所以y'=3ax2+2bx+6>0的解集为-2

　　也就是说,-2和3是方程3ax2+2bx+6=0的两根,即{■(12a"-" 4b+6=0"," @27a+6b+6=0"." )┤解得a=-1/3,b=1/2.

　　所以a,b的值分别为-1/3, 1/2.

能力提升

1若函数f(x)=x3-ax2-x+6在区间(0,1)内单调递减,则实数a的取值范围是(　　)

A.a≥1 B.a=1

C.a≤1 D.0

解析f'(x)=3x2-2ax-1. 学 ]

　　因为f(x)在区间(0,1)内单调递减,所以不等式3x2-2ax-1≤0在区间(0,1)内恒成立.

　　所以f'(0)≤0,f'(1)≤0.所以a≥1.故选A. 学 ]

答案A

2已知函数y=f(x)在定义域[-4,6]上可导,其图象如图,记y=f(x)的导函数为y=f'(x),则不等式f'(x)≤0的解集为(　　)

A.["-" 4/3 "," 1]∪[11/3 "," 6] B.[-3,0]∪[7/3 "," 5]

C.["-" 4",-" 4/3]∪[1"," 7/3] D.[-4,-3]∪[0,1]∪[5,6]

解析不等式f'(x)≤0的解集即函数y=f(x)的递减区间,由题图可知y=f(x)的递减区间为["-" 4/3 "," 1],[11/3 "," 6].故f'(x)≤0的解集为["-" 4/3 "," 1]∪[11/3 "," 6].

答案A

3若定义在R上的函数f(x)满足f(0)=-1,其导函数f'(x)满足f'(x)>k>1,则下列结论中一定错误的是0(　　)

A.f(1/k)<1/k B.f(1/k)>1/(k"-" 1)

C.f(1/(k"-" 1))<1/(k"-" 1) D.f(1/(k"-" 1))>k/(k"-" 1)

解析构造函数F(x)=f(x)-kx,

　　则F'(x)=f'(x)-k>0,

　　∴函数F(x)在R上为单调递增函数.

　　∵1/(k"-" 1)>0,∴F(1/(k"-" 1))>F(0).

　　∵F(0)=f(0)=-1,∴f(1/(k"-" 1))-k/(k"-" 1)>-1,

　　即f(1/(k"-" 1))>k/(k"-" 1)-1=1/(k"-" 1),

　　∴f(1/(k"-" 1))>1/(k"-" 1),故C错误.

答案C