20182019學(xué)年度學(xué)校第二次大考高一數(shù)學(xué)（答案）

1、湛江一中2019-2019學(xué)年第一學(xué)期“第2次大考高一級數(shù)學(xué)科試卷答案一、單項選擇題(共12題 ,每題只有一個正確選項 ,每題5分 ,共60分)題號123456789101112答案BBDABCDCADDB二、填空題共4題 ,每題5分 ,共20分13-2 14 15 (-,-1) 16三、解答題寫出詳細(xì)的解答過程 ,共6題 ,第17題10分 ,其余每題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, 即為2-22x-127 ,·····1分所以-2x-17 ,解得-1x8 ,所以A=x|-1x8·········&#

3、183;·3分因為對數(shù)函數(shù)y=log2x 在18,32上單調(diào)遞增 ,······························4分所以log218log2xlog232 ,即-3log2x5 ,所以B=y|-3y5·····6分2由1得A

4、B=x|-1x5···································7分當(dāng)C= 時 ,滿足CAB ,此時m+1>2m-1 ,解得m<2·······&

5、#183;····9分 當(dāng)C 時 ,由CAB得m+12m-1m+1-12m-15 ,解得2m3 ,··········11分綜上m3所以實數(shù)m的取值范圍是(-,3 ························

6、3;···12分1912分解：1當(dāng)時 ,函數(shù)為一次函數(shù) ,設(shè)其解析式為 ,點和在函數(shù)圖象上 ,解得························2分當(dāng)時 ,函數(shù)是二次函數(shù) ,設(shè)其解析式為 ,點在函數(shù)圖象上 ,解得·········

7、3;·······································4分綜上.··········

8、;·······························5分2由1得當(dāng)時 , ,。結(jié)合圖象可得假設(shè)方程有三個不同解 ,那么。實數(shù)的取值范圍.············

9、;··································8分3當(dāng)時 ,由得 解得 ；·············

10、83;····9分當(dāng)時 ,由得 ,整理得 解得或舍去·······················11分綜上得滿足的的取值集合是·················

11、;··········12分.20.12分解：1函數(shù)為奇函數(shù)證明如下：定義域為 ,又··············3分為奇函數(shù) ···················

12、83;························4分2函數(shù)在-1 ,1為單調(diào)增函數(shù)證明如下：任取 ,那么·····················

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

14、#183;···················8分故在-1 ,1上為增函數(shù)····························

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

16、83;············12分2112分解：1由圖可知A0,20 ,B25,45 ,C25,75 ,D30,70 ,設(shè)AB所在直線方程為P=kt+20 ,把B25,45代入P=kt+20得k=1 ,所以. lAB:P=t+20 ,··1分設(shè)CD所在的直線方程為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由題意 ,設(shè)Q=k1t+b ,把兩點5,35 ,15,25代入得5k1+b=3515k1+b=25 ,解得k

18、1=-1b=40所以Q=-t+40,把點20,20 ,30,10代入Q=-t+40也適合 , 6分所以Q=-t+400<t30. ······································

19、3;·····7分3設(shè)日銷售金額為y ,依題意得 ,當(dāng)0<x<25 時,y=t+20-t+40 ,配方整理得y=-t-102+900 ,······8分當(dāng)t=10 時 ,y在區(qū)間0,25上的最大值為900·······················

20、;········9分當(dāng)25x30 時 ,y=-t+100-t+40 ,配方整理得y=t-702-900 ,··10分所以當(dāng)t=25 時 ,y在區(qū)間上的最大值為1125. ···························&#

21、183;11分綜上可知日銷售金額最大值為1125元 ,此時t=25. ·························12分2212分解：1因為f(x)是定義域為R的奇函數(shù) ,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) , 因為f(1)>0 ,所以a-1a>0 ,又a>0且a1 ,所以a>1 , 所以fx=ax-1ax 在R上單調(diào)遞增 ,··············

23、·························3分 又f(x)是定義域為R的奇函數(shù) ,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 ,所以實數(shù)b的取值范圍為(-3,5). ···········································6分 3因為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)