Sets, Relation and Function
117354
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 \((-\infty,-6] \cup(-2, \infty)\)
Explanation:
A Given,
\(f(x)=y=\frac{x^2+2 x+1}{x^2+2 x-1}\)
\(y\left(x^2+2 x-1\right)=x^2+2 x+1\)
\(y x^2+2 x y-y-x^2-2 x-1=0\)
\(x^2(y-1)+2 x(y-1)-(y+1)=0\)
\(a=(y-1), b=2(y-1), c=-(y+1)\)
\(x \quad \frac{-b \pm \sqrt{(-b)^2-4 a c}}{2 a}\)
\(x=\frac{-2(y-1) \pm \sqrt{\{2(y-1)\}^2+4 \times(y-1)(y+1)}}{2(y-1)}\)
\(4(y-1)^2+4\left(y^2-1\right) \geq 0\)
\(\left(y^2+1-2 y+y^2-1\right) \geq 0\)
\(2 y^2-2 y \geq 0\)
\(2 y(y-1) \geq 0\)Range : \(\mathrm{y} \in(-\infty, 0] \cup[1, \infty)\)
