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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.

 May 4, 2022
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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.

 May 4, 2022

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