答案:A
5.函数f(x)=1/3x3-x2+a,函数g(x)=x2-3x,它们的定义域均为[1,+∞),并且函数f(x)的图像始终在函数g(x)图像的上方,则a的取值范围是( )
A.(0,+∞) B.(-∞,0)
C.("-" 4/3 "," +"∞" ) D.("-∞,-" 4/3)
解析:设h(x)=1/3x3-x2+a-x2+3x,则h'(x)=x2-4x+3=(x-3)·(x-1),所以当x∈(1,3)时,h(x)是减少的;当x∈(3,+∞)时,h(x)是增加的.当x=3时,函数h(x)取得最小值.
因为f(x)的图像始终在g(x)的图像上方,则有h(x)min>0,即h(3)=a>0,所以a的取值范围是(0,+∞).
答案:A
6.设x=1与x=2是函数f(x)=aln x+bx2+x的两个极值点,则常数a= .
解析:f'(x)=a/x+2bx+1,由题意得{■(a+2b+1=0"," @a/2+4b+1=0"," )┤
解得a=-2/3.
答案:-2/3
7.若函数f(x)=-1/2x2+bln(x+2)在(-1,+∞)上是减少的,则b的取值范围是 .
解析:∵函数f(x)在(-1,+∞)上是减少的,
∴f'(x)≤0在(-1,+∞)上恒成立.
∵f'(x)=-x+b/(x+2),∴-x+b/(x+2)≤0.
∵b≤x(x+2)在(-1,+∞)上恒成立,
∴b≤-1.
答案:(-∞,-1]
8.设函数f(x)=aln x-bx2,a,b∈R.若函数f(x)在x=1处与直线y=-1/2相切.
(1)求实数a,b的值;
(2)求函数f(x)在[1/e "," e]上的最大值.
解(1)f'(x)=a/x-2bx,∵函数f(x)在x=1处与直线y=-1/2相切,
∴{■(f"'(" 1")" =a"-" 2b=0"," @f"(" 1")" ="-" b="-" 1/2 "," )┤解得{■(a=1"," @b=1/2 "." )┤
(2)∵f(x)=ln x-1/2x2,
∴f'(x)=1/x-x=(1"-" x^2)/x.
当1/e≤x≤e时,令f'(x)>0,得1/e≤x<1;
令f'(x)<0,得1 ∴f(x)max=f(1)=-1/2. 9.已知函数f(x)=x3+ax2+bx(a,b∈R)的图像过点P(1,2),且在点P处的切线斜率为8. (1)求a,b的值. (2)求函数f(x)的单调区间. 解(1)∵函数f(x)的图像过点P(1,2), ∴f(1)=2,即a+b=1.0①