What values of $x$ satisfy $x^{2}<1$
(a) $x<1$
(b) $x>1$
(c) $-2<x<2$
(d) $x<-2$
(e) $x<0$
The correct option is (c).

Let us observe that if $x$ satisfies the inequality, then $-x$ also satisfies the inequality. Thus, the values that satisfy the inequality form a set symmetric with respect to the origin $\mathrm{e}$, so it is enough to check which $x \geq 0$ satisfy $x^{2}0$ for $x \geq 0$, we must have $x-2<0$, that is, $x<2$. By the symmetry observed, we conclude that $-2<x<2$ is the solution to the original inequality.