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17.解: ⑴ 2cosC?acosB?bcosA??c
由正弦定理得:2cosC?sinA?cosB?sinB?cosA??sinC
2cosC?sin?A?B??sinC
∵A?B?C?π,A、B、C??0,π? ∴sin?A?B??sinC?0 ∴2cosC?1,cosC?∵C??0,π?
1 2∴C?π 3⑵ 由余弦定理得:c2?a2?b2?2ab?cosC
17?a2?b2?2ab?
2?a?b?2?3ab?7
1333 S?ab?sinC?ab?242∴ab?6 ∴?a?b??18?7
2a?b?5
∴△ABC周长为a?b?c?5?7 18.解:(1) ∵ABEF为正方形 ∴AF?EF
∵?AFD?90? ∴AF?DF ∵DFEF=F
∴AF?面EFDC
AF?面ABEF
∴平面ABEF?平面EFDC ⑵ 由⑴知
?DFE??CEF?60?
∵AB∥EF
AB?平面EFDC EF?平面EFDC
∴AB∥平面ABCD
AB?平面ABCD
∵面ABCD面EFDC?CD ∴AB∥CD ∴CD∥EF
∴四边形EFDC为等腰梯形
以E为原点,如图建立坐标系,设FD?a
E?0,0,0??a3?,0,a?B?0,2a,0? C??2?2??A?2a,2a,?0
?a3?BC?,?2a,a?EB??0,2a,0?,??2?,AB???2a,0,0? 2??设面BEC法向量为m??x,y,z?.
?2a?y1?0??m?EB?0?1,即?a x1?3,y1?0,z1?? ?3a?z1?0???x1?2ay1??m?BC?0?22m??3,0,?1
?设面ABC法向量为n??x2,y2,z2?
?a3?az2?0?n?BC=0?x2?2ay2?.即?2 x2?0,y2?3,z2?4 ?2??2ax?0?n?AB?0?2n?0,3,4
??设二面角E?BC?A的大小为?. cos??m?nm?n??43?1?3?16??219 19219 1919解: ⑴ 每台机器更换的易损零件数为8,9,10,11
∴二面角E?BC?A的余弦值为?记事件Ai为第一台机器3年内换掉i?7个零件?i?1,2,3,4? 记事件Bi为第二台机器3年内换掉i?7个零件?i?1,2,3,4?
由题知P?A1??P?A3??P?A4??P?B1??P?B3??P?B4??0.2,P?A2??P?B2??0.4 设2台机器共需更换的易损零件数的随机变量为X,则X的可能的取值为16,17,18,19,20,21,22
P?X?16??P?A1?P?B1??0.2?0.2?0.04
P?X?17??P?A1?P?B2??P?A2?P?B1??0.2?0.4?0.4?0.2?0.16
P?X?18??P?A1?P?B3??P?A2?P?B2??P?A3?P?B1??0.2?0.2?0.2?0.2?0.4?0.4?0.24P?X?19??P?A1?P?B4??P?A2?P?B3??P?A3?P?B2??P?A4?P?B1??0.2?0.2?0.2?0.2?0.4?0.2?0.2?0.4?0.24
P?X?20??P?A2?P?B4??P?A3?P?B3??P?A4?P?B2??0.4?0.2?0.2?0.4?0.2?0.2?0.2P?x?21??P?A3?P?B4??P?A4?P?B3??0.2?0.2?0.2?0.2?0.08 P?x?22??P?A4?P?B4??0.2?0.2?0.04
17 18 19 20 21 22 X 16 P 0.04 0.16 0.24 0.24 0.2 0.08 0.04 ⑵ 要令P?x≤n?≥0.5,0.04?0.16?0.24?0.5,0.04?0.16?0.24?0.24≥0.5
则n的最小值为19
⑶ 购买零件所需费用含两部分,一部分为购买机器时购买零件的费用,另一部分为备件不足时额外购买的费用
当n?19时,费用的期望为19?200?500?0.2?1000?0.08?1500?0.04?4040 当n?20时,费用的期望为20?200?500?0.08?1000?0.04?4080 所以应选用n?19 20. (1)圆A整理为?x?1??y2?16,A坐标??1,0?,如图, 24BE∥AC,则∠C?∠EBD,由AC?AD,则∠D?∠C, 则EB?ED ?∠EBD?∠D,A32Cx1?AE?EB?AE?ED?AD?4
所以E的轨迹为一个椭圆,方程为x2y2⑵ C1:??1;设l:x?my?1,
43xy??1,(y?0); 432242B124E23D4P43因为PQ⊥l,设PQ:y??m?x?1?,联立l与椭圆C1 A21Nx?x?my?1?2得3m2?4y2?6my?9?0; ?xy2?1??3?442B124??MQ234则|MN|?1?m|yM?yN|?1?m|?m??1?1?|1?m22236m2?36?3m2?4?3m2?4|2m|1?m2?12?m2?1?3m2?4;
圆心A到PQ距离d?
?,
4m243m2?4?所以|PQ|?2|AQ|?d?216?,
21?m21?m22?SMPNQ21112?m?1?43m2?424m2?11?|MN|?|PQ|?????24??2?12,83 221223m?41?m3m?43?2m?1?xx21. (Ⅰ)f'(x)?(x?1)e?2a(x?1)?(x?1)(e?2a).
x(i)设a?0,则f(x)?(x?2)e,f(x)只有一个零点.
(ii)设a?0,则当x?(??,1)时,f'(x)?0;当x?(1,??)时,f'(x)?0.所以f(x)在(??,1)上单调递减,在(1,??)上单调递增.
又f(1)??e,f(2)?a,取b满足b?0且b?lna,则 2f(b)?a3(b?2)?a(b?1)2?a(b2?b)?0, 22故f(x)存在两个零点.
(iii)设a?0,由f'(x)?0得x?1或x?ln(?2a). 若a??e,则ln(?2a)?1,故当x?(1,??)时,f'(x)?0,因此f(x)在(1,??)上单调递增.又2当x?1时,f(x)?0,所以f(x)不存在两个零点. 若a??e,则ln(?2a)?1,故当x?(1,ln(?2a))时,f'(x)?0;当x?(ln(?2a),??)时,2f'(x)?0.因此f(x)在(1,ln(?2a))单调递减,在(ln(?2a),??)单调递增.又当x?1时,f(x)?0,所以f(x)不存在两个零点.
综上,a的取值范围为(0,??).
不妨设x1?x2,由(Ⅰ)知x1?(??,1),x2?(1,??),2?x2?(??,1),f(x)在(??,1)上(??)单调递减,所以x1?x2?2等价于f(x1)?f(2?x2),即f(2?x2)?0. 由于f(2?x2)??x2e2?x2?a(x2?1)2,而f(x2)?(x2?2)ex2?a(x2?1)2?0,所以
f(2?x2)??x2e2?x2?(x2?2)ex2.
设g(x)??xe2?x?(x?2)ex,则g?(x)?(x?1)(e2?x?ex).
所以当x?1时,g?(x)?0,而g(1)?0,故当x?1时,g(x)?0. 从而g(x2)?f(2?x2)?0,故x1?x2?2. 22.⑴ 设圆的半径为r,作OK?AB于K
∵OA?OB,?AOB?120?
∴OK?AB,?A?30?,OK?OA?sin30??∴AB与⊙O相切 ⑵ 方法一:
假设CD与AB不平行
OA?r 2CD与AB交于F
FK2?FC?FD① ∵A、B、C、D四点共圆
∴FC?FD?FA?FB??FK?AK??FK?BK? ∵AK?BK
∴FC?FD??FK?AK??FK?AK??FK2?AK2②由①②可知矛盾 ∴AB∥CD 方法二:
因为A,B,C,D四点共圆,不妨设圆心为T,因为所以O,T为AB的中垂线上,OA?OB,TA?TB,同理OC?OD,TC?TD,所以OT为CD的中垂线,所以AB∥CD.
?x?acost23.⑴ ? (t均为参数)
y?1?asint?∴x2??y?1??a2 ①
21?为圆心,a为半径的圆.方程为x2?y2?2y?1?a2?0 ∴C1为以?0,∵x2?y2??2,y??sin? ∴?2?2?sin??1?a2?0
即为C1的极坐标方程