Question

The sum of all possible solution(s) of the equation ||$x+2|-3|=\operatorname{sgn}\left(1-\left|\frac{(x-2)\left(x^2+10 x+24\right)}{\left(x^2+1\right)(x+4)\left(x^2+4 x-12\right)}\right|\right)$is
[Note : sgn (y) denotes the signum function of y.]

• A. 0
• B. -8
• C. -10
• D. not applicable

Answer 1

Answer: (D)

$|| x+2|-3|=\operatorname{sgn}\left(1-\left|\frac{(x-2)(x+6)(x+4)}{\left(x^2+1\right)(x+4)(x+6)(x-2)}\right|\right)$
$|| x+2|-3|= \operatorname{sgn}\left(1-\left|\frac{1}{x^2+1}\right|\right) x \neq 2,-4,-6$
$|| x+2|-3|= 1 \Rightarrow|x+2|-3= \pm 1 $
$ \Rightarrow|x+2|=4,2$
$ \Rightarrow x+2= \pm 4, \pm 2 $
$ \Rightarrow x=2,-4,0-6 .$