Question
The solutions of the equation $$2x - 2\left[ x \right] = 1,$$ where $$\left[ x \right] = $$ the greatest integer less than or equal to $$x,$$ are
A.
$$x = n + \frac{1}{2},n \in N$$
B.
$$x = n - \frac{1}{2},n \in N$$
C.
$$x = n + \frac{1}{2},n \in Z$$
D.
$$n < x < n + 1,n \in Z$$
Answer :
$$x = n + \frac{1}{2},n \in Z$$
Solution :
If $$x = n \in Z,$$ the equation is $$2n - 2\left[ n \right] = 1\,\,{\text{or 2}}n - 2n = 1\,\left( {{\text{impossible}}} \right).$$
If $$x = n + k,n \in Z,0 < k < 1$$ then the equation is $$2\left( {n + k} \right) - 2\left[ {n + k} \right] = 1$$
or $$2n + 2k - 2n = 1\,\,{\text{or, }}k = \frac{1}{2}\,\,{\text{and }}n \in Z.$$
Alternate Solution
$$\eqalign{
& 2x - 2\left[ x \right] = 1 \cr
& \Rightarrow x - \left[ x \right] = \frac{1}{2} \cr
& \Rightarrow \left\{ x \right\} = \frac{1}{2},\,{\text{where}}\left\{ {} \right\}{\text{is fractional part function}} \cr
& \therefore \,x = n + \frac{1}{2},\,n \in Z \cr} $$