Question

If $y=\frac{x^2+4x-6}{x^2+4x+6}$, for all real $x$, then number of integers in the range of $y$, is

• A. 7
• B. 6
• C. 5
• D. 4

Answer 1

Answer: (B)

$y=\frac{x^2+4x-6}{x^2+4x+6}$
$\Rightarrow y{x}^2+4yx+6y=x^2+4x-6$
$\Rightarrow (y-1)x^2+4(y-1)x+6(y+1)=0$
$\forall x\in R,D\ge 0$
$\Rightarrow 16(y-1{)}^2-4\cdot 6\cdot (y-1)(y+1)\ge 0$
$\Rightarrow 2\left(y^2-2y+1\right)-3\left(y^2-1\right)\ge 0$
$\Rightarrow y^2+4y-5\le 0$
$\Rightarrow (y+5)(y-1)\le 0$
$y\in [-5,1]$ But $y\ne 1$ (think)
$\therefore y\in [-5,1)$ number of integers $=6$.