故所抽取的两人所获得学习用品价值之差的绝对值不超过20元的概率为19．解析（Ⅰ）连接BD，交AC于O，则O是AC的中点，
∴OF∥CC1，CC1∥BB1，∴OF∥BB1，又OE∥AB， ∴平面OEF∥平面AA1B1B，又EF?平面OEF，
69． ···· 12分 95∴EF∥平面AA1B1B． ································································································· 4分
（Ⅱ）∵AD＝1，AA1＝2，A1D?3，∴△AA1D是直角三角形，且A1D⊥AD，∵侧面AD1⊥平面ABCD，∴A1D⊥平面ABCD，可知DA1、DA、DC两两垂直． ···················································· 6分
C(0,1,0)，，A(1,0,0)，A1(0,0,3)，
??????????????????????，DA1?(0,0,3)，AC，，C1(?1,1,3)，B(1,1,0)，则DB?(1,1,0)?(?1,1,0)AC?(?1,1,0)AA1?(?1,0,3)，11(0,0,)分别以DA1、DA、DC为x、y、z轴建立空间直角坐标系D－xyz，则D0 ············································································································································· 8分
??????????????????由DB?DA1?0，DB?AC11?0可得平面A1C1D的一个法向量????ACC1A1的法向量为n2?(x,y,z)， n1?DB?(1,1,，设平面0)?????n?AC??x?y?0,?由?2????取n2?(3,3,1)，······ 10分 ??n2?AA1??x?3z?0,则cos?n1,n2??n1?n22342??， n1n272?742． ·························································· 12分 7∴二面角C－A1C1－D的大小为arccos20．解析（Ⅰ）f?(x)?3x2?3a，由f(x)在x＝2处的切线方程为y＝9x－14，
?f(2)?4,?8?6a?b?4,?a?1,∴?则?解得?∴f(x)?x3?3x?2， ···························· 4分
?b?2,f(2)?9,12?3a?9,???则f?(x)?3x2?3?3(x?1)(x?1)，
由f?(x)?0，得x??1或x?1；由f?(x)?0，得?1?x?1．
故函数f(x)单调递减区间是(?1,1)；单调递增区间是(??,?1)，(1,??)． ·············· 6分 （Ⅱ）由（Ⅰ）知，函数f(x)在(0,1)单调递减，在(1,2)上单调递增， 又f(0)?2，f(2)?4，有f(0)?f(2)，
∴函数f(x)在区间[0，2]上的最大值f(x)max=f(2)=4． ··············································· 8分 又g(x)＝－x2＋2x＋k＝－( x－1)2＋k＋1
∴函数g(x)在[0，2]上的最大值为g(x)max=g(1)= k＋1． ············································ 10分 因为对任意x1?[0,2]，均存在x2?[0,2]，使成立， 所以有f(x)max＜g(x)max，则4＜k＋1，∴k＞3．
故实数k的取值范围是(3,＋∞)． ················································································ 12分 21．解析：（Ⅰ）由Sn?2an?2n?1，得Sn?1?2an?1?2n (n≥2).
两式相减，得an?2an?2an?1?2n，即an?2an?1?2n (n≥2)， ·································· 2分
∴
anan?1an，故数列··········································· 4分 {}是公差为1的等差数列， ·??12n2n2n?1ann，故． ············· 6分 a?(n?1)?2?2?(n?1)?n?1n2n又S1?2a1?22，则a1?4， ∴（Ⅱ）∵bn?log2不等式(1?an······································································· 7分 ?log22n?n， ·
n?111111)?(1?)???(1?)?m?b2n?1,即(1?1)?(1?)???(1?)?m?2n?1恒成立，也即b1b3b2n?132n?111(1?1)?(1?)???(1?)32n?1m?对任意正整数n都成立．··············································· 8分 2n?111111(1?1)?(1?)???(1?)(1?)(1?1)?(1?)???(1?)32n?12n?132n?1令f(n)?，知f(n?1)?,
2n?32n?1f(n?1)2n?24n2?8n?4∵???1，
2f(n)2n?1?2n?34n?8n?3∴当n∈N*时，f(n)单调递增， ················································································· 10分 ∴f(n)?f(1)?232323，则m?，故实数m的最大值为． ·························· 12分
333?2?a?b2?c2,?c2?,解得a?2，b?2， 22．解析（Ⅰ）由?e??a2??2a2?42,??cx2y2··············································································· 4分 ?1． ·故椭圆C的方程为?42?x2?2y2?4,（Ⅱ）联立?消去y，得(2k2?1)x2?4kmx?2m2?4?0，
?y?kx?m,则??16k2m2?4(2k2?1)(2m2?4)?32k2?8m2?16?0，又A(x1,y1)、B(x2,y2)
4km2m2?4，x1?x2?， ······································································· 6分 x1?x2??22k?12k2?1设直线MA：y?同理yQ?∵
y16y1(x?2)，则yP?， ······················································· 8分 x1?2x1?26y2， ······································································································· 9分 x2?2x?4x2?411x1?2x2?21111????0， ???，∴?，即1············· 10分
y1y26y16y26y16y2y1y2yPyQ∴(x1?4)y2?(x2?4)y1?0，∴(x1?4)(kx2?m)?(x2?4)(kx1?m)?0， 即2kx1x2?(m?4k)(x1?x2)?8m?0，
2m2?44km∴2k?2··························································· 12分 ?(m?4k)(?2)?8m?0， ·
2k?12k?1∴
?8k?8m··················································································· 13分 ?0，故k??m， ·
2k2?1故直线l方程为y?kx?k，可知该直线过定点(1,0)． 14分