∴原不等式的解集为{x|-7 (2)原不等式等价于两个不等式组: ①{■(3x"-" 4≥0"," @3x"-" 4>1+2x";" )┤ ②{■(3x"-" 4<0"," @"-(" 3x"-" 4")" >1+2x"." )┤ 由不等式组①,解得x>5. 由不等式组②,解得x<3/5. ∴原不等式的解集为{x├|x<3/5 "或" x>5┤}. 10.设函数f(x)=|x-1|+|x-a|.如果对任意x∈R,f(x)≥2恒成立,求实数a的取值范围. 解若a=1,则f(x)=2|x-1|,不满足题设条件; 若a<1,则f(x)={■("-" 2x+a+1"," x≤a"," @1"-" a"," a 所以f(x)的最小值为1-a. 若a>1,则f(x)={■("-" 2x+a+1"," x≤1"," @a"-" 1"," 1 所以f(x)的最小值为a-1. 所以对任意x∈R,f(x)≥2恒成立时,实数a需满足|a-1|≥2,从而a的取值范围为(-∞,-1]∪[3,+∞). ★11.已知f(x)=|ax+1|(a∈R),不等式f(x)≤3的解集为{x|-2≤x≤1}. (1)求a的值; (2)若|f"(" x")-" 2f(x/2)|≤k恒成立,求k的取值范围. 解(1)由|ax+1|≤3,得-4≤ax≤2. ∵f(x)≤3的解集为{x|-2≤x≤1}, ∴当a≤0时,不合题意;当a>0时,-4/a≤x≤2/a, ∴a=2. (2)记h(x)=f(x)-2f(x/2)=|2x+1|-|2x+2|, 则h(x)={■(1"," x≤"-" 1"," @"-" 4x"-" 3",-" 1 ∴|h(x)|≤1,因此k≥1.