课时作业5
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课时作业(五)1.在正方体ABCD -A 1B 1C 1D 1中,E 是上底面A 1B 1C 1D 1的中心,则AC 1与CE 的位置关系是( )A .重合B .垂直C .平行D .无法确定 答案 B2.已知斜三棱柱ABC -A 1B 1C 1中,底面ABC 是等腰直角三角形,AB =AC =2,CC 1=2,AA 1与AB ,AC 都成60°角,则异面直线AB 1与BC 1所成角的余弦值为( ) A.14 B.155 C.105 D.16 答案 D3.如图,正方体ABCD -A 1B 1C 1D 1的棱长为1,AM →=12MC 1→,点N 为B 1B 的中点,则|MN →|等于( )A.216 B.66 C.156 D.153 答案 A4.如图所示,在正方体ABCD -A 1B 1C 1D 1中,O 是底面正方形ABCD 的中心,M 是DD 1的中点,N 是A 1B 1的中点,则直线ON 与AM 的位置关系是( )A .平行B .垂直C .相交但不垂直D .无法判断答案 B5.对于空间任意一点O 和不共线的三点A ,B ,C ,有OP →=xOA →+yOB →+zOC →(x ,y ,z ∈R ),则“x =2,y =-3,z =2”是“P ,A ,B ,C 四点共面”的( ) A .必要不充分条件 B .充分不必要条件 C .充要条件 D .既不充分也不必要条件 答案 B解析 由题意,{OA →,OB →,OC →}构成空间的一个基底,OP →=xOA →+yOB →+zOC →(x ,y ,z ∈R ),则P ,A ,B ,C 四点共面等价于x +y +z =1.若x =2,y =-3,z =2,则x +y +z =1,所以P ,A ,B ,C 四点共面.若P ,A ,B ,C 四点共面,则x +y +z =1,不能得到“x =2,y =-3,z =2.所以“x =2,y =-3,z =2”是“P ,A ,B ,C 四点共面”的充分不必要条件. 6.如图,在大小为45°的二面角A -EF -D 中,四边形ABFE ,CDEF 都是边长为1的正方形,则B ,D 两点间的距离是( )A. 3B. 2C .1 D.3- 2 答案 D解析 {BF →,FE →,ED →}构成空间的一个基底,因为BD →=BF →+FE →+ED →,所以|BD →|2=|BF →|2+|FE→|2+|ED →|2+2BF →·FE →+2FE →·ED →+2BF →·ED →=1+1+1-2=3-2,故|BD →|=3- 2.7.已知空间四边形ABCD 的各边和对角线的长都等于a ,点M ,N 分别是AB ,CD 的中点,则MN ________AB (填“∥”或“⊥”). 答案 ⊥8.已知V 为矩形ABCD 所在平面外一点,且VA =VB =VC =VD ,若VP →=13VC →,VM →=23VB →,VN →=23VD →,则VA 与平面PMN 的位置关系是________.答案 平行解析 设VA →=a ,VB →=b ,VC →=c ,则{a ,b ,c }构成空间的一个基底,VD →=a +c -b .由题意知PM →=23b -13c ,PN →=23VD →-13VC →=23a -23b +13c .因此VA →=32PM →+32PN →,所以VA →,PM →,PN →共面.又VA ⊄平面PMN ,所以VA ∥平面PMN .9.如图,在正方体ABCD -A 1B 1C 1D 1中,P 是DD 1的中点,O 是底面ABCD 的中心.求证:B 1O ⊥平面P AC .解析 如图,连接BD ,则BD 过点O ,令AB →=a ,AD →=b ,AA 1→=c , 设|a |=|b |=|c |=1,∵AC →=AB →+AD →=a +b ,OB 1→=OB →+BB 1→=12DB →+BB 1→=12(AB →-AD →)+BB 1→=12a -12b +c .∴AC →·OB 1→=(a +b )·⎝⎛⎭⎫12a -12b +c =12|a |2+12a ·b -12a ·b -12|b |2+a ·c +b ·c =12-12=0. ∴AC →⊥OB 1→,即AC ⊥OB 1.又AP →=AD →+12DD 1→=b +12c ,∴OB 1→·AP →=⎝⎛⎭⎫12a -12b +c ·⎝⎛⎭⎫b +12c =12a ·b -12|b |2+c ·b +14a ·c -14b ·c +12|c |2=-12+12=0, ∴OB 1→⊥AP →, 即OB 1⊥AP .又AC ∩AP =A ,AC ,AP ⊂平面P AC ,∴OB 1⊥平面P AC .10.在正方体ABCD -A 1B 1C 1D 1中,已知E ,F ,G ,H 分别是CC 1,BC ,CD 和A 1C 1的中点.证明:(1)AB 1∥GE ,AB 1⊥EH ; (2)A 1G ⊥平面EFD .证明 (1)如图,设正方体棱长为1,AB →=i ,AD →=j ,AA 1→=k ,则{i ,j ,k }构成空间的一个单位正交基底.AB 1→=AB →+BB 1→=i +k ,GE →=GC →+CE →=12i +12k =12AB 1→,∴AB 1∥GE .EH →=EC 1→+C 1H →=12k +⎝⎛⎭⎫-12(i +j ) =-12i -12j +12k ,∵AB 1→·EH →=(i +k )·⎝⎛⎭⎫-12i -12j +12k =-12|i |2+12|k |2=0,∴AB 1⊥EH .(2)A 1G →=A 1A →+AD →+DG →=-k +j +12i .DF →=DC →+CF →=i -12j ,DE →=DC →+CE →=i +12k .∴A 1G →·DF →=⎝⎛⎭⎫-k +j +12i ·⎝⎛⎭⎫i -12j =-12|j |2+12|i |2=0,∴A 1G ⊥DF .A 1G →·DE →=⎝⎛⎭⎫-k +j +12i ·⎝⎛⎭⎫i +12k =-12|k |2+12|i |2=0,∴A 1G ⊥DE .又DE ∩DF =D ,DE ,DF ⊂平面EFD , ∴A 1G ⊥平面EFD .11.在棱长为1的正方体ABCD -A 1B 1C 1D 1中,M ,N ,H 分别在棱BB 1,BC ,BA 上,且满足BM →=34BB 1→,BN →=12BC →,BH →=12BA →,O 是平面B 1HN 、平面ACM 与平面B 1BDD 1的一个公共点,设BO →=xBH →+yBN →+zBM →,则x +y +3z =( ) A.105 B.125 C.145 D.165 答案 C解析 {BH →,BN →,BM →}构成空间的一个基底,如图,设Q 为AC 与BD 的交点,P 为BQ 的中点,连接MQ ,B 1P ,O 为MQ 与B 1P 的交点.过P 作PT ∥MQ 交BB 1于T .则T 为BM 的中点,所以MT =12BM =12×34BB 1=12×34×4MB 1=32MB 1.所以B 1O →=23OP →,因此BO→=35BB 1→+25BP →=35·43BM →+25·12(BH →+BN →)=45BM →+15BH →+15BN →,因为BO →=xBH →+yBN →+zBM →,所以z =45,x =15,y =15,所以x +y +3z =145.12.【多选题】如图,三棱柱ABC -A 1B 1C 1中,底面边长和侧棱长都等于1,∠BAA 1=∠CAA 1=60°.设AA 1→=a ,AB →=b ,AC →=c ,则( )A.BC 1→=a +b +cB .|BC 1→|= 2 C .BC 1⊥A 1B 1D .异面直线AB 1与BC 1所成角的正切值为 5 答案 BCD解析 以{a ,b ,c }为空间的一个基底,BC 1→=BB 1→+B 1C 1→=BB 1→+A 1C 1→-A 1B 1→=AA 1→+AC →-AB→=a +c -b ,A 不正确.因为a ·b =|a |·|b |cos ∠BAA 1=1×1×cos 60°=12,同理可得a ·c =b ·c =12,所以|BC 1→|=(a +c -b )2=a 2+c 2+b 2+2a ·c -2a ·b -2c ·b =1+1+1+1-1-1=2,B 正确.因为A 1B 1→=AB →=b ,所以A 1B 1→·BC 1→=b ·(a +c -b )=b ·a +b ·c -b 2=12+12-1=0,C 正确.因为AB 1→=a +b ,所以|AB 1→|=(a +b )2=a 2+b 2+2a ·b =1+1+1=3,因为AB 1→·BC 1→=(a +b )·(a +c -b )=a 2+a ·c -a ·b +b ·a +c ·b -b 2=1+12-12+12+12-1=1.所以cos〈AB 1→,BC 1→〉=AB 1→·BC 1→|AB 1→||BC 1→|=13×2=66,sin 〈AB 1→,BC 1→〉=1-⎝⎛⎭⎫662=306,tan 〈AB 1→,BC 1→〉= 5.所以异面直线AB 1与BC 1所成角的正切值为5,D 正确.13.【多选题】如图,已知斜三棱柱ABC -A 1B 1C 1中,∠BAC =π2,∠BAA 1=2π3,∠CAA 1=π3,AB =AC =1,AA 1=2,点O 是B 1C 与BC 1的交点.则下列结论正确的是( )A.AO →=12(AB →+AC →+AA 1→) B .|AO →|=32C .AO ⊥BCD .平面ABC ⊥平面B 1BCC 1 答案 AD解析 AO →=AB →+BO →=AB →+12(BC →+BB 1→)=AB →+12(AC →-AB →+AA 1→)=12(AB →+AC →+AA 1→),A 正确.设AB →=a ,AC →=b ,AA 1→=c ,则{a ,b ,c }构成空间的一个基底.|AO →|2=⎣⎡⎦⎤12(a +b +c )2=14(a 2+b 2+c 2+2a ·b +2b ·c +2a ·c )=14(1+1+4+0+2×1×2×cos π3+2×1×2×cos 2π3)=32,所以|AO →|=62,B 不正确.因为BC →=b -a ,所以AO →·BC →=12(a +b +c )·(b -a )=1≠0,C 不正确.如图,取BC 的中点E ,连接AE ,则AE →=12(AB →+AC →)=12(a +b ).因为AB =AC ,E 为BC 的中点,所以AE ⊥BC .又AE →·BB 1→=12(a +b )·c =12(1×2×cos 2π3+1×2×cos π3)=0,所以AE ⊥BB 1.因为BC ∩BB 1=B ,BB 1,BC ⊂平面B 1BCC 1,所以AE ⊥平面B 1BCC 1.又AE ⊂平面ABC ,所以平面ABC ⊥平面B 1BCC 1,D 正确.14.空间四边形OABC 的各边及对角线长均为2,E 是AB 的中点,F 在OC 上,且OF →=2FC →.(1)用{OA →,OB →,OC →}表示EF →=________;(2)向量OE →与向量BF →所成角的余弦值为________.答案 (1)-12OA →-12OB →+23OC → (2)-52142解析 (1){OA →,OB →,OC →}构成空间的一个基底,因为E 是AB 的中点,F 在OC 上,且OF →=2FC →,所以OE →=12(OA →+OB →),OF →=23OC →,于是EF →=OF →-OE →= 23OC →-12(OA →+OB →)=-12OA →-12OB →+23OC →.(2)由(1)得OE →=12(OA →+OB →),BF →=OF →-OB →=23OC →-OB →,因此|OE →|=12|OA →+OB →|=12×4+4+2×2×2×12=3,|BF →|=⎪⎪⎪⎪23OC →-OB →=49×4+4-43×2×2×12=273,又因为OE →·BF →=12(OA →+OB →)·⎝⎛⎭⎫23OC →-OB →=-53,所以向量OE →与向量BF →所成角的余弦值cos 〈OE →,BF →〉=OE →·BF →|OE →||BF →|=-533×273=-52142.15.在棱长为2的正四面体ABCD 中,点M 满足AM →=xAB →+yAC →-(x +y -1)AD →,点N 满足BN →=λBA →+(1-λ)BC →,当AM ,BN 最短时,AM →·MN →等于( )A .-43 B.43C .-13 D.13答案 A16.在如图所示的平行六面体ABCD -A 1B 1C 1D 1中,已知AB =AA 1=AD ,∠BAD =∠DAA 1=60°,∠BAA 1=30°,N 为A 1D 1上一点,且A 1N =λA 1D 1.若BD ⊥AN ,则λ的值为________.答案3-11.如图,直三棱柱ABC -A 1B 1C 1中,∠ABC =120°,AB =2,BC =CC 1=1,则异面直线AB 1与BC 1所成角的余弦值为( )A.32 B.155 C.105 D.33 答案 C 解析 设BA →=a ,BC →=b ,BB 1→=c ,则|a |=2,|b |=|c |=1, AB 1→=c -a ,BC 1→=b +c ,a ·b =|a |·|b |cos ∠ABC =2×1×cos 120°=-1,a ·c =b ·c =0, |AB 1→|=|c -a |=5,|BC 1→|=|b +c |=2, AB 1→·BC 1→=(c -a )·(b +c )=b ·c +c 2-a ·b -a ·c =0+1-(-1)-0=2,所以cos 〈AB 1→,BC 1→〉=AB 1→·BC 1→|AB 1→|·|BC 1→|=25×2=105.所以异面直线AB 1与BC 1所成角的余弦值为105. 2.如图,在空间四边形OABC 中,OA =8,AB =6,AC =4,BC =5,∠OAC =45°,∠OAB =60°,则OA 与BC 所成角的余弦值为( )A.3-225B.2-26C.12D.32 答案 A解析 因为OA →·BA →=8×6×cos 60°=24,OA →·AC →=8×4×cos 135°=-16 2.设异面直线OA与BC 的夹角为θ,则cos θ=|OA →·BC →||OA →||BC →|=|OA →·(BA →+AC →)||OA →||BC →|=24-1628×5=3-225.3.【多选题】如图,一个结晶体的形状为平行六面体ABCD -A 1B 1C 1D 1,其中,以顶点A 为端点的三条棱长均为6,且它们彼此的夹角都是60°,下列说法中正确的是( )A .AC 1=6 6B .AC 1⊥DBC .向量B 1C →与AA 1→的夹角是60° D .BD 1与AC 所成角的余弦值为63答案 AB解析 设AA 1→=a ,AB →=b ,AD →=c ,则{a ,b ,c }是空间的一个基底.因为以顶点A 为端点的三条棱长均为6,且它们彼此的夹角都是60°,所以AA 1→·AB →=AA 1→·AD →=AD →·AB →=a ·b =a ·c=c ·b =6×6×cos 60°=18,|AC 1→|2=(AA 1→+AB →+AD →)2=(a +b +c )2=a 2+b 2+c 2+2a ·b +2b ·c+2a ·c =36+36+36+3×2×18=216,则|AC 1→|=|a +b +c |=66,A 正确.AC 1→·DB →=(a +b +c )·(b -c )=a ·b -a ·c +b 2-b ·c +c ·b -c 2=0,B 正确.连接A 1D ,显然△AA 1D 为等边三角形,则∠AA 1D =60°.因为B 1C →=A 1D →,且向量A 1D →与AA 1→的夹角是120°,所以B 1C →与AA 1→的夹角是120°,C 不正确.因为BD 1→=c +a -b ,AC →=b +c ,所以|BD 1→|=(c +a -b )2=62,|AC →|=(b +c )2=63,BD 1→·AC →=(c +a -b )·(b +c )=36,所以cos 〈BD 1→,AC →〉=BD 1→·AC →|BD 1→||AC →|=3662×63=66,D 不正确.4.如图,在三棱柱ABC -A 1B 1C 1中,BC 1与B 1C 相交于点O ,∠A 1AB =∠A 1AC =60°,∠BAC =90°,A 1A =3,AB =AC =2,则线段AO 的长度为( )A.292 B.29 C.232 D.23 答案 A5.正四面体ABCD 中,M ,N 分别为棱BC ,AB 的中点,则异面直线DM 与CN 所成角的余弦值为________.答案 166.如图,在长方体ABCD -A 1B 1C 1D 1中,AB =1,BC =2,AA 1=3,E 为CC 1上的点,且CE =1,求异面直线AB 1,BE 所成角的余弦值.解析 AB 1→·BE →=(AB →+BB 1→)·(BC →+CE →) =AB →·BC →+AB →·CE →+BB 1→·BC →+BB 1→·CE →=0+0+0+3=3.依题意,易知|AB 1→|=10,|BE →|=5,所以cos 〈AB 1→,BE →〉=AB 1→·BE →|AB 1→|·|BE →|=310×5=3210.即异面直线AB 1,BE 所成角的余弦值为3210.。
课时作业(五)Unit 2Section ⅡLearning AboutLanguage层级一课时跟踪检测维度1单句语法填空1.This time tomorrow, we ________ (sit) in the cinema and watching a film.2.I feel so excited! At that time tomorrow morning, I ________ (fly) to Shanghai.3.In the past three years, Liu Mei ________ (make) great progress in her study.4.Daniel's family ________ (enjoy) their holiday in Huangshan this time next week.5.I ________ (fly) from Miami to New York to meet the other members of the group at this time tomorrow.6.At 7 o'clock this evening we ________ (celebrate) Tom's birthday.7.The car ________ (go) at the present speed until it reaches the foot of the mountain at about nine o'clock tonight.8.Will you ________ (work)on your report this time tomorrow?9.I ________ (tell) him the news when he comes back.10.Look at the dark clouds over there. I think it ________ (rain) soon.11.She ________ (go) to the zoo to see the lovely bear next week.12.Next Friday I will go to another concert.They ________ (play) some music by Mozart at that time.维度2语法与写作1.下个星期的这时候,我们将在那个工厂劳动。
课时作业(五) Writing Workshop, Viewing Workshop &Reading Clubs 1~2基础知识夯实Ⅰ.单词拼写根据汉语和首字母提示写单词。
1.When you are talking with him, you must pay attention to his r________ (回答;反应) to such a gesture.2.The f________ (灵活性) of distance learning would be suited to busy managers in particular.3.They help the body d________ (保卫) itself against some sorts of infections.4.He made no r________ (响应) to my question,and went on with his slides.5.The c________ (自行车运动员) should all have to be made to use the cycle lanes and wear helmets.6.Our original plan was o________ (突然发生) by events and we had to make a new one.7.John, a cyclist,got a bronze m________ (奖牌) in the race which took place last week.8.Nuclear energy may o________ (超过) oil as the main fuel in the future.Ⅱ.单句语法填空在空白处填入1个适当的单词或括号内单词的正确形式。
1.Not only you and I but Peter, the top student in our grade, ________ (be) not able to solve the problem.2.Wang Shu, a Chinese architect, won the 2012 Pritzker Architecture Prize, which ________ (refer) to as the Nobel Prize in architecture.3.A modern city has been set up in ________ was a wasteland ten years ago.4.Scientists have advanced many theories about why human beings shed tears, none of ________ has been proved.5.There is no way of knowing why one man makes an important discovery ________ another man, also intelligent, fails.6.The product was developed in ________ (respond) to customer demand.7.He is quite skilled and ________ (experience) though he has been in the factory for only a short time.8.Her bigcity friends were ________ (amaze) at the turn of events.9.In the distant past, friends hunted together and defended each other ________ dangerous animals or enemies.10.My readi ng was limited ________ stories by Englishspeaking authors.Ⅲ.根据汉语提示完成句子1.Parents ought to actively urge their children to ________(利用) the opportunity to join sports teams.2.The post has since ____________ (引起……的注意) social media users all over the world.3.Locke ________ (创立) this principle in the 17th century.4.The law was passed ____________________ (对……做出反应) public pressure.5.What does the underlined word “flexibility” in Paragraph 2 __________________(指的是)?6.I have some difficulty ____________________ (习惯) the senior high school life here.7.On the one hand, you have studied very hard, but ____________________ (另一方面), you don't have a good digestion of what you learned.8.Many audience packed into the gym ____________________ (为他们加油).Ⅳ.一句多译1.所谓网虫,就是上网上瘾的人。
课时作业(五) Unit1 PartⅤLesson3YourLifeIsWhatYouMakeItⅠ.单词拼写1.He has ________ (申请) for a position in a native company. 2.His ________ (自信的) leadership inspired his followers. 3.After we ________ (联系) with the charity organisation, we sent the orphan there. 4.We have lots of things in common ________ (除了) music. 5.The city council is ________ (负责的) for keeping the streets clean.6.The hotel has special ________ (设施) for welcoming disabled people.7.After ________ (毕业) from college, I took some time off to go travelling. 8.With his camera around his neck, he looked like a _______ _ (典型的) tourist.9.The thought suddenly ________ (闪过) through my mind that she didn't want to be here.10.As the world's population continues to grow, the ________ (供应) of food becomes more and more of a concern.Ⅱ.单句语法填空1.They are eager ________ (travel) abroad for further study. 2.They asked what his plans were after ________ (graduate).3.The government decided to supply the homeless ______ h ouses.4.Sophia intended to make a ________ (contribute) to charity all the time.5.Only when people overcome their fears do they live _____ ___ (confident).6.You have ________ (responsible) for collecting up the boo ks after the class.7.We aim to help disabled students to live and study ______ __ (independent).8.A career in the media is becoming increasingly ______ (att ract) to university graduates.9.________ (inspire) by her example, other zoologists (动物学家) have begun working with apes in the wild.10.The teacher, as well as his students, ________ (be) interv iewed shortly after the National Senior School English Compe tition for their perfect performance.11.The children are all ________ in the ________ story. (inter est)12.The film was so ________ that the audience were deeply ________.(move)13.Seeing the ________ result, we are all ________.(delight) 14.The ______ sound outside at night made her ______. (frig hten)15.I'm ________ that the ________ situation will go on. (worry )Ⅲ.完成句子1.I feel so confident that I can independently ________ (处理) all kinds of tough problems.2.Remember, you always ______________________________ __________ (有能力改变) your situation.3.Charlie Chaplin ________________________________ (对……做出了巨大贡献) the film industry.4.The book as well as the flowers _______________________ _ (是为……准备的) my mother, but my father thought they were for him and to ok them.5.Clearly and thoughtfully written, the book _______________ _________ students (激发了……的自信心) who wish to seek their own answers.Ⅳ.单句写作1.她这个人就是爱忘事。
2.3.3点到直线的距离公式一,选择题1.原点到直线x +2y -5=0的距离为( ) A .1 B. 3 C .2 D.52.已知点()(),20a a >到直线:30l x y -+=的距离为1,则a 等于( )11D.23.点(0,1)-到直线(1)y k x =+距离的最大值为( )A .1BCD .24.若点(),m n 在直线43100x y +-=上,则22m n +的最小值是( )A .2B .C .4D .5.点()sin P θθ到直线80x y ++=的距离的最小值为( )A.4B. C. D.6.设直线1:370l x y +-=与直线2:10l x y -+=的交点为P ,则P 到直线:20l x ay a ++-=的距离的最大值为( )B.4C.7.已知平面内两点(1,2),(2,2)A B -到直线l 的距离分別为23,,则满足条件 的直线l 的条数为( ) A.4B.3C.2D.18.直线l 过点A (3,4)且与点B (-3,2)的距离最远,那么l 的方程为( ) A .3x -y -13=0 B .3x -y +13=0 C .3x +y -13=0 D .3x +y +13=0二,填空题9.已知直线l 过点()3,4P 且与点()()2,2,4,2A B --等距离,则直线l 的方程为 . 10.若点(2,k )到直线5x -12y +6=0的距离是4,则k 的值是________.11.经过点P (-3,4),且与原点的距离等于3的直线l 的方程为________________. 三,简答题12.已知点()2,1P -.(1)求过点P 且与原点的距离为2的直线的方程.(2)是否存在过点P 且与原点的距离为6的直线?若存在,求出该直线的方程;若不存在,请说明理由.13.已知直线l 过两直线3450,2380x y x y +-=-+=的交点,且()()2,3,4,5A B -两点到直线l 的距离相等,求直线l 的方程.14.已知ABC 三边所在直线的方程分别为:3260,:23220AB AC l x y l x y -+=+-=,:340(,30)BC l x y m m m +-=∈≠R .(1)判断ABC 的形状;(2)当BC 边上的高为1时,求实数m 的值.15.已知直线l :x -2y +8=0和两点A (2,0),B (-2,-4). (1)在直线l 上求一点P ,使|P A |+|PB |最小; (2)在直线l 上求一点P ,使||PB |-|P A ||最大.参考答案1. D【解析】d =|0+2×0-5|12+22= 5.2. B【解析】由点到直线的距离公式,得1=,即|1|a +=0,1a a >∴,故选B. 3. B【解析】解法一:由点到直线的距离公式知,点(01)-,到直线(1)y kx =+的距离d ===当0k=时,1d =;当0k ≠时,d ,要使d 最大,需0k >且1k k +最小,∴当1k =时,max d = B.解法二:记点(01)A -,,直线(1)y k x =+恒过点(10)B -,,当AB 垂直于直线(1)y k x =+时,点(01)A-,到直线(1)y k x =+的距离最大,且最大值为||AB = B. 4. C【解析】因为(,)m n 在直线43100x y +-=上,所以43100m n +-=,则利用22m n +表示为直线上点到原点距离的平方的最小值来分析可知,为4 5. C【解析】点()sin ,3cos P θθ到直线80x y ++=的距离为d ==≥=故选C.6. A 【解析】由37010x y x y +-=⎧⎨-+=⎩,得12x y =⎧⎨=⎩,故()1,2P .直线l 的方程可整理为()210x a y ++-=,故直线l 过定点()2,1Q -.因为点P 到直线l的距离d PQ ≤,当且仅当l PQ ⊥时等号成立,所以max ||d PQ ===,故选A.7. B【解析】由题知满足题意的直线l 在线段AB 两侧各有1条,又因为||5AB =,所以还有1条为过线段AB 的一点且与AB 垂直 的直线,故共有3条直线满足题意.故选B. 8. C【解析】由已知可知,l 是过A 且与AB 垂直的直线,∵k AB =2-4-3-3=13,∴k l =-3, 由点斜式得,y -4=-3(x -3),即3x +y -13=0. 9. 23180x y +-=或220x y --=【解析】当直线斜率不存在时,直线方程为2x =-,不符合题意,设直线斜率为k ,则直线l 的方程为()34y k x =-+,整理得430kx y k -+-=,点A ,点B=,求得2k =或23-∴直线l 的方程为:23180x y +-=或220x y --=, 故为:23180x y +-=或220x y --=. 10. -3或173【解析】∵|5×2-12k +6|52+122=4,∴|16-12k |=52,∴k =-3或k =173.11. x =-3或7x +24y -75=0【解析】(1)当直线l 的斜率不存在时,原点到直线l :x =-3的距离等于3,满足题意; (2)当直线l 的斜率存在时,设直线l 的方程为y -4=k (x +3),即kx -y +3k +4=0. 原点到直线l 的距离d =|3k +4|k 2+(-1)2=3,解得k =-724.直线l 的方程为7x +24y -75=0.综上可知,直线l 的方程为x =-3或7x +24y -75=0.12.解:(1)①当直线的斜率不存在时,直线方程为2x =,符合题意; ②当直线的斜率存在时,设斜率为k ,则直线方程为()12y k x +=-,即210kx y k ---=.2=,解得34k =, 所以直线方程为34100x y --=.故符合题意的直线方程为20x -=或34100x y --=. (2)不存在.过点P 且与原点的距离最大的直线为过点P 且与OP 垂直的直线,此时最大距离为||OP而6>.13.解:由34502380x y x y +-=⎧⎨-+=⎩,得12x y =-⎧⎨=⎩,即交点为()1,2-.①当直线l 的斜率存在时,设直线l 的方程为()21y k x -=+,即20kx y k -++=.=,解得13k =-,所以直线l 的方程为()1213y x -=-+,即350x y +-=. ②当直线l 的斜率不存在时,直线l 的方程为1x =-,符合题意. 综上,可知所求直线l 的方程为350x y +-=或1x =-. 14.解:(1)直线AB 的斜率为32AB k =,直线AC 的斜率为23AC k =-,所以1AB AC k k ⋅=-,所以直线AB 与AC 互相垂直,因此ABC 为直角三角形. (2)由326023220x y x y -+=⎧⎨+-=⎩,得26x y =⎧⎨=⎩,即A 点坐标为()2,6.由点到直线的距离公式,得点A 到BC 边的距离即BC 边上的高为|30|5m d -==, 所以|30|15m -=,即|30|5m -=,解得25m =或35m =. 15.解:(1)设A 关于直线l 的对称点为A ′(m ,n ),则⎩⎪⎨⎪⎧n -0m -2=-2,m +22-2·n +02+8=0,解得⎩⎪⎨⎪⎧m =-2,n =8,故A ′(-2,8).因为P 为直线l 上的一点, 则|P A |+|PB |=|P A ′|+|PB |≥|A ′B |,当且仅当B ,P ,A ′三点共线时,|P A |+|PB |取得最小值,为|A ′B |,点P 即是直线A ′B 与直线l 的交点,则⎩⎪⎨⎪⎧ x =-2,x -2y +8=0,得⎩⎪⎨⎪⎧x =-2,y =3,故所求的点P 的坐标为(-2,3). (2)A ,B 两点在直线l 的同侧,P 是直线l 上的一点, 则||PB |-|P A ||≤|AB |,当且仅当A ,B ,P 三点共线时,||PB |-|P A ||取得最大值,为|AB |,点P 即是直线AB 与直线l 的交点,又直线AB 的方程为y =x -2,则⎩⎪⎨⎪⎧ y =x -2,x -2y +8=0,得⎩⎪⎨⎪⎧x =12,y =10,故所求的点P 的坐标为(12,10).。
课时作业(五) UNIT 2 Section ⅢDiscovering UsefulStructures层级一即时应用体验维度1 单句语法填空1.I noticed a thief ________ (steal) money from the old man's pocket when I passed by.2.With more and more farmers ________ (rush) into city,their children's education becomes a problem.3.________ (compare) with many other women, she leads a very happy life.4.Their car was caught in a traffic jam, ________ (cause) them to be late.5.________ (give) another five minutes, I can finish the work on time.维度2 用动词-ing形式完成语段A teacher once told each of her students to bring a clear plastic bag and a large bag of potatoes to school.1.____________________ (写下……名字)the person that the students refused to forgive in their life on a potato and put them in the plastic bag,they were then told to carry these bags with them everywhere for one week,2.________________________ (把他们放在床边)at night,on their car seats3.________________ (在开车的时候), and next to their desks at work.During this time,4.__________________________ (带着包到处走),they realized what a weight they were, and how they needed to pay attention all the time,so as not to forget them or leave them in embarrassing places.Of course,the potatoes were becoming rotten,5.________________ (闻起来很糟糕).维度3 句型转换1.As time passes by, we will have a better life.→____________, we will have a better life.2.I stood on the bridge and watched boats were passing by.→I stood on the bridge, ________________.3.I heard that Mary sang a song in the next room this time last night.→I heard Mary________________ in the next room this time last night.4.After he had eaten his dinner, the boy rushed out.→________________, the boy rushed out.5.When she turned around,she saw a car driving up.→________________,she saw a car driving up.6.As I did not know how to get there, I had to ask the way.→________________________, I had to ask the way.维度4 语法与语篇用适当的动词-ing形式完成下列短文Mary Smith looked at the beautiful ripe plums(梅子).They would make lovely jam. After she had finished 1.________ (cook),she filled all her empty jam jars 2.________ (leave) the rest of the jam in the pan. She would put it in the fridge when it was cooler.But just then the telephone rang.3.________________ (learn) that her mother was in hospital after a car accident,Mary picked up her bag and ran out of the house.Some days later, her husband,John,came home from a business trip.He had been travelling all day and felt like 4.________ (have) a drink and a piece of cake.5.________ (enter) the kitchen, he saw a pan with a dark red mess inside it.He lifted it up and smelled it.It smelled horrible.6.________ (think) Mary must have forgotten to clean this pan, he poured all the jam into the chicken yard and cleaned the pan.Then 7.________ (feel) comfortable, he began to eat a piece of cake.When Mary returned,she noticed the chickens 8.________ (behave) strangely.They were running round the yard as if they were sick. She saw the dark red mess on the ground and went closer.9.________ (see) a plum stone,she went into the kitchen.Her husband was at the table 10.________ (read) a newspaper. Angrily,Mary rushed up to him shouting “You threw away my jam!” Her husband said,“I'm sorry,but I thought it was porridge gone bad in the hot weather.”层级二主题阅读训练Ⅰ.阅读理解AWith the help of the car manufacturer (汽车制造商) Hyundai, a 13-year-old girlwas able to send a message to space, where her dad works on the International Space Station (ISS).“He gets to live and work in space and he is doing lots of experiments up there. He has to stay there for long periods of time... I miss him when he is gone, ” Stephanie, who is from Houston, Texas, said in a video. “I think if we could write a really big message he would be able to see it from space. ”Hyundai took her wish to heart and decided to take on the challenge. Using Nevada's Delamar Dry Lake as a canvas (画布), 11 drivers drove Hyundai cars, spelling out “Steph s you” across 2.14 square miles of desert. A promotional video, which Hyundai made to record the process, caught her father's response and showed the picture he took of her message from the ISS.“I am happy that he could see it and knows that we are think ing about him back home, ” Stephanie said. “He has seen so many things up there, but I hope that this message was the most special. ”There are many misunderstandings about what can and cannot be seen from space. Contrary to popular belief, the Great Wall of China cannot be seen from space, but many seemingly less important things can be seen.“A farmer from Louisiana could be burning wastes in his backyard, and it would make a big smoke trail that astronauts could see from space, ” said Mike Gentry, a photo researcher for NASA's human-tended vehicles. And with a special camera, they can see the earth in great detail and even keep an eye on their favorite sports teams.1.What can we know about Stephanie's father?A.He likes taking pictures.B.He spends little time at home.C.He works in a car company.D.He often sends messages to his daughter.2.How did Hyundai help the girl?A.By offering a free ride.B.By taking her to a guided tour.C.By replaying her father's response.D.By making a large picture with cars.3.Which of the following can be seen from space?A.A moving car.B.A standing farmer.C.The Great Wall of China.D.The smoke of burning rubbish.4.What is the best title for the text?A.Send Love to SpaceB.A Girl of Great TalentC.Observe the Earth Far AwayD.A New Way to Explore SpaceB[2022·福建师大附中高一期末]When it comes to who is happier, people with kids or those without, most research points to the latter. But a new study suggests that parents are happier than non-parents later in life, when their children move out and become sources of social enjoyment rather than stress.Most surveys of parental happiness have focused on those whose children still live at home. These tend to show that people with kids are less happy than their child-free peers because they have less free time, sleep and money.Christoph Becker at Heidelberg University in Germany and his colleagues wondered if the story might be different for parents whose kids have left home. To find out, they analyzed data from a European survey that asked 55,000 people aged 50 and older about their emotional well-being.They found that, in this older age group, people with children had greater life satisfaction and fewer symptoms of depression than people without children, but only if their kids had left home.“This may be because when children grow up and move out, they provide social enrichment to their parents minus (抵消) the day-to-day stress of looking after them,” says Becker. They may also give something back by providing care andfinancial sup port to their parents, he says. “Hence, children's role as caregivers, financial support or simply as social contact might outweigh negative aspects of parenthood,” he says.If parents disobey the idea of waiting for their kids to move out to maximize their potential happiness, they could move to a country with better childcare support, says Becker. A 2016 study of 22 countries found that parents with children at home were actually slightly happier than their child-free peers if they lived in places like Norway, Portugal and Sweden that have paid parental leave and generous childcare subsidies (补助).5.Why did Christoph Becker and his colleagues analyze data from a European survey?A.To show their opinions are different.B.To prove the earlier findings are wrong.C.To prove if parents can be happier under certain conditions.D.To figure out old people's emotional well-being.6.According to the new study, what is the key point of parental happiness?A.Whether the kids have moved out or not.B.Which country they choose to live in.C.Whether kids are to play roles of caregivers.D.Whether parents are willing to wait for kids' growth.7.How can people with kids at home achieve more happiness according to Becker?A.Moving to another country without trouble from kids.B.Asking their kids to move out as soon as possible.C.Living in a country with better policies on childcare.D.Paying for parental leave, generous childcare subsidies.8.What is the main idea of this passage?A.How people can achieve happiness in their life.B.People without children are happier than those with kids.C.Why people have greater life satisfaction in their older age.D.Parents whose children have left home feel happier than non-parents.Ⅱ.完形填空[2022·海南省琼海市嘉积中学高一下检测]Six years ago when I was told that my father suffered from dementia(痴呆), everything changed overnight. Until then, I was loved and lived with little __1__. My father became a child who needed round -the-clock __2__ for everyday activities. __3__, we learnt that the best way to deal with him was with patience and __4__.Our family started __5__ more time and activities together——Sundays were set aside for a drive, a park visit and dinner at a restaurant. Each __6__ gave us the chance to connect as a family.They were special to my father and gave him great __7__. It was also a time of __8__—— I found myself tested in different ways and found perseverance (坚毅) that I didn't __9__ I had.By December last year, my father's condition became more __10__. Unable to deal with his illness, we were __11__ to send him to a centre for dementia care.We brought my father home in February when he got __12__. The next month, the government advised us to stay home because of COVID-19. I was worried that he would miss his walks and would get __13__ if he had to stay home all day long, but when we explained the __14__ to him, he understood. Perhaps it was because he knew hisfamily would __15__ with him.,1.A.pay B.honourC.worry D.knowledge2.A.schedule B.purposeC.trust D.help3.A.Slowly B.ClearlyC.Similarly D.Usually4.A.power B.kindnessC.love D.bravery5.A.planning B.missingC.needing D.remembering6.A.game B.outingC.discussion D.visit7.A.luck B.joyC.pride D.surprise8.A.self-control B.self-studyC.self-discovery D.self-pity9.A.agree B.meanC.guess D.know10.A.serious B.commonC.important D.special11.A.ordered B.forcedC.trained D.wished12.A.better B.happierC.safer D.quicker13.A.injured B.lostC.concerned D.upset14.A.lesson B.changeC.story D.reason15.A.talk B.stayC.work D.waitⅢ.语法填空Loyalty is one of the most important 1.________ (quality) a person can have, a retired farmer Alastair Eckhoff said.Mr Eckhoff was a stock and station agent (代理商) for 38 years and owns a sheep farm in Moa Creek. When 2.________ (ask) what wisdom he would like to pass on to the next generation, he said,“The really major one is loyalty.”Loyalty proved its worth early in his career (职业), when he was just starting out as 3.________ young employee on the Wrightson's experimental farm near Lincoln.“I used to complain 4.________ the pay, as I was only getting 600 a year,”he said.His boss told 5.________ (he) that things would improve if he was loyal and did the job he was asked to do.Later, he was given a stock and station agent position in Omakau, which came with a high pay, a car and a house.He said it was also 6.________ (extreme) important to be loyal to a company and not always go after the best markets or the best deals.When he was younger, he 7.________ (help) into his farm by the original, older owner.He said he decided to lease (出租) his farm to a young couple 8.________ were just starting out earlier this year.“I was given the opportunity 9.________(go) farming,”he said.“Now, I am doing the same thing. I am 10.________ (excite) to pay it forward.”课时作业(五)层级一即时应用体验维度11.stealing 2.rushing pared 4.causing 5.Given维度21.Having written the name of 2.putting them beside their beds 3.while driving 4.carrying the bags around with them5.smelling very bad维度31.With time passing by 2.watching boats passing by3.singing a song 4.Having eaten his dinner5.Turning around 6.Not knowing how to get there维度41.cooking 2.leaving 3.Learning 4.having 5.Entering 6.Thinking 7.feeling 8.behaving 9.Seeing10.reading层级二主题阅读训练Ⅰ.阅读理解A【语篇解读】本文为记叙文。
课时作业(五)一、语言基础训练1.下列关于括号中属于何种戏剧语言的分析,正确的一项是()①周朴园(点着一支吕宋烟,看见桌上的雨衣,向侍萍)②鲁侍萍(哦,天哪,我觉得我像在做梦。
)③周朴园(我问,他现在在哪儿?)A.①对话②旁白③独白B.①舞台说明②独白③对话C.①独白②旁白③对话D.①背景介绍②对话③旁白答案 B解析戏剧语言包括台词和舞台说明。
①是舞台说明,说明演出时周朴园扮演者的动作。
②结合文意可知是鲁侍萍的独白。
③结合文意可知是周朴园与鲁侍萍的对话。
2.依次填入文中横线上的词语,最恰当的一组是() 人类时不时就会捕风捉影地寻找一些外星人存在的证据。
但这些所谓证据往往缺乏________考证,很难使人信服。
杨利伟作为中国进入太空第一人,关于他的航天故事许多人________,然而,鲜为人知的是,他进入太空的时候,还带了一把手枪。
因此,有谣言说,杨利伟带手枪是为了防止外星人的攻击。
每隔一段时间,就有这样一条谣言会在网上流传:美国联邦调查局(FBI)________了一批密件,其中一份证实了外星人的存在。
其实,FBI只是把已解密的材料放到官方网站上,用户可以________该网站,浏览下载。
只有极少数几篇提到对外星人和UFO(不明飞行物)的调查而已,却成了人们关注的焦点。
A.严谨的耳熟能详披露登录B.细致的耳濡目染披露登陆C.细致的耳熟能详揭露登陆D.严谨的耳濡目染揭露登录答案 A解析严谨:强调严密谨慎。
细致:强调精细周密。
文中强调谨慎的态度,应用“严谨的”。
耳熟能详:听得次数多了,熟悉得能详尽地说出来。
耳濡目染:形容听得多见得多了之后,无形之中受到影响。
由后文语境“然而,鲜为人知的是”可知,这里填“耳熟能详”更恰当。
披露:发表,公布。
揭露:使隐蔽的事物显露。
语境应用表示公布的“披露”。
登录:登记、注册。
登陆:渡过海洋或江河,登上陆地;商品打入某地市场。
网站应用“登录”。
3.(2019·天津)为纪念五四运动一百周年,某中学文学社团准备举办以某位作家为专题的展览。
课时作业(五)导数及其应用一、选择题1.(2013·浙江高考)已知函数y=f(x)的图象是下列四个图象之一,且其导函数y=f′(x)的图象如图1-5-1所示,则该函数的图象是()图1-5-1【解析】从导数的图象可看出,导函数值先增大后减小,当x=0时,f′(x)有最大值.结合导数的几何意义知,B正确.【答案】 B2.(2013·临沂模拟)已知f(x)=x3-92x2+6x-abc,a<b<c,且f(a)=f(b)=f(c)=0,现给出如下结论:①f(0)f(1)>0;②f(0)f(1)<0;③f(0)f(2)>0;④f(0)f(2)<0.其中正确结论的序号为()A.①③B.①④C.②④D.②③【解析】函数的导数为f′(x)=3x2-9x+6=3(x2-3x+2)=3(x-1)(x-2).则函数在x=1处取得极大值,在x=2处取得极小值,因为f(a)=f(b)=f(c)=0,所以函数有3个零点,则f(1)>0,f(2)<0,即⎩⎪⎨⎪⎧f (1)=1-92+6-abc >0,f (2)=23-92×22+6×2-abc <0,解得⎩⎪⎨⎪⎧abc <52,abc >2,即2<abc <52,所以f (0)=-abc <0,所以f (0)f (1)<0,f (0)f (2)>0.所以选D.【答案】 D3.(2013·青岛模拟)已知偶函数f (x )在R 上的任一取值都有导数,且f ′(1)=1,f (x +2)=f (x -2),则曲线y =f (x )在x =-5处的切线的斜率为( )A .2B .-2C .1D .-1【解析】 由f (x +2)=f (x -2)得,f (x +4)=f (x ),即4是函数f (x )的一个周期,又函数f (x )是偶函数,从而f (-5)=f (-1)=f (1),根据函数图象的对称性知,函数f (x )在x =-5处的切线的斜率k =f ′(-5)=f ′(-1)=-f ′(1)=-1.故选D.【答案】 D4.函数f (x )=ln xx 2的最大值为( ) A .e B .1e C .2eD .12e【解析】 函数f (x )的定义域为(0,+∞), 又f ′(x )=1x ·x 2-ln x ·2x x 4=1-2ln x x 3.令f ′(x )=0得x =e ,且当0<x <e 时,f ′(x )>0, 当x >e 时,f ′(x )<0,所以f (x )在x =e 处取得极大值,也就是函数在定义域上的最大值f (e)=12e . 【答案】 D5.(2013·天津高考)设函数f (x )=e x +x -2,g (x )=ln x +x 2-3.若实数a ,b 满足f (a )=0,g (b )=0,则( )A .g (a )<0<f (b )B .f (b )<0<g (a )C .0<g (a )<f (b )D .f (b )<g (a )<0【解析】 ∵f ′(x )=e x +1>0,∴f (x )是增函数.∵g (x )的定义域是(0,+∞),∴g ′(x )=1x +2x >0,∴g (x )是(0,+∞)上的增函数.∵f (0)=-1<0,f (1)=e -1>0,∴0<a <1.∵g (1)=-2<0,g (2)=ln 2+1>0,∴1<b <2,∴f (b )>0,g (a )<0.【答案】 A 二、填空题6.(2013·广东高考)若曲线y =kx +ln x 在点(1,k )处的切线平行于x 轴,则k =________.【解析】 y ′=k +1x ,由导数y ′|x =1=0,得k +1=0,则k =-1. 【答案】 -17.已知函数f (x )=12mx 2+ln x -2x 在定义域内是增函数,则实数m 的取值范围为________.【解析】 f ′(x )=mx +1x -2≥0对一切x >0恒成立, m ≥-(1x )2+2x ,令g (x )=-(1x )2+2x ,则当1x =1时,函数g (x )取最大值1,故m ≥1. 【答案】 [1,+∞)8.(2013·鄂州模拟)已知f (x )=x e x ,g (x )=-(x +1)2+a ,若∃x 1,x 2∈R ,使得f (x 2)≤g (x 1)成立,则实数a 的取值范围是________.【解析】 f ′(x )=e x +x e x =(1+x )e x ,当x >-1时,f ′(x )>0函数递增;当x <-1时,f ′(x )<0函数递减,所以当x =-1时f (x )取得极小值即最小值f (-1)=-1e .函数g (x )的最大值为a ,若∃x 1,x 2∈R ,使得f (x 2)≤g (x 1)成立,则有g (x )的最大值大于或等于f (x )的最小值,即a ≥-1e .【答案】 ⎣⎢⎡⎭⎪⎫-1e ,+∞三、解答题9.已知函数f(x)=ax2+1(a>0),g(x)=x3+bx.(1)若曲线y=f(x)与曲线y=g(x)在它们的交点(1,c)处具有公共切线,求a,b的值;(2)当a=3,b=-9时,若函数f(x)+g(x)在区间[k,2]上的最大值为28,求k 的取值范围.【解】(1)f′(x)=2ax,∴f′(1)=2a.又f(1)=a+1=c,∴f(x)在点(1,c)处的切线方程为y-c=2a(x-1),即y-2ax+a-1=0.又∵g′(x)=3x2+b,则g′(1)=3+b.又g(1)=1+b=c,∴g(x)在点(1,c)处的切线方程为y-(1+b)=(3+b)(x-1),即y-(3+b)x+2=0.依题意知3+b=2a,且a-1=2,即a=3,b=3.(2)记h(x)=f(x)+g(x).当a=3,b=-9时,h(x)=x3+3x2-9x+1,h′(x)=3x2+6x-9.令h′(x)=0,得x1=-3,x2=1.h(x)与h′(x)在(-∞,2]上的变化情况如下:当k≤-3时,函数h(x)在区间[k,2]上的最大值为h(-3)=28;当-3<k<2时,函数h(x)在区间[k,2]上的最大值小于28.因此,k的取值范围是(-∞,-3].10.(2013·济南模拟)已知函数f(x)=(ax2+x-1)e x,其中e是自然对数的底数,a∈R.(1)若a=1,求曲线f(x)在点(1,f(1))处的切线方程;(2)若a<0,求f(x)的单调区间;(3)若a=-1,函数f(x)的图象与函数g(x)=13x3+12x2+m的图象有3个不同的交点,求实数m的取值范围.【解】 (1)因为f (x )=(x 2+x -1)e x , 所以f ′(x )=(2x +1)e x +(x 2+x -1)e x =(x 2+3x )e x ,所以曲线f (x )在点(1,f (1))处的切线斜率为k =f ′(1)=4e. 又因为f (1)=e ,所以所求切线方程为y -e =4e(x -1), 即4e x -y -3e =0.(2)f ′(x )=(2ax +1)e x +(ax 2+x -1)e x =[ax 2+(2a +1)x ]e x , ①若-12<a <0,当x <0或x >-2a +1a 时,f ′(x )<0; 当0<x <-2a +1a 时,f ′(x )>0.所以f (x )的单调递减区间为(-∞,0],⎣⎢⎡⎭⎪⎫-2a +1a ,+∞; 单调递增区间为⎣⎢⎡⎦⎥⎤0,-2a +1a ②若a =-12,f ′(x )=-12x 2e x ≤0,所以f (x )的单调递减区间为(-∞,+∞). ③若a <-12,当x <-2a +1a 或x >0时,f ′(x )<0; 当-2a +1a <x <0时,f ′(x )>0.所以f (x )的单调递减区间为⎝ ⎛⎦⎥⎤-∞,-2a +1a ,[0,+∞); 单调递增区间为⎣⎢⎡⎦⎥⎤-2a +1a ,0 (3)由(2)知,f (x )=(-x 2+x -1)e x 在(-∞,-1]上单调递减,在[-1,0]单调递增,在[0,+∞)上单调递减,所以f (x )在x =-1处取得极小值f (-1)=-3e ,在x =0处取得极大值f (0)=-1.由g (x )=13x 3+12x 2+m ,得g ′(x )=x 2+x . 当x <-1或x >0时,g ′(x )>0;当-1<x <0时,g ′(x )<0.所以g (x )在(-∞,-1]上单调递增,在[-1,0]单调递减,在[0,+∞)上单调递增.故g (x )在x =-1处取得极大值g (-1)=16+m ,在x =0处取得极小值g (0)=m .因为函数f (x )与函数g (x )的图象有3个不同的交点. 所以⎩⎨⎧f (-1)<g (-1),f (0)>g (0).即⎩⎪⎨⎪⎧-3e <16+m ,-1>m .所以-3e -16<m <-111.(2013·新课标全国Ⅱ)已知函数f (x )=x 2e -x . (1)求f (x )的极小值和极大值.(2)当曲线y =f (x )的切线l 的斜率为负数时,求l 在x 轴上截距的取值范围. 【解】 (1)f (x )的定义域为(-∞,+∞), f ′(x )=-e -x x (x -2).①当x ∈(-∞,0)或x ∈(2,+∞)时,f ′(x )<0; 当x ∈(0,2)时,f ′(x )>0.所以f (x )在(-∞,0),(2,+∞)上单调递减,在(0,2)上单调递增. 故当x =0时,f (x )取得极小值,极小值为f (0)=0;当x =2时,f (x )取得极大值,极大值为f (2)=4e -2.(2)设切点为(t ,f (t )),则l 的方程为y =f ′(t )(x -t )+f (t ). 所以l 在x 轴上的截距为m (t )=t -f (t )f ′(t )=t +t t -2=t -2+2t -2+3.由已知和①得t ∈(-∞,0)∪(2,+∞).令h (x )=x +2x (x ≠0),则当x ∈(0,+∞)时,h (x )的取值范围为[22,+∞);当x ∈(-∞,-2)时,h (x )的取值范围是(-∞,-3).所以当t ∈(-∞,0)∪(2,+∞)时,m (t )的取值范围是(-∞,0)∪[22+3,+∞).综上,l 在x 轴上的截距的取值范围是(-∞,0)∪[22+3,+∞).。