Given the function f (x) = x2 alnx, G (x) = e ^ X - [x] (1) Proof: e ^ a > A (2) When a > 2E, discuss the number of zeros of function f (x) in the interval (1, e ^ a)

Given the function f (x) = x2 alnx, G (x) = e ^ X - [x] (1) Proof: e ^ a > A (2) When a > 2E, discuss the number of zeros of function f (x) in the interval (1, e ^ a)

prove:
Constructor f (x) = e ^ x-x
Then f '(x) = e ^ X-1
Ψ x > 0, f '(x) > 0, f (x) increasing
x