Question

已知二次函数g(x)对任意实数x都满足g(x-1)+g(1-x)=x2-2x-1,且g(1)=-1.令f(x)=g(x+

Answer 1

1.设g(x)=ax^+bx+c,

g(1)=a+b+c=-1,①

g(x-1)+g(1-x)

=a(x-1)^+b(x-1)+c

+a(1-x)^+b(1-x)+c

=2ax^-4ax+2a+2c

=x^-2x-1,

比较系数得2a=1,2a+2c=-1,

解得a=1/2,c=-1,

代入①,b=-1/2.

∴g(x)=(1/2)x^-(1/2)x-1.

2.存在x>0,使得f(x)=g(x+1/2)+mlnx+9/8=(1/2)x^+mlnx>=0,

x=1时上述不等式恒成立,

∴m的取值范围是R.

3.H(x)=f(x)-(m-1)x=(1/2)x^-(m-1)x+mlnx,x∈[m,1],m>0,

H'(x)=x-(m-1)+m/x

=[x^-（m-1)x+m]/x

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