VJesus12 wrote:If x≠0, is |x| < 1?

(1) x^2 < 1
(2) |x| < 1/x

The OA is the option D.

Why is sufficient the second statement? How can I prove it? Can someone help me? Thanks.

Hello Vjesus12.

Let's see your question.

(1) x^2 < 1

This statement implies that $$\sqrt{x}<\sqrt{1}\ \ \ \Leftrightarrow\ \ \ \left|x\right|<1.\ $$ Hence, this statement is sufficient.

(2) |x| < 1/x

Since x≠0, then |x|>0.

Now, if 0 < |x| < 1/x, then it implies that x must be positive.

If x>0, then |x|=x. Therefore, $$\left|x\right|<\frac{1}{x}\ \ \Rightarrow\ \ x<\frac{1}{x}\ \ \ \Rightarrow\ \ x\cdot x<1\ \ \ \Rightarrow\ \ \ x^2<1.$$ This last expression is the same as the first statement, therefore, |x|<1. This statement is also sufficient..

Therefore, the correct answer for this DS question is the option D.

I hope it helps.