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2022年2月19日 星期六

102年身障生升大學-數學甲詳解

102 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:$$\cases{(A)60^\circ={\pi\over 3} \Rightarrow r\theta=10\cdot {\pi\over 3}={10\pi\over 3} \gt 10\\(B)r\theta =5\cdot 2=10\\ (C)r\theta =\pi\cdot \pi = \pi^2 \lt 10\\ (D)r\theta =30\pi\cdot {1\over 10}=3\pi \lt 10} \Rightarrow {10\pi\over 3}最大,故選\bbox[red,2pt]{(A)}$$
解答:$$\cases{抽中1黑1黃的次數:C^1_1C^2_1=2 \\抽中1黃1白的次數: C^2_1C^3_1=6 \\抽中1黑1白的次數:C^1_1C^3_1=3 } \Rightarrow 兩球不同色的機率=(2+6+3)/C^6_2= 11/15,故選\bbox[red,2pt]{(C)}$$
解答:$$只需考慮x^2,x及常數項,也就是(x^2+2x)(7x^5+ \sqrt 2x^4-x^3+3x^2-3x+2)(\sqrt 3x^3+x+1) \\\xrightarrow{化簡}(x^2+2x)(3x^2-3x+2)(x+1) =(x+2)(3x^3-3x^2+2x)(x+1)\\ \xrightarrow{化簡}(x+2)(-3x^2+2x)(x+1) =(x+2)(-3x^3-x^2+2x) \\\xrightarrow{化簡}(x+2)(-x^2+2x),x^2係數=2-2=0,故選\bbox[red,2pt]{(A)}$$
解答:$$\overrightarrow{OP} \bot \overrightarrow{OQ} \Rightarrow \overrightarrow{OP} \cdot \overrightarrow{OQ}=0 \Rightarrow (-4+2t,2+t)\cdot (7+t,20-8t) = -6t^2+14t+12=0\\ \Rightarrow -2(3t+2)(t-3)=0 \Rightarrow t=3 \Rightarrow P=(-4+6,2+3)=(2,5),故選\bbox[red,2pt]{(D)}$$
解答:$$f(\bar z)=\overline{f(z)} \Rightarrow f(2+i)=f(\overline{2-i}) =\overline{f(2-i)} =\overline{1+3i} =1-3i\\ \Rightarrow (2-i)f(2+i)=(2-i)(1-3i)= -1-7i,故選\bbox[red,2pt]{(D)}$$
解答:$$\vec n= \overrightarrow{OA} \times \overrightarrow{OB} =(1,2,3)\times (3,1,4)=5(1,1,-1) \Rightarrow 平面E:x+y-z=0\\ P在E上 \Rightarrow 6-t-9+2t+15-4t=0 \Rightarrow t=4,故選\bbox[red,2pt]{(D)}$$
解答:$$7\lt r\lt \sqrt{(10-5)^2+(14-7)^2} =\sqrt{74},由於8\lt \sqrt{74} \lt 9,因此r可以為8,故選\bbox[red,2pt]{(A)}$$
解答:$${36\pi\over 7} =4\pi+{8\pi\over 7} \Rightarrow {8\pi\over 7}=\pi+{\pi\over 7}在第三象限\Rightarrow \tan {36\pi\over 7} \gt 0\gt \sin{36\pi\over 7} \gt \cos {36\pi\over 7} \\\Rightarrow \csc {36\pi\over 7} \lt \sec {36\pi\over 7}\lt 0 \Rightarrow \csc {36\pi\over 7}最小,故選\bbox[red,2pt]{(D)}$$
解答:$$P(X=k)= C^n_k/2^n \Rightarrow {5\over 4}P(X=7)= P(X=8) \Rightarrow {5\over 4}C^n_7 =C^n_8 \Rightarrow {5\cdot n!\over 7!(n-7)!} ={4\cdot n!\over (n-8)!8!} \\ \Rightarrow {5\over n-7}={4\over 8 }\Rightarrow n=17,故選\bbox[red,2pt]{(C)}$$
解答:$$\cases{A(5,0)\\ B(4,3)\\ C(1,2)} \Rightarrow \cases{\vec u=\overrightarrow{AB} =(-1,3)\\ \vec v=\overrightarrow{AC} =(-4,2)} \Rightarrow \cos  A ={\vec u\cdot \vec v\over |\vec u||\vec v|} ={10\over \sqrt{10}\cdot \sqrt{20}} ={1\over \sqrt 2} \\\Rightarrow \sin A={1\over \sqrt 2},故選\bbox[red,2pt]{(B)}$$
解答:$$\cases{3x+y+2z=7 \cdots(1)\\ x+2y-z=6 \cdots(2)\\2x-y+5z=1 \cdots(3)\\ 4x+3y+z=13 \cdots(4)} \Rightarrow (1)+(2)=(4) \Rightarrow 刪掉(3)後剩下三式就有無限多解,故選\bbox[red,2pt]{(C)}$$
解答:$$\cases{P(2,\log_a 2)\\ Q(8,\log_a 8)} \Rightarrow \overline{PQ} 斜率={\log_a8-\log_a 2\over 8-2} ={\log_a 4\over 6} ={1\over 10} \Rightarrow \log_a 4={3\over 5} \Rightarrow a^{3/5}= 4\\ \Rightarrow {3\over 5}\log a = \log 4= 2\times 0.301 =0.602 \Rightarrow \log a={5\over 3}\times 0.602 \approx 1 \Rightarrow a\approx 10,故選\bbox[red,2pt]{(D)}$$
解答:$$\cases{|k-7| \lt 8\sqrt 2 \Rightarrow 7-8\sqrt 2\lt k\lt 7+8\sqrt 2 \Rightarrow k=-4,-3,\dots,18\\[1ex] k^2-4k-13\gt 0 \Rightarrow (k-2)^2\gt 17 \Rightarrow k=7,8,\dots 或-3,-4,\dots} \\ \Rightarrow k=-4,-3,7,8,\dots,18,共14個整數,故選\bbox[red,2pt]{(B)}$$
解答:$$\overline{AB}=\overline{AC} \Rightarrow \angle B=\angle C \Rightarrow \sin B=\sin C={1\over 3} \Rightarrow \cos B=\cos C={2\sqrt 2\over 3} \\ \Rightarrow \sin 2B =2\sin B\cos B=2\cdot {1\over 3}\cdot {2\sqrt 2\over 3} ={4\sqrt 2\over 9} \Rightarrow   \sin A =\sin(\pi-2B) =\sin 2B= {4\sqrt 2\over 9}\\ ,故選\bbox[red,2pt]{(D)}$$
解答:$$\cases{O(0,0,0)\\ A(1/2,2/3,3/4)\\ B(1,1,1)} \Rightarrow \cases{\vec u=\overrightarrow{OA} =(1/2,2/3, 3/4)\\ \vec v= \overrightarrow{OB} =(1,1,1)} \Rightarrow \vec n= \vec u\times \vec v = {1\over 12}(-1,3,-2)\\\Rightarrow E:x-3y+2z=0\\ 由於點Q與直線L都在E上,因此d(P,E)=\min\{d(P,Q),d(P,L),d(P,E)\}\\ 現在只需要比較d(P,E)與d(P,xz平面)哪個比較短?\\\cases{d(P,E)= \left|{3-6+10 \over \sqrt{14}} \right| ={\sqrt{14}\over 2} \\ d(P,xz平面)= 2} \Rightarrow d(P,E) \lt d(P,xz平面),故選\bbox[red,2pt]{(C)}$$
解答:$$M=\begin{bmatrix} a & b\\ c & d\end{bmatrix} \Rightarrow \cases{M\begin{bmatrix}1\\ 2\end{bmatrix} =\begin{bmatrix} a +2 b\\ c +2 d\end{bmatrix} = \begin{bmatrix}1\\ 0\end{bmatrix} \\[1ex]M\begin{bmatrix}2\\ 0\end{bmatrix} =\begin{bmatrix} 2a\\ 2c  \end{bmatrix} = \begin{bmatrix}1\\ 1\end{bmatrix}} \Rightarrow \cases{a=1/2\\ b=1/4\\ c=1/2 \\d=-1/4} \\ \Rightarrow M= \begin{bmatrix} 1/2 & 1/4\\ 1/2 & -1/4\end{bmatrix} \Rightarrow M^{-1} =\begin{bmatrix} 1 & 1\\ 2 & -2 \end{bmatrix},故選\bbox[red,2pt]{(B)}$$
解答:$$令\cases{O(0,0)\\ P(-2,7)} \Rightarrow \overleftrightarrow{OP}斜率=-3.5,因此過P的直線L不經第三象限,\\其斜率m需滿足-3.5\le m \le 0 \Rightarrow m=0,-1,-2,-3,共四個整數值,故選\bbox[red,2pt]{(A)}$$
解答:$$\cases{出現兩正面,機率為p^2,期望值為100p^2 \\出現兩反面,機率為(1-p)^2,期望值為150(1-p)^2 \\出現一正一反面,機率為2p(1-p),期望值為400p(1-p) } \\ \Rightarrow 期望值= 100p^2+ 150(1-p)^2+400p(1-p) = -150(p^2-{2\over3}p-1) =-150((p-{1\over 3})^2-{10\over 9})\\ \Rightarrow p={1\over 3}有極大值,故選\bbox[red,2pt]{(B)}$$
解答:$$假設本金a \Rightarrow a(1+0.2)^{10}= a(1+p)^5 = 10000 \Rightarrow (1+0.2)^{10}= (1+p)^5 \\\Rightarrow 1+p=(1+0.2)^2 =1.44 \Rightarrow p=0.44,故選\bbox[red,2pt]{(C)}$$
解答:$$z^7 =7 = 7(\cos 2k\pi+i\sin 2k\pi),k\in \mathbb{Z} \Rightarrow z^3= 7^{3/7}(\cos {6k\pi\over 7}+i\sin {6k\pi\over 7}) 在第四象限\\ \Rightarrow  {3\pi\over 2}\lt {6k\pi\over 7} \lt 2\pi,取k=2 \Rightarrow z^7=7(\cos 4\pi+i\sin 4\pi) \Rightarrow z=7^{1/7}(\cos {4\pi \over 7}+i\sin {4\pi \over 7}) \\ \Rightarrow {4\pi\over 7}在第二象限,故選\bbox[red,2pt]{(B)}$$
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