Question

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

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

Answer 1

Answer: (D)

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