2022年新疆三校生数学答案
一、填空题:1.{1,2,3,4} 2.π3.\frac {1}{2} 4.\{ x \mid 1 \le x<2 \} 5.6π6.[2,+\infty)7.\sqrt {5} 9.\frac {1}{3} 10.(3,+\infty)11.±1 12.\frac {7}{2} 13.2 14.二、解答题:15.(1)设D(x,y).5分所以点D的坐标为(5,-6).6分(2)因为a= \overrightarrow {AB}=(2,-2)-(1,3)=(1,-5),b= \overrightarrow {BC}=(4,-1)-(2,-2)=(2,1), \cdots 8分所以ka-b=k(1,-5)-(2,1)=(k-2,-5k-1),a+3b=(1,-5)+3(2,1)=(7,-2).10分由ka-b 与a+3b 平行,得(k-2)\times(-2)-(-5k-1)\times 7=0, 12分所以k=- \frac {1}{3}.14分16.(1)\frac {3 \sin \alpha+2 \cos \alpha }{ \sin \alpha - \cos \alpha }= \frac {3 \tan \alpha+2}{ \tan \alpha -1} 2分= \frac {3 \times 2+2}{2-1}=8.10分(3)解法1:由\frac { \sin \alpha }{ \cos \alpha }= \tan \alpha =2, 得\sin \alpha =2 \cos \alpha , 又\sin ^{2} \alpha+\cos ^{2} \alpha =1, 故4 \cos ^{2} \alpha+\cos ^{2} \alpha =1, 即\cos ^{2} \alpha = \frac {1}{5}, 分因为α是第三象限角, \cos \alpha <0, 所以\cos \alpha =- \frac { \sqrt {5}}{5}.14分解法2:\cos ^{2} \alpha = \frac { \cos ^{2} \alpha }{ \cos ^{2} \alpha+\sin ^{2} \alpha }= \frac {1}{1+\tan ^{2} \alpha }= \frac {1}{1+2^{2}}= \frac { 2分因为α是第三象限角, \cos \alpha <0, 所以\cos \alpha =- \