2017届河南省南阳市第一中学高三上学期第一次月考数学试题

19.(12分)解:若p为真,则??a2?4?0,故a??2或a?2.

?当x?1时,h(x)有最小值,h(x)min?h(1)?e?e?a?a.

22??x?R,h(x)?0恒成立,?h(x)min?0,即a?0.

?“p?q”为真,?p为真且q为真.??从而所求实数a的取值范围为[2,??).

20.(12分)

?a??2或a?2, 解得a?2.

a?0,?ππTπ2ππ3

-?=,则=4×,∴ω=, (1)由图知A=2,=-?46?6?3ω323π3ππ

x+φ?.又f??=2sin?×+φ?=2sin?+φ?=2, ∴f(x)=2sin??2??6??26??4?πππ3πππ

∵0<φ<,<φ+<,∴φ+=,

244442

代点时优先代最值点,因为代零点时还要考虑上升还是下降段. 3π?π

即φ=,∴f(x)=2sin??2x+4?. 4

ππ3π3π

x-?=2sin??x-12?+?=2sin?x+?, (2)由(1)可得f??42??12??28????3x+π?1-cos4???x-π??2=4×?3x+π?, ∴g(x)=?f=2-2cos4???12???2

ππππ5π

-,?,∴-≤3x+≤, ∵x∈??63?444ππ

∴当3x+=π,即x=时,g(x)max=4.

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21.(12分)解:(Ⅰ)当a?0时,f?x??lnx?x2,其定义域为?0,???,f??x??所以f?x?在?1,e?上是增函数,当x?1时,f?x?min?f?1??1. 故函数f?x?在?1,e?上的最小值是1.

1?2x?0, x2x2?2ax?1,g?x??2x2?2ax?1. (Ⅱ)f??x??x(ⅰ)当a?0时,在?0,???上g?x??0恒成立, 此时f??x??0,函数f?x?无极值点;

(ⅱ)当a?0时,若??4a?8?0,即0?a?22时,

在?0,???上g?x??0恒成立,此时f??x??0,函数f?x?无极值点;

a?a2?2a?a2?2若??4a?8?0,即a?2时,易知当时,g?x??0,此时f??x??0; ?x?222a?a2?2a?a2?2当0?x?或x?时,g?x??0,此时f??x??0.

22a?a2?2a?a2?2所以当a?2时,是函数f?x?的极大值点,是函数f?x?的极小值点, x?x?22a?a2?2综上,当a?2时,函数f?x?无极值点;当a?2时,x?是函数f?x?的极大值点,

2a?a2?2是函数f?x?的极小值点. x?222.(12分)解:(Ⅰ)?f(x)?ln(1?x)?x?1?aax(a?0),?f'(x)? 2x?1(x?1)f'(1)?0 即a?2

(Ⅱ) ?f(x)?0在?0,???上恒成立, ?f(x)min?0

当0?a?1时,f(x)?0在?0,???上恒成立,即f(x)在?0,???上为增函数,

'?f(x)min?f(0)?0成立,即0?a?1

当a?1时,令f'(x)?0,则x?a?1,令f'(x)?0,则0?x?a?1,

即f(x)在[0,a?1)上为减函数,在(a?1,??)上为增函数,

?f(x)min?f(a?1)?0,又f(0)?0?f(a?1),则矛盾.

综上,a的取值范围为(0,1].

20162017120172017)?,只需证()?e 2017e2016201720171?1,即证ln?两边取自然对数得,2017ln, 2016201620172017111??0,即证ln(1?)??0, 即证ln2016201720161?2016x由(Ⅱ)知a?1时,f(x)?ln(1?x)?在?0,???单调递增.

x?11111?0,f(0)=0, 所以f()?ln(1?)??f(0)?0, 又

1?2016201620161?2016201620171)?成立. 所以(2017e(Ⅲ)要证(

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