The domain of `f(x)=sqrt(2{x}^2-3{x}+1),` where {.} denotes the fractional part in `[-1,1]` is `[-1,1]-(1/(2,1))` `[-1,-1/2]uu[(0,1)/2]uu{1}` `[-(1,1)/2]` (d) `[-1/2,1]`

Options

A. `[-1,1] ~((1)/(2),1)`
B. `[-1,-(1)/(2)] cup [0,(1)/(2)] cup {1}`
C. `[-1,(1)/(2)]`
D. `[-(1)/(2),1]`

Answer

B

Explanation

We must have
`2{x}^(2)-3{x}+1 ge 0, " i.e., " {x} ge 1 " or " {x} le 1//2.`
Thus, we have ` o le {x} le 1//2" or " x in [n,n+(1)/(2)], n in I.`