Question

The range of $f\left(x\right)=\frac{1}{-x^2+4x+5}$,$x\in R-\{-1,5\}$ is

• A. $(-\infty ,0]\cup [\frac{1}{9},\infty )$
• B. $(-\infty ,0)\cup [\frac{1}{9},\infty )$
• C. $(-\infty ,0]\cup (\frac{1}{9},\infty )$
• D. $(-\infty ,0)\cup (\frac{1}{9},\infty )$

Answer 1

The correct option is B $(-\infty ,0)\cup [\frac{1}{9},\infty )$

Let $y=f\left(x\right)$
$\therefore y=\frac{1}{-x^2+4x+5}\Rightarrow y{x}^2-\left(4y\right)x-5y+1=0$
Here, $y\ne 0$ because in quadratic equation coefficient of $x^2$ can not be $0.$

For $x\in R-\{-1,5\}$
$D\ge 0\Rightarrow 16y^2-4y(1-5y)\ge 0\Rightarrow 16y^2-4y+20y^2\ge 0\Rightarrow 9y^2-y\ge 0\Rightarrow y(9y-1)\ge 0\Rightarrow y\in (-\infty ,0]\cup [\frac{1}{9},\infty )$
But $y\ne 0$
$\Rightarrow y\in (-\infty ,0)\cup [\frac{1}{9},\infty )$