新课程高中数学测试题组(必修5)含答案(8)

2020-04-14 06:03

三、解答题

1. 解:Sn?3?2n,Sn?1?3?2n?1,an?Sn?Sn?1?2n?1(n?2)

?5,(n?1)而a1?S1?5,∴an??n?1

?2,(n?2)2. 解:设此数列的公比为q,(q?1),项数为2n,

1?(q)1?q22n则S奇?S偶S奇?85,S偶?2na2(1?q1?q2n2n)2?170,

?a2a1?q?2,1?21?4?85,2?256,2n?8,

∴q?2,项数为8

3. 解:an?3?(n?1)lg2,?an?是以3为首项,以?lg2为公差的等差数列,

Sn?n2[3?3?(n?1)lg2]??lg22*n?26?lg22n,

对称轴n?6?lg22lg2?10.47,n?N,10,11比较起来10更靠近对称轴

∴前10项和为最大。

?an?0另法:由?,得9.9?n?10.9

a?0?n?1?n?(?4),n为偶数???2n,n为偶数?24. 解:Sn?? ,Sn??,?2n?1,n为奇数?n?1?(?4)?4n?3,n为奇数??2 S15?29,S22??44S,31?6 1,S15?S22?S31??76

新课程高中数学训练题组参考答案(13976611338)

参考答案(数学5必修)第三章 [基础训练A组]

一、选择题

1.C ?2x?5x?2?0,(2x?1)(x?2)?0,212?x?2,

31

4x?4x?1?2x?2?2x?1?2x?2?2x?1?4?2x?3

22.B 对于A.2x?7,x?72,与 2x?x?7?x,0?x?72

对于C.x?3?1,x?3?1或x?3??1与x?3?1

对于D.(x?1)3?x3与

1x?1?1,

x当?1?x?0时,

1x?1?1

不成立

x1x?214?2x,x2?1?4?2x,x2?2x?3?0,?3?x?1,?y?2 ?2?()48114.C 对于A,B,倒数法则:a?b,ab?0??,要求a,b同号,

ab3.B 2x2?11?b??1?b?1,而a?1,对于a?2b的反例:a?1.1,a?1.21,b?0.8,2b?1.6

2225.B 设x?cos?,y?sin?,1?xy?1?2214sin2?

26.C 令f(x)?x2?(a2?1)x?a?2,则f(1)?0且f(?1)?0

2??a?a?0 即? ,?1?a?02??a?a?3?0二、填空题 1.1,?1222 ??4(m?12)?4m(3?m4n?n4?222 ?)20 2m?4mn?4n?2m?1?0,即(m?2n)?(m?1)?0

22而(m?2n)?(m?1)?0,即(m?2n)?(m?1)?0?m?1,且n??22212

2.13或24 设十位数为a,则个位数为a?2,

10a?a?2?30,a?2811,a?N?a?1或,2,即13或24

*3.????33111111?2,? ?x?x?0,??x?,递减则x??, ∴??x?

42222222?2244. ?1,大,1 y?x(2?x)??x?2x2??(x2?1),当?1x?1时,ym22ax?1

5. f(n)??(n)?g(n) f(n)?三、解答题

?x2?3?1或1. 解:(1)??2x?3?1

1n?1?n2g,n(?)1n?1?n2?,n(?)1n?n2 ?0?x2?3?1 得x?2或3?x?2, ??0?2x?3?132

?x?(3,2)?(2,??)

3?12x?x??42??123?2?x?2x?1?02(2)2?x?x??4,? ,?222?x?2x?5?0?1x2?x?3?2???22 ???x?2?1或x?????6?1?x?2?1,

6?16?1)

?x?(?6?1,?2?1)?(2?1,2. 解:?x2?8x?20?0恒成立,?mx2?2(m?1)x?9m?4?0须恒成立

当m?0时,2x?4?0并不恒成立;

?m?0当m?0时,则? 2???4(m?1)?4m(9m?4)?0?m?01?得?11 ?m??

2?m?,或m???423.解:(1)作出可行域 Zmax?3 ;(2)令x?5x',y?4y',

则(x')2?(y')2?1,z?10x'?4y',当直线z?10x'?4y'和圆(x')2?(y')2?1 相切时z?116,Zmax?116 4.证明:?log?a?1?a?loga?a?1??lgalg(a?1)?lg(a?1)lga?lga?lg(a?1)lg(a?1)lgalg(a?1)22

2?lg(a2?1)?lga2?lg(a?1)?lg(a?1)?2??()?lga 而lg(a?1)lg(a?1)?????222????2

即lg2a?lg(a?1)lg(a?1)?0,而a?2?lg(a?1)lga?0

?lga?lg(a?1)lg(a?1)lgalg(a?1)2?0,即log?a?1?a?loga?a?1??0

?log?a?1?a?loga?a?1?

参考答案(数学5必修)第三章 [综合训练B组]

一、选择题

33

1.D 方程ax2?bx?2?0的两个根为??12?1??b,?1?1?212和

13,

3a23a12x?112.B ?2,?0,x?,或x?0

xx2,a??12,b??2,a?b??14

3.B ?k2?2k?52?(k?1)?232?1,?x?1?x,x?12

1sinx4.D 对于A:不能保证x?0,对于B:不能保证sinx?2,

对于C:不能保证x?2?1x?22,

对于D:y?x?1x?1x?1?331?1?2

5.D 设x?cos?,y?sin?,3x?4y?3cos??4sin??5sin(???)?5

?a?b?c?3,a?c?2,c?2?a,0?2?a?1,1?a?2 6.B ?a?b?c?1?二、填空题

2221.???,?1???1,??? x2?2xy?y?y?1?1,(x?y)?1,x?y或?1x?y ??12.???,6?

a?b?ab?3?2ab,(ab)?2ab?3?0,2ab?3,ab?9,ab?3?6

A??x|x?a?b?ab?3,1a?,a?b?R??6,??,?CIA????,6?

3. 2 a?1?log1x?a,()?x?()21a?1221a?11,()?,a?2 224. 4 f(x)?1?cosx2?sinx28sixn?22cxo?s2sxin9xy?yx?28xsin?4tanx?cxos?102?9 1621?2xtan4? 4195. 16 x?y?(x?y)(??)xy1?02???1?x?3?x?2x?3?0???6.(1,3) ?2(???(x?2)x??x?x?2?0?x?3??1???,1?x?3 1?)0?x??10?三、解答题

34

1. 解:2x?2x?32?1?????2?3(x?1)?23?3x,x?x?6?0,?3?x?2,A???3,2?

2?9?x2?0?,?1?x?3,B?(?1,3),A?B?(?1,2) ?6?2x?0?2?9?x?6?2x方程x2?ax?b?0的两个根为?1和2,则a??1,b??2

?a?b??3

2. 解:y?x?5x?422?x?4?21x?42,令x2?4?t,(t?2)

y?t?1t在t??2,???上为增函数

12?52?当t?2时,ymin?2?

3. 解:y(x2?1)?mx2?43x?n,(y?m)x2?43x?y?n?0

显然y?m可以成立,当y?m时,方程(y?m)x2?43x?y?n?0 必然有实数根,???48?4(y?m)(y?n)?0, 即y2?(m?n)y?mn?12?0,而?1?y?7

??1和7是方程y?(m?n)y?mn?12?0的两个实数根

?m?n?6,m?1,n?5 则??mn?12??72?y?x?43x?5x?12x22 x2xx4. 解:?0?a?1,?a?2a?2?1,aa0,?x?2a?3?0 l oxx(?1?) (a?3)ax3,?a3 ?x?log a参考答案(数学5必修)第三章[提高训练C组]

一、选择题

35


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