QJEEM23Ma1e3c: JEE Main 2023 Mathematics - Class 11 JEEM23Ma1e3c Solution: The equation , where denotes the greatest integer function, has :

${x^2} - 4x + [x] + 3 = x[x]$

$\Rightarrow {x^2} - 4x + [x] + 3 - x[x] = 0$

$\Rightarrow (x - 1)(x - 3) - [x](x - 1) = 0$

$\Rightarrow (x - 1)(x - [x] - 3) = 0$

$\therefore$ $x = 1$

or

$x - [x] - 3 = 0$

$\Rightarrow \{ x\} - 3 = 0$ [As $\{ x\} = x - [x]$ ]

$\Rightarrow \{ x\} = 3$

But we know, 0 < \{ x\} < 1

$\therefore$ $\{ x\} \ne 3$

$\therefore$ $x$ has only one solution in ( $-\infty,\infty$ ) which is $x = 1$ .