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2022年2月9日 星期三

105年身心障礙學生四技二專甄試-數學(S)-詳解

105 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:$$將\cases{A(2,0)\\ B(-2,8)} 代入y=ax+b \Rightarrow \cases{2a+b=0\\ -2a+b=8} \Rightarrow \cases{a=-2\\ b=4} \Rightarrow y=-2x+4\\ \Rightarrow y-8=-2(x+2),故選\bbox[red,2pt]{(C)}$$
解答:$$重心=(A+B+C)\div 3 =((-2+2+3)\div 3,(1-1-3)\div 3)=(1,-1),故選\bbox[red,2pt]{(C)}$$
解答:$$此題相當於3個A、2個B、1個C排列數:{6!\over 3!2!} ={720\over 12} =60,故選\bbox[red,2pt]{(A)}$$
解答:$$2315\div 35\approx 66 \Rightarrow 23+66= 89,故選\bbox[red,2pt]{(C)}$$
解答:$$\cases{\vec u=(-3,1)\\ \vec v=(-2,-1)} \Rightarrow \cases{|\vec u|=\sqrt{10}\\ |\vec v|=\sqrt 5\\ \vec u\cdot \vec v=5} \Rightarrow \cos \theta ={\vec u\cdot \vec v\over |\vec u||\vec v|} ={5\over \sqrt{50}} ={1\over \sqrt 2} \Rightarrow \theta = 45^\circ,故選\bbox[red,2pt]{(D)}$$
解答:$$f(-1)=-2 \Rightarrow -2-a-3-7=-2 \Rightarrow a=-10,故選\bbox[red,2pt]{(B)}$$
解答:$$\sum_{k=0}^{10}(7k+2) = 7\sum_{k=0}^{10}k+ \sum_{k=0}^{10} 2 =7\times 55+11\times 2=407,故選\bbox[red,2pt]{(C)}$$
解答:$$\cases{\sin 833^\circ= \sin(360^\circ\times 2+113^\circ)= \sin 113^\circ \gt 0 \\ \sec 487^\circ = \sec(360^\circ +127^\circ)= \sec 127^\circ \lt 0} \Rightarrow P(\sin 833^\circ,\sec 487^\circ )在第四象限,故選\bbox[red,2pt]{(D)}$$
解答:$$將(2,-1)分別代入(A),(B),(D) ((C)的半徑不是5)\\(A)\times: (2-2)^2+(-1+1)^2= 0\ne 25 \\(B)\bigcirc:(2-2)^2 +(-1-4)^2= 25 \\(D)\times: (2+2)^2 +(-1-1)^2 = 20\ne 25,故選\bbox[red,2pt]{(B)}$$
解答:$${A_{k+1}-A_k\over A_k} =0.12 \Rightarrow A_{k+1}= 1.12A_k \Rightarrow A_k 為一等比級數,且公比為1.12 \Rightarrow A_{10} =1.12^9 A_1\\,故選\bbox[red,2pt]{(B)}$$
解答:$$\cases{P(2,5)\\ Q(a,b)\\ \overline{PQ}中點(3,7)} \Rightarrow \cases{3=(a+2)/2\\ 7=(b+5)/2} \Rightarrow \cases{a=4\\ b=9} \Rightarrow Q(4,9)\\ 令S(x,y),則\overrightarrow{SP} =\overrightarrow{RQ} \Rightarrow (2-x,5-y)=(4-5,9-10) =(-1,-1) \Rightarrow \cases{x=3\\y=6} \\ \Rightarrow S(3,6),故選\bbox[red,2pt]{(D)}$$
解答:$$ax^3+ bx^2-cx-9 = (x^2-2x+3)(3x-d)= 3x^3-(d+6)x^2 +(2d+9)x-3d\\ \Rightarrow \cases{a=3\\ b=-d-6\\ c=-2d-9\\ 9=3d} \Rightarrow \cases{a=3 \\b=\color{blue}{-9}\\ c=-15\\ d=3},故選\bbox[red,2pt]{(A)}$$
解答
$$等腰直角\triangle ABC \Rightarrow \overline{AB}=\overline{AC}(山高)=300\\ 又\angle D=30^\circ \Rightarrow \overline{AD} =\sqrt 3\cdot \overline{AC} =300\sqrt 3 \Rightarrow a=\overline{BD}=300\sqrt 3-300,故選\bbox[red,2pt]{(A)}$$
解答:$$(x-1)(x-3)=5 \Rightarrow x^2-4x-2=0 \Rightarrow x=2\pm \sqrt 6 \Rightarrow \alpha = 2+\sqrt 6 \Rightarrow \alpha-\sqrt 6=2,故選\bbox[red,2pt]{(C)}$$
解答:$$\left({13\over 17} \right)^{3x-1} =\left({289 \over 169} \right)^{2-x} = \left({17 \over 13} \right)^{4-2x} \Rightarrow 3x-1=2x-4 \Rightarrow x=-3,故選\bbox[red,2pt]{(A)}$$
解答:$$直線ax+2y=-6與坐標軸交於\cases{A(0,-3)\\ B(-6/a,0)} \Rightarrow 面積={1\over 2}\times |-3|\times |{-6\over a}|\lt 3 \\ \Rightarrow |{-6\over a}|\lt 2 \Rightarrow \cases{a=-4符合條件\\ a=-2,0,1皆不符合},故選\bbox[red,2pt]{(A)}\\本題的y\ge 0應該更正為y\le 0$$
解答
$$令\cases{直線L:3x-4y=25\\ C為\overline{AB}中點},則\cases{d(O,L)=5 \\\overline{AC}=24\div 2=12}\\ 直角\triangle OAC: r^2=5^2+ 12^2 =169 \Rightarrow r=13,故選\bbox[red,2pt]{(C)}$$
解答:$$\cases{抽中100元的機率=4x/100\\ 抽中1000元的機率=2x/100\\ 抽中10000元的機率=x/100} \Rightarrow 期望值=100\cdot {4x\over 100} +1000\cdot {2x\over 100}+ 10000\cdot {x\over 100} ={12400x\over 100} \\ 因此 800\le {12400x\over 100}\le 1100 \Rightarrow 80000\le 12400x \le 110000 \Rightarrow 6.45\le x\le 8.87,故選\bbox[red,2pt]{(B)}$$
解答:$$\log_2 (x^2+x+2)-\log_4 2=\log_2 (x+3)+\log_4 8 \\\Rightarrow \log_2 (x^2+x+2)-\log_2 \sqrt 2=\log_2 (x+3)+\log_2 \sqrt 8 \\\Rightarrow {x^2+x+2\over \sqrt 2} =\sqrt 8(x+3) \Rightarrow x^2+x+2 =4(x+3) \Rightarrow x^2-3x-10=0 \\ \Rightarrow (x-5)(x+2)=0 \Rightarrow x=5 (-2違反x為一正數),故選\bbox[red,2pt]{(D)}$$
解答:$$共有三場比賽:甲乙、乙丙、甲丙,每場結果有二種結果:勝或負,因此共有2^3=8種結果 \\ ,故選\bbox[red,2pt]{(D)}$$
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