The equation $\left|\right|x-1|+a|=4$ can have real solutions for x if ‘a’ belongs to the interval

We have

$\begin{array}{r}\left|\right|x-1|+a|=4\\ \Rightarrow |x-1|+a=+4\\ \Rightarrow |x-1|=\pm 4-a\end{array}$

For real solutions, we must have

$\pm 4-a>0$

$\Rightarrow 4-a\ge 0$ and $-4-a\ge 0$

$\Rightarrow a\le 4$ and $a\le -4$

$\Rightarrow a\le -4\Rightarrow a\in (-\mathrm{\infty },-4]$