6.已知函数f(x)=x3+2bx2+cx+1有两个极值点x1,x2,且x1∈[-2,-1],x2∈[1,2],则f(-1)的取值范围是(　　)

A.["-" 3/2 "," 3] B.[3/2 "," 6]

C.[3,12] D.["-" 3/2 "," 12]

解析:f'(x)=3x2+4bx+c,

　　由题意,得{■(f"'(-" 2")" =12"-" 8b+c≥0"," @f"'(-" 1")" =3"-" 4b+c≤0"," @f"'(" 1")" =3+4b+c≤0"," @f"'(" 2")" =12+8b+c≥0"," )┤作出平面区域如图中阴影部分.

　　f(-1)=2b-c,当直线过点A时,f(-1)取最小值3,当直线过点B时,取最大值12,故选C.

答案:C

7. 已知函数f(x)=x3-3x的图像与直线y=a有三个不同的交点,则a的取值范围是　　　　　.

答案: (-2,2)

8.函数f(x)=x3-3a2x+2a(a>0)的极大值为正数,极小值为负数,则a的取值范围是　　　　　.

解析:∵f'(x)=3x2-3a2=3(x-a)(x+a)(a>0),

　　∴f'(x)>0时,得x>a或x<-a,

　　f'(x)<0时,得-a

　　∴当x=a时,f(x)有极小值,x=-a时,f(x)有极大值.

　　由题意,得{■(a^3 "-" 3a^3+2a<0"," @"-" a^3+3a^3+2a>0"," @a>0"," )┤解得a>1.

答案:(1,+∞)

9.函数f(x)=aln x+bx2+3x的极值点为x1=1,x2=2,则a=　　　　　,b=　　　　　.

解析:f'(x)=a/x+2bx+3=(2bx^2+3x+a)/x,

　　∵函数的极值点为x1=1,x2=2,

　　∴x1=1,x2=2是方程f'(x)=(2bx^2+3x+a)/x=0的两根,即2bx2+3x+a=0的两根.

　　∴由根与系数的关系知{■(x_1+x_2="-" 3/2b=1+2"," @x_1 x_2=a/2b=1×2"," )┤

　　解得{■(a="-" 2"," @b="-" 1/2 "." )┤

答案:-2　-1/2

10.已知某质点的运动方程为s(t)=t3+bt2+ct+d,如图是其运动轨迹的一部分,若当t∈[1/2 "," 4)时,s(t)<3d2恒成立,试求d的取值范围.

解:∵质点的运动方程为s(t)=t3+bt2+ct+d,

∴s'(t)=3t2+2bt+c.