For real value of $x$, the range of $\frac{x^2+2 x+1}{x^2+2 x-1}$ is

For real value of $x$, the range of $\frac{x^2+2 x+1}{x^2+2 x-1}$ is
  1. $(-\infty, 0) \cup(1, \infty)$
  2. $\left[\frac{1}{2}, 2\right]$
  3. $\left(-\infty, \frac{-2}{9}\right] \cup(1, \infty)$
  4. None of these

Solution

$\begin{aligned} & \text { Let } y=\frac{x^2+2 x+1}{x^2+2 x-1} \\ & y\left(x^2+2 x-1\right)=x^2+2 x+1 \\ & y x^2+2 x y-y=x^2+2 x+1 \\ & y x^2-x^2+2 x y-2 x-y-1=0 \\ & (y-1) x^2+2(y-1) x-y-1=0 \end{aligned}$ For real values of $x, b^2-4 a c \geq 0$ $\begin{aligned} & {[2(y-1)]^2+4(y-1)(y+1) \geq 0} \\ & 4(y-1)^2+4(y-1)(y+1) \geq 0 \\ & 4(y-1)(y-1+y+1) \geq 0 \\ & 4(y-1)(2 y) \geq 0 \\ & 8 y(y-1) \geq 0 \\ & \text { At } y=0, x^2+2 x+1=0 \\ & (x+1)^2=0 \Rightarrow x=-1 \in \mathrm{R} \\ & \text { At } y=1, x^2+2 x+1=x^2+2 x-1 \\ & 1 \neq-1 \Rightarrow y \neq 1 \\ & y \in(-\infty, 0] \cup(1, \infty) \end{aligned}$

Asked in: AP EAMCET 2015

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