Question

The range of $f\left(x\right)=\frac{1}{2x^2-6x+7}$ is

• A. $(-\infty ,0)\cup [\frac{2}{5},\infty )$
• B. $(-\infty ,0]\cup [\frac{2}{5},\infty )$
• C. $(0,\frac{2}{5}]$
• D. $[0,\frac{2}{5}]$

Answer 1

The correct option is C $(0,\frac{2}{5}]$

$f\left(x\right)=\frac{1}{2x^2-6x+7}$
$2x^2-6x+7>0,\forall x\in R(\because D=36-56=-20<0)$
Now, let $y=f\left(x\right)$
$\Rightarrow y=\frac{1}{2x^2-6x+7}\Rightarrow \left(2y\right)x^2-\left(6y\right)x+7y-1=0$
for quadratic polynomial $y\ne 0$
Now, for $x\in R$
$D\ge 0\Rightarrow 36y^2-4\left(2y\right)(7y-1)\ge 0\Rightarrow 5y^2-2y\le 0\Rightarrow y\in [0,\frac{2}{5}]$
But $y\ne 0$
$\therefore$ Range is $(0,\frac{2}{5}]$