Correct Answer - A::B
`{x}=[x]x`
`{x}=x(x-{x})=x^(2)-x{x}`
`implies {x}=(x^(2))/(1+x)`
Now `0 le {x} lt 1`
`implies (x^(2))/(1+x) ge 0`
`implies 1+x gt 0`
`implies x gt -1`
`(x^(2))/(1+x) lt 1`
`implies x^(2) lt 1 +x`
`implies x^(2) -x-1 lt 0`
`implies (1-sqrt(5))/(2) lt x lt (1+sqrt(5))/(2)`
Given equation again can be written as
`x-[x]=x[x]`
`(x)/(1+x)=[x]=lambda`
`x=(lambda)/(1-lambda)`
`(1-sqrt(5))/(2) lt (lambda)/(1-lambda) lt (1+sqrt(5))/(2)`
` implies (-1-sqrt(5))/(2) lt lambda lt (sqrt(5)-1)/(2)`
`lambda=0 implies x=0`
`lambda= -1 implies x=(-1)/(2)`
