Squaring a function continuous at only one point

Detailed Solution

Since $f$ is continuous at $0$ and squaring is a continuous operation, the composition $g=f^2$ is guaranteed continuous at $0$. This rules out (C). To see that $g$ can be continuous well beyond just $x=0$, take the classical example $f(x)=x$ for rational $x$ and $f(x)=-x$ for irrational $x$, which is continuous only at $x=0$; here $g(x)=[f(x)]^2=x^2$ for every $x$ (rational or irrational), so $g$ is continuous on all of $\mathbb R$ — showing option (A)'s claim ("only at $x=0$") is not forced. Since $g$ is guaranteed continuous at $0$ and may (as this example shows) be continuous elsewhere too, (B) is the correct general statement. Differentiability at $0$, as claimed in (D), is not guaranteed in general by continuity and squaring alone. Thus the correct option is $\boxed{\mathrm{B}}$.