Answer
a. any value of $m$.
b. $m=2$.
Work Step by Step
a. For the function to be continuous at $x=0$, we need to evaluate the left and right limits and compare with the function value at this point. We have
$\lim_{x\to0^-}sin2x=0$, $\lim_{x\to0^+}mx=0$, and $f(0)=0$. Since these values are equal, we conclude that the function is continuous at $x=0$ for any value of $m$.
b. For the function to be differentiable at $x=0$, we need to evaluate the left and right derivatives at this point and let them equal. We have $\lim_{x\to0^-}f'(x)=\lim_{x\to0^-}2cos2x=2$ and $\lim_{x\to0^+}f'(x)=\lim_{x\to0^+}m=m$. Thus the function is differentiable at $x=0$ only if $m=2$.
