114年身障升大學-數學A詳解

114 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

解答: $$y=f(x)= ax^2+bx+5  \Rightarrow \cases{f(2)= 4a+2+5=1\\ f(4)= 16a+4b+5=5} \Rightarrow \cases{a=1\\ b=-4}\\ \Rightarrow y=f(x)= x^2-4x+5 =(x-2)^2+1 \Rightarrow 頂點坐標(2,1)，故選\bbox[red, 2pt]{(B)}$$

解答: $$\cases{A:100\pm 5\%\\ B:200\pm 11\%} \Rightarrow \cases{100(1-5\%)\le A\le 100(1+5\%) \\ 200(1-11\%) \le B\le 200(1+11\%)} \Rightarrow \cases{95\le A\le 105\\ 178\le B\le 222} \\ \Rightarrow 273\le A+B\le 327 \Rightarrow 300-27\le A+B\le 300+27 \Rightarrow {27\over 300}=9\% ，故選\bbox[red, 2pt]{(C)}$$

解答: $$\cases{a_1=-7/2\\ a_2= -2} \Rightarrow 公差d=a_2-a_1= {3\over 2} \Rightarrow a_3=-2+{3\over 2}=-{1\over 2} \Rightarrow a_4=-{1\over 2}+{3\over 2}=1 \\ \Rightarrow 公比r={a_4\over a_3} =-2 \Rightarrow a_5= a_4\cdot r=-2 \Rightarrow a_6=-2\cdot -2 =4 \Rightarrow \cases{a_3=-1/2\\ a_4=1\\ a_5=-2\\ a_6=4} \Rightarrow a_5\lt -1\\，故選\bbox[red, 2pt]{(C)}$$

解答: $$抽出卡片數字依序為a,b,c，\\b=6\Rightarrow (a,b,c)=(7,6,8-11),(8,6,\{7,9-11\}), (9,6,\{7,8,10,11\}),(10,6,\{7-9,11\}),\\\qquad (11,6,7-10), 共有5\times 4=20種\\ b=8,10,也各有20種，因此b為偶數共有60種\\ 其中遞增的情形:(a,b,c)=(\{6,7\}8,9-11), (6-9,10,11),共有6+4=10種，\\因此機率為{10\over 60}={1\over 6}，故選\bbox[red, 2pt]{(D)}$$

解答: $$取\cases{a=3/2\\ b=9} \Rightarrow \cases{\log a= \log 3-\log 2=0.4771-0.301=0.1761\\ \log b=\log 9= 2\times 0.4771=0.9542} \\ \Rightarrow \cases{(A) 3\log a+\log b= 1.4825\\ (B)\log a+2\log b= 2.0845\\ (C) \log_a b =\log b/\log a \approx 5.4\\ (D)\log_b a= \log a/\log b \approx 0.18} ，故選\bbox[red, 2pt]{(C)}$$

解答: $$假設\cases{兌換卡包x個\\ 兌換桌遊y個},則x+2y=5 \Rightarrow \cases{(x,y)=(1,2)有C^6_1C^5_2= 60種組合\\(x,y) =(3,1)有C^6_3C^5_1= 100種組合\\ (x,y)=(5,0) 有C^6_5C^5_0 =6種組合}\\ 共有60+100+6=166 種組合，故選\bbox[red, 2pt]{(D)}$$

解答:

$$假設圓心O(a,a),直線y=2x與圓交於A,B兩點，P為\overline{AB}中點，則\overline{OP}={|a|\over \sqrt 5} \\ \Rightarrow \overline{OA}^2= \overline{PA}^2+ \overline{OP}^2 \Rightarrow 3^2=1+{a^2\over 5} \Rightarrow 2\sqrt{10}，故選\bbox[red, 2pt]{(C)}$$

解答: $$\cases{\vec a=(\cos \theta, \sin \theta)\\ \vec b=(-\sin \theta, \cos \theta)} \Rightarrow 3\vec a+ 4\vec b=(3\cos \theta-4\sin \theta, 4\cos\theta+ 3\sin \theta) \\ \Rightarrow (1,2)\cdot (3\vec a+ 4\vec b) =(1,2) \cdot(3\cos \theta-4\sin \theta, 4\cos\theta+ 3\sin \theta) =11\cos \theta+ 2\sin \theta\\ \Rightarrow 最大值=\sqrt{11^2+2^2} =\sqrt{125} =5\sqrt 5，故選\bbox[red, 2pt]{(A)}$$

解答: $$\begin{bmatrix} a& b\\ c& d\end{bmatrix} =\begin{bmatrix} \cos \theta& -\sin \theta\\ \sin \theta& \cos \theta\end{bmatrix} \Rightarrow ad+bc=\cos^2 \theta-\sin^2\theta= 2\cos^2 \theta-1 ={7\over 25} \\ \Rightarrow \cos^2 \theta={16\over 25} \Rightarrow a= \cos \theta = \pm {4\over 5}，故選\bbox[red, 2pt]{(C)}$$

解答: $$\begin{array}{r} 區間& 範圍& 數量 & 累積數量\\\hline 1& a\le 50& 35 & 35\\ 2& 50\lt a\le 100& 30 &65\\ 3& 100\lt a\le 150& 20 & 85\\ 4& 150\lt a\le 200& 10& 95\\ 5& 200\lt a& 5 & 100\\\hline \end{array} \Rightarrow 甲所得130萬元屬於第3區間\\ 若k=77,也在第3區間，與甲所得差距不可能大於50萬元 ，故選\bbox[red, 2pt]{(C)}$$

解答: $$直線L:(1+t,t,t), t\in \mathbb R \Rightarrow 方向向量\vec u=(1,1,1)\Rightarrow A(1,0,0)在直線L上\\ P(1,2,4) \Rightarrow \overrightarrow{AP} =(0,2,4) \Rightarrow \cos \theta ={\overrightarrow{AP} \cdot \vec u\over |\overrightarrow{AP}||\vec u|} ={6\over 2\sqrt{15}} \Rightarrow \sin \theta =\sqrt{2\over 5}\\ \Rightarrow 圓半徑r=|\overrightarrow{AP}|\sin \theta \Rightarrow 2\sqrt{5} \cdot \sqrt{2\over 5}= 2\sqrt 2 \Rightarrow 圓面積=r^2\pi =8\pi，故選\bbox[red, 2pt]{(B)}$$

解答: $$f(x)=x^3 +ax^2 \Rightarrow f'(x)=3x^2+2ax \Rightarrow f''(x)=6x+2a\\ 若f''(x)=0 \Rightarrow x=-{a\over 3} \Rightarrow f'(-{a\over 3})=-12 \Rightarrow {a^2\over 3}-{2a^2\over 3}=-12 \Rightarrow a=\pm 6 \\ \Rightarrow -{a\over 3}=\pm 2，故選\bbox[red, 2pt]{(B)}$$

解答: $$假設\cases{x=2^a\\ y=2^b} \Rightarrow \cases{2^a=2^b-2\\ 4^a=4^b-12} \Rightarrow \cases{x=y-2\\ x^2=y^2-12} \Rightarrow \cases{x-y=-2\\ (x^2-y^2)=(x+y)(x-y) =-12} \\ \Rightarrow \cases{x-y=-2\\x+y=6} \Rightarrow \cases{x=2 =2^a\\ y=4=2^b} \Rightarrow \cases{a=1\\ b=2} \Rightarrow a-b=-1故選\bbox[red, 2pt]{(B)}$$

解答: $$1^2\times{100\over 500}+3^2\times{50\over 500}+ 16\times{50\over 500} ={1350\over 500}=2.7，故選\bbox[red, 2pt]{(C)}$$

解答: $$\cases{ax+by=1 \\ cx+dy=2} \Rightarrow {a\over c}={b\over d}={1\over 2} \Rightarrow \cases{c=2a\\d=2b} \Rightarrow \cases{ax+by+ 2z=1\\ cx+dy+z=1\\ x+y-z=0} \Rightarrow \cases{ax +by+ 2z=1\\ 2ax+ 2by+z=1\\ z=x+y} \\ \Rightarrow \cases{(a+2)x+ (b+2)y=1\\ (2a+1)x+ (2b+1)y=1} \Rightarrow {a+2\over 2a+1} ={b+2 \over 2b+1} ={1\over 1} \Rightarrow a=b=1，故選\bbox[red, 2pt]{(A)}$$

解答: $$Q=(a,b) \Rightarrow \cases{|\overrightarrow{OP}+ \overrightarrow{OQ}|= 3\\|\overrightarrow{OP}- \overrightarrow{OQ}|= 7} \Rightarrow \cases{(a+2)^2+ (b+3)^2 =3^2\\ (a-2)^2+ (b-3)^2=7^2} \\ \Rightarrow \cases{a^2+4a+4+b^2+6b+9=9\\ a^2-4a+4+b^2-6b+9=49}兩式相加\Rightarrow a^2+b^2=16\\ \Rightarrow |\overrightarrow{OQ}| =\sqrt{a^2+b^2} =4，故選\bbox[red, 2pt]{(A)}$$

解答:

$$\triangle BAD面積={1\over 2}\cdot \overline{AB}\cdot \overline{AD}\sin \angle BAD \Rightarrow {1\over 2}\cdot 5\cdot 5\sin \angle BAD= 12 \Rightarrow \sin \angle BAD ={24\over 25} \\ \Rightarrow \cos \angle BAD =-{7\over 25} ={\overline{AB}^2+ \overline{AD}^2-\overline{BD}^2 \over 2\cdot \overline{AB} \cdot \overline{AD}} ={50-\overline{BD}^2\over 50} \Rightarrow \overline{BD}=8 \\ \cos \angle BCD=-\cos \angle BAD ={7\over 25} ={\overline{BC}^2+ \overline{CD}^2-\overline{BD}^2 \over 2\cdot \overline{BC}\cdot \overline{CD}} ={\overline{CD}^2- 39\over 10\overline{CD}} \Rightarrow \overline{CD}={39\over 5}，故選\bbox[red, 2pt]{(A)}$$

解答: $$\overrightarrow{OA} 與L的方向向量(1,a,b)平行，即 \overrightarrow{OA}=k(1,a,b)=(k, ak,bk) \\ \Rightarrow \overrightarrow{OA}\cdot (-4,2,4)=-4k+2ak+4bk=3 \Rightarrow a+2b={3+4k\over 2k} =m為一定值\\ \Rightarrow 4k=2mk \Rightarrow m=2，故選\bbox[red, 2pt]{(B)}$$

解答: $$Q,R在x+y+z=1上,可假設\cases{Q(a,b,1-a-b)\\ R(c,d,1-c-d)}\\ \Rightarrow \cases{\overrightarrow{OQ} \cdot \overrightarrow{OP} =(a,b,1-a-b) \cdot (1,1,0)=a+b=0 \\ \overrightarrow{OR}\cdot \overrightarrow{OP} =(c,d,1-c-d)\cdot (1,1,0)=c+d=0} \\\Rightarrow \cases{Q(a,-a,1) \\ R(c,-c,1)} \Rightarrow \cases{\overline{OQ} =\sqrt{2a^2+1} =3\\ \overline{OR} =\sqrt{2c^2+1} =3} \Rightarrow \cases{a=\pm 2\\ c=\pm 2} \Rightarrow \cases{Q(2,-2,1)\\ R(-2,2,1)} 或\cases{Q(-2,2,1)\\ R(2,-2,1)} \\ \Rightarrow \overrightarrow{OQ} \cdot \overrightarrow{OR} =-4-4+1=-7，故選\bbox[red, 2pt]{(B)}$$

解答: $$\cases{a_1x+a_2y+a_3z=m_1\\ b_1x+b_2y +b_3z=m_2\\ c_1x+c_2y+c_3z=m_3} \Rightarrow (1,0,0),(1,1,0),(2,4,1),(3,6,2)均為c_1x+c_2y+c_3z=m_3的解\\ \Rightarrow \cases{c_1=m_3\\ c_2+c_2=m_3\\ 2c_1+ 4c_2+ c_3 =m_3\\ 3c_1 +6c_2+ 2c_3=m_3} \Rightarrow \cases{c_1=m_3\\ c_2=0\\ c_3=-m_3} \Rightarrow m_3x-m_3z=m_3 \Rightarrow x-z=1，故選\bbox[red, 2pt]{(D)}$$

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