114年身障生升四技二專-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：四技二專組-數學(B)

單選題，共 20 題，每題 5 分

解答: $$與L平行的直線方程式為2x-3y+k=0，故選\bbox[red, 2pt]{(C)}$$

解答: $$G=(A+B+C)/3 \Rightarrow (1,2)= \left({-1+2+y\over 3}, {x+3-1\over 3} \right) \Rightarrow \cases{x=4\\y=2}，故選\bbox[red, 2pt]{(D)}$$

解答: $$L:4x+3y+12=0 \Rightarrow \cases{x=0 \Rightarrow y=-4\\ y=0 \Rightarrow x=-3} \Rightarrow L與坐標軸交於\cases{A(0,-4) \\B(-3,0)} \\\Rightarrow \triangle OAB面積={1\over 2}\cdot 4\cdot 3 =6，故選\bbox[red, 2pt]{(D)}$$

解答: $$\log_2 4^{x+1} =\log_2 2^{2x+2} =(2x+2)\log_2 2 = 2x+2=x^2+3 \Rightarrow x^2-2x+1=0 \Rightarrow (x-1)^2=0\\ \Rightarrow x=1，故選\bbox[red, 2pt]{(A)}$$

解答: $$\cases{a=\sqrt[6]{64} =\sqrt[6]{2^6} =2\\ b=\sqrt[3]{27} =\sqrt[3]{3^3}=3\\ c=\log_9{27} ={\log 27\over \log 9} ={3\log 3\over 2\log 3} ={3\over 2} \\ d=\log_{1/2}{1\over 16} =4} \Rightarrow d\gt b\gt a\gt c，故選\bbox[red, 2pt]{(C)}$$

解答: $$1及-2為-2x^2+ax+b=0的兩根\Rightarrow \cases{-2+a+b=0\\ -8-2a+b=0} \Rightarrow \cases{a=-2\\ b=4} \\ \Rightarrow x^2-(a^2+1)x+(a-b) =0 \Rightarrow x^2-5x-6=0 \Rightarrow (x-6)(x+1) =0 \Rightarrow x=-1,6，故選\bbox[red, 2pt]{(A)}$$

解答: $$函數對稱x=-1, 又通過(0,1)代表也通過(-2,1) \Rightarrow \cases{f(0)=1\\ f(-2)=1} \Rightarrow \cases{b=1\\ -4-2a+b=1 \Rightarrow a=-2}\\ \Rightarrow a+b=-2+1=-1，故選\bbox[red, 2pt]{(A)}$$

解答: $$\cases{f(x)=p(x)(x-2)+3 \\ f(x)=q(x)(x+3)-2\\ f(x)=r(x)(x-2)(x+3)+ ax+b} \Rightarrow \cases{f(2)=3 \\ f(-3)=-2} \Rightarrow \cases{2a+b=3\\ -3a+b=-2} \Rightarrow \cases{a=1\\ b=1} \\ \Rightarrow 餘式:ax+b=x+1，故選\bbox[red, 2pt]{(A)}$$

解答: $$假設長方形的長為x, 寬為y, 則\cases{x-2=y+1\\ xy=(x-2)(y+1)} \Rightarrow (y+3)y =(y+1)^2 \Rightarrow y=1\\ \Rightarrow x=4 \Rightarrow 長方形周長=2(x+y)=10，故選\bbox[red, 2pt]{(B)}$$

解答:

$$所圍區域頂點為\cases{A(2,2) \\ B(4,-2) \\C(-2,0) \\D(0,0)\\ E(3,0)} \Rightarrow 所圍面積=\triangle ADE+ 梯形BCDE\\= {1\over 2}\cdot 3 \cdot 2+ (3+4)\cdot 2 / 2 =3+7= 10，故選\bbox[red, 2pt]{(B)}$$

解答: $$\sin 220^\circ =\sin(180^\circ+ 40^\circ) =-\sin 40^\circ =k \Rightarrow \sin 40^\circ =-k \Rightarrow \cos 50^\circ =-k \Rightarrow \tan 50^\circ =-{\sqrt{1-k^2} \over k}\\  \Rightarrow \tan 310^\circ =\tan(360^\circ- 50^\circ) = -\tan 50^\circ ={\sqrt{1-k^2}\over k}，故選\bbox[red, 2pt]{(A)}$$

解答: $$\cos \angle A={2^2+4^2-5^2 \over 2\cdot 2\cdot 4} =-{5\over 16} ={\overrightarrow{AB} \cdot \overrightarrow{AC} \over |\overrightarrow{AB} || \overrightarrow{AC}|}   ={\overrightarrow{AB} \cdot \overrightarrow{AC} \over 2\cdot 4} \Rightarrow \overrightarrow{AB} \cdot \overrightarrow{AC}=-{5\over 2}，故選\bbox[red, 2pt]{(B)}$$

解答: $$\cases{a= 6\cos \theta\\ b=6\sin \theta} \Rightarrow (a-5)^2+ (b+12)^2=36-10a+25+24b+144 =-10a+24b+205\\ =-60\cos \theta+144 \sin \theta+205 =\sqrt{60^2+ 144^2} \sin(\theta+ \alpha)+205 =156 \sin(\theta+\alpha)+205 \\ \Rightarrow 最小值=-156+205=49 \Rightarrow \sqrt{(a-5)^2+ (b+12)^2}的最小值= \sqrt{49}=7，故選\bbox[red, 2pt]{(C)}$$

解答: $$假設\cases{首項a_1\\ 公差d=3} \Rightarrow 前5項總和=5a_1+10d=A \Rightarrow a_1+2d =a_3={A\over 5} \\ \Rightarrow \cases{a_1=A/5-2d=A/5-6\\ a_{20} =a_3+17d={A/5+51}} \Rightarrow 級數總和={(A/5-6+A/5+51)}\times 10=({2A\over 5}+45)\times 10\\=4A+450，故選\bbox[red, 2pt]{(D)}$$

解答: $$\cases{\vec a=(2,-2) \\\vec b=(-3,7)} \Rightarrow t\vec a+\vec b= (2t-3,-2t+7)\Rightarrow |t\vec a+\vec b|^2 = (t\vec a+\vec b) \cdot (t\vec a+\vec b) =(2t-3)^2+ (-2t+7)^2 \\= 8t^2-40t+58 =8(t-{5\over 2}^2)+8 \Rightarrow 最小值為8 \Rightarrow |t\vec a+\vec b|的最小值=\sqrt 8 =2\sqrt 2，故選\bbox[red, 2pt]{(A)}$$

解答:

$$小明\bigcirc: {\overline{AC} \over \sin \angle B}={\overline{AB} \over \sin \angle C} \Rightarrow {2\over \sin 30^\circ} ={3\over \sin \angle C} \Rightarrow \sin \angle C={3 \over 4} \gt {1\over 2} =\sin 30^\circ \Rightarrow \angle C\gt 30^\circ \\ 阿美\bigcirc:\sin \angle C={3\over 4}\\ 大貴\times :\sin \angle C={3\over 4} \Rightarrow \cos \angle C=\pm {\sqrt 7\over 4}\\ 雄仔\bigcirc: {\overline{AC} \over \sin \angle B} =2R \Rightarrow 外接圓半徑=2 \\芊芊\times: \angle C不確定，因此內切圓半徑也不確定\\ 共三人說對，故選\bbox[red, 2pt]{(C)}$$

解答: $$假設球員選a位,啦啦隊選b 位,則(a,b)=(1,3), (2,2), (3,1),有三種情況，\\每種情況分別有C^6_1C^5_3=60, C^6_2C^5_2=150, C^6_3C^5_1=100 種選法，\\因此共有60+150+100 =310種選擇方法，故選\bbox[red, 2pt]{(D)}$$

解答:

$$九格編號依序為a,b,\dots, i，如上圖，\\符合要求的選法有(a,e,i), (a,f,g), (b, d ,i), (b, f,g), (c,e,g),(c,d,h)，共有六種情形，\\因此機率為{6\over C^9_6} ={1\over 14}，故選\bbox[red, 2pt]{(B)}$$

解答: $$調整前\cases{E(X)=45\\ Var(X)=9^2} \Rightarrow 調整後\cases{E(aX+b)=aE(X)+b=45a+b=63\\ Var(aX+b)=a^2Var(X)=a^2\cdot 9^2=15^2} \Rightarrow \cases{a=5/3\\ b=-12} \\ \Rightarrow 原始成績30分, 調整後變為{5\over 3}\cdot 30-12=38，故選\bbox[red, 2pt]{(D)}$$

解答:

$$兩人方向的夾角(\angle AOB)為120^\circ \Rightarrow \cos 120^\circ=-{1\over 2} ={50^2+30^2-x^2 \over 2\cdot 30\cdot 50} \Rightarrow x=70，故選\bbox[red, 2pt]{(D)}$$

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