Question

The solution of the equation $(x-2)\left[x\right]=\left\{x\right\}-1$ is ${}^{\prime }{x}^{\prime }$ such that $a\le x<b$ then $|a+b|=$
(where {.} represents fractional part function)

Answer 1

$\Rightarrow (x-2)(x-\{x\left\}\right)=\left\{x\right\}-1$
$\Rightarrow x^2-2x-\left\{x\right\}(x-2)=\left\{x\right\}-1$
$\Rightarrow x^2-2x+1=\left\{x\right\}(x-1)$
$\Rightarrow (x-1)(x-1)=\left\{x\right\}(x-1)$
$\Rightarrow x=1;\left\{x\right\}=x-1$
We know,
$0\le \left\{x\right\}<1$
$\Rightarrow 0\le x-1<1$
$\Rightarrow 1\le x<2$
$\therefore a+b=1+2=3$
$|a+b|=3$