The solution of x-1=(x-[x])(x- x ) (where [x] and x are the integral and fractional part of x is

Question

The solution of x-1=(x-[x])(x- x ) (where [x] and x are the integral and fractional part of x is (A) x ∈ R (B) x ∈ R ∼[1,2) (C) x ∈[1,2) (D) x ∈ R ∼[1,2]

Tardigrade Admin · Accepted Answer

∵(x-1)=(x-[x])(x- x ) ⇒ x=1+ x [x] ⇒ [x]+ x =1+ x [x ⇒ ( x -1)([x]-1)=0 ; x -1 ≠ 0 ∴ [x]-1=0 ⇒ [x]=1 ⇒ x ∈[1,2)

Q. The solution of $x - 1 = (x - [x]) (x - {x})$ (where $[x]$ and ${x}$ are the integral and fractional part of $x$ is

$x \in R$

$x \in R \sim [1, 2)$

$x \in [1, 2)$

$x \in R \sim [1, 2]$

$∵ (x - 1) = (x - [x]) (x - {x})$
$\Rightarrow x = 1 + {x} [x]$
$\Rightarrow [x] + {x} = 1 + {x} [x}$
$\Rightarrow ({x} - 1) ([x] - 1) = 0; {x} - 1 \neq = 0$
$∴ [x] - 1 = 0$
$\Rightarrow [x] = 1$
$\Rightarrow x \in [1, 2)$

Q. The solution of x−1=(x−[x])(x−{x}) (where [x] and {x} are the integral and fractional part of x is

x∈R

x∈R∼[1,2)

x∈[1,2)

x∈R∼[1,2]