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????????由已知得BA?AD,以A为坐标原点,AB的方向为x轴正方向,AB为单位长,建立如
图所示的空间直角坐标系A-xyz,则
1,0),P(0,则A(0,0,0),B(1,0,0),C(1,1,3),
????????PC?(1,0,?3),AB?(1,0,0)则
??????????BM?(x?1,y,z),PM?(x,y?1,z?3)
0,1)是底面ABCD的法向量,因为BM与底面ABCD所成的角为45°,而n?(0,所
以
?????cosBM,n?sin450,z(x?1)2?y2?z2?2 2即(x-1)2+y2-z2=0
?????????又M在棱PC上,设PM??PC,则
x??,y?1,z?3?3?
??2?x=1+?x=1-2????y=1(舍去),??y=1由①,②得???z??6?z???2??
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???????26?26?所以M?1-,1,?,从而AM??1-,1,?
??22?22?????设m=?x0,y0,z0?是平面ABM的法向量,则
????????m?AM?0?2-2x0?2y0?即?????????m?AB?0?x0?0??6z0?0
所以可取m=(0,-6,2).于是cosm,n?10 5m?n?mn10 5因此二面角M-AB-D的余弦值为20.解
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?????????(1)设P(x,y),M(x0,y0),设N(x0,0), NP??x?x0,y?,NM??0,y0?
?????????2NM得x0=x,y0?由NP?2y 2x2y2因为M(x0,y0)在C上,所以??1
22因此点P的轨迹方程为x2?y2?2
(2)由题意知F(-1,0).设Q(-3,t),P(m,n),则
????????????????OQ???3,t?,PF???1?m,?n?,OQ?PF?3?3m?tn, ????????OP??m,n?,PQ???3?m,t?n?,
????????由OP?PQ?1得-3m?m2?tn?n2?1,又由(1)知m2+n2=2,故
3+3m-tn=0
????????????????所以OQ?PF?0,即OQ?PF.又过点P存在唯一直线垂直于OQ,所以过点P且垂直于
OQ的直线l过C的左焦点F. 21.解:
(1)f?x?的定义域为?0,+??
设g?x?=ax-a-lnx,则f?x?=xg?x?,f?x??0等价于g?x??0 因为g?1?=0,g?x??0,故g'?1?=0,而g'?x??a?若a=1,则g'?x?=1?1,g'?1?=a?1,得a?1 x1.当0<x<1时,g'?x?<0,g?x?单调递减;当x>1时,g'?x?>0,xg?x?单调递增.所以x=1是g?x?的极小值点,故g?x??g?1?=0
综上,a=1
(2)由(1)知f?x??x2?x?xlnx,f'(x)?2x?2?lnx 设h?x??2x?2?lnx,则h'(x)?2???1?2??1?21x
????1?2?当x??0,?时,h'?x?<0;当x??,+??时,h'?x?>0,所以h?x?在?0,?单调递减,在?,+??单调递增
??1?2?又he?2>0,h??<0,h?1??0,所以h?x?在?0,?有唯一零点x0,在?,+??有唯一零
??
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???1??2??1?2??1?2?点1,且当x??0,x0?时,当x??x0,1?时,当x??1,+??时,h?x?>0;h?x?<0,h?x?>0. 因为f'?x??h?x?,所以x=x0是f(x)的唯一极大值点 由f'?x0??0得lnx0?2(x0?1),故f?x0?=x0(1?x0) 由x0??0,1?得f'?x0?<1 4因为x=x0是f(x)在(0,1)的最大值点,由e?1??0,1?,f'e?1?0得
??f?x0?>fe?1?e?2
所以e?2<f?x0?<2-2 22.解:
(1)设P的极坐标为
????,????>0?,M的极坐标为??,????>0?,由题设知
11OP=?,OM=?1=4 cos?由OM?OP=16得C2的极坐标方程?=4cos?因此C2的直角坐标方程为x?2(2)设点B的极坐标为
B??>0?
??2?y2?4?x?0?
B??,????>0?,由题设知
OA=2,?B=4cos?,于是△OAB面积
S=1OA??B?sin?AOB2????4cos??sin????3?????3?2sin?2????3?2??2?当?=-
3?12时,S取得最大值2+3 所以△OAB面积的最大值为2+3 23.解: (1)
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?a?b??a5?b5??a?ab?ab?b??a?b??2ab?ab?a?4?ab?a?b?6556332332224?b4?
?4(2)因为
?a?b?3?a3?3a2b?3ab2?b3?2?3ab?a+b??2+33?a+b?42?a+b??2?3?a+b?43
所以?a+b??8,因此a+b≤2.
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