Question

已知函数f(x)＝alnx－x2＋1.(1)若曲线y＝f(x)在x＝1处的切线方...

已知函数f(x)＝alnx－x2＋1.(1)若曲线y＝f(x)在x＝1处的切线方程为4x－y＋b＝0,求实数a和b的值； (2)求证：f(x)≤0对任意x>0恒成立的充要条件是 a＝2； (3)若a

Answer 1

已知函数f(x)＝alnx－x2＋1.(1)若曲线y＝f(x)在x＝1处的切线方程为4x－y＋b＝0,求实数a和b的值；(2)求证：f(x)≤0对任意x>0恒成立的充要条件是
a＝2；(3)若a0∴f’(x)＝a/x-2x==>
f’(1)＝a-2∵f(x)在x＝1处的切线方程为4x－y＋b＝0∴f’(1)＝a-2=4==>a=6,f(1)＝-1+1=0则切线方程y=4(x-1)==>b=-4∴a=6,b=-4(2)证明：充分性∵a=2==>
f(x)＝2lnx-x^2+1令f’(x)＝2/x-2x=0==>x=1f’’(x)＝-2/x^2-2==>当x>0时,f’’(x)0,f(x)≤0恒成立必要性∵函数f(x)＝alnx-x^2+1,其定义域为x>0令f’(x)＝a/x-2x=0==>x=√(2a)/2
(a>0)f’’(x)＝-a/x^2-2==>当x>0时,f’’(x)a=2∴对任意x>0,f(x)≤0恒成立,则a=2综上,a=2是f(x)≤0对任意x>0恒成立的充要条件(3)解析：∵函数f(x)＝alnx-x^2+1,其定义域为x>0当a|f(x1)-f(x2)|/|x1-x2|>=1==>|[f(x1)-f(x2)]/(x1-x2)|>=1由导数定义可知,即|f’(x)|>=1f’(x)＝a/x-2x>=1==>a>=x-2x^2==>
g(x)最大值为1/4（不合题意）f’(x)＝a/x-2xag(x)最小值为-1/4∴a的取值范围为a