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Q. The equation $x ^2-12 x +35=[ x ]+[- x ]$ has (where $[ x ]$ denotes largest integer less than or equal to $x$ )

Relations and Functions - Part 2

Solution:

$[x]+[-x]\begin{cases}0 ; & x \in I \\ -1 ; & x \notin I\end{cases}$
Case-I : When $x \in I$
$x^2-12 x+35=0 $
$(x-5)(x-7)=0$
$\therefore x=5, x=7$
Case-II : When $x \notin I$
$x^2-12 x+35=-1 $
$x^2-12 x+36=0 $
$(x-6)^2=0$
$x=6$
but $x \notin I$
$\therefore$ No solution from here