【领航密卷】江西省2014届高三第一次联考数学（文） PDF版含详解

大,最大值为2430 平方米. (13 分) ………………………… 21. 解:(1 )f (x )的定义域为(0 ,+∞).f ′(x )=x- a x = x-a x . 由f (x )在x=2 处的切线与直线3x-2y +1=0 平行,则f ′(2 )= 4-a 2 = 32 ,a=1. (4 分) ……………此时f (x )= 12 x-lnx ,f ′(x )= x-1 x . 令f ′(x )=0 ,得x=1. f (x )与f ′(x )的情况如下:x (0 ,1 ) 1 (1 ,+∞) f ′(x ) - 0 + f (x ) ↘ 12 ↗所以,f (x )的单调递减区间是(0 ,1 ),单调递增区间是(1 ,+∞). (7 分) …………………………………(2 )由f ′(x )=x- a x = x-a x ,由a>0 及定义域为(0 ,+∞),令f ′(x )=0 ,得x=a . ①若a ≤1 ,即0<a≤1 ,在(1 ,e )上,f ′(x )>0 ,f (x )在[1 ,e ]上单调递增,f (x )=f (1 )= 12 ; ②若1<a <e ,即1<a<e ,在(1 ,a )上,f ′(x )<0 ,f (x )单调递减;在(a ,e )上,f ′(x )>0 ,f (x )单调递增, 因此在[1 ,e ]上,f (x )=f (a )= 12 a (1-lna ); ③若a ≥e ,即a≥e ,在(1 ,e )上,f ′(x )<0 ,f (x )在[1 ,e ]上单调递减,f (x )=f (e )= 12 e-a. 综上,当0<a≤1 时,f (x )= 12 ;当1<a<e 时,f (x )= 12 a (1-lna ); 当a≥e 时,f (x )= 12 e-a. (13 分) ………………………………………………………………………