114年身障升大學-數學B詳解

114 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: $$f(x)=a(x-2)(x-b)+1 \Rightarrow \cases{f(3) =a(3-b)+1=1 \\ f(4) =2a(4-b)+1=2} \Rightarrow \cases{a(3-b) =0\\ a(4-b)=1/2} \\ \Rightarrow \cases{a=1/2\\ b=3} \Rightarrow f(x)={1\over 2}(x-2)(x-3)+1 \Rightarrow f(5) ={1\over 2}\cdot 3\cdot 2+1=4，故選\bbox[red, 2pt]{(B)}$$

解答: $$(\vec a+\vec b) \bot \vec a \Rightarrow (\vec a+\vec b) \cdot \vec a =0 \Rightarrow |\vec a|^2+\vec a\cdot \vec b=1+\vec a\cdot \vec b=0 \Rightarrow \vec a\cdot \vec b= -1\\ \Rightarrow (\vec a-\vec b) \cdot (\vec a+ 2\vec b) =|\vec a|^2+ \vec a\cdot \vec b-2|\vec b|^2 =1-1-2\cdot 4=-8，故選\bbox[red, 2pt]{(A)}$$

解答: $${A點向西移動\Rightarrow 經度小於123.9} ,再向北移動\Rightarrow 緯度大於19.8，故選\bbox[red, 2pt]{(A)}$$

解答: $$ABCD= \begin{bmatrix} 2\\3\end{bmatrix} \begin{bmatrix} -1& 2\end{bmatrix}  \begin{bmatrix} 0& 1\\ 1& 0 \end{bmatrix}  \begin{bmatrix} 1\\ 1\end{bmatrix} =   \begin{bmatrix} -2& 4\\-3 & 6\end{bmatrix}  \begin{bmatrix} 0& 1\\ 1& 0 \end{bmatrix}  \begin{bmatrix} 1\\ 1\end{bmatrix} =\begin{bmatrix} 4& -2\\ 6 & -3 \end{bmatrix}  \begin{bmatrix} 1\\ 1\end{bmatrix} = \begin{bmatrix} 2\\ 3\end{bmatrix}\\，故選\bbox[red, 2pt]{(A)}$$

解答: $$8\le |x-10|\le 14 \Rightarrow \cases{8\le x-10\le 14\\ 8\le 10-x\le 14} \Rightarrow \cases{18\le x\le 24\Rightarrow 7個解\\ -2\le -x\le 4 \Rightarrow -4\le x\le 2 \Rightarrow 7個解} \\ \Rightarrow 共14個整數解，故選\bbox[red, 2pt]{(D)}$$

解答: $$\cases{總和為4的情形:(1,1,2) \\總和為7的情形:(2,2,3) \\ 總和為16的情形:(1,7,8), (2,6,8), (2,7,7), (3,6,7) \\ 總和為23的情形:(7,7,8)} ，故選\bbox[red, 2pt]{(C)}$$

解答: $$\overline{DQ}: \overline{QC}=1:2 \Rightarrow \overrightarrow{PQ}={1\over 3} \overrightarrow{PC}+{2\over 3}\overrightarrow{PD} ={1\over 3} \left( \overrightarrow{PB}+\overrightarrow{BC} \right)+{2\over 3} \left( \overrightarrow{PA} +\overrightarrow{AD} \right) \\ ={1\over 3} \left({1\over 2}\overrightarrow{AB}+\overrightarrow{AD} \right)+{2\over 3} \left(-{1\over 2} \overrightarrow{AB} +\overrightarrow{AD} \right) =-{1\over 6}\overrightarrow{AB}+ \overrightarrow{AD} \\ \Rightarrow \cases{\alpha=-1/6\\ \beta=1} \Rightarrow \alpha+\beta={5\over 6}，故選\bbox[red, 2pt]{(B)}$$

解答: $$\sqrt{6}-\sqrt 5 \gt \sqrt{11}-\sqrt{10} \Rightarrow c\gt d且b\gt a\\ 又{1\over \sqrt{6}-\sqrt 5} =\sqrt{6}+\sqrt 5 \Rightarrow a\gt c , 因此b\gt a\gt c\gt d，故選\bbox[red, 2pt]{(C)}$$

解答: $$\langle a_n \rangle等比級數,其中\cases{a_1=16\\ a_9=a_1r^8=40} \Rightarrow r^8={40\over 16}={5\over 2} \\\Rightarrow a_{25}=a_1r^{24} = 16\cdot (r^8)^3=16\cdot {125\over 8} =250，故選\bbox[red, 2pt]{(D)}$$

解答: $${1\over 2}(0.1\times 1+0.5\times 5+0.8\times 8+ 1\times 6) ={3\over 4} = 75\% \\\Rightarrow 平均為原價的75\% \Rightarrow 20\times 75\%=15，故選\bbox[red, 2pt]{(B)}$$

解答: $$離消失點越近，視覺高度越短，其比值不相等，故選\bbox[red, 2pt]{(D)}$$

解答: $$a_1=1 \Rightarrow a_2=2a_1= 2 \Rightarrow a_3=a_2+3= 5 \Rightarrow a_4=10 \Rightarrow a_5=13\Rightarrow a_6=26 \Rightarrow a_7=29\\ \Rightarrow a_8=58 \Rightarrow a_9=61 \Rightarrow a_{10}=122 \Rightarrow k=10，故選\bbox[red, 2pt]{(B)}$$

解答: $$\cases{f(x)=x^2+ax+b \Rightarrow f(-1)=1-a+b=3 \Rightarrow a-b=-2 \cdots(1)\\ g(x)=3f(x)=3x^2+3ax+3b \Rightarrow g(-3) =27-9a+3b=9 \Rightarrow 3a-b=6 \cdots(2)} \\ \Rightarrow \cases{a =4\\ b=6} \Rightarrow f(x)=x^2+4x+6 \Rightarrow h(x)=2f(x)=2x^2+8x+12 \Rightarrow h(-2) =8-16+12=4\\，故選\bbox[red, 2pt]{(B)}$$

解答: $$600毫升飲料選a種、1200毫升飲料選b種 \Rightarrow (a,b) =(6,0),(4,1),(2,2), (0,3) \\ \Rightarrow 飲料選擇數=C^6_6C^5_0 +C^6_4C^5_1+C^6_2 C^5_2 +C^6_0C^5_3 =1+75+150+ 10= 236，故選\bbox[red, 2pt]{(C)}$$

解答:

$$假設地面距離\overline{OB}=x,由於\angle OPB=45^\circ \Rightarrow \overline{OP} =\overline{OB}=x \Rightarrow \tan \angle BAO={\overline{OB}\over \overline{IA}}\\ \Rightarrow \tan 30^\circ= {x\over x+9}  \Rightarrow {1\over \sqrt{3} +1} ={x\over x+9} \Rightarrow x={9\over \sqrt 3-1} ={{9\over 2}(\sqrt 3+1)}，故選\bbox[red, 2pt]{(D)}$$

解答: $$\cases{100\lt s\lt 1000 \Rightarrow 2\lt \log s\lt 3 \\ 0\lt r\lt 1/10  \Rightarrow 1\lt 10^r \lt \sqrt[10]{10} \lt 2} \Rightarrow 1\lt 10^r\lt \log s，故選\bbox[red, 2pt]{(B)}$$

解答: $$(x-1)^2+y^2=25 \Rightarrow \cases{圓心O(1,0) \\半徑r=5}, \\A(6,0)與圓心O同在x軸上\Rightarrow L_1:x=6\\(x-1)^2+y^2=25 \Rightarrow 2(x-1)+ 2yy'=0 \Rightarrow y'={1-x\over y} \Rightarrow y'(B)={1-(-2)\over 4} ={3\over 4} \\ \Rightarrow L_2斜率={3\over 4} \Rightarrow L_2: y={3\over 4}(x+2)+4 \Rightarrow C=L_1\cap L_2 =(6,10) \\ \Rightarrow \overline{CA}= \overline{CB}=10 \Rightarrow \overline{AB}= \sqrt{8^2+4^2} = 4\sqrt{5} \Rightarrow \cos \angle ACB ={10^2 +10^2-(4\sqrt 5)^2\over 2\cdot 10\cdot 10} ={3\over 5}\\，故選\bbox[red, 2pt]{(C)}$$

解答: $$A\begin{bmatrix} 2&-1\\ 4& 3\end{bmatrix}=\begin{bmatrix} 0&-1\\ 2& 0 \end{bmatrix} \Rightarrow A= \begin{bmatrix} 0&-1\\ 2& 0 \end{bmatrix} \begin{bmatrix} 2&-1\\ 4& 3\end{bmatrix}^{-1} = \begin{bmatrix} 0&-1\\ 2& 0 \end{bmatrix} \begin{bmatrix} \frac{3}{10} & \frac{1}{10} \\\frac{-2}{5} & \frac{1}{5} \end{bmatrix} = \begin{bmatrix} \frac{2}{5} & \frac{-1}{5} \\\frac{3}{5} & \frac{1}{5} \end{bmatrix} \\ \Rightarrow A^{-1}=\begin{bmatrix} 1 & 1 \\-3 & 2 \end{bmatrix}\Rightarrow A\begin{bmatrix}a\\b \end{bmatrix}= \begin{bmatrix}1\\1 \end{bmatrix} \Rightarrow \begin{bmatrix} a\\b\end{bmatrix} =A^{-1} \begin{bmatrix}1\\1 \end{bmatrix} =\begin{bmatrix}2\\-1 \end{bmatrix} \Rightarrow a+b=1，故選\bbox[red, 2pt]{(C)}$$

解答: $$直線L: y={3\over 4}x+a \Rightarrow \cases{L與y=0交於(-{4a\over 3},0)= Q(b,0)\\ L與x=4交於(4,3+a) =R(4,c)} \Rightarrow \cases{b=-4a/3\\ c=a+3} \\ \Rightarrow \triangle PQR面積= {1\over 2}\cdot \overline{PQ}\cdot \overline{PR} ={1\over 2}\cdot (4-b)\cdot c={1\over 2}(4+{4a\over 3})(a+3) =54 \\ \Rightarrow a^2+6a-72=0 \Rightarrow (a-6)(a+72) =0 \Rightarrow\cases{a=6 \Rightarrow b=-8,c=9 \\a=-72 \Rightarrow c=-69 \lt 0不在第一象限} \\ \Rightarrow a+b+c=6-8+9=7，故選\bbox[red, 2pt]{(A)}$$

解答: $$全班總分=na \Rightarrow \cases{na-100=(n-1)(a-2)\\ na=(n-1)(a+3)} \Rightarrow {na-100\over na}={a-2\over a+3} \\ \Rightarrow 1-{100\over na}=1-{5\over a+3} \Rightarrow {100\over na}={5\over a+3} \Rightarrow 5na=100a+300 \Rightarrow n=20+{60\over a} \\ \Rightarrow na=(n-1)(a+3)  \Rightarrow 20a+60=(19+{60\over a})(a+3) \Rightarrow a^2-57a-180=0 \\ \Rightarrow (a-60)(a+3) =0 \Rightarrow a=60 \Rightarrow n=20+{60\over 60}=21 \Rightarrow n+a=81，故選\bbox[red, 2pt]{(C)}$$

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