103年身心障礙學生四技二專甄試-數學(S)-詳解

103 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(S)

單選題，共 20 題，每題 5 分

解答：$$f(x)=(x^3+1)(x-1)(7x-5)+4(x-3)+6 =(x^2-1)p(x)+ax+b\\ \Rightarrow \cases{f(1)=-2=a+b\\ f(-1)=-10=-a+b} \Rightarrow \cases{ a=4 \\b=-6 } \Rightarrow 餘式為4x-6，故選\bbox[red,2pt]{(B)}$$

解答：$$a={1\over 2} +{1\over 4} +{1\over 8} + \cdots +{1\over 64} ={1\over 2}\cdot {1-1/64\over 1-1/2} ={63\over 64} \Rightarrow 1-a= {1\over 64}，故選\bbox[red,2pt]{(B)}$$

解答：$$\cases{A(1,3)\\ B(2,5)\\ C(3,1)\\ D(5,x)} \Rightarrow \cases{\overrightarrow{AB} =(1,2)\\ \overrightarrow{CD} =(2,x-1)}，因此\overleftrightarrow{AB}\parallel \overleftrightarrow{CD} \Rightarrow {1\over 2}={2\over x-1} \Rightarrow x-1=4 \Rightarrow x=5\\，故選\bbox[red,2pt]{(C)}$$

解答：$${1\over 25^x} ={1\over 5^{2x}} =5^{-2x}= 125   =5^3 \Rightarrow -2x=3 \Rightarrow x=-{3\over 2}，故選\bbox[red,2pt]{(A)}$$

解答：$${\log 2+\log 50\over \log \sqrt{10}} ={\log 2+\log {100\over 2}\over {1\over 2}\log 10} = {\log 2+2-\log 2\over {1\over 2}} =4，故選\bbox[red,2pt]{(D)}$$

解答：$$\cases{A(0,1)\\ B(2,3)\\ C(-1,-5)} \Rightarrow \cases{\overrightarrow{AB} =(2,2)\\ \overrightarrow{AC}= (-1,-6)} \Rightarrow 平行四邊形面積=\sqrt{|\overrightarrow{AB}|^2|\overrightarrow{AC}|^2 -(\overrightarrow{AB} \cdot \overrightarrow{AC})^2} \\=\sqrt{8\cdot 37-(-14)^2} = \sqrt{100}=10，故選\bbox[red,2pt]{(A)}$$

解答：$$\cases{A(-2,3)\\ B(1,7)} \Rightarrow \overline{AB}= 5，再加上\overline{AP}=2 \Rightarrow \overline{BP} =3 \Rightarrow  P(x,y)=(2B+3A)/5=(-{4\over 5},{23\over 5})\\ \Rightarrow x+y={19\over 5}，故選\bbox[red,2pt]{(A)}$$

解答：$$直線3x-y=2的斜率為3，因此可假設兩股長度為a及3a，因此9a^2+a^2=10^2 \Rightarrow a=\sqrt{10}\\ \Rightarrow 三角形面積={1\over 2}a\cdot 3a={3\over 2}a^2={3\over 2}\cdot 10=15，故選\bbox[red,2pt]{(A)}$$

解答：$$\sin 1000^\circ = \sin (360^\circ\times 3-80^\circ) = -\sin 80^\circ =-\cos 10^\circ = -\sqrt{1-k^2}，故選\bbox[red,2pt]{(B)}$$

解答：$$C^9_2\times 2= 72，故選\bbox[red,2pt]{(C)}$$

解答：$$\cases{S_1的平均值\bar x_1=(1+2+3+4)/4= 2.5\\ S_2的平均值\bar x_2=(-3-1+0+1+3)/4= 0\\ S_3的平均值\bar x_3=(7 +8+9+10)/4= 8.5\\ S_4的平均值\bar x_4=(10.2+10.3 +10.4+ 10.5)/4= 10.35 }\\ \Rightarrow \cases{ \sum (x_i-\bar x_1)^2 = 1.5^2+ 0.5^2+ 0.5^2+ 1.5^2  \\\sum (x_i-\bar x_2)^2 = 3^2+ 1^2+1^2+ 3^2  \\\sum (x_i-\bar x_3)^2 = 1.5^2+ 0.5^2+ 0.5^2+ 1.5^2  \\ \sum (x_i-\bar x_4)^2 =0.15^2+ 0.05^2 +0.05^2+ 0.15^2  } \Rightarrow S_2的變異最大，即標準差最大，故選\bbox[red,2pt]{(B)}$$

解答：$$f(x)= 2x^3-3x^2+3x-1 = x^2(2x-3)+3x-1\\ \Rightarrow \cases{f(-1)=-5-3-1\ne 0\\ f(-1/2)={1\over 4}\cdot (-4)-{3\over 2}-1\ne 0\\ f(1/2)= {1\over 4}\cdot (-2)+{3\over 2}-1=0\\ f(1)=-1+3-1\ne 0} \Rightarrow 2x-1為其因式，故選\bbox[red,2pt]{(C)}$$

解答：$$\cos 75^\circ = \sin 15^\circ \Rightarrow \cos^2 15^\circ +\cos^2 45^\circ + \cos^2 75^\circ =\cos^2 15^\circ +\cos^2 45^\circ + \sin^2 15^\circ = 1+\cos^2 45^\circ\\ =1+{1\over 2}= {3\over 2}，故選\bbox[red,2pt]{(D)}$$

解答：

$$令C為\overline{AB}的中點，則\overline{AC}=\overline{AB}\div =6\div 2=3 \Rightarrow \overline{AC}=\sqrt{4^2-3^2}=\sqrt 7，故選\bbox[red,2pt]{(A)}$$

解答：$$\angle C=180^\circ -\angle A-\angle B=180^\circ-105^\circ-45^\circ =30^\circ\\再由正弦定理: {\overline{AB}\over \sin \angle C} ={\overline{AC}\over \sin \angle B} \Rightarrow {\overline{AB} \over \overline{AC}} ={\sin \angle C\over \sin \angle B} ={\sin 30^\circ\over \sin 45^\circ} ={1/2\over 1/\sqrt 2} ={1\over \sqrt 2}，故選\bbox[red,2pt]{(B)}$$

解答：$$103+104+\cdots+113= (113+103)\times 11\div 2=1188 \\\Rightarrow 平均值={1188-106\over 10} ={1082\over 10}=108.2，故選\bbox[red,2pt]{(C)}$$

解答：$$恰1支中獎=2支會中獎的籤取1支及2支不會中獎的籤取1支 \Rightarrow 機率為{C^2_1C^2_1\over C^4_2} ={4\over 6} ={2\over 3}\\，故選\bbox[red,2pt]{(D)}$$

解答：$$\cases{A(-1,2)\\ B(3,4)\\ C(5,-1)} \Rightarrow \cases{\vec u=\overrightarrow{BA} =(-4,-2)\\ \vec v=\overrightarrow{BC}= (2,-5)} \Rightarrow \cos \angle ABC = {\vec u\cdot \vec v\over |\vec u||\vec v|} ={1\over \sqrt{145}} \Rightarrow \tan \angle ABC= 12\\，故選\bbox[red,2pt]{(D)}$$

解答：$$x(x-7)+ y(y+24)=0 \Rightarrow x^2-7x+y^2+24y=0 \\\Rightarrow x^2-7x+{49\over 4}+y^2+24y+144= {49\over 4}+144 \Rightarrow (x-{7\over 2})^2+(y+12)^2 = {625\over 4}\\ \Rightarrow 圓面積={625\over 4}\pi，故選\bbox[red,2pt]{(D)}$$

解答：

$$令\cases{\angle CAD=\theta\\ D在\overline{BC}上且\overline{AD}\bot \overline{BC}}，則\angle BCD=60^\circ-\theta \Rightarrow \overline{AD} =\overline{AC}\cos\theta = \overline{AB}\cos (60^\circ-\theta) \\ \Rightarrow 5\cos\theta = 8\cos(60^\circ-\theta) =8(\cos 60^\circ\cos\theta +\sin 60^\circ\sin \theta) =8({1\over 2}\cos\theta +{\sqrt 3\over 2}\sin \theta) \\ =4\cos\theta +4\sqrt 3\sin\theta \Rightarrow \cos\theta =4\sqrt 3\sin\theta  \Rightarrow \tan\theta ={1\over 4\sqrt 3} \Rightarrow \cos\theta ={4\sqrt 3\over 7} \\ \Rightarrow \overline{AD} =\overline{AC}\cos\theta = 5\cdot {4\sqrt 3\over 7} ={20\sqrt 3\over 7}，故選\bbox[red,2pt]{(D)}$$

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