Question

If $f\left(x\right)=\frac{x+2}{x^2-8x-4}$, then which of the following is correct?

• A. domain is $R-\{4-\sqrt{5},4+\sqrt{5}\}$
• B. range is $R$
• C. domain is $[4-\sqrt{5},4+\sqrt{5}]$
• D. range is $(-\infty ,-\frac{1}{4}]\cup [-\frac{1}{20},\infty )$

Answer 1

The correct option is D range is $(-\infty ,-\frac{1}{4}]\cup [-\frac{1}{20},\infty )$

For domain
$x^2-8x-4\ne 0$
$\Rightarrow x\ne 4+2\sqrt{5},4-2\sqrt{5}$
$\therefore$ Domain is $R-\{4-2\sqrt{5},4+2\sqrt{5}\}$

For range
$y=\frac{x+2}{x^2-8x-4}$
$\Rightarrow y{x}^2-(8y+1)x-(4y+2)=0$
Since, $x$ is real, $\Delta \ge 0,y\ne 0$
$\Rightarrow (8y+1{)}^2+4y(4y+2)\ge 0$
$\Rightarrow 80y^2+24y+1\ge 0$
$\Rightarrow (4y+1)(20y+1)\ge 0$
$\Rightarrow y\in (-\infty ,-\frac{1}{4}]\cup [-\frac{1}{20},\infty )-\left\{0\right\}$

But, for $x=-2,y=0$
Hence, range of $f$ is $(-\infty ,-\frac{1}{4}]\cup [-\frac{1}{20},\infty )$