Question

The equation $||x-2|+a|=4$ can have four distinct real solutions for $x$ if $a$ belongs to the interval

• A. $(-\infty ,-4)$
• B. $(-\infty ,0)$
• C. $[4,\infty )$
• D. None of these

Answer 1

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

$||x-2|+a|=4$
$\Rightarrow |x-2|+a=\pm 4$
$\Rightarrow |x-2|=\pm 4-a$
For four real roots,
$4-a>0$ or $-4-a>0$
or, $a\in (-\infty ,-4)$