内容发布更新时间 : 2025/2/22 18:45:30星期一 下面是文章的全部内容请认真阅读。

（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)