JEE MAIN - Mathematics (2023 - 24th January Morning Shift - No. 9)
The equation $${x^2} - 4x + [x] + 3 = x[x]$$, where $$[x]$$ denotes the greatest integer function, has :
exactly two solutions in ($$-\infty,\infty$$)
no solution
a unique solution in ($$-\infty,\infty$$)
a unique solution in ($$-\infty,1$$)
Explanation
$${x^2} - 4x + [x] + 3 = x[x]$$
$$ \Rightarrow {x^2} - 4x + [x] + 3 - x[x] = 0$$
$$ \Rightarrow (x - 1)(x - 3) - [x](x - 1) = 0$$
$$ \Rightarrow (x - 1)(x - [x] - 3) = 0$$
$$\therefore$$ $$x = 1$$
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 $$x = 1$$.
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