2023赣州二模文综

2023赣州二模文综试卷答案,我们目前收集并整理关于2023赣州二模文综得系列试题及其答案，更多试题答案请关注微信公众号：趣找答案/直接访问www.qzda.com(趣找答案)

2023赣州二模文综试卷答案，以下是该试卷的部分内容或者是答案亦或者啥也没有

forming8,000yearsago.Theyarehometoover2,000speciesoffish.GreenIslandisjustoneofthe900islandsoftheGreatBarrierReef,anditistheonlyonewitharainforest.Sherriisourparkguide,andshewilltakeyouthroughtheforestandtellyouallaboutthe120differentplantspecieswehave.Ifyou'reinterestedingettingwet,IsuggestyoufollowDamienandMarcus.They'reourdivingguides,andtheyhavethemostknowledgeaboutlifeontheisland.Anythingtodowithbeerorboats,youcanaskme.However,donotgoanywhereonyourownwithoutinformingusfirst.Wewouldn'twantanyonetogetbittenbyasharkorapoisonousspider.Justkidding,mates!

0得f(4)>04a+2b-1<0,(1)16a+4b-3>0,(2)f(-2)=4a-20-1)+1=4a-2+3由(1)(2)所表示的平面区域可求得4a-2b>0,故f(-2)=4a-2b+3>3所以f(2)的取值范围是(31+0)()方程ax2+(b-1)x+1=的两根为1,由根与系数的关系得x1+x222b由于不x2≠0,两式相除得b=∞(x1)由条件x2=+2可得x1+2,易知当∈(02)时,叫x是增函数当1∈0,2)时,a(x1)<(2)故b的取值范围是(4命题得证

5.D【解析】设△ABC的外接圆的半径为r,根据余弦定理可得,BC=AB2+AC2-2AB· ACcos120°=22+4-2×2×4×(-)=28可知BC=2/根据正弦定理可得BO2√Tin a sin 1202设三棱锥P一ABC的外接球的半径为R,则R2=r2+32=3,故外接球的表面积为S=4xR=4x×3=220r3故选D

19.解:(1)取A1B的中点P,连接PM,PN,因为M,P分别是AB,A1B的中点所以PM∥AA1,PM=2AA1又因为AA1∥CC1,且AA1=CC1,所以PM∥CN,且PM=CN,所以四边形PMCN为平行四边形,所以PN∥CM又因为CM平面A1BN,PNC平面A1BN,所以CM∥平面A1BN,6分)(2)取AC的中点O,连接BO,ON,M由题意知BO⊥AC,又因为平面A1ACC1⊥平面ABC,所以BO⊥平面A1ACC因为A1CC平面A1ACC1,所以BO⊥A:C,因为四边形A1ACC为菱形,所以A1C⊥AC1,又因为ON∥AC1,所以A1C⊥ON所以AC⊥平面BON,又BNC平面BON,所以A1C⊥BN.(12分)

2023赣州二模文综