求解一道数学题

2024-11-03 01:18:46
推荐回答(1个)
回答1:

n=∫[0:π/2][2sin(t/2)cos(t/2)t]dt
=∫[0:π/2]sintdt
=-cost|[0:π/2]
=-(cosπ/2-cos0)
=-(0-1)
=1
f(x)=a/x +lnx -1

对数有意义,x>0,分式有意义,x≠0
x>0,函数定义域为(0,+∞)
f'(x)=-a/x²+ 1/x=(x-a)/x²
令f'(x)≥0,得:(x-a)/x²≥0
x≥a,函数在(0,a]上单调递减,在[a,+∞)上单调递增
要f(x)在定义域上有零点,f(a)≤0
a/a +lna -1≤0
0a的取值范围为(0,1]