either the question is wrong or it is must be "for all positive integers number" . Because for real number there are examples where it is not true . For example x = 1/2 it gives 0 which is a even number . For all positive integers the statement is true by following proof :
f(x) = 2x-1
f(x)%2 = (2x-1)%2 = (2x-1+2)%2 = (2x+1)%2 = ((2*(x%2))%2 + 1)%2
now x%2 is either 1 or 0
for x%2 = 1
f(x)%2 = ((2*(1))%2 +1)%2 = (2%2+1)%2 = 1
for x%2 = 0
f(x)%2 = ((2*(0))%2 + 1)%2 = (0+1)%2 = 1
So in all cases f(x)%2 = 1 so f(x) must be odd for all positive integers
Here is another proof by mathematical induction :
let x = 1 be base condition then
f(1) = 1 so it is true for that
Lets assume f(x) is odd
then f(x+1) = 2(x+1)-1
f(x+1) = 2x+2-1
f(x+1) = 2x+1
f(x+1) = 2x-1 + 2
f(x+1) = f(x) + 2
f(x) = odd
so f(x) + 2 must be odd.