M08-11

Official Solution:


Is \(|x - 1| < 1\) ?

(1) \((x - 1)^2 \leq 1\).

Given that both sides of the inequality are non-negative, we can take the square root of both sides, resulting in \(|x-1| \le 1\). Consequently, \(|x-1|\) can be less than 1 (answer YES), as well as equal to 1 when \(x=2\) or \(x=0\) (answer NO). Hence, this statement is insufficient.

Notice here though that \(|x-1| \leq 1\) implies \(0 \leq x \leq 2\).

(2) \(x^2 - 1 > 0\).

Rearranging: \(x^2 > 1\). Again, since both sides of the inequality are non-negative, we can take the square root of both sides, leading to \(|x| > 1\). This means that \(x < -1\) or \(x > 1\). If \(x=1.5\), then the answer is YES, but if \(x=2\), then the answer is NO. This is also insufficient.

(1)+(2) From (1) we deduce that \(0 \leq x \leq 2\), and from (2) we infer that \(x < -1\) or \(x > 1\). As a result, \(1 < x \leq 2\). If \(x\) is any number between 1 and 2, not inclusive, the answer to the question is YES. However, if \(x=2\), the answer to the question is NO. Not sufficient.


Answer: E