Question

If $||2x-x^2+8|-|x^2+5||=|2x+13|$, then $x$ lies in

• A. $(-\infty ,-\frac{13}{2}]$
• B. $(-\infty ,-2]\cup [4,\infty )$
• C. $[4,\infty )$
• D. $(-\infty ,-\frac{13}{2}]\cup [4,\infty )$

Answer 1

The correct option is B $(-\infty ,-2]\cup [4,\infty )$

$||2x-x^2+8|-|x^2+5||=|2x+13|$
Let, $a=2x-x^2+8,b=x^2+5$
and
$a+b=2x+13$
Clearly, $||a|-|b||=|a+b|$ when $ab\le 0$
$\Rightarrow (2x-x^2+8)(x^2+5)\le 0$
$\Rightarrow (x^2-2x-8)(x^2+5)\ge 0$
$\Rightarrow (x-4)(x+2)(x^2+5)\ge 0$
$\because x^2+5$ is always positive
$\Rightarrow (x-4)(x+2)\ge 0$
$\Rightarrow x\in (-\infty ,-2]\cup [4,\infty )$