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2016年5月5日 星期四

105學年四技二專統測--數學(S)詳解

105 學年度科技校院四年制與專科學校二年制

統一入學測驗-數學(S)

解答:$$\cases{x+2=0\cdots(1)\\ x+y=0\cdots (2)},由(1)\Rightarrow x=-2代入(2)\Rightarrow -2+y=0 \Rightarrow y=2 \Rightarrow A(-2,2)\\ \Rightarrow \overline{AB} =\sqrt{(-3)^2+(-4)^2} =5,故選\bbox[red,2pt]{(B)}$$
解答:$$\cases{L_1與L_2有相同的y截距,即b_1=b_2 \gt 0\\ m_1\lt m_2且|m_1|\gt |m_2|} \Rightarrow |m_1|b_1 \gt |m_2|b_2,故選\bbox[red,2pt]{(D)}$$
解答:$$\cases{f(x)+g(x)=x^2-1 \cdots(1)\\ f(x)-g(x)= -x^2+2x+3\cdots(2)} \Rightarrow \cases{(1)+(2) \Rightarrow 2f(x)=2x+2\\ (1)-(2) \Rightarrow 2g(x)=2x^2-2x-4} \\ \Rightarrow \cases{f(x)= x+1\\ g(x)=x^2-x-2} \Rightarrow g(-1)=1+1-2=0,故選\bbox[red,2pt]{(A)}$$
解答:$$\cases{f(x)除以x-1之餘式為5 \\f(x)除以x^2之餘式為4} \Rightarrow \cases{f(1)=1+a+b+4=5 \Rightarrow a+b=0 \cdots(1)\\ f(x)=x^2(x^2+a)+bx+4 \Rightarrow bx+4=4 \cdots(2)}\\ 由(2)\Rightarrow b=0代入(1)\Rightarrow a=0 \Rightarrow f(x)=x^4+4 \Rightarrow f(2)= 16+4=20,故選\bbox[red,2pt]{(C)}$$
解答:$$假設直角三角形三邊長為a,b,25;又三邊長總和為60 \Rightarrow a+b+25=60 \Rightarrow a+b=35;\\再加上a^2+b^2=25^2 \Rightarrow (a+b)^2-2ab=25^2 \Rightarrow 35^2-2ab=25^2 \Rightarrow ab=300\\ \Rightarrow a(35-a)=300 \Rightarrow a^2-35a+300=0 \Rightarrow (a-20)(a-15)=0 \Rightarrow 短邊長度為15,故選\bbox[red,2pt]{(A)}$$
解答:$$\cases{(x+1)(x-3)\lt 0\\ (x-1)(x-5)\le 0} \Rightarrow \cases{-1\lt x\lt 3\\ 1\le x\le 5},兩式取交集可得1\le x\lt 3,故選\bbox[red,2pt]{(C)}$$
解答:$$將各頂點坐標代入\Rightarrow \cases{f(5,0)= 15\\ f(8,3)=27\\ f({10\over 3},10)=20\\ f(0,10)=10} \Rightarrow 最大值=27,故選\bbox[red,2pt]{(B)}$$
解答:$$\cases{a_5=2\\ a_6=a_5+4 =6} \Rightarrow 公比r=a_6/a_5= 6/2=3 \Rightarrow a_8 = a_6\cdot r^2=6\cdot 3^2=54,故選\bbox[red,2pt]{(D)}$$
解答:$$4^2\pi \cdot {\theta \over 2\pi} =\pi \Rightarrow \theta={\pi\over 8} \Rightarrow 弧長=1\cdot \theta ={\pi\over 8},故選\bbox[red,2pt]{(A)}$$
解答:$$960^\circ ={960\over 180}\pi ={16\over 3}\pi \Rightarrow 最大負同界角= {16\over 3}\pi-6\pi = -{2\over 3}\pi,故選\bbox[red,2pt]{(B)}$$
解答:$$等腰直角\triangle OAB,\overline{OA} =\overline{OB}=半徑=6 \Rightarrow \overline{AB}=6\sqrt 2 \Rightarrow \overline{PA} ={1\over 2}\overline{AB} =3\sqrt 2,故選\bbox[red,2pt]{(B)}$$
解答:$$\sin \theta +\cos \theta={\sqrt 3\over 2} \Rightarrow (\sin \theta +\cos \theta)^2= 1+2\sin\theta \cos\theta ={3\over 4} \Rightarrow \sin\theta \cos\theta =-{1\over 8}\\ \Rightarrow \sec \theta+\csc\theta ={1\over \cos\theta} +{1\over \sin \theta}= {\sin\theta +\cos\theta \over \sin\theta \cos\theta} ={\sqrt 3/2\over -1/8} =-4\sqrt 3,故選\bbox[red,2pt]{(B)}$$
解答
$$假設大樓最頂端為B,依題意\overline{OB}=508 \Rightarrow \overline{OA}=508\times \sqrt 3= 508\sqrt 3,故選\bbox[red,2pt]{(D)}$$
解答:$$\cases{A(2,-1)\\ B(-1,3)} \Rightarrow \overrightarrow{AB} =(-3,4) \Rightarrow \vec u={-{\overrightarrow{AB} \over |\overrightarrow{AB}|}} =-{(-3,4)\over \sqrt{(-3)^2+4^2}}= ({3\over 5},-{4\over 5}),故選\bbox[red,2pt]{(D)}$$
解答:$$\vec a\cdot \vec b=|\vec a||\vec b|\cos \theta  \Rightarrow {\vec a\cdot \vec b\over |\vec b|} =|\vec a|\cos \theta=4\times (-{1\over 3}) = -{4\over 3} \Rightarrow {|\vec a\cdot \vec b|\over |\vec b|} = {4\over 3},故選\bbox[red,2pt]{(C)}$$
解答:$$\cases{P(a,b)在x+3y-2=0上\\ d(P,4x+3y-2=0)=3} \Rightarrow \cases{a+3b-2=0 \cdots(1) \\ {|4a+3b-2|\over 5}=3 \cdots(2)},\\由(1)\Rightarrow 3b=2-a 代入(2) \Rightarrow {3a\over 5}=3\\ \Rightarrow a=5代回(1)\Rightarrow 5+3b-2=0 \Rightarrow b=-1 \Rightarrow a+b= 5-1=4,故選\bbox[red,2pt]{(D)}$$
解答:$$f(x)= ({6\over 0.2})^{-x} =30^{-x}= {1\over 30^x} \Rightarrow \cases{f(0)=1\\ f(x)\gt 0 \\ f(x)為遞減函數},故選\bbox[red,2pt]{(B)}$$
解答:$$\log_{10}0.375 = \log_{10} {375\over 1000} =\log_{10} 375-\log_{10} 1000 = \log_{10}(5^3\times 3)-3 = 3(1-\log_{10}2)+\log_{10} 3-3 \\ =3(1-0.301)+0.4771-3= -0.4259,故選\bbox[red,2pt]{(C)}$$
解答


$$圓心A(2,3)與水平線的距離=3+2=5 \Rightarrow C(2,3-5)=(2,-2) \Rightarrow 水平線為y=-2\\,故選\bbox[red,2pt]{(A)}$$
解答:$$直線與圓相切\Rightarrow 圓心至直線的距離=半徑\Rightarrow r=d((-3,-4),3x+4y=5)= {|-9-16-5|\over \sqrt{3^2+4^2}}\\ =6 \Rightarrow 圓方程式: (x+3)^2+(y+4)^2= 6^2,故選\bbox[red,2pt]{(A)}$$
解答:$$C^4_2=6,故選\bbox[red,2pt]{(C)}$$
解答:$$\cases{F,G,H,J任排有4!排法\\ F在最左邊有3!擺法} \Rightarrow F不能擺在最左邊的排法:4!-3!=18,故選\bbox[red,2pt]{(C)}$$
解答:$${4\times 4\over 6\times 6}={16\over 36}={4\over 9},故選\bbox[red,2pt]{(D)}$$
解答:$$出現3的機率皆為{1\over 6},故選\bbox[red,2pt]{(C)}$$
解答:$$\cases{甲+乙=92\\ 乙+丙=95\\ 丙+甲=83} ,三式相加\Rightarrow 2(甲+乙+丙)=270 \Rightarrow 甲+乙+丙=135\\ \Rightarrow {甲+乙+丙 \over 3}= {135\over 3}=45,故選\bbox[red,2pt]{(A)}$$
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