2018年普通高等学校招生全国统一考试数学试题理(全国卷3-含答案) 下载本文

∴MO?CD,VM(0,0,1)M?ABC最大.

urm?(x1,y1,z1),A(2,?1,0),B(2,1,0),C(0,1,0),D(0,?1,0),

,设面MCD的法

,MB?(2,1,?1),

,MD?(0,?1,?1),

设面MAB的法向量为

向量为

rn?(x2,y2,z2)uuurMA?(2,?1,?1)MC?(0,1,?1)r?2x1?y1?z1?0u?m?(1,0,2)?2x?y?z?0?111同理

rn?(1,0,0),,

,∴ sin??255.

∴cos??15?55

20.

解答:(1)设直线l方程为y?kx?t,设A(x,y),B(x,y),

1122?y?kx?t?2?xy2?1??3?4联立消y得(4k2?3)x2?8ktx?4t2?12?0,

则??64kt得4k1222?4(4t2?12)(3?4k2)?0,

6t?2m3?4k2?3?t2…①,

,y?y12且x?x

2??8kt?23?4k2?k(x1?x2)?2t?,

16

∵m?0,∴ t?0且k?0. 且

t?3?4k2?4k…②.

由①②得

4k2?3?(3?4k2)216k2,

∴k?12或k??12. ∵k?0,∴ k??12. (2)

uuFPr?uuFAr?uuFBr?r0,

uuFPr?2uuuFMr?r0,

∵M(1,m),F(1,0),∴P的坐标为(1,?2m). 由于P在椭圆上,∴ 144?m23?1,∴m?34,M(1,?32),又

x21y24?13?1,

x222y24?3?1,

两式相减可得y1?y2??3?x1?x2x?x4y,

121?y2又x1?x2?2,y1?y2?32,∴k??1,

直线l方程为y?34??(x?1), 即y??x?74, ???y??x?7∴

?4x22,

???4?y3?1消去y得28x2?56x?1?0,x14?3211,2?14,

17

uuruur|FA|?|FB|?(x1?1)2?y12?(x2?1)2?y22?3uur33|FP|?(1?1)2?(??0)2?22,

,

∴∴

uuuruuuruuur|FA|?|FB|?2|FP|uuurFA.

成等差数列,

21.∴d??328.

uuur

FP

uuurFB

uuuruuurccc2d?||FA|?|FB||?|a?x1?a?x2|??|x1?x2|aaa??111321(x1?x2)2?4x1x2??4???2271421. 解答:(1)若a?0时,f(x)?(2?x)ln(1?x)?2x(x??1),

11∴f?(x)?ln(1?x)?(2?x)1??2?ln(x?1)??1. xx?1?1, 令h(x)?ln(x?1)?x1?11?∴h?(x)?x1?1(x?1)?x(x?1)22.

∴当x?0时,h?(x)?0,h(x)在(0,??)上单调递增, 当?1?x?0时,h?(x)?0,h(x)在(?1,0)上单调递减. ∴h(x)min?h(0)?ln1?1?1?0,

∴f?(x)?0恒成立,

∴f(x)在(?1,??)上单调递增, 又f(0)?2ln1?0?0,

∴当?1?x?0时,f(x)?0;当x?0时,f(x)?0. (2)

1?ax2f?(x)?(2ax?1)ln(x?1)??1x?1,

18

2ax?12ax(x?1)?ax2?1f??(x)?2aln(x?1)???0x?1(x?1)22a(x?1)2ln(x?1)?(2ax?1)(x?1)?ax2?2ax?1?02a(x?1)2ln(x?1)?3ax2?4ax?x?0a[2(x?1)2ln(x?1)?3x2?4x]??x,

, ,

.

设h(x)?2(x?1)2ln(1?x)?3x2?4x∴h?(x)?4(x?1)ln(1?x)?2(x?1)?6x?4,h?(0)?6?0,h(0)?0,

h(x)?0,h(x)?0. ∴在x?0邻域内,x?0时,x?0时,

x?0时,a?2(x?1)?x2ln(1?x)?3x2?4x,由洛必达法则

得a??1, 6x?0时,a?2(x?1)?x2ln(1?x)?3x2?4x,由洛必达法则

得a??1, 6综上所述,a??1. 622. 解答:

x?cos?eO的普通(1)eO的参数方程为?,∴?y?sin??方程为x2?y2?1,当??90?时,直线:l:x?0与eO有两个

2|交点,当??90?时,设直线l的方程为y?xtan??2,由直线l与eO有两个交点有|0?0?1?tan2??1,得tan??1,∴

2 19