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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

AI Generated Solution

To find the domain of the function \( f(x) = \sqrt{2\{x\}^2 - 3\{x\} + 1} \), where \(\{x\}\) denotes the fractional part of \(x\), we need to ensure that the expression inside the square root is non-negative. The fractional part \(\{x\}\) is defined as \(x - \lfloor x \rfloor\), and it takes values in the interval \([0, 1)\). ### Step-by-Step Solution: 1. **Set up the inequality**: We need the expression inside the square root to be greater than or equal to zero: \[ 2\{x\}^2 - 3\{x\} + 1 \geq 0 \] ...
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