解法一:(Ⅰ)因B1C1?A,且B1C1?BB1,故B1C1?面A, 1B11ABB1从而B1C1?B1E,又B1E?DE,故B1E是异面直线B1C1与DE的公垂线. 设BD的长度为x,则四棱椎C?ABDA1的体积V1为
111V1?SABDA·BC?(DB?AA·)AB·BC?(x?2)·BC. 113661而直三棱柱ABC?A的体积为V?S·AA?AB·BC·AA1?BC. BCV2△ABC111122138由已知条件V1:V2?3:5,故(x?2)?,解之得x?.
65582从而B1D?B1B?DB?2??.
5529?2?22在直角三角形A1B1D中,A1D?A1B1?B1D?1????,
55??又因S△A1B1D?故B1E?211A1D·B1E?A1B·1B1D, 22A1B·2291B1D. ?A1D29A1 (Ⅱ)如答(19)图1,过B1作B1F?C1D,垂足为F,连接A1F,
B1 F
E D
C1
因
A1B1?B1C1,A1B1?B1D,故A1B1?面B1DC1.
由三垂线定理知C1D?A1F,故?A1FB1为所求二面角的平面角.
A B
C
答(19)图1
36?2?22在直角△C1B1D中,C1D?B1C1?B1D?2????,
55??又因S△C1B1D?故B1F?解法二:
211C1D·B1F?B1C1·B1D, 22B1C·AB33231B1D,所以tanA1FB1?11?. ?C1D9B1F20,2),A(0,(Ⅰ)如答(19)图2,以B点为坐标原点O建立空间直角坐标系O?xyz,则B(000,),,B1(0,1,0),
????????A1(01,,2),则AA1?(0,0,2),AB?(0,?10),. ?????0,2),则BC,0,0), 设C1(a,11?(a????又设E(0,y0,z0),则B1E?(0,y0,z0?2),
??????????????????从而BC. B1E?0,即B1E?BC11?11?????????又B1E?DA,所以B1E是异面直线B1C1与DE的公垂线. A11下面求点D的坐标.
z B1 F E D C1 ????设D(0,0,z),则BD(0,0,z).
因四棱锥C?ABDA1的体积V1为
y A x
C B(O) 答(19)图2
????1????????????????而直三棱柱ABC?AAB?BC?AA1?BC. 1?1B1C1的体积V2为V2?S△ABC?AA2由已知条件V1:V2?3:5,故
????1????????????????1V1?SABDA1?BC?(BD?AA1)?AB?BC
36????1?(z?2)?1?BC. 61388??(z?2)?,解得z?,即D?0,0,?. 6555??从而DB1?0,0,?,DA1??0,1,?,DE??0,y0,z0??.
???????2?5????????2?5???????8?5?接下来再求点E的坐标.
?????????2由B1E?DA1,有B1E?DA1?0,即y0?(z0?2)?0 (1)
5?????????y0又由DA?1∥DE得
1z0?2585. (2)
?????444810??448?联立(1),(2),解得y0?,z0?,即E??0,,?,得B1E??0,,??.
2929?2929??2929?22????229?4??10???故B1E??. ???29?29??29??????20,),过B1作B1F?C1D, (Ⅱ)由已知BC?2,则C1(2,0,2),从而DC1?(2,5垂足为F,连接A1F,
???????????????0,z1),则B1F?(x1,设F(x1,0,z1?2),因为B1F?DC1?0,故
2x1?24z1??0??????????????① 55
??????x8?????????因DF??x1,0,z1??且DF∥DC1得1?5?2?z1?2585,即
28x1?2z1?2?0??????????????② 55联立①②解得x1?24444??2,即F?2,z1?2,0,?.
272727??27??????2??210?????10?则A1F??2,?1,??,B1F??2,0,??.
27?27??27?27??????22??10?223. |B1F|?????27????9???27???????????21022?2?(?1)?0???0,故A1F?DC1, 又A1F?DC1?27275???????????????因此?A,?10),,从而A1B1?B1F?0, 1FB1为所求二面角的平面角.又A1B1?(0?????|A1B1|33??故A1B1?B1F,△A1B1F为直角三角形,所以tanA1FB1?????.
2|B1F|
(20)(本小题满分13分)已知函数f(x)?ax4lnx?bx4?c(x>0)在x = 1处
取得极值–3–c,其中a,b,c为常数。
(1)试确定a,b的值;(6分)
(2)讨论函数f(x)的单调区间;(4分)
(3)若对任意x>0,不等式f(x)??2c2恒成立,求c的取值范围。(3分) 解:(I)由题意知f(1)??3?c,因此b?c??3?c,从而b??3. 又对f(x)求导得f?(x)?4axlnx?ax??4bx?x3(4alnx?a?4b). 由题意f?(1)?0,因此a?4b?0,解得a?12.
(II)由(I)知f?(x)?48xlnx(x?0),令f?(x)?0,解得x?1. 当0?x?1时,f?(x)?0,此时f(x)为减函数; 当x?1时,f?(x)?0,此时f(x)为增函数.
33421x3
因此f(x)的单调递减区间为(0,1),而f(x)的单调递增区间为(1,∞?).
(III)由(II)知,f(x)在x?1处取得极小值f(1)??3?c,此极小值也是最小值,
要使f(x)≥?2c2(x?0)恒成立,只需?3?c≥?2c2.
即2c2?c?3≥0,从而(2c?3)(c?1)≥0,解得c≥所以c的取值范围为(??,?1]?
(21)(本小题满分12分)已知各项均为正数的数列{an}的前n项和满足S1?1,且
3或c≤?1. 2?3?,???. ??2?6Sn?(an?1)(an?2),n?N*
(1)求{an}的通项公式;(5分) (2)设数列{bn}满足an(2bn?1)?1,并记Tn为{bn}的前n项和,
求证:3Tn?1?log2(an?3),n?N*. (7分)
(I)解:由a1?S1?1(a1?1)(a1?2),解得a1?1或a1?2,由假设a1?S1?1,因此a1?2, 611又由an?1?Sn?1?Sn?(an?1?1)(an?1?2)?(an?1)(an?2),
66得(an?1?an)(an?1?an?3)?0,
即an?1?an?3?0或an?1??an,因an?0,故an?1??an不成立,舍去. 因此an?1?an?3,从而?an?是公差为3,首项为2的等差数列, 故?an?的通项为an?3n?1.
b(II)证法一:由an(2n?1)?1可解得bn?log2?1???1?3n?log; ?2a2?3n?1从而Tn?b1?b2???bn?log2??????36?253n??. 3n?1?33n?2?36因此3Tn?1?log2(an?3)?log2???. ????3n?1?3n?2?25
363n?2f(n?1)3n?2?3n?3?(3n?3)2?令f(n)????,则. ?????????23n?1?3n?2f(n)3n?5?3n?2?(3n?5)(3n?2)?25因(3n?3)3?(3n?5)(3n?2)2?9n?7?0,故f(n?1)?f(n). 特别地f(n)≥f(1)?3327?1,从而3Tn?1?log2(an?3)?log2f(n)?0. 20即3Tn?1?log2(an?3). 证法二:同证法一求得bn及Tn,
由二项式定理知,当c?0时,不等式(1?c)3?1?3c成立.
?1?由此不等式有3Tn?1?log22?1???2?31??1??1??1????? ?5??3n?1?33583n?23??3??3???log2···?·?log2(3n?2)?log2(an?3). ?log22?1???1????1?2?253n?1?2??5??3n?1?证法三:同证法一求得bn及Tn.
363n473n?1583n?2,Bn?··?··,Cn?··?.
253n?1363n473n?13n3n?13n?23n+22??因.因此An?AnBnCn?. 3n?13n3n?12·令An?··?3n??363从而3Tn?1?log22??? ???log2A2n?3n?1??253?log22AnBnCn?log2(3n?2)?log2(an?3).
证法四:同证法一求得bn及Tn.
下面用数学归纳法证明:3Tn?1?log2(an?3). 当n?1时,3T1?1?log227,log2(a1?3)?log25, 4因此3T1?1?log2(a1?3),结论成立.
假设结论当n?k时成立,即3Tk?1?log2(ak?3). 则当n?k?1时,
3Tk?1?1?log2(ak?1?3)?3Tk?1?3bk?1?log2(ak?1?3) ?log2(ak?3)?log2(ak?1?3)?3bk?1