103年身障生升大學-數學甲詳解

103 學年度身心障礙學生升學大專校院甄試
甄試類(群)組別：大學組數學甲

單選題，共 20 題，每題 5 分

解答：$$E、F獨立 \Rightarrow P(E\cap F)=P(E)P(F) \Rightarrow P(E\cup F)= P(E)+P(F)-P(E)P(F)\\ = {1\over 5}+{2\over 7}-{1\over 5}\times {2\over 7} ={17\over 35} -{2\over 35} ={15\over 35} ={3\over 7}，故選\bbox[red,2pt]{(B)}$$

解答：$${C^6_1C^4_1C^2_1\over C^{12}_3} ={48\over 220} ={12\over 55}，故選\bbox[red,2pt]{(B)}$$

解答：$$令M=\begin{bmatrix} a & b\\ c& d\end{bmatrix} \Rightarrow\cases{M\begin{bmatrix} 1\\ 0\end{bmatrix} =\begin{bmatrix} a & b\\ c& d\end{bmatrix} \begin{bmatrix} 1\\ 0\end{bmatrix}=\begin{bmatrix} a\\ c\end{bmatrix} =\begin{bmatrix} 3\\ 7\end{bmatrix} \\[1ex] M\begin{bmatrix} 0\\ 1\end{bmatrix} =\begin{bmatrix} a & b\\ c& d\end{bmatrix} \begin{bmatrix} 0\\ 1\end{bmatrix}=\begin{bmatrix} b\\ d\end{bmatrix} =\begin{bmatrix} 9\\ 21\end{bmatrix}} \Rightarrow M=\begin{bmatrix} 3 & 9\\ 7& 21\end{bmatrix} \\ \Rightarrow \det(M)=63-63=0，故選\bbox[red,2pt]{(A)}$$

解答：$$令L:y=mx+a，因此\cases{通過(2,3)\\L的x截距等於8 }\Rightarrow \cases{3=2m+a\\ -a/m=8 \Rightarrow a=-8m} \Rightarrow 3=2m-8m\\ \Rightarrow m=-{1\over 2}，故選\bbox[red,2pt]{(A)}$$

解答：$$\cases{P(3,2,1)\\ Q(4,a,b)\\ E:3x-4y+z=8} \Rightarrow \cases{\overrightarrow{PQ}= (1,a-2,b-1)\\ E的法向量\vec n=(3,-4,1)} \Rightarrow \overrightarrow{PQ} \parallel \vec n \Rightarrow {1\over 3} ={a-2\over -4} ={b-1\over 1} \\ \Rightarrow \cases{a=2/3 \\ b= 4/3} \Rightarrow a+b=2，故選\bbox[red,2pt]{(C)}$$

解答：$$\cases{8\lt \sqrt{79}\lt 9 \\ 28\lt \sqrt{791} \lt 29} \Rightarrow |x|=9,10,\dots,28，共20個整數 \Rightarrow x=\pm 9,\pm 10,\dots, \pm 28，共40個整數\\，故選\bbox[red,2pt]{(C)}$$

解答：$$756= 7\times 3^3\times 2^2 \Rightarrow \log 756000 =\log 7\times 3^3\times 2^2\times 1000 =\log 7+3\log 3+2\log 2+3\\ =0.8451+3\times 0.4771+2\times 0.301+ 3= 5.8784，故選\bbox[red,2pt]{(B)}$$

解答：$$\cases{2x+3y -z=4\\ 4x+5y-5z=-2\\ -5x-6y+7z= 5} \Rightarrow M=\left[\begin{array}{rrr|r}2 & 3 & -1 & 4\\4 & 5 & -5 & -2\\-5 & -6 & 7 & 5\end{array}\right] \xrightarrow{-2r_1+r_2,2.5r_1+r_3} \left[\begin{array}{rrr|r}2 & 3 & -1 & 4\\0 & -1 & -3 & -10\\0 & \frac{3}{2} & \frac{9}{2} & 15\end{array}\right] \\ \xrightarrow{-r_2, (3/2)r_3}\left[\begin{matrix}2 & 3 & -1 & 4\\0 & 1 & 3 & 10\\0 & 1 & 3 & 10\end{matrix}\right] \Rightarrow 無窮多組解，故選\bbox[red,2pt]{(D)}$$

解答：$$x^2+y^2 +2x-3y=17 \Rightarrow (x+1)^2+(y-{3\over 2})^2 =17+1+{9\over 4} ={81\over 4} \\ \Rightarrow \cases{圓心(-1,3/2) 在第二象限\\ 半徑r=9/2} \Rightarrow 第二象限的對角，即第四象限面積最小，故選\bbox[red,2pt]{(D)}$$

解答：$${\overrightarrow{QP} \cdot \overrightarrow{QR} \over \overrightarrow{RP} \cdot \overrightarrow{RQ}} ={\overrightarrow{QP} \cdot (\overrightarrow{QP} + \overrightarrow{PR})\over \overrightarrow{RP} \cdot (\overrightarrow{RP} +\overrightarrow{PQ})} = {\overline{QP}^2 +\overrightarrow{QP} \cdot\overrightarrow{PR}\over \overline{RP}^2 +  \overrightarrow{RP} \cdot \overrightarrow{PQ}} = {\overline{QP}^2 \over \overline{RP}^2} = {(2\overline{PR})^2 \over \overline{RP}^2} =4，故選\bbox[red,2pt]{(D)}$$

解答：$$P(X=k)= C^3_k/8, k=0,1,2,3 \Rightarrow E(X)= \sum_{k=0}^3 \left(k^2+(3-k)^2\right)C^3_k/8 \\ = {1\over 8}(9+5\times 3+5\times 3+9)= {48\over 8}=6，故選\bbox[red,2pt]{(B)}$$

解答：$$4^x =2^{2x}=3^{x+3} \Rightarrow 2x\log 2=(x+3)\log 3 \Rightarrow 2x\log 2-x\log 3=3\log 3\\ \Rightarrow x(2\log 2-\log 3)= 3\log 3 \Rightarrow x={3\log 3\over 2\log 2-\log 3 } ={3\times 0.4771\over 2\times 0.301-0.4771} ={1.4313\over 0.1249} \approx 11.5\\，故選\bbox[red,2pt]{(B)}$$

解答：$$\cos \theta \gt \sin \theta \gt \tan \theta  \Rightarrow {5\over 4}\pi \lt \theta \lt 2\pi，故選\bbox[red,2pt]{(D)}$$

解答：$$x^4+3x^3+1 = (x^3+2x^2+ax +b)(x+1)+(-a-2)x^2+(-b-a)x+(1-b)\\(A)\times: 0\Rightarrow \cases{a=-2\\ b=-a=2\\ b=1}，矛盾 \\(B)\bigcirc: 由(A)知:\cases{a=-2\\ b=2 }\Rightarrow 餘式為1-b=-1非零常數\\ (C)\bigcirc:  \cases{a=-2\\ b\ne 2} \Rightarrow 餘式為(2-b)x+(1-b) \\(D)\bigcirc: a\ne -1 \Rightarrow 餘式為(-a-2)x^2+(-b-a)x+(1-b)\\，故選\bbox[red,2pt]{(A)}$$

解答：$$圖形沒有通過第二象限，一定是凹向下，因此\cases{a\lt 0\\ y截距\le 0} \Rightarrow 兩根之積=b/a \ge 0，故選\bbox[red,2pt]{(C)}$$

解答：$$令Q在x+y=0上且\overline{PQ}\bot \overline{OQ}，則\triangle PQO三個角為30^\circ-90^\circ-60^\circ\\，因此\overline{PQ} =\overline{OP}\times{\sqrt 3\over 2} =4\sqrt 3，故選\bbox[red,2pt]{(C)}$$

解答：$$令\overline{PQ} =\overline{QR} =\overline{RP}=a，則\triangle PQR={\sqrt 3\over 4}a^2 = \sqrt 3 \Rightarrow a=2；\\ 又\triangle PQS={\sqrt 3\over 2}= {1\over 2}\triangle PQR，取S=\overline{QR}中點，則\angle PSQ=90^\circ \Rightarrow \sin \angle PSQ=1，故選\bbox[red,2pt]{(D)}$$

解答：$$假設正方體邊長為a，則任二頂點距離只有三種數據:a、\sqrt 2 a及\sqrt 3a；\\ 而\overline{PR} =\sqrt{2}\overline{PQ}，因此邊長為5\sqrt 2，兩頂點最長的距離=5\sqrt 2\times \sqrt 3=5\sqrt 6，故選\bbox[red,2pt]{(B)}$$

解答：$$z=1+2i \Rightarrow z^2=-3+4i \Rightarrow \cases{O(0,0)\\ P(z)=(1,2)\\ Q(z^2)=(-3,4)} \Rightarrow \cases{\vec u= \overrightarrow{OP}=(1,2)\\ \vec v= \overrightarrow{OQ}=(-3,4)} \\ \Rightarrow \cos \angle POQ = {\vec u\cdot \vec v\over |\vec u||\vec v|} ={5 \over 5\sqrt{5} } ={1\over \sqrt{5}} \Rightarrow \sin \angle POQ = {2\over \sqrt{ 5}} ={2\sqrt 5\over 5}，故選\bbox[red,2pt]{(A)}$$

解答：$$P(2,4),Q(4,3),R(x,y)共線\Rightarrow \overrightarrow{PQ}\parallel \overrightarrow{PR} \Rightarrow (2,-1)\parallel (x-2,y-4) \Rightarrow {2\over x-2}={-1\over y-4}\\ \Rightarrow x+2y=10 \Rightarrow R(10-2t,t),t\in \mathbb{R};\\又\overline{OQ}為\angle POR的角平分線 \Rightarrow {\overline{OP} \over \overline{OR}} ={\overline{PQ} \over \overline{QR}} \Rightarrow {2\sqrt 5\over \sqrt{(2t-10)^2+t^2}} ={\sqrt 5\over \sqrt{(2t-6)^2+(t-3)^2}} \\ \Rightarrow 4(2t-6)^2+4(t-3)^2 =(2t-10)^2 +t^2 \Rightarrow 3t^2-16t+16=0 \Rightarrow (3t-4)(t-4)=0\\ \Rightarrow t={4\over 3}(t=4 \Rightarrow R=P) \Rightarrow R的x坐標=10-2t =10-{8\over 3} ={22\over 3}，故選\bbox[red,2pt]{(C)}$$

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