Let f(x) = x + floor(x).
(a) Find all x >= 0 such that f(x) = 1.
(b) Find all x >= 0 such that f(x) = 3.
(c) Find all x >= 0 such that f(x) = 5.
(d) Find the number of possible values of f(x) for 0 <= x <= 10.
+9519 I will do (a) for you as an example.
x + floor(x) = 1
x = 1 - floor(x), which is an integer.
We conclude that x is an integer. Then floor(x) = x.
x + x = 1
2x = 1
x = 1/2 (rejected)
1/2 is rejected because it is not an integer.
There are no such x >= 0 such that f(x) = 1.
(b) and (c) are similar, and there are infinitely many possible values of f(x) for 0 <= x <= 10, if you consider all real numbers in that range.
