113年身障升大學-數學A詳解

113 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:$$頂點坐標(3,2) \Rightarrow y-2=(x-3)^2 \Rightarrow y=x^2-6x+11 =x^2+ax+b\\ \Rightarrow \cases{a=-6\\ b=11} \Rightarrow a+b=5，故選\bbox[red, 2pt]{(C)}$$

解答:$$低分成績加分或高分成績減分可使標準差變小，故選\bbox[red, 2pt]{(B)}$$

解答:$$\cases{ 100\times 80\% \le A\le 100\times 120\% \\ 600\times 80\% \le B\le 600\times 120\%} \Rightarrow \cases{80\le A\le 120\\ 480\le B\le 720} \\ \Rightarrow {80\over 80+720}\le  {A\over A+B} =x \le {120\over 120+480} \Rightarrow 10\%\le x\le 20\%，故選\bbox[red, 2pt]{(A)}$$

解答:$$\log_a b,\log_b a,c成等比 \Rightarrow (\log_b a)^2=c\log_a b \Rightarrow c={(\log_b a)^2 \over \log_a b} =\cfrac{\left({\log a\over \log b} \right)^2}{\log b\over \log a} =\left({\log a\over \log b} \right)^3\\ =\left(\log_b a \right)^3，故選\bbox[red, 2pt]{(D)}$$

解答:$$假設\cases{R:紅球\\ W:白球} \Rightarrow \cases{前2次都是RR 機率={2\over 3}\cdot {2\over 3}={4\over 9}\\ 前2次都是WW: 機率={1\over 3}\cdot {1\over 3}={1\over 9}\\ 後2次為RW:機率={2\over 3}\cdot {1\over 3}={2\over 9}\\ 後2次為WR:機率={1\over 3}\cdot {2\over 3}={2\over 9}} \Rightarrow 機率合計=1\\ 又前2次相同且後兩次相異情形\cases{RRRW: 機率＝({2\over 3})^3\cdot {1\over 3}={8 \over 81}\\ RRWR: 機率＝({2\over 3})^3\cdot {1\over 3}={8\over 81}\\ WWRW:機率＝({1\over 3})^3\cdot {2\over 3}={2\over 81}\\ WWWR:機率＝({1\over 3})^3\cdot {2\over 3}={2\over 81}} \Rightarrow 機率合計={20\over 81}\\ \Rightarrow 欲求之機率=1-{20\over 81}={61\over 80}，故選\bbox[red, 2pt]{(C)}$$

解答:$$\bbox[red, 2pt]{送分}$$

解答:$$假設三種套餐為A,B,C,兩天共四餐,因此可能為2A1B1C,1A2B1C,1A1B2C 三種情形\\ 考量2A1B1C的排列數為4!/2=12,需扣除AABC,AACB, BCAA,CBAA四種情形,剩下8種\\ 其他1A2B1C及1A1B2C也是各有8種情形,因此共有24種，故選\bbox[red, 2pt]{(B)}$$

解答:$$\triangle ABC面積={1\over 2}\overline{AB}\cdot \overline{AC} \sin \angle A= {1\over 2}\cdot 5\cdot 10\sin \angle A=25\sin \angle A=20 \Rightarrow \sin \angle A={4\over 5} \\ \Rightarrow \cos \angle A=-{3\over 5} ={\overrightarrow{AB} \cdot \overrightarrow{AC}\over \overline{AB} \cdot \overline{AC}} ={\overrightarrow{AB} \cdot \overrightarrow{AC}\over 5 \cdot 10} \Rightarrow \overrightarrow{AB} \cdot \overrightarrow{AC}=-30，故選\bbox[red, 2pt]{(B)}$$

解答:$$A=\begin{bmatrix}-1 & 4 \\3 & 2 \end{bmatrix} \Rightarrow 行列式|A|=-2-12=-14為最小 \Rightarrow |2A|=2^2\times (-14)=-56，故選\bbox[red, 2pt]{(C)}$$

解答:$$假設\overline{BC}=a \Rightarrow \cos \angle BAC={\overline{AC}^2+ \overline{AB}^2-b^2\over 2\overline{AC}\cdot \overline{AB}} ={8^2+5^2-a^2\over 2\cdot 8\cdot 5} ={89-a^2\over 80}={1\over 2} \\ \Rightarrow a=7 \Rightarrow \cos \angle ABC= {\overline{AB}^2+ b^2-\overline{AC}^2 \over 2b\cdot \overline{AB}} ={5^2+7^2-8^2\over 2\cdot 7\cdot 5} ={10\over 70}={1\over 7}，故選\bbox[red, 2pt]{(A)}$$

解答:$$\cases{\cos \theta=a \\ \sin \theta =b} \Rightarrow \cases{\cos(2\theta)= \cos^2\theta-\sin^2\theta =a^2-b^2\\ \sin(2\theta)=2\sin\theta \cos \theta=2ab} \Rightarrow \cases{\cos(-2\theta)=\cos(2\theta)=a^2-b^2\\ \sin(-2\theta)=-2\sin(2\theta)=-2ab} \\ \Rightarrow 逆時針旋轉-2\theta矩陣=\begin{bmatrix}\cos(-2\theta) & \sin(-2\theta) \\-\sin(2\theta) & \cos(-2\theta) \end{bmatrix} = \begin{bmatrix}a^2-b^2 & -2ab \\2ab &a^2-b^2 \end{bmatrix}，故選\bbox[red, 2pt]{(D)}$$

解答:$$\overline{AB}= \overline{AC} \Rightarrow \angle C=\angle B={\pi\over 11} \Rightarrow \angle A=\pi-{\pi\over 11}-{\pi\over 11}={9\over 11}\pi\\ \overline{AD}= \overline{AC} \cos \angle A \Rightarrow {\triangle ADC\over \triangle ABC} ={\overline{AD}\cdot \overline{AC}\sin \angle A \over \overline{AB}\cdot \overline{AC}\sin \angle A} ={\overline{AD}\over \overline{AB}} ={\overline{AD}\over \overline{AC}} ={\overline{AC} \cos \angle A \over \overline{AC}} \\=\cos \angle A=\cos {9\over 11}\pi =\cos {2\over 11}\pi，故選\bbox[red, 2pt]{(C)}$$

解答:$$\cases{平面x＋y＝1的法向量\vec n_1=(1,1,0)\\ 平面x＋z＝3的法向量\vec n_2=(1,0,1)} \Rightarrow 直線L的方向向量=\vec n_1 \times \vec n_2=(1,-1,-1) \\ \Rightarrow L:{x-1\over 1} ={y-1\over -1}={z-1\over -1}=-t \Rightarrow (x,y,z)=(-t+1,t+1,t+1)，故選\bbox[red, 2pt]{(D)}$$

解答:$$圓心在直線y＝ax上,因此假設圓心P(t,at) \Rightarrow P至x軸的距離＝P至直線y=-ax距離\\ \Rightarrow |at|={|2at|\over \sqrt{a^2+1}} \Rightarrow a^2t^2 ={4a^2t^2\over a^2+1} \Rightarrow {4\over a^2+1}=1 \Rightarrow a=\sqrt 3，故選\bbox[red, 2pt]{(C)}$$

解答:$$\triangle ABC面積={1\over 2}|\overrightarrow{AB}\times \overrightarrow{AC}| ={1\over 2}\sqrt{4+16+4} =\sqrt 6 ={1\over 2} \overline{AB}\cdot \overline{AC} \sin \theta \Rightarrow \sin \theta={2\sqrt 6\over  \overline{AB}\cdot \overline{AC} } \\ \cos\theta ={ \overrightarrow{AB}\cdot \overrightarrow{AC} \over \overline{AB} \cdot \overline{AC} } ={-5\over \overline{AB} \cdot \overline{AC}} \Rightarrow \cos^2\theta +\sin^2 \theta=1 \Rightarrow {49\over (\overline{AB} \cdot \overline{AC})^2} =1 \\ \Rightarrow \overline{AB} \cdot \overline{AC}=7，故選\bbox[red, 2pt]{(D)}$$

解答:$$\cases{(x,y,z)=(0,2,3) \Rightarrow \cases{2b+3c=r\\ 2a_2+3a_3=s \cdots(1)\\ 3a_4=t} \\ (x,y,z)= (-1,-2,3) \Rightarrow \cases{-a_1-2b+3c=r\\ -2a_2+3a_3=s \cdots(2)\\ 3a_4=t}}\\ 由(1)及(2)\Rightarrow 2a_2+3a_3=-2a_2+3a_3 \Rightarrow 4a_2=0 \Rightarrow a_2=0，故選\bbox[red, 2pt]{(B)}$$

解答:$$f(x)=(x^2-3x+2)P(x)+k=(x-3)Q(x)+k \Rightarrow f(x)-k=a(x-1)(x-2)(x-3) \\ \Rightarrow f(x) =a(x-1)(x-2)(x-3)+k=a(x^3-6x^2+11x-6)+k \\ \Rightarrow f'(x)=a(3x^2-12x+11) \Rightarrow f''(x)=a(6x-12)=0 \Rightarrow x=2 \\ \Rightarrow 對稱中心坐標(2,f(2))=(2,k)，故選\bbox[red, 2pt]{(B)}$$

解答:$$故選\bbox[red, 2pt]{(D)}$$

解答:$$f(x)=3\cos{x\over 2}-4\sin {x\over 2} \Rightarrow f'(x)=-{3\over 2}\sin {x\over 2}-2\cos {x\over 2} =0 \Rightarrow  \tan {x\over 2}=-{4\over 3}\\ \Rightarrow \cases{\sin {x\over 2}=\pm 4/5\\ \cos {x\over 2}=\mp 3/5} \Rightarrow \sin x=2\sin {x\over 2}\cos {x\over 2}= -{24\over 25}，故選\bbox[red, 2pt]{(D)}$$

解答:$$E:2(x-4)+(y-1)-2(z-5)=0 \Rightarrow 2x+y-2z+1=0\\ \Rightarrow \cases{d(A,E)=7/3\\ d(B,E)=4/3 \\ d(C,E)=0\\ d(D,E)=1/3} \Rightarrow d(A,E) 最大 \Rightarrow A的投影點離P最近，故選\bbox[red, 2pt]{(A)}$$

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解題僅供參考，其他歷年試題及詳解