Let $A = \{x∈R:-1≤x≤1\}$. Then, the mapping $f:A→A$ given by $f (x) = x|x|$, is

injective but not surjective surjective but not injective bijective none of these

bijective

We have, $f(x)=x|x|=\left\{\begin{matrix}x^2,&x≥0\\-x^2,&x<0\end{matrix}\right\}$ The graph f (x) is shown in Fig. Clearly, f is bijective.