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Q.
If $y=\frac{x^2+4 x-6}{x^2+4 x+6}$, for all real $x$, then number of integers in the range of $y$, is
Complex Numbers and Quadratic Equations
Solution:
$y=\frac{x^2+4 x-6}{x^2+4 x+6}$
$\Rightarrow yx ^2+4 yx +6 y = x ^2+4 x -6 $
$\Rightarrow ( y -1) x ^2+4( y -1) x +6( y +1)=0 $
$\forall x \in R , D \geq 0$
$\Rightarrow 16( y -1)^2-4 \cdot 6 \cdot( y -1)( y +1) \geq 0$
$ \Rightarrow 2\left( y ^2-2 y +1\right)-3\left( y ^2-1\right) \geq 0 $
$\Rightarrow y ^2+4 y -5 \leq 0 $
$ \Rightarrow ( y +5)( y -1) \leq 0 $
$y \in[-5,1]$ But $y \neq 1$ (think)
$\therefore y \in[-5,1)$ number of integers $=6$.