Question

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

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

Answer 1

Answer: (A), (B), (D)

$(-\infty ,-4]$
$(-\infty ,4]$
$[-4,4]$

Given equation $||x-1|+a|=4$
$\mid x-1\mid +a=\pm 4$
$\Rightarrow \mid x-1\mid =4-a$ or $-4-a.$
$\therefore$ for real $x$,
$4-a\ge 0$ or $-4-a\ge 0$.
$\Rightarrow a\le 4$ or $a\le -4$
Hence, option A , B and D satisfies above inequalities