内容发布更新时间 : 2024/12/23 8:28:24星期一 下面是文章的全部内容请认真阅读。
21. (1)由题意,得f'(x)?lnx?1, 故g(x)?ax2?(a?2)x?lnx?1, 故g'(x)?2ax?(a?2)?1(2x?1)(ax?1)?, xxx?0,a?0.
11,x2? 2a11① 当0?a?2时,?,
a211g'(x)?0?0?x?或x?;
2a11g'(x)?0??x?,
2a11a所以g(x)在x?处取极大值g()???ln2,
224111在x?处取极小值g()???lna.
aaa11②当a?2时,?,g'(x)?0恒成立,所以不存在极值;
a21111③当a?2时,?,g'(x)?0?0?x?或x?;
a2a211g'(x)?0??x?,
a2111所以g(x)在x?处取极大值g()???lna,
aaa11a在x?处取极小值g()???ln2.
2241a11综上,当0?a?2时,g(x)在x?处取极大值??ln2,在x?处取极小值??lna;
24aa111当a?2时,不存在极值;a?2时,g(x)在x?处取极大值??lna,在x?处取极
aa2a小值??ln2.
4x(2)F(x)?xlnx?x,定义域为x?(0,??),
ex?1F'(x)?1?lnx?x,而x?(1,2),
e令g'(x)?0,得x1?故F'(x)?0,即F(x)在区间(1,2)内单调递增 又F(1)??12?0,F(2)?2ln2?2?0, ee且F(x)在区间(1,2)内的图象连续不断,
故根据零点存在性定理,有F(x)在区间(1,2)内有且仅有唯一零点. 所以存在x0?(1,2),使得F(x0)?f(x0)?且当1?x?x0时,f(x)?当x?x0时,f(x)?x0?0, ex0x; exx, xe?xlnx,1?x?x0?所以m(x)??x
,x?x0??ex当1?x?x0时,m(x)?xlnx, 由m'(x)?1?lnx?0得m(x)单调递增; 当当x?x0时,m(x)?由m'(x)?x, ex1?x?0得m(x)单调递减; ex若m(x)?n在区间(1,??)内有两个不等实根x1,x2(x1?x2) 则x1?(1,x0),x2?(x0,??).
要证x1?x2?2x0,即证x2?2x0?x1
又2x0?x1?x0,而m(x)在区间(x0,??)内单调递减, 故可证m(x2)?m(2x0?x1), 又由m(x1)?m(x2), 即证m(x1)?m(2x0?x1),
2x0?x1 2x0?x1e2x?x记h(x)?xlnx?20x?x,1?x?x0,其中h(x0)?0
e0t1?t记?(t)?t,则?'(t)?t,
ee即x1lnx1?当t?(0,1)时,?'(t)?0;
当t?(1,??)时,?'(t)?0, 故?(t)max?1 e1, e而?(t)?0,故0??(t)?而2x0?x?1,
2x?x1??20x0?x?0, ee2x?x11因此h'(x)?1?lnx?2x?x?20x?x?1??0,
ee0e0所以?即h(x)单调递增,故当1?x?x0时,h(x)?h(x0)?0, 即x1lnx1?2x0?x1,故x1?x2?2x0,得证.
e2x0?x122. (1)因为圆C1:??x?cos??1(?为参数),
?y?sin?所以圆C1的普通方程是(x?1)2?y2?1 因为圆C2:??4sin?,
所以圆C2的直角坐标方程是x?y?4y?0. (2)因为圆C1:(x?1)?y?1, 圆C2:x?y?4y?0, 两式相减,得x?2y?0, 即公共弦所在直线为x?2y?0,
222222所以点(1,0)到x?2y?0的距离为
5, 5所以公共弦长为21?145, ?55所以S?AC1B?14552???. 255523.(1)(4a?1?4b?1)2?(1?4a?1?1?4b?1)2
?(12?12)?(4a?1?4b?1)
?2[4(a?b)?2]?2(4?1?2)?12,
当且仅当4a?1?4b?1,即a?b?故原式的最大值为12.
1时,取等号, 2ab11 ??b?2ab?2a1?2abab1212因为??(?)(a?b)
ababb2ab2a?1???2?3?(?)
abab(2)原式??3?2b2a??3?22, ab?b2a?a?2?1当且仅当?,即?时,取等号
ab??b?2?2所以原式?1?3?22,
3?22故原式的最大值为3?22.