Question

If the equation $||x-1|+a|=4$ has a real soltuion then $a$ belongs to

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

Answer 1

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

$||x-1|+a|=4$
$\Rightarrow |x-1|+a=4\text{or}|x-1|+a=-4$
$\Rightarrow |x-1|=4-a\text{or}|x-1|=-4-a$
As $|x-1|\ge 0$
So, $4-a\ge 0\text{or}-4-a\ge 0$
$\Rightarrow a\le 4\text{or}a\le -4$
$\Rightarrow a\le 4$
Hence, $a\in (-\infty ,4]$

Alternate:
$||x-1|+a|=4$
Since, $|x-1|\ge 0$ (always)
$\Rightarrow ||x-1|+a|\ge a$ (always)
For $||x-1|+a|=4$, the maximum value of $a$ can be $4$
$\therefore a\in (-\infty ,4]$