20182019学年度学校第二次大考高一数学（答案）

1、湛江一中2019-2019学年第一学期“第2次大考高一级数学科试卷答案一、单项选择题(共12题 ,每题只有一个正确选项 ,每题5分 ,共60分)题号123456789101112答案BBDABCDCADDB二、填空题共4题 ,每题5分 ,共20分13-2 14 15 (-,-1) 16三、解答题写出详细的解答过程 ,共6题 ,第17题10分 ,其余每题12分 ,共70分1710分解：1原式=1+49×94+(2-1)=2+1；·············&#

2、183;··········5分2原式=12log327+(lg25+lg4)-2=32+2-2=32. ·······10分1812分解：1不等式 142x-1128, 即为2-22x-127 ,·····1分所以-2x-17 ,解得-1x8 ,所以A=x|-1x8·········&#

3、183;·3分因为对数函数y=log2x 在18,32上单调递增 ,······························4分所以log218log2xlog232 ,即-3log2x5 ,所以B=y|-3y5·····6分2由1得A

4、B=x|-1x5···································7分当C= 时 ,满足CAB ,此时m+1>2m-1 ,解得m<2·······&

5、#183;····9分 当C 时 ,由CAB得m+12m-1m+1-12m-15 ,解得2m3 ,··········11分综上m3所以实数m的取值范围是(-,3 ························

6、3;···12分1912分解：1当时 ,函数为一次函数 ,设其解析式为 ,点和在函数图象上 ,解得························2分当时 ,函数是二次函数 ,设其解析式为 ,点在函数图象上 ,解得·········

7、3;·······································4分综上.··········

8、;·······························5分2由1得当时 , ,。结合图象可得假设方程有三个不同解 ,那么。实数的取值范围.············

9、;··································8分3当时 ,由得 解得 ；·············

10、83;····9分当时 ,由得 ,整理得 解得或舍去·······················11分综上得满足的的取值集合是·················

11、;··········12分.20.12分解：1函数为奇函数证明如下：定义域为 ,又··············3分为奇函数 ···················

12、83;························4分2函数在-1 ,1为单调增函数证明如下：任取 ,那么·····················

13、;·······6分 , ······························7分 ,即···········&

14、#183;···················8分故在-1 ,1上为增函数····························

15、;········9分3由1,2可得那么 解得： 原不等式的解集为x|0<x<13···································

16、83;············12分2112分解：1由图可知A0,20 ,B25,45 ,C25,75 ,D30,70 ,设AB所在直线方程为P=kt+20 ,把B25,45代入P=kt+20得k=1 ,所以. lAB:P=t+20 ,··1分设CD所在的直线方程为P-75=75-7025-30t-25 ,即P=-t+100 ,25x30 ,··2分所以P=t+20,0<x<25-t+100,25x30 ,·

17、83;········································4分2由题意 ,设Q=k1t+b ,把两点5,35 ,15,25代入得5k1+b=3515k1+b=25 ,解得k

18、1=-1b=40所以Q=-t+40,把点20,20 ,30,10代入Q=-t+40也适合 , 6分所以Q=-t+400<t30. ······································

19、3;·····7分3设日销售金额为y ,依题意得 ,当0<x<25 时,y=t+20-t+40 ,配方整理得y=-t-102+900 ,······8分当t=10 时 ,y在区间0,25上的最大值为900·······················

20、;········9分当25x30 时 ,y=-t+100-t+40 ,配方整理得y=t-702-900 ,··10分所以当t=25 时 ,y在区间上的最大值为1125. ···························&#

21、183;11分综上可知日销售金额最大值为1125元 ,此时t=25. ·························12分2212分解：1因为f(x)是定义域为R的奇函数 ,所以f(0)=0 ,所以1+(1-t)=0 ,所以t=2 , ·········&

22、#183;··················2分 2由1知：f(x)=ax-1ax(a>0,a1) , 因为f(1)>0 ,所以a-1a>0 ,又a>0且a1 ,所以a>1 , 所以fx=ax-1ax 在R上单调递增 ,··············

23、·························3分 又f(x)是定义域为R的奇函数 ,所以f(x2+bx)+f(4-x)>0f(x2+bx)>f(x-4)x2+bx>x-4······4分即x2+bx-x+4>0在xR上恒成立 ,所以=(b-1)2-16<

24、0 ,即-3<b<5 ,所以实数b的取值范围为(-3,5). ···········································6分 3因为f(1)=32 ,所以a-1a=32 ,解得 a=2 或 a=-12舍去 ,··············7分所以h(x)=22x+122x-2m(2x-12x)=(2x-12x)2-2m(2x-12x)+2 ,令u=f(x)