Question

设函数f(x)=|lgx|,a,b满足f(a)=f(b)=2f[(a+b)/2] 0

Answer 1

证:

由 |lga| = |lgb|

得 lga = lgb 或 lga = -lgb

得 a = b 或 a = 1/b

因为 0 1 ,所以 b² > 1 ,所以 1/b² ∈ (0,1)

又 2|lg[(a+b)/2] = |lgb| ,b > 1 ,a = 1/b

则 2|lg[(1/b + b)/2]| = lgb

由基本不等式得 (1/b + b)/2 > 2/2 = 1

则 2lg[(1/b + b)/2] = lgb

即 [(1/b + b)/2]² = b

即 1/b² + 2 + b² = 4b

则 2< 4b - b² = 2 + 1/b²