If $f(x)\left\{\begin{array}{l}x^a \sin \frac{1}{x} & x \neq 0 \\ 0 & x=0\end{array}\right.$ is continuous. At x = 0, then

$a \in(0, \infty)$ $a \in(1, \infty)$ $a \in(-1, \infty)$ $a \in(-\infty, 1)$

$a \in(0, \infty)$

f(x) is continuous at x = 0 hence $\lim\limits_{x \rightarrow 0} x^a \sin \frac{1}{x}=f(0)=0$ This is only possible when a > 0, thus the required set of values of a is (0, ∞) Hence (1) is the correct answer.