113年身障生升四技二專-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：四技二專組-數學(B)

單選題，共 20 題，每題 5 分

解答: $$(y-2)=3(x-0) \Rightarrow y=3x+2，故選\bbox[red, 2pt]{(A)}$$

解答: $$\cases{A(1,2)\\ B(2,1)\\ C(3,5)\\ D(4,k)} \Rightarrow \cases{\overrightarrow{AB} =(1,-1)\\ \overrightarrow{CD} =(1,k-5)} \Rightarrow \overline{AB} \bot \overline{CD} \Rightarrow \overrightarrow{AB} \cdot \overrightarrow{CD} =0 \\\Rightarrow (1,-1) \cdot (1,k-5)=1+5-k =0 \Rightarrow k=6，故選\bbox[red, 2pt]{(D)}$$

解答: $$f(x) = (x^2+2x+3)(x+7)+(2x+5) =x^3+9x^2+19x+26，故選\bbox[red, 2pt]{(B)}$$

解答: $$f(x) = x^{2023}+5x+6 \Rightarrow f(1)=1+5+6=12，故選\bbox[red, 2pt]{(C)}$$

解答: $$\cases{a_1=2\\ r=3} \Rightarrow \sum_{k=1}^6 a_k= {2\cdot 3^6-2\over 3-1}=3^6-1=728，故選\bbox[red, 2pt]{(B)}$$

解答: $${x+1\over 2}\lt 2-{x\over 3} \Rightarrow {x+1\over 2}+{x\over 3}\lt 2 \Rightarrow {5x+3\over 6}\lt 2 \Rightarrow 5x+3\lt 12\\ \Rightarrow 5x\lt 9 \Rightarrow x\lt {9\over 5}，故選\bbox[red, 2pt]{(A)}$$

解答: $$x^2+3x+5=0 \Rightarrow \cases{\alpha+\beta=-3 \\\alpha\beta=5} \Rightarrow {1\over \alpha}+{1\over \beta}={\alpha+ \beta \over \alpha\beta} ={-3\over 5}，故選\bbox[red, 2pt]{(B)}$$

解答: $$f(x,y)=x+y+1 \Rightarrow f(A)f(B)\lt 0 \Rightarrow (1+k+1)(2+2+1)\lt 0 \Rightarrow 5(k+2)\lt 0\\ \Rightarrow k\lt -2，故選\bbox[red, 2pt]{(D)}$$

解答: $$\log_2 6\cdot \log_6 8\cdot \log_8 16 ={\log 6\over \log 2}\cdot {\log 8\over \log 6} \cdot {\log 16\over \log 8} ={\log 16\over \log 2}=\log_2 16=4，故選\bbox[red, 2pt]{(A)}$$

解答: $$\cfrac{a^{3x}-a^{-3x}}{a^{x}-a^{-x}} = \cfrac{a^{4x}-a^{-2x}}{a^{2x}-1} = \cfrac{16-{1\over 4}}{4-1} =  \cfrac{63\over 4}{3} ={21 \over 4}，故選\bbox[red, 2pt]{(C)}$$

解答: $${54+640\over 2}=362 \Rightarrow 54-362\le x-362\le 670-362 \Rightarrow -308\le x-362\le 308\\ \Rightarrow |x-362| \le 308 \Rightarrow \cases{a=362\\ b=308} \Rightarrow {a\over 2}+{b\over 7}=181+44=225，故選\bbox[red, 2pt]{(D)}$$

解答: $$\cases{a\gt 0\\ b\lt 0} \Rightarrow \cases{a-b \gt 0 \\ b^4a \gt 0} \Rightarrow (a-b,b^4a) =(+,+)在第一象限，故選\bbox[red, 2pt]{(A)}$$

解答: $$\cos \angle C={a^2+b^2-c^2 \over 2ab} ={\sqrt 2 ab\over 2ab} ={\sqrt 2\over 2} \Rightarrow \angle C=45^\circ，故選\bbox[red, 2pt]{(B)}$$

解答: $$(\sin 45^\circ+\sin 15^\circ)^2+ (\cos 45^\circ-\cos 15^\circ)^2\\= \sin^2 45^\circ+2\sin 45^\circ \sin 15^\circ+ \sin^2 15^\circ +\cos^2 45^\circ-2\cos 45^\circ\cos 15^\circ+ \cos^2 15^\circ \\=(\sin^2 45^\circ+ \cos^2 45^\circ)+( \sin^2 15^\circ+ \cos^2 15^\circ)+ 2(\sin 45^\circ \sin 15^\circ-\cos 45^\circ\cos 15^\circ)\\ =1+1-2\cos(45^\circ+15^\circ)=2-2\cos 60^\circ=2-2\cdot {1\over 2}=1，故選\bbox[red, 2pt]{(A)}$$

解答: $$\cos \angle A={\overrightarrow{AB}\cdot  \overrightarrow{AC}\over \overline{AB} \cdot \overline{AC}} ={63 \over 5\cdot 13} ={65\over 63} ={\overline{AB}^2+ \overline{AC}^2-\overline{BC}^2 \over 2\overline{AB} \cdot \overline{AC}} ={194-\overline{BC}^2 \over 130} \\ \Rightarrow \overline{BC}^2=194-126=68 \Rightarrow \overline{BC}=\sqrt{68}，故選\bbox[red, 2pt]{(D)}$$

解答: $$\cases{\vec a=(2,4)\\ \vec b=(4,-k)} \Rightarrow \vec a+\vec b=(6,4-k)\\ \vec a\bot (\vec a+\vec b) \Rightarrow \vec a\cdot (\vec a+\vec b)=0 \Rightarrow (2,4)\cdot (6,4-k)=12+16-4k=0 \Rightarrow k=7，故選\bbox[red, 2pt]{(D)}$$

解答: $$有兩條切線 \Rightarrow \overline{OP} =5 \gt 圓半徑 r\\(A)\times: (x+1)^2+(y-1)^2=8 \Rightarrow 圓心(-1,1) \ne （1，－1） \\(B)\times: 圓心(-1,1) \ne （1，-1) \\ (C)\bigcirc: 圓心(1,-1)且半徑＝3\lt 5\\ (D)\times: 半徑5\not \lt 5\\，故選\bbox[red, 2pt]{(C)}$$

解答: $$四碼數字=AB3D \Rightarrow \cases{B\ne 3 \Rightarrow B有9種可能 \\D\ne 3 \Rightarrow D有9種可能 \\ A\ne B\Rightarrow A有9種可能 } \Rightarrow 共有9^3=729組數字，故選\bbox[red, 2pt]{(B)}$$

解答: $$\cases{Y=1.2X+10 \Rightarrow \bar Y =1.2\bar X+10=1.2\times 42.3+10=60.76\\ S_Y=1.2S_X=1.2\times 24=28.8 }\\\Rightarrow \bar Y-S_Y=60.76-28.8=31.96，故選\bbox[red, 2pt]{(C)}$$

解答: $$\cases{8/4=2 \Rightarrow Q_1=第2筆與第3筆資料的平均＝(2+3)\div2 =2.5\\ 8\times 3/4=6 \Rightarrow Q_3=第6筆與第7筆資料的平均=(8+9)\div2=8.5} \\ \Rightarrow 四分位距=Q_3-Q_1=8.5-2.5=6，故選\bbox[red, 2pt]{(C)}$$

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解題僅供參考，其他歷年試題及詳解