f(x) = ln(x)+x-x^2 => f'(x) = 1/x+1-2x
Searching for max/min by finding for which x we have f'(x) = 0. Hence1/x+1-2x = 0 => x = 1 and x = -1/2
But we're looking in the range [1/4,2] and x = -1/2 is not in that range. x = 1 are in the range, therefore x = 1 is a valid solution to the problem. Now you can check if this is a max or min by using the second derivative or by studying if the first derivative is positive or negative around x = 1.