Question

The range of the function $f\left(x\right)=\frac{8x^2+40x+28}{4x^2+20x+13},$ where $x\in R$ is

• A. $(-\infty ,\frac{11}{6}]\cup (2,\infty )$
• B. $(-\infty ,\frac{11}{6}]\cup [2,\infty )$
• C. $[\frac{11}{6},2)$
• D. $(\frac{11}{6},2)$

Answer 1

The correct option is A $(-\infty ,\frac{11}{6}]\cup (2,\infty )$

$f\left(x\right)=\frac{8x^2+40x+28}{4x^2+20x+13}$
$f\left(x\right)$ can be resolved as
$f\left(x\right)=2+\frac{2}{4x^2+20x+13}$

Let $y=2+\frac{2}{4x^2+20x+13}$
$\Rightarrow y-2=\frac{2}{4x^2+20x+13}$
$\Rightarrow 4x^2+20x+13=\frac{2}{y-2}$
$\Rightarrow 4x^2+20x+25-12=\frac{2}{y-2}$
$\Rightarrow (2x+5{)}^2=\frac{2}{y-2}+12$
$\Rightarrow (2x+5{)}^2=\frac{12y-22}{y-2}$
$\Rightarrow (2x+5)=\pm \sqrt{\frac{12y-22}{y-2}}$
Since $x\in R,$
$\frac{12y-22}{y-2}\ge 0$
For $y>2$ the inequation always holds true.

When $y<2$
$12y-22\le 0$
$\Rightarrow y\le \frac{11}{6}$

Hence, the range of the given function is $(-\infty ,\frac{11}{6}]\cup (2,\infty )$