The equation $|| x-1|+a|=4$ can have real solutions for x if 'a' belongs to the interval

(-∞, 4] (-∞, -4] (4, ∞) [-4, 4]

(-∞, -4]

We have,  $|| x-1|+a|=4$ $⇒ |x-1|+a=±4$ $⇒ |x-1|=±4-a$ For real solutions, we must have $±4-a>0$ $⇒ 4-a≥ 0$ and $-4-a≥0$ $⇒ a≤4$ and $a≤-4$ $⇒a≤-4⇒a∈(-∞, -4]$