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2022年2月20日 星期日

102年身心障礙學生四技二專甄試-數學(A)-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:四技二專組-數學(A)

單選題,共 20 題,每題 5 分

解答:$$\sin\theta-\cos \theta={3\over 4} \Rightarrow (\sin\theta-\cos \theta)^2 = 1-2\sin\theta \cos\theta ={9\over 16} \Rightarrow \sin\theta \cos\theta ={7\over 32},故選\bbox[red,2pt]{(B)}$$
解答:$$\cases{f(-2)=3\\ f(b)=-1} \Rightarrow \cases{-2a+1=3\\ ab+1= -1} \Rightarrow \cases{a=-1\\ b=2} \Rightarrow a+b=1,故選\bbox[red,2pt]{(C)}$$
解答:$$\cases{\vec a=(x,1)\\ \vec b=(3,-2)\\ \vec c=(-5,y)},因此\cases{\vec a\parallel \vec b\\ \vec b\bot \vec c} \Rightarrow \cases{{x\over 3} ={1\over -2} \\-15-2y=0} \Rightarrow \cases{x=-{3\over 2}\\[1ex] y=-{15\over 2}} \Rightarrow x+y=-9,故選\bbox[red,2pt]{(A)}$$
解答:$$\cases{f(-2)=0 \\ f(1)=3} \Rightarrow \cases{-8+4a-2b+4=0\\ 1+a+b+4=3} \Rightarrow \cases{2a-b=2\\ a+b=-2} \Rightarrow \cases{a=0\\ b=-2} \Rightarrow a-2b=4\\,故選\bbox[red,2pt]{(D)}$$
解答:$$4^x = {2\sqrt 2\over 8} \Rightarrow 2^{2x}={2^{3/2} \over 2^3} =2^{-3/2} \Rightarrow x=-{3\over 4},故選\bbox[red,2pt]{(B)}$$
解答:$$\cases{a= \tan(-236^\circ) = -\tan 236^\circ \lt 0\\ b=\sec(1206^\circ) = \sec(360^\circ\times 3+126^\circ) = \sec 126^\circ \lt 0} \Rightarrow (a,b)在第三象限,故選\bbox[red,2pt]{(C)}$$
解答:$$(\log_4 25-\log_8 5)(\log_5 4+\log_{25} 8) =({\log_5 25 \over \log_5 4}-{\log_5 5 \over \log_5 8})(\log_5 4+{\log_{5} 8 \over \log_5 25}) \\=({2  \over 2\log_5 2}-{1 \over 3\log_5 2})(2\log_5 2+{3\log_{5} 2 \over 2}) = \left({2\over 3}\cdot {1\over \log_5 2} \right)\left({7\over 2}\cdot \log_5 2\right) ={2\over 3} \cdot {7\over 2} ={7\over 3}\\,故選\bbox[red,2pt]{(B)}$$
解答:$$r\theta =s \Rightarrow r\cdot {\pi \over 12} ={\pi \over 4} \Rightarrow r=3 \Rightarrow 扇形面積= 3^2\pi \cdot {\pi/12\over 2\pi} ={3\over 8}\pi ,故選\bbox[red,2pt]{(D)}$$
解答:$$\alpha,\alpha+1 為9x^2+kx+4=0 的兩根\Rightarrow \cases{\alpha+\alpha+1 = -k/9\cdots(1) \\ \alpha(\alpha+1)= 4/9 \cdots(2)}\\ (2) \Rightarrow 9\alpha^2+9\alpha-4=0 \Rightarrow (3\alpha-1)(3\alpha+4)=0 \Rightarrow \alpha=1/3(\alpha \gt 0, -4/3不合)\\ 將\alpha=1/3代入(1) \Rightarrow {5\over 3} =-{k\over 9} \Rightarrow k=-15,故選\bbox[red,2pt]{(C)}$$
解答:$$兩直線\cases{3x-4y=2 \\ -6x+8y=3 \Rightarrow 3x-4y=-3/2}平行\\,因此兩平行線的距離即為高=\left|{2+3/2\over 5} \right| ={7\over 10},故選\bbox[red,2pt]{(A)}$$
解答:$$|x-1|\le 2 \Rightarrow -2\le x-1\le 2 \Rightarrow -1\le x\le 3 \Rightarrow x=-1,0,1,2,3,共五個整數解,故選\bbox[red,2pt]{(C)}$$
解答:$$x+2y+3=0的斜率為-1/2 \Rightarrow 與其垂直的斜率為2,故選\bbox[red,2pt]{(B)}$$

解答:$$x^2+y^2-6x+4y=0 \Rightarrow (x^2-6x+9)+(y^2+4y+4)=9+4 \Rightarrow (x-3)^2+(y+2)^2 = 13\\ \Rightarrow 圓心座標(3,-2),故選\bbox[red,2pt]{(D)}$$
解答:$$a_{10}=S(10)-S(9) = 2\cdot 10^2+3-(2\cdot 9^2+3)= 2\times 19=38,故選\bbox[red,2pt]{(A)}$$
解答:$$(-1)^7-2\times(-1)^5+ 3\times(-1)^4+4\times (-1)^3+5\times (-1)^2-1\\ =-1+2+3-4+5-1 =4,故選\bbox[red,2pt]{(D)}$$
解答:$$圖形為過原點直線y=2x的左半邊,因此不通過右下角的第四象限,故選\bbox[red,2pt]{(A)}$$
解答:$$進入有五種選擇,出來有四種選擇,因此有5\times 4=20種方法,故選\bbox[red,2pt]{(C)}$$
解答:$$五人任排有5!=120種排法,甲乙相鄰有4!\times 2=48種排法,\\因此甲乙不相鄰有120-48=72種,故選\bbox[red,2pt]{(D)}$$
解答:$$\cases{A=\{1,2,3\}\\ B=\{3,4,5,6\}} \Rightarrow A\cup B=\{1,2,3,4,5,6\}有六個元素,故選\bbox[red,2pt]{(C)}$$
解答:$$(60+40+ 62+78+83+61)\div 6= 384\div 6=64,故選\bbox[red,2pt]{(B)}$$
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