112年身障升大學-數學A詳解

112 學年度身心障礙學生升學大專校院甄試

甄試類(群)組別：大學組-數學 A

單選題，共 20 題，每題 5 分

解答：$$2,6, 10公差為4,即公比r=4 \Rightarrow a_3 = a_1r^2 = 2\cdot 4^2=32，故選\bbox[red,2pt]{(D)}$$

解答：$$\cases{y\gt x^2\\ y\le -3x} \Rightarrow x^2\lt y\le -3x \Rightarrow x^2\lt -3x \Rightarrow x(x+3)\lt 0 \Rightarrow -3\lt x\lt 0\\ \Rightarrow \cases{x=-1 \Rightarrow 1\lt y\le 3\\ x=-2 \Rightarrow 4\lt y \le 6} \Rightarrow (x,y)=(-1,2),(-1,3),(-2,5),(-2,6),共四點，故選\bbox[red,2pt]{(D)}$$

解答：$$\cases{ 125-15\le x\le 125+15\\100-15\le y\le 100+15  } \Rightarrow \cases{110\le x\le 140\\ 85\le y\le 115} \Rightarrow 110-115 \le x-y \le 140-85 \\ \Rightarrow -5\le x-y\le 55 \Rightarrow 0\le |x-y|\le 55，故選\bbox[red,2pt]{(A)}$$

解答：$$f(x)=(x-1)(x^2+x+1)p(x)+ x^2+bx+c\\=(x-1)(x^2+x+1)p(x)+ (x^2+x+1)+(b-1)x+(c-1) \\ =(x^2+x+1)((x-1)p(x)+1)+ (b-1)x+(c-1)\\ 由於f(x)除以x^2+x+1的餘式為2x+1 \Rightarrow (b-1)x+(c-1)=2x+1\\ \Rightarrow \cases{b=3\\ c=2} \Rightarrow f(1)=1+b+c=1+3+2= 6，故選\bbox[red,2pt]{(C)}$$

解答：$$餘弦定理：\cos \angle A={8^2+7^2-\overline{BC}^2 \over 2\cdot 8\cdot 7} ={113-\overline{BC}^2 \over 112}\\ 又60^\circ \lt \angle A\lt 120^\circ \Rightarrow -{1\over 2}\lt \cos \angle A\lt {1\over 2} \Rightarrow -{1\over 2}\lt {113-\overline{BC}^2 \over 112}\lt {1\over 2} \\ \Rightarrow -56\lt 113-\overline{BC}^2 \lt 56 \Rightarrow 57\lt \overline{BC}^2 \lt 169 \Rightarrow \sqrt{57}\lt \overline{BC}\lt 13 \Rightarrow 8\le \overline{BC}\lt 13 \\\Rightarrow \overline{BC}=8,9,10,11,12，共五個整數，故選\bbox[red,2pt]{(B)}$$

解答：$$\log_a b為整數\Rightarrow (a,b)=(2,4), (2,8),(2,16),(3,9),(4,16)，共五組\\ (a,b)共有20\times 19=380種組合，因此機率={5\over 380}={1\over 76}，故選\bbox[red,2pt]{(B)}$$

解答：$$\cases{\log x=1+2a\\ \log y=2+a} \Rightarrow a={\log x-1\over 2}= \log y-2 \Rightarrow \log x = 2\log y-3 \\\Rightarrow \log x=\log y^2-\log 10^3= \log{y^2\over 10^3} \Rightarrow x={y^2\over 10^3} \Rightarrow y^2 = 1000x，故選\bbox[red,2pt]{(D)}$$

解答：$$\cases{P(紅,紅)=1/16\\ P(白,白)=9/16} \Rightarrow \cases{兩球相同機率＝10/16\\ 兩球相異機率=6/16} \Rightarrow 期望值＝400\cdot {10\over 16}+ 200\cdot {6\over 16} \\= 250+75= 325，故選\bbox[red,2pt]{(C)}$$

解答：$$\cases{\overline{OP}= \sqrt{13} \\ \overrightarrow{OP} \cdot (0,1)=2} \Rightarrow \cases{a^2+b^2 =13\\ b=2} \Rightarrow a^2=9 \Rightarrow a=-3(a=3不合,P在第二象限)\\ \Rightarrow \overrightarrow{OP} \cdot (2,-1) =(-3,2) \cdot (2,-1) =-6-2=-8，故選\bbox[red,2pt]{(A)}$$

解答：$$A:\cases{x=n\\ y=10^n} \Rightarrow x越大則y越大\Rightarrow r_a\gt 0\\ B:\cases{x=n \\ y=\log 10^n= n} \Rightarrow x=y \Rightarrow r_b=1\\ C:\cases{x=n\\ y=1/7^n} \Rightarrow x越大y 越小 \Rightarrow r_c\lt 0\\ 因此r_c\lt r_a\lt r_b，故選\bbox[red,2pt]{(C)}$$

解答：$$(\cos\theta +1)^2 +(\sin \theta-2)^2 = 6 \Rightarrow \cos^2\theta +2\cos \theta+1 +\sin^2 \theta-4\sin \theta +4=6\\ \Rightarrow 2\cos \theta-4\sin \theta =0 \Rightarrow \tan \theta = {\sin \theta\over \cos\theta }={2\over 4} ={1\over 2}，故選\bbox[red,2pt]{(A)}$$

解答：$$假設上個月\cases{牛奶每瓶a元\\ 布丁每個b元} \Rightarrow 這個月\cases{第1間店：牛奶每瓶a+15元，布丁每個b元 \\第2間店：牛奶每瓶a元，布丁每個b+15元 \\第3間店：牛奶每瓶a+2元，布丁每個b+10元 } \\ \Rightarrow \cases{x(a+15)+by =519 \\ax+y(b+15)=504 \\x(a+2)+y(b+10)=500} \Rightarrow \cases{ax+by +15x=519 \cdots(1)\\ ax+by+ 15y=504 \cdots(2)\\ ax+by+2x+10y=500 \cdots(3)} \\ 因此\cases{(1)-(3) \Rightarrow 13x-10y=19\\ (2)-(3) \Rightarrow -2x+5y=4 } \Rightarrow \cases{x=3\\ y=2} \Rightarrow 2x+y = 6+2=8，故選\bbox[red,2pt]{(C)}$$

解答：$$通過(1,0),(0,-3)的直線L：y=3x-3\\ 圓C：x^2+y^2-6x+8y+k=0 \Rightarrow (x-3)^2 +(y+4)^2= 25-k \Rightarrow \cases{圓心O(3,-4) \\ 半徑r=\sqrt{25-k}}\\ 圓C與L相切\Rightarrow d(O,L)=r \Rightarrow \sqrt{10} =\sqrt{25-k} \Rightarrow k=15，故選\bbox[red,2pt]{(D)}$$

解答：$$(3,-1)為兩多項式的對稱中心\Rightarrow \cases{f(x)=a(x-3)^3+ b(x-3)-1\\ g(x)=c(x-3)^3+ d(x-3)-1}\\ 又\cases{f首項係數=2\\ g首項係數=-1} \Rightarrow \cases{a=2\\ b=-1} \Rightarrow \cases{f(x) =2(x-3)^3 +b(x-3)-1\\ g(x)=-(x-3)^3+d(x-3)-1} \\ 再加上皆通過(5,3) \Rightarrow f(5)=g(5)=3 \Rightarrow \cases{f(5)=16+2b-1=3\\ g(5)= -8+2d-1=3} \Rightarrow \cases{b= -6\\ d=6} \\ \Rightarrow \cases{f(x)=2(x-3)^3-6(x-3)-1\\ g(x)=-(x-3)^3+ 6(x-3)-1} \Rightarrow \\ (A) \times: \cases{f(0)=-54+18-1\lt 0\\ g(0)=27-18-1\gt} \Rightarrow f(0)\not \gt g(0) \\ (B)\times: \cases{f(1)=-16+12-1=-5\\ g(1)= 8-12-1=-5} \Rightarrow f(1)\not \gt g(1) \\(C)\bigcirc: \cases{f(2)= -2+6-1=3\\ g(2)=1-6-1=-6} \Rightarrow f(2)\gt g(2)\\ (D)\times: \cases{f(4)= 2-6-1=-5\\ g(4)=-1+6-1=4} \Rightarrow f(4) \not \gt g(4)\\，故選\bbox[red,2pt]{(C)}$$

解答：$$\begin{vmatrix}a & 0 &  1\\ b& \log 2 & \log 5 \\ 1& 0 & 1\end{vmatrix} =a\log 2-\log 2=(a-1)\log 2 = 0 \Rightarrow a=1，故選\bbox[red,2pt]{(B)}$$

解答：$$E與2x-2y+z=3不相交 \Rightarrow 相互平行 \Rightarrow E:2x-2y+z= k\\ \Rightarrow\cases{d((1,-2,3), E)= {|9-k|\over 3} \\ d((-5,4,1), E)= {|-17-k|\over 3}} \Rightarrow |9-k|=|-17-k| \Rightarrow  k=-4\\ \Rightarrow E:2x-2y+z+4=0 \Rightarrow d((3,-2,1),E) ={|6+4+1+4|\over 3} =5，故選\bbox[red,2pt]{(A)}$$

解答：$$\cos A= {\overline{AB}^2+\overline{AC}^2- \overline{BC}^2\over 2\overline{AB}\cdot \overline{AC}} ={1+1-1/4\over 2\cdot 1\cdot 1} ={7\over 8} \\ 又\cos A= {\overrightarrow{AB} \cdot \overrightarrow{AC} \over |\overrightarrow{AB} || \overrightarrow{AC}|} \Rightarrow {7\over 8}={\overrightarrow{AB} \cdot \overrightarrow{AC} \over 1\cdot 1} \Rightarrow \overrightarrow{AB} \cdot \overrightarrow{AC}={7\over 8}\\ \cases{\overrightarrow{DC} =\overrightarrow{DA} +\overrightarrow{AC}=-{1\over 2}\overrightarrow{AB}+ \overrightarrow{AC} \\ \overrightarrow{EB} =\overrightarrow{EA} +\overrightarrow{AB} =-{1\over 2}\overrightarrow{AC} +\overrightarrow{AB}} \Rightarrow \overrightarrow{DC} \cdot \overrightarrow{EB} =\left(-{1\over 2}\overrightarrow{AB}+ \overrightarrow{AC} \right)\cdot \left( -{1\over 2}\overrightarrow{AC} +\overrightarrow{AB}\right) \\={1\over 4} \overrightarrow{AB} \cdot \overrightarrow{AC} -{1\over 2}-{1\over 2}+\overrightarrow{AB} \cdot \overrightarrow{AC} ={5\over 4}\overrightarrow{AB} \cdot \overrightarrow{AC}-1 ={5\over 4}\cdot {7\over 8}-1 = {3\over 32}，故選\bbox[red,2pt]{(D)}$$

解答：$$A\begin{bmatrix} 1 \\0\end{bmatrix}= \begin{bmatrix} \sqrt 3/2 \\1/2 \end{bmatrix} = \begin{bmatrix} \cos (\pi/6) \\ \sin(\pi/6)\end{bmatrix} \Rightarrow A=\begin{bmatrix} \cos (\pi/6) & -\sin(\pi/6) \\ \sin (\pi/6) & \cos (\pi/6) \end{bmatrix} \Rightarrow A^6=\begin{bmatrix} \cos (\pi) & -\sin(\pi) \\ \sin (\pi) & \cos (\pi) \end{bmatrix} \\=\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \Rightarrow A^6\begin{bmatrix} -3 \\2 \end{bmatrix} =\begin{bmatrix} 3 \\-2 \end{bmatrix}=\begin{bmatrix} a \\b \end{bmatrix} \Rightarrow a+b=3-2=1，故選\bbox[red,2pt]{(B)}$$

解答：$$全部有C^4_2\cdot C^5_3=60種組合\\豆腐和波菜一起被選有:C^4_2 \cdot C^3_1=18種組合\\ 牛肉和韭菜一起被選有：C^3_1\cdot C^4_2=18種組合\\豆腐和波菜一起被選且牛肉和韭菜有：C^3_1=3種組合\\ 因此符合要求的選法有60-18-18+3=27種，故選\bbox[red,2pt]{(C)}$$

解答：$$\cases{\overrightarrow{AB}=(1,0,0)\\ E之法向量\vec n=(1,-1,\sqrt 6)} \Rightarrow \cos \theta ={\overrightarrow{AB} \cdot \vec n\over |\overrightarrow{AB}||\vec n|} =\sqrt{1\over 8} \Rightarrow \sin \theta =\sqrt{7\over 8} \\ \Rightarrow 投影長=|\overrightarrow{AB}|\sin \theta = 1\cdot \sqrt{7\over 8}=\sqrt{7\over 8}，故選\bbox[red,2pt]{(D)}$$
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