113年身障升大學-數學B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 B

解答: $$\cases{A=\begin{bmatrix}1 & 1 \\0 & 2 \end{bmatrix} \\[1ex] B=\begin{bmatrix}2 & 0 \\0 & 1 \end{bmatrix}} \Rightarrow \cases{AB= \begin{bmatrix}2 & 1 \\0 & 2 \end{bmatrix} \\[1ex]BA =\begin{bmatrix}2 & 2 \\0 & 2 \end{bmatrix}} \Rightarrow AB-BA= \begin{bmatrix}0 & -1 \\0 & 0 \end{bmatrix}，故選\bbox[red, 2pt]{(A)}$$

解答: $$a_{n+1}=1+ra_n \Rightarrow a_5=1+ra_4 =1 +r(1 +ra_3) =1+ r+r^2a_3 =1+r+r^2(1+ra_2) \\= 1+r +r^2+r^3a_2=1+r +r^2+r^3( 1+ra_1) =1+r +r^2+r^3+r^4，故選\bbox[red, 2pt]{(C)}$$

解答: $$f(x)=2x^3-12x^2+25x-23 \Rightarrow f'(x)=6x^2-24x+25 \Rightarrow f''(x)=12x-24\\ f''(x)=0 \Rightarrow x=2 \Rightarrow 對稱中心坐標(2,f(2))，故選\bbox[red, 2pt]{(A)}$$

解答: $$抽中3球中，紅球個數多於白球的情形:\cases{3紅:機率=C^5_3/C^8_3= 10/56\\2紅1白機率=C^5_2C^3_1/C^8_3 = 30/56} \\ \Rightarrow 紅球多於白球的機率={10\over 56}+{30\over 56}={40\over 56} ={5\over 7}，故選\bbox[red, 2pt]{(B)}$$

解答: $$\log_{1/2}8 +\log_2 \sqrt 2+\log_{16} 4+ \log_2 16+\log_4 {1\over 4} ={\log 8\over \log {1\over 2}} +{\log \sqrt 2\over \log 2} +{\log 4\over \log 16} +4+{\log {1\over 4}\over \log 4} \\={3\log 2\over -\log 2}+{{1\over 2}\log 2\over \log 2}+ {2\log 2\over 4\log 2}+4+{-\log 4\over \log 4} =-3+{1\over 2}+{1\over 2}+4-1=1，故選\bbox[red, 2pt]{(D)}$$

解答: $$x^2+y^2+4x-6y-8=0 \Rightarrow (x^2+4x+4)+(y^2-6y+9)-8-4-9=0 \\ \Rightarrow (x+2)^2+(y-3)^2=(\sqrt {21})^2 \Rightarrow \cases{圓心P(-2,3)在第二象限\\ 圓半徑r=\sqrt{21}} \\ \Rightarrow 圓心P至(0,0)距離=\sqrt{13}\lt r \Rightarrow (0,0)在圓內部，故選\bbox[red, 2pt]{(A)}$$

解答: $$\left|\sqrt2 x-\sqrt{18} \right| \le \sqrt{210} \Rightarrow (\sqrt 2x-\sqrt{18})^2 \le (\sqrt{210})^2 \Rightarrow 2x^2-12x+18 \le 210 \\ \Rightarrow 2x^2-12x-192=0 \Rightarrow 2(x-(3-\sqrt{105})) (x-(3+\sqrt{105})) \le 0 \\ \Rightarrow 3-\sqrt{105} \le x\le 3+\sqrt{105} \Rightarrow x=-7,-6,...,13,共21個整數，故選\bbox[red, 2pt]{(C)}$$

解答: $$\cases{\triangle ABC: \cos \angle B=\cfrac{\overline{AB}^2+ \overline{BC}^2-\overline{AC}^2}{2\cdot \overline{AB} \cdot \overline{BC}} =\cfrac{16+16-8}{2\cdot 4\cdot 4}\\ \triangle BAD: \cos \angle B=\cfrac{\overline{AB}^2 +\overline{BD}^2-\overline{AD}^2}{2\cdot \overline{AB}\cdot \overline{BD}} =\cfrac{16+9-\overline{AD}^2}  {2\cdot 4\cdot 3}} \\ \Rightarrow {24\over 32} ={25-\overline{AD}^2 \over 24} \Rightarrow 18=25-\overline{AD}^2 \Rightarrow \overline{AD}=\sqrt 7，故選\bbox[red, 2pt]{(D)}$$

解答: $$f(x)+g(x)=(x-2)(f(x)-g(x)) \Rightarrow (x-1)g(x)=(x-3)f(x) \\\Rightarrow g(x)=(x-3)\cdot {f(x)\over x-1} \Rightarrow x-3為g(x)的因式，故選\bbox[red, 2pt]{(D)}$$

解答: $$\cases{2^a=3 \\ 3^b=6} \Rightarrow \cases{a=\log_2 3\\ b=\log_3 6} \Rightarrow ab={\log 3\over \log 2}\cdot {\log 6\over \log 3} ={\log 6\over \log 2}=\log_2 6 =1+\log_2 3=1+a\\ \Rightarrow ab=a+1，故選\bbox[red, 2pt]{(C)}$$

解答: $$考慮數字變化較小,也就是離平均值較近的數字，故選\bbox[red, 2pt]{(B)}$$

解答: $$-{1\over 2}+{1\over 4}=-{1\over 4}\lt 0 \Rightarrow D在\triangle ABC外部,又-{1\over 2}\overrightarrow{BA}={1\over 2}\overrightarrow{AB}, 因此D在\angle A的內部，故選\bbox[red, 2pt]{(B)}$$

解答: $$\cases{A(1,4,2)\\ B(-2,2,-4)\\ C(5,-3, 3)\\ D(-4,0,-5)} \Rightarrow \cases{ A'(0,4,2)\\ B'(0,2,-4)\\ C'(0,-3, 3)\\ D'(0,0,-5)} \Rightarrow \cases{\overline{OA'}=\sqrt{20} \\\overline{OB'} =\sqrt{20} \\\overline{OC'} =\sqrt{18} \\\overline{OD'}=\sqrt{25} } \Rightarrow \overline{OC'}最短，故選\bbox[red, 2pt]{(C)}$$

解答: $$f(x)=\sin \left(\cfrac{5}{2}\pi x \right) \Rightarrow f({63\over 5}) =\sin{63\over 2}\pi=-1\\ (A) \times:f(0)=0 \\(B)\times: f(1)=\sin{5\over 2}\pi=1 \\ (C)\bigcirc: f(7)=\sin{35\over 2}\pi=-1\\ (D)\times: f(10)=\sin25\pi =0\\，故選\bbox[red, 2pt]{(C)}$$

解答: $$甲乙合併成A, 則A丙丁可排成A丁丙及丙丁A兩種排法;\\ 接著戊有4個位置可選擇,然後己有5個位置可選\\因此共有2\times 2\times 4\times 5=80種排法，故選\bbox[red, 2pt]{(B)}$$

解答: $$\Gamma：\cases{x+y\le 1 \cdots(1)\\ 2x-3y\le 4 \cdots(2)\\ 3x-2y\ge -4 \cdots(3)}\\(A)\times: (1,1)不符合(1)\\ (B)\times: (1,-1)不符合(2)\\ (C) \times: (-1,1)不符合(3) \\ (D)\bigcirc: 三式均符合，故選\bbox[red, 2pt]{(D)}$$

解答:

$$假設底面長方形頂點坐標為\cases{A(3,4,0)\\ B(-3,4,0)\\ C(-3,-4,0)\\ D(3,-4,0)}, 四角錐頂點坐標P(0,0,h)\\ 因此\overline{AP}=7 \Rightarrow \sqrt{3^2+4^2+h^2} =7 \Rightarrow h^2=24 \Rightarrow h=2\sqrt 6，故選\bbox[red, 2pt]{(A)}$$

解答: $$哥哥沒抽中3000元 \Rightarrow 哥哥可能抽到1200或600元，即\\\cases{哥哥抽到1200元,剩下5張鈔票 \Rightarrow 弟弟的期望值(抽中金額)= (3000+1200+ 600\times 3)/5= 1200 \\ 哥哥抽到600元,剩下5張鈔票 \Rightarrow 弟弟的期望值 =(3000+ 1200\times 2+ 600\times 2)/5=1320} \\ 也就是說\cases{哥哥有{2\over 5}的機率抽到1200元,此時弟弟抽到1200元\\\ 哥哥有{3\over 5}的機率抽到 600元,此時弟弟抽到1320元}\\ 因此弟弟壓歲錢的期望值= {2\over 5}\times 1200+ {3\over 5}\times 1320=1272元，故選\bbox[red, 2pt]{(D)}$$

解答: $$假設圓心O,圓半徑r=7\sqrt 3, 由於圓心角是圓周角的2倍,即\angle BOC=2\angle A \\\Rightarrow \cos \angle BOC = {r^2+r^2-\overline{BC}^2 \over 2r\cdot r} ={147+147-441\over 2\cdot 147} =-{1\over 2} \Rightarrow \angle BOC=120^\circ\\ \Rightarrow \angle A=60^\circ  \Rightarrow \cos \angle A={\overline{AB}^2+\overline{AC}^2-\overline{BC}^2 \over 2\overline{AB}\cdot \overline{AC}} \Rightarrow {1\over 2}={24^2+\overline{AC}^2-21^2 \over 2\cdot 24\cdot \overline{AC}} ={135+\overline{AC}^2\over 48\overline{AC}} \\ \Rightarrow \overline{AC}^2-24 \overline{AC}+135=0 \Rightarrow (\overline{AC}-9)(\overline{AC}-15)=0 \\\Rightarrow \overline{AC}=15(9不合,圓心需在三角形內部) \\ \Rightarrow \triangle ABC面積={1\over 2}\overline{AB} \cdot \overline{AC}\sin \angle A ={1\over 2}24\cdot 15\cdot {\sqrt 3\over 2} = 90\sqrt 3，故選\bbox[red, 2pt]{(B)}$$

解答: $$9=4+3+2,3+3+3,3+2+2+2有三種人數上的分組\\共有C^9_4C^5_3C^2_2 + {1\over 3!}C^9_3C^6_3C^3_3+ {1\over 3!}C^9_3C^6_2C^4_2C^2_2= 2800，故選\bbox[red, 2pt]{(A)}$$

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