Question

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

• A. $(-\infty ,4]$
• B. $(-\infty ,-4]$
• C. $[4,+\infty )$
• D. $[-4,4]$

Answer 1

The correct option is A $(-\infty ,4]$

1) When $x=1$,

$|a|=4a=\pm 4$

2) When $x>1$

$|x-1+a|=4$

$x-1+a=4x=5-a5-a>1a<4$

3) When $x<1$,

$-x+1-a=4x=-a-3-a-3>1a<-4$

Hence, option A is correct.