由题图可知,s(t)在t=1和t=3处取得极值.

　　则s'(1)=0,s'(3)=0,

　　即{■(3+2b+c=0"," @27+6b+c=0"," )┤∴{■(b="-" 6"," @c=9"." )┤

　　∴s'(t)=3t2-12t+9=3(t-1)(t-3).

　　当t∈(1/2 "," 1)时,s'(t)>0;

　　当t∈(1,3)时,s'(t)<0;

　　当t∈(3,4)时,s'(t)>0.

　　∴当t=1时,s(t)取得极大值4+d.

　　又s(4)=4+d,

　　∴当t∈[1/2 "," 4)时,s(t)的最大值为4+d.

　　∵当t∈[1/2 "," 4)时,s(t)<3d2恒成立,

　　∴4+d<3d2,即d>4/3或d<-1.

　　∴d的取值范围是(-∞,-1)∪(4/3 "," +"∞" ).

11.已知函数f(x)=ln x-1/2ax2-2x.

(1)若函数f(x)在x=2处取得极值,求实数a的值;

(2)若函数f(x)在定义域内递增,求实数a的取值范围;

(3)当a=-1/2时,关于x的方程f(x)=-1/2x+b在[1,4]上恰有两个不相等的实数根,求实数b的取值范围.

解:(1)f'(x)=-(ax^2+2x"-" 1)/x(x>0),

　　∵x=2时,f(x)取得极值,

　　∴f'(2)=0,解得a=-3/4,经检验知符合题意.

　　(2)函数f(x)的定义域为(0,+∞),

　　依题意f'(x)≥0在x>0时恒成立,

　　即ax2+2x-1≤0在x>0恒成立,

　　则a≤(1"-" 2x)/x^2 =(1/x "-" 1)^2-1在x>0恒成立,

　　即a≤[(1/x "-" 1)^2 "-" 1]_min(x>0),

　　当x=1时,(1/x "-" 1)^2-1取最小值-1,

　　∴a的取值范围是(-∞,-1].

　　(3)a=-1/2,f(x)=-1/2x+b,

　　即1/4x2-3/2x+ln x-b=0.

　　设g(x)=1/4x2-3/2x+ln x-b(x>0),

　　则g'(x)=("(" x"-" 2")(" x"-" 1")" )/2x.

　　列表:

x (0,1) 1 (1,2) 2 (2,4) g'(x) + 0 - 0 + g(x) ↗ 极大值 ↘ 极小值 ↗ ∴g(x)的极小值为g(2)=ln 2-b-2,极大值为g(1)=-b-5/4,

　　又g(4)=2ln 2-b-2,

　　∵方程g(x)=0在[1,4]上恰有两个不相等的实数根,则{■(g"(" 1")" ≥0"," @g"(" 2")" <0"," @g"(" 4")" ≥0"," )┤解得ln 2-2

12.已知函数f(x)=4x3-3x2cos θ+3/16cos θ(x∈R,θ∈[0,2π]).

(1)当cos θ=0时,判断函数f(x)是否有极值;