Question

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

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

Answer 1

Answer: (B)

$\left|x-1\right|+a\left|=4\Rightarrow \right|\left|x-1\right|+a=\pm 4$
$\Rightarrow \left|x-1\right|=-a\pm 4$
The above equation holds if $-a+4\ge 0$ or $-a-4\ge 0$
$\Rightarrow a\le 4$ or $a\le -4\Rightarrow a\in (-\infty ,-4]\cup (-\infty ,-4]$
$\Rightarrow a\in \left(-\infty ,4\right]$