2014年全国高考浙江省数学(理)试卷及答案[精校版] 下载本文

22.本题主要考查函数最大(最小)值的概念、利用导数研究函数的单调性等基础知识,同时考查推理论证、分类讨论、分析问题和解决问题等综合解题能力。满分14分。

?x3?3x?3a,(x?a)?3x2?3,(x?a)(I)因为f?x???3,所以f'?x???2,

?x?3x?3a,(x?a)?3x?3,(x?a)由于?1?x?1,

(i)当a??1时,有x?a,故f?x??x3?3x?3a,

此时f?x?在??1,1?上是增函数,因此M?a??f?1??4?3a,m?a??f??1???4?3a,

M?a??m?a??4?3a???4?3a??8

(ii)当?1?a?1时,

若x??a,1?,f?x??x?3x?3a,在?a,1?上是增函数,

3若x???1,a?,f?x??x?3x?3a,在??1,a?上是减函数,

3所以m?a??maxf??1?,f?1?,m?a??f?a??a,由于f?1??f??1???6a?2,

3??因此,当?1?a?11时,M?a??m?a???a3?3a?4,当?a?1时,33M?a??m?a???a3?3a?2,

3(iii)当a?1时,有x?a,故f?x??x?3x?3a,此时f?x?在??1,1?上是减函数,

因此M?a??f??1??2?3a,m?a??f?1???2?3a,故

M?a??m?a??2?3a??2?3a??4,综上8,?a??1?????a3?3a?4,??1?a?1????3???M?a??m?a???;

??a3?3a?2,?1?a?1?????3??4,?a?1????x3?3x?3a?b,(x?a)(II)令h?x??f?x??b,则h?x???3,

x?3x?3a?b,(x?a)??3x2?3,(x?a)h'?x???2,

?3x?3,(x?a)/

因为??f?x??b???4,对x???1,1?恒成立,即?2?h?x??2对x???1,1?恒成立,

2所以由(I)知,

(i)当a??1时,h?x?在??1,1?上是增函数,h?x?在??1,1?上的最大值是

2,且4?3a?b?2,h?1??4?3a?b,最小值是h??1???4?3a?b,则?4?3a?b??矛盾;

(ii)当?1?a?1时,h?x?在??1,1?上的最大值是h?1??4?3a?b,最小值是3h?a??a3?b,所以a3?b??2,4?3a?b?2,从而?2?a3?3a?3a?b?6a?2且

0?a?1?1?,令t?a???2?a3?3a,则t'?a??3?3a2?0,t?a?在?0,?上是增函数,3?3?故t?a??t?0???2,因此?2?3a?b?0, (iii)当

1?a?1时,h?x?在??1,1?上的最大值是h??1??3a?b?2,最小值是328h?a??a3?b,所以a3?b??2,3a?b?2?2,解得??3a?b?0,

27(iv)当a?1时,h?x?在??1,1?上的最大值是h??1??3a?b?2,最小值是

h?1???2?3a?b,所以3a?b?2?2,?2?3a?b??2,解得3a?b?0,综上3a?b的取值范围?2?3a?b?0.

/