Question

If the solution of the equation $|(x^4-9)-(x^2+3)|=|x^4-9|-|x^2+3|$ is $(-\infty ,p]\cup [q,\infty )$, then the value of $p+q$ is

• A. $0$
• B. $4$
• C. $1$
• D. $-1$

Answer 1

The correct option is A $0$

$|a-b|=|a|-|b|$

If $a$ and $b$ are of same sign and$|a|>|b|$
$\therefore |x^4-9|\ge |x^2+3|$
$⟹|(x^2+3)|(|x^2-3|-1)\ge 0$
$⟹x^2-3\ge 1$ or $x^2-3\le -1$
$⟹$$x$ belongs to $(-\infty ,-2)\cup (2,\infty )$.....(1)
and $(x^4-9)(x^2+3)\ge 0$
$⟹(x^2-3)(x^2+3{)}^2\ge 0$
$⟹x$ belongs to $(-\infty ,-\sqrt{3})\cup (\sqrt{3},\infty )$......(2)
From (1) and (2) $x$ belongs to $(-\infty ,-2)\cup (2,\infty )$.