow x^2+{1\over x^2}+2 =7+4\sqrt 3 \\ \Rightarrow x^2+{1\over x^2}=5+ 4\sqrt 3，故選\bbox[red, 2pt]{(D)}$$
解答:$$a={2+\sqrt 3\over 2-\sqrt 3} \Rightarrow a+{1\over a}= {2+\sqrt 3\over 2-\sqrt 3}+ {2-\sqrt 3 \over 2+\sqrt 3} = (2+\sqrt 3)^2+ (2-\sqrt 3)^2 =14，故選\bbox[red, 2pt]{(A)}$$
解答:$$\begin{vmatrix}7& x\\ 2& 1 \end{vmatrix} =5 \Rightarrow 7-2x=5 \Rightarrow x=1 \Rightarrow \begin{vmatrix}x+1& 1& 2\\ 3& x& 1\\ 2& 3& x-1 \end{vmatrix} = \begin{vmatrix}2 & 1& 2\\ 3& 1& 1\\ 2& 3& 0 \end{vmatrix}=10，故選\bbox[red, 2pt]{(D)}$$
解答:$$\log_3 x+ \log_3 y=2 \Rightarrow \log_3(xy)=2 \Rightarrow xy= 3^2=9 \Rightarrow (x+y)^2= x^2+y^2+2xy=103+18=121\\ \Rightarrow x+y=\sqrt{121}=11，故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{\vec a=(1,-2,3) \\ \vec b=(1,1,0)} \Rightarrow \cases{\vec a+k\vec b= (k+1,k-2,3) \\ \vec a-k\vec b =(-k+1,-k-2,3)} \\ (\vec a+k\vec b)\bot (\vec a-k\vec b) \Rightarrow (\vec a+k\vec b)\cdot (\vec a-k \vec b) =0 \Rightarrow (k+1,k-2,3) \cdot (-k+1,-k-2,3) =0\\ \Rightarrow 1-k^2+4-k^2+9=0 \Rightarrow k^2=7 \Rightarrow k=\sqrt 7，故選\bbox[red, 2pt]{(A)}$$
解答:$$\cases{弟弟今年a歲\\ 哥哥今年4a歲} \Rightarrow 四年後\cases{弟弟a+4歲\\ 哥哥4a+4歲} \Rightarrow 4a+4=2(a+4) \Rightarrow 2a=4 \Rightarrow a=2 \\ \Rightarrow 哥哥今年4\times 2=8歲 \Rightarrow 哥哥比弟弟大8-2=6歲，故選\bbox[red, 2pt]{(A)}$$
解答:$$\cases{\triangle ABC: \cos \angle B=\displaystyle {5^2+10^2-7^2\over 2\cdot 5\cdot 10}={76\over 100} \\ \triangle ABP: \cos \angle B=\displaystyle {5^2+5^2-\overline{AP}^2 \over 2\cdot 5\cdot 5} ={50-\overline{AP}^2\over 50}} \Rightarrow {76\over 100}={50-\overline{AP}^2 \over 50} \\\Rightarrow \overline{AP}^2=12 \Rightarrow \overline{AP}=2\sqrt 3，故選\bbox[red, 2pt]{(D)}$$
解答:$$(1+i)^2= 2i \Rightarrow (1+i)^{10} =(2i)^5 =32i，故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{L_1: x+y-7=0\\ L_2: 2x-y+4=0\\ L_3: y=0} \Rightarrow \cases{A=L_1\cap L_2 =(1,6) \\B=L_1\cap L_3 =(7,0) \\ C= L_2\cap L_3= (-2,0)} \Rightarrow \triangle ABC面積={1\over 2}\cdot \overline{BC}\cdot d(A,L_3) \\={1\over 2}\cdot 9\cdot 6 =27，故選\bbox[red, 2pt]{(A)}$$
解答:
$$假設樹的底部為C點及樹的高度為h，則\cases{\tan 30^\circ =h/\overline{AC} \\ \tan 60^\circ = h/\overline{BC}} \\ \Rightarrow \cases{\overline{AC}= \sqrt 3h \\ \overline{BC} =h/\sqrt 3} \Rightarrow \overline{AC}-\overline{BC}=20=\sqrt 3h-{h\over \sqrt 3} \Rightarrow h=10\sqrt 3，故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{\theta 為銳角\\ \sin\theta =1/4} \Rightarrow \cos \theta ={\sqrt{15} \over 4} \Rightarrow {1\over 1-\cos \theta}-{1\over 1+\cos \theta} ={4\over 4-\sqrt{15}}-{4\over 4+ \sqrt{15}} \\=4(4+\sqrt{15})-4(4-\sqrt {15}) =8\sqrt{15}，故選\bbox[red, 2pt]{(C)}$$
解答:
$$(x-5)^2+(y+6)^2=36 \Rightarrow \cases{圓心O(5,-6) \\ 圓半徑r=6}, 假設\overline{AB}的中點為P,則\overline{OP} =d(O,L) =2\sqrt 5\\ \Rightarrow \angle APO=90^\circ \Rightarrow{\overline{AP}}^2+ \overline{OP}^2 =\overline{OA}^2 \Rightarrow \overline{AP}^2+ 20= 36 \Rightarrow \overline{AP}=4 \Rightarrow \overline{AB}=2 \overline{AP}=8\\，故選\bbox[red, 2pt]{(D)}$$
解答:$$\cases{x=2\cos \theta\\ y=3\sin \theta} \Rightarrow 2x-y=4\cos \theta-3\sin \theta=5\left( {4\over 5} \cos \theta-{3\over 5}\sin \theta\right) =5 \sin(\alpha-\theta) \Rightarrow 最大值=5\\，故選\bbox[red, 2pt]{(C)}$$
解答:$$f(x)=\sqrt x \Rightarrow f'(x)={1\over 2\sqrt x} \Rightarrow 切線斜率=f'(4)={1\over 4} \\\Rightarrow 切線: y={1\over 4}(x-4)+2 \Rightarrow x-4y+4=0，故選\bbox[red, 2pt]{(B)}$$
解答:$$y=x^2與y=2x交於O(0,0)及P(2,4)兩點, 所圍面積=\int_0^2 (2x-x^2)\,dx = \left. \left[x^2-{1\over 3}x^3 \right] \right|_0^2 \\=4-{8\over 3} ={4\over 3}，故選\bbox[red, 2pt]{(C)}$$
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解答:$$\cases{\theta 為銳角\\ \sin\theta =1/4} \Rightarrow \cos \theta ={\sqrt{15} \over 4} \Rightarrow {1\over 1-\cos \theta}-{1\over 1+\cos \theta} ={4\over 4-\sqrt{15}}-{4\over 4+ \sqrt{15}} \\=4(4+\sqrt{15})-4(4-\sqrt {15}) =8\sqrt{15}，故選\bbox[red, 2pt]{(C)}$$
解答:
解答:$$\cases{x=2\cos \theta\\ y=3\sin \theta} \Rightarrow 2x-y=4\cos \theta-3\sin \theta=5\left( {4\over 5} \cos \theta-{3\over 5}\sin \theta\right) =5 \sin(\alpha-\theta) \Rightarrow 最大值=5\\，故選\bbox[red, 2pt]{(C)}$$
解答:$$f(x)=\sqrt x \Rightarrow f'(x)={1\over 2\sqrt x} \Rightarrow 切線斜率=f'(4)={1\over 4} \\\Rightarrow 切線: y={1\over 4}(x-4)+2 \Rightarrow x-4y+4=0，故選\bbox[red, 2pt]{(B)}$$
解答:$$y=x^2與y=2x交於O(0,0)及P(2,4)兩點, 所圍面積=\int_0^2 (2x-x^2)\,dx = \left. \left[x^2-{1\over 3}x^3 \right] \right|_0^2 \\=4-{8\over 3} ={4\over 3}，故選\bbox[red, 2pt]{(C)}$$
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