发布时间 : 星期四 文章2018年北京市海淀区高三一模理科数学试题及参考答案更新完毕开始阅读6eb60f9b541810a6f524ccbff121dd36a22dc478
（Ⅱ）由PO?平面ABC，OB?AC，如图建立空间
B(0,1,0)zPA B直角坐标系，则O(0,0,0)，C(1,0,0)，
A(?1,0,0)，
，P(0,0,1) 5
分
由OB?平面APC， 故平面APC的法向量为OB由BC?(1,?1,0)?(0,1,0)x ······ 6分
COy，PC?(1,0,?1)
?x?y?0 得：?x?z?0??n?BC?0设平面PBC的法向量为n?(x,y,z)，则由??n?PC?0········································· 7分 令x?1，得y?1，z?1，即n?(1,1,1)
cos?n,OB??n?OB|n|?|OB|?13?1?33 ·········································· 8分
由二面角A?PC?B是锐二面角， 所以二面角A?PC?B的余弦值为
33 ··········································· 9分
························································ 10分 （Ⅲ）设BN??BP，0???1，则 ·
BM?BC?CM?BC??CP?(1,?1,0)??(?1,0,1)?(1??,?1,?)
分
········ 11AN?AB?BN?AB??BP?(1,1,0)??(0,?1,1)?(1,1??,?) ·令BM?AN?0
············································· 12分 得(1??)?1?(?1)?(1??)?????0 ·即???1??12?1?11??······················ 13分 ，μ是关于λ的单调递增函数， ·
12当??[,]时，??[,]，
3345所以
BNBP?[145,2] ········································································ 14分
注：第（1）题四个得分点少一个扣1分；第（Ⅱ）题建系前的证明很简单，没写不扣分，若解答从点的坐标开始出错并且这个证明过程写了可以给1分；平面PBC的法向量不正确但方程组正确列出给1分；向量夹角余弦不正确但公式正确给1分；二面角是锐角没说明直接给出正确答案不扣分。第（Ⅲ）题写出BM?AN?0，而没有给出λ和μ的关系式，给1分；没有写“μ是关于λ的单调递增函数”而结论正确不扣分；最后正确求出μ的范围，
BNBP12而没写
?[45,]不扣分。
18. （本题满分13分）
（Ⅰ）当a?0时，f(x)?1lnxx······································ 1分 ，定义域为(0,??)
故
f'(x)?x?x?lnxx2?1?lnx ························································ 2分 x2 令f'(x)?0，得0?x?e
故f(x)的单调递增区间为(0,e) ······················································· 4分 （Ⅱ）法1：
································· 5分 因为a＞0，所以函数f(x)的定义域为(0,??)， ·
x?af'(x)?xaxax2?lnx21??ax?lnx2 ···················································· 6分
(x?a)(x?a) 令g(x)?1? 则g'(x)?? 由g(e)?ae?lnx?1x
x?ax2???0 ····················································· 7分
a?(1?a)?a?(1ea?1?0，g(ea?1)?1?ea?1?1)?0，
a?1 故存在x0?(e,e)，g(x0)?0························································ 9分
故当x?(0,x0)时，g(x)?0；当x?(x0,??)时，g(x)?0
x (0,x0)? x0 (x0,??) f'(x)f(x) 0? ↗ 极大值 ↘ ··································································································· 11分
a?f'(x0)?1??lnx0?02??x0??x0?e ········································ 13分 故?，解得?2??a?e?f(x)?lnx0?102?x0?ae? 故a的值为e2. （Ⅱ）法2：
································· 5分 因为a＞0，所以函数f(x)的定义域为(0,??)， ·
f(x)的最大值为
1e2的充要条件为对任意的x?(0,??)，
lnxx?a?1e2且存在
lnx0x0?a2x0?(0,??)，使得
?1e2，等价于对任意的x?(0,??)，a?e2lnx?x且存在
x0?(0,??)，使得a?elnx0?x0， ··················································· 8分
2等价于g(x)?elnx?x的最大值为a.
g'(x)?e2x··········································································· 9分 ?1， ·
令g'(x)?0，得x?e2.
x 2(0,e)?2 e2 (e,??)2 g'(x) 0? g(x)↗ 2222极大值 ↘ ······································································································ 11分 ·························· 13分 故g(x)的最大值为g(e)?elne?e?e，即a?e2. ·
注：第（Ⅱ）题法1中对函数f(x)求导若在第（Ⅰ）题中完成，则这1分计到本小题；方法1中找点g(e)、g(ea?1)，若用函数g(x)的变化趋势说明不扣分；
方法1中文字说明与列表有一个即可；
2方法2中“等价于g(x)?elnx?x的最大值为a”与最后结论“a?e2”出现一个即可。
（19）（本小题14分）
1?4??12?2ab??222··································································· 3分 （Ⅰ）由题意?a?b?c， ·
??e?c?3?a2?解得：a?22，b? 故椭圆C的标准方程为
2，c?x2················································· 4分 6 ·
···················································· 5分 ?1
8?y22（Ⅱ）假设直线TP或TQ的斜率不存在，则P点或Q点的坐标为(2，－1)，直线l的方程为y?1?12(x?2)，即y2?12x?2。
?xy??1??82 联立方程?，得x2?4x?4?0，
?y?1x?2??22
此时，直线l与椭圆C相切，不合题意。
························································· 6分 故直线TP和TQ的斜率存在。 ·
方法1：
设P(x1,y1)，Q(x2,y2)，则 直线TP:y?1?故|OM|?2?y1?1x1?2(x?2)，直线TQ:y?1?x2?2y2?1y2?1x2?2(x?2)
x1?2y1?1，|ON|?2? ·········································· 8分
由直线OT:y?2122x，设直线PQ:y?12x?t（t?0）························· 9分
?xy??1??82····························· 10分 联立方程?，得：x2?2tx?2t2?4?0 ·
1?y?x?t??22······································ 11分 当??0时，x1?x2??2t，x1?x2?2t?4 ·
|OM|?|ON|?4?(x1?2y1?1?x2?2y2?1)
?4?(x1?212x1?t?1?x2?212x2?t?1)
?4?x1x2?(t?2)(x1?x2)?4(t?1)14x1x2?212(t?1)(x1?x2)?(t?1)2 ························ 12分
?4?2t?4?(t?2)(?2t)?4(t?1)14(2t?4)?212(t?1)?(?2t)?(t?1)2 ·················· 13分
······································································· 14分 ?4 ·
方法2：
设P(x1,y1)，Q(x2,y2)，直线TP和TQ的斜率分别为k1和k2 由OT:y?122x，设直线PQ:y?212x?t（t?0） ······························· 7分
?xy??1??82····························· 8分 联立方程?，得：x2?2tx?2t2?4?0 ·
?y?1x?t??22······································ 9分 当??0时，x1?x2??2t，x1?x2?2t?4 ·
k1?k2?y1?1x1?2?y2?1x2?2
x2?t?1x2?21?2x1?t?1x1?2?12
?x1x2?(t?2)(x1?x2)?4(t?1)(x1?2)(x2?2)2 ············································· 10分 ·············································· 11分
?2t?4?(t?2)(?2t)?4(t?1)(x1?2)(x2?2)