Question

If $\text{sgn}\left(y\right)$ denotes the signum function of $y,$ then the number of solution(s) of the equation $||x+2|-3|=\text{sgn}(1-\left|\frac{(x-2)(x^2+10x+24)}{(x^2+1)(x+4)(x^2+4x-12)}\right|)$ is

• A. $0$
• B. $1$
• C. $3$
• D. $4$

Answer 1

The correct option is A $0$

$||x+2|-3|=\text{sgn}(1-\left|\frac{(x-2)(x+6)(x+4)}{(x^2+1)(x+4)(x+6)(x-2)}\right|)\Rightarrow ||x+2|-3|=\text{sgn}(1-\left|\frac{1}{x^2+1}\right|);x\ne 2,-4,-6$
$\Rightarrow ||x+2|-3|=\text{sgn}\left(\frac{x^2}{x^2+1}\right)\dots \left(1\right)$
We know $\frac{x^2}{x^2+1}\ge 0$ for all $x\in R$

Case $1:$
At $x=0,$
RHS of equation $\left(1\right)$ is $0$ but LHS $\ne 0$
Hence, $x=0$ is not the solution.

Case $2:$
For $x\ne 0$
$||x+2|-3|=1\Rightarrow |x+2|-3=\pm 1$
$\Rightarrow |x+2|=4,2\Rightarrow x+2=\pm 4,\pm 2\Rightarrow x=2,-4,0,-6$
But $x\ne -6,-4,0,2$
Hence, there is no solution.