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The domain of f(x)=sqrt(2{x}^2-3{x}+1), ...

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

A

`[-1,1] ~((1)/(2),1)`

B

`[-1,-(1)/(2)] cup [0,(1)/(2)] cup {1}`

C

`[-1,(1)/(2)]`

D

`[-(1)/(2),1]`

Text Solution

Verified by Experts

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.`
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