设EM=x(0

=(,3,0),|

|=

.

13. 设tanα=x,则=6,解得x=.

14.

因为(a+)5的展开式中的项为a2()3=,所以10a2=1,则|a|=.

15.-

易知曲线y=x2(x≥0)是抛物线C:x2=4y的右半部分,如图,其焦点为F(0,1),准线为y=-1.过A作AH⊥准线,垂足为

H,则|AH|=|AF|,因为|FB|=6|FA|,所以|AB|=5|AH|,tan∠ABH===,故直线l的斜率为-

.

16.(-∞,-3]∪(3,+∞) 设平行于直线y=-3x+1的切线的切点为(m,m3-am2),

∵y'=3x2-2ax,∴3m2-2am=-3,Δ=4a2-36≥0,解得a∈(-∞,-3]∪[3,+∞).

若切点在直线y=-3x+1上,则m3-am2=-3m+1,又3m2-2am=-3, 从而m3-3m+2=(m-1)2(m+2)=0,解得m=1或m=-2.

当m=1时,a=3,此时方程3m2-6m+3=0有两个相等的实根,曲线y=x3-ax2不存在平行于直线y=-3x+1的切线; 当m=-2时,a=-,此时方程2m2+5m+2=0有两个不等的实根,曲线y=x3-ax2仅存在一条平行于直线y=-3x+1的切线.

综上,a的取值范围为(-∞,-3]∪(3,+∞). 17.(1)证明:因为

-=1,

9

所以

+1=2(+1), ............................................................................................................................................................ 2分

又+1=2, ............................................................................................................................................................................... 3分

所以数列{+1}为等比数列,且首项为2,公比为2. .............................................................................................................. 4分

(2)解:由(1)知+1=2n, ........................................................................................................................................................... 6分

所以+2n=2n+2n-1. ............................................................................................................................................................. 7分

所以Sn=+

=2n+1+n2-2. ............................................................................................................................. 12分

18.(1)证明:因为AA1⊥平面ABC,BC?平面ABC,

所以AA1⊥BC. ....................................................................................................................................................................... 1分 因为AB=2

,AC=2BC=4,

所以AB2+BC2=AC2,所以BC⊥AB. ....................................................................................................................................... 3分 因为AB∩AA1=A,所以BC⊥平面ABB1A1. ............................................................................................................................ 4分 又A1D?平面ABB1A1,所以BC⊥A1D. .................................................................................................................................. 5分

(2)解:以B为坐标原点,建立空间直角坐标系B-xyz,如图所示, 则C(0,0,2),D(

,0,0),A1(2

,4,0). ....................................................................................................................................... 6分

设平面A1CD的法向量为n=(x,y,z), 则

.................................................................................................................................... 8分

令x=4,则n=(4,-,2

). ..................................................................................................................................................... 9分

易知平面BCC1B1的一个法向量为m=(1,0,0),.................................................................................................................... 10分

10

则cos=

=

. ...................................................................................................................................................... 11分

故所求锐二面角的余弦值为

. ...................................................................................................................................... 12分

19.解:(1)因为该厂只有2名维修工人,

所以要使工厂正常运行,最多只能出现2台大型机器出现故障, ........................................................................................ 1分 故该工厂能正常运行的概率为(1-)5+

××(1-)4+

()2(1-)3=. ............................................................................... 4分

(2)(ⅰ)X的可能取值为31,44, ............................................................................................................................................... 6分 P(X=31)=()5=

, .................................................................................................................................................................. 7分

P(X=44)=1-=则X的分布列为

, ................................................................................................................................................................. 8分

X P 31 44

9分

故EX=31×+44×=

. ............................................................................................................................................... 10分

(ⅱ)若该厂有5名维修工人,则该厂获利的数学期望为5×10-1.5×5=42.5万元, ............................................................ 11分 因为

>42.5,所以该厂不应再招聘1名维修工人. ........................................................................................................ 12分

20.(1)证明:依题意可得,解得

, ....................................................................................................... 2分

则c2=4,c=2,F1(-2,0),F2(2,0), .................................................................................................................................................. 3分 从而|PF2|=3,|F1F2|=4,|PF1|=5, ............................................................................................................................................. 4分 故|PF2|,|F1F2|,|PF1|成等差数列............................................................................................................................................. 5分 (2)解:因为直线PF1的斜率为,所以可设l的方程为x=-y+m. ......................................................................................... 6分

11

将l的方程代入+

=1消去x,得

y2-my+3m2-48=0, ................................................................................................... 7分

设A(x1,y1),B(x2,y2),所以y1+y2=

,y1y2=

, ................................................................................................................ 8分

则|y1-y2|==

, ............................................................................................................ 9分

所以四边形AF1BF2的面积S=|F1F2|·|y1-y2|==

, ....................................................................... 10分

解得m=0, ............................................................................................................................................................................. 11分 故l的方程为x=-y,即4x+3y=0. ....................................................................................................................................... 12分 21.解:(1)f'(x)=2e2x-3-2, ............................................................................................................................................................ 1分 令f'(x)=0,得x=; ................................................................................................................................................................... 2分

令f'(x)<0,得x<;令f'(x)>0,得x>. ...................................................................................................................................... 3分

故f(x)的单调递减区间为(-∞,),单调递增区间为(,+∞), .................................................................................................... 4分

从而f(x)min=f()=-2. ............................................................................................................................................................... 5分

(2)易证mn≤()2,则(x+y+1)(x-y-2)≤()2=,

当且仅当x+y+1=x-y-2,即y=-时,取等号. ........................................................................................................................... 7分

f(x)+2x=e2x-3,则e2x-3≤, ................................................................................................................................................ 8分

令t=2x-1(t>0),则et-2≤t2,即t-2≤2lnt-2ln2. ........................................................................................................................... 9分

12