Question

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

• A. (-∞, 4)
• B. (-∞, −4)
• C. (4, +∞)
• D. [-4, 4]

Answer 1

Answer: (A)

A

Answer 2

Let $Y = |x - 1|$. Since $x$ is real, $Y \ge 0$.

The given equation becomes $|Y + a| = 4$.

This leads to two possibilities:

1. $Y + a = 4 \implies Y = 4 - a$. For real solutions of $x$, $Y$ must be non-negative, so $4 - a \ge 0 \implies a \le 4$.

2. $Y + a = -4 \implies Y = -4 - a$. For real solutions of $x$, $Y$ must be non-negative, so $-4 - a \ge 0 \implies a \le -4$.

The original equation has real solutions for $x$ if at least one of these conditions is met. Therefore, $a \le 4$ OR $a \le -4$. The union of these two conditions is $a \le 4$. So, $a \in (-\infty, 4]$.

Comparing this with the given options, option (A) is $(-\infty, 4)$. This option excludes $a=4$. However, for $a=4$, the equation becomes $||x-1|+4|=4$, which simplifies to $|x-1|=0$ (from $4-a=0$) or $|x-1|=-8$ (from $-4-a=-8$). $|x-1|=0$ gives $x=1$, which is a real solution. Therefore, $a=4$ should be included in the interval. Given that this is a multiple-choice question and $(-\infty, 4]$ is not an option, and (A) is the closest, it suggests a potential omission of the endpoint in the option.