25【答案】C
【解析】如图所示,设BF?m,则AD?AF?3m,AG?3m,又 24AD?AG?2OF?2,∴m?,又CD?BE?318343?S?AOB??OF?CD?. ,233
26【答案】A 【解析】∵
||F1P?|2,c又∵F2P?5F2Q(F1P?F?0,∴|F1F2?1F2)?F2P,∴
|F2Q|?1a, 5121212a?4c2?a1112525∴|FQ, |?a?2a?a,在?F1F2Q中,cos?QF2F1?11552?a?2c51212122222a?4c?a222a?4c?4ca?4c?4c2525在?F1F2P中,cos?PF2F1?,∴?,
12?a?2c2?a?2c2?a?2c5?c2?
27【答案】A
b1522a,a?4b2,∴渐近线方程为y??x??x.
a24?1?2x,0?x?1??31【解析】根据题意得f(x)???x,1?x?2,分段函数图象分段画即可.
?445?51?x,2?x??422?
28【答案】C
【解析】由题意,得
a2?a1?1?0,a4?a3?1,a6?a5?1,,a60?a5,所以9?1k?1ka?aa??(?21),代入,得(k?2S奇?S偶.又a2k??2k1k?2k?212k?1a2k?ak??2??(22k1(k)?2,所以
a2?0,
a4?a2?21?(?1)2,
a6?a4?22?(?1)3,a8?a6?23?(?1)4,…,a2k?a2k?2?2k?1?(?1)k,将上式相加,
得
a2k?2?22?k?2k?1?(?1)2?(?1)3??(?1)k=
1?(?1)k?1k3?(?1)k?12?2??2?,
22所以S偶=(2?22?23?以S603021-21所-45=231?47,?229?230)?(15?2?15?4)=
21-2???2?231?47?=232?94.
29【答案】B
【解析】根据题意,原问题等价于曲线的最小值的平方.因为y'?2x?行且与曲线
y?x2?lnx上一点到直线x?y?2?0的距离
11,令2x??1,得x?1,可得与直线x?y?2?0平xxy?x2?lnx相切的切点为?1,1?,所以可得切线方程为x?y?0,所以直线
222?2,即曲线y?x?lnx上的点到直
x?y?0与直线x?y?2?0之间的距离为线x?y?2?0的距离的最小值为2,所以曲线y?x2?lnx上的点到直线
2x?y?2?0的距离的最小值的平方为2;所以(x1?x2)B.
30【答案】B
?(y1?y2)2的最小值为2,故选
【解析】设切点为Q?t,tlnt?,则切线斜率k?f??t?=1?lnt,所以切线方程为
y?tlnt??1?lnt??x?t?,把P?a,a?代入得a?tlnt??1?lnt??a?t?,整理得alnt?t,
显然a?0,所以
1lntlnt1?,设g?t??,则问题转化为直线y?与函数g?t?图象有两atta
1?lnt ,可得g?t?在?0,e?递增,?e,???递减,在x?e处取得极t2111大值,结合g?t?图象,可得0???a?e ,故选B.
eae个不同交点,由g??t??
二、填空题(4个小题)
31【答案】?3
m?n?(?1,?1),(m?n)?(m?n),??(2t?3)?3?0,【解析】m?n?(2t?3,3),解得t??3.
32【答案】68
【解析】回归直线过?x,y?,根据题意x?18?13?10???1??10,
4y?24?34?38?64?40,代入a?40???2??10?60,所以x??4时,
4y???2????4??60?68,所以用电量约为68度.
33【答案】1
【解析】f??x??x?8x?6,∵a1,a4031是函数f?x??213x?4x2?6x?3的极值点,32∴a1?a4031?6,又∵正项等比数列?an?,∴a2016?a1?a4031?6,
∴log
6a2016?log66?1.
34【答案】37 【解析】因为cos?ADB???272,所以sin?ADB?.又因为?CAD?,所以
41010?C??ADB??,所以sin?C?sin(?ADB?)?sin?ADBcos?cos?ADBsin
4444???ADAC722224?, ????.在?ADC中,由正弦定理得
sin?Csin?ADC1021025
74?AC?sin?CAC?sin?CAC?sin?C25故AD?????22.
sin?ADCsin(???ADB)sin?ADB7210又S?ABD?1172AD?BDsin?ADB??22?BD??7解得BD?5. 2210在?ADB中,由余弦定理得
AB2?AD2?BD2?2AD?BD?cos?ADB?8?25?2?22?5?(?三、解答题(14个小题)
35【答案】(1)an?3n?1;(2)10.
2)?37.10
【解析】:(1)设等差数列{an}的公差为d,由a2?1,a4?1,a8?1,得
(3?3d)2?(3?d)(3?7d),解得d?3或d?0(舍),
故an?a1?(n?1)d?2?3(n?1)?3n?1. (2)由(1)知bn?9113?3(?). ,bnbn?1?(3n?1)(3n?2)3n?13n?23n?111119n?)?3(?)?,3n?13n?223n?26n?4
1111b1b2?b2b3?...?bnbn?1?3(?+?+25589n45?解得n?10. 依题有
6n?432
36【答案】(1)
π;(2)3. 3122【解析】:(1)∵c(acosB?b)?a?b,由余弦定理
222222222得a?c?b?bc?2a?2b,a?b?c?bc.
222∵a?b?c?2bccosA,∴cosA?1. 2∵A??0,π?,∴A?π. 3(2)sinB?sinC?sinB?sin?A?B??sinB?sinAcosB?cosAsinB
33??sinB?cosB?3sin(B?). 226∵B??0,???5?????1???2??,∴,sinB?B??,??66????2,1?. ?663????????∴sinB?sinC的最大值为3.
37【答案】(1)
π;(2)2. 3【解答】:(1)由条件可知a(sinA?sinB)?bsinB?csinC,
a2?b2?c21?, 根据正弦定理得a?b?c?ab,又由余弦定理知cosC?2ab2222?0?C??,?C?(2)m?tanC(?3.
11sinCcosAcosB?)?(?) tanAtanBcosCsinAsinBsinCcosAsinB?cosBsinA2sin2C2c22(a2?b2?ab)????? cosCsinAsinBsinAsinBababab?2(??1)?2?(2?1)?2,当且仅当a?b即?ABC为正三角形时,
ba实数m的最小值为2.
n?2. n?22211【解答】:(1)由a1?2,2an?1?an得an?2?n?1?n?2,由题意知:
221当n?1时,b1?b2?1,故b2?2,当n?2时,bn?bn?1?bn,
nbb得n?1?n,所以bn?n. n?1n12nn(2)由(1)知 anbn?n?2.?Tn??1?0???n?2,
2222112nTn?0?1???n?1,两式相减得 222238【答案】(1)an?(2)Tn?8?,bn?n;n?21
王后雄高考关门卷·数文(含答案解析)
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